_ Ampacity Computation:

or Transmission, Distri
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GEORGE
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_ IEEE
PressPowerEngineering
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1149860
9
INUIT
RATING OF
ELECTRIC POWER CABLES
Ampacity Computations
for Transmission, Distribution,
and Industrial Applications

George J. Anders
Ontario Hydro Technologies

IEEE Power Engineering Society, Sponsor

IEEE Press Power Engineering Series

Dr. Paul M. Anderson, Series Editor

Pru Ga
McGRAW-HILL The Institute of Electrical
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10 g 8 #) 6 =) 4 3 Z 1

IEEE ISBN 07803-1177-9

McGraw-Hill ISBN 0-07-001791-3

Library of Congress Cataloging-in-Publication Data

Anders, George J. (date)

Rating of electric power cables: ampacity computations for
transmission, distribution, and industrial applications / George J.
Anders.
p. cm. -- (IEEE Press power engineering series)
Includes index.
ISBN 0-7803-1177-9
1. Electric cables--Standards. 2. Electric cables--Mathematical
models. 3. Electric currents--Measurements--Mathematics. I. Title.
Il. Series
TK3307.A69 1997
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 Contents

Preface xvii

Acknowledgments xxi

Symbols xxiii
PART! MODELING 1

Chapter 1 Cable Construction and Installations 3

1.1 Introduction 3
1.2 CableComponents 4
2 alee Conductorsmas
1.2.2 Insulation 6
1.2.3. Sheath/Concentric Neutral Wires 7
1.2.4 Armor 8
1.2.5 External Covering 8
1.3. Cable Installations 9
1.3.1 Laying Conditions 9
1.3.1.1 | Underground Installations 9
1.3.1.2 Cables Installed in Air 11
1.3.2 Bonding Arrangements 12
1.3.3 Forced Cooling of Cable Circuits 13
1.4 Heat Sources in Power Cables 13
1.4.1 Current-Dependent Losses 14
1.4.2 Voltage-Dependent Losses 16
1.5 Historical Note 16
References 20

ix
x Contents

Chapter 2 Modes of Heat Transfer and Energy Conservation

Equations 23
2.1 Introduction 23
2.2 Heat Transfer Mechanism in Power Cable Systems 23
2.2.1 Conduction 23
2.2.2 Convection 24
2.2.3 Radiation 25
2.2.4 Energy Balance Equations 26
2.3 Heat Transfer Equations 27
2.3.1 Underground Directly Buried Cables 27
Dis ee CablesmprAlies 0
2.4 Analyical versus Numerical Methods of Solving Heat Transfer
Equations 31
References 31

Chapter 3 Circuit Theory Network Analogs for Thermal Modeling 33

3.1 Introduction 33
3.2 Electrical and Thermal Analogy 34
3.2.1 Thermal Resistance 34
3.2.2. Thermal Capacitance 38
3.3 Construction of a Ladder Network of a Cable 40
3.3.1 Representation of Capacitances of the Dielectric 40
3.3.1.1 Van Wormer Coefficient for Long-Duration Transients 41
3.3.1.2. Van Wormer Coefficient for Short-Duration Transients 43
3.3.1.3 Van Wormer Coefficient for Transients Due
to Dielectric Loss 45
3.3.2 Reduction of a Ladder Network to a Two-Loop Circuit 46
3.3.3. Two-Section Ladder Networks for Some Selected Cable
Constructions 47
3.3.3.1 Solid and Self-contained Fluid-filled Paper-insulated
Cables, Extruded Cables and Thermally Similar
Constructions 48
3.3.3.2 High-pressure Fluid-filled Pipe-type (HFLF) Cables 50
3.3.3.3 Cables in Ducts 52
References 55

Chapter 4 Rating Equations—Steady State Conditions 57

4.1 Buried Cables with No Moisture Migration in the Soil 57
4.2 Buried Cables with Moisture Migration Taken into Account 60
4.2.1 Introduction 60
4.2.2 Development of Rating Equation 61
4.2.3 Determination of the Critical Temperature Rise A@x 62
4.3 CablesinAir 64
References 64

Chapter 5 Rating Equations—Transient Conditions 67

5.1 Introduction 67
5.2 Response to a Step Function 69
5.2.1 Preliminaries 69
Contents xi

5.2.2 Temperature Rise in the Internal Parts of the Cable 70

5.2.3. Second Part of the Thermal Circuit—Influence of the Soil 73
5.2.4 Second Part of the Thermal Circuit—Cables in Air 75
5.2.5 Representation of Dielectric Losses 75
5.2.6 Groups of Equally or Unequally Loaded Cables 76
5.2.7 Total Temperature Rise 77
my) Transient Temperature Rise Under Variable Loading 78
5.4 Diffusivity of Soil 78
ee) Conductor Resistance Variations During Transients 78
5.6 Calculation of Cyclic Ratings 87
5.6.1 Introduction 87
5.6.2 Cyclic Rating Computations 88
5.6.2.1 Single Cable 89
5.6.2.2 Groups of Identical, Equally Loaded Cables 92
5.6.3. Cyclic Rating with Partial Drying of the Surrounding Soil 97
5.6.3.1 Assumptions 97
5.6.3.2 Development of the Cyclic Rating Factor 98
a. Calculation of Emergency Ratings 100
5.7.1 Thermally Isolated Circuits 101
5.7.2. Groups of Circuits 103
References 104

PART Il EVALUATION OF PARAMETERS 107

Chapter 6 Dielectric Losses 109

References 113

Chapter 7 Joule Losses in the Conductor 115

i! Introduction 115
fy Resistance of Cable Conductor 116
7.2.1 DC Resistance of Stranded Conductors 117
7.2.2 AC Resistance of Conductors: Skin and Proximity Effects 118
7.2.3 Summary of ac Resistance Computations 124
ifsc, The Effect of Harmonics 129
References 135

Chapter 8 Joule Losses in Screens, Sheaths, Armor and Pipes 137

8.1 Introduction 137
8.2 Sheath Bonding Arrangements 139
8.3 Circulating Current Losses in Sheath and Armor 139
8.3.1 Internal Inductances 142
8.3.1.1 Hollow Conductor Internal Inductance 142
8.3.1.2. Conductor—Sheath Internal Inductance 143
8.3.1.3 Sheath-Sheath Internal Inductance 143
8.3.1.4 Internal Armor Inductances 143
8.3.2 Total Inductances 147
8.3.2.1 Conductor—Conductor Inductance 147
8.3.2.2 Conductor—Sheath Inductance 148
8.3.2.3 Sheath-Sheath Inductance 149
8.3.2.4 Armor Inductances 149
Xil Contents

8.3.3. Cable Impedances 150

8.3.4 Loss Factors 151
8.3.5 Circulating Current Losses in the Sheaths—Special Cases 156
8.3.5.1 Two Single-core Cables or Three Single-core Cables in Trefoil
Formation, Sheaths Bonded at Both Ends 156
8.3.5.2 Three Single-core Cables in Flat Formation, with Regular
Transposition, Sheaths Bonded at Both Ends 157
8.3.5.3 Three Single-core Cables without Transposition, Sheaths
Bonded at Both Ends 157
8.3.5.4 Variation of Spacing of Single-core Cables Between Sheath
Bonding Points 161
8.3.5.5 Effect of Unequal Section Lengths in Cross-bonded
Systems 161
8.3.5.6 Armored Cables with Each Core in a Separate
Lead Sheath (SL Type) 162
8.3.5.7. Losses in the Screens and Sheaths of Pipe-type Cables 162
8.3.5.8 Two-core or Three-core Armored Cables Having Extruded
Insulation and Copper Tape Screens 163
8.3.5.9 Circulating Currents in the Sheaths of Parallel Cables 163
8.3.6 Circulating Current Losses in the Armor—Special Cases 172
8.3.6.1 Armor Materials 172
8.3.6.2 Nonmagnetic Armor or Reinforcement 172
8.3.6.3 Magnetic Armor or Reinforcement 175
8.3.7. Losses in Steel Pipes 178
8.4 Sheath Eddy Current Losses 179
8.4.1 Overview 179
8.4.2 Loss Due to External Currents 180
8.4.3. Loss Due to Internal Current 183
8.4.4 Correction for Wall Thickness 184
8.4.5 Simplified Expressions for Three Single-core Cables in Flat
and Trefoil Formations 184
8.4.6 Two-core Unarmored Cables with Common Sheath 187
8.4.7. Three-core Unarmored Cables with aCommon Sheath 188
8.4.8 Two-core and Three-core Cables with Steel Tape Armor 188
8.4.9 Armored Three-core Cables with Each Core Having a Separate Lead
Sheath (SL Type) 189
8.4.10 Effect of Large Segmental-Type Conductors 189
8.5 General Method of Computation of Joule Loss Factors Using
Filament Heat Source Simulation Method 190
References 193

Chapter 9 Thermal Resistances and Capacitances 197

9.1 Introduction 197
9.2 Thermal Resistance Between One Conductor and Sheath Li oe
9.2.1 Single-core Cables 199
9.2.2 Three-core Cables 199
9.2.2.1 Overview 199
9.2.2.2 Two-core Belted Cables with Circular Conductors 200
Contents Xlll

9.2.2.3
Three-core Belted Cables with Circular and Oval
Conductors 200
9.2.2.4 Three-core Cables with Circular Conductors and Extruded
Insulation 201
9.2.2.5 Shaped Conductors 205
9.2.2.6 Three-core Cables with Metal Screens Around
Each Core 206
9.2.2.7 Fluid-filled Cables 207
9.2.2.8 SL Type Cables 209
9.3 Thermal Resistance Between Sheath and Armor T, 209
9.3.1 Single-core, Two-core, and Three-core Cables Having
a Common Metallic Sheath 209
9.3.2 SL Type Cables 210
9.4 Thermal Resistance of Outer Covering (Serving) T, 210
9.5. Pipe-type Cables 211
9.6 External Thermal Resistance 211
9.6.1 Single Buried Cable 212
9.6.2 Groups of Buried Cables (Not Touching) 214
9.6.2.1 Unequally Loaded Cables 214
9.6.2.2 Equally Loaded Identical Cables 214
9.6.3. Groups of Buried Cables (Touching) Equally Loaded 217
9.6.3.1 Overview 217
9.6.3.2 Two Single-core Cables in Flat Formation 217
9.6.3.3 Three Single-core Cables in Flat Formation 218
9.6.3.4 Three Single-core Cables in Trefoil Formation 219
9.6.4 Cables in Ducts and Pipes 222
9.6.4.1 Thermal Resistance Between Cable and
Duct (or Pipe) T, 222
9.6.4.2 Thermal Resistance of the Duct (or Pipe) Itself T; 228
9.6.4.3 External Thermal Resistance of the Duct (or Pipe) tT, 2228
9.6.5 Cables in Backfills and Duct Banks 229
9.6.5.1 The Neher—McGrath Approach 230
9.6.5.2 Extended Values of the Geometric Factor 232
9.6.5.3 Geometric Factor for Transient Computations 232
9.6.6 External Thermal Resistance of Cables Laid in Materials Having
Different Thermal Resistivities 234
9.6.7. The Neher—McGrath Modification of T,to Account for Cyclic
Loading 237
9.6.8 Cablesin Air 239
9.6.8.1 General Equation for the External Thermal Resistance 239
9.6.8.2 IEC Standard 287—Simple Configurations 240
9.6.8.3 IEC Standard 287—Derating Factors for Groups
of Cables 245
9.6.8.4 The Effect of Wind Velocity and Mixed Convection 250
9.6.8.5 Neher-—McGrath Approach 253
9.7. Thermal Capacitances 254
9.7.1 Oil in the Conductor 254
9.7.2 Conductor 255
9.7.3 Insulation 255
 Contents
XIV

9.7.4 Metallic Sheath or Any Other Concentric Layer 256

9.7.5 Reinforcing Tapes 256
9.7.6 Armor 256
9.7.7 Pipe-type Cables 256
Referencesa2. 7;
PART Ill ADVANCEDTOPICS 262

Chapter 10 Special Cable Installations 263

10.1 Introduction 263
10.2 Energy Conservation Equations 264
10.2.1 Energy Conservation Equation for the Cable Outside Surface 264
10.2.2 Energy Conservation Equation for the Wall Inside Surface 266
10.2.3. Energy Conservation Equation for the Wall Outside Surface 266
10.2.4 Energy Conservation Equations 267
10.3 Cables on Riser Poles 267
10.3.1 Introduction 267
10.3.2 Thermal Model 268
10.3.3. Intensity of Solar Radiation 270
10.3.4 Convection Coefficients 272
10.3.4.1 Riser Outside Surface 272
10.3.4.2. Convection in the Air Gap 272
10.4 Cables in Trays 277
10.4.1 Cables in Single Open-top Cable Tray 278
10.4.1.1 —Introduction 278
10.4.1.2. | Thermal Model 278
10.4.2 Cables in Covered Trays 280
10.4.2.1 —Cables in Fire-protection Wrapped Tray 282
10.5 Cables in Tunnels and Shafts 283
10.5.1 Horizontal Tunnels 283
10.5.1.1 Thermal Model 283
10.5.2 Vertical Shafts 288
10.6 Cables in Buried Troughs 288
10.6.1 Buried Troughs Filled with Sand 289
10.6.2 Unfilled Troughs of Any Type, with the Top Flush
with the Soil Surface and Exposed to Free Air 291
References 292

Chapter 11 Ampacity Computations Using Numerical Methods 295

I be Introduction 295
Lig General Characteristics of Numerical Methods 296
11.2.1 Selection of the Region to be Discretized 297
11.2.2 Representation of Cable Losses 297
11.2.3 Selection of a Time Step 297
1 IR The Finite-Element Method 298
11.3.1 Overview 298
11.3.2 Approximating Polynomials 300
11.3.3 Finite-Element Equations 303
11.3.4 Some Comments on Computer Implementation 309
11.4 The Finite-Difference Method 312
11.4.1 Overview 312
11.4.2 Finite Difference Approximations to Derivatives 312
Contents XV

11.4.3. Application of the Finite-Difference Method for the Calculation

of the Response of Single-Core Cables to a Step Function Thermal
Transient 319
11.4.3.1 General Characteristics of the Method 319
11.4.3.2 Description of the Method 319
1.4.3.3 Computation of Temperature Transient for a Single Cable 321
11.4.3.4 Mutual Heating from Other Cables 323
11.5 Modeling and Computational Issues 324
References 325

Chapter 12 Economic Selection of Conductor Cross Section 329

12.1 Introduction 329
12.2 Cost of Joule Losses 330
12.2.1 Constant Load Factor 331
12.2.2 Mean Conductor Temperature and Resistance 332
12.2.3 Growing Load Factor 336
12.3 Effect of Charging Current and Dielectric Losses 340
12.4 Selection of the Economic Conductor Size 343
12.5 Parameters Affecting Economic Selection of Cable Sizes 353
References 359

PART IV APPENDICES 361

APPENDIX A Model Cables 363

APPENDIX B An Algorithm to Calculate the Coefficients of the Transfer

Function Equation 371

APPENDIX C Digital Calculation of Quantities Given Graphically

in Figs. 9.1-9.7 373

APPENDIX D Properties of Air at Atmospheric Pressure 377

APPENDIX E Calculation Sheets for Steady-state Cable Ratings 379

APPENDIX F Differences between the Neher/McGrath

and IEC 287 Methods 417

Index 421

About the Author 427

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 Preface

The subject matter of this book is the computation of current ratings (also called current-
carrying capacities or ampacities) of electric power cables. Computations of cable am-
pacities are generally quite involved; therefore, cable engineers have traditionally used
published ampacity tables or performed approximate calculations to determine the cable
size and type required for a particular application or to assess the ratings of existing cables.
This practice could lead to the installation of oversized cables and increased installation
costs. The advent of inexpensive and powerful personal computers made the development of
fast, user-friendly computer programs for cable ampacity calculations feasible. These give
the engineer an opportunity to determine accurate cable ratings with an ease not available
before.
The widespread use of personal computers shifted the burden of ampacity computations
from a few experts to a wider group of cable engineers. Inevitably, without having the years
of experience required to perform cable thermal analysis, engineers ask questions about the
basis of the computations performed by the various cable rating programs. But even for the
experienced cable experts, the background of many of the computations may not always be
clear.
Currently, two major sources are used throughout the world as basic references about
computations of steady-state ratings: (1) the classic paper by Neher and McGrath (1957),
and (2) the IEC Standard 287 (1982, 1989, 1994, 1995). Equivalent sources for transient
analysis are Neher (1964) and the IEC Standards 853-1 (1985) and 853-2 (1989).! The IEC
standards, being the more recent documents, contain more up-to-date information. At the
writing of this book, a new, revised edition of the IEC Standard 287 is being prepared. Since
the method described in the Neher/McGrath (1957) paper is still widely used in the United
States, the differences between this method and the one outlined in the IEC Publication 287
(1994-1995) are summarized in Appendix F.

! See the list of references at the end of Chapters | and 2 for full details about these publications.

XVii
XVili Preface

Allof these sources contain hundreds of formulas, many of them developed empirically.
In the majority of cases, the equations are given without derivation or an explanation of
their origin and of the assumptions which may restrict their applicability. In addition, many
developments in cable rating computations which have taken place over the last few years,
such as the application of numerical methods, are not covered by international standards.
These facts clearly indicate that there is a need for a reference book which would help
cable engineers, researchers, and teachers to understand the theory behind the computation
of cable ratings. The present book was written with this need in mind. It is hoped that
by clearly describing different aspects of the theory and by providing numerical examples
which illustrate the concepts, the book will promote a consistent approach to applying
theory to the computations needed for standard and nonstandard cable installations.
The book is divided into three parts. In the first part, a general theory of heat transfer
is briefly described and, based on these theoretical principles, the steady-state and transient
rating equations are developed. In the second, computation of the parameters required in
rating equations is discussed. Whenever the equations describing parameter calculations
were developed empirically, they are simply reproduced in the text with the explanation of
their origin; if they are developed from basic principles, the full development is presented.
This is avoided only in a few cases where the theory is so complex that it would unnecessarily
cloud the presentation. Complete references are provided for all cases.
The third part contains specialized applications and advanced computational proce-
dures. In particular, cable installations in air requiring the solution of a set of heat transfer
equations are discussed. Also in Part III, an introduction to numerical methods for cable
rating computations is given, and an optimization problem is formulated for the selection
of the most economic conductor cross section.
The book contains a large number of numerical examples which explain the various
concepts discussed in the text. Each new concept is illustrated through examples based on
practical cable constructions and installations. To facilitate the computational tasks, I have
selected five model cables which will be used throughout the book. Three are transmission-
class, high-voltage cables, and two are distribution cables. The model cables were selected
to represent major constructions encountered in practice and are described in Appendix A.
Even though computer programs are now in common use for cable ampacity calcula-
tions, there are merits to performing some computations by hand, if only for the purpose
of checking sophisticated computer software. To this end, I have assembled, at the end
of the book, calculation sheets for steady-state ampacity computations. These sheets can
be used as templates for rating power cables in the most common installations. There is
considerable use of advanced mathematical derivations in the book. Therefore, in order to
emphasize the equations which are later assembled in the computation sheets and which
are important for the rating calculations, boxes are placed around some of the formulas in
the main text.
All equations use SI (metric) units since, with the exception of the United States, this
is common practice around the world (even in the United States, IEEE recommendations
suggest the use of the metric system in engineering computations). Therefore, even for the
cable system peculiar to North American installations (e.g., high-pressure oil-filled cables),
I converted imperial units into the metric system. A conversion table from imperial to
metric units is given at the end of the list of symbols.
Preface xix

While the book discusses a large number of subjects, there are still some topics which
are not addressed. In particular, forced cooling of the underground cables, ampacity com-
putations of dc cables, and rating of cryogenic cables are not considered. The first topic
is well covered in the book Thermal Design of Underground Systems by Weedy (Wiley,
1988). The remaining subjects are too specialized to be included in a general reference
book. The thermal analysis of cable joints is also not discussed, but the methods described
herein could be used for such analysis. The joints have, in the majority of cases, better
thermal characteristics than the other parts of the cable circuit, and thus do not require
special attention from the ampacity point of view. Cable heating during the short circuits
is not discussed in the book because it does not affect cable current-carrying capability in
normal and emergency operating conditions.
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 Acknowledgments

A large part of the material covered in this book was derived from various International
Electrotechnical Commission (IEC) standards and reports dealing with the thermal rating
of power cables. These publications are being prepared by Working Group 10 of the Study
Committee 20A (High Voltage Cables) of the IEC as an ongoing activity. The author is
particularly indebted to Mr. Mark Coates from ERA Technology in Britain, the Convener
of WG 10, who provided a substantial amount of background material used in Chapters
5, 8, and 9 of this book. Frequent discussions with the members of WG 10: Mr. J. Van
Eerde from NKF Kabel in Holland, Dr. B. Harjes of Felten & Guilleaume in Germany, Dr.
A. Orini of Pirelli Cavi in Italy, Mrs. S. Le Peurian of Electricite de France, Mrs. A. Van
Geertruyden from Laborelec in Belgium, and Dr. R. Wlodarski from BBJ-SEP in Poland, as
well as with the cable system designers in Ontario Hydro: Mr. D. J. Horrocks, Mr. J. Motlis,
and Mr. M. Foty, contributed greatly to the development of many procedures described in
the book. In addition, I could not have written this book without an involvement and close
association with several individuals who contributed their ideas and took the time to read the
manuscript. I am particularly indebted to Dr. G. L. Ford and Dr. J. Endrenyi from Ontario
Hydro Technologies (OHT) and Mr. D. J. Horrocks from Ontario Hydro Transmission
Design Department, who have reviewed the entire manuscript and provided several helpful
comments. Discussions with Dr. S. Barrett and Dr. B. Gu of OHT helped to clarify several
issues in the computation of joule losses and the analysis of cables on riser poles. All the
drawings were prepared by Mrs. G. Gostkowski from the drafting unit of OHT.
I would also like to acknowledge the personal encouragement of Mr. M. Bourassa
from the Standards Council of Canada (SCC) and Dr. G. L. Ford from OHT, as well as
the financial assistance of SCC and OHT in supporting my participation in the activities
of WG 10 of the IEC over many years. The Canadian Electrical Association (CEA) and
Ontario Hydro have contributed very substantial funds to the development of the Cable
Ampacity Program (CAP) used extensively in this book. Mr. Jacob Roiz from CEA was
overseeing the development of this software, now used by over 200 users in 33 countries on

xxi
xxii Acknowledgments

five continents. A close association with Dr. T. Rodolakis from CYME Int. helped me to
correct many problems in the program. Dr. Rodolakis also reviewed most of the manuscript.
Finally, but by no means last, I would like to thank my wife Justyna and my son Adam
who supported wholeheartedly this difficult endeavor.

George J. Anders
Toronto, Canada
 symbols

The symbols used in this book and the quantities which they represent are given in the list
below. Following the practice adopted in IEC Standard 287, all cable component diameters
are given in millimeters. When the formulas shown in the text require these dimensions to be
given in meters, an asterisk is added to the symbol. For example, D, denotes the external
diameter of the cable expressed in millimeters, whereas D? denotes the same diameter
expressed in meters.

Ag = cross-sectional area of the armor mm?

A = cost parameter related to condutor cross section (Chapter 13) $/m -mm?
Ay | = area of the wall inside surface for unit length riser m?
A = area of the jacket outside surface for unit length jacket m?
A, = area of the outside wall surface per unit length m?
Ags = equivalent area of the wall per unit length perpendicular to sunrays m7?
Asy = effective radiation area m?

Bo
By
By > = components of impedance due to the steel wires (Section 8.3.3) Q/m
Bs;
Bs
B = factor used in economic analysis of conductor size
CG = electrical capacitance per core F/m
CI = installation cost of a cable $
CL =ceost of losses $
CT =sum of the cost of installation and the cost of losses $
D = demand charge per year $/W - year
D!, = external diameter of armor mm
D, = diameter over armor bedding mm

Xxiii
XXIV Symbols

internal diameter of duct or pipe

external diameter of cable, or external diameter of pipe covering in
pipe-type cable
diameter over insulation
external diameter of duct or pipe
external diameter of metal sheath
diameter over the reinforcing tape
diameter of the imaginary coaxial cylinder which just touches the
crests of a corrugated sheath
diameter of the imaginary coaxial cylinder which would just touch
the outside surface of the troughs of a corrugated sheath = Dj; + 2t,
diameter of the imaginary cylinder which would just touch the in-
side surface of the crests of a corrugated sheath = D,,. —2t,
diameter over core screen
diameter of the imaginary cylinder which just touches the inside
surface of the troughs of a corrugated sheath
fictitious diameter at which effect of loss factor commences
constant used in Chapter 9
exponential integral function
mutual heating coefficient for a group of cables
factor for eddy currrent losses for large segmental conductors
(Chapter 8)
coefficient for belted cables defined in Chapter 9
coefficient for belted cables defined in Chapter 9
factor to take account of the length variations in a cross-bonded
system
auxiliary quantity in cost computations
i = 1, 2, or 3, auxiliary quantities in cost computations
thermal radiation shape factor between two long concentric
cylinders
aspect ratio, G= L/6d
geometric factor for belted cables
constant component of cost unaffected by size of cable (Chapter
13)
geometric factor for SL and SA type cables
Grashoff Number based on 6
intensity of solar radiation
magnetizing force (Section 8.3.1.4) ampere turns/m
thermal conductivity matrix in finite-element analysis
current in one conductor (rms value) A
maximum load on a cable during the first year (Chapter 13) A
constant current applied to cable prior to emergency load A
emergency load current which may subsequently be applied for
time ¢ so that the conductor temperature rise above ambient at the
end of the period of emergency load is @nax
rms conductor current
Symbols

= rms sheath current

= rms armor current PP
>
= charging current in a cable
= highest current of daily load cycle: used as denominator when
determining loss-load ordinates
= the magnitude of the nth harmonic current
= sustained (100% load factor) rated current to attain, but not exceed,
the permissible maximum conductor temperature
= current-carrying capacity for a maximum permitted temperature
rise of 9 —amp (Chapter 13)
= screening factor for the thermal resistance of screened cables
= diameter ratio, K= Dq/D.
= boundary conditions and heat source vector in finite-element
analysis
= coefficient used in Section 9.6.8.1
= depth of laying to cable axis or center of trefoil in steady-state rating mm

= axial depth of burial of a cable in transient rating and in Section

10.6.1
= length of the cable route in economic computations
= length of the riser
ce545
Sf
= distance from the soil surface to the center of a duct bank
= depth of tunnel centerline
= daily load factor
LF,, LF, = ultimate and present load factors (Chapter 13)
LEF = load-loss factor
= cyclic rating factor

= coefficients defined in Section 8.4.10

= coefficients used for calculating cable partial transient temperature

rise (Section 5.2.1)
= number of cables in a group
= economic life of a cable
= number of loaded cables in a duct bank (Section 9.6.5.1)
= period during which the load on a cable grows
= number of phase conductors per circuit
= number of circuits carrying the same value and type of load
= vector of shape coefficients in finite-element analysis
= average Nusselt Number
= cost of | watt hour of energy at the relevant voltage level
= Prandtl Number, one of the air thermal properties

= coefficients defined in Section 8.3.5.3

= total thermal capacitance of a cable

= thermal capacity matrix in finite-element analysis
= elements of two-part thermal circuit (see Section 3.3.2)
XXVi Symbols

thermal capacitance of armor J/m - K

thermal capacitance of conductor J/m -K

thermal capacitance of a duct J/m - K

thermal capacitance offilling in SL type cables, or offilling between
cores of a gas-pressure pipe-type cable J/m -

A
thermal capacitance of dielectric per conductor J/m -

thermal capacitance of first portion of insulation J/m -

thermal capacitance of second portion of insulation J/m -

thermal capacitance of a jacket J/m -

thermal capacitance of oil in pipe-type cable J/m -

thermal capacitance of a pipe J/m -

thermal capacitance of sheath and reinforcement J/m -

coefficients taking into account the increase in load, the increase in

cost of energy over the N years, and the discount rate
alternating current resistance of conductor at its maximum operat-
ing temperature
resistance of armor
resistance of sheath
dc resistance of conductor at maximum operating temperature
de resistance of conductor at 20°C
ac resistance of conductor before application of emergency current
ac resistance of conductor with sustained application of rated cur-
rent /p, 1.e., at standard maximum permissible temperature
ac resistance of conductor at end of period of emergency loading
mean ac resistance of conductor during economic life
resistance of reinforcing tape
modified Rayleigh Number
Reynolds Number
cross-sectional area of conductor
economic conductor cross section
total thermal resistance of a cable from conductor to outer surface
operating time at maximumjoule loss (Chapter 13)
YBeasa We elements of equivalent thermal circuit
thermal resistance per core between conductor and sheath
thermal resistance between sheet and armor
thermal resistance of external serving
thermal resistance of surrounding medium (ratio of cable surface
temperature rise above ambient to the losses per unit length)
external
thermal
thermal
thermal
resistance
resistance
resistance
between
of the duct
in free air, adjusted
cable and duct (or pipe)
(or pipe)
thermal resistance of the medium surrounding the duct (or pipe)
additional external thermal resistance caused by heating from other
for solar radiation

A
AK
cables in a group 2
apparent thermal resistances used to calculate cable partial transient
temperature rise (Section 5.2.1)
Symbols XXVii

a = thermal resistance of filling between cores of SL type cable and of

gas-pressure pipe-type cable
Is = thermal resistance of the oil in a pipe-type cable
THD = total harmonic distortion (Section 7.3)
Uo = voltage between conductor and screen or sheath

= constants used in Section 9.6.4.1

Vax = local air velocity

Wa = losses in armor per unit length
We = losses in conductor per unit length
WwW, = total J*R power loss of each cable
Win = amount of heat generated within the body
Wa = dielectric losses per unit length per phase
Ww, = losses dissipated by cable k
Ww, = losses dissipated in sheath per unit length
Woot = solar radiation absorbed by the riser surface per unit length
W(s+a) | = total losses in sheath and armor per unit length
Wror = total power dissipated in the trough per unit length
W, = total power dissipated in the cable per unit length
Weond,s—o = heat conduction rate from the wall inner surface to its outside sur-
face per unit length
Weonv,w = Natural convection heat transfer rate between the wall inside surface
and the air per unit length
Weonv.s = Natural convection heat transfer rate between the jacket outside
surface and the air per unit length
Weonv.o = natural convection heat transfer rate between the wall outside sur-
face and atmosphere air per unit length
Wiad.s—w = thermal radiation heat transfer rate between the wall inner surface
and the jacket outside surface per unit length
W,ad.o—sur
= thermal radiation heat transfer rate between wall outside surface
and surrounding objects per unit length
Xx = reactance of sheath (two-core cables and three-core cables in trefoil)
X| = reactance of sheath (cables in flat formation)
UG = mutual reactance between the sheath of one cable and the conduc-
tors of the other two when cables are in flat formation
s¢ = coefficient used in Section 9.6.4.1
Y,..Yo3. = scaled ordinate in (load) graph used in calculation of cyclic rating
factor
a = increase in load per year
a = length and width of square cable tunnel
do = shortest minor length in a cross-bonded electrical section having
unequal minor lengths
OMe = coefficients used for calculating cable partial transient temperature
rise l/s
b = increase in cost of energy per year, not including the effect of in-
flation
XXVili Symbols

¢ = demand charge escalation factor

Cc — distance between the axes of conductors and the axis of the cable for
three-core cables (= 0.55 r; +0.29 t for sector-shaped conductors) mm
c = specific volumetric heat capacity of the material (Chapters 2 and
10) J/m?
Cp = Specific heat at constant pressure, J/kg -K
d =mass density, kg/m?
d = mean diameter of sheath or screen mm
d = mean diameter of sheath and reinforcement mm
dy = dry density of soil kg/m?
d> = mean diameter of reinforcement mm
d, |= mean diameter of armor mm
d, = external diameter of belt insulation mm
d. = external diameter of conductor mm
d.; = external diameter of conductor shield mm
d-m = minor diameter of an oval conductor mm
d-m = major diameter of an oval conductor mm
d. = external diameter of equivalent round solid conductor having the
same central duct as a hollow conductor mm
d® =conductor diameter of equivalent single-core cable having the same
losses as the three-core cable (Section 3.3.1) mm
dy = diameter of a steel wire mm
d; |= internal diameter of hollow conductor mm
dy = major diameter of screen or sheath or an oval conductor mm
d,, = minor diameter of screen or sheath of an oval conductor mm
dpx = distance from center of pth cable whose rating is being determined
to an adjacent cable k mm
dig = distance from center of pth cable, whose rating is being determined,
to the image of an adjacent cable k (see Fig. 5.3) mm
drm = mean diameter of the reinforcing tape mm
d,. = diameter of an equivalent circular conductor having the same cross-
sectional area and degree of compactness as the shaped one mm
f =system frequency Hz
fo = factor by which thermal resistances are reduced for cables in trefoil
g = coefficient used in Section 9.6.8.1
g =acceleration of gravity m/s?
g, = coefficient used in Section 8.4.4
h = heat dissipation coefficient W/m2 - K>/4
hy = ratio 1, /TIp
Acony= total convection coefficient W/K - m?
h, =natural or forced convection coefficient at wall outside surface WI/K - m2
h, = radiation heat transfer coefficient W/K - m?
h; =natural convection coefficient at jacket outside surface W/K - m?
h, = natural convection coefficient at wall inside surface W/K -m
h® = element conductivity matrix in finite-element analysis
Symbols XXix
I = time h
i = discount rate
ko = factor used in the calculation of hysteresis losses in armor or rein-
forcement (Section 8.3.6.3)
k = ratio of cable, pipe, or duct external-surface temperature rise above
ambient to conductor temperature rise above ambient under steady
conditions
k® = element heat source vector in finite-element analysis
ky = value of k for a cable in a group
Kair ==air thermal conductivity W/K -m
k, | = lay-length factor of the wires in layer n of a stranded conductor
k, = factor used in calculating x, (proximity effect)
kee = factor used in calculating x, (skin effect)
l = length of a cable section (general symbol) m
la = length of lay of a stranded wire mm
€r = length of lay of a tape mm
In = natural logarithm (logarithm to base e)
m =(oiRp10w
n = number of conductors in a cable
idles = number of wires in layer n of stranded conductor
a = number of armor wires
ny = number of moisture barrier metallic tapes in pipe-type cable
p. = factor for apportioning the thermal capacitance of a dielectric (long-
duration transients—Section 3.3.1)
P = factor for computation of load loss factor (Chapter 13)
p* =factor for apportioning the thermal capacitance of a dielectric
(short-durations transients—Section 3.3.1)
Po = part of the perimeter of the cable trough which is effective for heat
dissipation (Section 10.6.2) m
p’ = factor for apportioning the thermal capacitance of cable coverings
Pa = factor for apportioning thermal capacitance of dielectric when cal-
culating transient caused by dielectric loss

P2 = coefficients used in Section 8.3.3.5

q2
q = heat flux W/m?
q’ =element capacity matrix in finite-element analysis
r = ratio Omax/Or(&)
ihe = mean radius of the armor mm
ee = external radius of the conductor mm
ike = mean radius of the sheath mm
rj = circumscribing radius of two- or three-sector shaped conductors mm
S = axial separation of conductors mm
S| = axial separation of two adjacent cables in a horizontal group of
three, not touching mm
t = time from start of application of heating, a general symbol for time,
usually in seconds (t = 36007) s
XXX Symbols

t = insulation thickness between conductors mm

t = insulation thickness between conductors and sheath mm
to = thickness of the bedding mm
b = thickness of the serving mm
ti = thickness of core insulation, including screening tapes plus half the
thickness of any nonmetallic tapes over the laid up cores mm
te = armor thickness mm
ty = thickness of the jacket mm
(hs = thickness of the sheath mm
ty = thickness of moisture barrier metallic tape in pipe-type cables mm
= (2L/D,) in Section 9.6.1
= (Le/7r;) in Section 9.6:5
Xp = argument of a Bessel function used to calculate proximity effect
xe = argument of a Bessel function used to calculate skin effect
ayy = sides of duct bank/backfill (y > x) (Section 9.6.5.1) mm
Yp = proximity effect factor
Vs = skin effect factor
a = temperature coefficient of electrical resistivity of conductor mate-
rial VK
a(t), B(t) = attainment factors for the conductor to cable surface and cable sur-
face to ambient temperature rises, respectively
A129 = temperature coefficient of electrical resistivity at 20°C per Kelvin I/K
ao = wall surface absorptivity to solar radiation
B = coefficient of volumetric expansion Ka
B = reciprocal of temperature coefficient of electrical resistivity atO°C K
B = angle between axis of armor wires and axis of cable
y = angular time delay
Vn = ratio of the nth harmonic current to the fundamental (Section 7.3)
Ve = temperature gradient vector in finite-element analysis
Ay i é :
ne = coefficients used in Section 8.4.5

AT = time step in transient analysis S

5 = soil thermal diffusivity m?/s
m) = derating factor for cable systems containing harmonics (Section
Du)
ny) = air gap thickness between the cable and the wall m
50 = equivalent thickness of armor or reinforcement mm
5 = thickness of metallic screens in screened-type cables mm
tan 6 = loss factor of insulation
€ = relative permittivity of insulation
Ew = emissivity of the wall inner surface
Ey = emissivity of the jacket outside surface
&5 = emissivity of the wall outside surface
n = moisture content of soil in percent of dry weight
0 = maximum operating temperature of conductor Cc
6; = conductor temperature at commencement of transient 1e:
Symbols XXxi

On = mean operating temperature of a conductor during economic life

6(t) = transient temperature rise of conductor above ambient, without cor-
rection for variation in conductor loss
6.(t) = transient temperature rise of conductor above the outer surface cable
O-a(t) = transient temperature rise of conductor above outer surface of the
cable due to dielectric loss
Oa(t) = transient temperature rise of conductor above ambient due to di-
electric loss
Oae(t) = transient temperature rise of cable surface above ambient due to
dielectric loss
0.(t) = transient temperature rise of outer surface of acable (or hottest cable
in a group of similarly loaded cables) above ambient temperature
O,(t) = transient average temperature rise of oil in a pipe-type cable above
the outer surface of the pipe
Or(t) = conductor temperature rise above ambient at time ¢ after application
of current Jz, neglecting variation in conductor resistance
Or(i) = 6(t) when the magnitude of the step function current is the sustained
(100% load factor) rated current, and i is expressed in hours
Or (OO) = value of Op in the steady state, i.e., the standard maximum permis-
sible temperature rise
8,(t) = conductor transient temperature rise above ambient corrected for
variation in conductor resistance with temperature
6(co) = conductor steady-state temperature rise above ambient
0. = temperature of the external surface of cable or duct
Oca, = air temperature in the gap (Chapter 10)
Omax = Maximum permissible temperature rise above ambient: 1) for a
daily cyclic load, 2) at end of period of emergency loading
On = mean temperature of medium between a cable and duct or pipe
A = average temperature at the wall outside surface (Chapter 10)
0s = average temperature of the jacket outside surface (Chapter 10)
0s = sheath temperature
Gy, = average temperature of the wall inner surface (Chapter 10)
Aé = permissible temperature rise of conductor above the ambient tem-
perature
A@i = factor to account for dielectric loss for calculating 74 for cables in
free air
A@,; = factor to account for both dielectric loss and direct solar radiation
for calculating 7;* for cables in free air
A@duct = difference between the mean temperature of air in a duct and am-
bient temperature
Aé@, = difference between the surface temperature of a cable in air and
ambient temperature
A@,, = temperature rise of the air in a cable trough
h1,A2 = ratio of the total losses in metallic sheaths and armor, respectively,
to the total conductor losses (or losses in one sheath or armor to the
losses in one conductor)
XXxii Symbols

dM, = ratio of the losses in one sheath caused by circulating currents in

the sheath to the losses in one conductor
i =ratio of the losses in one sheath caused by eddy currents to the
losses in one conductor
/ = loss factor for the middle cables
dim
ri/ = loss factor for the outer cable with the greater losses
/12 = loss factor for the outer cable with the least losses
= relative magnetic permeability of armor material
= loss-load factor of a load cycle
= longitudinal relative permeability
= relative permeability
= transverse relative permeability
= peremeability of nonmagnetic systems
= viscosity
= air kinematic viscosity
= conductor electrical resistivity at 20°C
= thermal resistivity of earth surrounding a duct bank/backfill
= electrical resistivity of metallic material
= thermal resistivity of concrete used for a duct bank
ROR
AR
ROK
= thermal resistivity of metallic screens on multicore cables
= mean ac resistivity of conductor during economic life
= thermal resistivity of soil
= thermal resistivity of material
= dielectric thermal resistivity
= absorption coefficient of solar radiation for the cable surface
= reflectivity of the wall inner surface
= reflectivity of the jacket outside surface
= angular frequency of system (27 f)
Symbols XXXiii

Table of North American Cable Sizes!

Cross-Sectional Area Cross-Sectional Area

Conductor Square Square Conductor Square Square

Size, AWG kemil2 inches millimeters Size, kcmil inches millimeters

8 16.51 0.01297 8.367 250 0.1964 126.7

6 26.24 0.02061 13.30 350 0.2749 hes)
4 41.74 0.03278 21.15 500 0.3927 253.4
2) 66.36 0.05212 33.62 750 0.5891 380.0
I 83.69 0.06573 42.4] 1000 0.7854 506.7
1/0 105.6 0.08291 53.49 1250 0.9818 633.4
2/0 133e1 0.1045 67.43 1500 1.178 760.0
3/0 167.8 0.1318 85.01 1750 1.374 886.7
4/0 211.6 0.1662 107.2 2000 1.571 1013
2250 1.767 1140
2500 1.964 1267
3000 2.356 1520
3500 2.749 1774
4000 3.142 2028

! Other useful conversions are : 1 in = 0.0254 m, | ft = 12 in = 0.3048 m.

2 Thousand circular mils (kcmil). A “mil” is a 1/1000 in, and circular mils represent the area
of an equivalent solid rod having a diameter expressed in mils. To convert mils to milli-
meters, the mil value is multiplied by 0.0254.
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7
PART | MODELING
 ‘
Cable Constructions
and Installations

1.1 INTRODUCTION

Underground cables are far more expensive to install and maintain than overhead lines. The
greater cost of underground installations reflects the high cost of the equipment, labor, and
time necessary to manufacture the cable, to excavate and backfill the trench, and to install
the cable. Because of the extra expense, most underground installations are constructed in
congested urban areas and as leads from generating plants and in substations. The large
capital cost associated with cable installations also makes it necessary that particular care
be applied in selecting the proper cable type and size to serve the load for the life of the
installation.
Information about the maximum current-carrying capacity which a cable can tolerate
throughout its life without risking deterioration or damage is extremely important in power
cable engineering and operation. Ampacity values are required for every new cable installa-
tion, as well as for cable systems in operation. With some underground transmission cable
circuits approaching the end of their design life, the development of a systematic method
for determining the feasibility of extending cable life and/or increasing current ratings is of
paramount importance.
The capacity of a transmission line is normally given in MVA (megavolt-amperes).
MVA is made up of two components, namely, the MW (megawatt) component, which
represents real power and is available to do work, and MVAR (megavolt-ampere-reactive),
the reactive component, which is present in the system due to its inductive and capacitive
elements, but cannot be utilized to produce work. Cable rating studies usually involve
computation of the permissible current flowing in the conductor for a specified maximum
operating temperature of the conductor. The current causes the cable to heat, and the limit of
loading capacity is determined from the acceptable conductor temperature. Occasionally,
4 Part I m»Modeling

the current value may be given, and the studies involve calculation of the temperature
distribution within the cable and in the surrounding environment.
The cable must be able to carry large amounts of current without overheating, and
must maintain an acceptable voltage profile. Cable heating presents one of the major
problems associated with underground lines. While it is relatively easy to dissipate the heat
generated by the flow of current through the conductor in overhead lines, the heat generated
by losses in underground systems must pass through the electrical insulation system to the
surrounding earth, and both the insulation system and the earth represent an obstacle to
heat dissipation. Because the maximum temperature at which conductors can operate is
limited by the conductor electrical insulation system and because these systems conduct
heat poorly, the result is a much larger conductor than would be required for an overhead
line of equal capacity.
The cable must carry the load currents without overheating, and also without producing
excessive voltage drop. This voltage drop is known as /Z drop after the formula used to
determine it, but in underground systems this is rarely a limiting factor.
In addition to the normal loads, a transmission system is customarily designed to carry
overloads due to equipment and line outages or other abnormal system conditions for limited
periods of time which may, nevertheless, be often 10 h or longer. Operation at higher than
normal temperatures is permitted during these overload periods, and the response of the
cable system to these overloads is evaluated in transient rating computations.
The rating of an electric power cable will depend on its construction and the method
of installation. There is a great variety of cable constructions currently used around the
world. Also, installation conditions vary widely. The Association of Edison Illuminating
Companies (AEIC) in the United States issues specifications and guides for various cable
constructions and installations. The current list of these specifications and guides is given
in the References section at the end of this chapter.
To help the reader understand the terminology and the computational process adopted
for cable ratings, in the following we will review briefly major cable components and
installation methods. This review is not exhaustive, and its aim is to focus on those cable
parameters and the installation requirements which affect thermal rating of the cable circuit.
We will close this chapter with a historical note on standardization efforts of cable ampacity
computations.

1.2 CABLE COMPONENTS

Every electric power cable is composed of at least two components: (1) an electrical con-
ductor, and (2) the conductor insulation which prevents direct contact or unsafe proximity
between conductor and other objects. The need to provide adequate electrical insulation
which will also permit heat to be conducted and dissipated poses technical challenges at
higher voltages. The problem of heat conduction is further aggravated by the fact that, in
the majority of power cables, the primary electrical insulation has to be protected against
mechanical, electromechanical, and chemical damage. This protection is provided by ad-
ditional concentric layers over the insulation. The most common form is a metallic sheath
which is often covered with a nonconducting material called an oversheath or jacket. Some
cables do not have a metallic sheath, but only a nonconducting jacket. Some cables have
concentric neutral wires instead of sheath. These wires serve mainly as a return path for
Chapter 1 = Cable Constructions and Installations 5

a neutral current or for short-circuit current during system faults. Submarine and special-
purpose cables usually have an additional metallic layer called armor. Armor may or may
not be further protected by a special outer covering. Each of these components is briefly
described below.

1.2.1 Conductors

Two materials are usually used in cable conductors: copper and aluminum. Since
the price of aluminum is less than 50% of the price of the copper, savings are possible
by utilizing aluminum. These savings are small since the cost of the conductor is only a
fraction of the installed cost of the system. Some of the savings in metal cost are offset
by the cost of additional insulation and accessories needed since the cross-sectional area
of aluminum required is approximately 1.5 times that of copper to carry the same load,
resulting in a cable with a larger diameter.
Conductor construction also plays an important role in rating computations. This
topic is discussed in Chapter 7, and we will restrict ourselves here to showing only typical
conductor constructions currently made by cable manufacturers (see Fig. 1.1).
An important parameter in rating computations is the conductor cross section. Con-
ductors are designed to conform to a range of nominal areas in graduated steps. In practice,
the following two ranges are used:

1. In North America and in some South American countries, cable cross section is
based on American Wire Gauge (AWG) up to 4/0, that is, 107 mm/?, and larger
sizes are in thousand circular mils (kcemil). A “mil” is a 1/1000 in, and circular
mils represent the area of an equivalent solid rod having a diameter expressed in
mils. To convert mils to millimeters, the mil value is multiplied by 0.0254.

2. Inall other countries, the metric system is used to define conductor area (mm7”) and
the dimensions specified in IEC Publication 228 (1978, 1982) are adopted.

In practice, the conductors are not manufactured to precise areas. Instead, manufac-
turers adjust their wire sizes and manufacturing processes to meet a specified maximum
resistance rather than area. The IEC Standard 228 (1982) specifies a single maximum dc
resistance for each size of conductor of a given material. American specifications adhere to
a pattern of a “nominal” resistance, together with a tolerance to provide a maximum resis-
tance for a single-core cable and a further tolerance for multicore cable. These tolerances
also vary with conductor classification.
The nominal values of resistances used in North America as well as those specified in
the IEC Standard 228 are quoted in Chapter 7.
In general, the larger the conductor cross section, the larger is the current-carrying
capability of the cable. For conventional cables with natural cooling, we can state that
doubling cable ampacity requires approximately a fourfold increase in conductor cross
section. For example, increasing transmission capacity from 300 to 600 MVA in a 230 kV
cable circuit requires a change of conductor cross section from 800 to 3200 mm? Cu or
4800 mm? Al. Since with the present technology the manufacture of a cable with a conductor
cross section larger than 3000 mm? is practically impossible, special cooling facilities or
several cables per phase are required when large transmission capacity 1s needed.
 Part I m Modeling

Figure 1.1 Typical conductor constructions contrasting compact designs (right) with con-
ventional stranding (left) (Barnes, 1964).

1.2.2 Insulation

The purpose of the electrical insulation is to prevent the flow of electricity from the
energized conductors to the ground or to an adjacent conductor. The insulation must be able
to withstand the electrical stresses produced by the alternating voltage and any superimposed
transient voltage stress on the conductor without dielectric failure and causing a short circuit.
There are many materials that have acceptable insulating properties. The most com-
monly used and the least expensive is the air surrounding the conductors on overhead
lines. There are, however, only a few materials available that can be used in underground
or enclosed applications to insulate conductors from ground and from each other. These
are: oil-impregnated paper tapes, solid insulations such as polyethylenes, and ethylene—
propylene rubber, and the more recently developed polypropylene (PPL) and compressed
gas insulation (such as sulfur hexafluoride or SF¢) systems.
When paper and the solid dielectric insulations are subjected to alternating voltage,
they act as large capacitors and charging currents flow in them. The energy required to
effect the realignment of electrons each time the voltage direction changes (i.e., 50 or 60
Chapter 1 m Cable Constructions and Installations I

times a second) produces heat and results in a loss of power which is called dielectric
loss, and should be distinguished from conductive loss. The magnitude of the required
charging currents is a function of the dielectric constant of the insulation, the cable length,
the dimensions of the cable, and the operating voltage. Charging current produces also a
resistive component of the losses in the insulation but, for ac applications, they are extremely
small compared to the capacitive component.
Various cable types used in power systems are often distinguished on the basis of the
type of insulation they have. For example, paper-insulated cables are classified as:

1. Mass-impregnated solid type where the insulation is saturated with impregnation

fluid before shipment.
2. Low-pressure fluid-filled (LPFF) with a filling of thin oil at a pressure of about
1 atm (see Appendix A for an example of such a cable); the oil pressure is maintained
by reservoirs feeding the cables along the route.
3. High-pressure fluid-filled (HPFF) cables, where the paper-insulated cores are in-
stalled in a steel pipe and the pipe is filled with insulating oil under a high pressure
of 1.38 MN/m? (200 psi). The pressure is maintained by a special pumping plant
consisting of an oil reservoir, controls, and a pump. The pump is activated auto-
matically when the oil pressure drops below a specified value.

Both LPFF and HPFF cables are impregnated during their manufacturing. A newer insulat-
ing material used in high-voltage cables is paper-polypropylene—paper (PPL). It combines
excellent insulating properties of paper with the low dielectric losses of polypropylene.
Other cable types are those using solid insulation (extruded cables) or using gas, usually
SF6, as an insulating medium.
The insulation type has a strong effect on cable rating. From a thermal point of view,
a good insulating material should have low thermal resistivity and should result in low
dielectric losses. A cross-linked polyethylene (XLPE) is a good example of such material.
It has primarily been used in low- and medium-voltage cables. However, recently, an
increasing number of manufacturers are producing high- and extra-high-voltage cables
with extruded insulation.
All modern electric power cables are constructed with semiconducting screens around
the conductor and around the insulation. For thermal calculations, these screens are con-
sidered to be a part of the insulation.

1.2.3 Sheath/Concentric Neutral Wires

Metallic sheaths are essential for paper-insulated cables to exclude water from the
insulation and to retain impregnating fluid in LPFF cables. For extruded insulations, there
is not such an obvious need; however, many early extruded cables without metallic sheaths
developed problems with integrity of the insulation. In addition, there are safety concerns
related to unshielded cables. Now IEC 502 (1983) requires that such cables of rated voltages
above | kV should have a metallic covering.
When a solid sheath is used in the cable construction, it is usually made of lead or
aluminum. Lead sheath, especially for large cables, may require a reinforcing metallic tape.
Aluminum sheaths, on the other hand, are lighter and, for additional flexibility, are often
corrugated. In some special constructions, a corrugated copper sheath may be used. In
 Part I m Modeling

addition to its insulation protection function, the sheath is used as the cable component to
carry neutral and/or fault current to earth in the event of an earth fault on the system. Some
cables may not have a solid sheath, but instead may be constructed with concentric neutral
wires to carry the fault current. The wires are usually made of copper and in some instances
of aluminum.
Because of safety considerations, metallic shields are always grounded in at least
one place. The nature of earthing of metallic shields has a profound effect on cable rat-
ing computations. For three-phase systems composed of single core cables with metallic
sheaths/concentric neutral wires, the bonding arrangement and the thermal resistivity of the
trench fill are the most important factors influencing cable rating which can be controlled
by the owner of the circuit. This topic is discussed briefly in Section 1.3.2 and in more
detail in Chapter 8.

1.2.4 Armor

Protective armor is usually made of steel wires or tapes. Steel construction, when
applied to single-core cables, may result in high magnetic hysteresis and circulating current
losses which reduce cable rating. In order to reduce magnetic losses, for these types of
cables, nonmagnetic materials such as aluminum or copper are preferably used.
The use of armor wires on cables with lead sheaths, installed in three-phase systems at
close spacing, causes additional sheath losses because the presence of armor wires reduces
sheath resistance (sheath and armor are connected in parallel), and the losses are largest when
the sheath-circuit resistance is equal to its reactance. Without armor wires, the reactance
of the sheath is always very much smaller than the resistance. To minimize this increase
in losses, armor wires made of high-resistance material such as copper-silicon—manganese
alloy are sometimes used. When the cables are spaced further apart, the reactance increases.
In this case, low-resistance armor can be used (e.g., aluminum alloy) because the combined
resistance is so much less than the reactance that the losses are reduced. Steel armor is
mostly used in submarine cables, and the rating of these cables is discussed in detail in
Section 8.3.

1.2.5 External Covering

Presently, most power cables are manufactured with external protective coverings.
Usually, these are extruded over the sheath or armor. Polyethylene (PE) or polyvinyl
chloride (PVC) are the materials used most often. On armored cables, compounded jute
or fibrous materials are sometimes used as an armor serving. Armored cables usually
have an additional nonconducting layer installed between the metallic sheath and the
armor. This layer, called armor bedding, is usually made of the same material as the
armor serving. The external covering provides additional restriction to the heat trans-
fer from the conductor, and therefore reduces the cable rating. The thermal resistance
of the cable external covering will depend on the material selected. Polyethylene has
the best thermal conductivity from all of the materials used for this purpose. Compu-
tation of the thermal resistances of various nonconductive layers is discussed in Chap-
ter 9.
Figure 1.2 shows an array of cross sections of various fluid-filled cables encountered
in practice.
Chapter 1 m Cable Constructions and Installations 9

(e) (f) (g)

Figure 1.2 Cross sections of low-pressure fluid-filled cables. (a)—(d) Single core cables.
(e)—(g) Three-core cables (from King and Halfter, 1983).

1.3 CABLEINSTALLATIONS
1.3.1 Laying Conditions

Insulated power cables are either installed underground or in air. It is important to

note that the ambient temperature used in the computation of cable ratings refers to climatic
conditions and the mode of cable installation—in the ground, in air, in an open space or in
a building, or under water. Typical cable laying conditions are briefly reviewed below.
1.3.1.1 Underground Installations. |The most common method for installing power
cables underground is to lay them directly in the soil at a depth of about | m (3-4 ft).
Typical installation configurations of a three-phase circuit composed of single-core cables
laid directly in the soil are shown in Fig. 1.3. The arrangement in Fig. 1.3a is called

& So RS ee
(a) (b) (c)

Figure 1.3 Typical installations of directly buried cables.

10 Part I » Modeling

a triangular or trefoil configuration, and the remaining two arrangements are called flat
configurations. In the arrangements shown in Fig. 1.3a and b, the cables are touching,
while in the last configuration, the phases are separated by a distance e. Separation of the
phases improves the heat dissipation process; however, in some cases, this arrangement
produces increased power losses, as discussed later in this section.
Occasionally, when it is important to achieve the highest possible current ratings, cables
installed underground are located in an envelope of a material characterized by better thermal
heat conduction than the native soil. This additional material is called thermal backfill. A
good backfill material can have a thermal conductivity two or more times greater than the
native soil. Figure 1.4 shows a typical installation of a cable circuit in a thermal backfill.
The effect of thermal backfill on cable ampacity is discussed in Chapter 9.

Native soil
Pg = 1.2 °C m/W

1.2m

0.65 m
Se
~
Seen
|EET
CF
a

<I
——__—_——— 0.75 m ———______®
Figure 1.4 Cables installed in a thermal backfill.

In urban areas, there is often a need to install a large number of cable circuits in one
trench. In such cases, a special concrete structure is built with uniformly spaced holes
to house the cables. Each hole is usually lined with a plastic tube, most often made of
polyethylene or PVC. Such a construction is called a duct bank. This construction permits
installation, and subsequent removal when necessary, of a large number of circuits. For
example, in Toronto, it is not uncommon to find 40 cables installed in one duct bank. Very
often, several cables may be installed in one duct. Figure 1.5 shows a typical installation
of two circuits in a duct bank. Several ducts are empty in this case, permitting installation
of additional circuits in the future.
The principal method of transmitting power at higher voltages in the United States is
through the use of pipe-type cable. This cable type makes use of three cables insulated by
layers ofoil-impregnated paper installed within a coated steel pipe which is then filled with
an insulating oil or gas. The heat generated within the cable is conducted by the oil or gas
and the steel pipe to the surrounding soil. The current rating of the pipe-type cable can
Chapter 1 m Cable Constructions and Installations 11

Groundsurface, ®amb=20icG
aaiae
Native soil
Pg beh oS m/W

Ground cable 2.03 m

Concrete ‘
Po = 0.6 °C m/W

0.1 m Transite A
see 0.71 m

Figure 1.5 Two circuits in a duct bank.

be increased by forced cooling, using heat exchangers and pumps to circulate and cool the
oil within the pipe. Pipe-type cables usually have a lower current-carrying capability than
directly buried cables with the same conductor size and operating at the same voltage level
because of the close proximity of the cores and the losses generated in the steel pipe. An
advantage of HPFF cable is the longer lengths of insulated cores that can be placed ona reel,
and therefore a fewer number of joints. There is also some advantage in using HPFF versus
self-contained cables in congested city areas. For the HPFF cable, only a short stretch of
roadway needs to be opened up at a time. After sections of a pipe are welded together, the
cables are pulled through the pipe at a later date. Figure 1.6 shows a typical cross section
of a cable circuit in a pipe.

1.3.1.2 Cables Installed in Air. Cables are often installed in air, outdoors and
indoors, and it is usually necessary to provide a cleating system to support such cables. For
example, power delivery systems frequently consist of acombination of overhead lines and
underground cables. In most lower voltage installations, the underground cable system is
connected to the overhead line through a short section of cable located in a protective riser.
The cable is secured to the pole, and the riser is provided for mechanical protection. The
current-carrying capacity of the composite system is limited by that segment of the system
that operates at the maximum temperature. Very often, the riser—pole portion of the cable
system will be the limiting segment.
Another type of cable installation in air can be found in an electric power generation
plant and in distribution systems where a typical arrangement could consist of a 3 in deep,
24 in wide metal trough or raceway containing anywhere from to 20 to 400 randomly
arranged cables ranging in size from #12 AWG to 750 kcmil. We refer to these installations
as cables in trays. This array of cables is usually secured along the cable tray to prevent
shifting should additional cables be pulled into the tray. In many cases, especially in nuclear
12 Part I mw
Modeling

Backfill

Coal-tar or plastic Insulating oil

protective coating Skid-wires (D-wires)
(steel, brass, zinc, etc.)

Steel Moisture seal

pipe
Screen
Oil-impregnated
kraft-paper wrapping
Inner protective
coating Conductivetape
Insulated-segment
compressed stranded conductor

Figure 1.6 Cross section of pipe-type cable in trench (Weedy, 1988).

power plants, the trays are covered with fire protection wrap around the raceway. Because
of the very strong mutual heating effects, the ampacity of cables in trays is usually lower
than that of cables installed in free air.
Yet another type of installation of cables surrounded by air are cables installed in
tunnels and shafts. Cables are sometimes installed in tunnels provided for other purposes.
In generating stations, short tunnels are often used to convey a large number of cable
circuits. Long tunnels are built or existing tunnels adapted solely for the purpose of carrying
major EHV transmission circuits which for various reasons cannot be carried overhead.
River crossings are obvious cases where tunnels would be used either for technical or
environmental reasons. The costs of such installations are considerable, and it is desirable
to optimize as far as possible the current-carrying capacity, groupings, and number of
circuits to be installed to meet a given transmission capacity.
Rating computations of cables in air are generally more complex than for underground
installations. Details of these computations are discussed in Chapters 9 and 10.

1.3.2 Bonding Arrangements

In a three-phase system, sheaths or concentric neutrals of power cables are always
bonded and grounded at least at one end. If the sheath is in the form of a metallic tube, eddy
Chapter 1 m Cable Constructions and Installations 13

currents will flow in it, producing additional heat which has to be dissipated from the cable
surface. If the sheaths are bonded at two or more points, circulating currents will flow in
them, producing additional losses and reducing cable ampacity. This current will increase
with increasing phase separation. Circulating current losses are usually much greater than
the losses induced by eddy currents. Therefore, from an ampacity point of view, single-point
bonded installations or installations utilizing special bonding arrangements are preferred.
These arrangements are referred to as single-point bonding or cross bonding.
In single-point bonded systems, the cable sheaths are bonded and directly grounded
at one end, with the remote end sheaths grounded through a voltage-limiting device. The
cable system must be designed to limit the sheath standing voltage to a locally permitted
level, and these values determine the maximum length of the cable circuit.
In cross-bonded systems, the cable run is divided into groups, each consisting of three
approximately equal sections. The cable sheaths are cross bonded so that induced voltages
are canceled, and are bonded together at the end of each group of three sections. In addition,
in some instances, the cables are physically transposed to enhance the cancellation of the
induced sheath voltages. However, it is an expensive procedure, and is therefore applied
mainly in installations with a conductor cross section above 500 mm?. Figure 1.7 shows
typical bonding and grounding arrangements of a transmission circuit composed of three
single-phase cables.
The computation of eddy and circulating current losses and the effect of bonding ar-
rangement are discussed in detail in Chapter 8. Various bonding arrangements are discussed
in the ANSI/IEEE Standard 575 (1988).

1.3.3 Forced Cooling of Cable Circuits

As mentioned earlier, in order to increase the capacity of cable circuits, special cooling
arrangements are sometimes made. The selection of the cooling method depends on the
cable type, length and trench profile, and limiting temperatures, and is related to the need to
provide adequate cooling of joints and terminations. The following are the most common
cooling arrangements: (1) flow of fluid in the central duct in the conductor (LPFF cables),
(2) forced cooling of high-pressure fluid-filled cables, (3) internal water cooling of a cable
(cable installed in a pipe filled with water), and (4) external water cooling (water pipes
installed externally to the cable). Figure 1.8 shows the permissible steady-state current
rating for a2500mm? copper conductor, fluid-filled cable system. The maximum conductor
temperature is 85°C, soil thermal resistivity is | K-m/W, and the cables are located 1 m
below the ground surface.
Computation of cable ratings in the presence of forced cooling is quite involved and
will not be discussed in this book. Some aspects of forced cooling are discussed by Weedy
(1988) and in several publications (EPRI, 1984; Aabo et al., 1995; Notaro and Webster,
1971; Flamand, 1966; Buckweitz and Pennell, 1976; CIGRE 1976, 1986).

1.4 HEAT SOURCES IN POWER CABLES

Depending on cable construction and installation conditions, there may be a number of

sources generating heat inside the cable. The heat produced inside a cable is referred to as
cable heat loss. In general, there are two types of losses generated in a cable:
14
Part I w Modeling

S
Installation 1
o—
-"
hex |(ta
ec
tae
tra
lle
ee
oe .
Installation 2 o—
—e
iy ee se
Installation3
—~¢
eH
e
ee M4
Installation 4

=e
Pe —e
Installation 5 a
$

& -
Installation 6
’
‘Ie pe
Figure 1.7 Installation of single-core cable circuits with bonding and grounding arrange-
ments of cable sheaths.

1. Current-dependent losses
2. Voltage-dependent losses.

The losses belonging to both groups are briefly discussed below and treated in detail in later
chapters of this book.

1.4.1 Current-dependent Losses

The current-dependent losses refer to the heat generated in metallic cable components,
namely, the conductor, sheath or screens, armor, and pipe. Sheath, armor, and pipe losses
are only generated, of course, when the cable is equipped with these components.
Chapter | m Cable Constructions and Installations 15

GVA
4

cide geniebse
Figure 1.8 The effect of forced cooling on rating
of a 2500 mm? LPFF cable. (a) Natural cooling
0200400600800
(a)
U
1000kV
(b) (C)
in the ground. (b) Forced cooling with the wa-
ter pipes located outside the cables. (c) Forced
cooing with water in pipes containing the cables. O O
isequal
The
to10°C
(from
Balletal.,1977).
temperature of water intake and ambient soil
SHES ever’ @® ©

Conductor Losses. All cables will have losses generated by the current flowing in
the conductor. These losses, denoted by W. (W/m), are often referred to as joule, or /*R
losses from the equation from which they are computed. Conductor losses are a function of
the load current, and in calculations of cyclic current rating, the losses are based on a load
factor representing the load variation within a specified time period. If the load is constant
at the maximum rating of the cable, the load factor is 100%.

Sheath Losses. Metallic parts of the cable other than the conductor may also be
a source of joule losses caused by the currents induced in them. There are two parts of
the cable which may generate these additional losses: (1) the sheath or screen (losses are
denoted by W,), and (2) the armor (the losses are denoted by W,).
In general, there are two types of losses generated in sheaths and screens. The first
type, called the sheath eddy loss, is caused by the induced eddy currents. It is known that,
whenever an alternating flux penetrates a piece of conducting material, eddy currents will
be produced therein. These currents circulate in the sheath. This loss occurs because no
point of any sheath is equidistant from all three of the current-carrying conductors.
The second type of sheath loss, known as the sheath circulating loss, which occurs in
single-core cable systems and in three-core SL cables (cables with lead sheath around each
core), is caused by induced currents flowing along the metallic sheath and returning through
the sheath of other cable phases or through earth. Sheath circulating loss only exists when
the sheaths of two or three single-core cables are bonded together at two different positions,
such as the ends of the cable route. In this case, the sheath eddy current is superimposed on
the sheath circulating current, and therefore the actual current flowing in the sheath will not
be uniformly distributed around the sheath circumference. However, in rating calculations,
eddy losses are neglected when the circulating losses are present.

Armor and Pipe Losses. Some cable installations, for example, submarine cables,
are constructed with protective armor. Since, in the majority of cases, this armor is mullti-
point grounded, circulating losses are produced therein. If the armor is made of a nonmag-
netic material, the armor and sheath losses are considered together in rating computations.
However, the majority of armored cables are constructed with magnetic steel wire or tape
and, in this case, the hysteresis losses have to be considered. For pipe-type cables, hysteresis
losses in the steel pipe are also included in rating computations.
16 Part I m Modeling

1.4.2 Voltage-dependentLosses
Two different types of losses belong to this group: (1) dielectric losses, and (2) losses
caused by the charging current. Both losses are always present when the cable is energized,
and they are dependent on the cable electrical capacitance.
Dielectric losses produced in the cable insulation are the result of the energy storage
capability of insulating materials subjected to alternating voltage. Cable acts as a large
capacitor subject to charging currents. The work required to charge this capacitor is called
dielectric loss (denoted by Wz).
The charging current produced by the cable capacitance generates ohmic losses in the
cable that are present any time the cable is energized. Therefore, these losses operate at
100% load factor as do dielectric losses. The magnitude of these losses is determined by
system reactive design and operation and is difficult to evaluate. The charging current must
be supplied by the system, but the charging current losses are not strictly losses within the
cable itself.
Dielectric losses can be neglected for distribution voltages, and the limiting voltage
levels for each insulation type can be those given in IEC Publication 287 (1982) and dis-
cussed in Chapter 6 of this book. Charging current losses are usually neglected in rating
computations.

1.5 HISTORICAL NOTE

Current rating techniques of electric power cables have as long a history as the cable itself.
Methods presented in some of the first publications on this subject (Kennelly, 1893; Mie,
1905) are still used in today’s standards. Over the last hundred years, many researchers and
engineers have worked on various aspects of cable ratings. There are too many of them to
even mention their names; therefore, only a few selected developments are highlighted here,
with the emphasis on the work which has had the greatest influence on the standardization
efforts! The major contributors to the development of the cable rating techniques are
mentioned throughout the book when the technical issues are discussed.
The first rating tables were issued in Britain in the early 1920s. As a matter of fact, the
history of the calculation of current ratings in the United Kingdom can be traced through the
reports of one company that is still active in the field of cable ratings. This company is ERA
Technology Ltd., formerly the Electrical Research Association. The early involvement
of ERA can best be seen by quoting the preface from a 1924 ERA Report, “Permissible
Current Loading of British Standard Impregnated Paper Insulated Electric Cables.” The
preface reads as follows:

The work herein described was initiated in 1913 by the Institution of Electrical Engineers,
continued under the auspices of the Electrical Research Committee, and finally transferred to
the Electrical Research Association on its formation in 1921.
There has been little change in the personnel of the working Committee since the commence-
ment. The new research work herein described has been carried out by Mr S. W. Melsom of

! The account of the history of standardization of cable rating computations in the United Kingdom is based
on a private communiqué of Mr. Mark Coates, and that in the United States is based on the presentations by
M. A. Martin (1989) and W. Z. Black (1989) during th IEEE/PES T&D Conference in New Orleans, LA.
Chapter 1 m Cable Constructions and Installations 17

the National Physical Laboratory, and Mr E. Fawssett, of the Newcastle-upon-Tyne Electric

Supply Co., who are members of the directing Committee.

The committee structure, then in place for setting out standard methods of calculating,
appears very similar to that currently operating in the IEC with its technical committees
and working groups.
The first ERA report on the subject, published in 1923, set out a standard method of
calculating current ratings and provided tabulated ratings. Work continued and calculation
methods improved in many areas with papers being published by a number of workers; most
notable is the work by Arnold on joule losses. The next major U.K. document was prepared
by Whitehead and Hutchings and published in 1939. This report was limited to methods
of calculation, with tabulated ratings being published in a separate document. In the next
two decades, many reports were published which improved one aspect or another in the
calculation of current ratings. These improvements were brought together by Goldenberg in
a 1958 report numbered F/T 187. This was the last U.K. document to give a comprehensive
guide to current rating calculations before the publication of IEC 287.
In the United States, the National Electric Light Association (NELA) adopted in 1931
a set of standard constants and reference conditions for the calculation of load capabilities
of cables. This, in turn, resulted in the publication of the first set of current-carrying
capacity tables for paper-insulated, lead-covered cables in underground ducts or in air.
This publication was issued in March 1932 and was entitled, “Determination of Rating for
Underground and Aerial Cables—NELA Publication #28.”
In 1933, the tables were expanded and republished by the Edison Electric Institute
(EEI), “Current-Carrying Capacity of Impregnated Paper—Insulated Cable” (EEI Publi-
cation A-14-1933). These tables lasted for ten years when, in 1943, after considerable
improvements in cable insulation materials and growth in underground systems, the Insu-
lated Power Cable Engineering Association (IPCEA) published three documents, P-29-226,
P19-102, and P20-161, for impregnated paper, rubber, and varnished cambric insulated ca-
bles, respectively.
In 1951, W. A. Del Mar of Phelps Dodge Wire & Cable Company coined the word
“ampacity” to replace “current-carrying capacity.”
Concurrent with the development of power cable ampacity calculations over the years,
the National Electrical Code (NEC) in the United States has published ampacity tables
for building wires since the early 1900s. These ampacity tables have dealt primarily with
cables rated 600 V and below. The technical basis for the development of the tables issued
by NEC was developed by a committee chaired by S. J. Rosch, and dealt primarily with
low-voltage, code-grade, rubber-insulated cables. The committee’s final report was issued
in June 1938. Unfortunately, the work completed by Rosch was very crude by today’s
engineering standards and, in fact, it was in error in several fundamental technical areas.
In the 1930s, the field of heat transfer was in its infancy, and Rosch did not appear
to be aware of some of the heat transfer correlations that were first beginning to become
available from early heat transfer studies conducted mostly in the United Kingdom. Lacking
any heat transfer foundations and unaware of other similar work, he and his committee
resorted to a series of experimental tests to determine the relationship between the current
in a conductor and the conductor temperature. The committee’s tests were limited to a
relatively small number of all copper conductors insulated with code grade rubber and
18 Part I » Modeling

suspended horizontally in air. No buried cables were considered, and shield losses were not
included. Rosch neglected the variation of the heat transfer coefficient with the cable surface
temperature, which is an error that can be quite significant. He did, however, make two
additional assumptions that result in a conservative value for the recommended ampacity:
the air surrounding the cable is still and radiation from the surface to the environment is
neglected.
Rosch recognized some of the shortcomings of his ampacity model, and in 1938 he
published a paper that detailed a series of experimental test on cables placed in horizontal
conduits. Temperatures of cable sizes between 12 AWG and | 000 000 cmils were measured
for a number of different current levels. Using these data, he was able to calculate a Q
value that was to be incorporated in the NEMA tables, so that ampacity values predicted
by the model were brought into line with the experimental data. In fact, the value of Q was
a correction factor that pointed out that the correlation suggested by the ampacity model
was actually in error; its use was necessary to correct this error. If correct heat transfer
correlations had been used in the first place, the application of a correction factor would not
have been necessary.
Ampacity work progressed steadily from the time of Rosch’s final report until 1957
when Neher and McGrath published their paper which successfully summarized most of
the important ampacity work by extending an earlier comprehensive work of Simmons.
Their paper actually introduced no new advancements in the area of ampacity; it simply
and effectively put all of the ampacity principles into a single, all-encompassing paper.
Naturally, this is a tall order when we consider the complexity of the ampacity problem.
Due to Neher and McGrath’s successful summary paper, most engineers in North America
refer to the calculation procedure used to determine ampacity values as the Neher—McGrath
method. Actually, the technique that they described is based on a simple model of energy
balance in the conductor, and on an analogy between the flow of electric current and the
flow of heat. Both of these principles were well known long before 1957. Nonetheless, the
Neher—McGrath paper is credited as the paper which forms the basis for modern ampacity
standards.
The Neher—McGrath Model, as opposed to the Rosch Model, is based on a technically
correct set of equations. It effectively accounts for a much greater diversity of cable designs
and installation geometries. It considers heat generated in the shield material resulting
from induced shield currents and it also accounts for dielectric heating which can become
significant in high voltage cables. It uses more accurate and more technically correct heat
transfer coefficients than the Rosch Model. It describes the equations that should be used
for underground installations, as well as cables oriented vertically and horizontally in air.
It describes a procedure that can be used to derate cables when the heat generated by one
cable influences other cables. In other words, it permits ampacity calculations for multiple
cable installations where mutual heating between cables can be significant. It also includes
a technique for calculating the thermal resistance of a duct bank that may contain multiple
cables.
Although the Neher—McGrath Model is technically correct, it does have some weak-
nesses, mainly in dealing with cables in air. Since the method is based on the analogy
between the flow of heat and current, one must know the thermal resistances on the circuit
before the ampacity value can be calculated. Unfortunately, convective heat transfer coeffi-
cients are temperature-dependent, and therefore the temperature at local points in the thermal
Chapter 1 m Cable Constructions and Installations 19

circuit must be known before the problem is solved. Neher and McGrath use assumed tem-
perature values to solve this dilemma. In some instances, these assigned temperatures can
lead to unacceptable errors. Also, the Neher/McGrath Model uses experimental constants
to aid in the computation of the thermal resistance of fluid layers in the thermal circuit. As
Black (1989) pointed out, the accuracy of these constants has not been thoroughly verified.
And finally, the model accounts for only a single value of the thermal resistance of the soil
layer for buried installations. Changes in soil resistivity can occur when the soil adjacent
to the cable—earth interface begins to dry. This change is known to be very influential in
the cable ampacity because the soil thermal resistance is the largest single resistance in the
composite circuit. Such changes in the resistance of the soil can lead to thermal runaway
of the cable temperature resulting from a phenomenon referred to as the thermal instability
of the soil.
After several dissenting views were resolved at a symposium devoted to the subject,
the electric power industry adopted the Neher—-McGrath methods for the calculation of
power cable ampacity. In 1962, the American Institute of Electrical Engineers (AIEE)
and IPCEA jointly published new ampacity tables based on the Neher/McGrath paper.
These tables were published in two volumes, one for copper and one for aluminum cables
(AIEE S-135-1 and S-135-2), providing ratings for impregnated paper, varnished cloth,
rubber, thermoplastic polyethylene- and asbestos-insulated cables. Subsequently, these
tables became known in industry circles as the black books. In 1967, the IPCEA published
a revised version of ampacities for impregnated paper-insulated cables (IPCEA #48-426)
because the Association of Edison Illuminating Companies (AEIC) revised upward their
specification for conductor operating temperatures.
After the publication of Goldenberg’s report in the United Kingdom and the Neher-—
McGrath paper in the United States, it was felt that sufficient methodology has been de-
veloped to warrant the issuing of an international standard. Such a standard was prepared
by the International Electrotechnical Commission (IEC 287, 1969, 1982). The immediate
predecessor of IEC 287 was a CIGRE report presented in Paris in 1964. The countries that
participated in preparing the CIGRE document were the United Kingdom, the Netherlands,
France, Germany, Italy, and the United States. The major contributions to the development
of this standard were made by such well-known workers in this field as R. G. Parr and
H. Goldenberg of the United Kingdom, A. Morello of Italy, M. McGrath of the United
States, H. Brekelmann of Germany, and R. Wlodarski of Poland. It is the same countries,
with the addition of Canada and Belgium, that are active in the continued development of
IEC 287.
The CIGRE report was adopted by IEC in 1969, and after a number of amendments, a
second edition was published in 1982. For the third edition, IEC 287 is being divided into
a number of parts and sections, each of which covers a different aspect of the calculation
of cable ratings. At the time of writing this book, the IEC working group responsible for
current ratings is preparing tabulated current ratings for publication by IEC. This takes us
back to 1923 when both methods of calculating current ratings and tabulated ratings were
published in the United Kingdom.
Meanwhile, in the United States, after the publication of IPCEA tables in 1967, the
work on any new ampacity tables remained dormant for approximately ten years. In 1972,
due to a proliferation of shielded single-conductor cables and the absence of ratings for
cables with circulating currents in the shields, a working group was formed within the
20 Part I m»Modeling

Insulated Conductors Committee (ICC). This group reviewed the industry needs and rec-
ommended that additional ampacity tables be published to supplement the “black books”
(AIEE S-135-1, S-135-2).
During the late 1970s, the Insulated Conductors Committee requested and received
a project authorization from IEEE to revise the “black books.” This project was necessi-
tated by the outdated parameters and cable constants and the major changes in heat trans-
fer technology associated with modern cable systems. It was later decided to include in
the revision the documents AIEE $135-1 and $135-2, IPCEA P-48-426, P-53-426, and
P-54-440 (Cable Tray Ampacities).
To accomplish this task, another working group was organized in ICC, and in 1981
it published a set of new parameters for the new ampacity tables. The parameters were
to be used to develop tables for extruded and laminar dielectric underground power cables
of 600 V-500 kV. The tables, published in 1994 (IEEE, 1994), address new issues such
as cable/earth interface temperature and limiting heat flux, cables in vented and nonvented
risers, and cables in open and covered trays. Over 3000 ampacity tables for extruded
dielectric power cables rated through 138 kV and laminar dielecric power cables rated
through 500 kV are provided. The development of these tables has been spearheaded by
M. A. Martin, Jr. of M& E Technology Inc.
The work on refining cable ampacity computations is being continued. It proceeds
in two directions: (1) experimental studies are being performed to fine-tune some of the
computational formulas and adjust the value of constants, and (2) numerical methods are
being applied to overcome limitations inherent in the analytical approach. From the perusal
of the book, the reader will be able to learn about some of the most important historical
developments in power cable rating computations, and will also be able to follow the latest
developments in the application of numerical methods in this field.

REFERENCES

Aabo, T., Lawson, W. G., and Pancholi, S. V. (1995), “Upgrading the ampacity of HPFF
pipe-type cable circuits,’ JEEE Trans. Power Delivery, vol. 10, no. 1, pp. 3-8.
AEIC CS1-90 (1990), “Specifications for impregnated paper-insulated, metallic-sheathed
cable, solid type.”
AEIC CS2-90 (1990), “Specifications for impregnated paper and laminated paper polypropy-
lene insulated cable, high pressure pipe type.” |
AEIC CS3-90 (1990), “Specifications for impregnated paper-insulated, metallic-sheathed
cable, low-pressure gas-filled type.”
AEIC CS4-93 (1993), “Specifications for impregnated-paper-insulated low and medium
pressure self contained liquid filled cable.”
AEIC CS5-94 (1994), “Specifications for cross-linked polyethylene insulated shielded
power cables rated 5 through 46 kV.”
AEIC CS6-87 (1987), “Specifications for ethylene propylene rubber insulated shielded
power cables rated 5 through 69 kV.”
AEIC CS7-93 (1993), “Specifications for cross-linked polyethylene insulated shielded
power cables rated 69 through 138 kV.”
Chapter | m Cable Constructions and Installations 21

AEIC CS31-84 (1984), “Specifications for electrically insulated low viscosity pipe filling
liquids for high-pressure pipe-type cable.”
AEIC G1-68 (1968), “Guide for application of AEIC maximum insulation temperatures at
the conductor for impregnated-paper-insulated cables.”
AEIC G2-72 (1972), “Guide for electrical tests of cable joints 138 kV and above.”
AEIC G4-90 (1990), “Guide for installation of extruded dielectric insulated power cable
systems rated 69 kV through 138 kV.”
AEIC G5-90 (1990), “Underground extruded power cable pulling guide.”
AEIC G7-90 (1990), “Guide for replacement and life extension of extruded dielectric
5-35 kV underground distribution cables.”
ANSI/IEEE Standard 575 (1988), “Application of sheath-bonding methods for single con-
ductor cables and the calculation of induced voltages and currents in cable sheaths.”
Ball, E. H., Endacott, J. D., and Skipper, D. J., (1977), “U. K. Requirements and future
prospects for forced-cooled cable systems,” Proc. IEE, part C, no. 5.
Barnes, C. C. (1964), Electric Cables. London: Sir Isaac Pitman & Sons.
Black, W. Z. (1989), “The ampacity table dilemma: plotting a future course,” presented at
the llth IEEE/PES Transmission and Distribution Conf. and Exposition, New Orleans,
LA, Apr. 2-7, 1989.
Buckweitz, D. B., and Pennell, D. B. (Apr. 1976), “Forced cooling of UG lines,” Transmis-
sion and Distribution, pp. 51-58.
CIGRE (1976), “The calculation of continous ratings of forced cooled cables,’ WG 08 of
CIGRE SC 21, Electra, no. 66, pp. 59-84. Erratum in Electra, no. 76, p. 48, May 1981.
CIGRE (1986), “Forced cooled cables. Calculation of thermal transients and cyclic loads,”
WG 08 of CIGRE SC 21, Electra, no. 104, pp. 23-38.
Flamand, C. A. (1966), “Forced cooling of high-voltage feeders,’ JEEE Trans. Power App.
Syst., vol. PAS-85, no. 9, pp. 980-986.
IEC Standard 228 (1978), “Conductors of insulated cables,” 2nd ed., Ist suppl., 1982.
IEC Standard 287 (1969, 1982), “Calculation of the continuous current rating of cables
(100% ldad factor),” Ist ed. 1969, 2nd ed. 1982, 3rd ed. 1994-1995.
IEC Standard 502 (1983), “Extruded solid dielectric insulated power cables for rated volt-
ages from.1 kV to 30 kV.”

IEEE (1994) Standard 835, “IEEE Standard—Power cable ampacity tables.”

EPRI (1984), “Designer handbook for forced-cooled high-pressure oil-filled pipe-type cable
systems,” EPRI EL-3624, Project 7801-5, Report July 1984.
Kennelly, A. E. (1893), “Minutes, ninth annual meeting, Association of Edison Illuminating
Companies,” New York, NY.
King, S. Y., and Halfter, N. A. (1983), Underground Power Cables. London: Longman.
Martin, Jr., M. A. (1989), “Historical perspective on power cable ampacity ratings,” pre-
sented at the 11th IEEE/PES Transmission and Distribution Conf. and Exposition,
New Orleans, LA, Apr. 2—7, 1989.
Mie, G. (1905), “Uber die Warmeleitung in einem verseilten Kable,” Electrotechnische
Zeitschrift, p. 137.
22 Part I » Modeling

Notaro, J., and Webster, D. J. (1971), “Thermal analysis of forced cooled cables,” EEE
Trans. Power Syst., vol. PAS-71, no. 3, pp. 1225-1231.
Weedy, B. M. (1988), Thermal Design of Underground Systems. Chichester, England:
Wiley.
 2
Modes of Heat Transfer
and Energy Conservation
Equations

2.1 INTRODUCTION

Ampacity computations of power cables require solution of the heat transfer equations
which define a functional relationship between the conductor current and the temperature
within the cable and in its surroundings. In the previous chapter, we discussed how the heat
is generated within the cable. In this chapter, we will analyze how the heat is dissipated to
the environment. We will also develop the basic heat transfer equations, and discuss how
these equations are solved, thus laying the groundwork for cable rating calculations.

2.2 HEAT TRANSFER MECHANISM IN POWER CABLE SYSTEMS

The two most important tasks in cable ampacity calculations are the determination of the
conductor temperature for a given current loading, or conversely, determination of the tol-
erable load current for a given conductor temperature. In order to perform these tasks,
the heat generated within the cable and the rate of its dissipation away from the conduc-
tor for a given conductor material and given load must be calculated. The ability of the
surrounding medium to dissipate heat plays a very important role in these determinations,
and varies widely because of several factors such as soil composition, moisture content,
ambient temperature, and wind conditions. The heat is transferred through the cable and
its surroundings in several ways, and these are described in the following sections.

2.2.1 Conduction

For underground installations, the heat is transferred by conduction from the conductor
and other metallic parts as well as from the insulation. It is possible to quantify heat transfer

23
24 Part I m Modeling

processes in terms of appropriate rate equations. These equations may be used to compute
the amount of energy being transferred per unit time. For heat conduction, the rate equation
is known as Fourier’s law. For a wall having a temperature distribution 6(x), the rate
equation is expressed as

The heat flux g (W/m*) is the heat transfer rate in the x direction per unit area perpen-
dicular to the direction of transfer, and is proportional to the temperature gradient d0/dx
in this direction. The proportionality constant p is a transport property known as thermal
resistivity (K -m/W) and is a characteristic of the material. The minus sign is a consequence
of the fact that heat is transferred in the direction of decreasing temperature (see Fig. 2.1).

Unit area

Figure 2.1 Illustration of Fourier’s law.

2.2.2 Convection

For cables installed in air, convection and radiation are important heat transfer mech-
anisms from the surface of the cable to the surrounding air. Convection heat transfer may
be classified according to the nature of the flow. We speak of forced convection when the
flow is caused by external means, such as by wind, pump, or fan. In contrast, for free (or
natural) convection, the flow is induced by buoyancy forces which arise from density dif-
ferences caused by temperature variations in the air. In order to be somewhat conservative,
in cable rating computations, we usually assume that only natural convection takes place at
Chapter 2 » Modes of Heat Transfer and Energy Conservation Equations 25

the outside surface of the cable (see Chapter 9). However, both convection modes will be
considered in Chapter 10.
Regardless of the particular nature of the convection heat transfer process, the appro-
priate rate equation is of the form

= h(@s —Oamb) (Cz)

where q, the convective heat flux (W/m), is proportional to the difference between the
surface temperature and the ambient air temperature, 6, and Oamp,respectively. This ex-
pression is called Newton’s law of cooling and the proportionality constant h (W/m? - K)
is referred to as the convection heat transfer coefficient. Determination of the heat convec-
tion coefficient is perhaps the most important task in computation of ratings of cables in
air. The value of this coefficient varies between 2 and 25 W/m? - K for free convection and
between 25 and 250 W/m’ - K for forced convection.

2.2.38 Radiation

Thermal radiation is energy emitted by cable or duct surface. The heat flux emitted
by a cable surface is given by the Stefan—Boltzmann law:

q = €0;0" (2.3)
where 0* is the absolute temperature (K) of the surface,! og is the Stefan—Boltzmann
constant (og = 5.67-10~8W/m* - K*), and € is a radiative property of the surface called the
emissivity. This property, whose value is in the range 0 < € < 1, indicates how efficiently
the surface emits compared to an ideal radiator. Conversely, if radiation is incident upon
a surface, a portion will be absorbed, and the rate at which energy is absorbed per unit
surface area may be evaluated from the knowledge of surface radiative property known as
absorptivity a; that is,

dabs = %Ginc (2.4)

where 0 < a < 1. Equations (2.3) and (2.4) determine the rate at which radiant energy
is emitted and absorbed, respectively, at a surface. Since the cable both emits and absorbs
radiation, radiative heat exchange can be modeled as an interaction between two surfaces.
Determination of the net rate at which radiation is exchanged between two surfaces is
generally quite complicated. However, for cable rating computations, we may assume that
a cable surface is small and the other surface is remote and much larger. Assuming this
surface is one for which a@= € (a gray surface), the net rate of radiation exchange between
the cable and its surroundings, expressed per unit area of the cable surface, is

q = €ap(0**—0%.) amb (2.5)

A variation of equation (2.5) will be used in the book for rating of cables on riser poles and
in tunnels (see Chapter 10).
Throughout this book, we will use a notion of heat rate rather than heat flux. The
heat transfer rate is obtained by multiplying heat flux by the area. Thus, the heat rate for

I Throughout this book, the temperature with an asterisk will denote absolute value in degrees Kelvin.
26 Part I m Modeling

radiative heat transfer will be given by the following equation:

Wead
= €opAsr(a4 —6%.) | (2.6)

where A,, (m7) is the effective radiation area per meter length.
In power cable installed in air, the cable surface within the surroundings will simulta-
neously transfer heat by convection and radiation to the adjoining air. The total Bis of heat
transfer from the cable surface is the sum of the heat rates due to the two modes.* That is,

WShAv@)= Onh) reAgan(Ge ars) (2.7)

where A, (m*) is the convective area per meter length.

For some special cable installations, the ambient temperature used for heat convec-
tion can be different from the one used for heat transfer by radiation. The appropriate
temperatures to be used are described in Chapter 10.

2.2.4 Energy Balance Equations

In the analysis of heat transfer in a cable system, the law of conservation of energy
plays an important role. We will formulate this law on a rate basis; that is, at any instant,
there must be a balance between all energy rates, as measured in joules per second (W).
The energy conservation law can be expressed by the following equation:

Went= Wint= WoutaleAWs, (2.8)

Wentis the rate of energy entering the cable. This energy may be generated by other cables
located in the vicinity of the given cable or by solar radiation. Wj, is the rate of heat
generated internally in the cable by joule or dielectric losses, and A W,, is the rate of change
of energy stored within the cable. The value of W,y, corresponds to the rate at which energy
is dissipated by conduction, convection, and radiation. For underground installations, the
cable system will also include the surrounding soil.
In words, this relation says that the amount of energy inflow and generation act to
increase the amount of energy stored within the cable, whereas outflow acts to decrease
the stored energy. The inflow and outflow terms Wen and Woy are surface phenomena,
and these rates are proportional to the surface area. The thermal energy generation rate
Wint iS associated with the rate of conversion of electrical energy to thermal energy and is
proportional to the volume. The energy storage is also a volumetric phenomenon, but it is
simply associated with an increase (AW,, > O) or decrease (AW,, < 0) in the energy of
the cable. Under steady-state conditions, there is, of course, no change in energy storage
CAW
= 0):
In analysis of cables installed in air, we will often apply the conservation of energy
requirement at the surface of the cable. In this special case, we will only deal with the
surface phenomena, and the conservation requirement becomes

Went — Wour = 0 (2.9)

: Throughout the book, symbol W will be used for heat transfer rate.
~ The heat conduction in air is often neglected in cable rating computations (see, however, Section 9.6.8).
Chapter 2 » Modes of Heat Transfer and Energy Conservation Equations 27

Even though thermal energy generation will be occurring within the cable, the process will
not affect the energy balance at the cable surface. Moreover, this conservation requirement
holds for both steady-state and transient conditions.
We will use the fundamental equations described in this chapter to develop rating
equations throughout the reminder of the book.

2.3 HEAT TRANSFER EQUATIONS

As we mentioned earlier, current flowing in the cable conductor generates heat which
is dissipated through the insulation, metal sheath, and cable servings into the surrounding
medium. The cable ampacity depends mainly upon the efficiency of this dissipation process
and the limits imposed on the insulation temperature. To understand the nature of the heat
dissipation process, we need to develop the relevant heat transfer equations. Energy balance
equation (2.8) will be the focal point in our investigations.

2.3.1 Underground Directly Buried Cables

Let us consider an underground cable located in a homogeneous soil. In such cable,
the heat is transferred by conduction through cable components and the soil. Since the
length of the cable is much greater than its diameter, end effects can be disregarded, and
the heat transfer problem can be formulated in two dimensions only as shown in Fig. ike
The terms in equation (2.8) are defined as follows. When heat passes through the body
shown in Fig. 2.2, it meets certain thermal resistance in its path. Applying Fourier’s law of

MA

SVVVVVVVV VV VV VV VV VV VV V VV VV VV VV VV VV VS
—
Soil
Wy+ dy

aioedy a
Wy
Figure 2.2 Illustration of a heat conduction problem.

4 Because end effects are neglected, all thermal parameters will be expressed in this book on a per-unit
length basis.
28 Part I » Modeling

heat conduction [equation (2.1)], we have

Weaebeees (2.10)

where W,. = heat transfer rate through the area S in the x direction, W
p = thermal resistivity, K - m/W
S = surface area perpendicular to heat flow, m?
0 ay Ai rey
aaa temperature gradient in x direction.
Xx
Consider the small element dx dy in Fig. 2.2. If there are temperature gradients,
conduction heat transfer will occur across each of the surfaces. The conduction heat rates
perpendicular to each of the surfaces at the x and y coordinate locations are indicated by
terms W, and W,, respectively. The conduction heat rates at the opposite surfaces can then
be expressed as a Taylor series expansion where, neglecting higher order terms,

Wyr+dx
= W,+ aw: dx
an (2.11)
Wrigley
agi “o ogh8
Within the element dx dy, there may also be an energy source term associated with the rate
of thermal energy generation. This term is represented as

W, = Windx dy

where W;,; is the rate at which energy is generated per unit volume of the body by resistive
and capacitive currents (W/m?/m). In addition, there may occur changes in the amount of
internal energy stored by the material in the small body dx dy. These changes are related
to the capacitive nature of cable insulation. On a rate basis, we express this energy storage
term as
00
AW,, = c—dx dy
or
where c is the volumetric thermal capacity of the material. Recognizing that the conduction
rates constitute the energy inflow and outflow and there are no other energy transfer modes,>
the energy balance equation (2.8) for this body can be written as

00
W, + Ww,+ Windx dy — Wy4ax — Wy+ay = pepe dy

Observing that, in this case, S = dx dy, we can rewrite the last equation by substituting
from equations (2.10) and (2.11) to obtain:

~(G E)+s 1 00 LW a0
ax \p ax dy \p dy intrr (2512)

5 Since the case of an underground cable is considered here, heat is transferred by conduction only.
Chapter 2 m Modes of Heat Transfer and Energy Conservation Equations 29

For a cable buried in soil, equation (2.12) is solved with the boundary conditions
usually specified at the soil surface. These boundary conditions can be expressed in two
different forms. If the temperature is known along a portion of the boundary, then

0 = O(s) (2.13)

where 6 1s the boundary temperature that may be a function of the surface length s. If heat
is gained or lost at the boundary due to convection h(@ — Oamp)or a heat flux q, then
2 GES Ieee 0 (2.14)
p on
where n is the direction of the normal to the boundary surface, h is a convection coefficient,
and @is an unknown boundary temperature.
Occasionally, it may be advantageous to express the heat transfer equation in cylindrical
coordinates. In two dimensions, they become

ikx) PalNe 1 00 Fea (2.15

PoOreNep
an r2d¢\p ad oereee ny
In cable rating computation, the temperature of the conductor is usually given, and the
maximum current flowing in the conductor is sought. Thus, when the conductor heat loss
is the only energy source in the cable, we have Wi,, = 1*R, and equation (2.12) or (2.15)
is used to solve for J with the specified boundary conditions.
The challenge in solving equations (2.12) and (2.15) analytically stems mostly from
the difficulty of computing the temperature distribution in the soil surrounding the cable.
An analytical solution can be obtained when a cable is represented as a line source placed
in an infinite homogenous surrounding. Since this is not a practical assumption for cable
installations, another assumption is often used, namely, that the earth surface is an isotherm.
In practical cases, the depth of burial of the cables is on the order of ten times their external
diameter, and for the usual temperature range reached by such cables, the assumption of an
isothermal earth surface is a reasonable one. In cases where this hypothesis does not hold,
namely, for large cable diameters and cables located close to the earth surface, a correction
to the solution equation has to be used or numerical methods should be applied. Both are
discussed later in this book.

EXAMPLE 2.1
In order to illustrate the solution process for the heat transfer equation, we will determine the tem-
perature distribution within the insulation of a single core cable. Let as assume that the conductor
temperature is 9, and the temperature at the outer surface of the insulation is 6;. The cable has a
conductor radius r, and a radius over insulation r;. Steady-state conditions are assumed.
For steady-state conditions with no heat generation within the insulation, equation (2.15), with
the assumption of radial symmetry, reduces to

Ue¢ =)= (2.16)

Assuming the value of p to be constant, equation (2.16) may be integrated twice to obtain the general
solution
6(r) =C, Inr + Cp (MN
30 Part I » Modeling

To obtain the constants of integration C, and C2, we use the specified boundary conditions at
the inner and outer surface of the insulation. Substituting these conditions into equation (2.17) and
solving for C, and C2, we obtain

6. = 6; Fr.
Or) = In a (2.18)
Wess Yj
Yj

We observe that the temperature distribution in the insulation associated with radial conduction
through a cylinder is logarithmic.

An additional difficulty in finding the analytical solution to equation (2.12) arises in

the case of cables being located in a nonuniform medium, for example, in a backfill or duct
bank. In this case, the thermal resistivity is not constant. Here again, either correction
factors or numerical techniques are applied, and these are discussed in detail in Chapters 9
and 11.

2.3.2 Cables in Air

For an insulated power cable installed in air, several modes of heat transfer have to
be considered. Conduction is the main heat transfer mechanism inside the cable. Suppose
that the heat generated inside the cable (due to joule, ferromagnetic, and dielectric losses) is
W, (W/m). Another source of heat energy can be provided by the sun if the cable surface is
exposed to solar radiation. Energy outflow is caused by convection and net radiation from
the cable surface. Therefore, the energy balance equation (2.9) at the surface of the cable
can be written as

W, a Wool= Weonvaa Wrad = 0 (7.19)

where W,,) is the heat gain per unit length caused by solar heating, and Weonyand W,aqare
the heat losses due to convection and radiation, respectively. Computation of the losses
generated inside the cable is discussed in detail in Chapters 6-8. Substituting appropriate
formulas for the remaining heat gains and losses [equation (2.7)], the following form of the
heat balance equation is obtained:

W,+ oD3H —1Dth (63—6%)

amb —1Dios (054—0**,)=0
amb (2.20)
where 6* = cable surface temperature, K
o = solar absorption coefficient
H = intensity of solar radiation, W/m?
o, = Stefan—Boltzmann constant, equal to 5.67 - 10-8 W/m2- K4
€ = emissivity of the cable outer covering
D* = cable external diameter,® m
oxamb — ambient temperature, K

© We recall that the dimension symbols with an asterisk refer to the length in meters and without it to the
length in millimeters.
Chapter 2 m Modes of Heat Transfer and Energy Conservation Equations 31

This equation is usually solved iteratively. In steady-state rating computations, the

effect of heat gain by solar radiation and heat loss caused by convection are taken into account
by suitably modifying the value of the external thermal resistance of the cable. Computation
of the global solar radiation H and the convection coefficient h can be quite involved
(Morgan, 1991). This topic is discussed in Chapters 9 and 10 of the book. Recommended
values are also discussed in the same chapters.

2.4 ANALYTICAL VERSUS NUMERICAL METHODS OF SOLVING

HEAT TRANSFER EQUATIONS

Equations (2.12) and (2.20) can be solved either analytically, with some simplifying
assumptions, or numerically. Analytical methods have the advantage of producing cur-
rent rating equations in a closed formulation, whereas numerical methods require iterative
approaches to find cable ampacity. However, numerical methods provide much greater flex-
ibility in the analysis of complex cable systems and allow representation of more realistic
boundary conditions. In practice, analytical methods have found much wider application
than the numerical approaches. There are several reasons for this situation. Probably the
most important one is historical: cable engineers have been using analytical solutions based
on either Neher/McGrath (1957) formalism or IEC Publication 287 (1982) for a long time.
Computations for a simple cable system could often be performed using pencil and paper or
with the help of a hand-held calculator. Numerical approaches, on the other hand, require
extensive manipulation of large matrices, and have only become popular with an advent of
powerful computers. Both approaches will be described in this book; analytical methods
are discussed in Parts I and II, whereas the numerical approaches are dealt with in Part III.

REFERENCES

IEC Standard 287 (1982), “Calculation of the continuous current rating of cables (100%
load factor),” 2nd ed., 3rd amendment, 1993.
Neher, J. H., and McGrath, M. H. (Oct. 1957), “The calculation of the temperature rise and
load capability of cable systems,” AJEE Trans., vol. 76, part 3, pp. 752-772.
 ieownapel
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 3
Circuit Theory
Network Analogs
for Thermal Modeling

3.1 INTRODUCTION

Analytical solutions to the heat transfer equations are available only for simple cable con-
structions and simple laying conditions. In trying to solve the cable heat dissipation prob-
lem, electrical engineers noticed a fundamental similarity between the heat flow due to the
temperature difference between the conductor and its surrounding medium and the flow
of electrical current caused by a difference of potential (Pashkis and Baker, 1942). Using
their familiarity with the lumped parameter method to solve differential equations repre-
senting current flow in a material subjected to potential difference, they adopted the same
method to tackle the heat conduction problem. The method begins by dividing the physical
object into a number of volumes, each of which is represented by a thermal resistance and
a capacitance. The thermal resistance is defined as the material’s ability to impede heat
flow. Similarly, the thermal capacitance is defined as the material’s ability to store heat.
The thermal circuit is then modeled by an analogous electrical circuit in which voltages are
equivalent to temperatures and currents to heat flows. If the thermal characteristics do not
change with temperature, the equivalent circuit is linear and the superposition principle is
applicable for solving any form of heat flow problem.
In a thermal circuit, charge corresponds to heat; thus, Ohm’s law is analogous to
Fourier’s law. The thermal analogy uses the same formulation for thermal resistances and
capacitances as in electrical networks for electrical resistances and capacitances. Note that
there is no thermal analogy to inductance and in steady-state analysis; only resistance will
appear in the network. The analogy between electrical and thermal networks is explored in
the next section.

55
34 Part I m Modeling

Since the lumped parameter representation of the thermal network offers a simple
means for analyzing even complex cable constructions, it has been widely used in thermal
analysis of cable systems. A full thermal network of a cable for transient analysis may
consist of several loops. Before the advent of digital computers, the solution of the network
equations was a formidable numerical task. Therefore, simplified cable representations were
adopted and methods to reduce a multiloop network to a two-loop circuit were developed. A
two-loop representation of a cable circuit turned out to be quite accurate for most practical
applications and, consequently, was adopted in international standards. In this chapter,
we will explain how the thermal circuit of a cable is constructed, and we will show how
the required parameters are computed. We will also explain how full network equations
are solved, and present several examples of the two-loop representation of major cable
constructions.

3.2 ELECTRICAL AND THERMAL ANALOGY

3.2.1 Thermal Resistance

All nonconducting materials in the cable will impede heat flow away from the cables
(the thermal resistance of the metallic parts in the cable, even though not equal to zero, is so
small that it is usually neglected in rating computations). Thus, we can talk about material
resistance to heat flow. In order to explain the concept of thermal resistance, let us consider
a cylindrical nonconducting layer with constant thermal resistivity o,;,. Cable insulation is
a good example of such a layer. If the internal and external radii of this layer are r; and rz,
respectively, then the temperature distribution inside this layer is given by equation (2.18)
as

SVPeeds70) G1)

where 6) and 6 are given temperatures at locations corresponding to r; and rz, respectively.
We may use this temperature distribution in Fourier’s law, equation (2.10), to determine the
conduction heat rate. Taking the derivative in equation (3.1), we obtain

The area of the cylinder per unit length is equal to 27r. Substituting this into equation
(2.10) yields the following rate equation:

Ding
W= —— (1 - &) (3.2)
PthIneg
|
From equation (3.2), we can observe an analogy between the diffusion of heat and
electrical charge mentioned in the Introduction. Just as an electrical resistance is associated
with the conduction of electricity, a thermal resistance may be associated with the conduction
of heat. Defining resistance as the ratio of a driving potential to the corresponding transfer
Chapter 3 = Circuit Theory Network Analogs for Thermal Modeling 35

rate, it follows from equation (3.2) that the thermal resistance for conduction of a cylindrical
layer per unit length is

je :
ie” (3.3)
yg iA

For a rectangular wall, we have

where = p;, = thermal resistivity of a material, K -m/W

S = cross section area of the body, m?
/ = thickness of the body, m

Similarly, for electrical conduction in the same system, Ohm’s law provides an electrical
resistance of the form

OAV Ve l
R
T = Pes (3.5)
The analogy between equations (3.4) and (3.5) is obvious. We also can write that

(3.6)

which is the thermal equivalent of Ohm’s law.

A thermal resistance may also be associated with heat transfer by convection at a
surface. From Newton’s law of cooling [equation (2.2)],

W= héonvAs (0. ae Gamb) (3.7)

where A, is the area of the outside surface of the cable for unit length, heony is the cable
surface convection coefficient, and 9, is the cable surface temperature.
The thermal resistance for convection is then

Oe al Oamb ]
Tag ee 3.8a
a WwW heonvAs

Yet another resistance may be pertinent for a cable installed in air. In particular,
radiation exchange between the cable surface and its surroundings may be important. It
follows that a thermal resistance for radiation may be defined as

62-65, 1
Trad = (3.8b)
Wrad - hrat
where A,, is the area of the cable surface effective for heat radiation for unit length of
the cable and 6;,, is the temperature of the air surrounding the cable which, when cable
36 Part I m Modeling

is installed in free air, is equal to the ambient temperature, 0%,. 1, is the radiation heat
transfer coefficient obtained from expression (2.6) for radiation heat transfer rate:

Wrad
=€OBAsr
(0:4ne4)
=copAsy
(05—O35) (or+05,.)
(03?
+ioeNa (6:- 6.5)
Hence,
hy=€08(02+Oss)
(5?+632) (3.9)
The
total
heat
transfer
coefficient
foracable
inairisgiven
by
hy = Aeony+ h, (3.10)

EXAMPLE 3.1
Determine an expression for the external thermal resistance for a cable in air. The external diameter
of the cable is D? (m).!
The external thermal resistance for the cable is obtained from equation (3.6):

0. - amb
W,

where W, is the total heat loss per unit length, and is equal to

W, = 1 D*h, (8. — Pam) (3.10a)

Hence,
1
lige (3.10b)
‘ mDth;,

In the remainder of this book, we will use symbol / to denote the heat transfer coeffi-
cient. In the majority of cases, this transfer will be purely convective. In Chapter 9, we will
show the relationship between coefficients / defined in IEC 287 and h, as defined above.
Circuit representations provide a useful tool for both conceptualizing and quantifying
heat transfer problems. The equivalent thermal circuit for the composite cylindrical wall
is shown in Fig. 3.1. The heat transfer rate can be determined separately considering each
element in the network. That is,
LORD eS MEO WhGg—Opoma OO
~ paln(r2/ri) ~~ pa ln(r3/r2) pc In(r4/r3) I
TG 2m 20 2mrg4h
where all the radii are expressed in meters.
In terms of the overall temperature difference 0, —Qam»and the total thermal resistance
Tio, the heat transfer rate can also be expressed as

0; aa Oamb
W=
Trot

! We will denote diameters expressed in meters with an asterisk.

Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling Sy

UPL RRB As
1 In(to/t4) In(r3/t,) In(r,/r.
; 3) 1
h,2nr, 2mk, 2rk, 2nk, h,2nr,

Figure 3.1 Temperature distribution for a composite cylindrical wall.

Because the conduction and convection resistances are in series and may be added up,
we have
Tne CAMSae
Sey oa Napea is be eel r 1
De APA ag ify Pg igs Daigle

EXAMPLE 3.2
Consider cable model No. | (see Appendix A) installed in air. The possible existence of an optimum
insulation thickness for a cable (from a thermal point of view) is suggested by the presence of
the competing effects: on the one hand, the conduction resistance decreases with the reduction of
insulation thickness, and on the other, a reduction in the overall cable diameter increases the external
thermal resistance. Hence, an insulation thickness may exist that maximizes the heat transfer rate. The
optimal insulation thickness must, of course, be sufficient to satisfy voltage insulation requirements.
Let the overall heat transfer coefficient be equal to 10 W/m”: K. To facilitate the computations,
we will make the following two assumptions: (1) the presence of concentric neutral wires will be
neglected (they have negligible effect on cable’s thermal resistance), and (2) the thermal resistance of
the jacket remains constant as the diameters change. The last assumption implies that the ratio of the
jacket thickness to the diameter under the jacket is constant, that is, 1;/D; = k. We will also assume
38 Part I a Modeling

that the value of k is equal to the rt, /D; ratio obtained for nominal conditions for this cable which are
specified in Table Al. The insulation screen is treated as a part of the insulation.
Neglecting the thermal resistance of concentric wires, the total thermal resistance of the cable

is equalto ae ate
Tigfee F D; ate eOe
Dig ihe, Pons D; m(D; + ty)h
SO Dia Di ain
Or dou mDi(A+kjyh
The optimum insulation thickness would be associated with the value of D; that minimizes this
resistance. This value can be obtained by solving the following equation:

DTrot= Pi 1000 me
dD; 2xD; whD?(1+k) —
For the cable model No. 1, the jacket has a thickness of 3.2 mm and the diameter under the
jacket is 30.1 mm. The PVC jacket has a thermal resistivity of 3.5 K - m/W; hence,

2000 2000
= —_—_ = = 1.7, mm:
ph +k) arCFAKO
i? ou( Wile
* S
0.0301
e saat
It remains to be verified that the thermal resistance is minimum at this point. This is achieved by
computing the second derivative of T,, . The second derivative is positive at D; = 51.7 mm; hence,
the task is accomplished.
We observe that this value is much greater than the actual diameter over the insulation for cable
model No. 1. Therefore, in order to improve the heat dissipation process for this cable, we could
increase the thickness of the insulation almost four times. This is, of course, not practical because
of cost and weight considerations. Nevertheless, any increase of this thickness, up to the value of
51.7 mm, will result in an improvement of the heat dissipation process for this cable under assumed
conditions.

3.2.2 Thermal Capacitance

Many cable rating problems are time-dependent. Consider, for example, two cable
circuits operating in parallel under steady-state conditions. When one of the circuits is
suddenly switched off, the second circuit will carry additional current. This sudden change
of loading will cause slower changes in the temperature distribution within the cable and in
the surrounding medium.
To determine the time dependence of the temperature distribution within the cable
and its surroundings, we could begin by solving the appropriate form of the heat transfer
equation, for example, equation (2.12). In the majority of practical cases, it is very difficult
to obtain analytical solutions of this equation and, where possible, a simpler approach is
preferred. One such approach may be used where temperature gradients within the cable
components are small. It is termed the lumped capacitance method. Since this method
offers the only hope for solving the heat equations analytically for a cable system, it is
typically applied in practical cable rating computations. In order to satisfy the requirement
that the temperature gradient within the body must be small, some components of the cable
system, for example, the insulation and surrounding soil, must be subdivided into smaller
entities. In Section 3.3, we discuss how this subdivision is performed for a cable’s insulation
and jacket. Treatment of the external cable environment is discussed in Chapter 5.
Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling 39

We will explain the concept of heat capacitance by considering heat transfer in the
jacket of a cable installed in air. At time t = 0, the current is switched off and the cooling
process begins. The essence of the lumped capacitance method is the assumption that the
temperature of the solid is spatially uniform at any instant during the transient process. This
assumption implies that the temperature gradients within the body are negligible.
From Fourier’s law, heat conduction in the absence of a temperature gradient implies
the existence of infinite thermal conductivity. Such a condition is clearly impossible. How-
ever, this condition can be closely approximated if the resistance to conduction within the
jacket is small compared with the resistance to heat transfer between the cable and the
surrounding air. Let us assume for a moment that this is, in fact, the case.
Applying equations (2.8) and (3.10a), the energy balance equation for the cable jacket
with volume V now takes the form

meeLUout
Wst
or
do
—h,A (0 Pa amb) = Vco——
dt
where _ A = total area of the jacket exposed to heat transfer by convection and
radiation, m*
V = volume of the body, mm
c = volumetric specific heat of the material, J/m? - °C

Assuming that the initial jacket temperature is 0(0) = 6, the solution to this equation is

O(t)
=Oamb
+(90
—
Gamb)
EXP
|-(Ve)|
h,A
This result indicates that the difference between jacket and ambient temperatures must
decay exponentially to zero as t approaches infinity. From the last equation, it is also
evident that the quantity (Vc/h,A) may be interpreted as a thermal time constant. Utilizing
equations (3.10b), this time constant can be written as

C=(=) vo=T-0 GzLL

where Q is the lumped thermal capacitance of the jacket.
We can immediately observe that the foregoing behavior is analogous to the voltage
decay that occurs when a capacitor is discharged through a resistor in an electrical RC
circuit. This analogy suggests that RC electrical circuits may be used to determine the
transient behavior of thermal systems. In fact, before the advent of digital computers, RC
circuits were widely used to simulate the transient thermal behavior of cables. Even today,
this is the preferred method for such analysis, and we will explore it further in Chapter 5.
Meanwhile, we observe that just as we found in the case of Ohm’s law, an analogy
between thermal and electrical networks exists, which is the following:

Electrical:
AV =£ Gaw)
Thermal: A@
_ Wn
Orn
40 Part I m Modeling

where C = electrical capacity, F

QO; = thermal capacity, J/°C
tO | electrical charge stored in C, C
W,, = heat stored in Q,p, J
AV = voltage rise across C due to Q, V
A@ = temperature rise in Q,, due to W,;, °C

As mentioned above, an equivalent thermal network will contain only thermal resis-
tances T and thermal capacitances Q,;. The thermal capacitance Q;, can be defined as the
“ability to store the heat,” and from equation (3.11) we have

As an illustration, the formula for the thermal capacitance is established for a coaxial
configuration with internal and external diameters Df and D3 (m), respectively, which may
represent, for example, a cylindrical insulation.
The thermal capacitance is calculated from equation (3.13):

Os TU
ar AU *2 *2
Bea | 14
(3.14)

Thermal capacitances and resistances are used to construct a thermal ladder network to
obtain the temperature distribution within the cable and its surroundings as a function of
time. This topic is discussed in the next section.

3.3 CONSTRUCTION OF A LADDER NETWORK OF A CABLE

The electrical and thermal analogy discussed in Section 3.2 allows the solution of many
thermal problems by applying mathematical tools well known to electrical engineers. An
ability to construct a ladder network is particularly useful in transient computations. To
build a ladder network, the cable is considered to extend as far as the inner surface of the
soil for buried cables, and to free air for cables in air.
In constructing ladder networks dielectric losses, which are described in Chapter 6,
require special attention. Although the dielectric losses are distributed throughout the
insulation, it may be shown that for a single-conductor cable and also for multicore, shielded
cables with round conductors, the correct temperature rise is obtained by considering for
transients and steady state that all of the dielectric loss occurs at the middle of the thermal
resistance between conductor and the sheath. For multicore belted cables, dielectric losses
can generally be neglected, but if they are represented, the conductors are taken as the source
of dielectric loss (Neher and McGrath, 1957).
Thermal capacitances of the metallic parts are placed as lumped quantities correspond-
ing to their physical position in the cable. The thermal capacitances of materials with high
thermal resistivity and possibly large temperature gradients across them (e.g., insulation
and coverings) are allocated by the technique described in Section 3.3.1 below.

3.3.1 Representation of Capacitances of the Dielectric

In the early days of transient rating computations, the thermal capacity of the insu-
lation was typically divided equally between the conductor and the sheath (Buller, 1951).
Chapter 3 = Circuit Theory Network Analogs for Thermal Modeling 41

However, the thermal capacity of the insulation is not a linear function of the thickness of
the dielectric. To improve the accuracy of the approximate solution using lumped constants,
Van Wormer (1955) proposed a simple method for allocating the thermal capacity of the
insulation between the conductor and the sheath so that the total heat stored in the insulation
is represented. An assumption made in the derivation is that the temperature distribution in
the insulation follows a steady-state logarithmic distribution for the period of the transient.
We will present the derivation of the Van Wormer factor for a long-duration transient and
simply state the result for the short-duration conditions. Whether the transient is long or
short depends on the cable construction. For the purpose of transient rating computations,
long-duration transients are those lasting longer than iz T- XQ, where XT and &Q are the
internal cable thermal resistance and capacitance, respectively. The methods for computing
the values of T and Q are discussed in Chapter 9.
3.3.1.1 Van Wormer Coefficient for Long-duration Transients. |The dielectric is
represented by lumped thermal constants. The total thermal capacity of the dielectric (Q;)
is divided between the conductor and the sheath as shown in Fig. 3.2.
Conductor Ty
nnn
Somaagah en | Outside
diameter
ofinsulation
=—Q
Fane Q
fae T (1-p)Q
|meeee
ae
ae
Figure 3.2 Representation of the dielectric for times greater than 407 - XQ. T; = total
thermal resistance of dielectric per conductor (or equivalent single-core con-
ductor of a three-core cable; see below).
where (Q; = total thermal capacitance of dielectric per conductor (or
equivalent single-core conductor of a three-core cable)
Q, = thermal capacitance of conductor (or equivalent single-core
conductor of a three-core cable).
(IEC Standard 853-2, 1989)

Note: When screening layers are present, metallic tapes are considered to be part of
the conductor or sheath, while semiconducting layers (including metallized carbon paper
tapes) are considered part of the insulation in thermal calculations.
The Van Wormer coefficient p is derived as follows.
Consider a part of the cable extending to the cable sheath as shown in Fig. 3.3. As
indicated earlier, we will consider a cable 1 m long. Let g; denote the capacitance of the
insulation expressed per unit area, that is, Q; = A-4q;. If p represents the part of the
insulation thermal capacity to be placed at conductor temperature 6, and (1 —p) the part to
be placed at sheath temperature 6,, the total heat stored in the insulation may be accounted
for as follows:
D;/2
pAgi9- + (1 —p)Agi9s = qi / 6.24 dr (Gals)
d,/2
where __D;= external diameter of dielectric

d. = external diameter of conductor.

From Fig. 3.3,

Afi
A= 4
—(D;
—d2) (3.16)
42 Part I » Modeling

Figure 3.3 Cable conductor, insulation, and

sheath.

In steady state, the temperature in the insulation at a distance r from the conductor is
obtained from equation (3.2) as

Pi if
6. —6, = W.—
In In —
dk Sle
Sew

and in particular,

A Si hesgnila i D;
ee (3.18)
Ww Ge

where W, = heat generated in the conductor

(; = thermal resistivity of the insulation.

Combining equations (3.15)—(3.18) results in

pAgi(@-:—6;) - Agi@, / D;/2

=274; (4— i In=)rdr
W.—
Afterperformingintegration,weobtain
d./2 20
Bod braid a| 6, (D?—
8 d?) pase
Diet Leet 16
—de)
wetnD
ey
A i 6. = 6, +A 105= Dir i - : W.. a — ! <

Combining this equation with equations (3.15) and (3.17), the following solution for
p 1s obtained:

nS l 1
or D; Dee (3.19)
2In | — (3) an]
di a,

Equation (3.19) is also used to allocate the thermal capacitance of the outer covering
in a similar manner to that used for the dielectric. In this case, the Van Wormer factor is
given by

(3.20)

where D, and D, are the outer and inner diameters of the covering.
Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling 43

For long-duration transients and cyclic factor computations, the three-core cable is re-
placed by an equivalent single-core construction dissipating the same total conductor losses
(Wollaston, 1949). The diameter d* of the equivalent single-core conductor is obtained on
the assumption that new cable will have the same thermal resistance of the insulation as the
thermal resistance of a single core of the three-core cable; that is,

where Dj’ is the same value of diameter over dielectric (under the sheath) as for the three-
core cable, and 7; is the thermal resistance of the three-conductor cable as given in Chap-
ter 9; p; is the thermal resistivity of the dielectric. Hence, we have

(aed)

Thermal capacitances are calculated on the following assumptions (see Example 3.8):

1. The actual conductors are considered to be completely inside the diameter of the
equivalent single conductor, the remainder of the equivalent conductor being oc-
cupied by insulation.
2. The space between the equivalent conductor and the sheath is considered to be
completely occupied by insulation (for fluid-filled cable, this space is filled partly
by the total volume of oil in the ducts and the remainder is oil-impregnated paper).

Factor p is then calculated using the dimensions of the equivalent single-core cable, and is
applied to the thermal capacitance of the insulation based on assumption (2) above.
Van Wormer has also suggested that when the insulation is thick, the thermal capacity
of the insulation should be placed in part at the conductor, in part halfway through the
insulation thermal resistance, and in part at the sheath.

3.3.1.2 Van Wormer Coefficient for Short-duration Transients. Short-duration

transients last usually between 10 min and about | h. In general, for a given cable con-
struction, the formula for the Van Wormer coefficient developed in this section applies
when the duration of the transient is not greater than ;2T - £Q. The heating process for
short-duration transients can be regarded as if the insulation were thick.
The method is the same as for long-duration transients except that the cable insulation
is divided at diameter d, = /D; - d,, giving two portions having equal thermal resistances,
as shown in Fig. 3.4.

1
ls 1
=i
Conductor _2 4 dy 21
‘© Outside diameter

is ofinsulatio
Q,+p’,
a - | iad Oe (1-p")Qin
.
O
Figure 3.4 Representation of the dielectric for times less than or equal to t DT -xuQ: TEC
Standard 853-2, 1989)
44 Part I » Modeling

The thermal capacitances Q;; and Qj are defined in Chapter 9.

The Van Wormer coefficient is given by

anes Leead
il | (3.22)
@) B=
PRs clieeeeie:
Sereventtalites.,
U e e e e
EXAMPLE3.3
Constructa laddernetworkfor modelcableNo.4 in AppendixA for a short-durationtransient.
This network is shown in Fig. 3.5.
As shown in Fig. 3.5, the insulation thermal resistance is divided into two equal parts, the
insulation capacitance into four parts, and the capacitance of the cable serving into two parts.

@ D £L@>)
a.
rep 5 o i=
co) a
>
5 3 oan 8 8
E ame S lec) iB amelie $ 8
= w D fe) c
2 oe & 5 Che Celie a= gs 2
ernie E (= £2 6s < Sia is
£¢& ——~— £ = 6 --JVE SS —
Physical
aa 6 OG
OMOve 2 T2 s
2 €@
fc B§ ie << Ts
, ESS pkey ° at , Se
SO

Son pas+ LR a ae ee ee SRSyeres)yp

Qo [Qe|Qix |Qi2)/
Pig |Qig|Qs|, Qb1 |Qb2Qa Qe4

——_» uM—_,—Y Ve SY pornet carmen#

Electrical Cy Ry Co Ro Cy R3 Cy Ry
Figure 3.5 Thermal network for model cable No. 4 with electrical analogy.

A three-core cable is represented as an equivalent single-core cable as described above

for durations of about 4 T - ©Q or longer (the quantities XTand XQ refer to the whole
cable). However, for very short transients (i.e., for durations up to the value of the product
XT -2Q where XT and © Q now refer to the single core), the mutual heating of the cores
is neglected, and a three-core cable is treated as a single-core cable with the dimensions
comesponding to the one core. For durations between these two limits, ©T - ©Q for one
core and +DT - ©Q for the whole cable, the transient is assumed to be given by a straight
line aeiearslany in a diagram with axes of linear temperature rise and logarithmic times.

EXAMPLE 3.4
Consider cable No. 2 (Appendix A) located in a PVC duct in air shaded from solar radiation. Duct
internal and external diameters are 100 and 102 mm, respectively. The external thermal resistance
Chapter 3 = Circuit Theory Network Analogs for Thermal Modeling 45

of this cable is composed of: (1) resistance of the air inside the duct (equal to 0.4167 K - m/W), (2)
resistance of the duct (equal to 0.0216 K - m/W), and (3) resistance of the air outside the duct (equal
to 0.259 K - m/W) (see Chapter 9). The external resistance per core of the air inside the duct and the
duct itself is therefore

T, + T, = 0.4167+ 0.0216= 0.4383K-m/W

The thermal capacitances on a per-core basis are: (1) conductor 1035 J/K - m, (2) insulation
915.6 J/K - m, (3) screen 4 J/K - m, (4) jacket 432.4 J/K - m, (5) air inside the duct 4.8 J/K - m, and (6)
duct 179.8 J/K - m (see Chapter 9 for the derivation of these values). The total thermal capacitance
of a single core of the cable is the sum of these values and is equal to 2571.6 J/K - m.
The time constant of a single core of cable No. 2 installed in a duct is therefore equal to
0.59 h. With the insulation and jacket thermal resistances equal to 0.307 and 0.078 K - m/W, respec-
tively, the time constant of the entire three-core cable installed in the duct is equal to

xT -XQ = [0.307+3:- (0.078 +0.4383)]-[3-(1035+915.6+4+432.4+4.8+ 179.8) ]/3600 = 4.0h

Hence, long-duration transients for this cable are those lasting about 2 h or more and short-
duration transient last between 10 and 36 min. For durations between 36 min and 2 h, the transient is
assumed to be given by straight line interpolation on linear temperature rise/logarithmic time axes.

3.3.1.3 Van Wormer Coefficient for Transients Due to Dielectric Loss. In the
preceding sections, it has been assumed that the temperature rise of the conductor due to
dielectric loss has reached its steady state, and that the total temperature at any time during
the transient can be obtained simply by adding the constant temperature value due to the
dielectric loss to the transient value due to the load current.
If achange in load current and system voltage occur at the same time, then an additional
transient temperature rise due to the dielectric loss has to be calculated (Morello, 1958).
For cables at voltages up to and including 275 kV, it is sufficient to assume that half of
the dielectric loss is produced at the conductor and the other half at the insulation screen
or sheath. The cable thermal circuit is derived by the method given above with the Van
Wormer coefficient computed from equations (3.19) and (3.22) for long- and short-duration
transients, respectively.
For paper-insulated cables operating at voltages higher than 275 kV, the dielectric loss
is an important fraction of the total loss and the Van Wormer coefficient is calculated by
(IEC 853-2, 1989)

(yet) (323)

= (2) -1][o(2)]
In practical calculations for all voltage levels for which dielectric losses are important,
half of the dielectric loss is added to the conductor loss and half to the sheath loss; therefore,
the loss coefficients (1 + 4) and (1 + A; + A2) used to evaluate thermal resistances and
capacitances are set equal to 2 (see Example 5.4).

EXAMPLE 3.5
To illustrate formula 3.23, we will compute the Van Wormer coefficient for dielectric losses for cable
No. 3 described in Appendix A. From Table Al, we have D; = 67.26 mm and d, = 41.45 mm.
Hence,
 PartI »Modeling
67.26oa 67.26 Mi67.26“aa 2p os
(Gras) "\ 41.45 41.45 2| \41.45 = 0.585
i 67.26\
2)a) 67.26\
41.45 1 °
|e (ON
Len(meee
An example of the transient analysis with the voltage applied simultaneously with the current
is given in Example 5.4.

3.3.2 Reduction of a Ladder Network to a Two-loop Circuit

CIGRE (1972, 1976) and later IEC (1985, 1989) introduced computational proce-
dures for transient rating calculations employing a two-loop network with the intention of
simplifying calculations and with the objective of standardizing the procedure for basic
cable types. Even though with the advent and wide availability of fast desktop computers
the advantage of simple computations is no longer so pronounced, there is some merit in
performing some computations by hand, if only for the purpose of checking sophisticated
computer programs. To perform hand computations for the transient response of a cable to
a variable load, the cable ladder network has to be reduced to two sections. The procedure
to perform this reduction is described below.
Consider a ladder network composed of v resistances and (v+ 1) capacitances as shown
in Fig. 3.6. If the last component of the network is a capacitance, the last capacitance Q,,+ is
short circuited. An equivalent network, which represents the cable with sufficient accuracy,
is derived with two sections T4 Q, and Tg Qz as shown in Fig. 3.7.

LPs, Ts Ty aL
v-1
T
V

5aE I
rent ae
+Oa ibQs diQy | Qy.1| Q, Qya4

| Figure 3.7 Two-loop equivalent network.

The first section of the derived network is made up of T, = T, and Q4 = Q, without

modification in order to maintain the correct response for relatively short durations.
The second section Tz Q, of the derived network is made up from the remaining
sections of the original circuit by equating the thermal impedance of the second derived
section to the total impedance of the multiple sections. The Laplace transform of the thermal
impedance of the circuit Tz Q, to T,Q, is
Chapter 3 m=Circuit Theory Network Analogs for Thermal Modeling 47

Z(s)=1
SOs
Tz+1
sQ,+1 (3.24)
Tes
ey PU
NOwii:
and the corresponding operational equation of the simple equivalent network is

253s)
=pe Ue 225)

The total thermal resistance must be the same for each case; therefore,

Tp Sie sl ee aT, (3.26)

Equating equations (3.24) and (3.25), an approximation for the equivalent capacitance is
obtained by comparing the first degree terms in s and neglecting terms of higher degree:

T,+05
DitOait +---+T,
Gaerne \?
rsAr Gri)

+(Ree EYon+ +( ua Jo
PRTOok pe PyeTp
eT, opsen
Even though formulas (3.26)—(3.27) are straightforward, a great deal of care is required
when the equivalent thermal resistances and capacitances are computed in the case when
sheath, armor, and pipe losses are present (IEC, 1985). This is because the location of these
losses inside the original network has to be carefully taken into account. The following
section discusses several examples which illustrate this point.

3.3.3 Two-section Ladder Networks for Some Selected Cable Constructions

As indicated above, in order to make computations reasonably convenient and uniform

throughout the industry, CIGRE (1972) proposed that two-section networks be used in
transient rating computations. Earlier researchers (Buller, 1951; Van Wormer, 1955; Neher,
1964) used two-section networks for transient analysis as well since they afforded a relatively
simple mathematical solution. In this section, we will derive several such networks for
common cable constructions. It should be noted that, in a cable, all thermal resistances,
capacitances, and losses are to be calculated as quantities per conductor, or per equivalent
conductor in the case of multicore cables.
In a two-section network representing the cable, the first section always includes the
thermal capacitance of the conductor and the inner portion of the dielectric along with the
thermal resistance of the dielectric (or a part thereof for short-duration transients). The
second section includes thermal resistances and capacitances of the remaining components
of the cable.
48 Part I m Modeling

EXAMPLE 3.6

3.3.3.1 Solid and Self-contained Fluid-filled Paper-insulated Cables, Extruded Ca-

bles, and Thermally Similar Constructions. We will construct a two-loop equivalent network
for model cable No. | assuming: (1) a short-duration transient, and (2) a long-duration transient.
(1) Short-duration transient
From Table Al, we observe that, for this cable, short-duration transients are those lasting half
an hour or less. The diagram of the full network for a short-duration transient is shown in Fig. 3.8.

|1 ilkey
pie. qs! 3

Figure 3.8 Network diagram for cable No. | for short-duration transients.

The method of dividing insulation and jacket capacitances into parts is discussed in Sec-
tion 3.3.1. Before we apply the reduction procedure, we combine parallel capacitances into four
equivalent capacitances. In the equivalent network, only conductor losses are represented. There-
fore, to account for the presence of sheath losses, the thermal resistances beyond the sheath must be
multiplied, and the thermal capacitances divided by the ratio of the losses in the conductor and the
sheath to the conductor losses. By performing these multiplications and divisions, the time constants
of the thermal circuits involved are not changed. Thus,

OV —0.-- ppOn Os — (ep On tp On

Q;-— Qj Qj 3.28)
Q3 = (l—"p")
pyQ Oyo 04 = Ses
ase QO:
— rae --

To compute numerical values, we will require expressions for Q;; and Q;7. These expressions
are given in Chapter 9. The numerical values are as follows: Q;,; = 763 J/K - mand Q;. =
453.9 J/K - m. With these values and with the additional numerical values in Table Al, D; = 30.1
mm, d. = 20.5 mm, D, = 35.8 mm, and D, = 31.4 mm, we have

| | l |
— = = 0.468
D; D; 30.1 30.1
In | — —)-1 In —— —|
d.. d. 20.5 20.5

‘ | | l l
2 = 0.478
20 (2)
D. De\,3 epee a4)
35, BB" 2 ge
D, De 31.4 31.4
Q; = 1035+ 0.468- 763= 1392.1J/K-m
Q2 = (1 — 0.468)763 + 0.468 - 453.9 = 618.3 J/K- m
Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling 49

4 + 0.478 - 394.8
O3 = (1 —0.468)453.9 = 241.5 J/K-m, Q4= oe ae 176.8 J/K-m

_ (1 = 0.478)394.8
Q5= ite = 189.1 J/K-m

The final capacitance Qs is omitted in further analysis because the transient for the cable response is
calculated on the assumption that the output terminals on the right-hand side are short circuited.
Since the first section of the network in Fig. 3.8 represents the conductor, and in rating com-
putations the conductor temperature is of interest, the equivalent network will have the first section
equal to the first section of the full network; that is,

Tied ond | O, = 0; (3.29)

Fromequation(3.26),wehave
Tp= 311+ (1+A)T3 (3.30)
Thermal capacitance of the second part is obtained by applying equation (3.27):

(1+A))T3 E
QO, = =Q2}+ Bese
Oe ( Q3 SI
+ Qs)
4 (3.31)
3351

The sheath loss factor and thermal resistances for this cable are given in Table Al as 4, = 0.09,
T; = 0.214 K - m/W, and 73 = 0.104 K - m/W. Substituting numerical values in equations (3.29)—
(3.31), we obtain

Ti OO Kem Wi Oe —139251Kee
Tg= 0.107+ 1.09- 0.104= 0.220 K- m/W
1.09 - 0.104
OF—
Oso.( 0.107 + 1.09 - 0.104 ) (241.5
+176.8)
=729.4
J/K-m
(2) Long-duration transients
Long-duration transient for this cable are those lasting longer than 0.5 h. The appropriate
diagram is shown in Fig. 3.9.
In this case, we have

The insulation and jacket are split into two parts with the Van Wormer coefficients given by equations
(3.19) and (3.20), respectively. Since the last part of the jacket capacitance is short circuited (see
Section 3.3.2), Q,4 and Qz are simply obtained as the sums of relevant capacitances:

T3

diesoogl
cae
Figure 3.9 Network diagram for a long-duration transient for model cable No. 1.
50 Part I mwModeling

Qa = OQ,+ pQi QO,=(1-—p)Qi+ eet

Substituting numerical values, we obtain

l 1 = 0.437
PpSS2in (2)
D; 5 as
DiY\ =sey 2n (=)
30.1 Ss.
a M $4_,
30.1)
hi ZL. 205) (205

T, =0.214K-m/W, Q4 = 1035+0.437- 915.6= 1435.1/K +m

Tz= 1.09-0.104 = 0.113 K- m/W
4+ 0.478 - 394.8
Qp = (1 — 0.437)915.6 + el = 692.3 J/K-m

EXAMPLE 3.7

3.3.3.2 High-pressure Fluid-filled Pipe-type (HPF F) Cables. | Consider model cable

No. 3 described in Appendix A. The parameters of this cable are: D; = 67.26 mm, d, = 41.45 mm,
D,. = 244.48 mm, and D, = 219.08 mm, A; = 0.010, 42 = 0.311, T; = 0.422 K - M/W, Th =
0.082 K . mW, and 7; = 0.017 K - m/W.
The thermal circuits are quite different for short- and long-duration transients.
(1) Short-duration transients
Short-duration transients for this cable are those lasting up to 6 h. The thermal circuit for short
duration transients is shown in Fig. 3.10.

dp)
oO
(oy
Ss
ze
a 2
5 dp)
@ C=]
©
— — (a)

+—}—_
Conductor 3
B
£
3
;
(7p)
;
a.
o
:
.
[at

o_O,
1/27,
}— = ae
flige :li ts
5 D © it F
e ©
.
Q,
t—|— 5

Figure 3.10 Ladder network for an HPLF cable for short-duration transients.

The first section of the equivalent network is composed ofthe conductor and half of the insulation;
that is,

T, = iT; Q4=(Q. + p*Qii)/3 (3.34)

The thermal resistance of the second part is simply the summation of the remaining thermal
resistances with appropriate multiplying factors to take account of losses in skid wires and screens as
well as those in the pipe:

Tg =2T, + (1+1)T + (1 +a, +A2)73 (3.35)

Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling 51

To compute Q,, we apply equation (3.27):

er
ps eeOt On tee
he ees
2) la-pyona+
3SPHie Tp 1+,

ais( (+a; +4)T3\?

) ( 42, vee’0
2 peat Oy (3.36)
Tp 1+, L+A,+A,
where Q;, and Q;» are defined in Chapter 9, Tg is given by (3.26), Q,, Qo, Q,, and Q; denote skid
wire, oil, pipe, and pipe covering capacitances, respectively.
The numerical values of Q;; and Q;> are 1680.5 (J/K - m) and 2726.9 (J/K - m), respectively.
Capacitance of the oil is equal to 38570 (J/K - m) (see Section 9.7.1). Therefore,

] | ]
Fy ieiweyeDi D, intho7 2mm GT.06 = 0.460
In | — —)-1 Ih—— ——-
41.45 41.45
]
= — — eee
Dp 2In( D. —
DA 2eefmehIn244.48 panne 2eS
D, D, 219.08 219.08
T, = $1,/3 = 40.422/3= 0.07K- mW,
QO, = (3484.5 + 0.46 - 1680.5) /3 = 1419.2 /K-m

Tp, =-0.07 + 1.01 - 0.082 +.1.321 0.017 = 0.175 K. m/W

Op = (1 —0.46) 1680.5/3 + 0.46 - 2726.9/3

( 0.175 )‘( 04021200

=0107 Be12/3);
0S868:570
) 1.01

(=;0.175
aaa (=-38570
rie 16126.4
+(221
0.482-15
720.9
) = pee8381.2
edt
/K-m.
ae
(2) Long-duration transient
For a long-duration transient, oil thermal resistance is split in half as shown in Fig. 3.11.

T (a eI

(Pulled ieee L
lies oes. aed |
Figure 3.11 Ladder network for an HPLF cable for long-duration transients.

The parameters of an equivalent network are obtained from

te NEES OV (Oop Oi)/3 (Geo)

Tp Xl Apostle Ay Aa) 3 (3.38)

52 Part I » Modeling

Oz ee
= ( = pyOwe + Q, [20
+anh+ +e+h ri Qo)
3(1
+A1) Tp I+, (3.39)
ri (14 ApAd) *0,+p'Q;
Tp Vea aya!
Using the numerical values obtained from Table Al, we have

a
P= 2In +)
Senn Camere ee se 2;
A pey
D, LN\2 ast) cepa67.26 (uaa
d, d. 41.45 41.45
T, = 0.422/3 = 0.141 K-m/W, Qa = (3484.5 + 0.421 - 1458.4)/3 = 1366.2 J/K -m

Tg = 1.01 - 0.082 + 1.321 - 0.017 = 0.105 K - m/W

12 0.5 - 1.01-0.082+ 1.321-0.017\” (38570

Qz = (1 —0.421)1458.4/3+ Se SSS
3-1.01 0.105 1.01
Usdd 2G1
e,O1L\GeOdar0.482 9O20.)eite nae Kee
0.105 1.321

EXAMPLE 3.8

3.3.3.3 Cables in Ducts. Consider cable No. 2 (Appendix A) installed in a PVC duct
as described in Example 3.4. The parameters of this cable are: D; = 30.1 mm, d. = 20.5 mm,
D, = 72.9 mm, and D, = 65.9 mm, A; = 0.022, T; = 0.307 K - m/W, 73 = 0.078 K - m/W, and the
thermal resistances of the air in the duct and the duct itself are equal to 0.4167 and 0.0216 K - m/W,
respectively.
The following formulas, derived similarly as in previous examples, apply in this case:
(1) Short-duration transients
The diagram in Fig. 3.12 represents the ladder network for this case.

= _—
s 2 3
& oO D a)

+
Conductor
3ae
is
127,
(dp)
aryoe
5
2
®
s)
=
epee
x
eS
<x (a)
ay oie
&
s
~
S

Speeeaeee ere ence

Da sigsa
sel eifhadigeeluepaical
as
Figure 3.12 Ladder network for an equivalent single-core cable for cable No. 2 in duct for
short-duration transient.

™=1%.. Os=0-+P*On (3.40)

Te = 3T+ +4)(B+T%+T%) (3.41)

To compute Qx, we apply equation (3.27):

Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling 53

Ox =( Pp)On4p 20+| (Gye

T, i i { P )Qn+ Oen0;
Tage
Si Ona One.| eA =p \On
(3.42)

[cee ems)
“8 Tp Ih(uae) =ol Ou
se
Jb Tp TAs
where Ty is given by equation (3.41), iL, and ‘Gi are the thermal resistances of the air space in duct
and the duct itself, respectively, and Q, is the capacitance of the duct.
To obtain numerical values, we recall that for short-duration transients, the mutual heating of
the cores is neglected, and equations (3.40)—(3.42) apply to a single core of the cable. The time
limit for the short-duration transient is equal to ©T - £Qwhere UT and YQ refer to one core only.
Short-duration transients for cable No. 2 located in duct in air last less then 36 min. The long-duration
transients are those longer than 2 h (see Example 3.4). For transients having durations between these
two values, a network diagram for short-duration transients must be solved. The numerical values
required for this case are as follows:

I 1 1 1
S= = = 0.468
P ae D, D, ie patel
ea etha rye tg et |
a a 20.5 20.5
The Van Wormer coefficient for the cable jacket is equal to

i = : pele | 1 area FE
56483
| Saale a
De
2In( — De 2 Seen GORD
51 fem! Aa 2 age
Ds De oh EE)

For the duct dimensions as specified above, the values of T, and T, are equal to 0.4167 and
0.0216 K - m/W, respectively. The thermal capacitances of the first and second part of the insulation
are as follows (see Chapter 9 for the derivation of these values):

With these values and the remaining parameters specified in Table Al, we obtain

T, = 0.307/6 = 0.051 K-m/W Qy, = (1035 + 0.468 - 371)/3 = 402.9 J/K -m

ip = 5 -0.307/3 + 1.022(0.078 + 0.4167 + 0.0216) = 0.580 K - m/W

On=(1—
=i Ie0.468)371/3
+0.468- 544.7/3
468- 544.7/3
4 ( 0.580
—
aeBee)
+ c —0.468)544.7/3 + ——____——
4 + 0.483 - 1297.3
1.022

1.022(0.4167 + 0.0216) ]*(1 — 0.483) 1297.3

0.580 O22,

1.022 - 0.0216 \*539.4

= 1134.4 J/K-m
0.580 1.022
(2) Long-duration transients
Long-duration transients for this cable last 2h or more. The diagram in Fig. 3.13 represents the
network diagram for this case.
 Part I m Modeling

T, T, T,’ ivy
i (est ===fl il festdr sh 5 I dc S
==Q, | pQ; ==(1-p)Q; T Q, TH ih (1-p')Qi
i QairT Qy -O

Figure
3.13Ladder
network
foranequivalent
single-core
cable
forcable
No.2induct
for
long-duration
transient.
Ty,
= T, Q,= 2. +pQi (3.43)
Ty=(1+Ai)(T3+T,
+T,) (3.44)

(1 DO pe Q.+p0;
ee ee | +a
ee +T%)
| (A-p)Q;
Qzg=(1-p)Qi+fsa ty Ts 1+),
| ees (3.45)
Tp 1+),
To obtain numerical values, we will represent the three-conductor cable as an equivalent single-
core cable. Thermal resistance of a single cable is equal to one third of the value given in Table Al.
The equivalent conductor diameter is obtained from equation (3.21) as

2nT, 270.307
a= Die = 65.9e° 335 3
= 54.8 mm

The capacitance of the equivalent conductor is equal to the sum of the capacitances of the three
conductors plus the capacitance of that portion of the insulation which is enclosed within the perimeter
of the equivalent conductor.
The insulation area enclosed within the equivalent conductor is equal to

Sins= 1 - (54.87—3 -20.5")/4 = 1368.4mm?

Hence, from equation (3.13) and the material properties as given in Table 9.1, the capacitance of the
equivalent conductor is given by

Q. = 3- 1035 + 2.4 - 1368.4 = 6389.2 J/K -m

The equivalent cable has a diameter over insulation equal to 65.9 mm. The capacitance of the
insulation is obtained from equation (3.14) as

a = (65.9" —54.87)2.4 = 2525.4 /K-m

The Van Wormer coefficient is thus equal to

] l 1 l
p= 5 = = 0.469
D;
2In(3) \2 wal
Dy\* eels:85.9)
at 6:8 2Woe
s2aa
d- d. 54.8 54.8
Chapter 3 m Circuit Theory Network Analogs for Thermal Modeling 55

The equivalent network parameters are now obtained from equations (3.43)—(3.45):

T, = 0.307/3 = 0.102 K-m/W Q, = (6368 + 0.469 - 2525.4) = 7552.4 J/K-m

Tg = 1.022(0.078 + 0.4167 + 0.0216) = 0.5277 K - m/W

0.483 - 1297.3 1.022(0.4167 + 0.0216) ]*(1 — 0.483) 1297.3

Op = (1 —0.469)2525.4 +
1.022 0.5277 1.022
1.022 - 0.0216 \* 539.4
= 2427.9 J/K-m
0.5277 1.022

The evaluation of the losses, thermal resistances and capacitances shown in the ladder
network is discussed in detail in Chapters 6-9. Once all the parameters for the lad-
der network are determined, the network equations are solved as with electrical RC net-
works. Solution of the network equations yields the temperatures at various nodes of the
network corresponding to points in the cable. This task is fairly straightforward for steady-
state computations and somewhat more laborious for transient analysis.
In closing this chapter, we note that the lumped parameter network is not the only
model that can be applied to solve heat conduction equations. In some publications, fully
distributed models using Fourier integrals have been used. In the fully distributed model,
the temperature is a function of time and the radial distance from the cable conductor. The
lumped parameter model allows the computation of the temperature of each cable region
as a function of time only. Since this book is focused on the most common methods used
by cable engineers, we will consider the lumped parameter model only. Readers interested
in an example where Fourier integrals are used for transient cable ratings are referred to
Bernath and Olfe (1986).

REFERENCES

Bernath, A., and Olfe, D. B., (July 1986), “Cyclic temperature calculations and mea-
surements for underground power cables,” JEEE Trans. Power Delivery, vol. PWRD-1,
pp. 13-21.
Bernath, A., Olfe, D. B., and Martin, F. (July 1986), “Short term transient temperature
calculations and measurements for underground power cables,” JEEE Trans. Power De-
livery, vol. PWRD-1, pp. 22-27.
Buller, EF.H. (1951), “Thermal transients on buried cables,” Trans. Amer. Inst. Elect. Eng.,
vol. 70, pp. 45-52.
CIGRE (Oct. 1972), “Current ratings of cables for cyclic and emergency loads. Part 1.
Cyclic ratings (load factor less than 100%) and response to a step function,” Electra,
no. 24, pp. 63-96.
CIGRE (Jan. 1976), “Current ratings of cables for cyclic and emergency loads. Part 2.
Emergency ratings and short duration response to a step function,” Electra, no. 44,
pp. 3-16.
IEC Standard (1985), “Calculation of the cyclic and emergency current ratings of cables.
Part 1: Cyclic rating factor for cables up to and including 18/30 (36) kV,” Publication
853-1.
56 Part I » Modeling

IEC Standard (1989), “Calculation of the cyclic and emergency current ratings of cables.
Part 2: Cyclic rating factor of cables greater than 18/30 (36) kV and emergency ratings
for cables of all voltages,” Publication 853-2.
Neher, J. H., and McGrath, M. H. (1957), “Calculation of the temperature rise and load
capability of cable systems,” AJEE Trans., vol. 76, part 3, pp. 755-772.
Neher, J. H. (1964), “The transient temperature rise of buried power cable systems,” JEEE
Trans., vol. PAS-83, pp. 102-111.
Pashkis, V., and Baker, H. D. (1942), “A method for determining the unsteady-state heat
transfer by means of an electrical analogy,’ ASME Trans., vol. 104, pp. 105-110.
Van Wormer, F. C. (1955), “An improved approximate technique for calculating cable
temperature transients,” Trans. Amer. Inst. Elect. Eng, vol. 74, part 3, pp. 277-280.
Wollaston, F. O. (1949), “Transient temperature phenomena of 3 conductor cables,’ AJEE
Trans., vol. 68, part 2, pp. 1248-1297.
 4

Rating Equations—
oteady-State Conditions

The current-carrying capability of a cable system will depend on several parameters. The
most important of these are (1) the number of cables and the different cable types in the
installation under study, (2) the cable construction and materials used for the different cable
types, (3) the medium in which the cables are installed, (4) cable locations with respect to
each other and with respect to the earth surface, and (5) the cable bonding arrangement. For
some cable constructions, the operating voltage may also be of significant importance. All
of the above issues are taken into account; some of them explicitly, the others implicitly, in
the rating equations presented in this chapter. The lumped parameter network representation
of the cable system discussed in Chapter 3 is used for the development of steady-state and
transient rating equations. These equations are developed for a single cable either with one
core or with multiple cores. However, they can be applied to multicable installations, for
both equally and unequally loaded cables, by suitably selecting the value of the external
thermal resistance as shown in Section 9.6.
The development of cable rating equations is quite different for steady-state and tran-
sient conditions. Therefore, we will discuss these issues in two separate chapters, Chap-
ter 4 and Chapter 5.

4.1 BURIED CABLES WITH NO MOISTURE MIGRATION IN THE SOIL

Steady-state rating computations involve solving the equation for the ladder network shown
in Fig. 3.1 with the thermal capacitances neglected. The resulting diagram which also
includes the external thermal resistance is shown in Fig. 4.1.
The unknown quantity is either the conductor current / or its operating temperature 6.
In the first case, the maximum operating conductor temperature is given, and in the second

57
58 Part I mModeling

Figure 4.1 The ladder diagram for steady-state rating computations. (a) Single-core cable.
(b) Three-core cable.

case, the conductor current is specified. Since losses occur at several positions in the cable
system (for this lumped parameter network), the heat flow in the thermal circuit shown in
Fig. 4.1 will increase in steps. Thus, the total joule loss W; in a cable can be expressed as

W, = W. + Ws + Wa = We + Ai + Az) (4.1)

where W,, W,, and W, are conductor, sheath, and armor losses, respectively. The quantity
A, is called the sheath loss factor, and is equal to the ratio of the total losses in the metallic
sheath to the total conductor losses. Similarly, A2 is called the armor loss factor, and is
equal to the ratio of the total losses in the metallic armor to the total conductor losses.
Incidentally, it is convenient to express all heat flows caused by the joule losses in the cable
in terms of the loss per meter of the conductor.
Referring now to the diagram in Fig. 4.1, and remembering the analogy between the
electrical and thermal circuits, we can write the following expression for A@, the conductor
temperature rise above the ambient temperature:

Ad = (W. + 4Wa)T) +[We +41) + WalnTe

4 (4.2)
+ LWel + Ay Aa) Wa ln(T3 + Ta)

where W,, A, and A> are defined above, and n is the number of load-carrying conductors
in the cable (conductors of equal size and carrying the same load). W, represents dielectric
losses of which the evaluation is discussed in Chapter 6. The ambient temperature is the
temperature of the surrounding medium under normal conditions at the location where the
cables are installed, or are to be installed, including any local sources of heat, but not the
Chapter 4 # Rating Equations—Steady-State Conditions 39

increase of temperature in the immediate neighborhood of the cable due to the heat arising
therefrom. 7), T>, T3, and Ty are the thermal resistances where 7; is the thermal resistance
per unit length between one conductor and the sheath, 7> is the thermal resistance per unit
length of the bedding between sheath and armor, 73 is the thermal resistance per unit length
of the external serving of the cable, and T, is the thermal resistance per unit length between
the cable surface and the surrounding medium.
The permissible current rating is obtained from equation (4.2). Remembering that
W.. = I? R, we have

a AGaWalOSian 1a -elagt 14)] i (4.3)

RT +nRO +Ai)T2 +nRd + Ay +A2) (73 + Ta)

where R is the ac resistance per unit length of the conductor at maximum operating tem-
perature.

EXAMPLE 4.1
Determine the steady-state rating of a cable system composed of three single-core cables in flat
formation using cable model No.1 in Appendix A. The parameters of this system are specified in
Table Al as A, = 0.09, 7; = 0.214 K-m/W, 73 = 0.104 K-m/W, and Ty = 1.933 K-m/W. Since for
this cable dielectric losses are negligible, applying equation (4.3) results in

AS Ue
ir — 629A
PRT. 10° (0.21421 (le 0109) “0 1 1+ 0.09) 0.1044 1.933)

Equation (4.2) is often written in a simpler form which clearly distinguishes between
internal and external heat transfers in the cable. Denoting

ii
T=— +1 +a)h+ th +ATs
(4.4)
T)
Tg = —+1,4+T;
2n
equation (4.2) becomes

A@ =n(Wel + Wielait Wala) (4.5)

where W, are the total losses generated in the cable defined by

W, = W, + Wa = We +A, +A2) + Wa (4.6)

and JT computed from (4.4) is an equivalent cable thermal resistance. This is an internal
thermal resistance of the cable which depends only on the cable construction. The external
thermal resistance, on the other hand, will depend on the properties of the surrounding
medium as well as on the overall cable diameter, as explained in Chapter 9.
The last term in (4.5) is the temperature rise caused by dielectric losses. Denoting it
by Aézq,
Ady = nWa Ty (4.7)
60 Part I » Modeling

4.2 BURIED CABLES WITH MOISTURE MIGRATION TAKEN INTO ACCOUNT

4.2.1 Introduction

The current-carrying capacity of buried power cables depends to a large extent on the
thermal conductivity of the surrounding medium. In fact, results reported by El-Kady (1985)
indicate that the sensitivity of cable temperature to variations in thermal conductivities of the
surrounding medium is at least an order of magnitude greater than sensitivity to variations in
other parameters such as ambient temperature, heat convection coefficient, or cable current.
Soil thermal conductivity is not a constant, but is highly dependent on its moisture
content (Mochlinski, 1976). Under unfavorable conditions, the heat flux from the cable
entering the soil may cause significant migration of moisture away from the cable. A dried-
out zone may develop around the cable, in which the thermal conductivity is reduced by a
factor of three or more over the conductivity of the bulk. This, in turn, may cause an abrupt
rise in temperature of the cable sheath which may lead to damage to the cable insulation.
Studies of this problem have made it apparent that a need exists for a well-formulated
procedure for calculating cable ampacities taking into account heat and moisture migration
in the soil. The problem of thermal runaway has been studied by Groeneveld et al. (1984),
Black et al. (1982), and Arman et al. (1964). In the first two cases, an analytical solution
to the moisture migration problem was proposed and the governing equations were solved
as single-dimensional approximations using the finite-difference method.
However, in typical engineering practice adopted by power utilities, the current ratings
of cables are established on the basis of an assumed thermal conductivity of the surrounding
medium. One of the reasons for this is that strict mathematical explanations and physical
models describing moisture migration phenomena are very complicated, and adequate eval-
uations of the quantities concerned have not yet been made. In order to give some guidance
on the effect of moisture migration on cable ratings, CIGRE (1986) has proposed a simple
two-zone model for the soil surrounding loaded power cables, resulting in a minor mod-
ification of equation (4.3). Subsequently, this model has been adopted by the IEC as an
international standard (IEC, 1993).
The concept on which the method proposed by CIGRE relies can be summarized as
follows. Moist soil is assumed to have a uniform thermal resistivity; but if the heat dissipated
from a cable and its surface temperature are raised above certain critical limits, the soil will
dry out, resulting in a zone which is assumed to have a uniform thermal resistivity higher
than the original one. The critical conditions, that is, the conditions for the onset of drying,
are dependent on the type of soil, its original moisture content, and temperature.
Given the appropriate conditions, it is assumed that, when the surface of a cable
exceeds the critical temperature rise above ambient, a dry zone will form around it. The
outer boundary of the zone is on the isotherm related to that particular temperature rise (see
Fig. 4.2). An additional assumption states that the development of such a dry zone does not
change the shape of the isothermal pattern from what it was when all the soil was moist,
only that the numerical values of some isotherms change. Within the dry zone, the soil has
a uniformly high value of thermal resistivity, corresponding to its value when the soil is
“oven dried” at not more than 105°C. Outside the dry zone, the soil has uniform thermal
resistivity corresponding to the site moisture content. The essential advantages of these
assumptions are that the resistivity is uniform over each zone, and that the values are both
convenient and sufficiently accurate for practical purposes.
Chapter 4 m Rating Equations—Steady-State Conditions 61

Ground surface

Cable

Py
moist

Figure 4.2 Illustration of the concept of dry zone within an isothermal circular boundary.

The method presented below assumes that the entire region surrounding a cable or
cables has uniform thermal characteristics prior to drying out, the only nonuniformity being
that caused by drying. As a consequence, the method should not be applied without further
consideration to installations where special backfills, having properties different from the
site soil, are used.

4.2.2 Development of Rating Equation

Let 8, be the cable surface temperature corresponding to the moist soil thermal resis-
tivity ;. Then, without moisture migration, we obtain the following relations by applying
equation (3.6) and remembering that soil thermal resistance is directly proportional to the
value of resistivity:

= 6, —Ogmbni (6. —Ox) aig(6, —6amb) (4.8)

Ww,
Hisee T; Te

and

0, —0,
nWy Cay (4.9)

where C is aconstant and n is the number of cores in the cable. 6., Qamp,and 0, are cable
surface temperature, ambient temperature, and the temperature of an isotherm at distance
x, respectively. The total losses are given by equation (4.6)
If we now assume that the region between the cable and the 9, isotherm dries out so
that its resistivity becomes 2, and that the power losses W, remain unchanged, we have

6! — 6,
———r
nw, CE (4.10)

where @/is the cable surface temperature when moisture migration has taken place.
Combining equations (4.9) and (4.10), we obtain the following form of equation (4.8):

nw, (4.11)
r a
62 Part | m Modeling

We can now define A@, = 6, — Oampas the critical temperature rise of the boundary
between the moist and dry zones above ambient temperature. From equation (4.5), the
cable surface temperature is equal to

Substituting (4.12) into (4.11)

and recalling that W. = I? RandW, = W.(1+A,+A2)+ Wa,
we obtain the following equation for the current rating with moisture migration taken into
account :

= E — W,[0.57; + n(72 + T3 + vTy)] + (v - Des.) “

RT; +nR( +A))TIo +nRCU+A, +A2)(T3 + vT4)

where v = p2/; and the remaining variables are defined above. Note that 7, is the external
thermal resistance of the cable when it is laid in soil having a uniform resistivity p1.
We can observe that equation (4.3) has been modified by the addition of the term
(v — 1)A@, in the numerator, and the substitution of v74 for 7, in both the numerator and
the denominator.

EXAMPLE 4.2
We will illustrate the effect of moisture migration on the rating of model cable No.1 located in standard
conditions described in Appendix A. All the numerical values required in equation (4.13) are given
in Table Al as A; = 0.09, T; = 0.214 K-m/W, 7; = 0.104 K-m/W, and 7, = 1.933 K-m/W. In
addition, we assume the values p2 = 2.5 K-m/W for the dry soil thermal resistivity and A6, = 35°C
for the critical temperature rise. Then, applying equation (4.13),

ia 75+(2.5—1)+35 a Sage
~ [7.81
-10-5(0.214
+1-(1+0.09)-0+1-(1+0.09)(0.104
+2.5-1.933)
]
Thus, when the moisture migration is taken into account a 14% reduction in cable ampacity is required
in comparison with the result obtained in Example 4.1.

The question of determination of A@,, the critical temperature rise above ambient, has
been addressed by Donnazi et al. (1979). In practice, values between 35 and 50°C are
used in most countries for the critical isotherm temperature value 6,. The following section
discusses this subject in more detail.

4.2.3 Determination of the Critical Temperature Rise Aé,

There is not a great deal of direct information on the practical behavior of moisture
in soil under the influence of a varying temperature distribution. Determination of the
temperature, and hence the position of the critical isotherm, is a complicated matter, but
examples of theoretical and experimental derivations are given in the literature (Donnazi
etal., 1979; Brakelmann, 1984; Groeneveld er al., 1984; Black et al., 1982). The following
method is an adaptation of a work published by Donnazi et al. (1979), (CIGRE, 1992). It is
the only practical method known at present which provides values of critical temperature and
resistivity (apart from empirical values adopted in some countries), and which is relatively
simple and backed up by experimental evidence.
Chapter 4 m Rating Equations—Steady-State Conditions 63

The method relies on two experimentally determined quantities: (1) a critical moisture
content expressed as a critical degree of saturation, and (2) a migration parameter. The
critical degree of saturation can be determined by the use of a migration test or, for most
sandy materials, from thermal resistivity/moisture content measurements. The migration
parameter is obtained from a migration experiment. In the description of the parameters
and their use, it is assumed that the soil surrounding the cable is homogeneous except for
the moisture distribution derived from the two zones.
The critical temperature rise of a soil is then related to the critical degree of saturation
Scr (expressed as a fraction) and a parameter n(K~!), as follows:

Ab, = —ja
1 SS) 1 — Scr I , 2
re 5 a ¥. Sn) ae (Sa a. Ser) sa 250) (4.14)
n I Sa

where s, is the degree of saturation of the soil controlled by the ambient moisture at
the site.
An experimental method of deriving the values of s,, and n is described by Donnazi
et al. (1979) and in CIGRE (1992). Table 4.1 (CIGRE, 1992) gives the values of the
parameters s,, and 7 and the critical temperature rise A@, as a function of s,, for some
selected soils!:

TABLE 4.1 Values of Critical Temperature Rise (K)

Type of Soil

Sand/Clay Sandy Crushed Rock Selected Sands

Porosity 0.47 0.42 0.24 0.25,

Ge 0.5 0.3 0.18 0.25
nee 2.104 f104 0.6-104 0.6:10-+
y, Critical Temperature Rise
0.25 BS)
0.3 13 1
0.35 0.6 40 8
0.4 5 90 30
0.45 19 74
0.5 49
0.55 0.4
0.6 4
0.65 15
0.7 3)
0.75 85

In closing this section, we would like to point out that the crucial assumption used in
the above developments, that the critical temperature rise is independent of the heat flux
at the surface of the cable, may not be valid when the soil becomes thermally unstable. In
fact, developments presented by Hartley and Black (1982) suggest that the heat flux at the

! The soil descriptions given in Table 4.1 are not very precise, but they give a good indication of the desired
thermal properties of soils. Crushed rock has the best thermal properties since its critical temperatures are high at
low moisture contents. Selected sands follow closely.
64 Part I mModeling

surface of the cable is an important factor in determining the time required for a soil to
become unstable. Therefore, equation (4.10) should be used with caution, and only ina
completely static case where drying out occurs and soil reaches an equilibrium condition.

4.3 CABLES IN AIR

When cables are installed in free air, the same ladder network is used as discussed in Sec-
tion 4.1. However, the external thermal resistance now accounts for the radiative and con-
vective heat loss. For cables exposed to solar radiation, there is an additional temperature
rise caused by the heat absorbed by cable external covering. The heat gain by solar absorp-
tion is equal to 0D,H, with the meaning of the variables defined below. In this case, the
external thermal resistance is different than for shaded cables in air, and the current rating
is computed from the following modification of equation (4.3):

i‘ AO —W,[0.5T, + n(T + T3+ Tj)] — 0DHT; i a)

LRT, +n2RU +A) Io +nRO +A, + A2)(T3 + Tf)

where o = absorption coefficient of solar radiation for the cable surface; the values
of this coefficient are given in Table 9.2

H = intensity of solar radiation; if this value is unknown locally, it could be

assumed to be equal to 1000 W/m?
T,* = external thermal resistance of the cable in free air, adjusted to take
account of solar radiation (see Section 9.6.8).

We will illustrate the application of equation (4.13) in Section 9.6.8.1.

Computation of the values of several quantities appearing in equations (4.3) and (4.15)
is discussed in detail in Chapters 6 through 9. Computation of dielectric losses is discussed
in Chapter 6. Chapter 7 discusses computation of joule losses in the conductor, giving
formulas for the computation of the conductor resistance R. Chapter 8 describes evaluation
of sheath and armor loss factors, and Chapter 9 deals with the computation of thermal
resistances.

REFERENCES

Arman, A. N., Cherry, D. M., Gosland, L., and Hollingsworth, P. M. (1964), “Influence
of soil moisture migration on power rating of cables in H.V. transmission,” Proc. EE,
vol. 111, pp. 1000-1016.
Black, W. Z. et al. (Sept. 1982), “Thermal stability of soils adjacent to underground
transmission power cables,” Final Report of EPRI Research Project 7883.
Brakelmann, H. (1984), “Physical principles and calculation methods of moisture and heat
transfer in cable trenches,” ETZ Report 19. Berlin: VDE Verlag.
CIGRE (1986), “Current ratings of cables buried in partially dried out soil. Part 1: Simpli-
fied method that can be used with minimal soil information: (100% load factor),” Electra,
no. 104, pp. 11-22.
Chapter 4 m Rating Equations—Steady-State Conditions 65

CIGRE (Dec. 1992), “Determination of a value of critical temperature rise for a cable
backfill material,’ Electra, vol. 145, pp. 15-29.
Donnazi, F., Occhini, E., and Seppi, A. (1979), “Soil thermal and hydrological characteris-
tics in designing underground cables,” Proc. IEE, vol. 126, no. 6.
El-Kady, M. A. (Aug. 1985), “Calculation of the sensitivity of power cable ampacity to
variations of design and environmental parameters,’ JEEE Trans. Power App. Syst.,
vol. PAS-103, no. 8, pp. 2043-2050.
Groeneveld, G. J., Snijders, A. L., Koopmans, G., and Vermeer, J. (Mar. 1984), “Improved
method to calculate the critical conditions for drying out sandy soils around power cables,”
Proc. IEE, vol. 131, part C, no. 2, pp. 42-53.
Hartley, J. G., and Black, W. Z. (May 1981), “Transient simultaneous heat and mass transfer
in moist unsaturated soils,’ ASME Trans., vol. 103, pp. 376-382.
IEC (1993), “Amendment 3 to IEC publication 287: Calculation of the continuous current
rating of cables (100% load factor).”
Mochlinski, K. (1976), “Assessment of the influence of soil thermal resistivity on the ratings
of distribution cables,” Proc. IEE, vol. 123, no. 1, pp. 60-72.
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 S

Rating Equations—
Transient Conditions

9.1 INTRODUCTION

J. H. Neher began his famous 1964 paper (Neher, 1964) by stating: “The calculation of the
transient temperature rise of buried cable systems, that is, the determination of the curve of
conductor temperature rise versus time after the application of a constant conductor current,
has intrigued the more mathematically minded cable engineers for many years.” Thirty
years have passed since this paper was presented in Toronto, and the fascination with the
subject by mathematically minded cable engineers has not abated. On the contrary, judging
by the number of recent publications dealing with this topic, one might conclude that either
the number of mathematically inclined cable engineers must have risen dramatically in the
past 30 years, or that the subject has increased significantly in prominence after steady-state
computations were more or less standardized or, possibly both.
In transient rating computations, the following three questions are of particular impor-
tance:
1. Given current operating conditions (i.e., conductor temperature), what will be the
final component temperature if loading is increased by a given amount and if the
new loading is maintained for a specified period of time?
2. What is the maximum current a cable can carry for a given period of time if the
temperature of the conductor cannot exceed a specified limit?
3. Given current operating conditions, how long can a new higher loading be applied
without exceeding a specified temperature limit?

The first question requires computation of the temperature of the component of interest
after a specified time period (including the steady-state conditions). This can be achieved
by the methods described in Sections 5.2 and 5.3.

67
68 Part I m Modeling

The second question relates to the classical problem of transient ratings which also
includes cyclic conditions. The initial operating temperatures, which represent specified
preloading conditions, are given. These conditions can be obtained either from steady-state
computations, or from on-line monitoring. The time period of interest can vary between 10
min and infinity. In the latter case, one obtains the steady-state cyclic rating of the cables.
The load curves, possibly different for each cable circuit, could represent the same curves
which existed before the onset of transient conditions, but scaled appropriately to reflect
changes in the loading. Alternatively, special curves may apply for the duration of the
transient (e.g., a step function could be applied to some cables).
The third question reflects operating concerns when an emergency condition exists in
the system and the system operator needs to know how long this condition can be tolerated.
To answer the first question, computations of component temperature variation with
time are required. These computations are described by the mathematical model presented
in this chapter. Special attention will be paid to: (1) the selection of an appropriate time
step, (2) proper handling of the mutual heating of cables, and (3) the necessity to account
for the variation of electrical resistances of metallic components with temperature. The
treatment of the last two items is discussed in Sections 5.2 and 5.5. The selection of the
time step depends on the procedure adopted for the discretization of the load curve. A brief
discussion of this topic is presented in Section 11.2.3.
In general, computation of transient ratings (question 2 above) requires an iterative
procedure. The main iteration loop involves the adjustment of cable loadings. At each
step of iteration, the loading of each cable is selected and the temperature of the desired
component is calculated (see Fig. 5.1). This step requires the same procedure as that of
question |.
To answer the third question, the main iteration loop must allow positive or negative
increments in time. Once the time horizon is set for a particular iteration, the block com-
puting the temperatures is called. If the temperature of the component of interest is within
the specified tolerance from the desired temperature, the process is stopped; otherwise, a
new time horizon is set and the algorithm is repeated.
Based on the above discussion, the procedure to evaluate temperatures is the main
computational block in all three cases being considered. This block requires a fairly complex
programming procedure to take into account self and mutual heating, and to make suitable
adjustments in the loss calculations to reflect changes in the conductor resistance with
temperature.
Transient rating of power cables requires the solution of the equations for the network
in Fig. 3.5. The unknown quantity in this case is the variation of the conductor temperature
rise with time,! 6(t). Unlike in the steady-state analysis, this temperature is not a simple
function of the conductor current / (t). Therefore, the process for determining the maximum
value of / (rt) so that the maximum operating conductor temperature is not exceeded requires
an iterative procedure. An exception is the simple case of identical cables carrying equal
current located in a uniform medium. Approximations have been proposed for this case,
and explicit rating equations developed. We will discuss this case later in this chapter.

| Unless otherwise stated, in this chapter, we will follow the notation in (IEC, 1989), and we will use the
symbol @ to denote temperature rise and not A@ as in Chapter 4 and (IEC, 1982).
Chapter 5 m Rating Equations—Transient Conditions 69

Read/enter initial conditions.

Compute time-independent parameters.

Select the values of currents.

Temperature Computation Block

¢ Perform temperature computations

in time increments.
* Iterate within each time step adjusting
resistances of metallic parts and
recomputing Joule losses.

a Computed - specified
temperature < tolerance

NO

Adjust ratings up or down as required.

Figure 5.1 Flowchart for rating computations in transient analysis.

5.2 RESPONSE TO A STEP FUNCTION

5.2.1 Preliminaries

Whether we consider the simple cable systems mentioned above, or a more general
case of several cable circuits in a backfill or duct bank, the starting point of the analysis is
the solution of the equations for the network in Fig. 3.5. Our aim is to develop a procedure
to evaluate temperature changes with time for the various cable components. As observed
by Neher (1964), the transient temperature rise under variable loading may be obtained by
dividing the loading curve at the conductor into a sufficient number of time intervals, during
any one of which the loading may be assumed to be constant. Therefore, the response of a
cable to a step change in current loading will be considered first.
This response depends on the combination of thermal capacitances and resistances
formed by the constituent parts of the cable itself and its surroundings. The relative impor-
70 Part I » Modeling

tance of the various parts depends on the duration of the transient being considered. For
example, for a cable laid directly in the ground, the thermal capacitances of the cable, and
the way in which they are taken into account, are important for short-duration transients,
but can be neglected when the response for long times is required. The contribution of the
surrounding soil is, on the other hand, negligible for short times, but has to be taken into
account for long transients. This follows from the fact that the time constant of the cable
itself is much shorter than the time constant of the surrounding soil.
The thermal network considered in this work is a derivation of the lumped parameter
ladder network introduced early in the history of transient rating computations (Buller, 1951;
Van Wormer, 1955; Neher, 1964; CIGRE, 1972; IEC 1985, 1989). For computational
purposes, Baudoux (1962) and then Neher (1964) proposed to represent a cable in just
two loops. Baudoux provided procedures for combining several loops, to obtain a two-
section network which was latter adopted by CIGRE WG 02 and published in Electra
(CIGRE, 1972). This topic is addressed in Section 3.3.2. However, transformation of a
multiloop network into a two-loop equivalent not only requires substantial manual work
before the actual transient computations can be performed, but also inhibits the computation
of temperatures at parts of the cable other than the conductor. A procedure is given below for
analytical solution of the entire network. Generally, the network will be somewhat different
for short- and long-duration transients, and usually, the limiting duration to distinguish these
two cases can be taken to be | h. Short transients are assumed to last at least 10 min. A
more detailed time division between short and long transients can be found in Section 3.3.
The temperature rise of a cable component (e.g., conductor, sheath, jacket, etc.) can
be represented by the sum of two components: the temperature rise inside and outside the
cable. The method of combining these two components, introduced by Morello (Morello,
1958; CIGRE, 1972; IEC, 1985, 1989), makes allowance for the heat which accumulates
in the first part of the thermal circuit and which results in a corresponding reduction in
the heat entering the second part during the transient. The reduction factor, known as the
attainment factor a(t), of the first part of the thermal circuit is computed as a ratio of the
temperature rise across the first part at time ¢ during the transient to the temperature rise
across the same part in the steady state. Then, the temperature transient of the second part
of the thermal circuit is composed of its response to a step function of heat input multiplied
by a reduction coefficient (variable in time) equal to the attainment factor of the first part.
Evaluation of these temperatures is discussed below. The validity of Morello’s hypothesis
was demonstrated in experiments carried out by Wlodarski and Cabiac (1966).

5.2.2 Temperature Rise in the Internal Parts of the Cable

The internal parts of the cable encompass the complete cable including its outermost
serving or anti-corrosion protection. If the cable is located in a duct or pipe, the duct and
pipe (including pipe protective covering) are also included. For cables in air, the cable
extends as far as the free air.
Analysis of linear networks, such as the one in Fig. 3.5, involves the determination of
the expression for the response function caused by the application of a forcing function.
In our case, the forcing function is the conductor heat loss, and the response sought is the
temperature rise above the cable surface at node i. This is accomplished by utilizing a
mathematical quantity called the transfer function of the network. It turns out that this
transfer function is the Fourier transform of the unit-impulse response of the network. The
Chapter 5 m Rating Equations—Transient Conditions 71

Laplace transform of the network’s transfer function is given by a ratio

| PCs)
H(s) (S21)
O(s)
P(s) and Q(s) are polynomials, their forms depending on the number of loops in the
network. Node i can be the conductor or any other layer of the cable. In terms of time, the
response of this network is expressed as (Van Valkenburg, 1964)?

n
6:(t) = We Ti (1—e”") (522)
j=!

where _ 6; (t) = temperature rise at node i at time r, °C

W.. = conductor losses including skin and proximity effects, W/m
T;; = coefficient, °C - m/W
P; = time constant, s~!
t = time from the beginning of the step, s
n = number of loops in the network
i = node index
J = index from | ton

The coefficients 7;; and the time constants P; are obtained from the poles and zeros
of the equivalent network transfer function given by equation (5.1). Poles and zeros of the
function H(s) are obtained by solving equations Q(s) = 0 and P(s) = 0, respectively.
From the circuit theory, the coefficients 7;; are given by

Tj; = -— (5.3)

where = d(,_;); = Coefficient of the numerator equation of the transfer function

b, = first coefficient of the denominator equation of the transfer function
Z, = zeros of the transfer function
P; = poles of the transfer function

An algorithm for the computation of the coefficients of the transfer function equation is
given in Appendix B.
Before desktop computers became widely available, several researchers presented
simplified versions of equations (5.2) and (5.3). An interesting review of the work conducted
in this area by Baudoux (1962), Van Wormer (1955), Morello (1958), and Wlodarski (1963)

2 Unless otherwise stated, in the remainder of this chapter, all the temperature rises are caused by the joule
losses in the cable.
72 Part I » Modeling

can be found in Wlodarski and Cabiac (1966). They also present the results of experiments
carried out to verify various models. The results of their analysis were later adopted by
CIGRE and IEC and form the basis of today’s standards.

EXAMPLE 5.1
A simple expression of equation (5.2) is obtained for the case of n = 2. Construction of such a
network is discussed in Section 3.3.2 and the network is shown in Fig. 5.2.

T eA i QB
Figure5.2Anequivalent
thermal
network
com-
OP POSER
OL OLOODSs
In this simple case, the time-dependent solution for the conductor temperature can easily be ob-
tained directly. However, to illustrate the procedure outlined above, we will compute this temperature
from equations (5.1)—(5.3).
The transfer function for this network is given by

oe (T,
Ih se Thy)aeollash
B) A BOe (5.4)
1+s(T4Q4 + TgQp + TgQa) + 5*TsQaTeQp

Since we are interested in obtaining conductor temperature, i = 1 and j = 1,2. To simplify the
notation, we will use the following substitutions:

T =T1, Ty=Ti2, Mo =0.5(T4Q4+TgQe3+TpQ4). No =TaQaTsQp (5.5)

The zeros and poles of the transfer function (5.4) are easily obtained as

Li T,: + T,z P, = —a, P,; =—b

T,TzQp

where

My + (M2 — No ; Mo — J M3 —No
a= er No ntA = _._.___.__.
No (5.6)

From equation (5.4),

da-11 = T4TgQp, b2 = TATgQAQz.
Thus,
a\\ = l
by - On
From equation (5.3), we have

T, + Tp
=e.
ee
q Q, —a(—b+a) a—b\Q, aTsTgQaQp

but
l
ab = ——
TsTgpQaQp
Chapter 5 # Rating Equations—Transient Conditions 73

Hence,

jie 1 |=| -bit,+T)|and7=

a—b| Qa , %,+T,
—T, (5.7)
Finally, the conductor temperature as a function of time is obtained from equation (5.2):

6.(t) = W. [T, (1-e"“) + 7 (1-e*)] (5.8)

where W, is the power loss per unit length in a conductor based on the maximum conductor temperature
attained. The power loss is assumed to be constant during the step of the transient. Further,

6.(t)
1) =WT,+Ts) oo)
Because the solution of network equations for a two-loop network is quite simple, IEC
publications 853-1 (1985) and 853-2 (1989) recommend that this form be used in transient
analysis. Examples of conversion of multiloop networks to their two-section equivalents
are given in Section 3.3.3. The two-loop computational procedure was published at a time
when access to fast computers was very limited (Wlodarski, 1966; CIGRE, 1972). Today,
this limitation is no longer a problem, and a full network representation is recommended
in transient analysis computations. This recommendation is particularly applicable in the
case when temperatures of cable components, other than the conductor, are of interest.

5.2.3 Second Part of the Thermal Circuit—Influence of the Soil

The transient temperature rise 6.(t) of the outer surface of the cable can be evaluated
exactly in the case when the cable is represented by a line source located in ahomogeneous,
infinite medium with uniform initial temperature. Under these assumptions, equation (2.15)
can be written as
B70 oN 28 apne
prey the be ESL
OE
where (Ps;= soil thermalresistivity,K -m/W
6 = 1/p,c = soil thermal diffusivity (see Section 5.4), m?/s.

Integrating this equation, we obtain

a= we | eS“enim
| 2 du
ne dpe
and making the change of variables,

6.(t) p ©
=De[i di
(-=)|
[o,@)
where —Ei(—x) = [ £~ dv is called the exponential integral. The value of the exponen-
x v

tial integral can be developed in the series

xP res
—Fi(—x) = —0.577 —Inx +x — Ped3x3 vee ge4
5.9a
74 Part I m Modeling

When x < 0.1,

—FEi(—x) = —0.577 — Inx +x

to within 1%accuracy. For largex,

ae! ae 5ea 3
Ei(—x) =—- l= ee
x XK PRESx
The National Bureau of Standards published in 1940 Tables of Exponential Integrals,
Vol. 1, in which values of —Ei(—x) can be found. IEC has also published nomograms
from which —Ei(—x) can be obtained (IEC 853-2, 1989).
The mathematical solution obtained so far is valid under the assumption that the cable
is treated as a line source with the internal thermal resistivities equal to that of the sur-
rounding infinite soil. This result is valid for very short times and very deep cable locations
only. However, for practical applications, we have to use another hypothesis, namely, the
hypothesis of Kennelly, which assumes that the earth surface must be an isotherm. Under
this hypothesis, the temperature rise at any point M in the soil is, at any time, the sum
of the temperature rises caused by the heat source W, and by its fictitious image placed
symmetrically with the earth surface as the axis of symmetry and emitting heat —W, (see
Fig, 5.3).
In this case, we have

ake
Sut) Ps
ral sfgakSa
i(
Say Deen am)*f
i( pae
az)
es ein(os |
Placing the point M at the surface of the cable and assuming identical material in the
inside and outside of the cable, we obtain

) | Ps ~
i( (ne
25;)
ORa pegpee di + Bi
( Ext
3)] 10)

\ Soil surface

oe
N,
=—
Ss
<

Figure 5.3 Illustration of Kennelly’s hypothesis.

Chapter 5 # Rating Equations—Transient Conditions 75

where D* = external surface diameter of cable, m

L* = axial depth of burial of the cable, m

Under steady-state conditions, t > oo and x approaches zero. In this case, the terms
after the logarithm in (5.9a) can be neglected. Summing now the exponential integrals in
equation (5.10), we obtain

In De +1 ne Zit bes
— [CSiog n Os oe WeeehDE

Hence,in this case,equation(5.10)becomes

Ps , 4L*
6.(00)
e (00)= W,—1
tae n D*
(es

Several researchers proposed modifications of equation (5.10) to remove the assump-

tion that the cable is a line source (Whitehead and Hutchings, 1938; Goldenberg, 1964).
However, in experiments carried out to test various formulas, Wlodarski and Cabiac (1966)
have shown that the results obtained with equation (5.10) are very close to the measured
values for a wide range of times, and therefore this formula has been adopted by CIGRE
and IEC as the standard equation.

5.2.4 Second Part of the Thermal Circuit—Cables in Air

For cables in air, it is unnecessary to calculate a separate response for the cable en-
vironment. The complete transient 6(t) is obtained from equation (5.1), but the external
thermal resistance 74, computed as described in Chapter 9, is included in the cable network.

EXAMPLE 5.2
For a cable in Example 5.1, obtain the transient conductor temperature if the cable is located in air.
In this case, the transient response of the internal part of the cable is given by equation (5.8). It
is sufficient to modify the value of Tz to obtain the required answer. Thus, according to the discussion
related to Example 5.1, we have

l
Tz = 5h al spainMCa2tha)

5.2.5 Representation of Dielectric Losses

The temperature rise caused by dielectric losses is of significance only for high-voltage
cables because, as explained in Chapter 6, they are strongly voltage dependent. If the
transient analysis is performed with a starting conductor temperature different from the
ambient, the dielectric losses are assumed to be included in the temperature rise obtained at
the onset of the transient conditions. If the application of load current and system voltage
occur at the same time, then an additional transient temperature rise due to the dielectric
loss has to be calculated. Assuming that the dielectric power factor is constant and equal
to an appropriate value in the temperature range of interest, dielectric losses can be taken
into account in transient conductor temperature rise computations as follows.
76 Part I » Modeling

For cables at voltages up to and including 275 ky, it is sufficient to assume that half of
the dielectric loss is produced at the conductor and the other half at the insulation screen or
sheath. The cable thermal circuit is derived by the method given in Chapter 3 with the Van
Wormer coefficient computed from equations (3.19) and (3.22) for long- and short-duration
transients, respectively.
For cables operating at voltages higher than 275 kV, the dielectric loss is an important
fraction of the total loss (i.e., for paper-insulated cables) and the appropriate Van Wormer
coefficient is given by equation (3.23). Also, in this case, it is sufficient to assume that half
the dielectric loss is produced at the conductor and the other half at the sheath.
Hence,

Wa [47 +n(I) +T73+T4)| if cable energized before t = 0 ae

Ga(t= a f
, Oca(t) + g(t) Gea(t) if cable energized at t = 0

where the subscript d refers to the circuit in which only dielectric losses are represented.
This subject is illustrated in Example 5.4 below.

5.2.6 Groups of Equallyor Unequally Loaded Cables

In a typical installation, several power cables are laid in a trench. The mutual heating
effect reduces the current-carrying capacity of the cables, and this effect must be taken into
account in rating computations. A major limitation of the methods presented by Neher
(1964), Morello (1958), CIGRE (1972, 1976), and IEC (1985, 1989) is that they apply only
in the case of identical, equally loaded, directly buried cables. However, one of the most
common cases of interest is that associated with the loss of one circuit and a corresponding
increase in loading in the remaining circuits. A method to account for the presence of other
cables is discussed below (Anders and El-Kady, 1992; Affolter, 1987).
When there is more than one cable, for example, in a three-phase circuit composed of
single-core cables, each will have its own thermal diagram and equivalent network from
which the transient temperature rises caused by its own losses are evaluated. If the duration
of the transient is sufficiently long for mutual heating (thermal coupling) from other cables
to be significant, this must be taken into account when calculating the true temperatures
used to update cable losses at each time step.
The temperature for each cable is obtained at each point in time by adding to its own
temperature the temperature rise caused by other nearby cables. To achieve better accuracy
in calculations with multiple time steps, the effect of other cables should be added at each
time step so that their effect can be included with that caused by the temperature rise of the
cable itself. Thus, the temperature rise in the cable of interest “p” due to one other adjacent
cable “k” can be computed from

pad)“ an
Ds
Gt)=Wiest;( ‘a
beat
|SBb | we
(
le iim )
d*?
pooa | bt
in which W7;,are the total joule losses in cable k, and ak and d*,
pk (Mm)denote the distance
from the center of cable p to the center of cable k and its image, respectively, as shown
in Fig. 5.4.
Chapter 5 m Rating Equations—Transient Conditions ad

iO) |

Air /
TINNNENOS
ONIONS
NTENJEN
ON
NJNTONTENTO
ST”
NV NLENTOINE
NINN
NENTS
NS
Soil |
|
O |
‘i
Mh 1 kp
|
p
k “pk

Figure 5.4 Example of cable configuration and image cables.

The temperature rise in the cable of interest caused by the dielectric losses of cable
“k” is obtained from

d*2
Wakps In oe if cable energized before t = 0
pk
Opar(t) = ss Me
Wee —Ei _ Ak + Ei ei if cable energized at t = 0
4n 4dt ASt
(5.13)

5.2.7 Total Temperature Rise

The total transient temperature rise of a cable at any time is the sum of the rise due
to its own losses, given by its own network, and the rises due to mutual heating given by
the networks of other cables and image sources, as appropriate. Thus, the final temperature
rise at any layer of the cable of interest (that is, at any node of the equivalent network) at
time ¢ after the beginning of the load step is obtained from

N-1

Oprot
(t)= O;(t)+ a(t) e(t)+ Oa(t)+ e(t) ¥° [Opx(t)
+ Opan(t)] (5.14)
k=)
where a(t) is the attainment factor for the transient temperature rise between the conductor
and outside surface of the cable and N is the number of cables. The temperature rise
Onak(t) in equation (5.14) is multiplied by a(t) only if cable k is energized at time ¢ = 0.
78 Part I » Modeling

The internal temperature rise 0,(t) is given by equation (5.11). In equation (5.14), Aproxis
defined for any layer of the cable, and in the above formulation, only 6;(t) is different for
each layer.
The attainment factor varies in time, and a reasonable approach for obtaining a(t) is
to use (Morello, 1958)
tempertaure across cable at time ¢
= (5.15)
steady-state temperature rise across the cable
The conductor attainment factor is computed from this definition using equations (5.2)
and (4.5):

WadeTey(1 citi) mamietatleera)

oli = O.(t)
(00)
jel W.T
eel
= T
(5.16)

The attainment factor associated with dielectric losses is obtained from a similar equa-
tion with the network parameters reflecting the presence of dielectric loss only.

9.3 TRANSIENT TEMPERATURE RISE UNDER VARIABLE LOADING

In order to perform computations for variable loading, a daily load curve is divided into a
series of steps of constant magnitude. For different successive steps, the computations are
done repeatedly, and the final result is obtained using the principle of superposition. The
diagram in Fig. 5.5 illustrates how the temperature rise caused by an application of a single
step of current lasting | h is computed.
From the diagram in Fig. 5.5, the temperature rise above ambient at time t can be
represented as

6(r) —O(t —1) (3:17)

5.4 DIFFUSIVITY OF SOIL

In the majority of cases, the soil diffusivity is not known. In such a case, a diffusivity
value of 0.5 - 10~° m?/s can be used. This value is based on a soil thermal resistivity of 1.0
K - m/W and a moisture content of about 7% of dry weight. When the density, moisture
content, and thermal resistivity are known, the diffusivity of the soil can be computed from
an empirical equation (IEC, 1985):
% Loa
(5.18)
~~ pydy(0.82 + 0.0427)

where dy = dry density of soil, kg/m*

7)= moisture content, % of dry weight.

If only the thermal resistivity is known, then the diffusivity can be obtained from
another empirical formula proposed by Neher (1964):
4.68
pA
} =, 10-7 (5.19)
Chapter 5 m Rating Equations—Transient Conditions 79

Step of current

Equivalent positive
and negative steps

Resulting temperature

Figure 5.5 Temperature rise due to one step of

current.

9.9 CONDUCTOR RESISTANCE VARIATIONS DURING TRANSIENTS

Since the conductor electrical resistance, as well as the resistance of other metallic parts
of the cable, changes with temperature, the effect of these changes should be taken into
account when computing conductor and sheath losses. Goldenberg (1967, 1971) developed
a technique for obtaining arbitrarily close upper and lower bounds for the temperature rise
of the conductor, taking into account the changes of the resistance of metallic parts with
temperature. His upper bound formula has been adopted by CIGRE (1972, 1976) and
IEC (1989). The derivation of this formula is very complex; therefore, it is simply quoted
below:

~ O(t)
“N71 + A19(00) = OO) ae

where @(t) = conductor transient temperature rise above ambient without correction
for variation in conductor loss, based on the conductor resistance at the
end of the transient

8(oo) = conductor steady-state temperature rise above ambient

80 Part I » Modeling

a = temperature coefficient of electrical resistivity of the conductor

material at the start of the transient. a = 1/[B + 6(0)] with B being
a reciprocal of temperature coefficient at 0°C and @(0) the conductor
temperature at the start of the transient.

EXAMPLE 5.3
In this example, we present calculations for a circuit of three cables (cable model No. 1) in flat
formation, not touching. The parameters of this cable are given in Appendix A as A; = 0.09,
T, = 0.214 K - m/W, T; = 0.104 K - m/W, and 7; = 1.933 K - m/W. Laying conditions are as
follows: cables are located | m below the ground in a flat configuration. Uniform soil properties
are assumed throughout. Spacing between cables is equal to one cable diameter (spacing between
centers equal to two cable diameters). Ambient soil temperature is 15°C. The thermal resistivity of
the soil is equal to 1.0 K - m/W. The cables are solidly bonded and not transposed. We will compute
the transient temperature response of the center cable for the first 24 h after the application of a step
function of rated current. The temperature rise due to the dielectric loss is negligible for this cable.
Since we will perform computations without the help of a computer, we will use a two-section
equivalent network.
(1) The thermal circuit of the cable
The thermal circuit for this cable has been derived in Example 3.6. The following values were
obtained there:
T, =0.214K- m/W, Q, = 1435.1 J/K- m

Tes 0213 Kee Wee © pe—109223.0/

Koen

(2) Calculation of the response of the cable circuit

The response of the cable circuit to a step function of rated current is computed from equations
given in Example 5.1:

Mo= 0.5(T,Q4 + TgQp+TpQa), No=TsQaTeQp

My+ ,/M2 —No Mo —,/ Ma—No

a ee AE PEBe eae A ane
No No
T, = a ] Al Di
l Bae |
a al iypsi y and 7, =1T,+Tp
= —Ty
ro

Substituting numerical values, we obtain

My = 5(0.214 - 1435.1 + 0.113 +692.3 + 0.113 - 1435.1) = 273.8 s

No = 0.214 - 1435.1 - 0.113 - 692.3 = 24025.3 s?

1 = 213.8 + V273B = 240253 Ryne:

=a 240253 Sa San

p — 223.8 —¥273.8? = 24025.3 Aon i

Ps 24035.3 Paes

a
~0.0208
—
| 0.0020
las — 0.0020(0.214 + 0.1 13) = 0.0023 K - m/W

T, = 0.214 + 0.113
—0.0023 = 0.3247 K - m/W
Chapter 5 m=Rating Equations—Transient Conditions 81

The transient temperature rise for each step is determined from equation (5.8); it is calculated
for 1 h (3600 s) below; the results for other times are tabulated in Table 5.2.

O-(1)= W.[T,(1—e~*)+ T;(1—e~*)]

= 30.82[00023{17 "e-9.008
250) 193247 (1= 20900) 10.1K
The conductor to cable surface attainment factor is determined for each time step from equation
(5.9); it is calculated for 1 h below; results for the other times are shown in Table 5.2.

6.(t) 10.1
«)iS=Wr, 47s)~30820214
= 40.113)
=0.9)
We can observe that because of the small value of the cable time constant, the conductor tem-
perature rise above the cable surface temperature, reaches its steady state value within 1 h.
(3) Calculation of the response of cable environment
The response of the cable environment is given by equation (5.12). A sample calculation for 1
h is presented below. The other values are given in Table 5.1.

d*, = 0.072m

ie, anyO722 2 —)200 1m

6 =0.5-10~ m2/s
On a term-by-term basis for 1 h (3600 s), we have

= D:*2 = pDs38 2 = 0.0445, —Ei(—x) = 2.579

16t6 16 - 3600 - 0.5 - 10-°

#2 2:
td 3600 - 0.5 - 10-°

dr 0.072?

yen)os = es0012 =556.1,—Fi(—k)eS0

416 4-3600-0.5-10-
6,(1)=a 3810.579— 0)+2(0.365
—0)]=8.8K
The second terms in equations (5.10) and (5.12) represent the effect of the image sources, and are
usually negligible at normal depths of laying and for durations of less than 24 h.

TABLE 5.1 Components for Cable Environment Partial Transient

i
Time(h)
x Item
pane)ceri)Sareeepet 1 2 3
itp otismest adreesrinterrey 4 5
et AE TF fie RE ttre 6 12
a5) sit a 24

De x 0.045 0.022 0.015 0.011 0.009 0.007 0.004 0.002

1616 -E(-x) 2.579 3.25 3.648 3.932 4.153 4.334 5.024 5715
de x 0.720 0.36 0.24 0.18 0.144 0.12 0.06 0.03
ae -E(-x) 0.365 0.775 1.076 1.31 1.50 1.66 2.295 2.959
82 Part I m»Modeling

(4) Complete temperature rise ;

The complete transient is calculated from equation (5.14). A sample calculation for 1 his shown
below. The values for this and other times are summarized in column 5 of Table 5.2.

6(1) = 10.1 + 0.999 - 8.8 = 18.9K

Correcting for variation of conductor losses using equation (5.20), we obtain

A(t) 18.9
64(1) = 154K

| + 534.5410: ss)
The corrected values of conductor temperature rise are summarized in column 6 of Table 5.2,
along with the actual conductor temperature in column 7.

TABLE 5.2 Summary of Temperature Transient Components

ee
Actual
Conductor
Time 6.(t) 6 .(t) A(t) 6,(t) Temperature
(h) (K) a(t) (K) (K) (K) CC)

(1) (2) (3) (4) (5) (6) (7)

1 10.1 0.999 8.8 18.9 15.4 30.4

2 10.1 i 12.8 22.9 18.9 33.9
3 10.1 1 15.4 aD Pa es! 36.3
4 10.1 1 17.4 PAS 23.1 38.1
5 10.1 1 19.0 29.1 24.6 39.6
6 10.1 1 20.3 30.4 25.8 40.8
12 10.1 1 PS) 35.6 30.8 45.8
24 10.1 1 30.9 41.0 36.1 oy!Al

EXAMPLE 5.4
In this example, we present calculations for a system of two cables (cable model No. 2 and cable
model No. 3 described in Appendix A) buried underground. Laying conditions are shown in Fig. 5.6.
We will compute the transient temperature response of both cables for the first 6 h after the application
of a step function of current. The losses in each conductor of the pipe-type cable are assumed to be
equal to 9 W/m and in each conductor of the three-core cable 22 W/m. Total joule losses in the
pipe-type cable and the three-core cable are 40 and 72 W/m, respectively. The dielectric loss in the
pipe-type cable is 4.83 W/m per cable. System frequency is 60 Hz. We will assume that the voltage
is applied simultaneously with the current.
Since the response of the system in | h steps for the first 6 h is required, long-duration transient
conditions can be assumed for the three-core cable and short-duration transient conditions for the pipe-
type cable (see Examples 3.8 and 3.7). The temperature rise due to the dielectric loss is negligible
for the three-core cable, but has to be evaluated for the pipe-type cable. Therefore, we will start by
constructing an equivalent network for the pipe-type cable for the computation of the temperature
rises caused by dielectric loss.
(1) Thermal network for dielectric loss of pipe-type cable
Parameters ofthis cable are given in Table Al as A; = 0.010, A2 = 0.311, 7, = 0.422 K - m/W,
T, = T, = 0.082 K - m/W, and 7; = 0.017 K - m/W. The Van Wormer coefficient for dielectric loss
Chapter5 mRating-Equations—Transient
Conditions 83

im

102)
er SE

+— 03m—>

Figure 5.6 Laying conditions for two cables in Example 5.4.

for this cable has been computed in Example 3.5 and is equal to 0.585. Therefore, we have

Tg ely, Ox = (ORE: 07) /3

Te = 4M +0 +4) +0 +A) +273

penn Ores That.) Wyfraise

>
Qn=a es +(454 (ips) Ory3itn
fe oeBPR
+(¢ Pe a oon
Ts\
1+ A2)) 1 os Os ‘0;
5oO 4 Let PQ; (5.21)
ilps (ee aes
Substituting numerical values, we obtain

i $0.422/3 = 0.07K:-m/W, Qa = (3484.5 + 0.585 - 1680.5)/3 = 1489.2 J/K -m

Tg = 0.07 + 2 - (0.082 + 0.017) = 0.268 K - m/W
Qe = (1 — 0.585) 1680.5/3 + 0.585 - 2726.9/3

(2268 B=
(0.07.
0.268 )\o
( 1— ) hoe
0.585)2726.9/3 12/3+a)
(; 0. 2;16126.4+
~ 6126.4 + 0.482- 15720.9
0.482 15720.9
+ 38570/2
5)
/ —6734.2 /K-m
0.268 2

The remaining cable parameters are obtained from equations (5.5)—(5.7) and are equal to

a=0.012s"', b=0.00045s"', T, =0.046K-m/W, 1, = 0.292 K- m/W

The transient temperature rise caused by dielectric loss for each step is determined from equation
(5.8). Half of the dielectric losses are located at the conductor and half at the sheath. The calculation
84 Part I m Modeling

for 1h (3600s)isgivenbelow;thevaluesforthisandtheothertimesaresummarized
inTable5.4.
Od Wally(ae ei ee) |

zs 3 -4.83[0.046(1ane
es) ae0.292(1eeoo eae) | DOK
2
The conductor to cable surface attainment factor is determined at each time step from equation
(5.9); the calculation for | h is given below. The values for this and for the other times are shown in
Table 5.4.

=on ae eee
wa)=7945(0.07 =(817
40.268),
(2) Thermal circuit for the temperature rise caused by joule losses, and calculation of the
response of the cable circuit
The thermal circuit for the pipe-type cable is obtained from the data in Table A.1 and Example

3.7. The following values are obtained:

a=0.0119s"', b=0.00058s"', T,=0.0498K-m/W, T, = 0.1952 K-m/W

The transient temperature rise for each step is determined again from equation (5.8); the calcu-
lations for 1 h (3600 s) is given below; the values for this and the other times are assembled in Table
S55:
8-(1)= 27[0.0498(1 —.e—001191300)
+ 0.1952(1 —e790 130M))—6.0K
The conductor to cable surface attainment factor is determined at each time step from equation
(5.9); the value for 1 h is computed below; otherwise, see Table 5.4.

6.(t) 6.0
CXA0)= SSS SS See SY
WAT a lin) 27 (0:07 2 0al75)

For the three-core cable, new values of 74, Tg, Q4, and Qg have to be computed because the
system frequency for this example is 60 Hz and the values in Table A.1 and Example 3.8 were obtained
for 50 Hz. The only difference between these two cases is the value of the sheath loss factor. This
factor is computed by a method discussed in Section 8.3. For a system frequency of 60 Hz, it is equal
to 0.103. For illustration purposes, we will assume a single set of capacitance values. Proceeding
now similarly as in Example 3.8 and remembering that the cable is directly buried, we obtain the
following network parameters:

T, = 0.104 K-m/W, QO,= 743.1J/K-m

Tz = 0.086 K-m/W, Qs = 1018.7K-m
and

a = 0.0286s"', b=0.0052s7!, T, = 0.0156 K-m/W, T, = 0.1744 K - m/W

The transient temperature rise for each step is determined again from equation (5.8); the sample
calculation for | h (3600 s) is given below; the values for other times are assembled in Table 5.5.

A.(1)= 66[0.0156(1 —e°9°8919) + 0.1744(1 —7 0.005213600))

_ 19.5K

Theconductorto cablesurfaceattainmentfactorfor f= 1h is equalto

1225
aA ee Sa Se
66(0.104 + 0.086)
Chapter 5 # Rating Equations—Transient Conditions 85

The remaining values are given in Table 5.5.

(3) Calculation of the response of the environment for the pipe-type cable
The response of the cable environment is given by equation (5.10). A sample calculation is
given for | h. The other values are assembled in Table 5.4.

Se. 0.2445? “hae ov

Wt “WELV pee ee ee

L* 12
ees + =555.6, —Ei(—x)=0
~ #3 3600:0.5 107

So
6.(1) Ds
ral
=W,— i(
| —Hi D*
2) + i( L*
ry
)]
|——£E|i— =40— 1
4m
-0.044=0.1K
The external temperature rise due to dielectric loss is equal to

1- 14.49

(4) Calculation of the response of the environment for the three-core cable
The response of the cable environment is given again by equation (5.10). A sample calculation
is given for | h. The other values are assembled in Table 5.5.

De 0.07297
x= — = —______ = 01845, —Ei(—x) = 1.289
16¢5 16 - 3600 - 0.5 - 10-®

L* 12

1-72
6.(1) = —— - 1.289 =7.4K
Ar

(5) Calculation of the mutual heating effect

Since we have only two cables located at the same depth, the exponential integrals will be the
same for the mutual heating effect for both cables.
From Fig. 5.6, we obtain

d= OSmirds 4 08 + 252.02
The values of the exponential integral for += 1 h are computed below, and for other values are shown
in Table 5.3.

Pie
SR418 0,33
SEpat4-3600-0.5-
Getasie 10-6
al 98 55 Feyp
e ingy
AL6
Ae)
a). 2.022
(ee opel te = 566.7, —Ei(—x) =0
4té 4. 3600 - 0.5 - 10-6
The temperature rise in the pipe-type cable due to joule losses in the three-core cable is obtained
from equation (5.12). As can be seen from Table 5.3, the mutual heating effect will take place from
t = 5hon. Computations for 6 h are shown below, with the remaining values summarized in Table
5.4.
Heating of the pipe-type cable by the three-core cable

i
O54(6). = 72 » — - 0.044 = 0.3 K
Ant
86 Part I m Modeling

TABLE 5.3 Components for Cable Environment Partial Transient

Dueto MutualHeating
— eid, quelle Se: okt Seen ee ee
Time

x Item 1 2 3 4 5 6

d yk
— 12.5 6.25 4.17 4.12 25 2.08
Ais -E(-x) 0 0 0 0 0.025 0.044

TABLE 5.4 Summary of Temperature Transient Components as a Function

of Time for the Pipe-Type Cable

Actual
Cond.

Time 0,(t) 0,(t) @,(t)” 8, 4(t) @(t) 0,(t) Temp.

(h) (K) a(t) (K) (K) (K) (K) (K) (°C)

(1) (2) (3) (4) (5) (6) (7) (8) (9)

1 6.0 0.907 0.1 24 0 Se 7.0 27.0

) 6.5 0.988 0.7 a5 0 9.7 8.4 28.4
3 6.6 0.998 heed 2.4 0 10.5 9.1 29.1
4 6.6 1.0 1.7 28 0 11.1 9.6 29.6
5 6.6 1.0 ae) 2.9 0.1 11.8 10.3 30.3
6 6.6 1.0 2.6 3.0 0.3 12.5 10.9 30.9

“This column contains total temperature rise due to dielectric as computed from equation (5.11).

Heating of the three-core cable by the pipe-type cable

For the computation of the mutual heating effect, the total losses in the pipe-type cables are
entered in equation (5.12). Thus, we have
1
6(6) = 54.49 - — -0.044 = 0.2K
Ant

1
Onak(6) = 14.49 -— - 0.044 = 0.05 K
An

(6) Complete temperature rise

The complete transient is calculated by using equation (5.14). A sample calculation for 1 h is
shown below; the values for other times are summarized in columns 7 of Tables 5.4 and 5.5 for the
pipe-type and the three-core cable, respectively.
Pipe-type cable

6(1) = 6.0 + 0.907 - 0.1 + 2.1 + 0.817 - 0.05 + 0.901 -0 = 8.2K

Correcting for the variation of conductor losses using equation (5.20), we obtain

8.2
641) = = /.0)
priite
ohowe a
a 234.5+ 20(Ot)
Three-core cable

(1) = 12.54 1.0-7.44+1.0-(0+0) =19.9K

Chapter5 mRatingEquations—Transient
Conditions 87

TABLE5.5 Summaryof TemperatureTransientComponentsas a Function

of TimefortheThree-CoreCable
ee ee ee ee eee ee
Actual
Cond.
Time 6.(t) 6,(t) Bpag(t) 8,4(t) 0(t) 6,(t) Temp.
(h)
a (K) aa(t) (kK) (K)
ge (kK)
ee ee (K) ee (K) CC)
() (2) (3) (4) (5) (6) (7) (8) (9)
Sreeeee te SACS St? saosin) erienalil:pxtriten priors le ett
1 12.5 1.0 7.4 0 0 19.9 16.6 36.6
y) 12.5 1.0 10.9 0 0 23.4 19.8 39.8
3 ee 1.0 13.0 0 0 25.6 21.8 41.8
4 12.5 1.0 14.6 0 0 al 222 43.2
5 12.5 1.0 15.8 0.03 0.1 28.4 24.4 44.4
6 125 1.0 16.8 0.05 0.3 29.8 257 45.7
“This column contains total temperature rise due to dielectric as computed from equation (5.13).

Correcting for the variation of conductor losses using equation (5.20), we obtain

19.9
yyy = Sg a ae 16.6K
eei (IK)
234.5 + 20 = IO)

The corrected values of conductor temperature rise are summarized in columns 8 of Table 5.4
and 5.5, along with the actual conductor temperature in columns 9.

Goldenberg’s approximate formulas pertain to the application of a step function, and

would have to be repeated for each new time step considered. However, taking advantage
of fast desktop computers which are now in wide use, one can perform the computations
iteratively, adjusting the resistances at each computational step. This is performed in the
following way. At each time step, the temperature at the end of the step is computed based
on the losses at the beginning of the step. New resistance values are then obtained and the
computations for this time step are repeated. The process is carried out until convergence
is obtained. As pointed out by Thomann ef al. (1991), seldom are more than two iterations
required to achieve the desired accuracy.

5.6 CALCULATION OF CYCLIC RATINGS

5.6.1 Introduction

The complexity of cyclic rating computations varies depending on the shape of the
load curve and the amount of detail known for the load cycle. If only the load-loss factor or
a daily load factor is known, a method proposed by Neher and McGrath (1957) can be used.
This method involves modification of the cable external thermal resistance as discussed in
Section 9.6.7. This modified value is then used in equation (4.3). Another possibility is to
make the assumption that the load curve has a flat top lasting for at least 6 h and use the
method described in Example 5.6 in this section.
If a more detailed analysis is required, the algorithm described in Sections 5.2 and 5.3
can be used to handle different cable types with different loading patterns (Anders et al.,
88 Part I mModeling

1990). For the case of a single cable or a group of identical cables, the simplified approach
described below gives satisfactory results.

5.6.2 CyclicRating Computations

The general algorithm described in Sections 5.2 and 5.3 is applicable to a group of
equally or unequally loaded cables with each circuit carrying a different cyclic load. Since
the algorithm requires intensive computations, simplified approaches have evolved which
are quite suitable for hand calculations. The first simplified approach was proposed by
McGrath (1964) in response to a complex method involving application of the Fourier
integral introduced by Neher (1964). This simplified approach uses the steady-state rating
equation (4.3), but requires modification of the external thermal resistance of the cable.
McGrath’s approach continues to be the basis for the majority of cyclic loading computations
performed in North America. Since it only requires modification of the value of 74, we
introduce it in Section 9.6.7 where thermal resistances are discussed.
Another simplified approach was introduced by Goldenberg (1957) and later adopted
by the IEC (1985, 1989). This approach, applicable to a single cable or a cable system
composed of identical, equally loaded cables located in a uniform medium, is presented for
both cases below.
The cyclic rating of a single three-core cable or a group of equally loaded identical
cables located in a uniform soil requires computation of a cyclic rating factor M by which
the permissible steady-state rated current (100% load factor) may be multiplied to obtain
the permissible peak value of current during a daily (24 h) cycle such that the conductor
temperature attains, but does not exceed, the standard permissible maximum temperature
during the cycle. A factor derived in this way uses the steady-state temperature, which
is usually the permitted maximum temperature, as its reference. The cyclic rating factor
depends only on the shape of the daily cycle, and is independent of the actual magnitudes
of the current.
Goldenberg (1956) derived an expression for M taking into account the form of the
whole cycle. Later, he showed that it is sufficient to consider only the loss-load factor of
the cycle and the detail of the load current for the 6 h prior to the maximum conductor
temperature (Goldenberg, 1957). Earlier values can be represented with sufficient accu-
racy by using an average. The loss-load factor jz provides this average (computation of
this factor is illustrated in Example 5.5 below). Location of the time of maximum tem-
perature is done by inspection, bearing in mind that although it usually occurs at the end
of the period of maximum current, this may not always be the case. The loss-load factor
4 of the daily current cycle is determined by decomposing the cycle into hourly rect-
angular pulses. The temperature response to the complete cycle of losses of the cable
and the soil is found by adding together the responses to each hourly rectangular pulse,
taking into account the time period between each pulse and the time to maximum temper-
ature.

EXAMPLE 5.5
Determine a load-loss factor jx for a daily load curve shown in Fig. 5.7 (IEC, 1989).
The daily load cycle is given as a fraction of the maximum current in Table 5.6 (IEC, 1989) (the
last column will be used in Example 5.6 later in this section).
The load-loss factor is then equal to
Chapter 5 m Rating Equations—Transient Conditions 89

1.0
e . z e
0.8 NM POs Note that 1(8) is
ee H
@e e highest value
Ele 0.6 : }

Time of day (h)

ORR pete Oe BOa0 12 214 TIGA1890 M0) 24. O° AG

Time of day (h)

@ 2 @& & Fe We dW Wt ae i 2) ee at pp» @ «&

(b)

Figure 5.7 (a) Cyclic load divided by highest load; (b) load-loss graph. (IEC Standard
853-2, 1989)

@22)

For the numerical values of the load curve in Table 5.6, the load-loss factor is equal to

0.091 + 0.061 + 0.052 + --. +0.360

og 24 = 0.504

5.6.2.1 Single cable. The developments presented in this section apply to a single-
conductor single cable located in a uniform, semi-infinite medium. The three-core cable is
replaced in cyclic computations by an equivalent single-core construction dissipating the
same total conductor losses as described in Section 3.3.1, with the thermal capacitances of
an equivalent cable calculated with the assumptions specified there.
To obtain an expression for the conductor temperature rise at an arbitrary time T, we
will make an assumption that the heat dissipated in the cable is directly proportional to the
square of the applied current. This simplifying assumption gives quite accurate results, as
confirmed by the author by performing finite-element studies. The required temperature
is then obtained as an algebraic sum of the hourly temperature rises as given by expres-
sion (5.17).
90 Part I » Modeling

TABLE 5.6 Details of the Load Cycle in Fig. 5.7

EEUU UEEEEE
EISEN
Time Load
(h) (p.u.) Value of Y; Order of Y;

0 0.302 0.091 Yi7

| 0.247 0.061 Yi6
2 0.227 0.052 Y15
3 0.232 0.054 Yi4
4 0.235 0.056 Y13
5 0.246 0.061 Y1>
6 0.290 0.084 Yi
i 0.600 0.360 Yi9
8 1.000 1.000 Yo
9 0.950 0.902 Y,
10 0.940 0.884 Y;
11 0.910 0.828 Y¢
12 0.892 0.796 Ys
13 0.770 0.593 Y4
14 0.772 0.596 Y;
15 0.800 0.640 Y;
16 0.853 0.728 Y,
17 0.996 0.992 Yo
18 0.853 0.728 Y53
19 0.790 0.624 Y55
20 0.740 0.548 ¥>,
21 0.740 0.548 Yo
jy 0.722 0.521 Yio9
23 0.600 0.360 Yig

Let the conductor temperature rise at time i after the application of a step function of
losses corresponding to the rated current Jp be Og(i). The temperature rise corresponding
to the cyclic current with a maximum value / will be denoted by 6(7). The maximum
conductor temperature rise at time t after an application of a rectangular current pulse of
1 h duration is given by expression (5.17). Since the temperature rise is assumed to be
proportional to the square of the current, the temperature rise at time tf = 0 is equal to
5

I
4(0) = —HOR(00) (5.23)
Tr

Equation (5.23) assumes that a uniform current ,/y J has been applied for a long time
before t = 0. At the end of time ¢ = r, the temperature rise caused by this average current
is equal to
>

<4 [0R(00) —Oe(t)] (5.24)

Ik
If at tr= O, a current pulse with magnitude /p is applied, the maximum conductor
temperature rise at time f = | is obtained from (5.17) and (5.24) as

2
Ornax
= 7 {H[8x(00) —Or(1)] + Yo[@r(1)—Ar(0)]}
R
where Yo is the ratio 13/1? and 6p(0) = 0.
Chapter 5 # Rating Equations—Transient Conditions 91

Similarly, when a cyclic load is applied to the conductor, at the end of t hours, the
maximum conductor temperature rise is equal to

Omax
= [2 [124000—URC) =
t-1Y;[Or@
+ 1)—mio}
TR r—()

=TOR
(00)
ip l! ji Ort)
ea s|emxeee
i” ye 2
0x
( 1)
© OR)
00)Heo ao)
525 i=0

where each of the Ys is expressed as a fraction of their maximum value, and Y; is a measure
of the equivalent square current between i and (i + 1) hours prior to the expected time of
maximum conductor temperature (see Example 5.5).
The ratio 6(t)/Ar (Oo) is computed as follows. Consider a step current Jp applied at
time t = 0. Conductor temperature rise above ambient is obtained from (5.14):

Or(t) = G-(t) + a(t) - O.(t) (5.26)

From the definitions of the attainment factors a(t) and f(t) of the cable conductor and
the cable outer surface, respectively, we have

6-(t) =a(t)O.(co) and 6,(t) = B(t)@-(co) (5.27)

The steady-state temperature rises, on the other hand, are obtained from (4.5) as fol-
lows:

Orx(CO)= WoT +Wils, 98.(0) =W.T, 8.(00) = W7T4 (5.28)

where W, is the total joule loss in the cable.

Combining now (5.26)—(5.28), the required ratio is obtained as

Or(t) :
9p(00) = [l1—k+kB(t)la(t)
The fori > 1 (5.29)

Or(0) = 0

where

0. (00) WT
= eee (5.30)
Ba(oo)h moWeTGW
Ts

Whitehead and Hutchings (1939) computed the attainment factor for the cable surface
temperature assuming that a cable can be represented by a thin cylindrical constant source
of heat in a semi-infinite volume of soil. The source was located along the circumference of
the cable. The thermal properties of the cable were assumed to be those of the surrounding
soil. Goldenberg (1967) has shown that a better approximation of the value of A(t) can be
obtained by applying the exponential-integral formula. Thus, from equations (5.10) and

from
(3.6),
we
obtain
6 "7 W[i (-75)yw(=té)|
D* L*
e(t)=Ort
elee oer eeeOe
PY S 6.(00) WT (5.31)
92 Part I m Modeling

As will be shown in Chapter 9, the external thermal resistance of a single cable buried
in a uniform soil is equal to (p,/277) In(4L*/D*), where L* is the depth of cable burial
and D* is its external diameter, both in meters. Substituting this into the last equation, we
obtain
-Ei(- D*=) + 8i(- L* )
1616 td (5.32)
B(t) = 4L*
2 In
De

For standard cyclic-rating-factor calculations, the maximum conductor temperature

rise reached during cyclic load is assumed to be equal to the maximum conductor tempera-
ture rise reached for steady-state (100% load factor) current. The cyclic rating factor M is
defined so that J] = M/p. Therefore, from equation (5.25), we obtain

(533)

1—fi
ste al y,ES
“hao +1)
brarang
|———- Or(i)
ipa
i=0

with the temperature ratios given by (5.29) and Tt is usually taken to equal 6 h.
Calculations are simplified considerably when the conductor attainment factor can be
assumed to be equal to one. In this case, equation (5.33) takes the form
]
GYfhe ree ee Oks (5.34)
VU —k)Y¥o + k{B + “fl —B(r)]}

where

t-l

B= UY, ®n = B(m + 1) — Bm) (5.35)

i=0
IEC (1989) identifies the following cases when the internal cable capacitances can be
neglected. If the period from the initiation of the thermal transient is longer than
1. 12h for all cables,
2. the product XT - &OQwhen dealing with fluid-pressure pipe-type cables and all
types of self-contained cables where the product ©T - XQ < 2h, and
3. the product 2- XT - &Qwhen dealing with gas-pressure pipe-type cables and all
types of self-contained cables where the product ©T- XQ> 2h

where XT and &Q are the total internal thermal resistance (simple sum of all resistances)
and capacitance (simple sum of all capacitances), respectively, of the cable.
Table 5.7, based on design values commonly used at present for the determination of
cable dimensions, shows when cases 2 and 3 apply (IEC, 1989).

5.6.2.2 Groups of Identical, Equally Loaded Cables. In this section, we consider

groups of N cables with equal losses where the cables or ducts do not touch. In this case,
Chapter 5 m Rating Equations—Transient Conditions 93

TABLE 5.7 Cases When the Attainment Factor Can Be Assumed

to Be Equal to 1

Type of Cable Case 2 Case 3

Fluid-filled cables 1. All voltages < 220 kV 1. 220 kV: sections > 150 mm?
2. 220 kV: sections <150 mm? 2. All voltages > 220 kV
Pipe-type, fluid-pressure cables 1. All voltages < 220 kV 1. 220 kV: sections > 800 mm?
2. 220 kV: sections < 800 mm? 2. All voltages > 220 kV
Pipe-type, gas-pressure cables Vr 220 KV;
2. Sections < 1000 mm?
Cables with extruded insulation 1. All voltages < 60 kV 1. 60 kV: sections > 150 mm?
2. 60 kV: sections < 150 mm? 2. All voltages > 60 kV

Or(i) is the conductor temperature rise of the hottest cable in the group. The external
thermal resistance of the hottest cable in equations (5.30) and (5.31) will now include the
effect of the other (NV— 1) cables and will be denoted by 7, + A7y. Applying equation
(5.12), we obtain the following new form of equation (5.31):

D* L* N he a
i( a3)+i( =)+>|( Ars
StFe | eee
et\ ays jel = ey te yl eee

Ps
ae) — Ar
k#p (T4+ AT)
(5.36)
The valueof AT, is obtainedin Section9.6.2.1as
IN Psalne
in
20
where ;
ew oe Gp pk pw GER
dp -dpr +++++adpx»dpn
with factor ae /dpp excluded, leaving (N — 1) factors in (5.37a). The distances de) and
dpp are defined in Fig. 5.4.
Introducing the notation
4L*
a5 = Faw)’ (5.37b)
equation (5.36) can be approximated by

Ei
( D*
oe)+Bi
Oyeeg (
Dn L*
=)
fees+0 1| (
E d
1605
ae ii
qjEl
E ==|
* |
Bi(t) = aL F
2 In
Dy
(3.38)

Also, equation (5.30) becomes

fs W,1(14
(74+
+ AT.AT4) (5.39)
tase W nATs)
94 Part I m Modeling

The cyclic rating factor is given by equation (5.33) with

Or(t)
= [1—k, + ki Bi (t)Jo(t) (5.40)
AR(0O)

EXAMPLE 5.6
Determine the cyclic rating factor of the cable system analyzed in Example 5.3. First, we have to
identify the time during the 24 h period at which we expect that the conductor temperature rise will
reach its maximum value. From analysis of the values in Table 5.6, we select i = 18 as the hour at
which this maximum occurs. Note that this is not the time of maximum load, and our selection is
based on engineering judgment. The six preceding hours are underlined in Table 5.6 and the values
of Y reordered accordingly. The new order is shown in the last column of Table 5.6.
Much of the work for the determination of the cyclic rating factor for the cable system under
consideration has already been performed in Example 5.3. In particular, the conductor attainment
factors a(t) and the exponential integral values have been computed for all hours. The load-loss factor
has been evaluated in Example 5.5. The next step is to evaluate the cable surface attainment factors
B,(t). Asample computation is performed below for t = 1 h with the remaining values summarized
in column 3 of Table 5.8.

Di =00358me LA=1.0mpedp.= 0072

dj = O.0T2 2.0 = 2.001 m, “3= 0.5 210° m’/s aN = 3

The auxiliary variables defined in equations (5.37a) and (5.37b) are equal to

dy =n _ 2.001:2.001_ eu Ee ©W a210 = 0.144

dpi-dp2 0.072-0.072 MOAR ALESISOg pe
From equation (5.38) and Tables 5.1 and 5.2, we obtain

Ei( 1616
oe)+Ei Diwi
5 va |rpager|
peated
bare
i( f—)
ue
Bi(1)= *
>In4L*F
D:
_ (2.579
—0)+2-(0.365
—0)_
Sea
Ir eee
2.9 0.0358
Next, from equation (5.39), we compute factor k, with the value of 7, + AT, given above
(the external thermal resistance given in Table Al already includes the mutual heating effect) and
equivalent thermal resistance 7, and Tg from Example 5.3. We have

W,(74+ AT,) Syehvoteys

IA2)8%)
ke' W.T + Wi(T, + ATs) = 30.82(0.214 + 0.113) + 33.38 - 1.933 = 0.865

The ratios 0p(i)/@z (00) are computed from equation (5.40). As an example, we will calculate
this ratio fori = 1. The remaining values, corresponding to the underlined Y values in Table 5.6, are
shown in column 5 in Table 5.8.

3 If no other information is available, the time of the highest loading should be selected.
Chapter
5 = RatingEquations—Transient
Conditions 95
TABLE 5.8 Evaluation of Cyclic Rating Factor
——
Time

(h) Y; a(t) Bit) 0g(i)/ Op(~)

(1) (2) (3) (4) (5)
0 0.992 0.999 0.146 0
1 0.728 1 O2TT 0.261
2 0.640 1 0.255 0.318
3 0.596 1 0.288 0.356
4 0.593 1 0.315 0.384
5 0.796 1 0.337 0.407
6 —_— 1 0.356 0.426

Or(1)
= [1 —k, +k, B)(t1)] a(t) = (1 — 0.865 + 0865 - 0.146) - 1.0 = 0.261
Ar(OO)

The cyclic rating factor is now computed from equation (5.33) as follows:

1
Vie

p(T)
“|| —
ales —,[OrG
9x
—— y,|———(+1) Ori)
00)ae
- i=0

= [0.504(1 — 0.426) + 0.992(0.261) + 0.728(0.318 — 0.261) + 0.640(0.356 — 0.318)

+ 0.596(0.384 — 0.356) + 0.593(0.407 — 0.384) + 0.796(0.426 — 0.407)]'/? = 1.23

Thus, the permissible peak value of the cyclic load current is

1.23 -629=774A

EXAMPLE 5.7
Determine a cyclic rating factor for a single three-core cable with the load curve having the following
characteristics: (1) a sustained maximum current lasts for a minimum of 6 h, and (2) there are no
restrictions on the shape of the reminder of the cycle, except that the maximum conductor temperature
rise occurs at the end of the duration of the sustained high current.
Since in this case Yo = Y; = --- = Yo = 1, equation (5.33) simplifies to

M ee ] ee, (5.41)
é Or (6) Or (6)
Or(CO) Op(Co)

Substituting now equation (5.29), we obtain

1
a (5.42)
w+ —pyll—k + kB 6)]a(6)]

EXAMPLE 5.8
Assume that a cyclic rating factor M has been derived for the cable system located in a soil with
thermal resistivity p,. We will determine a new cyclic factor for the same cable system with the soil
96 Part I » Modeling

thermal resistivity equal to p,. We will then use the resulting formula to determine the cyclic rating
of the system in Example 5.6 with a soil thermal resistivity of 0.85 K . m/W.
The factors k and k, [see equations (5.30) and (5.39)] are dependent on the depth of laying and
soil resistivity. We will first rewrite equation (5.30) as follows:

WT, W_T, W_T; T;

k= = — (5.43)
Wel + W, 14 W, ik +W)T; Wiad Wily + T.aa
“Tet
1+, +A2

where T. is the
€
cable internal thermal resistance as if all the joule losses were generated at the
conductor. Now, if the laying conditions “a,” for which a value of k is known, change to conditions
“b. then

pe
k(a) = 4 (5.44)
ace dive

and

T?
k(b) = & ; (5.45)
Dome)

where 7; and T/ are the external thermal resistances for conditions “a” and “b,” respectively.
Computing 7. from (5.44) and substituting this in (5.45), we obtain

(5.46)

The new value of k, is obtained in the same way.

In our example, the depth is not changed. Therefore,

ihe ve 1
— = — = — =1,176
yiee ios PES

and
k= : 0.845
fe LP
TR eed
1+ 1.176 pensGibos en
0.965
The new ratio 6r(1)/Ar (co) = 1 — 0.845 + 0.845 - 0.211 = 0.333. The remaining ratios are as
follows:

0.333 0.370 0.398 0.421 0.440 0.456

The revised cyclic loading factor is thus equal to

M = [0.504(1 — 0.456) + 0.992(0.333) + 0.728(0.370 — 0.333) + 0.640(0.398 — 0.370)

+0.596(0.421 — 0.398) + 0.593(0.440 — 0.421) + 0.796(0.456 — 0.440)]-'/? = 1.21

The steady-state ampacity of this cable system is now larger than before. The new steady-state rating
is equal to 674 A. Therefore, the new cyclic rating is equal to 1.21 -674 = 815 A.
ae ae ea
Chapter 5 m Rating Equations—Transient Conditions i

9.6.3 CyclicRating with Partial Drying of the Surrounding Soil

5.6.3.1 Assumptions. The method to be presented in this section, based on CIGRE
(1992), applies a two-zone soil model and is an extension of the techniques described
previously for calculating the cyclic rating factor for a cable in uniform soil. We should
stress again that there are several assumptions associated with an application of a two-zone
model. These assumptions were reviewed in Section 4.2. Additional assumptions, related
to the application of the method for cyclic factor computations, are listed below.
© Once the soil around a cable has dried out, it will not rewet while there is any heat
flux flowing in a direction away from the cable.

e The position of the dry zone boundary will be where the peak soil temperature during
a load cycle just attains the critical value.
¢ Cyclic load has persisted for a duration long enough so that the temperature variations
in the soil have attained their final values.

Based on the results of numerous calculations on different types of cables, Parr (1987,
1988) has observed that the cyclic rating factor is essentially independent of the value of
the critical temperature or the size of the dry zone. For a given conductor temperature,
the relationship between the sizes of the dry zones for both cyclic and sustained load
varies in such a way that, although the steady-state rating changes substantially with the
critical temperature, the cyclic rating factor multiplying that steady-state rating is largely
unaffected.
Further, calculations for a wide range of soil and load characteristics indicate that a
factor derived for cyclic conditions where the critical temperature of the soil is just equal to
the peak cyclic temperature of the cable surface (that is, a dry zone has just not developed)
will be on the safe side by not more than 5% from the cases where there is a substantial
dry zone (Parr, 1987, 1988; CIGRE, 1992), and in most cases the error is not greater than
2%. Such an error is well within the accuracy with which soil and load characteristics are
usually known.
This feature makes it possible to considerably simplify the computations of cyclic
rating factors when soil dry out is expected. In the developments presented below, it is
first assumed that the soil critical temperature is equal to the peak cyclic value of the cable
surface temperature; that is, a dry zone is at the point of occurring. At this point, the soil
surrounding the cable has uniform properties appropriate to its wet, in situ, state. The cyclic
rating factor for these conditions is derived by equation (5.33). The factor is then adjusted
so that it applies to the steady-state rating for the same assumed value of critical temperature
when there is a dried out zone.
In general, the size of a dry zone where the boundary just achieves a certain critical
temperature rise with cyclic loading is smaller than the zone which will form for the same
critical temperature rise with steady-state loading. We also observe that the size of dry zone,
and hence the cable external thermal resistance, changes with the type of loading. The last
observation has an important implication for the computational procedure adopted in this
section. As indicated above, the procedure is to make an adjustment to the cyclic rating
factor computed for uniform soil when the external thermal resistance is the same for cyclic
and the steady-state conditions. Therefore, for cyclic rating computations, an adjustment
is made by simply using the ratio of the appropriate external thermal resistances. This is
shown below.
98 Part I » Modeling

5.6.3.2 Development of the Cyclic Rating Factor. The computation of a cyclic

rating factor is performed as follows. The thermal response of the cable and its environment,
including the effect of its internal thermal capacitance, is obtained by using the appropriate
formulas in Section 5.2. According to the convention adopted at the beginning of this
chapter, the temperature rises referred to here are those due to the joule losses only, and the
equations quoted take this into account at the appropriate stages. The factor M is adjusted
for the presence of a dry zone under steady-state loading conditions; Parr (1987) developed
the following procedure to find the new cyclic factor M, for two-zone soils. Let / be the
maximum current permitted by applying the factor from equation (5.33) to the steady-state
rating based on uniform soil around the cable. Then

I= MIp (5.47)

where Jp is the rating based on an external thermal resistance with no dry zone. A load
equal to Jp can be carried only if the cable surface temperature under steady-state loading
does not exceed the critical temperature for the soil immediately around the cable. In this
case, factor M is applicable without correction.
If such is not the case, a dry zone is presumed to form, and the cable external thermal
resistance will be increased. Let Ii denote the steady-state rating current when there is a
dry zone. This current is obtained from equation (4.13). In order to maintain the same peak
value of current 7, we should have

I = MIrp = Mi Ip (5.48)

From equation (5.48), we obtain

TR
Tp

Using the special notation adopted in this chapter, the rated current /p is obtained from
(4.3):

n= | 6 (oo) —W,dl [0.5Tptn(h+T3+ a)

% Loe TAP (5.50)
Re (Te + T4)

where
R. =nR(1 +A, +2)

T
Fane SR Res pC NNRp
IGWp
T.=
LE Ay ng
and 6(0o) represents permitted conductor temperature rise (due to all losses). T,.represents
the internal thermal resistance of the cable computed under the assumption that all the joule
losses are produced at the conductor. The external thermal resistance 7; is that of the wet
soil.
The rated current with moisture migration taken into account is given by equation
(4.13), rewritten below with the special notation adopted in this chapter:

ib Ee — Wz [0.5T; +n(T) + Ts + v+T4)] + (v — 196, 7°

‘ RATe + V+ Ts) (S61)
Chapter5 # RatingEquations—Transient
Conditions 99

where

Vs = G2)
Pi
and p, and p2 are the thermal resistivities of the wet and dry soil, respectively.
We also have
Topi dmade da telly
Cre (ee
=1+k(v—1) (5.53)
because,from equation (5.43),
petals
lates
Dividing (5.50) by (5.51) with the ratio of thermal resistances given by (5.53), we
obtain the required expression for M. In developing this ratio, we are taking into account
the fact mentioned above, that a factor derived for cyclic conditions where the critical
temperature of the soil is just equal to the peak cyclic temperature of the cable surface (that
is, a dry zone has just not developed) will be on the safe side by a small margin. This means
that we make the following substitution:
OF = 0.(t) -nWala

Hence, M, can be expressed as

eels Sah)
M,=M (5.54)
8.(T)
1.4 ites) wane

where 6,(t) is the peak cable surface temperature rise occurring at time tf = t. Also,
in accordance with the notation adopted in this chapter, all 6s represent temperature rises
above ambient due to joule losses.
Equation (5.54) is applicable only when the total cable surface temperature rise
[9.(t) +nW,T4] is greater than the critical temperature rise of the soil; otherwise, there
will be no drying and the cyclic rating factor is M without any correction. Thus, in order
to determine which rating factor to use, the value of [0.(t) + W474] has to be determined
first. This is accomplished as follows.
From the definition of the cyclic rating factor
Omax= Or(00) (3,95)

or
Oc(T) + A(T) = Or(O0)
On the other hand, assuming that the conductor-above-cable-surface temperature rise
is proportional to the square of the current, we have from equation (5.25)

eae 176.(00)O(t)
2 feel | | Sy[ei+)
R
a)
8.(00)(00) 0)

= M76.(00){u[1 —a(t)] + A} = M*OR(Co)(1 —k){u[ —o(t)] + A}

(5.56)
100 Part I m»Modeling

because
M’I?=I; and6-(00)
=(1—
k)Or() (5.57)
where
dil fait) Oi) ]_ S ;
cance sre: = d Y;Ciao)1)- batt! Ope
5.58
Substituting (5.56) into (5.55), we obtain

O.(t)=Og(00)
{1—M*(1
—k){A’
+wl —
a(z)]}} (5.59)

EXAMPLE 5.9
Determine the cyclic rating factor for the cable system examined in Example 5.6, taking moisture
migration into account. Assume a dry soil thermal resistivity value equal to 2.5 K - m/W and a critical
soil temperature rise of 35 K.
First we compute from equation (5.58) the value of A’ with the temperature ratios and attainment
factors given in Table 5.8:

Me Ostia 50.0)
i=0
= 0.992(1.0) + 0.728(1.0 — 1.0) + ent — 1.0)

+ 0.596(1.0 — 1.0) + 0.593(1.0 — 1.0) + 0.796(1.0 — 1.0) = 0.992

The conductor temperature rise due to joule losses is equal to 90 — 15 = 75 K because the
dielectric losses are negligible. From Example 5.6, M = 1.23, k = 0.856, and a(6) = 1. The loss
factor was computed in Example 5.5 and « = 0.504. Therefore, the peak temperature rise of the
cable surface is equal to

6¢(6) = 8e(t) = x(00) {1 — M?(1 —KA! + wll —a())}}

= 75- {1 — 1.237(1 —0.865)[0.992 + 0.504(1 — 1.0)]} = 59.8 K

This is greater than the critical value of 35 K; hence, drying can be expected. The corrected
rating factor for this condition is, from equation (5.54), equal to

M,=M eo
— ee ) 1.93 I1 +0.865(2.5
—
i reeae i—Iea) 1+ ——-(2.5—-
rere 1)
The steady-state rating with drying out when the critical temperature rise is 35 K was obtained in
Example4.2andis equalto 541A;hence,thepeakcurrentis 1.26-541= 682A.
ee ee a Se ie ee ae a

9.7 CALCULATION OF EMERGENCY RATINGS

During emergency conditions (e.g., loss of acompanion circuit in a two-circuit transmission

line), power cables may be required to carry substantially higher currents than permitted
Chapter 5 » Rating Equations—Transient Conditions 101

by the steady-state rating. Emergency conditions usually last only a few hours, and the
conductor temperature is often permitted to reach a higher value than that allowed in a
steady-state operation. In this section, we will develop formulas for calculating the short
time rating of a single circuit based on knowledge of the conductor temperature transient
as derived in Section 5.2.

5.7.1 ThermallyIsolated Circuits

Consider an isolated buried circuit carrying a constant current J; applied for a suffi-
ciently long time so that steady-state conditions are effectively reached. If acyclic load with
a peak value of current equal to J amperes has been applied for a long time, then J; = /u/
where ju is the load-loss factor of the cyclic load. Subsequently, from a time defined by
t = 0, an emergency load current /> (greater than /,) is applied. If J, is applied for any
given time f, the question is how large may J be so that conductor temperature does not
exceed a specified value, taking into account the variation of the electrical resistivity of the
conductor with temperature. The effect of dielectric loss is first neglected, but is taken into
account at the end of this discussion.
In the following developments, we will assume that the heat generation per unit volume
of the conductor for the time t of emergency loading is constant and equal to its value Winax
at the end of the period of emergency loading, at which time the conductor temperature
rise above ambient is 64x. Goldenberg (1971) has shown that with this assumption, a safe
value of emergency current can be obtained. We can write

Omax- OR(0) i} Wmaxts Wo

(5.60)
Or(t) Wr

where Wo = I? R; and Wr = [7 Rp are the heat generation per unit volume of conductor at
time t = Oand during the steady-state, respectively. /p is the steady-state rated current and
the conductor ac resistance corresponding to this current is denoted in this section by Rr.
Substituting now the definition of the conductor losses, equation (5.60) can be rewritten as

i? .
r —h{Ri/Rr z Ip i ve hi Ri (5.61)
Or(t)/Or(CO) Rr Rr
Omax
where i
AR(0O)
hy =— I
TR
Or(co) = steady-statetemperaturerise correspondingto current [pr
Omax = Maximumpermissibletemperaturerise aboveambientat the end
of the emergency period
Or(t) = conductor temperature rise above ambient at time ¢ after application
of current Jp neglecting the variation of conductor resistance with
temperature from t = 0
4r(0) = steady-state conductor temperature rise above ambient following
application of current /
102 Part I » Modeling

The emergency rating current is obtained by solving equation (5.61) for 1):

h2R, R: [r= 18R I2

aay
epee )4*1 maxae R (5.62)
Ar(00)
The ratio Op(t)/Ap (oo) is obtained by computing both temperatures rises separately.
The steady-state conductor temperature rise above ambient is simply

AR(OO)= 0 33Oambaa 64 (5.63)

with 4,mb and 6, denoting the ambient temperature and the temperature rise caused by
the dielectric loss, respectively. 6. is the maximum steady-state conductor temperature
expressed in °C. The value of z(t) is obtained by applying equation (5.26) while neglecting
dielectric losses.
The effect of heating due to dielectric loss is taken into account by calculating from
equation (5.11) the steady-state conductor temperature rise 6g due to the dielectric loss and
subtracting this value from Onax, Or(t), and Or(co). Any change in dielectric loss with
temperature is neglected. The calculation of /> then proceeds as above, using the modified
values Of Omax,Or(t), and Or(oo). The values of Ro, Rmax, R;, and R> are not altered.
Goldenberg (1971) has shown that the emergency rating current obtained from equation
(5.62) is on the safe side with an error not exceeding 2% for an emergency duration of 3 h
or more, 3% for an emergency duration of 2 h, and 1-5% for an emergency duration of 1 h.

EXAMPLE 5.10
Assume that the cable system examined in Example 5.3 has been operating at 550 A continuously,
which gives a steady-state temperature of 70°C (the ampacity of this circuit at 90°C is given in
Appendix A as J, = 629 A). Determine the emergency current level that can be carried for 6 h
without exceeding the maximum operating temperature of 105°C.
From the statement of the problem, we have

Oran Ashes? Sito wy ey 6 h (21 600 h

i Osteo) =a ———_
agO ens = ], . = (zw habla
‘ = ee
cig 629 ve
R, = 7.338 - 10-°Q/m (ac resistance at 70°C)

Rmax = 8.151 - 10-°&/m (ac resistance at 105°C)

Rr = 7.810 - 10-°2/m (ac resistance at 90°C)

Temperature rise due to dielectric loss = 0 K. Thus, from equation (5.63)

Or (CO) = 0, — Gam — Og = 90 — 15 = 75 K

Ox = 30.4 (see Table 5.2, column 5) for application of a step function of rated current.
The emergency current is obtained from equation (5.62):
Chapter 5 a Rating Equations—Transient Conditions 103

bh=I1
on ae Ont)
Or(CO)
LoBhs 0.8742 a 7.810
-7.338 8.151[129-087457.338]
0 ) 2 on
- S151 30.4 .
hS

5.7.2 Groups of Circuits

Where groups of circuits which are not thermally independent carry loads behaving
in the same way, the methods for calculating the emergency ratings given in Section 5.7.1
can be applied.
When the loads do not behave in the same way, for example, the first circuit has an
increased load while the second goes off load, the matter requires careful consideration.
In most cases of buried cables, the time lag associated with the mutual heating between
circuits is so great that any worthwhile reduction in temperature of the first circuit due to
the second going off load does not occur for many hours.
Figure 5.8 (CIGRE, 1978) shows the time taken for a 1 K reduction in temperature of
the first circuit of a double-circuit 1935 mm7, 400 kV installation when the current in the
second circuit is reduced to zero. Thus, any increase in load for the first circuit during this
period will depend entirely on what increase in conductor temperature can be permitted
above the value the first circuit had attained before the change in load.

Osos UpteG
p——

(Axial
Sseparation
m) 0.6

0.4

0.2

20 40 60 80 100 120 140 160 (h)

Time (h)

Figure 5.8 Time for 1 K reduction in temperature at the axis of the central cable of a first
circuit due to current reduced to zero in a second circuit (IEC, 1989).
104 Part I m Modeling

REFERENCES

Affolter, J. F. (1987), “An improved methodology for transient analysis of underground

power cables using electrical network analogy,” M.E. Thesis, McMaster University,
Hamilton, Canada.
Anders, G. J., Moshref, A., and Roiz, J. (July 1990), “Advanced computer programs
for power cable ampacity calculations,” JEEE Comput. Appl. in Power, vol. 3, no. 3,
pp. 42-45.
Anders, G. J., and El-Kady, M. A. (Oct. 1992), “Transient ratings of buried power cables—
Part 1: Historical perspective and mathematical model,” JEEE Trans. Power Delivery,
vol. 7, no. 4, pp. 1724-1734.
Baudoux, A., Verbisselet, J., and Fremineur, A. (1962), “Etude des regimes thermiques
des machines electriques par des modeles analogiques, application aux cables et aux
transformateurs,” Rapport CIGRE, No. 126.
Buller, F. H. (1951), “Thermal transients on buried cables,” Trans. Amer. Inst. Elect. Eng.,
vol. 70, pp. 45-52.
CIGRE (Oct. 1972), “Current ratings of cables for cyclic and emergency loads. Part 1.
Cyclic ratings (load factor less than 100%) and response to a step function,” Electra,
No. 24, pp. 63-96.
CIGRE (Jan. 1976), “Current ratings of cables for cyclic and emergency loads. Part 2.
Emergency ratings and short duration response to a step function,” Electra, no. 44,
pp. 3-16.
CIGRE (Dec. 1992), “Methods for calculating cyclic ratings for buried cables with partial
drying of the surrounding soil,” Electra, no. 145, pp. 33-67.
Goldenberg, H. (1957), “The calculation of cyclic rating factors for cables laid direct or in
ducts,” Proc. IEE, vol. 104, part C, pp. 154-166.
Goldenberg, H. (1958), “The calculation of cyclic rating factors and emergency loading for
one or more cables laid direct or in ducts,” Proc. IEE, vol. 105, part C, pp. 46-54.
Goldenberg, H. (1964), “Cyclic loading and transient temperatures of power cables,” Work-
ing Paper of CIGRE WG02 of Study Committee 21.
Goldenberg, H. (1967), “Thermal transients in linear systems with heat generation linearly
temperature-dependent: application to buried cables,” Proc. IEE, vol. 114, pp. 375-377.
Goldenberg, H. (1971), “Emergency loading of buried cable with temperature-dependent
conductor resistance,” Proc. IEE, vol. 118, pp. 1807-1810.
IEC Standard (1985), “Calculation of the cyclic and emergency current ratings of cables.
Part 1: Cyclic rating factor for cables up to and including 18/30 (36) kV,” Publication
853-1.
IEC Standard (1989), “Calculation of the cyclic and emergency current ratings of cables.
Part 2: Cyclic rating factor of cables greater than 18/30 (36) kV and emergency ratings
for cables of all voltages,” Publication 853-2.
King, S. Y., and Halfter, N. A. (1982), Underground Power Cables. New York: Longman.
McGrath, M. H. (1964), Discussion contribution to Neher, 1964, IEEE Trans. Power App.
Syst., vol. 83, p. 113.
Chapter 5 m Rating Equations—Transient Conditions 105

Morello, A. (1958), “Variazioni transitorie die temperatura nei cavi per energia,” Elettrotec-
nica, vol. 45, pp. 213-222.
Neher, J. H. (1964), “The transient temperature rise of buried power cable systems,” JEEE
Trans. Power App. Syst., vol. PAS-83, pp. 102-111.
Parr, R. G. (1987), “Cyclic ratings for cables. A simple adaptation for partly dried soil,”
ERA Report No. 87-0265.
Parr, R. G. (1988), “Cyclic ratings for cables. A simple adaptation for partly dried soil.
Supplement to Report No. 87-0265,” ERA Report No. 88-0127.
Thomann, G. C., Aabo, T., Ghafurian, R., McKernan, T., and Bascom, E. C. (1991),
“A Fourier transform technique for calculating cable and pipe temperatures for periodic
and transient conditions,’ JEEE Trans. Power Delivery, vol. 6, no. 4, pp. 1345-1351.
Van Valkenburg, M. E. (1964), Network Analysis. Englewood Cliffs, NJ: Prentice-Hall.
Van Wormer, F. C. (1955), “An improved approximate technique for calculating cable
temperature transients,” Trans. Amer. Inst. Elect. Eng., vol. 74, part 3, pp. 277—280.
Whitehead, S., and Hutchings, E. E. (1938), “Current ratings of cables for transmission and
distribution,” J. IEE, vol. 38, pp. 517-557.
Wlodarski, R. (1963), “Echauffement de cables souterrains de transport d’énergie,” Edition
de l’Ecole Polytechnique de Varsovie.
Wlodarski, R., and Cabiac, M. (1966), Etudes et Experiences Récentes Concernant la
Détermination de l’Echauffement Transitoire des Cables Enterrés. Warsaw: Panst-
wowe Wydawnictwo Naukowe.
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 6

Dielectric Losses

When paper and solid dielectric insulations are subjected to alternating voltage, they act
as large capacitors and charging currents flow in them. The work required to effect the
realignment of electrons each time the voltage direction changes (i.e., 50 or 60 times a
second) produces heat and results in a loss of real power which is called dielectric loss, and
which should be distinguished from reactive loss. For a unit length of a cable, the magnitude
of the required charging current is a function of the dielectric constant of the insulation,
the dimensions of the cable, and the operating voltage. For some cable constructions,
notably for high-voltage, paper-insulated cables, this loss can have a significant effect on
the cable rating. In this chapter, we will develop formulas for calculating the dielectric
losses and examine the effect of cable construction on the value of these losses. Before
doing so, however, we will review a few basic concepts related to the behavior of dielectrics
subjected to ac voltages.
Cable insulation is a material whose dielectric response is a result of its capacitive
nature (ability to store charge) and its conductive nature (ability to pass charge). The
material can be represented by a resistor and capacitor in parallel (Fig. 6.1).
When a voltage Uo is applied to this circuit, the current 7 will form an angle g with
the voltage as shown in Fig. 6.1. This current is composed of two components: capacitive
(charging) current J, and the resistive (leakage) current /,. Since, in the case of good
insulating materials, the magnitude of the leakage current vector is much smaller than that
of the capacitance current vector, the loss angle 6 is very small. The charging and leakage
currents are equal to

Ia jog wanasy1; = 2 (6.1)

i

109
 Part II » Evaluation of Parameters
110

Uo

Figure 6.1 Representation of a cable insulation.

where C is the capacitance of the insulation and w= 27f, f being the system frequency
and j = /—1. To compute C, we observe that the effect of dielectric material has been
traditionally described by introducing the concept of relative permittivity, denoted by ¢ and
efinedas
defined C
é
ae
where Cp is the capacitance of identical size and construction capacitor with vacuum as the
dielectric. The quantity ¢ is often referred to as the static or low-frequency value of the
permittivity or the dielectric constant. Then,

(6.2)

where D; is the external diameter of the insulation excluding screen and d, is the diameter
of the conductor, including screen. The same formula can be used for oval conductors if
the geometric mean of the appropriate major and minor diameters is substituted for D; and
de
Another measure of a dielectric is the dissipation factor denoted by tané and often
referred to as the loss factor of the insulation at power frequency. From Fig. 6.1, we can
see that

7, | Uo l
ifhal@y
= = ——_—_—= —— (6.3)
\I.| R;CwUp R;Cw

Evidently, the smaller the value of tan 4, the more the dielectric material approaches
the condition of a perfect insulator. For a given set of electric field and system frequency,
the loss factor will undergo change with temperature. This relationship is shown in Fig. 6.2
(Westinghouse, 1957).
In practice, € and tan 6dare assumed constant in computations of cable ratings. Their
values are given in Table 6.1 (IEC 287, 1989; IEC 287-1-1, 1994).
The dielectric loss per unit length in each phase is then obtained from equation (6.4)
as
Chapter 6 m Dielectric Losses 111

0.006
Oil-filled cable

|
6Tan 0.004 Solid-type
| cabl
My if
0.002 is
Gas-pressurized cable

10522077309 407750) 605.70 80. 90) 100

Temperature °C

Figure 6.2 Variation of tan d with temperature.

TABLE 6.1 Values of ¢ and tan 64 for the Insulation of Power Cables
at Power Frequency

Type of Cable é tan 6

Cables insulated with impregnated paper 4 0.01

Solid type, fully impregnated, preimpregnated
or mass-impregnated nondraining
Fluid-filled low-pressure
up to Up = 36 kV 3.6 0.0035
up to Up = 87 kV 3.6 0.0033
up to Up = 160 kV 35 0.0030
up to Up = 220 kV 3.5 0.0028
Fluid-pressure, pipe-type Bu 0.0045
Internal gas-pressure 3.4 0.0045
External gas-pressure 3.6 0.0040
Cables with other kinds of insulation
Butyl rubber 4 0.050
EPR—up to 18/30 kV 3 0.020
EPR—above 18/30 kV 3} 0.005
PVC 8 0.1
PE (HD and LD) eis} 0.001
XLPE up to and including 18/30 (36) kV—unfilled PES) 0.004
XLPE above 18/30 (36) kV—unfilled 25 0.001
XLPE above 18/30 (36) kV—filled 3 0.005
Paper—polypropylene—paper (PPL)" 2.8 0.001

“The dielectric constant and the loss factor of PPL insulation have not been
standardized yet.

Us f
Wy= R = wCU, tand (6.4)
l

Dielectric loss is voltage dependent, and thus only becomes important at higher voltage
levels. Table 6.2 (IEC 287-1-1, 1994) gives, for the insulation materials in common use, the
112 Part II » Evaluation of Parameters

value of Up at which the dielectric loss should be taken into account if three-core screened or
single-core cables are used. It is not necessary to calculate the dielectric loss for unscreened
multicore or de cables.

TABLE 6.2 Phase-to-Ground Voltage above Which

Dielectic Losses Must Be Calculated

Up
Type of Cable (kV)

Cables insulated with solid paper 38

Cables insulated with fluid/gas 63.5
Butyl rubber 18
EPR 63.5
PVC 6
PE (HD and LD) 127
XLPE unfilled 127
XLPE filled 63.5
PPIEP 63.5

EXAMPLE 6.1
We will determine dielectric losses of the model cable model No. 3 (pipe-type cable) and compare
them with the dielectric losses of model cable model No. 5 (400 kV PPL cable) assuming that both
cables operate at 60 Hz. The parameters of the pipe-type cable are (see Appendix A) D; = 67.1 mm,
d. = 41.45 mm, and ¢ = 3.5. The parameters of the cable model No. 5 are D; = 94.6 mm, d.. = 58.6
mm, and € = 2.8.
The capacitances of cables 3 and 5 are

& 3.5
Cas = er apa 10° = Sara Tacs 10-° = 0.4036-10°° F/m
ide heyy sie
d. 41.45
€ 2.8
Ceabs = ———>—~ : 10°” -10-? = 0.3248- 107° F/m

18in
(2 181n
gee
Dielectric losses
d.
are obtained from equation
58.6 (6.4) with the loss factor from Table 6.1:

U2 A 13
Wacab3= —* = wCU? tand = 2760 -0.4267 - 107° - ( : 7) -0.005 = 4.83 W/m
R; J3

UZ ; 4 °
Wacabs
=is =wCUR
tand
= 2760
-0.3248
-10-°.( a) -0.0001
=6.53W/m
We can observe that in spite of a voltage level about three times higher, the dielectric losses in
the PPL cable are almost the same as the corresponding losses in the pipe-type cable with standard
paper insulation for these examples.
Chapter 6 m Dielectric Losses 113

REFERENCES

IEC 287 (1982), “Calculation of the continuous current rating of cables (100%) load factor,”
IEC Publication 287.
IEC 287-1-1 (1994), “Electric cables—Calculation of the current rating—Part 1: Current
rating equations (100% load factor) and calculation of losses. Section 1: General,” IEC
Publication 287.
Westinghouse (1957), Underground Systems Handbook.
 BWGAFIS ° we net oor he

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A P
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2 Ree ier :
Ta anal A
mura”)
1P= yurOnamae
th
onataatutailipn
enie
10)"emis| nine OeSnoan aria bd 0 sr AMA
T AG7.
| valle
dhotrwand
inept Enaniab ae noe : ‘e i
ee a>« ov ns ae ae
i a , o _
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> ood
ae :
ap =
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== <a S

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2 7 oi! @—hecP) 1D Gare Apquees Ty
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: i>of Mie,; : ae)
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a =. W@ou Cedebie0" ah oe wae
ar ©
Coenen y 9 el Vw he:
meee: :
zl a Nal jor ae > & a)

6 - UI oe } 7 ive at . ! oe7 } joe = ‘a

; 7 i , ae 1
x a y iy : n ax ae a
e.. L al | , ona » £6040 7 ee ( —— OD) = ¢
@ ‘ i
7 ot! Ds : , ae hw i
OPva OHNNG Bia |= Np o> 9 GegnGea ikelirrihiy@ i
> Vii,'sins ee Wilane 4CRRA I @
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 7

Joule Losses
in the Conductor

7.1 INTRODUCTION

The majority of losses in the power system are the result of a natural physical characteristic
of electrical conductors, referred to as resistance. The resistance causes electrical energy to
be converted to heat energy whenever current flows. Although this conversion process can
be harnessed as is done in electric stoves, clothes dryers, electric heaters, and so on, such
conversions in the conductors of a power system are largely wasted or “lost” since the heat
is usually dissipated into the atmosphere or to the ground. These losses, often referred to
as joule losses and denoted by W, (W/m), are computed from the equation

W.=I°R

where / is the conductor current and R is its ac resistance at the operating temperature of
the conductor.
In power cable installations, the heat generated by these losses has to be dissipated
through the surrounding soil or through air. Assuming that the current to be transmitted
through a cable is given, joule losses can be reduced by reducing the ac resistance of the
conductor. Alternating current (ac) resistance is always higher than the direct current (dc)
resistance mainly because of the presence of skin and proximity effects which are described
in Section 7.2.2. Therefore, the ac resistance should be kept as low as possible by a careful
design of the construction of the conductor. In this chapter, we will review how the conductor
resistance is computed for rating calculations.

115
116 Part II » Evaluation of Parameters

7.2 RESISTANCE OF CABLE CONDUCTOR

The relation between the dc resistance per unit length at 20°C, R20, and the cross section S
of a solid conductor is expressed by the well-known equation

(7.1)

where 29 is the electrical resistivity of the conductor material at 20°C. The conductivity of
copper used as a cable conductor material was standardized in 1913 by the IEC. Table 7.1
gives the values of (29 for conductors and sheaths listed in the IEC Standard 287 (1982).
For stranded conductors, we will assume that S is an effective area equal to the area of an
equivalent solid conductor.
With changes of temperature, copper and aluminum change their dimensions and
resistivity. The latter can affect the thermal design of cables. For the practical temperature
range between —40 and 125°C, a linear relationship holds approximately between the
resistivity and temperature; hence,

R’ = Ry[1 + a20)6 — 20)] Cie)

where @9 is the temperature coefficient of resistance at 20°C and @ is the actual conductor
temperature (°C). Besides temperature, the temperature coefficient of resistance varies with
annealing and purity. The values given in Table 7.1 (IEC 287, 1982) are recommended in
rating computations.
Equations (7.1) and (7.2) give reasonably accurate values for the dc resistance of a
solid conductor. However, the vast majority of conductors in electric power cables are of
stranded construction, and the resistance of stranded conductors is not so accurately known.
It is commonly assumed that the current is confined to the individual strands, and does not
transfer from strand to strand in the direction parallel to the axis of the conductor. To take
account of an increased resistance of stranded conductors, the resistance computed from
equation (7.1) is in practice multiplied by an empirical factor of 1.02.
We will review the computation of stranded conductor resistances in the following
sections.
TABLE 7.1 Thermal Resistivities and Temperature Coefficients of Metals Used
in Cable Construction

Resistivity (39) -10° Temperature Coefficient (9) - 10

Material Q.-m at 20°C per K at 20°C

Conductors
Copper 1.7241 3.93
Aluminum 2.8264 4.03
Sheaths and armor
Lead or lead alloy 21.4 4.0
Steel 13.8 4.5
Bronze 3.5) 3.0
Stainless steel 70 Negligible
Aluminum 2.84 4.03
eee
Chapter 7 m Joule Losses in the Conductor 117

7.2.1 DC Resistance of Stranded Conductors

Because of the additional length due to stranding, the dc resistance of stranded con-
ductors is larger than that of solid conductors of equal area. The resistance at 20°C, R%,of
the wires of diameter d,, in layer n is given by

Lage md?ny, (73)

R5o 4 p20kn ,
where n,, is the number of wires in layer n and k, is the lay-length factor of the wires in

layer
n,given
by
kn=i1+(dy,
? )- (7.3a)
where £,, is the lay length of layer n and d,, is the mean diameter of layer n. The quantity
£,/d, 18known as the lay ratio of layer n. The total resistance of the conductor with N
layers is

1 |
wae >is- (7.4)
n
n=1 20

EXAMPLE 7.1
We will compute the dc resistance of the parallel combination of the skid wire and tape for the model
cable No. 3. The cable shield consists of a mylar tape intercalated with a 7/8(0.003) in bronze tape—1
in lay, and a single 0.1 (0.2) in D-shaped bronze skid wire—1.5 in lay.! The diameter over the tape is
equal to 2.648 in. The resistances of the tape and skid wire at an operating temperature of 60°C are
obtained as follows.
The mean diameter of the tape equals”

a — 000726 10/627 10°* = 0.0672 m

The cross-sectional area of the tape is obtained from

Ar = 0.762 - 10~* - 0.0222. 1 = 0.169 - 10-° m”

Tape resistance at 60°C is obtained by combining equations (7.1)—(7.3a), and is equal to

kp | 20)

eat
ee Ae, m0.0254
-0.0672°
) ae
[1 + 0.003(60 — 20)] = 0.194 Q2/m
0.169- 10-5
' 1 in = 0.0254 m.
2 We recall that throughout the book, the distance symbols with an asterisk represent values in meters.
118 Part IJ » Evaluation of Parameters

Similarly, for the D-shaped bronze skid wire, we obtain

d* = 0.06759 — 1 -0.005/2 = 0.0651 m

wm

x - 0.005? - 1/4
area = = = 0.101710) im

ee m - 0.0651 \? i
R ae Su [1 + 0.003(60 —20)] = 0.0211 Q/
= , = == U8 m
2 O10T 104

The combined resistance of skid wire and the reinforcing tape is

0.194 - 0.0211
Rat"~ee0.194+0.0211 = 0.0190 2/m

7.2.2 AC Resistance of Conductors: Skin and Proximity Effects

The resistance of a conductor when carrying an alternating current is higher than that
of the conductor when carrying a direct current. The principal reasons for the increase are:
skin effect, proximity effect, hysteresis, and eddy current losses in nearby ferromagnetic
materials, and induced losses in short-circuited nonferromagnetic materials nearby. The
degree of complexity of the calculations that can economically be justified varies consid-
erably. Except in very high-voltage cables consisting of large segmental conductors, it is
common to consider only skin effect, proximity effect, and in some cases, an approximation
of the effect of metallic sheath and/or conduit. Figure 7.1 illustrates the flow of eddy-type
skin and proximity currents which are superimposed onto the load current, resulting in an
uneven current density in the conductor.
The ac resistance R with the skin and proximity effects is given by

— Rk skkpr ~ R' (kx ae Koz—1)

with R’ defined in equation (7.2).

Load current

Proximity
eddy current

Skin
eddy current Figure 7.1 Illustration of the skin effect in
electrical conductors.
Chapter 7 m Joule Losses in the Conductor 119

Denoting y, = kx — 1 and y, = kyr — 1 (the notation used in IEC 287, 1982), we

have

R=R(1+ys+yp) (7.5)
The factors kx, ys, kp, and y, are called skin and proximity effect factors, respectively.
The ratio R/R’ is usually close to unity at power frequency, and may be a few percent above
unity for large diameter conductors. The derivation of these factors is very complex and
involves the use of Bessel functions. Therefore, in the following sections, only the applicable
final results are presented with the references to the full mathematical treatment provided
as appropriate.
7.2.2.1 Skin Effect. Since not all the magnetic flux due to filaments of current near
the center of a homogeneous conductor cuts the whole conductor, the inductance per unit
area will decrease towards the surface; hence, the current per unit area will increase towards
the surface (see Fig. 7.1).
The skin effect phenomenon was investigated by such people as Maxwell, Heaviside,
Rayleigh, and Russell. The following approximate formulas for the skin effect factor are
due to Arnold (1946) and Goldenberg (1961). Let

m= V2rU,bof/p (7.6)

where /; is the electrical resistivity at the operating temperature, ju, is the relative perme-
ability of the conductor (44, = | for copper and aluminum conductors), {49is the permeabil-
ity of the free space (49 = 47 - 107’), and f is the frequency. m is ./2 times the reciprocal
of the skin (penetration) depth. Let us consider a circular conductor with the diameter d,.
We will use the following notation:

= (=) 2 — BRS 19-7 (7.7)

To take into account the stranding and treatment of a conductor, a factor k, < 1 is
introduced in equation (7.7) to yield

enti inal
8 (7.8)
R’
The values of k, are tabulated in Table 7.2. The skin effect factor is obtained as follows.
For QO yas 2:8

x4
see en (7.9)

For2.8 <,253:8
ys = —0.136—0.0177x,+ 0.0563x? (7.10)
For3:s =X,
Septet ucyeyth (7.11)
120 Part II » Evaluation of Parameters

In the absence of alternative formulas, IEC 287 (1982) recommends that the same
expressions should be used for sector and oval-shaped conductors. Since for the majority
of practical cases x, < 2.8, equation (7.9) is used in the standards.
For a tubular conductor with inner and outer diameters d; and d,, respectively, the IEC
standard (IEC 287, 1982) gives the following approximate value of the constant k; to be
used in conjunction with equation (7.8):

Y } esate
Pe CWRU / BN (7.12)
d' +d;\ di+d;
where d/ is the diameter of an equivalent solid conductor having the same central duct.
Equations for the skin effect factor for a tubular conductor were developed by Dwight
(1922). The rigorous solution of the problem of skin effect involves a Bessel equation for
the determination of current distribution. Dwight’s equations were rearranged by Lewis
and Tuttle (1959) to permit convenient programming. To avoid the laborious evaluation of
the Bessel functions, Arnold (1936) proposed the following remarkably accurate approxi-
mations. Let
d;
eee eeendz =0.25m?k,(d-
—d;)? (7.13)
Then
je ra
Ys=a(z)|1—Bat 2
5 —Bb) (7.14)
where,
for0<z<5,
Tz? 56
Ge Siang habe” = OTe?
For 5 < z < 30, the values of a(z) and b(z) are calculated by the following poly-
nomials (useful for computer programming):

a(z) =0.19701—0.1546295z+ 0.073796z?—9.02854- 1073z3

+6.27032- 10-424
—2.69028- 10-°z?+7.0647-10-72—1.04301 - 10-8z?+ 6.62315-10-!!z8
b(z)= 0.5356—0.21030734z+ 6.495563
- 10~2z2—1.089373- 107223
+ 1.03728739
- 107324
—5.8238557 - 1075z>
+1.91099645- 10-66
—
3.38936767- 10-8z’+ 2.509622- 107!z8 (7.15)

Forz > 30
a(z) = f2z/2-—il einyalHO) = ee
4./2z —5
Dwight (1922) gave curves for the skin effect ratio k,,. These were later recomputed
and extended by Lewis and Tuttle (1959) and are shown in Fig. 7.2.

7.2.2.2 Skin Effect of Large Segmental Conductors. —Special conductor construc-

tions have been applied for underground cable systems with large conductor cross sections.
Chapter 7 # Joule Losses in the Conductor 121

Skin effect curves for

tubular conductors

R/R'
d

f = frequency in Hz
R' = ohms per mile

Figure 7.2 Skin effect resistance ratio for circular cylindrical conductors. R’ is in ohms
per mile (Lewis and Tuttle, 1959).

Conventional conductors are not suitable for large cross sections as their current-carrying
capacity would be seriously reduced by skin and proximity effects. Asegmental conductor
design (British Patent 412 017), which is commonly referred to as the “Milliken” or “type
M” conductor, has lightly insulated segments which reduce the magnitude of these effects.
Figure 7.3 shows the reduction of ac resistance achieved with the Milliken conductor in
comparison with conventional conductor construction (Ball and Maschio, 1968).
Recent work conducted in Germany and France resulted in the following empirical
expression for constant k, for Milliken conductors (IEC WG10, 1991):

k, = —0.4 + 0.145 In(S) (7.16)

where S (mm?) is the nominal cross-sectional area of the conductor. This expression applies
to conductors up to 2000 mm” having uninsulated wires with four, five, or six segments,
irrespective of the direction of lay of the wires. Since no experimental data are available
for larger conductors, equation (7.16) can be used for conductors with the cross-sectional
122 Part II m Evaluation of Parameters

R/R'

0 100 200 300 400 500

Spacing (mm)

Figure 7.3 Comparison of ac to dc resistance ratio for segmented and nonsegmented con-
ductors. (a) Annular conductor. (b) Measured values for segmental conductor.
(c) Calculated values for segmental conductor. (Bell and Maschio, 1968)

area exceeding 2000 mm’. This subject continues to be reviewed and developed; therefore,
equation (7.16) should be used with caution.>
Large aluminum conductors are constructed with peripheral strands around the seg-
ments to make the conductor round. The constant k, is then computed from the following
formula (IEC WG10, 1991):

k, = {12(1 —b) (fai —b) —0.57 + [a1 —b) —0.5](W — a) (1 — b)

+ 0.33(y —a)?(1 —b)?) + b(3 —6b + 4b7)}°° (7.17)

where b is the ratio of the total cross-sectional area of peripheral strands to the total cross-
sectional area of the conductor,

c 1 _ 2ni/n+2/3
te (1 + sinz/n)?’ Vite 2(1+7/n) he)

and n is the number of segments. Equation (7.17) is applicable to aluminum conductors up

to 1600 mm”. If the total cross-sectional area of the peripheral strands exceeds 30% of the
total area of the conductor, then k, may be regarded as unity.

EXAMPLE 7.2
Compute the skin effect factor for a 1200 mm? aluminum conductor with two layers of peripheral
strands. The ratio of the total cross-sectional area of peripheral strands to the total cross-sectional
area of the conductor equals 27%.
From equation (7.18),

a= —___
(1 + sin 2/2)?
= ODS
oe he7407
Qn /2+2/3
2(1+7/2)

and from equation (7.17),

3 Preliminary results from the experiments carried out in the United Kingdom indicate that the value of ks
computed from equation (7.16) may be too large.
Chapter 7 m Joule Losses in the Conductor 123

ks = {12(1 — 0.27)((0.25(1 — .27) —0.5]* + [0.25(1 —0.27) —0.5](0.7407 —0.25)(1 —0.27)

+ 0.33(0.7407 —0.25)?(1 —0.27)?) + 0.27(3 —6 - 0.27 + 4 - 0.272)}95 = 0.84

7.2.2.3 Proximity Effect. | When two conductors carrying alternating currents are
parallel and close to each other, the current densities on the sides facing each other are
decreased, and those on the remote sides are increased because of the difference in magnetic
flux densities. This results in an increase in the conductor ac resistance and is called the
proximity effect. The skin and proximity effects are seldom separable in cable work, and the
combined effects are not directly cumulative. The proximity effect in three-conductor cables
ordinarily is to slightly reduce the effect of skin effect alone. However, for computational
convenience, these effects are considered separately.
To make the foregoing equations applicable to stranded conductors, the empirical
transverse conductance factor k, has been introduced (Neher and McGrath, 1957). This
coefficient plays a role in proximity effect computations similar to that which the coefficient
k, plays in skin effect calculations. In analogy to equation (7.7), we have

8x f
eo hake e 10a kp (7.19)
In the majority of practical applications, x, < 2.8. In this case, the following approx-
imate formulas are given in IEC 287 (1982):
For two-core cables and two single-core cables,

and for three core-cables and for three single-core cables,

PnBray(0. 7 PGs
(ioaioy 1.18
eee (7.21
)

where é
WE xpBee
192200 d.
is

and s is the spacing between conductor centers.

Just as with the skin effect factor, the exact expressions for the proximity factor require
the solution of Bessel equations. Arnold (1941) proposed the following approximations for
various ranges of xp:
For two-core cables and two circular single-core cables,

2G
Vi oe coll
at A; (7.22)
= AG v=vB(Xp)

For three-core cables and for circular three single-core cables,

2
gJaiAe) (7.23)
ale mean
StePIERS
P 2- Sy?H(xp)
124 Part II m Evaluation of Parameters

For three stranded segmental conductors carrying three-phase current,

2.5y’G(xp) (7.24)
\p = SSS
“P 2 —Sy? (xp)
where,for0< x»)< DEB.
0.042+ 0.012x4
AGS) Trp haere ele
i B (7.25)
Lx} eray ees 0.0283x%
Ce) srg zepoe whlue Si eames wig 004I
Fon 2. Siaky 133.0,

A(xp) = —0.223 + 0.237xp —0.0154x2, B(x) = 0

G(xp) = —1.04 + 0.72x, —0.08x;, (7.26)
and H(x,) = 0.095 +0.119xp + 0.0384x2

For3.8< Xp,
A(xp)= 0.75 1.128x5=) Bp) 10.094 —0:376x7
1
2x» —4.69 (7.27)
G(x»)a
=—:--
4/2 8
l andpAaeonsATS
The values of the constant k, are shown in Table 7.2.

7.2.2.4 Skin and Proximity Effects in Pipe-type Cables. For pipe-type cables, the
skin and proximity effects calculated by the above formulas are increased by a factor of 1.5
(Silver and Seman, 1982). For these cables,

R = R'[1+1.5(y; + yp)] (7.28)

This is an empirical relation obtained for cables operating at voltages of up to

345 kV. Some cable designers use a value of 1.7 for this factor as recommended by Neher
and McGrath (1957) and the IEC 287 (1982). The latter value can be used if we prefer a
conservative design.

7.2.3 Summary of ac Resistance Computations

In rating computations, an engineer usually specifies the required conductor cross

section and its construction. Since computation of the de resistance of stranded conductors
from this information is not very straightforward, a common practice is to use the tabulated
values for Roo in equation (7.2). Tables 7.3 and 7.4 give the appropriate values as proposed
in IEC 228 (1982) and IPCEA (1978), respectively. When the conductor size is outside the
range covered by these references, the value of R29 may be chosen by agreement between
the manufacturer and purchaser. Otherwise, it is computed from either equation (7.1) or
(7.3) for solid and stranded conductors, respectively.
The procedure for computation of the ac resistance of cable conductor can be summa-
rized as follows:
Chapter 7 m Joule Losses in the Conductor 125

TABLE7.2 SkinandProximityEffects
rt ee
Whether Dried
Type of Conductor and Impregnated or Not k, ky

Copper
Round, stranded Yes ] 0.8
Round, stranded No 1 1
Round, compact Yes 0.8
Round, compact No 1 l
Round, segmental 0.435 0.37
Hollow, helical stranded Yes Eq.(712) 0.8
Sector-shaped Yes 1 0.8
Sector-shaped No 1 |
Aluminum .
Round, stranded Either 1
Round, four segment Either 0.28
Round, five segment Either 0.19
Round, six segment Either 0.12
Segmental with periperal
strands Either Bqi@el7)

“Since there are no accepted experimental results dealing specifically with aluminum stranded
conductors, IEC 287 recommends that the values of k, given in Table 7.2 for copper conductors
also be applied to aluminum stranded conductor of similar design to copper conductors.

1. Read the dc resistance at 20°C from Tables 7.3/7.4 or use equations (7.1) or (7.4).

2. For given conductor operating temperature 9 (°C), compute the de resistance R’

from equation (7.2).
3. Compute the skin effect factor y;(y; = ks, — 1) from equations (7.7)—(7.11) with
the constant k, obtained from Table 7.2. For segmental hollow core conductors,
equation (7.12) should be used to obtain k, or, alternatively, equations (7.16)—(7.18)
can be applied. For more accurate calculations of k;,, equations (7.13)-(7.14)
and/or Fig. 7.2 can be used. Note that IEC 287 lists equations (7.9) and (7.12) only.
4. Compute the proximity effect factor y,(yp, = kp, —1) from equation (7.20) for two-
core or two single-core cables or from equation (7.21) for three-core or three single-
core cables. Constant k, is read from Table 7.2. For more precise computations,
equations (7.22)-(7.27) may be used.
5. Compute the ac resistance from equation (7.5) or (7.28).

EXAMPLE 7.3
We will compute the ac resistance (at 90°C) of model cable No. 1 using the IEC 287 method (assume
that the cable is not dried or impregnated).
The dc resistance at 20°C of a 300 mm? conductor is read from Table 7.3 and the de resistance at
90°C is computed from equation (7.2): Roo = 6.01-10-> Q/mand R’ = 6.01-10°5(1+0.00393-70) =
7.663 -10~> Q/m. With the factor k, = 1, the value of x, is obtained from equation (7.8):

ee 8x f
eae ; ea ee 8750-1077 - 1 1981
a are ahs 7,663 - 10-5
126 Part II m Evaluation of Parameters

TABLE 7.3. Maximum Resistance of Stranded Conductors at 20°C

Copper Conductor Aluminum Conductor

Plain, Metal-Coated,
Nominal Cross-
Plain Wires Metal-Coated Wires or Metal-Clad Wires
Sectional Area
Q/m- 10% Q/m:- 10-7 Q/m- 10-7
(mm’)

“TEC 228 (1978) includes similar tables for solid and flexible conductors.

The skin effect factor is obtained from equation (7.9) as

xe
Ss
West?
Ys Se 0.0138
192 0:8x% LO2-FOs e231

The proximity effect factor is obtained from equation (7.21). Since a = 0.0138 (a = y, in this case)
and y = 20.5/71.6 = 0.286, we have

1.18 1.18
Yp= ay” (0.3129° +
a+0.27 )=0.0138
-0.2867
(0312
102867 prevekscr eg
0.0138 + 0.27
)=0.004
Thus, the ac resistance is equal to

R = R'(1 + ys + yp) = 7.663 - 10-5(1 + 0.0138 + 0.0047) = 7.81 - 1075 Q/m

=eEssSesSSMSSeSeSSSSSSSSSSSSSSFssese
Chapter 7 » Joule Losses in the Conductor 127

TABLE 7.4 Maximum Resistance of Stranded Conductors at 20°C

Copper Conductor Aluminum Conductor

Conductor Size Q /m- 107 Q/m- 10%
Hollow Hollow
Core No Core Any Type Core No Core Any Type

162.8 85
211.6 107.2
250 127
300 ilsy
350 177
400 203
450 228
500 Ue)
550 279
600 304
650 329
700 355
750 380
800 405
900 456
1000 507
1250 633
1500 760
1750 887
2000 1013
2250 1140
2500 1267
2750 1393
3000 1520
3500 1773
4000 2027

EXAMPLE 7.4

Compute the ac resistance at 90°C of a six-segment, copper Milliken conductor with a cross section
of 2500 mm’, external diameter of 69.1 mm, and hollow core diameter of 25 mm. Three single-
conductor cables are in flat formation. Assume that the two neighboring cables are spaced 27 cm
apart. The dc resistance at 20°C is equal to Ry) = 6.896 - 10-° Q/m.
The decresistance at 90°C is given by

R’ = 6.896 - 10~°[1 + 0.00393(90 — 20)] = 8.794 - 10° Q/m

We will compute the ac resistance using two approaches. In one case, we will apply the formulas
recommended by the IEC 287 (Approach 1). In the second case, we will use more precise formulas
presented in this chapter (Approach 2).

Approach 1. The constant k, is obtained from equation (7.12). The equivalent diameter d/
is obtained by solving the following equation:

252. d?.
500 2 =;
128 Part II m» Evaluation of Parameters

which yields d’ = 71.7 mm. Thus,

oF re /
eee _61.7ss
— oe
25 .=)
een di+d,) 61.7—25\61.7+252Seti
The skin effect factor is computed from equations (7.8) and (7.9):

“at10-7k,
=xk,= Ts aa-./0.7026
=3.78
-0.8382
=3.168
,=— 405 4 = 0.370
= 192+0.8x4 192 + 0.8- 3.1684

The proximity effect is computed from equation (7.21) with k, = 0.8 obtained from Table 7.2.
When the proximity factor is computed from equation (7.21), we first need the values of x,, y, and
a. These are equal to

= xifiy = 3.78 08=3.381, ¢ = ee x4 eee 3.3814 = 007

maps ae “* 192+0.8x4 192 +0.8- 3.3814

d. 69.1
oee859
Rareto76

The proximity factor is now obtained as

1.18
sade ( ep 1.18
= ay*(0.312y? )=0.4407
-0.25597
;(0.312
0.2559
;+0.4407
Se
ee +0.27
)=0.048
The ac resistance
a+0.27
in this case is obtained from equation (7.5) and is equal to

R = 8.794 - 10-°(1 + 0.370 + 0.0485) = 1.247 - 10-°Q/m

Approach 2. The constant k, is obtained from equation (7.16):

k, = —0.4 + 0.145 In($) = —0.4 + 0.145 In(2500) = 0.7345

The skin effect factor y, is computed from equations (7.13) and (7.14):

pm elonpony means:
a. i te
One,
Byeeddies ee ee 81250 -1077 ee oe gate
p 1.7241-10-8({1 +0.00393(90 — 20)}
= 0.25m7k,(d. —d;)? = 0.25 - 17 958 - 0.7345(0.0691 —0.025)? - 107° = 6.413

Since 5 < z < 30, the values of a(z) and b(z) are computed from equation (7.15). Thus, we
have

a(z) = 0.19701 —0.156295z + 0.073796z? —9.02854 - 107323 + 6.27032 - 10-424

— 2.69028 - 10-525 + 7.0647 - 10-7z® — 1.04301 - 10-82? + 6.62315 - 107!!z8
b(z) = 0.5356 —0.21030734z + 6.495563 - 10-7z? — 1.089373 - 10-2z3 + 1.03728739 - 10-3z4
—5.8238557 - 10-°z° + 1.91099645 - 10-62° —3.38936767 - 10-82? + 2.509622 - 107!%z8
Chapter 7 m Joule Losses in the Conductor 129

Substituting the value of z, we obtain

a(z) = 0.6619, b(z) = 0.2264

ys=a(z)
I =in ak p.2)| = 0.6619(1 —0.6382/2 —0.6382? -0.2264) = 0.3896

We will compute the proximity factor using equation (7.24). From equation (7.19), we have

pat 8eef LO, (ES 10-7-0.8= 3.381

; R’ i 8.794- 10-6
Since 2.8 < x, < 3.8, from equation (7.26), we have

G(xp) = —1.04+0.72x, —0.08.x- = —1.04 + 0.72 - 3.381 —0.08 - 3.3817 = 0.4798

A (x,) = 0.095 + 0.119x, + 0.03847, = 0.095 + 0.119 - 3.381 + 0.0384 - 3.3817 = 0.9363

The value of y is simply y = d,./s = 69.1/270 = 0.2559. Thus, from equation (7.24), we have

2.5y°G(xp) 2.5-0.2559?
-0.4798
i eee Aeneagma
aae oh,le eS
DEESOH(xp 2 012559"70.9363
The ac resistance in this case is equal to

R = 8.794 - 10-°(1 + 0.3896 + 0.0398) = 1.257 - 10-5 Q/m

Thus, in this example, both approaches give similar results.

7.3 THE EFFECT OF HARMONICS

Because the resistance ratios, and hence the system losses, increase dramatically at higher
frequencies, consideration of the effect of harmonics on cable losses is clearly justified.
Three recent publications address this subject (Hiranandani, 1992; Meliopoulos and Martin,
1992; Palmer et al., 1993).
Since the harmonic currents may appear not only in the conductor, but also in the
screens and pipes, we will write the total joule losses in the cable at a given frequency as

W, =I°R,
where R, is the apparent ac resistance of a conductor per unit length, taking into account
both the skin and proximity effects and losses in metal screens, armor, and pipe. The value
of R; can be obtained from

Ry= PR(Lbs yp) at A2) (7.29)

where y, and y, are the skin and proximity effect factors and 4, and 2 are the loss factors
defined in Chapter 8.
Due to the orthogonality of sinusoids having different frequencies, the effect of har-
monics on losses may be calculated at each frequency and summed, that is,

We) spike) >) Er = (7.30)

n= n=! n
130 Part II » Evaluation of Parameters

where n is the index of the harmonic. If , is taken as the ratio of the nth harmonic current
to the fundamental, equation (7.30) becomes

[o.@)
W/=RTYR;
2(Z) (7.3
so that the effective ac/dc resistance ratio, with the ac resistance including the effect of

harmonics,canbegivenas
(=)
R =\\y= (z=)
R (7.32)
R dist SS R n
This effective resistance ratio is a function of the magnitude of each harmonic flow in
the conductor of the cable. Thus, the losses are dependent not only on the total harmonic
distortion (THD), but also on the magnitude of each harmonic current. Because of this
dependence, IEEE Standard 519 (1993) recommends limitations on both the total harmonic
distortion and the distortion caused by any single harmonic. The total harmonic distortion
is calculated as

yiieaw!
Fr
Ce (7.33)
1
and the limits dictated by IEEE Standard 519 are shown in Tables 7.5—7.7.
Since harmonic currents produce additional joule losses in the cable, the ampacity of
a cable will be lower than in the case when only the fundamental frequency is considered.
The derating factor is obtained from

([deratea)”
#. R ")($=)
R; = (rated) o R’ (z)R; (7.34)
R dist R fund

where

Iderateda d 7 5) Trated

and the distorted resistance ratio is calculated by equation (7.32). This cable derating factor
6 assumes that when currents are measured on a system containing harmonics, only the
fundamental is measured, that is, the derating is applied against the fundamental current.
The current derating factor 5 further assumes that an acceptable current rating has been
determined for a system free of harmonics, and a harmonics profile was later detected on
that system. From equation (7.34), the factor is calculated as

(7233)
Chapter 7 m Joule Losses in the Conductor 131

TABLE7.5 CurrentDistortionLimitsforGeneralDistributionSystems
(120-69000V)
a a ee es eee
MaximumHarmonicCurrentDistortion
in Percentof J;
IndividualHarmonicOrder(OddHarmonics)
I,/T, n<ll Mil<n<al7 17<n<23 23<n<35 35<n THD

<20* 4.0 2.0 is) 0.6 0.3 5.0

20<50 7.0 3.5 25 1.0 0.5 8.0
50<100 10.0 4.5 4.0 Ss 0.7 12.0
100<1000 12.0 25) 5.0 2.0 1.0 15.0
>1000 15.0 7.0 6.0 DSS) 1.4 20.0

Even Harmonics are limited to 25% of the odd harmonic limits above.

Current distortions that result in a dc offset, e.g., half-wave converters, are not allowed.

“All power generation equipment is limited to these values of current distortion, regardless
of actual /,.//,.
/,.= maximum short-circuit current at point of common coupling.
J; =maximum demand current (fundamental frequency component) at point of common
coupling.

TABLE 7.6 Current Distortion Limits for General Subtransmission Systems

(60 001 V—161 000 V)

Maximum Harmonic Current Distortion

in Percent of J;
Individual Harmonic Order (Odd Harmonics)

Le, n<ll el <r ai Ke iaKe3} 23S <35 35<n THD

<20* 2.0 1.0 0.75 0.3 0.15 DPS)

20<50 3.5 eS) 1.25 0.5 0.25 4.0
50<100 5.0 225 2.0 0.75 0.35 6.0
100<1000 6.0 QS Dis 1.0 0.5 Ts
>1000 7.5 She) 3.0 LOS 0.7 10.0

Even Harmonics are limited to 25% of the odd harmonic limits above.

Current distortions that result in a de offset, e.g., half-wave converters, are not allowed.

“All power generation equipment is limited to these values of current distortion, regardless
of actual /,./1;.
/,.= maximum short-circuit current at point of common coupling.
I, =maximum demand current (fundamental frequency component) at point of common
coupling.

EXAMPLE 7.5
Consider cable model No. | with laying conditions specified in Appendix A. Let us assume that the
system load has a third harmonic which constitutes 20% of the fundamental current. We will compute
the derating factor for this cable system.
We start by computing the conductor resistance and loss factors for the third harmonic. From

equation (7.7), we have

132 Part IT m Evaluation of Parameters

TABLE 7.7 Current Distortion Limits for General Transmission Systems

(> 161 kV), Dispersed Generation and Cogeneration
CEU UU
UE
EEE EEE
Individual Harmonic Order (Odd Harmonics)
rn Ee i ee
Leelin n<ll cere Nes 17<n<23 23<n<35 35<n THD
a er a a EE ll
<50* 2.0 1.0 0.75 0.3 0.15 2
>50 3.0 ES 1.15 0.45 0.22 cee)
Even Harmonics are limited to 25% of the odd harmonic limits above.

Current distortions that result in a de offset, e.g., half-wave converters, are not allowed.

/,.= maximum short-circuit current at point of common coupling.

],, =maximum demand current (fundamental frequency component) at point of common

coupling.
ae2 8 (imme 8x-1 cnr gp satSr aL
-150 i em
no R’ 0.0601
- 10-3(1
+ 0.00393
-70) 0.07663-10-3
Since (x,)3; < 2.8, the skin effect factor is obtained from equation (7.9) as

4
Se 4.922 0.1
(¥s)3 = 192+0.8x4 192 +0.8 - 4.92?

To compute the proximity effect factor, we observe that, since the cable is assumed to be neither
dried nor impregnated, by equation (7.19), (x;,), = (x7), = 4.92. Thus, from equation (7.25), we
obtain
Wes 11 - 4.92?
G(xp) = SS= = SSS = 0.224
704 + 20x°, 704 + 20 - 4.922
1 1+0.0283x4 1 140.0283 -4.92?
H(x,) = =: apps eae ket a lH)
3 140.0042x4 3 140.0042 -4.922
The proximity effect factor is obtained from equation (7.23):

atde 205“Dene.
camelome:2 ema
2.5y*G(xp) 2.5-0.286?
-0.224
~2- SH) 2-2 -0288-0510

Hence, the ac resistance of the conductor for the third harmonic is equal to

(R)3 = 0.07663 - 10 7(1 + 0.011 + 0.023) = 0.0792 - 107382/m

When the conductor reaches an operating temperature of 90°C, the temperature of the concentric
neutral wires computed using internal thermal resistances is equal to 83°C. Thus, the resistance of
the screen is equal to

R, = 0.759 - 10-°(1 + 0.00393 - 63) = 0.947 - 103 Q2/m

The circulating current loss factors are computed in Example 8.2 and are equal to

(Ai), = 1.48, (Ain),= 1.82, (Ai), = 0.195

Chapter 7 » Joule Losses in the Conductor 133

The ac resistance of all three cables for the third harmonic is thus

(R7)3 = 0.0792 - 10~° - (3 + 1.48 + 1.82 + 0.195) = 0.514. 1073 Q2/m

The de resistance was computed in Example 7.3 as 0.0766 - 10-3 Q/m. The skin and proximity
factors were also computed in the same example, and are equal to 0.0139 and 0.0047, respectively.
The concentric wires loss factor for this cable is equal to 0.09 (see Section 8.3.5.3). Hence, the
effective distorted ac/dc resistance ratio, computed from equation (7.32), is

=)
(;sal dist
=D ae (3)
d, Yale,
RY; ;
| =16-C.+0.0139 af +.0.09)
+0.0047)(1 0.514
- 10-3
+0.2?.3-0.0766
-10-3= 1.2
——__—__.

Finally, the ampacity reduction factor is obtained from equation (7.35) as*

» men A1 + 0.0139+ 0.0047)(1+

a Oe 0.09) _ 39%

EXAMPLE 7.6>
Consider a 2000 kemil (1010 mm’) pipe-type cable with the following parameters:

where) Sd —1k59in
R’ = 5.24 wQ/ft
1 ==UNCOyeysii
(DY3 22 hh
gS 250 im
ky S O35

k, = 0.33 for cradled configuration and k, = 0.15 for triangular configuration.

The conductors have enameled strands, and the inside diameter of the pipe is 8.125 in. We will
consider two cable arrangements inside the pipe: cradle and triangular. The only difference is the
value of the proximity effect coefficient as shown above.
Sample harmonic scenarios are taken from IEEE Standard 519 (1993) and are shown in Table
7.8. The first scenario, designated “A,” in the table, is the profile given for a 12-pulse converter as
described in Table 13.1 of the Standard. The second scenario takes the values for the first scenario
and attenuates all harmonics so that the magnitude of THD limitations of the Standard are met. The
third and fourth scenarios are taken from Table 13.7 of the IEEE Standard representing unfiltered and
most filtered harmonic distribution. Case “E” is the profile used by Meliopoulos and Martin (1992)
as an example applied to a distribution system. For all of these cases, the triple harmonics above the
third and all even harmonics are negligible, as is common in power transmission systems. Therefore,
only frequencies with nonzero harmonics are shown.
These harmonic scenarios were applied to the cable described above. The cable derating factors
were computed for each case taking into account the effect of harmonics on the conductor resistance
and losses in screens and pipe. The values of the derating factor for the above cable and harmonic

4 We assume that the screen loss factor for the fundamental frequency is the same for all three cables.

5 This example is extracted from Palmer er al. (1993).

134 Part IJ # Evaluation of Parameters

TABLE 7.8 Harmonics Profile (Percent)

n A B Cc D E
yee ee eee ee ee eee
Fundamental 100 100 100 100 100
3 0.00 0.00 0.00 0.00 22.85
5 19.20 5.00 9.59 0.00 3.43
v 13.20 5.00 6.60 2:33 3.43
11 7.30 3.50 3.66 1.70 0.00

13 5.70 3.50 2.85 37, 0.00

17 3.50 3.00 Ps 0.87 0.00
19 2.70 2.70 1.35 0.68 0.00

2 2.00 2S) 1.0 0.51 0.00

25 1.60 125 0.80 0.41 0.00
29 1.40 1.25 0.70 0.36 0.00

Bil 1.20 1.20 0.60 0.31 0.00

35 1.10 0.70 0.55 0.28 0.00
37 1.00 0.70 0.00 0.00 0.00
41 0.90 0.70 0.00 0.00 0.00
43 0.80 0.70 0.00 0.00 0.00
47 0.80 0.70 0.00 0.00 0.00
49 0.70 0.70 0.00 0.00 0.00

THD 25.7% 10% 12.8% 3.49% 23.3%

scenarios are shown in Table 7.9. These results indicate that harmonics may create a very significant
increase in losses. Scenario “A,” which was a 12-pulse converter with no filtering, required a derating
of as much as 16.25%. Comparison with scenario “B” indicates the very strong influence of even
minimal filtering. The benefit of complying with the IEEE Standard are also clearly seen as the
derating decreases from 16.25 to 5.29% when the limits are impressed on the harmonic scenario
created by the 12-pulse converter.

TABLE 7.9 R,/R’ and Derating Factor for Pipe-Type Cable and Harmonics

A B ( D 5

Cradle R/R’ 2.251 1.760 1.738 1.598 e792

5 (%) 16.25 a) 4.71 0.62 6.13

Triangular R/R’ 1.811 1.466 1.455 1.356 1.500

5 (%) 13.90 4.29 1.46 0.50 5.37

A comparison of scenarios “B” and “C” clearly indicates that losses are not dependent on THD
alone. While scenario “B” had a THD of 10%, the losses were greater than for scenario “C” which
had a THD of 12.8%. The reason for this is the existence of higher harmonics in case “B.” Hence,
the losses are dependent upon both the individual harmonic level and the total harmonic distortion.
Additionally, it is clear that the loss increases are dependent on the cable configuration inside the pipe,
resulting in deratings that vary by as much as 3% between the triangular and cradled configurations.
In observing scenario “D,” we can see that for systems with a high level of filtering, the effect
of harmonics on cable rating is minimal.
Scenario “E” requires a derating of slightly over 6%. This can be compared with the derating
of 11% calculated by Meliopoulos and Martin (1992) for a given distribution system.
pr a ead
Chapter 7 # Joule Losses in the Conductor 135

REFERENCES

AEIC CS4-93 (1993), “Specifications for impregnated-paper-insulated low and medium

pressure self contained liquid filled cable,” Association of Edison Illuminating Compa-
nies, New York.
Arnold, A. H. M. (1936), “The alternating current resistance of tubular conductors,” J. JEE,
no. 78, pp. 580-593.
Arnold, A. H. M. (1941), “Proximity effect in solid and hollow round conductors,” J. JEE,
no. 88 II, pp. 349-359.
Arnold, A. H. M. (1942), “Eddy current losses in single-conductor, paper-insulated, lead-
covered unarmoured cables of a single phase system,” J. JEE, no. 89 II, pp. 639-647.
Arnold, A. H. M. (1946), The Alternating Current Resistance of Non-Magnetic Conductors.
London: (DSIR) HMSO.
Ball, E. H., and Maschio, G. (1968), “The A.C. resistance of segmental conductors as used
in power cables,” JEEE Trans. Power App. Syst., vol. PAS-87, no. 4, pp. 1143-1148.
Dwight, H. B. (1922), “Skin effect and proximity effect in tubular conductors,” AJEE Trans.,
no. 41, pp. 189-198.
Goldenberg, H. (1961), “Some approximations to Arnold’s formulae for skin and proximity
effect factors for circular and shaped conductors,” Brit. Elec. Res. Assoc. Report F/T203.
IEC 228 (1978), “Conductors of insulated cables,’ IEC Publication 228.
IEC 287 (1982), “Calculation of the continuous current rating of cables (100%) load factor,”
IEC Publication 287.
IEC WG10 (1991), “Skin and proximity effects,’ Document of WG 10 of IEC SC20A—
High Voltage Cables.
IEEE Standard 519 (1993), “IEEE recommended practices and requirements for harmonic
control in electric power systems,” IEEE Ind. Appl. Soc./Power Eng. Soc., Apr. 1993.
Lewis, W. A., and Tuttle, P. D. (1959), “The resistance and reactance of ACSR,” AIEE
Trans., vol. 77 Ill, pp. 1189-1217.
Meliopoulos, A. P. S., and Martin, M. A., Jr. (Apr. 1992), “Calculation of secondary cable
losses and ampacity in the presence of harmonics,” JEEE Trans. Power Delivery, vol. 7,
no. 2, pp. 451-457.
Palmer, J. A., Degeneff, R. C., McKernan, T. M., and Halleran, T. M. (Oct. 1993), “Pipe-
type cable ampacities in the presence of harmonics,” JEEE Trans. Power Delivery, vol.
8, no. 4, pp. 1689-1695.
Silver, D. A., and Seman, G. W. (1982), “Investigation of A.C./D.C. resistance ratios of
various designs of pipe-type cable systems,” JEEE Trans. Power App. Syst., vol. PAS-101,
no. 9, pp. 3481-3497.
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 8
Joule Losses in Screens,
sheaths, Armor, and Pipes

8.1 INTRODUCTION

Dielectric losses and joule losses from conductors have already been considered in Chapters
6 and 7. The third source of losses, considered in this chapter, are those from metallic cable
screens and coverings such as sheaths and armoring. These losses are generated by currents
induced in these components by the current flowing in the conductors.
Sheath losses are current dependent, and can be divided into two categories according to
the type of bonding. These are losses due to circulating currents which flow in the sheaths of
single-core cables if the sheaths are bonded together at two points, and losses due to eddy
currents, which circulate radially (skin effect) and azimuthally (proximity effect). Eddy
current losses occur in both three-core and single-core cables, irrespective of the method of
bonding. Eddy current losses in the sheaths of single-core cables which are solidly bonded
are considerably smaller than circulating current losses, and are ignored except for cables
with large segmental conductors.
Losses in protective armoring also fall into several categories depending on the cable
type, the material of the armor, and installation methods. Armored single-core cables
without a metallic sheath generally have a nonmagnetic armor because the losses in steel-
wire or tape armor would be unacceptably high. For cables with nonmagnetic armor, the
armor loss is calculated as if it were a cable sheath, and the calculation method depends on
whether the armor is single-point bonded or solidly bonded. For cables having a metallic
sheath and nonmagnetic armor, the losses are calculated as for sheath losses, but using the
combined resistance of the sheath and armor in parallel and a mean diameter equal to the
rms value or the armor and sheath diameters. The same procedure applies to two- and
three-core cables having a metallic sheath and nonmagnetic armor. For two- and three-core

137
138 Part I] » Evaluation of Parameters

cables having metallic sheath and magnetic wire armor, eddy current losses in the armor
must be considered. For two- and three-core cables having steel tape armor, both eddy
current losses and hysteresis losses in the tape must be considered together with the effect
of armor on sheath losses.
Submarine cables require special consideration. Single-core ac cables for submarine
power connections differ in many respects from underground, buried directly or in ducts; in
fact, submarine cables are generally armored, can be manufactured in very long lengths, and
are laid with a very large distance between them. For these reasons, calculation methods
described in the IEC 287 (1982) must be supplemented and modified in some points. In this
chapter, we will introduce required tools to compute sheath and armor losses in single-core
cables with metallic sheath and magnetic armor representing common submarine cable
construction. However, we will ignore the presence of the sea in our calculations. This
implies that the theory presented here is applicable to cables laid in air, buried underground,
or in ducts or cables in shallow waters near landing points. The reader interested in calcu-
lations involving consideration of the sea impedance is referred to the paper by Bianchi and
Luoni (1976).

8.2 SHEATH BONDING ARRANGEMENTS

Sheath losses in single-core cables depend on a number of factors, one of which is the
sheath bonding arrangement. In fact, the bonding arrangement is the second most important
parameter in cable ampacity computations after the external thermal resistance of the cable.
For safety reasons, cable sheaths must be earthed, and hence bonded, at least at one point
inarun. There are three basic options for bonding sheaths of single core cables. These are:
single-point bonding, solid bonding, and cross bonding (ANSI/IEEE, 1988).
In a single-point-bonded system, the considerable heating effect of circulating currents
is avoided, but voltages will be induced along the length of the cable. These voltages are
proportional to the conductor current and length of run, and increase as the cable spacing
increases. Particular care must be taken to insulate and provide surge protection at the free
end of the sheath to avoid danger from the induced voltages.
One way of eliminating the induced voltages is to bond the sheath at both ends of the
run (solid bonding). The disadvantage of this is that the circulating currents which then
flow in the sheaths reduce the current-carrying capacity of the cable.
Cross bonding of single-core cable sheaths is a method of avoiding circulating currents
and excessive sheath voltages while permitting increased cable spacing and long run lengths.
The increase in cable spacing increases the thermal independence of each cable, and hence
increases its current-carrying capacity. The cross bonding divides the cable run into three
sections, and cross connects the sheaths in such a manner that the induced voltages cancel.
One disadvantage of this system is that it is very expensive, and therefore is applied mostly
in high-voltage installations.
Figure 8.1 gives a diagrammatic representation of the cross connections.
The cable route is divided into three equal lengths, and the sheath continuity is broken
at each joint. The induced sheath voltages in each section of each phase are equal in
magnitude and 120° out of phase. When the sheaths are cross connected, as shown in Fig.
8.1, each sheath circuit contains one section from each phase such that the total voltage in
each sheath circuit sums to zero. If the sheaths are then bonded and earthed at the end of
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 139

Cable conductors Cable sheaths Cable sheaths

Total voltage in each sheath circuit = e + ae + a2e = 0 =

(where a indicates phase rotation through 120°)

(a)

Sheath sectionalizing insulators

Joints Transposition of cables

& Sheaths solidly bonded and earthed +

(b)

Figure 8.1 Diagrammatic representation of a cross-bonded cable system. (a) Cables are
not transposed. (b) Cables are transposed.

the run, the net voltage in the loop and the circulating currents will be zero, and the only
sheath losses will be these caused by eddy currents.
This method of bonding allows the cables to be spaced to take advantage of improved
heat dissipation without incurring the penalty of increased circulating current losses. In
practice, the lengths and cable spacings in each section may not be identical, and therefore
some circulating currents will be present. The length of each section and cable spacings
are limited by the voltages which exist between the sheaths and between the sheaths and
earth at each cross-bonding position. For long runs, the route is divided into a number of
lengths, each of which is divided into three sections. Cross bonding as described above can
be applied to each length independently.
The cross-bonding scheme described above assumes that the cables are arranged sym-
metrically, that is, in trefoil. It is usual that single-core cables are laid in a flat configuration.
In this case, it is acommon practice in long-cable circuits or heavily loaded cable lines to
140 Part IJ m»Evaluation of Parameters

transpose the cables as shown in Fig. 8.1b so that each cable occupies each position for a
third of the run.
A number of practical points must be considered before adopting cross bonding, the
most important of which are the high voltages which can occur on the sheaths and across
sheath insulating joints during switching surges or other transients. Experimental work
by Gosland (1940) has shown that voltages as high as the full service voltage can appear
across insulating glands under transient conditions, even when there are only a few meters
of cable in the circuit. These voltages cannot be avoided, but use of suitable surge diverters
will prevent damage to the cable system. Other practical points relate to the voltages on
the sheath under normal service or fault conditions and the need to ensure that the sheath is
effectively insulated from earth for the life of the system.

8.3 CIRCULATING CURRENT LOSSES IN SHEATH AND ARMOR

The basic equations for calculating circulating current losses were developed by a number
of authors in the 1920s: Morgan et al. (1927), Arnold (1929), and Carter (1927, 1928). In
some cases, the effects of eddy currents were included in the equations developed; others
concluded that the effects of eddy currents were insignificant compared with circulating
current losses, and hence could be ignored. The equations presented in IEC Publication
287 (1982) and in the Neher—McGrath paper (1957) are taken from the work by Arnold
(1929), and ignore eddy current losses except for the case of cables with large segmental
conductors.
All of the equations for sheath losses given in this section assume that the phase
currents are balanced. The equations also require a knowledge of the temperature of the
sheath, which cannot be calculated until the cable rating is known, and therefore an iterative
process is required. For the first calculation, the sheath temperature must be estimated; this
estimate can be checked later after the current rating has been calculated. If necessary, the
sheath losses, and hence the current rating, must be recalculated with the revised sheath
temperature.
As discussed above, the power loss in the sheath or screen (A; ) consists of losses caused
by circulating currents (4) and eddy currents (/). Thus,

fySayeeeayoh (8.1)
The loss factor in armor is also composed of two components: that due to circulating
currents (4) and, for magnetic armor, that caused by hysteresis (43). Thus,

Ag=a t+ay (8.2)

As mentioned above, for single-core cables with sheaths bonded at both ends of an
electrical section, only losses caused by circulating currents are considered. An electrical
section is defined as a portion of the route between points at which the sheaths or screens of
all cables are solidly bonded. Circulating current losses are much greater than eddy current
losses, and they completely dominate the calculations. Of course, there are no circulating
currents when the sheaths are isolated or bonded at one point only. The formulas for the
circulating current losses in sheath and armor are developed below.
Consider a three-phase cable circuit. The complex currents flowing in the conductor,
sheath, and armor are denoted by /,, /;, and J,, respectively. Then, the sheath and armor
Chapter 8 m=Joule Losses in Screens, Sheaths, Armor, and Pipes 141

loss factors due to circulating currents are defined as

poll TaleRs
Wiley: (8.3)

/ WaleRe
Ky = LER (8.4)

where R ({2/m) denotes the ac resistance of the conductor at operating temperature and the
subscripts s and a represent sheath and armor, respectively.
As can be seen from the above equations, in order to compute the loss factors, the
sheath and armor currents have to be expressed as functions of the conductor current. In
order to compute /; and J,, we observe that the voltages along the conductor, sheath, and
armor are related to the currents by

V. Lee Zics Zea I,

Vs — Lise Ze. Zsa Is (8.5)
Ve Lise Zas Zaa Iq

where Z;; is an impedance between element i and j. Calculation of these impedances is

discussed in the following subsections. The approach to deriving inductances will be to
compute “flux linkages,” except for the inductances resulting from magnetic fluxes within
the thickness of the armor. In that case, an energy approach turns out to be more suitable.
We will use Fig. 8.2 to illustrate the flux linkage concept. This figure shows three
cables in trefoil configuration, where each cable is composed of a conductor, sheath, and
magnetic armor. Also illustrated is magnetic flux due to current in the bottom conductor.
Consider, for example, the conductor—conductor impedance Z,- which will be evaluated
for the bottom conductor in Fig. 8.2. This impedance consists of the conductor resistance,
the conductor’s self-reactance (jw times the self-inductance), and the conductor’s mutual
reactance with the other two conductors.
The conductor’s self-inductance is the linkage of the lower conductor’s flux with its
own current. The mutual inductance is the linkage of the lower conductor’s flux with the
current in the other conductors. (Flux linkage, generally, is the spatial integral of flux due
to a unit current times the proportion of current that it links.)

Figure 8.2 Three cables in trefoil configuration.

142 Part IJ » Evaluation of Parameters

The evaluation of conductor—conductor inductance, including both the self- and mutual
inductances, consists of integrating all of the conductor flux linkage from the center of the
conductor to the dashed line passing through the centers of the other two conductors in Fig.
8.2. The reason that this integration ends halfway through the upper two conductors is that
flux entering the upper two conductors near their bottom edges, as illustrated by the inner
dashed line, links only a small proportion of the current in those conductors. On the other
hand, flux at the upper edge, as illustrated by the outer dashed line, links nearly all of the
current. The total flux linkage from the lower surface of the two conductors to the upper
surface (linking, on average half of the current) is approximately the same as the integral
of flux to the halfway point, linking all of the conductor current.
The inductances due to flux within the conductor and within the thicknesses of the
sheath and armor walls are derived in the following subsections. We will begin by evaluating
all internal cable inductances.

8.3.1 Internal Inductances!

8.3.1.1 Hollow Conductor Internal Inductance. The self-inductance of a hollow

conductor is computed in the following manner. For unit total current, unit outer radius,
and an inner radius of a (see Fig. 8.3), the fraction of conductor current within radius r is
given by

po)
le, = —— (8.6)
l =a“
and the flux within dr is given by

jig! ap
doop = ——d
er ry (8.7)

The internal inductance of a hollow conductor is thus given by

vece—int= iof 1Z,dr

ito[Po eee)
On Jan car Qn (1 —a2)2 (88)

where a is the ratio of inner to outer radius.

For a solid conductor (a = 0), this reduces to (49/277) -(1/4). Forathin shell (a > 1),
it converges to (9/27) - (1/3)(1 —a) = (0/27) - CLySyGs/ Te)

Figure 8.3 Hollow-core conductor with unit

outer radius.

1 In the development ‘
of inductances, : :
all linear dimensions should be expressed in meters. However, since
the final results contain only the ratios of these dimensions, the formulas contain the symbols for the thicknesses
the spacing, and the armor length of lay without an asterisk, so the dimensions in millimeters can be substituted
in the final expressions.
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 143

where —_r,
= mean diameter of the sheath, mm

ts mean thickness of the sheath, mm.

8.3.1.2 Conductor—Sheath Internal Inductance. For mean sheath radius r; and

thickness f,, the conductor flux density for unit conductor current is approximately given
by

B. © uo/ (2r?) (8.9)

The proportion of sheath current at radius less than r is given approximately by

fey Cpr be)2) be (8.10)

The internal inductance is then given by the flux linkage

Leen Lo rs+t,/2 es
f=—Ts, ty /2 [pS Mo Sgfs Lo ,[eae+ts/2
rs 8.11
Se oar i t Dar). In a ere Boe

8.3.1.3 Sheath—-Sheath Internal Inductance. For mean sheath radius r, and thick-
ness f,, and unit total current in the sheath, the fraction of sheath current within a distance
x of the inner radius is equal to /-, = x/t,. The flux d@ within dx encircling this current
is given by

db ae Tex (8.12)
Pang
ie

The radius has been approximated by ,, at all values of x. The inductance is given by
the integral of the flux times the proportion of current that it surrounds, so that the internal
self-inductance? is approximately given by

Iyeaa: = Ho if : Pedy ==Ho Mey

1 stse HoIn Ts+ts/2
5 t,/2 (8.13)
QTY Go LI gOarmedit rs+t,/6
8.3.1.4 Internal Armor Inductances. Magnetic flux within the thickness of the
armor wall affects all of the inductances. The internal inductances resulting from the flux
in the armor tapes or wire are fairly complicated. The armor may be treated as a coil
surrounded by circular flux as a result of conductor and sheath currents and the armor’s
own current. In addition, the armor surrounds solenoidal flux as a result of its own current.
This solenoidal flux was not considered by Bosone (1931), and is therefore not included
in the IEC 287 (1994) Standard. The magnetic fields within the thickness of the armor are
a superposition of the axial and circular fields. The circular fields from the conductor and
sheath are nearly uniform between the inner and outer surfaces of thin armor and are treated
as uniform. The fields from the armor’s own current, however, are nonuniform. The axial
field is maximum at the inner surface of the armor and decreases to zero at the outer surface

2 In solving the field equations, the voltage measured at the outer surface of the sheath is equal to Jsurface-
To second order, this is the sum of the resistive voltage due to the total sheath current (or average current density)
and the voltage from internal inductance. The latter therefore depends on the difference between the surface
current density and average current density. The resulting self-inductance, to second order, however, is the same
as that obtained with a uniform-density flux-linkage calculation.
144 Part II =» Evaluation of Parameters

because the axial field at a given radius is caused by current at larger radii. Conversely,
the circular field is maximum at the outer surface of the armor and decreases to zero at
the inner surface because the circular field at a given radius is caused by current at smaller
radii. The situation is complicated by the fact that the magnetic permeability of the wires is
different, parallel and perpendicular to the wires. The circular and axial fields in the armor
wall must therefore be resolved into directions parallel and perpendicular to the wires (see
Fig. 8.4). In deriving inductance formulas, the helical geometry of the currents and fluxes
leads to flux linkages that are sometimes difficult to visualize. Therefore, to avoid errors,
the derivations presented here are based on the integral of complex power over the volume
of the armor material. These integrals are identical to those encountered in flux-linkage
calculations. The total electromagnetic power is given by Poynting’s Theorem (Jackson,
1962):

p+jo-=/ff (E+BHAV] (8.14

1 RSPR
Tad USE LOUPatt: WA

// 3)
/
hed
ey!
rt a)
Hindfu}

(
fig,lay8 at
Ai=Aire+Haxit
Poa nee) 1=Aoirea
+ ax
circ
ere

etre
Figure 8.4 Resolution of magnetic field in parallel and perpendicular components.
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 145

where J =current density, A/m?

E = electric field, V/m
B = magnetic flux density, Tesla
H = magnetic field, A-turns/m

and the integral is over all space. The J - E term results in the J*R power loss. The second
term, integrated only over the volume of the armor material, will be used to obtain internal
armor inductances, as well as the contribution that internal armor flux makes to all of the
other cable inductances.

The complex power (real is dissipated; imaginary is stored) is also given by

P+jQ=) (h-1) Zi (8.15)

iJ
where Z;; = mutual (or self fori = j) impedances

felipe) vied) imnU)smGi)= il; el 1) f2 (8.16)

Equating these two expressions allows the identification of jwLj;;(part of Z;;) in the
power obtained from the first expression.
All currents, voltages, and fields are rms values. The subscripts may represent the
conductor, sheath, or armor. The losses are due to currents in these elements, but take place
within the thickness of the armor. Because the longitudinal permeability of the wires is
complex and is included in all of the impedances, there is loss corresponding to all of the
impedances, not just the diagonal ones.
To simplify the computations, it will be assumed that the armor is composed of tapes
rather than wires. The error in the resulting formulas when used for round wires is negligible.
The mean radius and the thickness of the armor are denoted by r, and fz, respectively. The
length of lay of armor wires or tapes is denoted by @,. The fields within the tape (or
wire) are obtained by resolving the solenoidal field (1 — x)Jq/€q and the circular field
(I. +1, +x1I,)/(2rq) into components parallel and perpendicular to the tape. The parallel
and perpendicular components of the magnetic field at position x within the armor wall
(assuming uniform current density in the armor) are given by

Fee ree, $21) Biel tye ie 17)

and
sin(B) cos(f)
dial<=—(1a oar ar le orAfotxJ) nrg
1
= [(- + Jy)cos(8) + Ia(x —sin?(8))|_———- (8.18)
2mrq cos(B)
146 Part II » Evaluation of Parameters

where I. = rms conductor current, A

7, = rms sheath current, A
I, = rms armor current, A
x = distance from inner surface/armor thickness
(x = 0 at inner surface; x = | at outer surface)
B = helical lay angle with respect to the cable axis
€, = helical lay length of the armor, mm
rq = mean radius of the armor, mm.

The identity 277r,/€, = sin(B)/cos(B) was used above and below. The first term of
H, has —(1 —x) because that term is from the axial field of the armor, and its perpendicular
component in the tape is in the opposite direction to the perpendicular components of all
the circular fields.
Considering now the second term on the right-hand-side of equation (8.14), the com-
plex parallel power is given by

(P + jQ)) = jorote HyAy perunitvolume

m af |
= aaa + I, + I,|? cos?(B) perunitvolume
a
J@OLoMeAa 2 2 :
= re, +1; + Iq|\°cos“(B) perunittapelength
a
J@OLoMeAa 2 §
= cagapiaanese +I1,+I,|" cos(B) —perunitcablelength
a
= JOMLoMeAa on F
= 0 aim +1,+1,|°sin(B) perunitcablelength (8.19)
aa
where = 4g = complex relative longitudinal magnetic permeability
Aq = sum of the wire or tape cross-sectional areas (mm?)
Vae-Pulabodale = ldol tal del lola clas lee ole Ll, el, «ee

From this expression and equations (8.15) and (8.16), it can be seen that
(HoleAq/2mralq) sin(B) contributes equally to all of the cable inductances L.., Les, Lea,
Les Lees Disa: Dae: Las; and Daan
The complex perpendicular power is given by
(P+ jQ)1 = jopou,H, At per unit volume

jopons | 1[HL Hf] (2mrata)dx per unitcable

length
= ——
0
JOMoLr tq a) 2
Tbr, —| |/, + 1,|"
| cos“(B)
7)
(B) + |Ta|
le (se
1
in?
3 c0s%(B) sin*(B)

+ (UW.
+15) +Ig (5l —sin?(6))
3 per unit cable length (8.20)

where 4, = complex relative transverse magnetic permeability.

Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 147

For round wires, 1; in the IEC 287 Standard has been adjusted so that the transverse
inductance for a cylinder may still be used. Although the formula above permits ju, to be
complex, the standard treats it as real. From this point on, we will also treat it as real since
it is small compared with the longitudinal permeability, and therefore has little associated
hysteresis loss. For touching steel wire armor, 4; = 10; for nontouching steel wires,
Mr = 1.
From the expression for perpendicular power and equations (8.15) and (8.16), it can
be seen that

[(Holr)/(27)] (ta/Ta)cos?(B)contributes
equallyto Lec,Les,Lsc,andLss
[(HoMr)/(27)] (ta/Ta)[1/ (3cos*(B))—sin*(B)]contributestoLag
[(Moltr)
/(270)](ta/Tra)
[1/2—sin’(B)| contributesequallyto fbr Lepeeaescand)
Ly.)
The total contribution of internal armor flux to each cable inductance is given by the
sum of the parallel and perpendicular power contributions.

8.3.2 Total Inductances

With the internal inductances computed in the preceding section, we can now derive
the expressions for the total inductances of all cable components.

8.3.2.1 Conductor—Conductor Inductance. Integration of the flux linkage from

the center of the conductor to its outer surface yields the internal inductance L.._jint given
by equation (8.8). It may be assumed that the permeability of the sheath is jo, and so
the presence of the sheath makes no difference in the amount of conductor flux within the
thickness of the sheath walls. The presence of the sheath does, however, make a difference
in the total flux because the conductor flux induces currents in the sheath, which are taken
into account by conductor—sheath inductance. If the presence of the armor is temporarily
ignored, the external conductor-conductor inductance is given by the integral of conductor
flux (linking all of the conductor current) from the conductor surface r, to distance s,
yielding

Lext—no
armor
=fefTL
-dr=GeIn(=)Cie) pet 21 Le (8.21)
where 40 = magnetic permeability of free space, 42 x 10-7 H/m
re = outer radius of the conductor, mm
s = axial distance between conductors, mm.

Magnetic armor makes a difference to the conductor—conductor inductance because of

its high magnetic permeability. The inductance from conductor flux within the thickness
of the armor is derived in Section 8.3.1.4. However, before adding it to Lext above, it is
necessary to subtract the contribution that has been included in equation (8.21) for the space
that the armor occupies, which is given by

Tatta /2 1
Larmor
Space
lanso f -dr ®me<=
0 Ho f
(8.22)
Tvra—ta/2
y ae Ya
148 Part I] » Evaluation of Parameters

The total conductor—conductor inductance is obtained by combining equations (8.8),

(8.21), 8.22) and the contribution of the armor as discussed in Section 8.3.1 A:

Lec= Hoed bie.

Ho in( s )
eT d ft beeMofa
sie i Ho¢
fa.
Teich[141
cos?(B)]+ HoleAa
ena Y (B) (8.23)
asin 8.23

1/4- a* +.a*(3/4 —Ina)

f= (1 —a?) ( 8.24 )

and a is the ratio of inner to outer radius of a hollow conductor.

Also,
B = helical lay angle with respect to the cable axis (see Fig. 8.4)

€, = helical lay length of the armor, mm

rq = mean radius of the armor, mm

t, = armor thickness, mm
[te = complex relative longitudinal magnetic permeability; the
imaginary part of it describes the hysteresis loss of the magnetic material
[4; = complex relative transverse magnetic permeability
Aq = sum of the wire or tape cross-sectional areas, mm”.

The first two terms may be combined into one by defining an effective conductor radius
ar, where a = exp(—/). For a solid conductor, a = exp(—1/4) = 0.7788.
The total conductor—conductor inductance then reduces to

Lo n( S ) + ae
Lo ta [Mrcos’(B)
5 ]+ aye ESsin(B)
HobeAa
a (8.25)
Lien—os Ih ee 4 a 1 :

For nonmagnetic armor, the last two terms are ignored. For nontouching magnetic armor
wires, the second term is ignored.

8.3.2.2 Conductor-Sheath Inductance. The derivation of conductor—sheath induc-

tance is similar to that of conductor—conductor inductance. The flux linkage is between
the flux from the bottom conductor in Fig. 8.2 and all three sheath currents. The internal
conductor—sheath inductance L,,—jn; (the first term below) is given by equation (8.11). The
external conductor-sheath flux linkage is obtained by integrating from the outer sheath
radius to distance s and then taking into account the presence of the armor as before. The
total conductor—sheath inductance is thus given by

Lo
20 ( rst+
Les= —In|————t,/2
re ) Ho
af21P€ 5
+t;
—Ip{pe /2 Lot
ikQFrz[ym
S eay
LaSat 2B)— 1]
A eA sin(B)
4 bestia .
adele

ate[A in(=) + po ta a comehyge

) | tatesefhacn) (8.26)
2mYa Qnrala
la
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 149

Where rr, = mean radius of the sheath, mm

t; = sheath thickness, mm.

It can be seen that, within the thickness of the sheath, the conductor flux links ap-
proximately half of the sheath current. The internal and external logarithm terms can thus
be combined, to a good approximation, as the integral of flux from the mean radius of the
sheath to distance s.
For nonmagnetic armor, the last two terms are ignored. For nontouching magnetic
armor wires, the second term is ignored.
8.3.2.3 Sheath-Sheath Inductance. The sheath-sheath inductance is the linkage of
the flux from the bottom sheath in Fig. 8.2 with the currents in all three sheaths. The internal
sheath—sheath inductance L,5—jn; (the second term below) is given by equation (8.13).
The external sheath—sheath flux linkage is obtained by integrating from the outer sheath
radius to distance s and then taking into account the presence of the armor as before.

Lee# in(—*_)ofae (227)

Ssts2 + ~tqLee
cos”
(6) 1]
20 rs+t,/2 r,;+t;/6
HobeA
Lae a nant)
SeQnn(-
Lo eeeaaa
isiee seoseapgle
Hofa ccs(Bl |Nl [LobeA
ee oe ae
aante) ee
8.27
We can observe that, within the thickness of the sheath, the sheath’s own flux links
approximately one third of the sheath current. The internal and external logarithm terms
can thus be combined, to a good approximation, as the integral of flux from r,; + t, /6 of the
sheath to distance s. For manual computations, the t,/6 may be ignored with little loss of
accuracy.
For nonmagnetic armor, the last two terms are ignored. For nontouching magnetic
armor wires, the second term is ignored.
8.3.2.4 Armor Inductances. The conductor—armor, sheath—armor, and armor—armor
internal inductances have been derived in Section 8.3.1.4. The external portion of the armor
inductances is obtained by integrating from the outer surface of the armor rg +1, /2 to distance
s, yielding

Hida
Ho(=
Lga—ext
=——=
| mei]
Ss
wee eee!
8.28
The following armor-related inductances are given as the sum of the external inductance
and the internal inductances for both parallel and perpendicular power (power associated
with flux parallel and perpendicular to the armor tapes or wires), derived in Section 8.3.1.4.
The conductor—armor and sheath—armor inductances are given by
Lo s Mol: ta|1 . » MobeAa
Pea
=Usa
20:(-as
= ) Dieta
E (p)|
+Qnrala
wa
= =—In|—————|-—-—sin“sin(B)P (8.29)
For nonmagnetic armor, the second term becomes (ota /27 rq)(1/2) and can be absorbed
into the first term, to a good approximation, by dropping the “+1,/2” in the denominator
in the logarithm term. The last term is ignored for nonmagnetic armor.
150 Part II » Evaluation of Parameters

For nontouching magnetic armor wires, 4; is assumed to be 1. If the sin’(B) term is

ignored, the first and second terms may then be combined (by ignoring the “+7, /2” in the
denominator in the logarithm term) as before. The last term remains unchanged.
The armor—armor inductance is given by

Se, Ho n(— sos)

s ha Holtr
ange ta | a ] a | ai
HolteAa. Gin .
Laa = — In | ———— — _—| ——~— —-sin* > sin 8.30)

For nonmagnetic armor, the second term becomes ({iofg¢/27ra)(1/3) and can be absorbed
into the first term, to a good approximation, by changing the “rg + t)/2” to “rg + tq/6” in
the denominator in the logarithm term since tg/(3rg) © In[(ra + ta/2)/(ra + ta/6))]. The
last term is ignored for nonmagnetic armor.
If the sin?(f) term is ignored, the first and second terms may the combined (by changing
the “r, + t,/2” to “rg + t,/6” in the denominator in the logarithm term as before. The last
term remains unchanged.

8.3.3 Cable Impedances

In order to compute the loss factors, the impedances will be separated into real and
imaginary components. Every cable inductance includes the longitudinal internal armor
term given by

= JOLolLe Aa:
a
JiX nr, sin(B) (8.31)

where @ = angular frequency

/e = complex relative longitudinal magnetic permeability; the
imaginary part of it describes the hysteresis loss of the magnetic material.

Equation (8.31) may be split into the imaginary and real parts, as shown below.
The complex longitudinal magnetic permeability {1. of the wire may be written as

He = |Mel|(cosy —j siny) (8.32)

where y is the angular time delay of the magnetic flux density B w.r.t. the field H.
Real and imaginary components in equation (8.31) are

WMo|MelAa .
Bo=S8 x7) = ————
0 = Sm(jXz) eer sin(B) cos(y) (8.33)

and

By2 = ;Re(jX,)= ———_*

Wo|Le| Ag 5 z
e(jX_z) att, sin(B) sin(y) (8.34)

From equation (8.25), the conductor—conductor impedance is given by

J@Lo s J@MLota -
Lec = ( (R+ Byase ca l (—) Pe) BOT 4 [u:cos*(B)—1]
27a) — (8.35)
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 151

where R = conductor ac resistance, Q/m

5 = distance between cables in trefoil configuration or the geometric mean of
the three spacing between cables in flat formation, mm
ar, = effective radius of the conductor (a = 0.7788 for solid conductor), mm.

From equation (8.26), the conductor—sheath impedance is given by

Les = Lsc = By+ JB (8.36)

where
Bp a) S
te a (=) WOLot a
+ By+ <—* = fu, cos2(B)—1] (8.37)
Oar rs 20 Tay
The conductor—armorand sheath—-armor
impedancesobtainedfrom equation (8.29),
are given by
Lea = Lac = Zsa = Zas = Bo+ J B3 (8.38)
where
ogresWL n(—— S + ORTies
OL =u;
ta 1 sin
2 | (8.39)
B3 = ——In | ————~ Bo + —— — —~—s 3.39

The sheath—sheath impedance is obtained from equation (8.27). Without significant

loss of accuracy, we can ignore the term ¢,/6 in the equation, yielding

Las ~ (Rs St Bz) ar JB (8.40)

From equation (8.30), the armor—armor impedance is given by

Lad ~ (R, + Bo) + j Ba (8.41)

where
a ieedi Mu(- + =) + Bo oMo
Sa ta Mr Eas 1 uM,| (8.42)
si mieozs By + — sin? 8.42

where R, = armor resistance, including the stranding increment |/ cos B, Q/m

Zaa = the only impedance that would differ for round wires as compared to
tapes. For a solid cylinder without helical lay, the internal flux linkage
contains the factor 1/3 (the square bracket of Bs when B = 0). For
round wires without helical lay, this factor becomes 0.356.

8.3.4 Loss Factors

The voltages V, and V,, in equation (8.5) are zero when both ends of the sheath and
armor are grounded, and so the sheath and armor voltages are described by the following
matrix equation:

(Ret Bye ie By + jB3 } \[ Ike B -| (Bo + jBi)l. (8.43)

Bo + j B3 (R, + Bz) + j Ba) Ip (B2 + j B3)I.
152 Part IT » Evaluation of Parameters

The solutions for 7, and J, in terms of J, are given by

he (Bo+ 7B) (Ra+ Bz)+ j Ba]—(Bo+ j Bs)? i (8.44)

© (CR,+ Bo)+ 7Bi][(Ra+ Bz)+ j Ba]—(Bo+ j Bs)?
and

= (B2 + j B3)Rs
le =a es ae ee (8.45)
[((R; + Bo) + j Bi] ((Ra + Bo) + j Ba] — (Bo + j Bs)

Define

Y; = Bo(Ra + Br) — B, By — BS + BS

Y> = By(Ra + Bz) + Bo Bs — 2B2B3

Y3 = R;(Rq + Bo)
(8.46)
Y= RB;
Y; = Re B>
Ye= KR,Bs

Collecting the real and imaginary parts of equations (8.44) and (8.45) together, the
solutions become

Y,+ jY2
1, = ————+__!——__ /. (8.47)
(Y; + Y¥3)+ j (Yo + Yo)

lL,
Ys+ jY¢
= -————_——__ |. (8.48)
: (Y; + ¥3) + j(Yo + Ya)

Define Y = (Y; + Y3)*+ (Y + Y4)?.Then

un Atl
[Zs|?= + 1, |? Vlad
Ve (8.49)
BALLS i Y
(Lal?= =——*|1, Faroe
|?
NZ
The circulating loss factors for the sheath and armor are thus given by

yt WeEReReVE +3 Ae
Caanaren= Ge oy: 5.20)
and

Alallil’ Res pRede

Ao = =~ 56. (8.51)
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 153

The hysteresis loss factor for the armor is given by

pete leet Tal. myBy (Yo3.Y5)2-e(Ya o)e

A, = (8.52)
[Z.|2R pate Y

The armor loss factor is the sum of the loss factors given by equations (8.51) and (8.52).

EXAMPLE 8.1
Consider a cable model No. 4, and assume that the armor is composed of touching steel wires. The
parameters of this cable are: 1) lead sheath: tr, = 3.2 mm, D, = 75.4 mm; 2) three copper tapes:
Dy; = 78 mm, tr = 0.13 mm, wr = 0.13 mm, 3) steel armor: 59 = 5.189 mm, £, = 121.8 mm,
D, = 98.4 mm, ng = 51. The circuit is partially located under water and partially underground. In
its land portion, the cables are placed in a triangular configuration as shown in Fig. 8.5.

p, = 1 K.m/W
O00

One

OLS

Figure 8.5 Three submarine cables in triangular

formation. se Ooo Tae

All of the dimensions will be expressed in meters. We will begin by computing the resistances
of the metallic parts assuming the frequency of the system equal to 50 Hz. The resistances of the
conductors, sheaths, and reinforcing copper tapes are slightly different in the upper cable since it
operates at slightly lower temperature. However, the differences are very small, and we can assume
that the values of the lower cable apply to all cables. We will assume that the sheath operates at 70°C
and the armor at 68°C. The conductor ac resistance at the operating temperatures of 85°C is equal to

R. = 0.356: 10°* Q/m

The resistance of the lead sheath at 8, = 70°C is obtained from

Dear?
‘ = (0.0754 — 0.0032) - 0.0032 1 =+ 0.004(70 —20)] = 0.000354 &2/m
/

The resistance of the copper reinforcing tape is computed as an equivalent tube resistance. The
mean diameter of the tape is equal to

d; = 0.078 —3 - 0.0013 = 0.0741 m

The cross-sectional area of the tapes is given by

Ar = 3- 0.0013 - 0.025 = 0.975 - 10-* m?

154 Part II m» Evaluation of Parameters

The equivalent tube resistance is obtained taking into account the length of lay 7 of the tapes

Ry; = —————__ [1 + a (8, —20)]

Ar

GI fn 007A
OMe ON cM Rerere roel |
= [1 + 0.00393(70 —20)] = 0.477 - 10° Q/m
0.975 - 10-4
The resistance of the parallel combination of the sheath and the reinforcing tape at 70°C will continue
to be denoted by R, and is equal to
0.354 - 1073 - 0.477 - 10-3
R, = = 0.203 - 10°? Q/m
0.354 - 10-3 + 0.477 - 10-3
The rms diameter of the equivalent sheath is equal to

pe [2.0758
s CUB ery
The cross-sectional area of the armor wires is equal to

A, = - (0.00519)? - 51/4 = 0.00108 m?

The mean diameter of the armor is given by

d* = 0.0984 —0.00519 = 0.0932 m

The resistance of armor is obtained taking into account the length of lay of the wires
0.138 - 10°°. (eee - 0.0932 \° “sy

R, = —————_—__—_————_
0.00108 [1
Lie+ 0.0045(68
( —20)]
O) == O-40Ses10m
0.405- 1073
Om

Since this is a relatively large cable, the ac resistance of the armor wires is 1.4 times greater than the
dc resistance computed above; that is,
Ra = 1.4 -0.405 - 105° = 0.567- 10>? Q/m

In order to compute the remaining impedance, we need to calculate the angle between the axis
of each armor wire and the axis of the cable. This angle is obtained from a knowledge of the mean
diameter of the armor and the length of lay of the wires (see Fig. 8.4):
yee,
B = tan = (oft
a

The longitudinal relative permeability of steel wires is equal to 400 and y = 45°. From equations
(8.33) and (8.34), we have
Bo = @HolHelAa
a) éeAg sin(B)
. cos(y)
2m Wafa

(= -50-47 - 107’-400-51 -2 -0.005192

) sin 68 cos 45 = 0.00313 Q/m
2m-0.0932/2-0.1218-4
By= OHolHel
w alAAasin(B)
sin(y)
anae;
eitaca
(= -50-47 - 10-7-400-51 - 2 -0.00519?
II
) sin 68 sin45 = 0.00313 Q/m
2 -0.0932/2- 0.1218- 4
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 155

The value of B, is obtained from equation (8.37) with the value of uw,= 10

@) Ss @) te
By= et In(=) + Bo+ ett a3[1u,cos*(B)
—1]
lies PE Tp
Oe Se Oray 20S 0.00313
= 20 n 0.0767ib f
nt 2m -50-42 - 107’ 0.00519 -2
[10 cos? (68) — 1] = 0.00329 Q/m
20 0.0932

The constants B; and By from equations (8.39) and (8.42), respectively, are

265;
WLo
oy— ] n(—_)+s Dip =u,|>
Who
t a 1
B palaDeeape oe Ra|
te
DitSOA
LENS fTrolEE Om! —___________—
Ons }+0.00313
20 0.0932/2
+0.00519/2
Rf——————————
geSeOn 2geAO!——__—
O0.0932
CU?_.-10]E=1— auz |
sin“(68) i
| =0.00326
Q
é WLLo
20 i
Be= ein ee(~ s
=) Bidet
+ Wota
QTsne
| By ee ee eon 1 et 2
me |
eae!. aeale
o Mr? 0.5 oe | atone
aly[petehaoa
20 0.0932/2
+0.00519/2
£50.
sAares
a 27aUDAWi 1057 2
gOS se95 :(68)=0.00338
- sin? 2/m
20 0.0932 3cos?(68)
Constants Y are obtained from equations (8.46)

Y, = By(R, + Bo) — B, By — Bs + B}
= [3.13(0.567 + 3.13) — 3.29 - 3.38 — 3.13? + 3.267] - 10-° = 1.28 - 10-®

Y, = By(R, + Bz) + BoB, — 2B) Bs;

= [3.29(0.567 + 3.13) + 3.13 - 3.38 — 2 - 3.13.- 3.26] -10-* = 233-10 °

Y; = R,(Rz + Bo) = [0.203(0.567 + 3.13)] - 10-° = 0.750= 10°

Y, = R, Bs = [0.203 - 3.38] - 10-° = 0.686 - 10°
Y; = R,B, = [0.203 - 3.13] - 10-° = 0.635 - 10-°
Ys = R, B3 = [0.203 - 3.26] - 10-° = 0.662 - 10-°
Y = (¥, 4+ Ys)? + (Y2 + ¥4)* = [(1.28 + 0.750)? + (2.33 + 0.686)7] - 10-2 = 13.2. 10-?

The loss factors are now computed from equations (8.50)—(8.52)

ALY ery O20 ANDY?

42 2Be
pes ee 2 = = 3.05
VER Y 0.0356 13.2
2
RAs Re Vata 2 sa 0.567 0.635?2 + 0.6627
t 2? _ ne
2” AR % 0.0356 13

, By (3 —Ys)? + (Ys —¥%)? 3.13 (0.750 —0.635)? + (0.686 —0.662)?

Aoi Op Y ~ 0.0356 [32
156 Part IJ » Evaluation of Parameters

The rating of the hottest (bottom) cable is given by equation (4.3)

[c= Pam) — Wa [0.57 + nT + Ts + Ta) |

~ LRT) tn +A) Tp+n. + Ar +2 + Ts)I

Dielectric losses for the cable considered here are equal to 6.62 W/m. The ambient temperature
is equal to 20°C. With the cable construction given in Fig. A.4 and circuit location shown in Fig. 8.5,
the thermal resistances of the thermal circuit are as follows (see Chapter 9):

T, = 0.568 K-m/W, 7=0.082K-m/W, 7; =0.066K-m/W, Ty,= 0.789K- m/W

where the external thermal resistance is that of the lower cables.
The sheath loss factor is equal to 3.05 since the eddy current losses are neglected and the armor
loss factor is equal to 1.1. The rating of the cable is given by

i= 85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 0.789) 0.5

ae = 549A
0.356 - 10-4(0.568 + (1 + 3.05) - 0.082 + (1 + 3.05 + 1.1) - (0.066 + en |

8.3.5 Circulating Current Losses in the Sheaths—Special Cases

The equations for the sheath and armor circulating currents loss factors derived in this
section are very general and are applicable to cables in flat and trefoil configurations with
the spacings between cables properly taken into account. The sheath and armor loss factors
are computed simultaneously and the calculations are quite involved. The computations
can be much simplified, with a small loss of accuracy, if we consider sheath and armor
losses separately following the practice presented in the standards. This is illustrated in the
following subsections, where we will consider several cable constructions and typical cable
arrangements.
8.3.5.1 Two Single-core Cables or Three Single-core Cables in Trefoil Formation,
Sheaths Bonded at Both Ends. Since the sheaths are bonded at both ends of an electrical
section, A/ = 0, except for cables having large conductors of segmental construction, in
which case 41 is calculated by the method given in Section 8.4.10.
Let us consider only the sheath of a cable. In this case, the last two terms of equation
(8.37) will disappear and the reactance B), usually denoted by X, takes the form
my _ @Lo 2s > tehe 25

where —_s
= distance between conductor axis in the electrical section being
considered, mm

d = mean diameter of the sheath, mm

—for oval-shaped cores d is given by /dy «dm, where dy and d,, are the
major and minor mean sheath diameters, respectively
—for corrugated sheaths d is given by (D,. + Dj;)/2.

Equation (8.49) reduces in this case to

Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 157

and the sheath loss factor is given by:

vo TE Rete
Ki XG R, 1 ey
ERR OE ORR. ELoe, me R.\?
1+ (+) P28)
X

Equation (8.54) illustrates that A) is increased by a decrease in the conductor resistance

R and an increase in X*, which means by the use of larger conductors and wider spacing.
By computing the first derivative of A, with respect to R, and equating it to zero, we find that
the sheath loss factor attains its maximum when R, = X. Under typical conditions when
R; > X,a decrease in R, increases i. This explains why, in general, aluminum-sheathed
cables have considerably greater sheath losses than lead-sheathed cables.

8.3.5.2 Three Single-core Cables in Flat Formation, with Regular Transposition,

Sheaths Bonded at Both Ends. Equations were also developed by Arnold for the calcu-
lation of sheath losses of cables laid in a flat formation with the middle cable equidistant
from the two outer cables. For a regular transposition condition, the voltage in each sheath
at the third transition is equal to the vector average of the voltages induced in each of the
three sections. This average voltage V; is given by

Wsi x+78)
aon 3
where X,, = mutual reactance between the sheath of the outer cable and the conduc-
tors of the other two, (2/m

= 21077 - In(2), 2/m.

For convenience, the term (X + (X,,,/3)) is defined as X,, and from equation (8.53),
it is given by

epee E:v2(=)] (8.55)

The sheath loss factor is given again by equation (8.54) with X replaced by X;; that is

R, 1
: G +)
] eh

Also, Me= 0, except for cables having large conductors of segmental construction
when Aj is calculated by the method given in Section 8.4.10.

8.3.5.3 Three Single-core Cables without Transposition, Sheaths Bonded at Both

Ends. As before, only balanced currents are considered. When the cables are laid in flat
formation with the middle cable equidistant from the two outer ones, the mutual inductance
between the middle cable and each of the outer cables is given by twice the value given by
158 Part I » Evaluation of Parameters

equation (8.53). However, since the axial spacing between the two outer cables is 2s, the
mutual reactance between them is given by

p) 2s
P = 20-10-TIn2- = = 210-7 In2 + 20 10-7 In = = Xm + X

Let
Lee
h=h\-=+j—
(8.57)
ree (-5 pert
-i) a8

Clearly,

Ts)ta I59+ 153= 0

(8.58)
E51 = Es: = E33 = Eo

where Ep (V/m) is a residual voltage along the cable sheaths, which usually does not exceed
50 V and could be zero when both ends of the cables are grounded.
Applying equations (8.57) and (8.58), the residual voltage computed for each sheath
separately is equal to

| ae
By — Eo — Pahs = JX) = ane - 2G) 3 In(X at Nas a JiaAm
2
Boe By Mes
(Rete fd ie fd x (8.59)
1 3
E.3 = Eo = 153(Rs + JX) — Pile ce — Xm) + 2 In(X =FXm)=aJ1s1Xm

Solving equations (8.58) and (8.59) for the sheath currents, we obtain

ae|h Q: aes
Pat
oe ee7 9 st Reser
5 7
se 7RO s/3P
2)|
ny 7
Re-- OQ? R2 + P?
Q? Bea? )
a52 edie
2(=rama
+ 0 se ere
R2+ O? (8.60)
Fiaong Q? J3R.P ig Pe is 3 P?
(= 13. | , Rae
2|R2+02 Renee ee ape. eT
R24P2 "1\ R242 * R24P?
where P andQ are definedby

Pre iXpepoX iO aX eS (8.61)

Taking the magnitudes of the sheath currents in equations (8.60) and remembering that
in a balanced system |/| = |/|, we obtain the following expressions for the sheath loss
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 159

factors:

| Ld 1 ae a 2R,POXn |
Mh lenin
RY)RETO?” oe +P? ae /3(R? + O°)(R?+ P?) intheleadingphase

Ry Ae
Baa
Im Bae
R es
R2+Q2 : ;
inthemiddlecable

Eyer 1 s)
Aa 2R,;POXn
Mo—eens Gohl a oe inthelaggingphase
RR ATOS REV 3( R260?)
(RA+P2)
(8.62)
Observe from equations (8.62) that the three loss factors are different.
When single-core cables are installed in flat formation without cross bonding or trans-
position, the sheath losses increase as the cable spacing increases, but not linearly. Also,
the external thermal resistance decreases as the cable spacing increases. Therefore, the
ideal spacing is a balance between these two factors. For the rating of cables in air, the
loss factor for the outer cable carrying the lagging phase should be used since it has the
highest loss factor, provided that the spacing is sufficient to ensure thermal independence.
For spaced buried cables, the loss for all three cables is used in the calculation of external
thermal resistance.
We will illustrate the computation of the loss factors by again examining Example 7.5.

EXAMPLE 8.2
Consider cable model No. 1 with laying conditions specified in Appendix A. We assume that the
system load has third harmonic components which constitute 20% of the fundamental current. We
compute the loss factors for the fundamental frequency and for the third harmonic for this cable
system. The sheath diameter and the cable spacing are d = 31.2 mm and s = 71.6 mm, respectively.
The required resistances were computed in Example 7.5.
The auxiliary quantities required to compute the value of the screen loss factor are obtained
from equation (8.61). For the fundamental frequency, we have

) = 0.957 - 10°-*Q/m
2= =450-107In 2-71.6
X=e-2-10-7In ae
Xm = 40-50-1077 - In(2) = 0.436 - 10-* Q/m
PH xX. + xX= 0957-107 +0.436 10 20.1392 10° Q/m

0.436-10-4
Q=X—-X,,/3= 0.957-107*—ano ogame 0.812-10-4Q/m
When the conductor reaches an operating temperature of 90°C, the temperature of the concentric
neutral wires is 83°C. Thus, the resistance of the screen is equal to

R, = 0.759 - 1073(1 + 0.00393 - 63) = 0.947 - 107? Q/m

The circulating current loss factor for the outer cable carrying the leading phase is obtained from
equation (8.62)
160 Part II » Evaluation of Parameters

R? + P? = (0.947 - 10-3)? + (0.139 - 107°)? = 0.916 - 10°°

R? + Q? = (0.947 - 10-3)? + (0.0812 - 10-7) = 0.903 - 10-°

2 3 2
Re leg? a 2R,POX
~ R|RE+Q?
K+P? /3(R2
+0?)(RE
+P?)
0.947-10-3| 5-0.0812?
-10-°7 2.0.139?-10-°
~ 0.0781
- 10%| 90.903 310 0.916-10-6
2 - 0.947 - 0.139 - 0.0812 - 0.0436 - 107!?
= 0.206
J3-916- 0.903- 10-!2
The loss factor for the other outer cable is also obtained from equation (8.62) and is given by

ee
hat ed inne ras 2 2R;POXn
12a|
© OR he
(PR?+Q? R24 P*” /3(R2 + O7)(R?+P?)
0.947
-1073
: -0.0812?-
= 0.0781 - 10-3
——____
10- -
—3.0,139?
-10-6
0.903 - 10-° 0.916 - 10-°
2 -0.947 -0.139 - 0.0812 - 0.0436 - 10-!
0D V3 - 916 -0.903 - 10-2

For the middle cable, the loss factor is equal to

RineeO2 0.947- 10-7 0.8127. 10°

ye = oe See LUC WF",
R R2+Q? —0.0781-10-3 0.903 10-6 ita

Using a similar approach for the third harmonic, we have

(X)y
=4-150:
10-7
In Ze
75)
116=0.287.
10-3
Q/m
(Xm)3 = 4 - 150-107 - In(2) = 0.131 - 107? Q/m
P = 0.287- 107? + 0.131 - 107? = 0.418 - 10° Q/m

0.131 - 10-3
Q = 0.287- 1073 — — = 0.243 - 107° Q/m

R? + P? = (0.947 - 10-7)? + (0.418 - 1077)?= 1.07 - 10-°

R? + Q? = (0.947 - 10-7)? + (0.243 - 1077)?= 0.956 - 10-8

0. Ye0.947-10-3| 4 -0.2437-10-6 3. 0.418?- 10-6

ae 0.0792.10— 0.956- 10-6 1.07- 10-6
2 -0.947-0.418- 0.243-0.131- 10-2
/3 +1.07- 0.956- 10-!2
= 1,48
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 161

; 0.947=10-901 0,243 10 193150:4182 710-°

Qi), = : ue
“2"* 0.0792¢10 | 90,9567 10-4 1.07-10-6
2-0.947-0.418-0.243-0.131.ne]
V3- 1.07-0.956«10-!2
= lew

0947-1072. .0.1227a.105°
(Mim)= 0.0792:102 0912-106 ~ 9!

8.3.5.4 Variation of Spacing of Single-core Cables Between Sheath Bonding Points.

We found that for single-core cables spaced uniformly and with sheaths solidly bonded at
both ends and possibly at intermediate points, the circulating currents, and the consequent
losses, increase as the spacing increases. However, it is not always possible to install cables
with constant spacing along aroute. When the distance between cables varies in an electrical
section, the mutual reactance will vary, and an average value for this reactance should be
used in the formulas in the preceding sections. Note again that a section is defined as a
portion of the route between points at which sheaths of all cables are solidly bonded.
When the spacing along a section is not constant but its various values are known, the
average value of X is given by

ya Xat loXn
tot hXn 8.63
Leal, ares,

where la, lp, ... , l, = lengths with different spacings along an electrical section
Xa, Xp,..- , Xn = the reactances per unit length of cable, where the
appropriate spacings Sq, Sp,... , Sn are used.

When, in any section, the spacing between cables and its variation along the route
are not known and cannot be anticipated, IEC 287 recommends that the losses in that
section, calculated from the design spacing, should be arbitrarily increased by 25%. This
increase, which is not applicable to cross-bonded and single point bonded systems, has
been determined through experience with lead-sheathed high-voltage cables. Where the
section includes a spread-out end, the allowance of 25% may not be sufficient and it is
recommended that an estimate of the probable spacing be made and the loss calculated
employing formula (8.63).
If the cable ends are widely spaced compared to the main part of the route and form
a significant part of the route length, the sheath losses should be based on the average
reactance X, as calculated by equation (8.63).
8.3.5.5 Effect of Unequal Section Lengths in Cross-bonded Systems. The ideal
cross-bonded system will have equal lengths and spacings in each of the three sections. If
the section lengths are different, the induced voltages will not sum to zero and circulating
currents will be present. These circulating currents are taken into account by calculating
the circulating current loss factor, A, assuming the cables were not cross-bonded and
multiplying this value by a factor to take into account length variations. This factor, Fe, is
given by

pee Pata a2 2 (8.64)

in pha
162 Part II » Evaluation of Parameters

where pa = length of the longest section

qna = length of the second longest section
a = length of the shortest section.

Where lengths of the minor sections are not known, IEC 287-2-1 (1994) recommends
that the value for 4, based on experience with carefully installed circuits, be:

i, = 0.03 for cables laid directly in the ground

d, = 0.05 for cables installed in ducts. (8.65)
8.3.5.6 Armored Cables with Each Core in a Separate Lead Sheath (SLType). From
the point of view of circulating currents, SL type cables behave like single core lead-sheathed
cables in trefoil, except that the magnetic armoring increases the circulating current losses.
These losses are now given by the loss factor?

(8.66)

2
where X = reactance of sheath, Q/m = 2107’ - In (>) ,Q/m

In this case, s = distance between conductor axes, mm.

8.3.5.7 Losses in the Screens and Sheaths of Pipe-Type Cables. | Nonmagneticcore

screens, copper or lead, of pipe-type cables will carry circulating currents induced in the
same manner as in the sheaths of solidly bonded single core cables. As with SL type cables,
there is an increase in screen losses due to the presence of the steel pipe. The loss factor is
given by [see footnote to equation (8.66)]

(8.67)

If additional reinforcement is applied over the core screens, the above formula is
applied to the combination of sheath and reinforcement. In this case, Rs is replaced by
the resistance of the parallel combination of sheath and reinforcement, and the diameter is
taken as the rms diameter d’, where

ace S (8.68)

with d = mean diameter of the screen or sheath, mm

d> = mean diameter of the reinforcement, mm.

3 An amendment to the IEC 287 to be introduced in 1997 will propose to change the factor 1.7 to 1.5
according to the recommendations made by Silver and Seman (1982).
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 163

EXAMPLE 8.3
Compute the loss factor A, for a pipe-type cable described by cable model No. 3 in Appendix A.
The parameters of this cable are D;, = 67.26 mmand D, = 67.59 mm. The combined resistance
of the skid wire and the bronze reinforcing tape at operating temperature was computed in Example
7.1 as R, = 0.019 Q2/m. The equivalent diameter is obtained from

D2
Sn +D2 [67.26+
eee 67.592
mae eR

The reactance of the tape and skid wire is

X= 20-107 Dee = 47-60-1077

-In—— In DANES)
ae = 0.525510°
- 1074
*:2/m

Taking the conductor resistance from Table Al (R = 0.245-10~4 Q/m) and applying equation (8.67),
we obtain

: Rs 17 0.0190 le7/
A Ay = = 5 = 0.010
R ‘a (#) 0.245 - 10 ee 0.0190
xe OS25e 10m

8.3.5.8 Two-core or Three-core Armored Cables Having Extruded Insulation and

Copper Tape Screens. By considering the similarity between an SL type cable and an
armored cable having copper tape screens around each core, it is clear that the same equations
are applicable, that is [see footnote to equation (8.66)],

R ee ine Ses
ay (8.69)
Ribas
Rs
\*
alee
In the above equation, it is recommended that the screen resistance is taken as that of
the equivalent tube. In practice, the effective resistance will be somewhat higher because
there will be contact resistance between each turn of the copper tape and a portion of the
induced current will follow the helical path followed by the tape. The use of an equivalent
tube resistance will produce a result which errs on the side of safety.

8.3.5.9 Circulating Currents in the Sheaths of Parallel Cables. |The equations set
out in the preceding sections have been derived by solving a set of simultaneous equations for
the three unknown sheath currents. In all of the above cases, it has been reasonable to assume
a balanced system in which the conductor currents are equal. When it is necessary to install
a number of cables per phase in one circuit, the reactances of the sheaths and conductors
are functions of their spacings from all the other sheaths and conductors. Because of this,
not only will the impedance of the sheaths vary, but also the impedance of each phase
conductor may vary, depending on the relative positions of the cables. Hence, for cables in
parallel, the current flowing in each conductor may be different. This leads to the need to
solve simultaneous equations for both the conductor and sheath currents. For example, for
two cables per phase in a three-phase system, the six conductor currents and the six sheath
164 Part II # Evaluation of Parameters

currents must be found. In this case, a set of 12 simultaneous equations with 12 unknowns
has to be solved, each equation having a real and an imaginary component. In general terms,
for n cables per phase, 6n simultaneous equations must be solved. Additional equations
may be required to set up the boundary conditions for voltages and currents. This is not a
task for manual calculations.
Equations for the voltage drop can be generalized to represent three-phase multicable
installation. In what follows, the term “conductor” refers either to the conductor or to the
metallic sheath of a cable. When the cables have nonmagnetic armor, the armor and the
sheath are combined into an equivalent sheath as illustrated in Example 8.4 below. Cables
with magnetic armor can also be treated by the method discussed in this section; however,
the matrix G would have to be constructed in such a way as to take into account the lay
angle of the tapes or wires. In this case, the relative magnetic permeability of the armor
material should also be considered.
The longitudinal voltage drop E in a conductor is given by the sum of the resistive and
reactive voltage drops (from the fundamental equations set out by Arnold and others):

E=[Ri+
i5-G|aT (8.70
where Ry =n Xn matrix with vector [R,, Ro,--- , R,] of conductor resistances
in the diagonal and zeros outside the diagonal
E = vector of voltage drops
I = vector of currents in all conductors

l 1
In— In— -;:-- In—
Sq] S12 Sin
G=

] 1
i . Ses in—
Sn] Snn
Sj = geometric mean radius of conductor, m

S;j = geometric mean distance between conductors i and /, m.

Since both E and I are complex quantities, their separate components must be de-
termined. In effect, therefore, there are 2n equations and 4n unknown quantities to be
found.
The solution of equations (8.70) for conductor currents is based on an application of
Kirchhoff’s laws: (1) the sum of all the currents flowing into a connection or node is zero,
and (2) the potentials between the ends of conductors connected in parallel are equal.
Application of the second law means that the values of the voltage drop for all con-
ductors in parallel in the same phase are equal. It is therefore possible to eliminate the
impedance voltages as shown below. The required equations have been developed by Parr
(1988) and are set out in IEC 287 (to be published). In the developments presented below,
we will consider cables with one additional metallic layer in addition to the conductor, and
we will refer to it as the sheath. A more general case will be treated in Section 8.5.
For a three-phase system with p phase conductors per phase, there are a total of n
conductors, where n = 6p. The phase conductors are assigned even numbers and the
sheaths odd numbers so that the phase conductor currents become Jp, I4, Ig, +++ , I, and the
sheath currents J), 13,-++ , J,—1.
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 165

To set up the equations eliminating the voltage drop in equations (8.70), the parallel
conductors are considered in pairs:
For the sheath conductors 1 and 3,

O= N(R; + jXe —fX31) + hCGX12 —fX32) + (i X13— Rs —Xo)

rol 4( MidemepXaa)or “= 1, (J Xan = J X3n)

For the sheath conductors 3 and 5,

O= NG Xs1 —JX51) + bhGX32 —fXs52) + (Ry + [Xe — j X53)

Ta) Spaeeeoes i Xas Remi Xe) ke ln (J Xan — J ASn)

For the phase conductors 2 and 4,

0= hQGXa —jXa1) + h(R + [Xe — jXa2) + bj X23 —jfX43)

; ; : (8.71
+ Ig Xa —R — jfXe) + 15(j X05—X45) + + Ln Xm —Xan)

In general terms, for conductors f and f + 2,

OSTA JA pea ee a Ap Ree

TA — JA feos) Tey xpest —JX p43
3p)
+o Xpp = Re J Xe) In IX fa JA Fn)
(8.72)

Also, since the sum of the currents in each set of phase conductors must equal the phase
currents Jp, Is, and /7, we have

Ip{l + jOl Sloe Lge eth lop

Is[—0.5 —j0.866] = Inp42 + Tapa + ++++ Lap (8.73)

and the sum of the sheath currents must equal zero:

04+ j70=)4+h+15+---+)h-1 (8.74)

where, from equation (8.71) and the definition of matrix G,

R, = conductor resistance at operating temperature (sheath or phase), 2/m

X= 20107) In (—)1 (8.75)
58
_4 2 be 2
Kas 2010, gin aE: for phase conductors and X, = 2w10°' In FF
ad* ie
for the sheath

d* = conductor diameter, m

d* = mean diameter of the sheath, m

a = coefficient depending on the conductor construction (ad*/2 is often

called the geometric mean radius of the conductor).
166
Part II » Evaluation of Parameters

The terms X, and X fg are treated as if they were reactances, but they do not represent
true physical quantities. In reality, these terms must exist in pairs as a complete loop. Values
of w were estimated by Dwight (1923) for multiwire conductors and are given in Table 8.1.

TABLE 8.1 Coefficients a for Computation

of Conductor Self-Reactance

Number of Wires Value of a

i 0.779
3 0.678
7 0.726
19 0.758
37 0.768
61 O72
91 0.774
ey 0.776
Compacted conductors 0.779

The values of a for hollow core conductors are dependent on the inner and outer diam-
eter of the conductor as well as the number of strands, and should be computed separately.
The required calculations are described by equation (8.24).
The equations above will give n simultaneous equations with n unknowns /,. These
can be written in a matrix form as

where the matrix Q contains the left-hand side of the above equations, Z contains the
coefficients of 7, from the right-hand side of the above equations, and I contains the un-
known currents /,. To solve the equations for J,, the inverse of the matrix Z is found, and
equation (8.76) is rewritten as

I=Z'xQ (8.77)
The circulating current loss factor for cable a is then given by

i/ ( fin
i ) . R;
R (8.78)
is | ee

where /,, = circulating current in the sheath of cable a, A

Iq = current in the conductor of cable a, A.

EXAMPLE 8.4
We will consider a circuit composed of model cables No. 4 with two cables per phase in the arrangement
shown in Fig. 8.6.
Assuming that the total conductor current is equal to 1600 A per phase, we will compute the
loss factors for armor and sheath of each cable and the current split between the two cables in each
phase.
The conductor resistance at the operating temperatures of 85°C is equal to R = 0.356-10~4 Q/m.
Assuming, again, that the sheath and armor temperatures are 70 and 68°C, respectively, we will use
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 167

Cable 1 Cable 3 Cable 5 Cable 6 Cable 4 Cable 2

<— 0.5 m—<—

0.5 m—|< 0.5 m—'<— 0.5 m—|< 0.5 m—

R1 Sil ila Wz S2 R2
Conductor 1 2 5 6 9 Oma 12. Sieo 4
Figure 8.6 Cable arrangement for Example 8.4.

the equivalent resistance of the sheath and the reinforcing tape computed in Example 8.1, that is
R, = 0.203 - 10-7 Q/m. The rms diameter of the equivalent sheath is equal to d* = 0.0767 m.
However, the resistance of the copper armor wires has to be recalculated since steel wire armor
was used in Example 8.1. The cross-sectional area of the wires and the mean armor diameter were
computed in Example 8.1 as 0.00108 m? and 0.0932 m, respectively. Thus,

(avon)
O1724-10- 0.1218
[1 + 0.00393(68 — 20)] = 0.0494 - 107? Q/m
R,= 0.00108

This multiplied by 1.4 gives R, = 0.0692 02/km. Even though equations (8.76) can be set up
for both sheath and armor conductors, for the purpose of this example, we will combine sheath and
armor into an equivalent sheath. The resistance of the parallel combination of sheath, reinforcing
tape, and armor is
0.203 - 0.0692
Reo 0.203 + 0.0692 = 0516. 105 iy

The rms diameter of this combination is equal to

fia | 0.07672 — 2 —0.0853 m

The cable coordinates are entered into the array

Ley
oY @ Cable 1, phase R
is) AY Cable 2, phase R
la: 20 Cable 3, phase S
ws 2: One0 Cable 4, phase S$
LO =50 Cable 5, phase T
to Cable 6, phase T

The zero coordinates can be fixed at any point; it is convenient to take this point at the center of the

leftmost cable.
The cable spacings, in meters, are calculated using the following equation:

ee roe pe A DY, = Per Gn = 1s 56

168 Part I] m» Evaluation of Parameters

The spacings are given in the following array:

OF 25 9051e2:0 OO ees
25 > (07) 247-0508 CoD
0:5 2.0L. lS O10
DO 05. iat 0. bas
0°. oer” lO aa S
L310 7 Oe OS aye

Clearly, this array is symmetrical about its diagonal, and it is not necessary to calculate the spacing
between cables m and n and again between n and m.
The effective reactances X, and Xm»,are calculated using the following equations:
(1) For the phase conductors,

l
Fom2n
=ifc: =0/2 120807
ln ste Smin
i a2
S10"
(2) For the sheath,

] 2
Pon lan = if es > 0, V—-1-2@- 1G In ,V-1:-2@- 10 In =)
mon

Fon 130) = TF (sn a 4f—1-205 10. 10 1 V/=—1-20-1077 In =)

2
’
Smin

Iya lap low = 0. V—1 70710 it

Smon
l V=1-
20-107
n=)
2
NOTE: the last three equations are identical; three equations are used only to fill the array. The
calculated values are given in the array F with sample computations shown below.
The phase conductors are compacted hollow core. To compute the coefficient a, we apply
equation (8.24). Let a = d;/d. = 17.5/33.8 = 0.518. From equation (8.24),

As
Ps 1/4 —a? +.
a 4(3/4—Ina) - 1/4 —0.518"+0.5184(3/4
— In0.518)
(1 —a)? (1 —0.5182)? ak

The coefficient @ is then

a=el =e! ! — 0.86

The effective self-reactance of the conductor is obtained from equation (8.75):

2 2
X, = 20107 In (=) =4-m-50-1077In = 0.266 - 10-3 2/m
0.86 - 0.0338

The effective self-reactance of the equivalent sheath, obtained from equation (8.75), is equal to

X, = @) 2
2w107/In{
— })=4-2-50-10771 2
n (=) 1 n 0853 =).0.1986-sh107°Q/m
0

From equation (8.75), the effective reactance between the sheath of cable | and the phase
conductor of cable 2 is

X14
= 201077 | 7 Os10°*
In(—}=4-n-50-107’In
pre Q/m
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 169

Continuing the same way, the following matrix is obtained:

SaIO3; =195 O576 0.576 —0.436 —0.436 0.436

INOS. Seon OSS 0.576 —0.436 —0.436 0.436
0.576 0576 —1.98 —I1.98 0.436 0.436 —0.436
0.576 0.576 —-1.98 -—2.66 0.436 0.436 —0.436
—0.436 —0.436 0.436 0.436 1S 1 O55
P= 27.104 —0.436 —0.436 0.436 743 Oi 9 Se 200 0255
J 0.436 0.436 —0.436 —0.436 0.255 0.255 —1.98
0.436 0.436 —0.436 —0.436 0.255 0259 —1.98
0 0 0.255 0.255 —0.436 —0.436 0
0 0 0.255 0.255 —0.436 —0.436 0
0.255 0.255 0 0 0 0 —0.436
0.255 0.255 0 0 0 0 —0.436

0.436 0) 0 0.255 0.255

0.436 0 0 O55 (0),2595)
—0.436 0.255 0.255 0 0
—0.436 0.255 0255 0 0
0.255 —0.436 —0.436 0 0
0.255 —0.436 —0.436 0 0
—1.98 0 0 —0.436 —0.436
—2.66 0 0 —0.436 —0.436
—1.98 -—1.98 —0.436 —0.436
—1.98 -—2.66 —0.436 —0.436
—0.436 —0.436 —0.436 —-1.98 —1.98
—0.436 —0.436 —0.436 -—1.98 —2.66

The components of matrix Z in equation (8.76) are calculated as follows:

(1) For the phase conductor loops,

Phase R Phase S Phase T

Z2p = Frp —Fap Z6,p = Fo.p —Fe.p Z10,p = Fio,p —Fi2,p p= LSZee Le
222 = Za2+R Z66 = Zo6+R Z\0,10 = Zi0,10 + R
Z214=Z24—-R 263 =Zog—R Z10,12 = Z10,12 — R.

(2) For the sheath loop,

Di Uh pe ee a ee a eee lo On1D

Ze Leet
Zine = Zenza — Rs

We can observe that when constructing matrix Z for this example, the rows 4, 8, 11, and 12
will be filled with zeros. In order to save space, we will move the last four rows representing the
coefficients of the currents J on the right-hand-side of equations (8.73)—(8.74) into rows 4, 8, 11, and
12. The matrix H of the current coefficients in equations (8.73)-(8.74) is given by

Omi OMe
erOeLOGL0
) Om
00-0.0.80%)phase.R
COMOOme
Om0mateel Ore
O 5
F 0"
1 0 phase
Haag go207mOnOOo Om0” 1 liphase’T S
Ome eee Oto eo teen emt| o-Sheath
170 Part II m Evaluation of Parameters

Substituting numerical values for variables Z;,; and combining them with the matrix H in rows
4, 8, 10, and 12, we obtain

0.516 + j2.55 72.55 PU SiG —7255 = 72.55

2.55 0.356 + j3.2 S255 20.356 —72:2
—j1.01 27101 0.516 + j2.42 j2.42
0 1 0 I
j0.876 j0.876 —j0.876 —j0.876
Cee j0.876 j0.876 —j0.876 —0.876
= —j0.436 —j0.436 j0.691 j0.691
0 0 0 0
0.255 j0.255 —j0.255 —j0.255
70.255 j0.255 —j0.255 —0.255
0 0 0 0
1 0 1 0

j0.876 j08.76 —j0.876 —j0.876

j0.876 j0.876 —j0.876 —j0.876
—0.516 —j2.42 —j2.42 j0.691 j0.691
0 0 0 0
0.516 + j2.24 j2.24 —0.516 —j2.24 —72.24
j2.24 0.356 + j2.91 =j204 2(1356-—~ 72.91
—j0.691 —j0.691 0.516 + j1.98 71.98
0 1 0 1
0.436 0.436 —j0.436 —j0.436
0.436 j0.436 —j0.436 —j0.436
0 0 0 0
1 0 1 0

70.255 70.255 —j0.255 —j0.255

j0.255 70.255 —0.255 —70.255
~j0.691 —j0.691 0 0
0 0 0 0
j0.436 j0.436 —j0.436 —j0.436 j0.436
j0.436 0.436 —j0.436 —j0.436
—0.516 — 1.98 —j1.98 0 0
0 0 0 0
0.516 + j1.54 1.54 —0.516 —j1.54 =f 154
1.54 0.356 + j2.22 ~ 71.54 =0.356— 72.22
0 0 I
! 0 0

The array Q in equation (8.76) is given by

Q=[0001000 —0.5 —j0.86600 —0.5 + j0.866 0}!

where ¢ denotes transposition (Q is a column vector).

Inverting matrix Z and substituting into equation (8.77), we obtain the following vector for the
currents and their magnitudes:
Chapter 8 # Joule Losses in Screens, Sheaths, Armor, and Pipes 171

—0.4875 — j0.1404 0.5073

0.5000 — 70.0000 0.5000
—0.4875 — j0.1404 0.5073
0.5000 + j 0.0000 0.5000
0.3523 + j0.3014 0.4637
I= —0.2500 — 70.4330 iq] = 0.5000 A
0.3523 + j0.3014 1) 90.4637
—0.2500 — 70.4330 0.5000
0.1305 — j0.4408 0.4597
—0.2500 + j0.4330 0.5000
0.1305 — j0.4408 0.4597
—0.2500 + 70.4330 0.5000

The magnitudes of the phase conductor and sheath currents are given in Table 8.2, assuming a
total phase current of 1600 A. The equivalent sheath loss factors are computed from equation (8.78)
and are given in column 4 of Table 8.2.

TABLE 8.2 Results of Example 8.4

Phase
Conductor Sheath
Current Current EquivalentSheathLoss SheathLoss ArmorLoss
(A) (A) Factor Factor Factor

| Lom
| -1600 on | -1600 ( | L2m-1|-1600)?-R,
) sa
Sia
( | Zam
| -1600)”-R
Cable 1, phase R 800 812 1.49 0.379 eda
Cable 2, phase R 800 813 1.49 0.379 Heil
Cable 3, phase S$ 800 742 P25 0.318 0.932
Cable 4, phase S 800. 742 Ds 0.318 0.932
Cable 5, phase T 800 736 123 0.313 0.917
Cable 6, phase T 800 139 123 0.313 0.917

To separate the equivalent sheath loss factor into sheath (with tape) and armor loss factors, we
consider each cable separately and observe that

X;Xaqa x? T=XG Na
I, =
x? ia X5Xaa J(XsRa aleXqRs)
—jXaRs
l=
xa X;5Xqcup(Xs RaSitXqR;)
From the definition of the loss factors, we thus have

/ i (Xs ~ Xq)2 an X7Ra

di, 2 (8.79)
A2 R;Ra X?Rs
When X, * X,, which usually is a good assumption, equation (8.79) leads to

/ Rg
Je once aig a oD (8.80)
Ry + Ra

Using equation (8.80), the results summarized in the last two columns of Table 8.2 are obtained.
172 Part IJ » Evaluation of Parameters

8.3.6 Circulating Current Losses in the Armor—Special Cases

8.3.6.1 Armor Materials. | Armored single-core cables for general use in ac systems
usually have nonmagnetic armor. This is because of the very high losses which would
occur in closely spaced single-core cables with magnetic armor. On the other hand, when
magnetic armor is used, losses due to eddy currents and due to hysteresis in the steel must
be considered. A method of calculating these losses is given by Bosone (1931), and the
results agree with those obtained in the limited experimental work reported by Whitehead
and Hutchings (1939). The latter work demonstrated that the losses in the sheath and
armor combination could be several times the conductor losses, depending on the bonding
arrangements of the sheaths and armor (see Fig. 8.7).

1100
Sheath and armor bonded

oSheath
loss
and
conductor
%
aarmor
as
f
—o——_—

1000 }-
Sheaths only bonded
Eee
900

75 mm spacing no bonding
800 fF =

150 mm between sheaths

700 Se ee a

600 }- 25 mm between sheaths

ae SNS

500

400
100 150 200 250 300 350 400
Conductor current, A

Figure 8.7 Effect of steel wire armor on cable losses.

Several other authors addressed the subject of armor losses. Acomprehensive account
of this subject is given by Carter (1954), Schurig et al. (1929), and Schelkunoff (1934).
More recent publications include the works by Bianchi and Luoni (1976), Kawasaki et al.
(1981), and Weedy (1986).
The armoring or reinforcement on two-core or three-core cables can be either magnetic
or nonmagnetic. These cases are treated separately in the next two subsections. Steel wires
or tapes are generally used for magnetic armor.

8.3.6.2 Nonmagnetic Armor or Reinforcement. | When nonmagnetic armor is used,

the losses are calculated as a combination of sheath and armor losses. The equations set out
above for sheath losses are applied, but the resistance used is that of the parallel combination
of sheath and armor, and the sheath diameter is replaced by the rms value of the mean armor
and sheath diameters.
For nonmagnetic tape reinforcement where the tapes do not overlap, the resistance of
the reinforcement is a function of the lay length of the tape. The advice given in IEC 287
to deal with this is as follows.
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 173

1. If the tapes have a very long lay length, i.e., are almost longitudinal tapes, the
resistance taken is that of the equivalent tube, that is, atube having the same mass
per unit length and the same internal diameter as the tapes.
2. If the tapes are wound at about 54° to the axis of the cable, the resistance is taken
to be twice the equivalent tube resistance.

3. If the tapes are wound with a very short lay, the resistance is assumed to be infinite;
hence, the reinforcement has no effect on the losses.
4. If there are two or more layers of tape in contact with each other and having a very
short lay, the resistance is taken to be twice the equivalent tube resistance. This is
intended to take account of the effect of the contact resistance between the tapes.

EXAMPLE 8.5
We will compute loss factors and ratings of a three-phase circuit composed of cable model No. 4
with all necessary parameters as given in Appendix A and with the resistances of the metallic parts
at operating temperature as computed in Example 8.4. From this example, R, = 0.203 - 10-* Q/m
and R, = 0.0692 - 10-3 Q/m. The equivalent sheath/tape/armor resistance is 0.516 - 10-4 Q/m
and the mean diameter is d = 85.3 mm. The conductor ac resistance is given in Table Al as
R = 0.356 - 10-4 Q/m. We will consider two laying conditions: (1) when the cables are transposed,
and (2) when they are not transposed. In both cases, the cables are in a flat formation.
(1) Cables regularly transposed
The mutual reactance between one equivalent sheath and a conductor of the neighboring cable
is given by equation (8.55):

500
x)=20-107
-n[2.
V3(5)]=4-2-50-107-1n]
2.3(FE
)]=1.69-
107-4
Q/m
The sheath loss factor is given by equation (8.56):

4! R, ] =0.516
-10+ ] meee
ae eeRON?
0856-10 O516410-\.
1+(ro)
The sheath and armor loss factors are obtained from equation (8.80):

Ni Rye . 0.0692
Gs = 0.338
= PARSER, Boe 0.203 010692
Ky = lo — 0) 33he) == OLB

The rating of the middle cable is obtained from equation (4.3) with the thermal resistances and the
dielectric losses given in Appendix A as 7, = 0.568 K - m/W, 7; = 0.082 K - m/W, 73 = 0.066 K -
m/W, and 7; = 0.814 K - m/W. Wz = 6.62 W/m. Thus, we have

: KO WAST ene oh)! ik

~ | RT, tnR(L+ A1)TotnR(L +A, + A2)(T3
+ Ts)
85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 0.814) i
7 aa - 10-4[0.568 + (1 + 0.338) - 0.082 + (1 + 0.338 + 0.992) - (0.066 + 0.814)]

==HIM
174 Part IJ » Evaluation of Parameters

(2) Cables are not transposed

The values of P and Q are given by equation (8.61)

Xm =4-2-50- 1077 In2 = 0.436- 10-4 Q/m

2 2 - 500 i
X =20- 107 n= =4-7-50- Oneness = 1.55-10°* Q/m

P=X,+X =0.436- 10 + 1.55 - 10-* = 1.99 10° &2/m

QO=X — Xm/3 = 1.55 - 10-4 —0,436 - 1074/3 = 1.40: 10-* 2/m

The rating of the circuit is computed for the hottest cable. The loss factor for the middle cable
is obtained from equations (8.62), with the factor 10~* cancelled in all the fractions

pene Ky G2 _ 0.516 1.40? ning

Imsa ~~
R R2+Q? 0.356 0.5162+ 1.402 ~

The sheath and armor loss factors are obtained from equation (8.80)

Paphos ie
re 0.0692 peor
7s| = os
eck5 Ry Ey RTTOTO aTP Be
ho = 1.28 — 0.325 = 0.955

The rating of the middle cable is obtained from equation (4.3)

= 85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 0.814) i

~ | 0.356 - 10-4[0.568 + (1 + 0.325) - 0.082 + (1 + 0.325 + 0.955) - (0.066 + 0.814)]
= IN.

We observe that these loss factors are almost equal to the ones obtained in part (1); hence, the
ampacities are almost the same in both cases.
The magnitudes of combined sheath and armor currents can be computed from equations (8.60)
or from the loss factors formulas (8.62)

=i Ps ae 2 —-
Shee Sea 5 aa
Ri+Q? R}+P? ./3(R2 + Q?)(R2 + P?)

Tweet Lae heals

ieWea oa 2RsPOXn
3= So
Ri+Q? aeRi+P? ./3(R2
+ Q2)(R2
+P?)
Substituting numerical values, we have

Tint = 109 Agcorksgesande tolvendaa, ah.

0.0692
Is) = - 709 = 242 A, 14; = 709 — 242 = 467 A
0.203

Similarly,
Ig = 246A, Ig2 =477A, 1,3 = 263A, 1,3 = 508A
Chapter 8 m Joule’Losses in Screens, Sheaths, Armor, and Pipes 175

We can observe that combined sheath and armor current is in the same order of magnitude as the
conductor current, and that most of this current flows in the armor.

8.3.6.3 Magnetic Armor or Reinforcement.

8.3.6.3.1 SINGLE-CORE LEAD-SHEATHED CABLES WITH STEEL WIRE ARMOR. The
armor losses are lowest when the armor and sheath are bonded together at both ends of a run;
thus, this condition is selected for the calculations below. The method, derived by Bosone
(1931), gives a combined sheath and armor loss for cables which are very widely spaced
(greater than 10 m). It has been applied for submarine cables where the cable spacing may
be very wide, and there is a need for the mechanical protection provided by the steel wire
armor. The equations contained in the IEC Standard 287 can be also derived by simplifying
equations (8.43). The simplification involves replacement of the sheath and armor by a
single resistance equal to

Ree 8.81
ul Rataks ean

where _R,= sheath resistance with the cable at its maximum operating tempera-
ture, (2/m

Rq = armor resistance with the cable at its maximum operating tempera-

ture, (2/m.

The ac resistance of the armor wires will vary between about 1.2 and 1.4 times its
dc resistance, depending on the wire diameter, but this variation is not critical because the
sheath resistance is generally considerably lower than that of the armor wires. The use
of the resistance R, assumes that the current split between sheath and armor is inversely
proportional to the resistances. This would be true only if sheath and armor had the same
self- and mutual inductances. Even a relatively small difference in inductances can cause
the current split (and hence the losses) to be quite different; the actual split will be controlled
by inductances.
Two important factors make the sheath and armor inductances different:

1. the gap between the sheath and armor,

2. the longitudinal inductance caused by the lay of the armor.

Next, the impedances B3 and By,appearing in equations (8.50)—(8.52) are both re-
placed by B,. Summing now the loss factors given by equations (8.50) —(8.52), we obtain
the following expression for the total loss factor

/ :
R. 7+
Dy BP
a+eR.Bp
sinOe | (8.82)
ys =
(Re+ Bo)?+ Bi
IEC Standard 287 (1982) suggests that the sheath and armor loss factors have the same
value, namely, one half of the value obtained from equation (8.82). Alternatively, equation
(8.80) can be used to obtain these loss factors.
Replacing B3;and By by B, in equations (8.50)-(8.52) assumes that sheath and armor
have the same inductances, which is approximately correct because the Bo term of Bs,
176 Part II » Evaluation of Parameters

By, and B, completely dominates the other terms. The last terms of B3 and By must be
replaced, incorrectly, with the last term of B,. The reason that these terms do not converge
towards the IEC Standard is that circular fields were assumed to have 100% flux linkage in
the armor, and solenoidal fields in the armor were not considered at all in the original paper
by Bosone (1931). These errors are fairly minor because of the dominance of the Bo and
B> terms resulting from the large longitudinal flux in the steel tapes or wires. Thus, it turns
out that the approximation given by equation (8.82) does not introduce any significant error
in cable rating (see, for instance, Example 8.6).

EXAMPLE 8.6
We will return to the case examined in Example 8.1, but this time we will apply equation (8.82) to
compute the sheath and armor loss factors.
The resistance of the parallel combination of sheath, reinforcing tape, and armor is

_ 0.203 - 0.567
- 1073 = 0.149. 1077 Q/m
“~~ 0.203 + 0.567

Applyingequation(8.82),sheathand armorlossfactorsare equalto

R, (2 + BP+ as)
Ap hy 2R \(R. + Bs)? + Bi

vtediin0. 149. 10-3 0.003137 + 0.00329" + 0.00313 - 0.149 - 10~?

= 2.05
~ 2 [0.0356 - 10-3 (0.149 - 10-3 + 0.00313)? + 0.003292

The rating of the hottest (bottom) cable is given by

ns 85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 0.789) }

0.356 - 10-4(0.568 + (1 + 2.05) - 0.082 + (1 + 2.05 + 2.05)) - (0.066 + 0.789))
== O3)0)7

The ampacities computed using equations (8.50)-(8.52) and equation (8.82) are in very good
agreement. In numerous studies performed by the author, the difference in the cable ratings obtained
using both approaches did not exceed 10%.

8.3.6.3.2. TWO-CORE CABLES WITH STEEL WIRE ARMOR. When calculating armor
losses for cables with steel wire, account must be taken of both armor losses and hysteresis
losses. Equations for such losses were developed by Arnold (1941) and Whitehead (1939)
for both two- and three-core cables. The equation given in IEC 287 for two-core cables is
derived from the work by Whitehead, and is intended for cables with shaped conductors, but
it is also sufficiently accurate for cables with circular conductors. The relative permeability
of the armor wires is assumed to be 300. The armor loss factor is given by

1
0.62w2107!4
2A" 1.48r;
RR, +t] +
R
=
d2 +95.7A,
(8.83)

where —-R,
= ac resistance of the armor at maximum cable operating temperature, Q/m
d, = mean diameter of the armor, mm?
Aa total cross-sectional area of the armor, mm

r\ radius of the circle which circumscribes the conductors, mm

t insulation thickness between conductors, mm.

Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 177

8.3.6.3.3 THREE-CORECABLESWITHSTEELWIREARMORANDROUNDCONDUCTORS.
The equations generally used for three-core cables are derived from those developed by
Arnold:

Ia [fXEz——_____.,—
ie-Sn0se*(=) 1 (8.84)
R \da (a)
= })+]
WM
where _c = distance between the axis of the conductor and the center of the cable, mm.

8.3.6.3.4 THREE-CORE AND FOUR-CORE CABLES WITH STEEL WIRE ARMOR AND
SECTOR-SHAPED CONDUCTORS. The equations used for this type of cable are derived
from those developed by Arnold. For three-core cables,

earn RMaryNe
(=) ta 1 ao (8.85)
R\d, Gaol
eee el
W
where _—r;
= radius of the circle circumscribing the shaped conductors, mm.

For four-core cables, the coefficient 0.358 is replaced by 1.09.

8.3.6.3.5 SL TYPE CABLES. The armor loss in SL type cables is reduced by the
screening effect of the sheath currents. It is calculated as for three-core cables with circular
conductors and then multiplied by the factor (1 — 4), where Aj is the circulating current
loss factor for SL type cables.

8.3.6.3.6 THREE-CORECABLESWITHSTEELTAPE ARMORORREINFORCEMENT. For

two- or three-core cables with steel tape armor, it is necessary to consider the hysteresis
and eddy current losses separately. Equations for the hysteresis loss in tapes of a thickness
between 0.3 and 1 mm have been developed by Whitehead (1939). It is recognized that
this equation overestimates the hysteresis loss for tapes less than 0.3 mm in thickness. A
suitable approach has been presented in Section 8.3.4.
The equation for hysteresis loss developed by Whitehead has been used to derive the
equation given in IEC 287. The armor loss factor A2 is given by the sum of the hysteresis
loss 4, and the eddy current loss factor (5. The hysteresis loss factor is given by

peseki
O
hoesRd, y (8.86)
h ieee
Beaiacsieer a
aveindsta) NSO
Lt
s = distance between conductor axes, mm
f = supply frequency, Hz
yt = relative permeability of the steel, usually taken as 300
59 = equivalent thickness of the armor = A,/7d,, mm.
178 Part II m» Evaluation of Parameters

The eddy current loss factor is given by

2.25s7k75910-*
"
oie
(8.87)
Rd,
The total armor loss factor is then

8.3.7 Losses in Steel Pipes

Empirical equations for the losses in steel pipes were developed by Meyerhoff (1952).
These equations are used by Neher and McGrath (1957) and in IEC 287. The equations
were derived for the size of pipe and type of steel commonly used in the United States at
that time, and should be treated with caution for other sizes of pipe or types of steel. Three
equations are given by Neher and McGrath (1957) depending on the cable configuration
within the pipe.
For a three-core cable,

Ao
=cen aeeeaee
0.0199s
- — 10->
0.001485, R (8.89
where Ss= axial spacing of adjacent conductors, mm
D 4 = internal diameter of the pipe, mm
R = ac conductor resistance at maximum operating temperature, Q/m.

If the three cores have a flat wire armor applied over them, the pipe losses are ignored
and the armor losses calculated as for an SL type cable.
For three single cores in trefoil,

0.0115s —0.001485 Dy or
Ao.= | ———————__ 1 10 ~ (8.90)
R

For three single cores in cradle formation on the bottom of the pipe,

i — ae0.00438s + 0.00226D, 10> (8.91)

In practice, the three single cores will lie in a formation somewhere between trefoil and
cradle. For practical cases, the losses should be calculated as the mean value between the
trefoil and cradle formations using the following empirical equation:

ho = —orwr)
0.00794s + 0.00039D, Log (8.92)
R
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 179

The above equations apply to systems operating at 60 Hz. For systems operating at 50 Hz,
the calculated loss factor should be multiplied by 0.76.

EXAMPLE 8.7
We will compute the pipe loss factor for a pipe-type cable (cable model No. 3). The parameters of
this cable obtained from Appendix A are s = 67.6 mm, R = 0.245-10~4 Q/m, and Dy = 206.4 mm.
Since the cables are assumed to be in the cradle configuration, we apply equation (8.91) to obtain

ae os ae—)
R
Ge= =— -67.6
+0.00226
-206.4 0.245- 10-4 ) 10> = ae0:

8.4 SHEATH EDDY CURRENT LOSSES

8.4.1 Overview

Sheath eddy current losses must be included in the equations for calculating current rat-
ings for three-core cables, two-core cables, and single-core cables when circulating current
losses are eliminated by choosing a suitable method of bonding.
In this section, we will consider cables with a continuous tubular sheath. Cables with
corrugated sheaths are treated as having extruded sheaths with the outer diameter replaced
by the geometric mean of inner and outer diameters. Cables with concentric neutral wires
when bonded at one end will have no eddy current losses.
Early investigations by Carter (1927), Dwight (1923), and Arnold (1929) form the
basis of the formulas used in today’s standards. Accurate calculation of eddy current losses
for single-core cable sheaths is very complicated, and analytical simplifications leading to
semi-empirical equations have been developed by a number of workers. The most notable
of these are Carter (1927), Miller (1929), Goldenberg (1958), and Morello (1959). The
methods developed by these authors were accurate enough for the lead-sheathed cables in
general use at the time. The equations developed by Miller were included in the first edition
of IEC 287 (1969).
The increasing use of aluminum sheaths, whose electrical resistance is an order of
magnitude lower than the equivalent lead sheath, has led to reconsideration of the equations
for eddy current loss. Work by Heyda et al. (1973) provided a good general approach to
the problem, yet the solutions could only be expressed as complicated mathematical series
unsuitable for general use. The work by Heyda et al. (1973) was examined by Parr (1979)
in order to derive semi-empirical equations for general use. These equations are included
in the second edition of IEC 287 (1982). They contain correction factors for eddy current
losses in thick sheaths where additional losses due to currents in other conductors and sheath
losses due to the conductor current of the cable itself must be included.
The developments presented in this section are based on the work by Jackson (1975).
The eddy currents flowing in a tubular sheath are composed of many components. The first-
order eddy current in a sheath is caused by the combined effect of its own axial conductor
current and the currents of neighboring cables. These two eddy currents can be considered
separately, and the resulting losses added (Jackson, 1975). In most practical cases, the
self-induced losses are negligible. The second-order eddy current arises from the effect
of the magnetic field of first-order eddies in other tubes. The process is repeated until the
180 Part II » Evaluation of Parameters

successive order eddies have a negligible effect on the total eddy current distribution in the
tube under consideration.
The original formulas developed by Carter, Dwight, and Arnold were made amenable
for hand calculations by considering approximations to the series expressions in the equa-
tions. Even though their application may result in an error reaching about 20% in sheath
loss factor computations (Jackson, 1975), the convenience of the formulas and a small effect
of eddy current losses on cable rating could justify their continued use provided that the
range of application is clearly defined. We will consider this subject next.
In the developments presented below we will first assume that the sheath thickness
is small compared with its radius, so that eddy currents can be assumed to be uniformly
distributed across the wall and to act at the mean sheath radius. The correction to be applied
to thick sheaths is considered separately. The resulting geometry and nomenclature for
analysis are shown in Fig. 8.8. In this figure, two cables, A and X, separated by a distance
Sax are considered. Currents J, and Jy flow in conductor A and X, respectively. The
sheath thickness is denoted by f, and the outer radius of the sheath is r4. A magnetic flux
at the point P is considered.

Figure 8.8 Nomenclature for analysis of eddy

currents (Jackson, 1975).

8.4.2 Loss Due to External Currents

First-Order Eddy Currents: The eddy current density in any one sheath is deter-
mined by first considering the influence of the magnetic field caused by the neighboring
conductor currents. A current so induced is the first-order eddy current. The basic induction
equation linking the magnetic flux ©with current density i, is given by

M Ol, ; t, oD
An = JOU |aHore ap lie —sS10 (8.93)

where eg te A
Lo = 4x - 10~’, H/m = permeability of nonmagnetic systems
i, = relative permeability, equal to 1 for nonmagnetic materials
t; = sheath thickness, m
¢ = sheath material electrical resistivity, Q/m
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 181

Solving equation (8.93) with the notation in Fig. 8.8, we obtain (Jackson, 1975)

lA, == Fis
= = cosné
+F’,,
sinn@) (8.94)
where

Fan—(Fn)xoat
»E (Sax) n
exp(J Wx)COSnaax

q
Los= CA xsati
S, (8ax)”
exp(j Wy) sin naax (8.95)

J2rama
hh. (8.96)
a MOA
HLLow
= 4m Rs ( 8.97 )

X =a dummy variable and the summation }~4_ ,, , represents the

summation of all conductors X exterior to A. The analysis does
not assume identical tubes, but each tube may be completely
specified by its mean radius and the frequency/resistance ratio m.
Myx and yy = the magnitude and phase angle of the current /y defined with
respect to an arbitrary reference current /; that is,

Ix = Mx exp(jvx)1 (8.98)

From equation (8.94), we obtain the following expression for the eddy current sheath
lossfactor:
CTAi iA 2
te Jo=0i|2
1aa =
I2R =Svea
rt:, sit ae[? = 2+|Fanl’)(8.99)
TR(Fan 12
R;=
=5K 2+|Finl)
De(Fae! PD
Substituting equations (8.95) into (8.99), we obtain

nM!=a Rs [o-e) pot M4 2 q M,

ieee R ) A n + m2 (sax)”
= A X=A+1 AX
(8.100)
My
cosn(a4x — ay) cos(Wax — Way)
ONDA Mes O ae
YZA
Equation (8.100) gives a general expression for the first-order eddy current loss factor
as a function of the system geometry. Computations can become tedious when several
182 Part II » Evaluation of Parameters

cables are involved. For the following common arrangements, further simplifications are
possible.
(1) Single-phase go-and-return circuit. Let s be the distance between the cables in
millimeters. Since there are only two cables, the second summation term in equation (8. 100)
is equal to zero. Also, My = 1; hence, equation (8.100) becomes

I eget d\*" m2
oe es DE == ae 8.101

with m = ma defined by equation (8.97) and d is the mean sheath diameter, mm.
(2) Cables in flat formation with three-phase balanced currents. For three-phase
balanced currents, we have

Ma=Mzp=Mc=1,wWa=0O,
Ua 4a 20
Ve ray (8.102)
The eddy current losses will be different in all three phases. For the center cable,

aa
yt | lee
Rec)
=AS Ga)map
d 2nae Leh)
ue hho) (8.10
For the outer cables, we have

a. ye n+
1 RDe (=)
n=1
iem2 banat
( aQ2n
Q)n (8.10
(3) Cables in trefoil-touching formation with three-phase balanced currents. Substi-
tuting (8.102) into (8.100), we obtain in this case

oe Rs3 3(ar m- (2 nit

R 2s n+m2 pina
=i (8.105
Higher Order Currents: The procedure for determining the second-order eddy
current ¢4, caused by the magnetic field set up by the first-order eddy currents flowing
in adjacent tubes follows the same development as that for finding i4,. Thus, the total
second-order current in sheath A caused by all adjacent g tubes is

if (o,@)
ia, = bah De (Gancosné +G',, sinnd) (8.106)
n=]
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 183

where

(=) nk icy
Ge = | ae
2; tS (n —1)!k!
q k
ry
Gath [Fxecos(n+ k)aax + Fy, sin(n+ Koay]
. X=A-+1
Be : a (8.107)
Gay
r = (=)n pal (n +k —1)!(—1)*
hI
or iy ikt
q pe kSo. [ Fxxsin(n+ k)aaxy—Fy,cos(n+k)a |
ico (sax rte AX Xk AX

Similar expressions can be written for other sheaths replacing the suffix A by the
appropriate legend. Higher order currents follow a similar pattern. For example, the third-
order current 74, is obtained from an equation similar to (8.106) with variable G replaced
by variable H. Variables H and H’ are computed from equations (8.107) with Fy, and
Fy, replaced by Gy, and G'‘,,, respectively. Jackson (1975) has shown that the effect of
fourth- and higher order currents can usually be neglected.

Total Eddy Current and Loss Factor Due to External Currents: The total eddy
current density in sheath A is

I
BA TA la he ae Yo[(Fan+ Gan+++-)
cosnd + (Fy, + Gy, +--+) sinnd]
(8.108)
Usingthenotation
Cai Ss
alb,tknels (8.109)
Cy = Fin + Gy to

and repeating the developments which lead to equation (8.99), the following expression for
the total eddy current loss factor due to external currents is obtained:

ste, A (ICanl?+ Ci
ie 2Rn=l n n ‘) (8.110)
8.4.3 Loss Due to Internal Current

The eddy current in a tube induced by its own coaxial conductor current circulates in
the wall thickness, and is constant for all angular positions 6 assuming a uniform density
of the current in the coaxial conductor. Thus, there is only a single loss term equivalent to
n = 0 and the loss can be added directly to the external loss. Since the loss is in general
very small, it is permissible to ignore the magnetic field caused by the eddy current itself.
The exact expression for eddy current internal loss factor was developed by Imai (1968)
 Part IJ » Evaluation of Parameters
184

and is equal to
R,; ) (Bits)? =Wai
EY Rebkuatlo
MA Ee SAK)a) (8.111)

where
pe 4 w
1 Vi107%
and t, (mm) is the sheath thickness.
The total loss factor is obtained by summing expressions (8.110) and (8.111):

= 3+|Canl”)
ate= b»(ICanl? M? Ss
+aS
n=
. 10— (8.112
8.4.4 Correction for Wall Thickness

The preceding analysis has assumed that the eddy currents are uniformly distributed
across the sheath wall. For thicker than usual sheaths, Jackson (1975) has developed
a correction factor by which equation (8.110) is multiplied. This correction factor gs, is
plotted in Fig. 8.9 against the nondimensional parameter 8; D*, where D* (m) is the external
diameter of the sheath for various ratios of the sheath thickness versus the outside diameter.
Analytical expressions to approximate the curves in Fig. 8.9 have been developed and
are given by the following expression:

t 1.74
s=1+(4) (B,D,- 10-*—1.6) (8.113)
No correction for wall thickness is required for the losses caused by the internal current.

Ons
ts/Ds =0.2,7 0.10.08 0.06 0.05 0.04

factor
correction
tube
Thick
gg

0 10 20 30 40 50
Figure 8.9 Correction to external loss factor
B1Ds (Jackson,1975).

8.4.5 Simplified Expressions for Three Single-core Cables in Flat and TrefoilFormations

It is evident that the computation of the loss factor from equation (8.112) is fairly
complex because it involves a summation of a series within a series. Equation (8.112) can
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 185

be evaluated by suitable computer programs; however, in applications published in cable

rating standards, significant simplifications of this expression are used for the purpose.
Two types of simplifications are employed for the external component of the eddy sheath
loss factor: 1) the first term in equations (8.101), (8.103)-(8.105) is used to approximate
first-order eddies, and 2) an analytical expression is used for approximating the remaining
terms in equation (8.112). The relevant expressions are as follows. Equation (8.112) is
approximated by

M=aR,estoSEIN
SEIN)
Ree
e ot10-8]
‘ (8.114)
where g, is given by equation (8.113). The formulas for Ao (the first term of the first-order
eddy current loss) A; and A> are given below.
(1) Three single-core cables in trefoil formation.

Dh Sita size
Meo (=) Ge
i obtained by substituting n = | in (8.105)
2S

A, = (1.14m** + 0.33) (=) 0.92m+1.66 (8.115)

A, =0

where m is defined by the same equation as my, in (8.97).

(2) Three single-core cables, flat formation.
(a) center cable:

dine
Ao = (=) i obtained by substituting n = | in (8.103)
S
A; = 0.86m> (=)d \'4m+0.7 (8.116)
2s
A, =0

(b) outer cable leading phase:

a) : (=)d\? 1 +m?m?
= 1.5{—

qd \0-16m+2
Ay = 49m"! (8.117)
2s

csi ft (5)Se 1.47m+5.06

Ay = 21m>?
186 Part IJ » Evaluation of Parameters

(c) outer cable lagging phase:

si dy
ee) )\amuitis
(2s) 14m?obtained
bysubstituting
n=1in(8.104)
MR _0.740n+2)m0.5(<.) m+1 (8.118)
2+(m—0.3)2\2s
A>
=0.92m>’
(=)
IAS
In many practical cases, only the first-order eddy currents will be of interest. Jackson
(1975) has shown that the approximate formulas (8.101), (8.103)-(8.105) are good approxi-
mations (maximum error of 10%) for all cases where d/2s < 0.2,m < 1.0. For the majority
of power cables, m values are typically in the range 0.3—-0.7with spacing factors less than
(0.2, and here the use of the first-order solution is permissible. In addition, for lead-sheathed
cables, g, can be taken as unity, and the term (6)f,)*/12 - 10~'? can be neglected. For
aluminum-sheathed cables, both terms may have to be evaluated when the sheath diameter
is greater than about 70 mm or the sheath is thicker than usual.

EXAMPLE 8.8
We will find the sheath loss factor for the center cable in the circuit composed of three cables (model
No. 5) in flat formation. The sheaths in this cable circuit are cross bonded and the lengths of minor
sections are not known. The spacing between cable centers is 0.5 m. The required parameters of this
circuit are taken from Appendix A: D,, = 113 mm, D;, = 102 mm, and t, = 6 mm.
To compute the cross-sectional area of the sheath, we will take the diameter of the equivalent
noncorrugated sheath and multiply it by the sheath thickness and by 7. The mean diameter of the
equivalent solid sheath is equal to

—Doc+ Dit 102+ 113

be ee
Assuming sheath operating temperature of 75°C, the sheath resistance is given by

ees ra gs
Ss [1 + 0.00403(75 — 20)] = 1.71. 10°° Q/m
~ 1107.5 -6- 10-8

The frequency-to-resistance ratio is given by equation (8.97):

_ How _ 22-60-1077
= Soh?
4 R, Al o Oe

With the sheath temperature assumed to be 75°C, the component of the loss factor caused by the
conductor current in the same cable is obtained from equation (8.111) as

4rw 4n -2n - 60
Bi = = = = INKops
10’¢ 10’ - 2.84(1 + 0.00403 - 55) - 10-8

MM= Res
—Mi (Bite) -10°?oved-ih-
= 107
———_...] . (116.8-6)*
———_—... 1 0° ®=
PTS Cea 1.26- 10-5 12 a ee
Theoutsidediameterof theequivalentsheathis givenby
Doe# Dj 102+ 113
D, = ~—— +1, =———_ + 6 = 113.5mm
2 2
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 187

The correction factor for thick sheaths is given by equation (8.113) as

e=1+(4) i 1.74
(41D, 10-16) =1+ (72.6 1.74
(129.2. 113.5.10°?—1.6)= 1.08

The approximate equations (8.116) for eddy current loss factor yield

3 (=)d\* m?
1+m? (=)TOPS 29?
1+ 2.2? pO
NO ee = 6 | ee)

A, = 0.86m?8 (=)d 1.4m+0.7= 0/86 2.2°° (5)107. 1.4-2,2+0,7

= 0.002
2 - 500
A>=1()

AY= R,
— |] eA +A (Bit;)*
———. 19°”
| R E Od ss 4M)ar AND)
ae 5) 10

ee We [1.08- 0.0575- (1+ 0.002 OL027 FON

pp nip eae: -(1 + 0.002)]+ 0. = (0),

Since the lengths of the minor sections are unknown, the value of the circulating current loss is
assigned by equation (8.65), that is, 4, = 0.03, and the total sheath loss factor is equal to

vie Ose 003i Ones

8.4.6 Two-core Unarmored Cables with Common Sheath

The equations used for the calculation of eddy current losses in two-core cable with
round or oval conductors are approximations made by Whitehead (1939) of the equations
developed by Carter (1927). The equations for sector-shaped conductors are further devel-
opments of Carter’s work.
For two-core cables with round or oval conductors, the loss factor is given by

pe saat eyf eg(5) (8.119)

and for sector-shaped conductors,

es 10.82
-107!6
/ 1.48r,
ee+t\? (e487,
tN
where c = distance between the axis of one conductor and the axis of the cable, mm

t = insulation thickness between the conductors, mm

r, = radius of the circle circumscribing the two sector-shaped conductors, mm

d = mean diameter of the sheath, mm; for oval sheaths, d = /dy - dj, where
dy and d,, denote the major and minor diameters of the sheath; for
corrugated sheaths, d = (Doc + Dj; /2) where Do is the external diameter
over the crests of the sheath and Dj, is the internal diameter of the roots
of the corrugations.
188 Part II » Evaluation of Parameters

8.4.7 Three-core Unarmored Cables with a Common Sheath

The equations used for the calculation of eddy current losses in three-core cable with
round or oval conductors are based on the equations developed by Carter (1927). For higher
resistance sheaths, the equations are those developed by Whitehead (1939). For lower
resistance sheaths, the equation has been extended by one term in the series to improve
accuracy. The equations for sector-shaped conductors are further developments of Carter’s
work.
For three-core cables with round or oval conductors and with a sheath resistance
< 100uQ/m, the loss factor is given by

—_.+(%) 2c\47 1re (8.121)

lee R,107
\* d L+4 R,10’\*
wW
@W
For round or oval conductors and with a sheath resistance greater than 100;.02/m, the
loss factor is given by

ee Bi De\
(=) 10-14 (8.122)

Forsector-shaped
conductors
atanyvalueofR,,
P
aoa Rat ee \aBS l eee (8.123)
R d 1+(=RANE)
@
where —__r;
= radius of the circle circumscribing the three sector-shaped conductors, mm
~ insulation thickness between the conductors, mm
d = mean diameter of the sheath, mm.

8.4.8 Two-core and Three-core Cables with Steel Tape Armor

The addition of a magnetic tape armor to multicore cables will increase the eddy current
losses in the sheath. The equation which has been developed for the increase is only known
to apply to tapes between 0.3 and 1.0 mm in thickness and installed in a single layer. The
factor F, by which the eddy current loss factor calculated from the above formulas must be
multiplied is given by
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 189

so that

Ne = Ay (8.124)

where dq = mean diameter of armor, mm

4 = relative permeability of the steel tape (usually taken as 300)
69 = equivalent thickness of the armor = A,/2d,, mm
Aq = cross-sectional area of armor, mm*.

8.4.9 Armored Three-core Cables with Each Core Having

a Separate Lead Sheath (SL Type)
For this type of cable, the eddy current losses are zero and sheath losses are restricted
to circulating current losses. The calculation of circulating current losses is covered in
Section 8.3.2.6.

8.4.10 Effect of Large Segmental-type Conductors

In this section, we consider cables with large conductors of insulated segmental con-
struction bonded at both ends. Where the conductor is designed to have a reduced proximity
effect, such as in the cables considered in this section, sheath eddy current losses must be
considered in addition to circulating current losses. When cable sheaths are single-point
bonded, eddy currents losses are considered to be due only to the electromagnetic effects
of the conductor currents. In solidly bonded systems, sheath eddy current losses are a
function of both conductor currents and sheath circulating currents. Accurate calculation
of eddy current losses under these conditions is extremely complex and unnecessary when
it is considered that the eddy current losses have only a small effect on the current rating
of solidly bonded single-core cables. A suitable approximation for the manner in which
circulating currents modify the eddy current loss was developed by Miller (1929). This
approximation is used in IEC 287 where the eddy current losses are calculated by assuming
only conductor currents, and then are multiplied by a factor F given by the formula (IEC
287,1982)
4M?N?
(i pec ee + (M +N)?
acne (8/125)
A(M*= 1)(N*+1)
where
MN Ry
xX forcablesintrefoilformation
and
pees
rolNi
R forcablesinflatformation
withequidistant
spacing
Nees :
Dn
190 Part II » Evaluation of Parameters

If the spacing along a section is not constant, the value of X is calculated from equa-
tion (8.63).

8.5 GENERAL METHOD OF COMPUTATION OF JOULE LOSS FACTORS

USING FILAMENT HEAT SOURCE SIMULATION METHOD

A general configuration for a closely spaced cable system with random separations is illus-
trated in Fig. 8.10.

@©
©) Figure 8.10 General system of n parallel con-
ductors.

Such systems can conveniently be analyzed with the application of the filament heat
source simulation (f.h.s.s.) method (Ford, 1970). We will use, again, the term conductor
to denote any metallic component of the cable. Applying this method, the conductors are
replaced by a large number of smaller cylindrical subconductors or filaments. The number
of filaments should be large enough so that the current density can be assumed uniform
throughout each filament cross section. The size of the filaments is calculated such that the
sum total cross-sectional area of the filaments equals the total conductor cross-sectional area.
Helically wound wires are replaced by tubes with equivalent resistances. The longitudinal
magnetic fluxes are not indicated in the presentation below, but are included in the computer
software mentioned earlier (Anders et al., 1990).
The expressions describing the electrical connections of the filaments are as follows:

P= Be (8.126)

E = B'.E‘ (8.127)

where the superscript ¢ denotes transpositions and

I = [7;, 15,--- , Iyc]' = column vector of total conductor currents

I= [/,, h,--- , J,]' = column vector of filament currents
Ef = [E{, E5,-+- , Eyc]' = column vector of conductor longitudinal
voltage drop
E = [E), E2,--- , En)! = column vector of filament longitudinal
voltage drop
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 191

‘i LaeOy 5094 3y Sane(60 OAR EO

One i 4 1 © . @
0 1

B= ‘ ‘ = connection matrix

; 0
0 1

where NC = number of conductors

n = number of filaments

c = superscript referring to conductor quantities

The value of NC will depend on the number of cables per phase and on whether or
not the cables have metallic armor or sheath. The presence of neutral cables and earth
conductors will further increase NC.
We can see from equations (8.126) and (8.127), the connection matrix B is such that
the sum of the filament currents in each conductor equals the total conductor current, and
the longitudinal voltage drops in each filament of a conductor are equal to the conductor
longitudinal voltage drop.
Our aim is to express filament currents as a function of phase conductor currents and
the geometry of the system. The longitudinal voltage drop in a filament is given by equation
(8.70), and is repeated here with an inclusion of the relative permeability of the material
and the assumption of tubular sheath and armor:

E=(Ri+j—“G].1 (8.128)
20
where R, =n Xn matrix with vector [R,, Ro,--- , R,] of filament
resistances in the diagonal and zeros outside the diagonal.
[to = 4m - 107’ = permeability of air, H/m
4, = relative permeability of the material (4, = 1 for
nonmagnetic materials)

1 1 1
In— In— ... In—
S11 S12 Sin

G=

1 i!
In — In —
Sn] Snn

where _ 5;; = geometric mean radius of filament, m

sij = geometric mean distance between filaments i and 7, m

Since both E and I are complex quantities, their separate components must be deter-
mined. In effect, therefore, there are 2n equations and 4n unknown quantities to be found.
192 Part II m Evaluation of Parameters

By equating E in equations (8.128) and (8.127), we obtain

B'ES
=[Ra
+j<*G]
I
20
or
1=[Ri+js“c]i BES (8.129
Digs
where j4 = [lo/,. Premultiplying by B and substituting for B - I in equation (8.128), we
obtain
‘=B [Ra“ js*6]efBES
2m
or (8.130
—1

Ee
=E[Ru
+j<"6]
B'|Ic
—1
20
Premultiplying the last equation by B' and substituting for B'E* in equation (8.129), we
obtain

<I
r=|RI+7-—G|
BB[Ry Baa
+56] B'|Ic (8.13
For systems where all the conductor currents are known, evaluation of the above
equations represents the required solution. For systems in which total conductor currents
are not known, calculations must be performed to determine the unknown values of the
currents.
Equations (8.131) can now be used to determine the sheath and armor loss factors by
suitably specifying the matrix boundary conditions. If the sheaths are solidly bonded, the
sheath and armor filaments are solidly bonded. Equation (8.131) yields both the circulating
and approximate eddy currents after observing the boundary conditions that the voltage
drops in all sheath filaments are equal and the sum of all sheath filament currents is zero. If
the sheaths are bonded at one end only, the filaments representing the sheath of each cable
are bonded together, but not those belonging to different cables. The boundary conditions
now require the sum of sheath filament currents in each cable be equal to zero. Thus, the
eddy currents and standing voltages are computed from equations (8.131) and (8.128).
For solidly bonded systems, for example, we proceed as follows. Let us suppose that
the first i — | entries in the vector I® represent known cable conductor currents. From
Kirchhoff’s first and second laws, we have

: ; (8.132)
E; = E;,, =-++:= Eyc = Eo = aconstant
where Eo is the sheath longitudinal voltage drop. Combining equations (8.130) and (8.132)
and letting

C=[B[Re
+S46)w']
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 193

weobtain

Ey Cris gCt2 se Gis I;

. i

ES Ie
Ne ea eet (8.133)

Eo Cnc Cne.nc :
0 1S OR Blk wpinyesy ih

This constitutes a set of NC+ 1 equations in NC+ 1 unknowns. Some reduction in

computational effort can be obtained by noting that the longitudinal voltage drops of the
central conductors are not of interest (see Section 8.3.5.9).
Equations similar to (8.133) can be set up for single-point bonded systems. The sheath
voltages in this case are different and are to be computed.
The loss factor for a particular conductor (a sheath, armor, or pipe) composed of
filaments k to m is equal to

YOUR
ee pe
ieee
i=k (8.134)
DOWER,
j
where / is the index of central conductor filaments belonging to the same cable as the sheath
or armor. The currents represent the rms values.

REFERENCES

ANSI/IEEE Standard 575 (1988), “Application of sheath-bonding methods for single con-
ductor cables and the calculation of induced voltages and currents in cable sheaths.”
Anders, G. J., Moshref, A., and Roiz, J. (July 1990), “Advanced computer programs
for power cable ampacity calculations,’ IEEE Comput. Appl. in Power, vol. 3, no. 3,
pp. 42-45 (1996 revision).
Arnold, A. H. M. (1929), “Theory of sheath losses in single-conductor lead-covered cables,”
J. IEE, no. 67, pp. 69-89.
Arnold, A. H. M. (1941), “Eddy current losses in multi-core paper-insulated lead-covered
cables, armored and unarmored, carrying balanced 3-phase current,” J. JEE, no. 88,
part II, pp. 52-63.
Bianchi, G. and Luoni, G. (1976), “Induced currents and losses in single-core submarine
cables,” JEEE Trans. Power App. Syst., vol. PAS-95, no. 1, pp. 49-58.
Bosone, L. (1931), “Contributo allo studio delle perdite e dell’ autoinduzione dei cavi unipo-
lari armati con fili di ferro,” /’Elettrotecnica, vol. 18, pp. 2-8.
194 Part II » Evaluation of Parameters

Carter, F. W. (1927), “Eddy currents in thin cylinders of uniform conductivity due to peri-
odically changing magnetic fields, in two dimensions,” Proc. Cambridge Philosophical
Soc., vol. 23, pp. 901-906.
Carter, F. W. (1928), “Note on losses in cable sheaths,’ Proc. Cambridge Philosophical
Soc., vol. 24, pp. 65-73.
Dwight, H. B. (1923), “Proximity effect in wires and thin tubes,” AJEE Trans., no. 42,
pp. 850-859.
Ford, G. L. (1970), “Calculation and measurement of current distributions in closely spaced
bus system,” Ontario Hydro Technologies, Private Communication.
Goldenberg, H. (1958), “The calculation of continuous current ratings and rating factors
for transmission and distribution cables,’ ERA Report F/T 187.
Gosland, L. (1940), “Transient sparkover voltages at insulating glands on lead-sheathed
single-core cables—Suggestions for their reduction or elimination,” ERA Report G/T113.
Heyda, P. G., Kitchie, G. E., and Taylor, J. E. (1973), “Computation of eddy current losses
in cable sheaths and bus enclosures,” Proc. IEE, vol. 120, no. 4, pp. 447-452.
IEC 287 (1969), “Calculation of the continuous current rating of cables (100% load factor),”
Ist ed., IEC Publication 287.
IEC 287 (1982), “Calculation of the continuous current rating of cables (100% load factor),”
2nd ed., IEC Publication 287.
IEC 287-1-1 (1994), “Electric cables—calculation of the current rating—Part 1: Current
rating equations (100% load factor) and calculation of losses—Section |: General,” IEC
Publication 287.
Imai, T. (1968), “Exact equation for calculation of sheath proximity loss of single-conductor
cables,” Proc. IEE, vol. 56, no. 7, pp. 1172-1181.
Jackson, R. L. (1975), “Eddy-current losses in unbonded tubes,” Proc. IEE, vol. 122, no. 5,
pp. 551-557.
Jackson, W. D. (1962), Classical Electrodynamics. New York: John Wiley & Sons Inc.,
1962.
Kaniuk, G. (1984), “Calculation of sheath eddy current losses for a double circuit installa-
tion,’ ERA Report 84-0189.
Kawasaki, K., Inami, M., and Ishikawa, T. (1981), “Theoretical considerations of eddy
current losses in non-magnetic and magnetic pipes for power transmission systems,”
IEEE Trans. Power App. Syst., vol. PAS-100, no. 2, pp. 474-484.
King, S. Y., and Halfter, N. A. (1983), Underground Power Cables. London: Longman.
Meyerhoff, L. (1952), “A.C. resistance of pipe-cable systems with segmental conductors,”
AIEE Trans., vol. 71, part II, pp. 393-414.
Miller, K. W. (1929), “Sheath currents, sheath losses, induced sheath voltages and apparent
conductor impedances of metal sheathed cables carrying alternating currents,” Thesis in
Electrical Engineering, University of Illinois, Urbana, IL.
Morgan, P. D., Wedmore, S., and Whitehead, E. B. (1927), “A critical study of a three-phase
system of unarmored single-conductor cables, from the standpoint of the power losses,
line constants and interference with communication circuits,” ERA Report F/T22.
Chapter 8 m Joule Losses in Screens, Sheaths, Armor, and Pipes 195

Morello, A. (1959), “Calculation of the current ratings for power cables,” /’Elettrotecnica,
vol. 46, pp. 2-17.
Neher, J. H., and McGrath, M. H. (1957), “The calculation of the temperature rise and load
capability of cable systems,” AJEE Trans., vol. 76, part III, pp. 752-772.
Parr, R. G. (1979), “Formulae for eddy current loss factors in single-point or cross-bonded
cable sheaths,” ERA Report 79-97.
Parr, R. G., and Coates, M. W. (1988), “Current sharing between armored single core cables
in parallel,’ ERA Report 88-0393.
Schelkunoff, S.A. (1934), “The electromagnetic theory of coaxial transmission lines and
cylindrical shields,” Bell Syst. Tech. J., pp. 532-575.
Schurig, O. R., Kuehni, H. P., and Buller, F. H. (1929), “Losses in armored single-conductor
lead-covered A.C. cables,” AJEE Trans., vol. 48, pp. 417-435.
Silver, D. A., and Seman, G. W. (1982), “Investigation of A.C./D.C. resistance ratios of
various designs of pipe-type cable systems,” JEEE Trans. Power App. Syst., vol. PAS-101,
no. 9, pp. 3481-3497.
Weedy, B. M. (1986), “Prediction of return currents and losses in underwater single-core
armored AC cables with large spacings,” Elec. Power Syst. Res., vol. 10, pp. 77-85.
Whitehead, S., and Hutchings, E. E. (1939), “Current rating of cables for transmission and
distribution,’ ERA Report F/T131.
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Thermal Resistances
and Capacitances

9.1 INTRODUCTION

In this chapter, we will discuss the computation of thermal resistances and capacitances
associated with the components and the environment of cables. The internal thermal resis-
tances and capacitances are characteristics of a given cable construction and were defined
in Section 3.2. Without loss of accuracy, we will assume that these quantities are constant
and independent of the component temperature. Where screening layers are present, we
will also assume that for thermal calculations, metallic tapes are part of the conductor or
sheath, while semiconducting layers (including metallized carbon paper tapes) are part of
insulation.
We recall from Chapter 4 that the current rating of a cable is a function of the fol-
lowing thermal resistances: 1) 7; = thermal resistance between conductor and sheath, 2)
T) = thermal resistance between sheath and armor, 3) 7; = thermal resistance of external
covering, and 4) 7, = thermal resistance of cable external environment. The units of the
thermal resistance are K/W for a specified length. Since the length considered here is | m,
the thermal resistance of a cable component is expressed in K/W per meter, which is most
often written as K - m/W. This should be distinguished from the unit of thermal resistivity
which is also expressed as K - m/W. In transient computations discussed in Chapter 5,
thermal capacitances associated with the same parts of the cable were identified. The unit
of thermal capacitance is J/K - m and the unit of the specific heat of the material is J/K-
m°. The thermal resistivities and specific heats of materials used for insulation and for
protective coverings are given in Table 9.1.
The thermal resistances of the insulation and the external environment of a cable have
the greatest influence on cable rating. In fact, for the majority of buried cables, the external

197
198
Part II » Evaluation of Parameters

TABLE 9.1 Thermal Resistivities and Capacities of Materials

(Based on IEC 287-2-1, 1994)
i S
Thermal Specific
Resistivity Heat
(p) (c: 10°)
Material (K-m/W) [J/(m?-K)]
nn EEE
Insulating materials”
Paper insulation in solid type cables 6.0 2.0
Paper insulation in fluid-filled cables 5.0 2.0
Paper insulation in cables with external gas pressure 5.5 2.0
Paper insulation in cables with internal gas pressure
preimpregnated 6.5 2.0
mass impregnated 6.0 2.0
PE 3:5 2.4
XLPE 3.5 2.4
Polyvinyl chloride
up to and including 3 kV cables 5.0 1.7
greater than 3 kV cables 6.0 ilae/
EPR
up to and including 3 kV cables 3)9) 2.0
greater than 3 kV cables 5.0 2.0
Butyl rubber 5.0 2.0
Rubber 5.0 2.0
Paper-polypropylene-paper (PPL) 6.5 2.0
Protective coverings
Compounded jute and fibrous materials 6.0 2.0
Rubber sandwich protection 6.0 2.0
Polychloroprene 5.5 2.0
PY C
up to and including 35 kV cables 5.0 U7
greater than 35 kV cables 6.0 1)
PVC/bitumen on corrugated aluminum sheaths 6.0 1.7
PE 3 2.4
Material for duct installations
Concrete 1.0 1.9
Fiber 4.8 2.0
Asbestos 2.0 2.0
Earthenware lipo? re?
PVC 6.0 te)
PE 3:5) 2.4

"For the purpose of current rating computations, semiconducting screening materials

are assumed to have the same thermal properties as the adjacent dielectric materials.

thermal resistance accounts for more than 70% of the temperature rise of the conductor.
For cables in air, the external thermal resistance has a smaller effect on cable rating than in
the case of buried cables.
The calculation of thermal resistances of the internal components of cables for single-
core cables, whether based on rigorous mathematical computations or empirical investi-
gations, is straightforward. For three-core cables, the calculations are somewhat more
involved. Also, the calculation of the external thermal resistance requires particular atten-
tion. These topics are discussed next.
Chapter 9 m Thermal Resistances and Capacitances 199

9.2 THERMAL RESISTANCE BETWEEN ONE CONDUCTOR AND SHEATH 7,

9.2.1 Single-core Cables

The thermal resistance between one conductor and the sheath is computed from equa-
tion (3.3):

1 anp n( ae 2t1
d ) (9.1)
T; = —In{ 1+ —

where p = thermal resistivity of insulation, K - m/W

d, = diameter of conductor, mm

t; = thickness of insulation between conductor and sheath, mm

T, = thermal resistance of the insulation, K - m/W.

For corrugated sheaths, t; is based on the mean internal diameterof the sheath which
is equal to
(= ts Dec )
——SS Ue
2
The dimensions of the cable occur in

n(a 2t}
Injil+—)},
and therefore this expression plays the role of a geometric factor or shape modulus, and has
been defined as the geometric factor.

EXAMPLE 9.1
We will compute the value of 7; for model cable No. 5. The required parameters of this circuit are
taken from Appendix A: d, = 58 mm, D,. = 113 mm, D;,; = 102 mm, and t, = 6 mm.
First, we compute the mean internal diameter of the sheath:

_ Dit Doo, _ 13+102

d a= —6= 101.5 mm
2) 2
From equation (9.1), we obtain

beipain(+7)
Tipe ewes
(alte Onn(58
Vere 101.5
geenye
=|=0.579
Kee
as
9.2.2 Three-core Cables

9.2.2.1 Overview. The computation of the internal thermal resistance of three-core

cables is more complicated than for the single-core case. Rigorous mathematical formulas
cannot be determined, although mathematical expressions to fit the conditions have been
derived either experimentally or numerically. The general method of computation employs
geometric factor (G) in place of the logarithmic term in equation (9.1); that is,

(oz)
200 Part II m»Evaluation of Parameters

Several methods of determining such factors have been devised. The first paper on
this subject dates from 1905 (Mie, 1905). Further work on determining the value of 7
for various types of three-core cables was done by Russel (1914), Atkinson (1924), and
Simmons (1923, 1932). The values of the geometric factor published in IEC 287 are based
on empirical investigations carried out at E.R.A. in the United Kingdom in the 1930s. These
values are, in turn, based on measurements of electrical resistance performed in a series
of tests on models comprising copper-tube electrodes soldered to resistance-alloy sheets
to represent the dielectric. The results were further checked using the graphical method
devised by Wedmore (E.R.A., 1923) and Simmons (1923, 1932).
The difficulty in analytically solving the problem of a three-core cable has been over-
come by employing numerical methods such as the integral-equation method, the filament
heat source simulation (f.h.s.s.) method (King and Halfter, 1982), and the finite-element
method (Van Geertruyden, 1994; Anders, et al. 1997). By the integral-equation method, the
actual conductor surfaces, including the sheath, are regarded as boundaries. The thermal
field within the domain defined by the boundaries, which, in this case, is the region of cable
insulation, is then formulated and solved by numerical means.
In the f.h.s.s. method, instead of taking the conductor surfaces as boundaries, the
surfaces are simulated with the aid of filament heat sources (see also Section 8.5). This
approach is similar to Mie’s method (Mie, 1905), but a larger number of heat sources is
employed. Once the positions and magnitudes of the heat sources are fixed, the thermal
field can be determined.
In the finite-element method, the thermal resistance of the insulation is computed
directly, assuming that the conductor and sheath boundaries are isothermal. With given
losses dissipated by each conductor per unit length W.. (W/m) and the temperature 6, of the
sheath, the finite-element method (see Chapter 11) is used to compute the temperature of
the conductors 6,. The value of 7, is then obtained from

ai 0. =a0s
T, (9.3)
W.

The values of the geometric factor for various cable constructions as standardized
in IEC Publication 287-2-1 (1994), as well as those obtained by numerical methods, are
discussed in the following sections.

9.2.2.2 Two-core Belted Cables with Circular Conductors. The curves defining
the value of the geometric factor are shown in Fig. 9.1 and analytical expressions fitting the
curves are given in Appendix Cl.

9.2.2.3 Three-core Belted Cables with Circular and Oval Conductors. The curves
defining the value of the geometric factor are shown in Fig. 9.2 and analytical expressions
fitting the curves are given in Appendix C2.
Simmons (1923) also proposed the following empirical formula to evaluate the geo-
metric factor for each core of a three-core belted cable:

o=foasooa(of(er-@))()u] 6s
where f and ft; (mm) are the thicknesses of the insulation between conductors and between
one conductor and the sheath, respectively.
Chapter 9 m=Thermal Resistances and Capacitances 201

G
3.0

7215)

2.0

1.5 t = thickness of insulation

between conductors

t1 = thickness of insulation
between conductor
and sheath

dc = diameter of conductor
(circular)

Figure 9.1 Geometric factor for two-core belted

cables with circular conductors (IEC 287-2-1,
1994).

King and Halfter (1982) developed the empirical formulas given in Table 9.2 using an
integral-equation approach.
Cables with oval conductors are treated as cables with an equivalent circular conductor
with an equivalent diameter d. = /d-udcem (mm) where d.y and d.», are the major and
minor diameters of the oval conductor, respectively.

9.2.2.4 Three-core Cables with Circular Conductors and Extruded Insulation. To-
day’s standards are based on work performed by Simmons (1932) and Whitehead and
Hutchings (1938). Over 60 years have passed since the method for computation of 7;
used in today’s standards was developed. Since then, many new insulating materials have
appeared in three-core cable constructions; cross-linked polyethylene (XLPE) insulation in
particular has been a material of choice in newer three-core cable designs. All three-core
cables require fillers to fill the space between insulated cores and the belt insulation or a
sheath. In the past, when impregnated paper was used to insulate the conductors, the resis-
tivity of the filler material very closely matched that of the paper (around 6 K - m/W). With
polyethylene insulation having much lower thermal resistivity (3.5 K - m/W), the higher
thermal resistivity of the filler may have a significant influence on the overall value of 7,
and hence on the cable rating. To give an indication of the effect of the resistivity on the
internal thermal resistance of the cable, finite-element studies were conducted for a range of
values of the thermal resistivities of the insulation and the filler. From these investigations,
 Part I] » Evaluation of Parameters

t = thickness of insulation
between conductors

t1 = thickness of insulation
between conductor
and sheath

dc = diameter of conductor
(circular)

Figure 9.2 Geometric factor for three-core belted

cables with circular conductors (IEC 287-2-1,
1994).

TABLE 9.2 Geometric Factor for Three-Core Unscreened Cables

Cable GeometricFactor—
Configuration EmpiricalFormula
(2t,-t)/t for 0.5 <x <2 with t,/d.=x

0.332 + 3.029x —1.954x2 + 0.774x7 —0.126x4

oon
Nn
0.344 + 2.985x —2.028x2 + 0.827x° —0.137x4
0.358 + 2.914x —2.13x + 0.91329 —0.157x4

thefollowingapproximating
formulaemerged(Andersefal., 1997):
rile —6 + 0.031(py —pj)e%™/% (9.5)
;
where py and p; are the thermal resistivities of filler and insulation, respectively, and G is
the geometric factor obtained from Fig. 9.2 assuming pr = (;.
When the values of 7|, computed from equation (9.5) and obtained using the finite-
element method for a wide range of cable designs were plotted and regression analysis was
performed, the regression curve had a slope of 0.998 with a standard error of 0.8% (Anders
et al., 1997).
Chapter 9 m Thermal Resistances and Capacitances 203

1.6
° Filler TR = 10

12 Filler TR =8

o Filler TR=7

(Kem/W)
T, ; Filler TR =5

0.8

0.6

0.4L:
Onis

Figure 9.3 7) as a function of filler thermal resistivity for various cable parameters.

Figure 9.3 shows the relationships between 7; and the ratio t, /d, for various values
ofpr.
Equation (9.5) was developed for unscreened cable constructions. Modern three-core
cables usually have a metallic screen around each core. However, in ampacity computations
for such cables, a two-step procedure is followed: first, the value of T; for unscreened cable
is computed [e.g., using equation (9.2) or (9.5)]; then, this resistance is modified as described
in Section 9.2.2.6.

EXAMPLE 9.2
In order to demonstrate the effect of the thermal resistivity of the filler on cable ampacity, we will
consider modified cable model No. 2, ignoring the presence of the metallic screen around each core
and adding a metallic sheath over the laid-up cores. The parameters of this cable are as follows:

conductor diameter d, = 20.5 mm

diameter over conductor shield Ga) eamin
insulation thickness t, = 3.4mm
diameter over semiconducting insulation screen Dj;, = 30.1 mm
diameter over the three cores D3, = 64.7 mm
diameter over the lead sheath D, = 68.7 mm
diameter over the PVC jacket D, = 74.7 mm

The cable is installed in air, away from a wall, with no direct solar radiation. The ambient air
temperature is 25°C.
With the above parameters, the following values of thermal resistances and loss factors are
obtained using standard calculation procedures:
204 Part II » Evaluation of Parameters

thermal resistance of the insulation T; = 0.518 K - m/W

thermal resistance of the jacket T; = 0.067 K - m/W

external thermal resistance T; = 0.346 K - m/W
conductor resistance at 90°C (60 Hz) R = 0.798 - 10-* Q/m
sheath loss factor 0026

Computing the thermal resistance of the insulation, the thicknesses of semiconducting screens
over the conductor and over the insulation are added to the thickness of the insulation.
We will vary the thermal resistivity of the filler between 3.5 and 10 K - m/W. The values of 7;
are summarized in Table 9.3 and a sample computation for p¢ = 6 K - m/W is given below.

TABLE 9.3. Thermal Resistance of the Insulation and the Rating

of the Three-Core Cable Examined in Example 9.2

py (K-m/W) She) 6 8 10

T\(K-m/W) 0.518 0.609 0.666 0.733

I(A) 675 658 648 638

From Fig. 9.2, G= 0.93 and the thermal resistance of the insulation is obtained from equation
(9.5) as

i BE) :
T= ate + 0.031(oy —p)) e997" = 50.93 + 0.031(6—3.5)e0°™48/5= 0.609 K -m/W
IT ‘ a
The rating of the cable is obtained from equation (4.3) as

pea (6. = Oamb)— Wa [0.5T) +n (Tz + Ts + Ts)] e

RT; +nR (1 +A\) Io +nR(1 +A) + A2) + M4)

For the cable under consideration, we have values of 7; = 0, Wy = 0, and A> = 0. Hence,

fee
FPAoem,-10-4[0.609
+3-1.026
BOofteSheet-(0.067
+al
abetted 9025
0.5 Bratt

The effect ofthe filler resistivity is quite noticeable. In this example, for the highest value of p;,
the ampacity is reduced by about 6% in comparison with the value computed from equation (9.2).
Finally, we will compare the value of the geometric factor obtained from Fig. 9.2 with the
values given by Simmons [equation (9.4)] and by King and Halfter (Table 9.2). With the ratio of the
insulation thickness between conductor and sheath to the insulation thickness between conductors
equal to t;/t = 0.5, the geometric factor obtained from equation (9.4) is equal to

(hejo.+02(2+ei')|In|(s32.2(=ra'))(+)»:1
t t d.
=[0.85
+.0.2(2-0.5—
1]Intc ~2.2(2-0.5—1))
(Ss)a1~0.92
4.8
From Table 9.2, for x = 4.8/20.5 = 0.234, we obtain

G = 0.358 + 2.914x — 2.13x? + 0.913x? —0.157x4

= 0.358 + 2.914 - 0.234 — 2.13 - 0.2347 + 0.913 - 0.2343 — 0.157 - 0.2344 = 0.93
Chapter 9 m Thermal Resistances and Capacitances 205

The values obtained from equation (9.4), Table 9.2, and Fig. 9.2 are identical, and they also agree
with the computations performed using the finite-element method.

9.2.2.5 Shaped Conductors. The use of shaped conductors reduces the thermal
resistance of a cable, the precise effect depending on the configuration of the conductor.
Some tests on this aspect were carried by Atkinson, and sector correction factors published
(Atkinson, 1924). Later tests carried out by E.R.A. covered twin-, three-, and four-core
models having a wide range of dimensions. The results of these tests are used in today’s
standards.
9.2.2.5.1 TWO-CORE BELTED CABLES WITH SECTOR-SHAPED CONDUCTORS. The ge-
ometric factor is given by

=Fin(s*)
G=F,1da (9.6)
9.6
4.4t
Where F,; =2+

d, = external diameter of the belt insulation, mm

r,; = radius of the circle circumscribing the conductors, mm

d, = diameter of a circular conductor having the same cross-sectional area

and degree of compaction as the shaped one, mm
t = insulation thickness between conductors, mm

9.2.2.5.2 THREE-COREBELTEDCABLESWITHSECTOR-SHAPED CONDUCTORS. The

geometric factor is given by equation (9.6) with the coefficient F; defined as

Or
[ieee ui Sanne
i seks 9.7
Qn(d,+t) —t ae

King and Halfter (1982) developed the equations for the geometric factor using the
f.h.s.s. method. These are shown in Table 9.4.
When compared with the case of round conductors, a reduction of about 15% in
geometric factor in the case of (2t, — t)/t = 0.5 is observed.

TABLE 9.4 Geometric Factor for Three-Core Unscreened Cables

with Sector-Shaped Conductors

Cable Geometric Factor—

Configuration Empirical Formula
(2t,-t)/t for 0.5 <x <2 with t)/de =x

| 0.2 + 2.53x —1.032x2 + 0.183x°

0.5 0.2 + 2.58x —1.248x2 + 0.3x° + 0.0163x4
0 0.23 + 2.529x —1.5472? + 0.599x> —0.098x4
i
206 Part II » Evaluation of Parameters

9.2.2.6 Three-core Cables with Metal Screens Around Each Core.

9.2.2.6.1 CIRCULAROR OVAL CONDUCTORS. Screening reduces the thermal resis-

tance of a cable by providing additional heat paths along the screening material of high
thermal conductivity, in parallel with the path through the dielectric. The thermal resis-
tance of the insulation is thus obtained in two steps. First, the cables of this type are
considered as belted cables for which rt;/t = 0.5. Then, in order to take account of the ther-
mal conductivity of the metallic screens, the results are multiplied by a factor K, called the
screening factor, values of which are obtained from Fig. 9.4, and the analytical expressions
fitting the curves are given in Appendix C3. Thus, we have

T,=K£-G (9.8)
20

Screening
factor K
1.0 8, = thickness of metallic screen
on core

Pz; = thermal resistivity of insulation

d, = diameter of conductor(circular)
0.9 t, = thickness of insulation between
conductor and screen

Pp, = thermal resistivity of screening

material:
27 x 10 K.m/W for copper
08 48 x 10° K.m/W for aluminum

0.7

0.6

34 XPT
0 Sar lnlOi| 15° 820 “258 TS02
cXPm

Figure 9.4 Thermal resistance of three-core screened cables with circular conductors as
compare to that of a corresponding unscreened cable (IEC 287-2-1, 1994).

The values of factor K were obtained in the experiments carried out at E.R.A. (White-
head and Hutchings, 1938). The models previously described were provided with an ad-
Chapter 9 m Thermal Resistances and Capacitances 207

ditional copper strip soldered to the resistance alloy and surrounding the core electrode to
simulate the screen. The screen was so dimensioned as to have the correct relative conduc-
tance as determined by the parameter tp; /(27rp,,.), where t is the thickness of the screen, r
the radius or equivalent radius of conductor, p; the thermal resistivity of the dielectric, and
sc the thermal resistivity of the screen material.
As discussed in Section 9.2.2.4, filler resistivity may have a significant influence on
the value of 7; for plastic-insulated cables. At present, an equation similar to (9.5) is being
developed for screened cables with fillers. However, until this work is completed, it is
recommended that the thermal resistance of such cables be first computed from equation
(9.5) and then multiplied by the screening factor K obtained from Fig. 9.4.

EXAMPLE 9.3
We will compute the value of 7, for model cable No. 2.
First, we determine the value of the screening factor. For this cable, t,/d. = 4.8/20.5 = 0.234
and the ratio 5; - p;/(d- - Pm) = 0.2 - 3.5/(20.5 - 27. 10-4) = 12.65. From Fig. 9.4, we obtain
iS = O59)
To obtain the geometric factor G, we first assume t,/t = 0.5, and from Fig. 9.2, G= 0.93. The
thermal resistance of the insulation is obtained from equation (9.8):

a5
Te KeenG= 0:59="0,94= 0.306ReaalW
20 Qn

This value is somewhat different from the one given in Table Al because the latter was computed
with the aid of formulas in Appendix C.

Cables with oval conductors are treated as an equivalent circular conductor with an
equivalent diameter d. = /d-mdem-
9.2.2.6.2 SHAPEDCONDUCTORS. For these cables, 7) is calculated the same way as
for belted cables with sector-shaped conductors, but d, is taken as the diameter of a circle
which circumscribes the core assembly. The result is multiplied by a screening factor given
in Fig. 9.5 or computed as described in Appendix C4.

9.2.2.7 Fluid-filled Cables.

9.2.2.7.1 THREE-CORE CABLES WITH CIRCULAR CONDUCTORS AND METALLIZED
PAPER CORE SCREENS AND CIRCULAR OIL DUCTS BETWEEN THE CoRES. In cables of
this construction, oil ducts are provided by laying up an open spiral duct of metal strip in
each filler space as shown in Fig. 9.6.
The expression for the thermal resistance between one core and the sheath was obtained
experimentally:

2t;

where ¢; (mm) is the thickness of core insulation including carbon black and metallized
paper tapes plus half of any nonmetallic tapes over the three laid-up cores. Equation (9.9)
 Part I. » Evaluation of Parameters
208

Screening
factor K
1.0
8, = thicknessof screen
Pz = thermal resistivity of insulation
d, = diameter of circularconductor
havingthe same section and
0.9 compaction
t, = thickness of insulation between
1 conductor and screen

P,, = thermal resistivity of screening

material:
08 27 x 10° K.m/W for copper
‘ 48 x 10° K.m/W for aluminum

OR

0.6

he 34 XPr
0 5 10 15 20 25 30
dy X Pry

Figure 9.5 Thermal resistance of three-core screened cables with sector-shaped conductors
as compared to that of a corresponding unscreened cable (IEC 287-2-1, 1994).

Outer jacket
Insulation

Sheath

Oil duct

Conductor
Figure 9.6 Three-core oil-filled cable.

assumes that the space occupied by the metal ducts and the oil inside them has a very high
thermal conductance compared with that of the insulation; the equation therefore applies
irrespective of the metal used to form the duct or its thickness.
9.2.2.7.2 THREE-CORE CABLES WITH CIRCULAR CONDUCTORS AND METAL TAPE CORE
SCREENS AND CIRCULAR OIL DUCTS BETWEEN THE CORES. The thermal resistance be-
Chapter 9 m Thermal Resistances and Capacitances 209

tween one conductor and the sheath is

|T,=0.35p
p((0.923
—pee
=
2t;) (9.10)
where ¢; (mm) is the thickness of core insulation including the metal screening tapes plus
half of any nonmetallic tapes over the three laid-up cores.
9.2.2.7.3. THREE-CORE CABLES WITH CIRCULAR CONDUCTORS, METAL TAPE CORE
SCREENS, WITHOUT FILLERS AND OIL DUCTS, HAVINGA COPPER WOVEN FABRIC TAPE
BINDING THE CORES TOGETHER AND A CORRUGATEDALUMINUM SHEATH._ Studies carried
out in Germany (Brakelmann er al., 1991) have shown that the air gap between the cable
insulation and corrugated sheath forms additional thermal resistance when the corrugated
sheath is not in direct contact with the underlying layer. The thermal resistance 7; of these
cables was obtained recently in experiments carried out in Britain, and is given by the
following equation (IEC 287-2-1, 1994):

ATS
i=pin
jy es E
(2)N
2 fo n( Deas
d.
rir | ee ) oe
9.11
where
1,105|(Fs)
D;Dee —2.16
|
D, = diameter of a core over its metallic screen tapes, mm

6; = average nominal clearance between the core metallic screen tapes

and the average inside diameter of the sheath, mm.

Equation (9.11) is independent of the metal used for the screen tapes.

9.2.2.8 SL Type Cables. In SL type cables, the lead sheath around each core may be
assumed isothermal. The thermal resistance 7; is calculated from equation (9.1) the same
way as for single-core cables.

9.3 THERMAL RESISTANCE BETWEENSHEATHAND ARMOR 7,

9.3.1 Single-core, Two-core, and Three-core Cables Having a Common Metallic Sheath

The thermal resistance between sheath and armor is obtained from equation (3.3)
representing thermal resistance of any concentric layer. With the notation applicable to this
part of the cable, we have

(S212)

where p = thermal resistivity of the armor bedding, K - m/W

D, = external diameter of the sheath, mm
t> = thickness of the bedding, mm
210 Part I m Evaluation of Parameters

9.3.2 SL TypeCables
In these cables, the thermal resistance of the fillers between sheaths and armoring has
been obtained using the graphical method of Wedmore (1929). The thermal resistance of
fillers and bedding under the armor is given by

Toteé (9.13)

Ge)

0.7
Sheaths touching

0.6

0.5

0.4

0.3

Equal thicknesses of material

between sheaths and between
0.2 sheaths and armor

0.1

0
0 0.05 ye oe Figure 9.7 Geometric factor for obtaining the
Thicknesses of material between sheaths and thermal resistances ofthe filling material between
armor expressed as a fraction of the outer the sheaths and armor of SL-type cables (IEC 287-
diameter of the sheath 1-1, 1994).

9.4 THERMAL RESISTANCE OF OUTER COVERING (SERVING) 7;

The external servings are generally in the form of concentric layers, and the thermal resis-
tance 73 is given by
Chapter 9 m Thermal Resistances and Capacitances 211

n=2in(i+22) (9.14)
where p = thermal resistivity of the serving, K - m/W
D/, = external diameter of the armor, mm; for unarmored cables, D’, is
taken as the external diameter of the component immediately beneath
it, that is, sheath, screen or bedding

t; = thickness of serving, mm

For corrugated sheaths,

p Docap
2
:SA (2s. 2) t3 C2)
2 Tis
9.5 PIPE-TYPE CABLES

A basic construction of a pipe-type cable is shown in Fig. 1.6.

For these three-core cables, the following computational rules apply:
1. The thermal resistance 7) of the insulation of each core between the conductor and
the screen is calculated from equation (9.1) for single-core cables.
2. The thermal resistance 7> is made up of two parts:

(a) The thermal resistance of any serving over the screen or sheath of each core.
The value to be substituted for part of 7> in the rating equation (4.3) is the value
per cable, that is, the value for a three-core cable is one third of the value of a
single core. The value per core is calculated by the method given in Section
9.3.1 for the bedding of single-core cables. For oval cores, the geometric mean
of the major and minor diameters /dyd,, is used in place of the diameter for
a circular core assembly.
(b —The thermal resistance of the gas or liquid between the surface of the cores and
the pipe. This resistance is calculated in the same way as that part Ty which is
between a cable and the internal surface of a duct, as given in Section 9.6.4.1.
The value calculated will be per cable and should be added to the quantity
calculated in a) above before substituting for 7> in the rating equation (4.3).

3. The thermal resistance 7; of any external covering on the pipe is dealt with as in
Section 9.4. The thermal resistance of the metallic pipe itself is negligible.

We will illustrate computation of the thermal resistances of the pipe-type cables in

Example 9.8 after we review computations for cables in ducts.

9.6 EXTERNAL THERMAL RESISTANCE

The current-carrying capability of cables depends to a large extent on the thermal resistance
of the medium surrounding the cable. For a cable laid underground, this resistance accounts
212 Part II » Evaluation of Parameters

for more than 70% of the temperature rise of the conductor. For underground installations,
the external thermal resistance depends on the thermal characteristics of the soil, the diameter
of the cable, the depth of laying, mode of installation (e.g., directly buried, in thermal
backfill, in pipe or duct, etc.), and on the thermal field generated by neighboring cables.
For cables in air, the external thermal resistance has a smaller effect on the cable rating.
For aerial cables, the effect of installation conditions (e.g., indoors or outdoors, proximity
of walls and other cables, etc.) is an important factor in the computation of the external
thermal resistance. In the following sections, we will describe how the external thermal
resistance of buried and aerial cables is computed.

9.6.1 Single Buried Cable

When an analytical expression is sought for the value of the external thermal resistance,
it is necessary to consider the thermal resistivity of the soil as being unaffected by the
temperature which may be attained at various points in the general field. We further idealize
the conditions by assuming that the thermal resistivity is constant throughout the field.
Under this condition, the superposition theorem becomes applicable, that is, the temperature
change existing at any point in the general heat field becomes equal to the sum of the
temperature changes produced at that point by each of the heat fields by itself.
We will first consider a single cable laid directly in a uniform soil. If the diameter of
the cable is small compared with the depth of burial, it will be reasonable to represent the
cable as a filament heat source laid in an infinite medium. Under steady-state conditions,
equation (2.15) now simplifies to

W;=0 (9.16)

The temperature rise at any point M located at a distance d from the center of the cable
is obtained by integrating equation (9.16) between the limits r = co and r = d. Thus,
dm Vs Ps
AOé=] _ W,dr = ——W, Ind (9.17)
Poe 5 hg 20
As explained in Section 5.2.2, in order to avoid the assumption of an infinite uniform
medium, we have to use another hypothesis, namely, the hypothesis of Kennelly referenced
in Chapter 5, which requires the assumption that the earth surface is an isotherm. Under
this hypothesis, the temperature rise at any point M in the soil is, at any time, the sum
of the temperature rises caused by the heat source W, and by its fictitious image placed
symmetrically with respect to the earth surface and emitting heat —W, (see Fig. 9.8). If
these two heat flows operate simultaneously, the resulting temperature is obtained by the
superposition theorem, adding to equation (9.17) a term corresponding to the fictitious heat
source located at the distance d’ from point M (see Fig. 9.8):
/

AG ee AY t Vn
nc AY ,Ind / Se S Win
; In 9
( .18)

If the point M is placed at the surface of the cable, and expressing the values of d and
d' in terms of L and D,, the depth of the center of the cable and its diameter, equation (9.18)
can be written as

Ps
So EA In(2u) (9719)
Chapter 9 » Thermal Resistancesand Capacitances 213

1'-Wt

O
Ow : p
Cable No. 1

Figure 9.8 Illustration of the development of an equation for the external thermal resistance
of a single cable buried under an isothermal plane.

where ip=2L/ Ds and

(Ps = soil thermal resistivity, K - m/W

D,. = external diameter of the cable, mm
L = depth of burial of the center of the cable, mm
W, = total losses inside the cable (W/m).

Equation (9.19) assumes that the heat flow lines emerge from the geometric center of
the cable. Strictly speaking, the heat flow lines terminate at a small distance vertically above
the geometric center of the heat source. The magnitude of this displacement or eccentricity

isgiven
by
EIDE (uSpe
ud &1)
Thus,
equation
(9.19)
takes
theform
(known
asalongform
oftheKennelly
formula)
Age= W,In(unaAe -1) (9.19a)
From equations (9.19a) and (3.6), the external thermal resistance can be obtained as

Me= In(uTW)=1) (9.20)

When the depth of burial is much greater than the cable diameter, u becomes much
greater than one, and equation (9.20) reduces to
 Part I » Evaluation of Parameters
214

ps , 4b
pb liknel
PM pllyeer
tay (9.21)

The magnitude of the error involved in using equation (9.21) in place of equation (9.20)
depends on the value of wuand reaches the high value of 15% for u = 1.5. In practice, u 1s
seldom less than 10, and the short form of the Kennelly formula [equation (9.21)] can be
used.

9.6.2 Groups of Buried Cables (Not Touching)

For several loaded cables placed underground, we must deal with superimposed heat
fields. The principle of superposition is applicable if we assume that each cable acts as
a line source and does not distort the heat field of the other cables. Therefore, in the
following subsections, we will assume that the cables are spaced sufficiently apart so that
this assumption is approximately valid. The axial separation of the cables should be at
least two cable diameters. The case when the superposition principle is not applicable is
discussed in Section 9.6.3.

9.6.2.1 Unequally Loaded Cables. |The method suggested for the calculation for
ratings of a group of cables set apart is to calculate the temperature rise at the surface of
the cable under consideration caused by the other cables of the group, and to subtract this
rise from the value of A@ used in the equation (4.3) for the rated current. An estimate of
the power dissipated per unit length of each cable must be made beforehand, and this can
be subsequently amended as a result of the calculation where it becomes necessary.
From equation (9.18), the temperature rise A@;, at the surface of the cable p produced
by the power W; watt per unit length dissipated in cable k is equal to

Savant ode
Abe = oe W, In 2 (9.22)

The distances of d,, and d/,, are measured from the center of the pth cable to the center
of cable k, and the center of the reflection of cable k in the ground—air surface, respectively
(see Fig. 9.8). Thus, the temperature rise A@, above ambient at the surface of the pth cable,
whose rating is being determined, caused by the power dissipated by the other (q — 1)
cables in the group, is given by

AO, = AOip + AOoy+--+ AOky+--+ + ADgy (9.23)

with the term A@,, excluded from the summation.

The value of A@in equation (4.3) for the rated current is then reduced by the amount
of A@,,,and the rating of the pth cable is determined using a value of 7; corresponding to
an isolated cable at position p. This calculation is performed for all cables in the group,
and is repeated where necessary to avoid the possibility of overheating any of the cables.
Let 4.» denote the external temperature of cable p in isolation. Substituting equation
(9.22) into the right-hand side of equation (9.23) and applying equation (9.20), the following
Chapter 9 » ThermalResistancesand Capacitances 215

general expression for the external thermal resistance of cable p is obtained:

Ta
4 Ben
ee+AB,—
a
W, eenOamb
Ds a ( yee S| Nee
esees
Bt niu+ vu )+ W,aa as
Any
ak.
InDe (9.24)
k#p

EXAMPLE 9.4
Consider a cable model No. 2 and a cable model No. 3 located horizontally 1 m below the ground
and the centers spaced 50 cm apart. The thermal resistivity of the soil is | K - m/W and the ambient
temperature is 15°C. We will compute the rating of cable model No. 2 assuming that the pipe-type
cable (cable model No. 3) carries the rated current as specified in Table A1.
From Table A1, the parameters of the three-core cable are

thermal resistance of the insulation T, = 0.307 K - m/W

thermal resistance of the jacket T; = 0.078 K - m/W

conductor resistance at 90°C (50 Hz) R =0.798-10-* Q/m
sheath loss factor A, = 0.0218
external diameter D,. = 72.9 mm

The external thermal resistance of the three-core distribution cable is obtained from equation
(9.20):

ioe aen(u+Vw?
1) =sn (2744.
V8 —1)=0.637
K-m/W
The temperature rise of the three-core cable due to the heat dissipated in the pipe-type cable
(3 - 31.15 W) is obtained from equation (9.22):

a
ODN Ds
Wal
a RIAdit
e
eeeee |
eeea diel a)
5Ing2
Ne
ers 2 0.52
From equation (4.3), the rating of the three-core cable is

| 0.5
CLO) they 2208
~ 10.798 - 10-4(0.307 + 3 - 1.0218 - (0.078 + 0.637 =520A
The single three-core cable without the presence of the pipe-type cable would have an ampacity
))
of 613 A.

9.6.2.2 Equally Loaded Identical Cables. Whena group of identical, equally loaded
cables is considered, the computations can be much simplified. In this type of grouping, the
rating of the group is determined by the ampacity of the hottest cable. It is usually possible
to decide from the configuration of the installation which cable will be the hottest, and to
calculate the rating for this one. In cases of difficulty, a further calculation for another
cable may be necessary. The method is to calculate amodified value of 7, which takes into
account the mutual heating of the group and to leave unaltered the value of A@ used in the
rating equation (4.3).
216 Part II » Evaluation of Parameters

When the losses in the group of cables are equal, equation (9.24) simplifies to

t= tn} Vie=1)-
(2)(#)-(4)-@)] (9.2
There are (q — 1) factors in square brackets, with the term aoe /dpp excluded.

EXAMPLE 9.5
We will compute the rating of a circuit composed of three single-core cables of the model cable No.
4. The location of the cables is shown in Fig. A5 with the backfill removed. The native soil thermal
resistivity is | K - m/W and the ambient temperature is 20°C. The external diameter of this cable is
D,. = 105 mm. The loss factors for this cable were computed in Example 8.5. They are A; = 0.325
and A, = 0.955. From Appendix A, 7; = 0.568 K - m/W, T) = 0.082 K - m/W, and T; = 0.066 K -
m/W.
The external thermal resistance of the middle cable, which is usually the hottest, is obtained

from
equation
(9.25):
ra (|) Ty = — In
5] 5}
(9.26)

=find
fa a=a}-1+
(2)|}
where s; is the axial spacing between two adjacent cables (mm). With the factor [ + Ju? — |
approximated by 2u and substituting numerical values, we obtain

r Le on) ie 2-1000\7 rent

a = Ose
=—— iN 105 500 ——
+ Wei - m/W

The rating of this circuit is therefore equal to

os | 85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 1.03) 0.5

= =700 A
0.356 - 10-4 (0.568 + (1 + 0.325) - 0.082 + (1 + 0.325 + 0.955) - (0.066 + ml

When the losses in the sheaths of single-core cables laid in a horizontal plane are
appreciable, and the sheaths are laid without transposition and/or sheaths are bonded at all
joints, their inequality affects the external thermal resistance of the hottest cable. In such
cases, the value of 7 to be used in the numerator of the rating equation (4.3) is as given
by equation (9.26), but a modified value of 7, must be used in the denominator. This value
is obtained from equation (9.24) remembering that the conductor losses are assumed to be
equal, but the sheath losses are different. The total joule losses in the cable are expressed
as W,; = 1*R(1+A4,) where the subscript i equals 1 or 2 for the outer cables and equals
m for the middle Chble Substituting this in equation (9.24), we obtain

Tya i [utVue
1]+| ee a) eIn
[si() || (9.2 Im 1
Chapter 9 m Thermal Resistances and Capacitances 217

This assumes that the center cable is the hottest cable. The value of 4, to be used in
rating equation (4.3) is that for the center cable. Equation (9.27) is modified slightly if the
cable has both sheath and armor as illustrated in Example 9.6 in Section 9.6.3.3.

9.6.3 Groups of Buried Cables (Touching)EquallyLoaded

9.6.3.1 Overview. When the cables are touching or are laid in a close proximity to
each other, the thermal field of a cable will be distorted by the thermal fields of the cables
located nearby. The principle of superposition is not applicable in this case. Goldenberg
(1969a, 1969b) has shown that the minimal axial separation of the cables when equation
(9.25) can be safely used is equal to two cable diameters.
The derivation of the formulas for the external thermal resistance of cables in flat
and trefoil touching formations was carried out by Symm (1969) and Goldenberg (1969a,
1969b). Symm considered the general case of touching cables and used the integral-equation
method for potential theory. The equations governing the distribution of an electric field
around a current-carrying conductor are very similar to the heat conduction equations dis-
cussed in Chapter 2. Goldenberg solved the heat conduction problem by developing several
formulas for the external thermal resistance of cables in flat and trefoil formations. In
the case of two buried cables, Poritsky’s (1931) formula was applied to give the potential
distribution due to two parallel conducting cylinders of equal radius, in infinite space, with
equal charges of the same sign each. For cables in trefoil-touching formation, Goldenberg
employed the technique of conformal transformation, and the formulas were derived using
restricted application of the principle of superposition.
Other authors also have tried to solve the problem. Cronin and Conangla (1971) ex-
amined the relationship between heat flux and cable sheath temperature for closely spaced
buried power cables using both a numerical method and an electrolytic tank analog. King
and Halfter (1977) used the method of successive images to determine the external thermal
resistance of groups of equally or unequally loaded touching cables. More recently, numer-
ical studies for three cables in flat and trefoil formations and two cables in flat formation
were performed by Van Geertruyden (1992, 1993) using the finite-element method.
All authors using analytical techniques assumed that the cable surfaces are isothermal.
In all cases, the derivations are very involved and will not be repeated here. Instead, we will
quote the approximations used in today’s standards and discuss their validity. The formulas
developed by Van Geertruyden are also reported below.
9.6.3.2 Two Single-core Cables in Flat Formation. _Using Poritsky’s (1931) expres-
sion for the potential distribution due to two parallel conducting cylinders of equal radius,
Goldenberg (1969a) developed analytically the following formula for the external thermal
resistance of two cables touching in flat formation:

Tz = [in(coth
oll (9.28)
where u is defined in equation (9.19). This formula can be simplified by using a series
expansion for coth x. If u > 5, equation (9.28) can be replaced by

(pee hig 4eee ) (9.29)

It IU 12u
with an error of less than 0.01%.
218 Part IJ » Evaluation of Parameters

Van Geertruyden (1993), using finite-element analysis, developed the following for-
mulas for the external thermal resistance of two cables touching in flat formation.
For metallic sheathed cables, with the sheath assumed to have sufficient thermal con-
ductance to provide an isotherm at the cable surface,

T;= —[In(2u)
—0.451] (9.30)
XN
When the external surfaces of the cables cannot be assumed to be isothermal, we have

Ty = -“[In(2u) — 0.295) (9.31)

A

This formula applies for nonmetallic sheathed cables having a copper wire screen
and for the external thermal resistance of touching ducts. The value of 7, computed form
equation (9.31) differs by less than 0.4% from the value computed by the finite-element
method.
9.6.3.3 Three Single-core Cables in Flat Formation. The expression used in today’s
standards for the external thermal resistance of three touching cables in flat formation is
based on Symm’s (1969) paper. The following equation, not in the quoted paper, was
derived empirically by the members of WG10 of SC20A of the IEC, to fit the calculated
results given in Table 2 of the paper:

Ts = ps [0.475 In(2u) — 0.346] foru > 5 (9.32)

Performing numerical studies using finite-element analysis, Van Geertruyden (1992)

concluded that equation (9.32) gives temperature values corresponding to the mean temper-
ature of the outer cables. This means that when using this equation, the external temperature
of the central cable, and thus the conductor temperature, will be underestimated. On the
other hand, using equation (9.25) for the same case, she found that the temperatures will
be slightly overestimated. She concluded that equation (9.25) is better suited for the com-
putation of the rating of the hottest cable in this configuration than is equation (9.32). In
addition, she proposed the following equation, derived from numerical studies, when the
cable surfaces are nonisothermal (e.g., for cables in ducts):
1.505
T, = [In(2u) —0.297] (9.33)
T

EXAMPLE 9.6
We will reconsider Example 9.5, but will place the cables in flat touching formation. We will compute
the rating of the hottest cable applying equations (9.25), (9.32), and (9.33). We will recall that the
cables are located | m underground and the external diameter of this cable is 105 mm.
(1) Isothermal cable surface; principle ofsuperposition assumed.
Since the cable has both the sheath and an armor, the total joule losses are obtained from

W,= PR(1 +A),+A2)

Substituting this into equation (9.24) and remembering that sheath and armor are combined in
cable No. 4, we obtain

n=9(n[u+
s Je=i]+[
115pate‘ | nfs(24)
2]t I ot ri msA Ss]
Chapter 9 m Thermal Resistances and Capacitances 219

The loss coefficients for this new arrangement have to be recomputed because the spacing of
the cables has changed. Employing the same procedure as used in part 2 of Example 8.5, we obtain
the following values of the loss factors: 4, = 0.159, Aj, = 0.295, A/,, = 0.089, and Ar = 0.865.
Since u = 2 - 1000/105 = 19.05, we have

T= 5| {In[19.05
+ 19.05"
—1]+ 1+0.
+0.865
+.0.5
-(0.159
5 (0! +0.295)
1 + 0.089 + 0.865

nae 31000\7

105

= 1.58 K- m/W
Since the sheath loss factors for the outer cables are almost the same as for the center cable,
the coefficient in front of the second logarithm is almost equal to one. Therefore, the same value of
T, will be used in the numerator and denominator of equation (4.3). The rating of the center cable
becomes

al 85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 1.58) “

~ [0.356 - 10-4[0.568 + (1 + 0.089) - 0.082 + (1 + 0.089 + 0.865) - (0.066 + 1.58)]

ao OA

(2) Isothermal cable surface; Symm’s equation (9.32).

Ts = p;[0.475 In(2u) — 0.346] = 1 - [0.475 In(2 - 19.05) —0.346] = 1.38 K - m/W

ie (0, meOrmp)raeWa(0.57; ainn(Tyar T; ae T4)] )

~ ERT)nR (1 44,) DBPaRG +4) + 2) $7)
85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 1.38) li
~ | 0.356 - 10-4[0.568 + (1 + 0.089) - 0.082 + (1 + 0.089 + 0.865) - (0.066 + 1.38)]

= 654A

(3) Nonisothermal cable surface; finite-element equation (9.33).

ea a
7, = 12"! nu) — 0.297] = [In(2 - 19.05) — 0.297] = 1.60 K- m/W
TC ath
4 85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 1.60) i
~ | 0.356 - 10-4 (0.568 + (1 + 0.089) - 0.082 + (1 + 0.089 + 0.865) - (0.066 + 1.60))

= 608A

We can observe that, in this example, the external thermal resistance increases by about 16% and
the rating is reduced by 7% between two extreme cases. If we accept the fact that the finite-element
equation (9.33) best represents the present situation, Van Geertruyden’s conclusion that equation
(9.25) is better suited for the computation of 7; than equation given in the IEC Standard 287 (1982)
can be seen to be justified in this case.

9.6.3.4 Three Single-core Cables in Trefoil Formation. —For this configuration, L

is measured to the center of the trefoil group and D, is the diameter of one cable. The
external thermal resistance 7, is that of any of the cables, and the configuration may be with
the apex either at the top or at the bottom of the group.
220 Part I] » Evaluation of Parameters

As in the case of touching cables in flat formation, the early work used in today’s
standards was performed by Goldenberg (1969b) and Symm (1969). Goldenberg used
restricted application of superposition to derive the following equation in the case when the
external surfaces of the cables can be assumed to be isothermal (Goldenberg, 1969b; IEC
287-2-1, 1994):

[ieeeSeye
e UCA0] (9.34)
IE
Symm (1969), using the integral-equation method of potential theory, confirmed that
equation (9.34) is valid for u > 4.
If we neglect the effect of circumferential heat conduction of metallic layers for touch-
ing cables in trefoil formation, the thermal resistance of the insulation and the outer covering
is increased because dissipation of the heat is obstructed. Thus,

Trsfeli Ua fols (9.35)

where 7; and 73 are calculated by the methods in Sections 9.2 and 9.4, respectively. The
value of f, represents the fraction of the cable circumference which is obstructed by the
neighboring cables. This fraction will depend on the angle g shown in Fig. 9.9. Referring
to this figure,

fo = (9.36)

Figure 9.9 Three cables in trefoil-touching formation.

Chapter 9 m Thermal Resistances and Capacitances 221

We can also see that

_2 tape A Wee, | eae

Oia 7 i 6 a 180sin (=) (9.37)

For the standard range of cable dimensions, equations (9.36) and (9.37) give the value
of f, between 1.27 and 1.42. IEC 287 recommends that the value of fi, = 1.6 be used in
the computation of the thermal resistance of the outer covering 7;. Van Geertruyden (1995)
has performed several numerical studies using the finite-element method to empirically
determine the values of f, for cables in touching trefoil and flat formations. She concluded
that the value of f, = 1.6 suggested in IEC 287 (1982) is too large, but its magnitude has
a negligible effect on the cable rating (see also Example 9.7).
When the cable is only partially metal-covered (where helically laid armor or screen
wires cover from 20 to 50% of the cable circumference), equation (9.34) applies for wires
having a long lay (15 times the diameter under the wire screen), 0.7 mm diameter individual
copper wires having a total cross-sectional area of between 15 and 25 mm”. In this case,
the factor f, will take the following values for the standard cable dimensions:
1, : . fg = 1.07 for cables up to 35. kV
fo = 1.16 for cables from 35 to 110 kV (9.38)
f kiitos yl ep

Cables for voltage above 110 kV are usually not buried in trefoil-touching formation.
But if this were the case, equations (9.36) and (9.37) should be used in conjunction with
equations (9.35).
In the nonisothermal case, the IEC 287 (1982) document proposes the following for-
mula:

Tye tinue 2inGge)] (9.39)

20
Performing finite-element studies, Van Geertruyden (1992) concluded that equations (9.34)
and (9.39) are well suited for cables in trefoil formation.

EXAMPLE 9.7
Even though high-voltage cables are not normally placed in touching formation, we will reconsider
Example 9.6 and assume that the cables are in trefoil. We will consider both cases where the cable
surface is isothermal or nonisothermal.
(1) Isothermal cable surface; equation (9.34)
The external thermal resistance is equal to

5s 1.5
7, = 125 nu) — 0.630] = —[In(2 - 19.05) — 0.630] = 1.437 K- m/W
"4 TT
Since cable model No. 4 used in this example has a metallic sheath, there is no reduction in the
thermal resistance of nonmetallic layers. However, in order to illustrate the computational procedure,
we will compute the factor f, given by equations (9.36) and (9.37). Since the same considerations
apply for the armor bedding, the value of 7) is also increased by the factor f,. The reduction factor
is given by

Pane oer
mseen
/4OE!)
NORHeMaran)
895.5
Voees
Cert BaeRASHe ()-F+ am Date
222 Part II » Evaluation of Parameters

We observe that this value is lower than that recommended by IEC 287 (1982). The internal
thermal resistances are now obtained from equation (9.35):

T* = 1.280.568 =0.727, Ty = 1.28-0.082=0.105, Tf = 1.28 - 0.066 = 0.0845

The sheath and armor loss factors are computed from equations (8.56) and (8.80) and are equal
to A, = 0.206 and A = 0.604. The cable rating becomes

i 85 — 20 — 6.62(0.5 - 0.727 + 0.105 + 0.0845 + 1.437) iF

~ | 0.356 - 10-4[0.727 + (1 + 0.206) - 0.105 + (1 + 0.206 + 0.604) - (0.0845 + 1.437)]

(63315)
4
The rating is comparable to the case of the flat touching formation examined in Example 9.6, but
is much smaller in comparison with the case when the cables are separated as examined in Example
oS:

9.6.4 Cables in Ducts and Pipes

This section deals with the external thermal resistance of cables in ducts or pipes filled
with air or a liquid. Cables in ducts which have been completely filled with a pumpable
material having a thermal resistivity not exceeding that of the surrounding soil, either in
the dry state or when sealed to preserve the moisture content of the filling material, may be
treated as directly buried cables.
The external thermal resistance of a cable in duct or pipe consists of three parts:

1. The thermal resistance of the air or liquid between the cable surface and the duct
internal surface, T;.
2. The thermal resistance of the duct itself, 7,’. The thermal resistance of a metal pipe
is negligible.
3. The external thermal resistance of the duct, 7”.

The value of 74 to be substituted in the current rating equation (4.3) will be the sum
of the individual parts; that is,

Te leet, el (9.40)
9.6.4.1 Thermal Resistance Between Cable and Duct (or Pipe) T;. The develop-
ment of a rigorous equation for this thermal resistance is quite involved, and the expression
depends on the cable surface temperature. This equation, even though amenable to computer
implementation, is not suitable for standardization. Therefore, after developing a general
expression for 7;, we will simplify it in several stages until we arrive at the equation given
by Neher and McGrath (1957) and in IEC 287-2-1 (1994).
In the development presented below, we will assume that the inner surface of the duct
or pipe is isothermal. This assumption is usually valid for metallic conduits. For ducts
made from materials having poor heat transfer properties, the average temperature inside
the duct will be assumed.
Considering the outside surface of the jacket under steady-state conditions, the con-
duction heat flux from its inner surface is equal to the heat loss through conduction, free
convection, and thermal radiation. The energy balance equation (2.8) takes the form

W,—Weonv,s
aeWeond
ar Wrads—w (9.41)
Chapter 9 m Thermal Resistances and Capacitances 223

where Weonv,s= natural convection heat transfer rate between the cable outside
surface and the surrounding medium per unit length, W/m
Weond= Conductive heat transfer rate in the medium surrounding the cable,
W/m
Wrad,s—w= thermal radiation heat transfer rate between the duct (pipe) inner
surface and the cable outside surface, per unit length, W/m

W, = total energy per unit length generated within the cable, W/m. Its
value is given by equation (4.6).

Free convection heat transfer in the annular space between long, horizontal concentric
cylinders has been considered by Raithby and Hollands (1975). The heat transfer rate per
unit length of the duct may be obtained from equation (2.2):

Woonv,s= hs (0, = Cad As (9.42)

where h s = natural convection coefficient at the surface of the cable, W/K - m?

6. = average
a temperature of the cable outside surface, °C

0 , = temperature of the duct/pipe inner surface,! °C

iJ

A, = area effective for convective heat transfer, (m7), for unit length.

The value of A, reflects the series connection of two thermal resistances corresponding
to the outer surface of the cable and the inner surface of the duct wall and is equal to
(Incropera and de Witt, 1990)?

Ae eeD (9.43)
¥
In a
é
The convective heat transfer coefficient is obtained by assuming that the cable and the
conduit are concentric cylinders. This assumption is almost always violated in practical
installations since cables are usually placed at the bottom of the conduit. The topic is
discussed in detail in Anders et al. (1987) where the heat flux emanating from various parts
of the duct is examined. However, since the thermal resistance of the gas/liquid surrounding
a cable in duct or pipe constitutes a small portion of the total external thermal resistance of
the cable, the proposed simplifications have a very small effect on the accuracy of the final
results. The heat transfer coefficient represents in this case the effective thermal conductivity
of the fluid (gas or oil). The empirical correlation is given by Raithby and Hollands (1985):

erat eiro 1 (srr) P 1/4(Ra)!/4

op \0.861
+ Pr
«7 p*)\]4
Ra= [In(Di/D3)| eB, Tel6,)d?pcp (9.44)
Cai Saver alan
—Sap eeseeee 3

! To be more precise, the average temperature of the medium surrounding the cable should be used in
equation (9.42). However, in order to facilitate further simplifications, the temperature of the inner wall of the
duct is used here with only a small loss of accuracy.
2 Diameters with an asterisk denote dimensions in meters.
224 Part I] = Evaluation of Parameters

where Ra = Rayleigh Number

B = volumetric thermal expansion coefficient, K =
Cp = Specific heat at constant pressure, J/kg - K
d = mass density, kg/m?
g = acceleration due to gravity mm/s”
jl = viscosity, kg/s - m
p = thermal resistivity of the fluid, K -m/W
Pr = Prandtl Number
D*, = inside diameter of the conduit, m
D* = external diameter of the cable, m

When the formula is used for a group of cables in a conduit, D? becomes the equivalent
diameter of the group as follows:

e twocables: D* = 1.65 times the outside diameter of one cable, m

e three cables: | D* = 2.15 times the outside diameter of one cable, m
e four cables: | D* = 2.50 times the outside diameter of one cable, m

Equation (9.44) may be used in the range 10? < Ra < 10’. For Ra < 100,h,; = 1/p.
Denoting by D7, the factor representing the cable-duct geometry and substituting (9.43) and
(9.44) into (9.42), we obtain

conv, Tv (saa Pr
=) alt feh | Mchdecalee
up3 ( ) (9.45)
Woe = or 60386| ee eee De pee G26 yt
with

:
Bogayht Bin23/5\75/3
an acl) (9.46)
If the medium between the cable and the enclosure wall is air at atmospheric pressure,
which is usually the case for cables in ducts, the values of the physical constants can be
obtained from the formulas in Appendix D. For other gases and fluids, the material properties
can be obtained from the tables found in most of the books on heat transfer, for example,
in Incropera and de Witt (1990).
From the theoretical standpoint, the expression for the conduction component should
take into account any eccentricity between cylindrical radiation and the enveloping isother-
mal enclosure. But as in the case of convection, we can assume without much loss of
accuracy that the cable and the conduit are concentric cylinders.* The conductive heat loss
is thus given by the formula

Weond
= wane (9.47)
In—2
pln,

3 Whitehead and Hutchings (1939) give an expression to account for the eccentricity of the cable and conduit.
Chapter 9 m Thermal Resistances and Capacitances 225

The net radiation heat transfer rate between the cable and the inside surface of the
conduit is based on the radiative exchange between two surfaces:4

Wrad,s—w
= As,FswOR
(0%4
faadee) (9.48)
where og = Stefan—Boltzmannconstant, equal to 5.67 x 1078 W/(m2- K4)
Fw = thermal radiation shape factor, its value depending on the geometry of
the system
As, = area of the cable surface effective for heat radiation, (m7), for unit
length of the cable.

This equation is applicable to the case of gas-filled ducts or pipes. Computation of the
radiation shape factor F, ,, and the effective radiation area is discussed in detail in Chapter
10. The thermal properties which occur in the above equations are temperature dependent.
However, for approximate calculations, they can be assumed constant over the small range
of expected operating temperatures. For some of the parameters in equation (9.45), Sellers
and Black (1995) proposed the values listed in Table 9.5.

TABLE 9.5 Thermal Properties for the Intervening Media

Air Gas
Quantity (1 atm) (200psi) Oil
p—thermal resistivity 355) 39.0 WAS
(K-m/W)
B—thermal coefficient 0.00308 0.00310 0.00068
of expansion (1/K)
Pr—Prandtl Number 0.715 0.7567 1126.1
v—kinematic viscosity" 1.88-10° 1.303-10~° 8.278 -10°°
(m/s)

“ The following relation holds between kinematic viscosity v and viscosity LU:
v=,/d, where d is the mass density of the liquid.

The thermal resistance between the cable surface and the inner surface of the duct
(pipe) is obtained by dividing the temperature drop across the duct (pipe) gap by the total
heat emanating from the cable surface. Therefore, by equation (9.41),

0, vvOwSe 6, agOy
C= (9.49)
i Ww, Weonv,s
==Weond
ai Wrad,s—w
When the first attempts to determine the thermal resistance between the cable and the
duct wall were made (Whitehead and Hutchings, 1939; Buller and Neher, 1950), finding
the solution of equation (9.49) appeared to be a formidable task. The value of T; depends
on the unknown cable surface and inner wall temperatures, and the material parameters are
also dependent on the mean temperature of the medium. Several iterations are required to
compute 7. In the absence of digital computers, it became apparent that several simplifi-
cations were required. The approach proposed by Buller and Neher (1950) provided such

4 We recall that the temperature symbols with an asterisk denote absolute temperatures.
226 Part IJ » Evaluation of Parameters

simplifications. The same approach was later adopted by Neher and McGrath (1957), and
became the basis for North American and IEC standards.
The first approximation concerns the representation of cable/conduit geometry. The
effective diameter given in equation (9.46) is approximated by

3/4_
(Ds) =~(Dz (9.50

Next, the following assumptions were made.

1. In the case of inert gas, the physical properties of the medium were assumed to
be substantially independent of temperature over the working range, but it was
noted that the density is a direct function of the pressure. Thus, if P represents
the pressure (in atmospheres), with appropriate numerical values for the material
constants, we obtain from equations (9.45) and (9.50)

W, eu
conv,
s(gas)SFGhig:= A :Pp}/2
:Aes (9.51)
Absw es wa £ |
Di
where AO, = 6, — Oy:
2. When mineral oil is employed as the pipe-filling medium, it was assumed that the
physical constants are substantially independent of pressure with the exception of
viscosity which, for the type of oil commonly employed, was taken as varying
inversely as the cube of the temperature. This leads to the following form of
equation (9.45):

Weonv.s il)
ad
AOsw
2.733
OU eats . ADE3/4COs
* m
OAD
sw
(9.52
1.39 +
Dj
where 6,,, is the mean oil temperature in °C.
3. The radiation component with gas as the medium is assumed to be given with
sufficient accuracy by the following expression:

Wrad,s
—w
Ad (gas) = 13.21 - DF - e, -(1+ 0.0167 - 6,,) (53)

where &, is the emissivity of the cable outside surface. The radiation component is
ignored for pipes filled with oil. Substituting (9.47) and (9.51)—(9.53) into equation
(9.49) with appropriate values of the thermal resistivities, we obtain

. (DEA 0.5279
=T,(gas)i fea Se Di . pi? . Ael4 ds : wi:
vege: Dj inod
Bg
+13,21-) D* . 6. 14-0 0167 4.0.,) (9.54)
Chapter 9 » Thermal Resistances and Capacitances 227

ee fo 0.8763
—(oil)= 2.733
- - 03!" Ags 4 ; (9.55)
T; igo ee
Dj De

Next, Buller and Neher (1950) proposed to linearize equations (9.54) and (9.55). First,
they assumed that the second term in both equations and the radiation term in equation (9.54)
are constant. Considering equation (9.54), the conduction term constitutes about 14% of the
total in the case of a typical cable in duct installation, and about 8% for a typical gas-filled
pipe-type installation at 200 psi. The corresponding values for the radiation term are 63
and 43%, respectively. Normal variation of D,/D, may produce considerable variation
in the conduction term, but the overall effect is small because conduction is such a small
part of the total heat flow. In addition, the variation of this ratio has opposite effects on the
convection and conduction terms. Buller and Neher concluded that a minimum error should
therefore prevail when the conduction term is treated as a constant if the denominator of
the convection term is also treated as a constant.
Variation of 6,,, can affect the radiation term by as much as 20% over a sufficiently wide
operating range; however, when calculating a cable rating with fixed conductor temperature
on the order of 70—80°C, the range of this variable is very small, and an inaccuracy on the
order of 3—5% may be expected.
In the case of equation (9.55), the conduction term constitutes about 24% of the total
for a typical oil-filled pipe installation. Variation of @,, is more important than in the case
of gas-filled pipe cables, but is still within tolerable limits.
Under the above assumptions, equations (9.54) and (9.55) can be rewritten as (Neher
and McGrath, 1957)

i (gas) (9.56)

TO
. ee
0.120-(D363.AG.) 4 + 0.183 (9.57)
m

The constants a, b, and c in equation (9.56) were established empirically from data in
Buller and Neher (1950) for cables in pipe, and from data in Greebler and Barnett (1950)
for cables in fiber and transite ducts. Their values are given in Table 9.6. The constants in
equation (9.57) were also determined empirically.
By further restricting the value of A@,,, to 20°C for cables in ducts and to 10°C for
gas-filled pipe-type cables, and restricting the range of D, to 25-100 mm for aus in ducts
and to 75-125 mm for three cores in pipe, equation (9.56) can be reduced to

Ree u (9.58)
T,eae
= 0.1(V + Y0,,)D,

5 It should be noted that in equation (9.58), the cable external diameter D- is expressed in mm.
 Part II » Evaluation of Parameters
228

TABLE 9.6 Values of Constants a, b, c and U, V, Y

InstallationCondition a b G U V Ve
eg pon a ee ee ie See
In metallicconduit 11.41 15.63 0.2196 52 1.4 0.011
In fiberduct in air 11.41 4.65 0.1163 32 0.83 0.006
In fiberduct in concrete 11.41 Sep) 0.1806 5.2 0.91 0.010
In asbestoscement
duct in air 11.41 11.11 0.1033 522, 12 0.006
duct in concrete 11.41 10.20 0.2067 a pH ih) 0.011
Gas pressure cable in pipe 11.41 15.63 0.2196 0.95 0.46 0.0021
Oil pressure pipe-type cable — — 0.26 0.0 0.0026
Earthenware ducts = — — 1.87 0.28 0.0036

in which the values of the constants U, V, and Y depend on the installation and are given
in Table 9.6.
In the case of oil-filled pipe cable, assuming the average value of A6,,, = 7°C anda
range of 3810-8890 mm -°C for D.,,, equation (9.57) reduces to equation (9.58) with the
values of U, V, and Y given in Table 9.6.

9.6.4.2 Thermal Resistance of the Duct (or Pipe) Itself T,{. This resistance is
obtained by a direct application of equation (3.3):

(9.59)

where p is the thermal resistivity of the material and D, (mm) is the outside diameter of
the duct.

9.6.4.3 External Thermal Resistance of the Duct (or Pipe) T;’. For a single-way
duct not embedded in concrete, this thermal resistance is calculated in the same way as for
a cable, using the appropriate formulas from Sections 9.6.1, 9.6.2, or 9.6.3 with the external
radius of the duct or pipe including any protective covering thereon replacing the external
radius of the cable. Ducts embedded in concrete are treated as described in Section 9.6.5.
The external thermal resistance of buried pipes for pipe-type cables is calculated as
in the case of ordinary cables, using equation (9.20). In this case, the depth of laying L is
measured to the center of the pipe and D, is the external diameter of the pipe, including
anti-corrosion coating.

EXAMPLE 9.8
We will determine the value of the thermal resistances of the oil, pipe covering, and the external
thermal resistance for the cable model No. 3 (the pipe-type cable from the Neher—McGrath (1957)
paper) and compute the percentage ofthe heat transfer rate in the oil attributed to the two modes of heat
transfer. The parameters of this cable are given in Appendix A as D; = 67.26 mm, d. = 41.45 mm,
D, = 67.59 mm, D, = 244.48 mm, D, = 219.08 mm, Dy = 206.38 mm, A; = 0.010, A> = 0.311,
T, = 0.422 K - m/W, T, = 0.082 K - m/W, and T; = 0.017 K - m/W, W, = 19.93 W/m, W, = 4.83
W/m.
Since the oil temperature is unknown, we will initially assume that its average value is 60°C.
The external diameter appearing in equation (9.58) is that over the skid wire D,. Applying equation
(9.58) with the constants taken from Table 9.6 and remembering that we have three cores in a pipe,
Chapter 9 = Thermal Resistances and Capacitances 229

we obtain (Note: for pipe-type cables Tj = T>and Tj = T3: see Section 9.5)
U 0.26
1)Pike
eS =e oda
ee eye ne eS eon
ee 80EE00i6
ee 602eee1567s = eer '

Let us employ the computed parameters for this cable as given in Table Al. The temperatures
of the cable surface and inner wall are equal to

6, = 0. —3- (W. +T,/3 + Wy - T,/6) = 70 — 3 - (19.93 - 0.422/3 + 4.83 - 0.422/6) = 60.6°C

Oy = 8; — 3[W-(1 +. A1) + Wa] To = 60.6 — 3[19.93(1 + 0.01) + 4.83] - 0.082 = 54.5°C

Thus, the mean temperature of the oil is equal to (60.6 + 54.5) /2 = 57.6°C. With this tempera-
ture, 7, is recalculated. Equation (9.58) yields Tj = 0.082 K - m/W, and the temperature of the inner
wall of the pipe is equal to 54.4°C. The resulting value of Tj = 0.082 K - m/W, and the process has
converged.
The heat transferred by convection is obtained from equation (9.52):

Weony.s
(O11)
=27133
wha*\3/4OG) * AO anes,
3h) oFDi2

Woon.<(0il) = 2.733.
(2.15-0.06759)3/4 -57.5°/4 . (60.6 —54.4)°/4 = 62.75 W/m
Soe
2.15-0.06759
0.20638
The conduction heat transfer is obtained from the second term in equation (9.55):

AO» 0.8763 +AG, 0.8763 - (60.6 — 54.4)

Weona
(Oil)= = 15.49W/m
fi Do 20638
Ds 2.15 -0.06759
Thus, the conductive heat transfer constitutes 15.49/(62.75 + 15.49) = 0.20 or 20% of the total heat
flow. Using the computed values of temperatures and heat flow rates, the thermal resistance of the oil
becomes
6, — 6, 60.6 — 54.4
Ti = = 0.079 K- m/W
Weonv.sa5 Weonda 62.75 aia 15.49
which gives an error of about 3% in comparison with the value obtained using equation (9.58).
The thermal resistance of the pipe can be neglected, but the losses in it must be taken into
account. The relevant loss factor was computed in Example 8.7 and is equal to 0.311. The thermal
resistance of the pipe coating is calculated from equation (9.59) and is equal to

ene ND.eal 024448

~ On
Dire 021908 = 0.017 K- m/W
ih PEEae
The external thermal resistance of the pipe is computed from equation (9.20):

Ds - 0.8 2-051 22001 \7

Cee
— peas (u a 21) ) SS an 1"| 0.24448 * V \:0.24448 —| = 0.343 K - m/W

9.6.5 Cables in Backfills and Duct Banks

In many North American cities, medium- and low-voltage cables are often located
in duct banks in order to allow a large number of circuits to be laid in the same trench.
230 Part II » Evaluation of Parameters

The ducts are first installed in layers with the aid of distance pieces, and then a bedding of
filler material is compacted after each layer is positioned. Concrete is the material most
often used as a filler. High- and extra-high-voltage cables are, on the other hand, often
placed in an envelope of well-conducting backfill to improve heat dissipation. What both
methods of installation have in common is the presence of a material which has a different
thermal resistivity from that of the native soil. The first attempt to model the presence of
a duct bank or a backfill was presented by Neher and McGrath (1957) and later adopted in
IEC Standard 287 (1982). In later works by El-Kady and others (El-Kady and Horrocks,
1985; El-Kady ef al. 1988, Tarasiewicz et al. 1987; Sellers and Black, 1995), the basic
method of Neher and McGrath was extended to take into account backfills and duct banks
of elongated rectangular shapes, and to remove the assumption that the external perimeter
of the rectangle is isothermal.
In the following subsections, we will review the computation of the external thermal
resistances of cables laid in backfills and duct banks, neglecting the effect of moisture
migration. The subject of moisture migration in the vicinity of backfills and duct banks is
treated briefly in Heinhold (1990).

9.6.5.1 The Neher-McGrath Approach. When the cable system is contained within
an envelope of thermal resistivity p,, the effect of thermal resistivity of the concrete or
backfill envelope being different from that of the surrounding soil p, is handled by first
assuming that the thermal resistivity of the medium is p, throughout. A correction is then
added algebraically to account for the difference in the thermal resistivities of the envelope
and the native soil. The correction to the thermal resistance is given by

N
3°" = (pe —pe) In (u + Vu? = 1) (9.60)
20
where N = number of loaded cables in the envelope
Le¢
“= —
rb
Log = depth of laying to center of duct bank or backfill, mm
rp = equivalent radius of the envelope, mm.

The equivalent radius of the thermal envelope is obtained as follows. Considering

the surface of the duct bank to be an isothermal circle of radius r,, the thermal resistance
between the duct bank and the earth’s surface will be a logarithmic function of Lg and rp.
In order to evaluate r; in terms of the dimensions x and y (x < y) ofa rectangular thermal
envelope, consider two circles: one inscribed and outside the envelope. The radius of the
circle inscribed in the envelope is

r| = kp2.
and the radius of a larger circle touching the four corners is

dediTSE
Let us assume that the circle of radius r;, lies between these circles, and that the
magnitude of r; is such that it divides the thermal resistance between r; and rp in the ratio
Chapter 9 m=Thermal Resistances and Capacitances 231

of the portions of the heat field between r; and rz occupied and unoccupied by the envelope.

Thus,
i xy— mr?
ya cat Sla (in) iy AT Jie
a te
Taym(r3—r?) r\ rp n(rz—r?) |
from which

35
ine (F-F)
axeeAlnex(14)
leeen) a y? 3 55
‘d
or

rp=exE =(- a) (14% +n 9.61

2= P --{-—--
dey i Byn x2
— oy = 8)
Equation (9.61) is only valid for ratios of y/x less than 3.

EXAMPLE 9.9
We will reconsider Example 9.5 with the cables located in the thermal backfill as shown in Fig. 9.10.

[\

1m Soil p = 1 Km/W

A
Backfill p = 0.6

1m

eat SA Poe

—~—!0.5 m l=
Ss HOTA

Figure 9.10 Cable model No. 4 in a thermal backfill.

The external thermal resistance of the middle cable, assuming a uniform soil thermal resistivity
of 0.6 K - m/W, is obtained from the short form of equation (9.26):

Cars
nf
Ds +(#)|
— clhaePy In”\\
baN:
0.6
(eae=—|] 500
4-
1 000
1 | S09
set =0.618
K-M/W
For the installation in Fig. 9.10, x = 1000 mm, y = 1500 mm, and Lg = 1000 mm. From
equation (9.61), we have

Mexe West
3yeyfeas
cds
1 »(i)y?
-—~—{——-—]JInfl+—)]+In- 8
"3
exp|
Beer 11000/4
ee (-ee 1aa)
| 51500 000
Te 150021000
n(BF
apo)
| +ein
ay) a
==|634.6mum.
232
Part II » Evaluation of Parameters

Thus, uw= 1000/634.6 = 1.576, and from equation (9.60) the correction factor becomes

Tot= (pe p.)In(u+ fu -1)=2 —( —0.6)

In(1.576+1.5762—ie= 0.196 K-m/W
Theexternal
thermal
resistance
ofthemiddle
cableis
T;= 0.618+0.196
= 0.814
K-m/W
The loss factors were computed in Example 8.5 and are equal to 4; = 0.325 and Az = 0.955. The
rating of the hottest cable is obtained from equation (4.3):

7— [Ge= Pam)= Wa(0.57)+ (Ta + Ts+ T)] i?

WMRTME
ARUP R(t Cay)
85 — 20 — 6.62(0.5 - 0.568 + 0.082 + 0.066 + 0.814) |e
~ | 0.356 - 10-4[0.568 + (1 + 0.325) - 0.082 + (1 + 0.325 + 0.955) - (0.066 + 0.814)]

= VAS

We can observe that the addition of a thermal backfill resulted in a 10% increase in cable rating in
comparison with the results in Example 9.5.

9.6.5.2 Extended Values of the Geometric Factor. The approximation of a rectan-

gular duct bank or backfill by an isothermal circle with a radius given by equation (9.61) is
only valid if the height/width ratio is in the range of 1/3 to 3. To overcome this restriction,
El-Kady and Horrocks (1985) used the finite-element method to develop values for the ge-
ometric factor if the thermal envelopes have ratios beyond the above range. Their method
is described in Example 11.5, and the resulting values of the geometric factor are shown
in Table 9.7. The values of the geometric factor are displayed in terms of the height/width
(h/w) and depth/height (Lg/h) ratios.
Since the geometric factor G, = In2Lg/rp, the equivalent radius is equal to
2LG
rh = (9.62)
e%

where Gy, is obtained from Table 9.7.

El-Kady and Horrocks (1985) compared the values in Table 9.7 with these computed
by the Neher—McGrath formula [the logarithmic factor in equation (9.60)]. The comparison
had shown that for a given value of (Lg/h) and in the range of h/w > 1.0, the geometric
factor calculated using the Neher-McGrath method tends to decrease as the ratio h/w
increases. This contradicts the more exact results obtained by the finite-element method
(Table 9.7). This contradiction can be explained by the fact that the Neher-McGrath method
does not differentiate between cases where h > w and cases where h < w. On the other
hand, the finite-element method recognizes the difference between the heat flux patterns
associated with these two cases.
Sellers and Black (1995) used conformal transformation to calculate the external ther-
mal resistance of cables in duct banks and backfills. However, as explained in Section 9.6.6,
this method is not easy to use in a general case.

9.6.5.3 Geometric Factor for Transient Computations. |The method of Neher and
McGrath to calculate the steady-state temperature rise of cables in duct banks and backfills,
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234
Part IJ » Evaluation of Parameters

given in Section 9.6.5.1, permits the introduction of a thermal resistivity for the duct bank
which may differ from that used for the earth. This is compensated for by a correction
factor. Recently, El-Kady and Affolter (Affolter, 1987) proposed a simple correction factor
to take into account the presence of a duct bank or backfill in transient calculations.
In analogy to equation (9.20), the external thermal resistance in transient analysis is
obtained from equation (5.10) as

De
alt)ral ee:
Ei
( a)
dupealinnee Bal
+Bi
(
(eue: a kes
iv) ee
9.63
where 5 = soil thermal diffusivity (m*/s) and ¢ is the time in seconds.
In analogy to equation (9.60), the following formula can be applied for the correction
term:
1, Ce adoa hcrp + Ei| —Ly (9.64)
3 An Ast St
This correction is then added to (9.63) in the same way as was done in the steady-state
analysis.

9.6.6 External Thermal Resistance of Cables Laid in Materials

Having Different Thermal Resistivities

The approach for the computation of the external thermal resistance discussed so far
assumes that the heat path between the cable and the ground surface is composed of a
region having uniform thermal resistivity. Even in the case of a backfill or duct bank, the
same assumption was made initially, and a correction factor was applied later to account
for different thermal resistivities. In practice, several layers of different thermal resistivities
may be present between the cable surface and the ground/air interface. To deal with a
general case of varying thermal resistivities of the soil, CIGRE WG 02 proposed a method
using conformal transformation to compute the value of the external thermal resistance
(CIGRE, 1985). This method is briefly described in the following.
We will start by writing the temperature rise of the surface of cable i in a group of N
cables in the following form:

N
D6; = >> Rik Wr (9.65)
k=]
where W,, = total power loss per unit length of cable k, W/m
R;; = self thermal resistance of ith cable, K - m/W
Rix (i A k) = mutual thermal resistance between cable i and cable k, K - m/W.

The matrix with elements Rj, plays the same role as the external thermal resistance Ty
for one cable as given in IEC Publication 287 (1982), but covers the case where there are
several cables, each of which may have different losses.
To explain the concept of the particular conformal transformation, we will consider
a single cable in a uniform soil. The field can be drawn for this situation, resulting in the
isothermal lines taking the form of eccentric circles as shown in Fig. 9.11, one of which is
the cable surface.
Chapter 9 a Thermal Resistances and Capacitances 235

i9 Ground surface
+X

yy

|_—|sotherms
Cable surface

L = axial depth of cable

y,= depth of equivalent line source

(a)

T
Ground surface 0

Cable surface

(b)

Figure 9.11 (a) Kennelly assumption for thermal field around a cable; (b) conformal map
of field around a cable. (CIGRE, 1985)

The conformal transform used to redraw this simple case is given by the function

at
w=06)=In| ZF 11
(9.66)

where z =x + jy = complex coordinates of a point being considered in the

original plane of Fig. 9.1la
w =u + jv = complex coordinates of the corresponding point in the
transformed plane represented in Fig. 9.11b
0+ jy; = complex coordinates of the line source corresponding to an
isothermal surface of the reference cable in Fig. 9.1 1a.
236 Part IT » Evaluation of Parameters

The function ® has been chosen so that the new diagram (Fig. 9.11b) has rectangular
finite boundaries, the upper and lower ones representing the ground and cable surfaces,
respectively. The two side boundaries both correspond to a particular flux line; in the case
of Fig. 9.1 1a, it is the flux line cr.ad on the-y axis passing through the center of the cable.
The differential equation (2.12) of heat conduction is invariant under a conformal
transform, so that the temperatures at corresponding locations in the w and in the z planes
are the same when the boundary conditions are the same. The advantage of the transformed
plane is that the isotherms become straight lines parallel to the line representing the isother-
mal ground surface, and the flux lines can be represented by a second set of parallel lines
perpendicular to the isotherms. The thermal resistance along any tube of flux lines is the
same in both planes, but is more readily computed in the rectangular geometry of the w
plane. The external thermal resistance is given directly by the height u of the transformed
domain.
If we now consider an installation where several cables are present, similar transfor-
mations may be applied, using each cable in turn as the reference cable, so as to obtain the
external thermal resistance R;; of each cable separately. In so doing, the presence of other
cables is ignored in accord with the restricted application of superposition. The determina-
tion of the temperatures at the location of the centers of the remaining cables in the w plane
will yield the “mutual resistance” terms Rj, (i # k).
If, finally, we include variations in soil resistivity, the boundaries between regions of
different resistivity may be mapped, point by point, on the w plane using the same function
®. The only difference from the case with uniform soil in Fig. 9.1 1b is that, when imposing
a temperature difference between the lines representing the soil surface (u = 0) and the
cable surface (u = u-,), the resulting flux lines are not necessarily parallel to the u axis due
to refraction at the boundaries between regions of different resistivity.
Evaluation of the temperatures in the thermal field of the w plane is achieved by ap-
proximating the composite medium represented in the w plane with an analogous resistance
network covering the same rectangular area. The equations of the network may be solved
by any standard method of numerical analysis.
There are several drawbacks to the method outlined above. The major one is that the
equations describing the transformed network are equivalent to finite-difference equations
obtained by discretizing the heat equation in the w plane, and hence the complexity of a
numerical solution of the heat conduction problem is not avoided. Another drawback is that
both the earth and cable surfaces are assumed to be isothermal. In addition, transformation
of the boundaries between regions with different resistivities point by point is very laborious,
and the resulting computer software cannot efficiently handle more than four cables in one
installation.
All of these limitations can easily be overcome by the application of the finite-element
method. The nonuniform soil conditions and nonisothermal boundaries are handled natu-
rally by this method. The computational efficiency of this approach is also quite satisfying.
With presently available personal computers, calculations involving networks with several
thousand nodes can be performed in a matter of minutes.© The finite-element method is
described in Chapter 11.

© The author performed computations on a network of 7000 nodes in less than 5 min using a 66 MHz
486 PC.
Chapter 9 m Thermal Resistances and Capacitances 237

9.6.7 The Neher-McGrath Modification of T, to Account for CyclicLoading

As discussed in Section 5.6, the computation of a cyclic rating factor requires an

evaluation of the cable capacitances and conductor temperatures at several time points.
Neher and McGrath (1957) proposed an alternative, simple method to account for the
cyclic loading of a cable. Their approach requires a modification of the external thermal
resistance of the cable. The underlying principles are discussed below.
In order to evaluate the effect of a cyclic load upon the maximum temperature rise of
a cable system, Neher (1953) observed that one can look upon a heating effect of a cyclical
load as a wave front which progresses alternately outwardly and inwardly in respect to
the conductor during the cycle. He further assumed that the heat flow during the loss
cycle is represented by a steady component of magnitude j W, plus a transient component
(1 —2)W, which operates for a period of time t during each cycle. The transient component
of the heat flow will penetrate the earth only to a limited distance from the cable, thus the
corresponding thermal resistance 74,; will be smaller than its counterpart 74,, which pertains
during steady-state conditions.
Assuming that the temperature rise over the internal thermal cable resistance is com-
plete by the end of the transient period t, the maximum temperature rise at the conductor
may be expressed as

Ad = W, [T + LT4s5 + CL— 1) Ter] (9.67)

where T is the internal thermal resistance of the cable, 7,, is the external thermal resistance
with constant load, and 74, is the effective transient thermal resistance in the earth. Further,
Neher (1953) assumed that the last thermal resistance may be represented with sufficient
accuracy by an expression of the general form

B65 -T
Ther = Aps In
é

in which constants A and B were evaluated empirically to best fit the temperature rises
calculated over a range of cable sizes. Using measured data, Neher (1953) obtained the
following values for the constants A = 1/27 and B = 61 200 when Tt is expressed in hours
and 6 is expressed in m?/s.
Introducing the notation

D, = 61200,/6(lengthofcycleinhours) (9.68)
theexternalthermalresistanceinequation(9.67)canbewrittenas
Fee! hn a yn? |
Ts== UT4s5
+ CL
—1b)Ter= Bate it D,
—— = pI SSS
(9.69a)
hadpeg wh
yA
pi
iw | eee TDs
The last equation can be generalized for a group of equally loaded identical cables taking
into account equation (9.25)
238 Part IJ » Evaluation of Parameters

r=ladcuflee} ((%#)(2)-(2)--(
(9.69b)
where u = 4L/D,. When the cables are laid in a flat formation without transpositions, and
when the sheaths are solidly bonded, equation (9.27) rather than (9.25) is substituted into
(9.69a) (see Example 9.10).
From equation (9.68), we observe that the fictitious diameter D, (mm) at which the
effect of loss factor commences is a function of the diffusivity of the medium 6 and the
length of the loss cycle.
In the majority of cases, the soil diffusivity will not be known. In such cases, a value
of 0.5 - 10~° m?/s can be used. This value is based on a soil thermal resistivity of 1.0 K -
m/W and a moisture content of about 7% of dry weight (see Section 5.4 for more details on
this subject). The value of D, for a load cycle lasting 24 h and with a representative soil
diffusivity of 0.5 - 10~° m/s is 211 mm (or 8.3 in).
Alternative expressions for D, are given by Heinhold (1990):

205
D, = —— for sinusoidalload variation (9.70)
AT ies

493./ se
1D),= Ree for rectilinear load variation (9.71)
hwpo

— Me
103+eased
246 for other load variations (9.72)
Jw.
wherew is the numberof loadcyclesin a 24h period.

EXAMPLE 9.10
We will compute the rating of the cable circuit studied in Example 5.6 using the Neher-McGrath
approach. The parameters of this circuit are 4, = 0.206, 4}, = 222, A}, = 0.089, T, = 0.214 K -
m/W, and 73 = 0.104 K - m/W. Cables are spaced two cable diameters apart.
The fictitious diameter D, is first obtained from equation (9.68) and is equal to 211 mm. The
external thermal resistance of the middle cable is obtained from equation (9.69b) with the loss factor
computed in Example 5.5 equal to 0.504. In this case, equation (9.27) is used for the external thermal
resistance with unity load factor, and with the sheath loss factors computed in Example 8.2:

: |bs}, Ds
D. M4 4L
D, | 14+0.5(a4,
mane+49) i 1+ (=)
Lye
ie Tl
In —+ iS ah||)=e
|
ies

1 21140,504: fin4.1000a 1+0.5(0.2

= —(In=—— eee 2 eee eee| ?
2x\ 35.8 211 1+ 0.088 2.35.8
II A

Assuming the sheath loss factor for the middle cable, the rating is equal to

re 90 —15
0.781 - 10-4[0.214 + (1 + 0.09) - (0.104 + 1.11) 40.5
=790A
Chapter 9 = Thermal Resistances and Capacitances 239

If we use Heinhold’s equation (9.72), the fictitious diameter D, is equal to

p. — 103+246v0.504
—=Tf {owiantan
x eK : jo4

The external thermal resistance and the rating of the cable become

nim
a1(enage
+0804.
[n , 140.5(0.206
277.6 4.1000
20 +0an]i+
(3ee)I}
1+0.5(0.206
+0.2
3550 4 277.6 1+ 0.089 IESBIS
Ve)
= 1.14K-m/W

ae 90 —15

~ [0.781 - 10-4[0.214 + (1 + 0.09) - (0.104 + 1.14) j 0.5

=782
A
We can observe that these values are not very different form the rating obtained with the more precise
method in Example 5.6, with the Heinhold approximation giving the more accurate result.

9.6.8 Cables in Air

9.6.8.1 General Equation for the External Thermal Resistance. In this section, we
will consider cables either installed horizontally on noncontinuous brackets, ladder supports,
or cleats or clipped to a vertical wall. Heat transfer phenomena are more complex for
cables installed in free air than for those located underground, and proper handling of these
situations requires the solution of a set of energy balance equations. The relevant energy
balance equation for the surface of the cable was formulated in Section 2.3.2 [equation
(2.20)] and, for a cable with n conductors, is given by

nI?R(L+A,+2) + Wa+oDEH—1Décony(02—6%)—7Dieson(0 —0%.)=0

(9:73)

where —/cony
= Convective heat transfer coefficient, W/m? - K~!

0* = cable surface temperature, K

o = solar absorption coefficient (see Table 9.10)
H = intensity of solar radiation, W/m?
og = Stefan—Boltzmann constant, equal to 5.67 - 10-8 W/m?. K*

€, = emissivity of the cable outer covering

D* = cable external diameter, m

6*» = ambienttemperature,
K
There are two unknown variables in equation (9.73), the conductor current J and the
cable surface temperature 0;. The second equation linking these two variables is obtained
from the following relationship between conductor and cable surface temperatures [see
equation (4.5)]:

6*—0*=n(I°R-T + Ada)
Cc
(9.74)
240 Part IJ » Evaluation of Parameters

where 7, the internal thermal resistance of the cable, is defined in equation (4.4) and the
temperature rise caused by dielectric losses, A@z, is given by equation (4.7). The constant
n represents the number of the conductors in the cable.
To obtain an expression for the external thermal resistance of a cable in air, we first
rewrite equation (9.73) using the notation in equation (3.10a),

W,= a Doh,AO, (Jah

where AO, = 07 — O_, is the cable surface temperature rise above ambient, h, is the total
heat transfer coefficient defined in equation (3.10), and W, includes the heat gained by solar
radiation. From equation (9.75), the external thermal resistance of the cable is given by

(9.76)

Before the external thermal resistance is computed from equation (9.76), the value of
the heat transfer coefficient h, has to be determined first. This coefficient is a nonlinear
function of the cable surface temperature (see Table 10.2 in Section 10.3.4.2). However, for
standardization purposes, a simpler method is required for the determination of Ty. Such a
method is discussed next.

9.6.8.2 IEC Standard 287—Simple Configurations. From a large number of tests

on various cables in various configurations, carried out in the United Kingdom in the 1930s,
Whitehead and Hutchings (1939) deduced that the total thermal dissipation from the surface
of cable in air may conveniently be expressed as

W,= 1D*h(A0,)>" (9.77)

where h (W/m? - K°/*) is the heat transfer coefficient embodying convection, radiation,
conduction, and mutual heating. From equation (9.77), the following expression for the
external thermal resistance of cables in free air is obtained:

l
Le (9.78)
x D*h(A0,)'/4

From equations (9.76) and (9.78), we can observe that for a single cable,

h,=hael (9.79)

The values of the heat transfer coefficient were obtained experimentally and plotted as
a function of the cable diameter for various cable arrangements (Whitehead and Hutchings,
1939; IEC 287-2-1, 1994) The curves of Whitehead and Hutchings were later fitted with
the following analytical expression (IEC 287, 1982):

eae (9.80)

where constants Z, E, and g are given in Table 9.8.

Chapter 9 m Thermal Resistances and Capacitances 241

TABLE 9.8 Values for Constants Z, E, and g for Black Surfaces of Cables
in Free Air

Installation i, E g Mode

1 Singlecable* 0.21 3.94 0.60 Sat ere oe De

2 Twocablestouching,horizontal 0.21 2735, 0.50 =| aS 0S Dz

3 Threecablesin trefoil 0.96 1.25 0.20 (4-205 Da

|
4 Three cables touching, horizontal 0.62 E95 0.25

>| (~s
205
Da
coo
S Twocablestouching,vertical 1.42 0.86 0.25 PO.) Ds

6 TwocablesspacedD>, vertical 0.75 2.80 0.30 Pit 210:5pe

| Oy
Chive
7 Threecablestouching,vertical 1.61 0.43 0.20 | << > iL)Di

8 ThreecablesspacedDe. vertical i133 2.00 0.20 P4210 Dz

/ Od. :

9 Single cable 1.69 0.63 0.25 b

10 Three cables in trefoil 0.94 0.79 0.20 jee

*Values for a “single cable” also apply to each cable of a group when they are spaced horizontally
with a clearance between cables of at least 0.75 times the cable overall diameter.
242 Part II » Evaluation of Parameters

Served cables and cables having a nonmetallic surface are considered to have a black
surface. Unserved cables, either plain lead or armored, should be given a value of h equal
to 88% of the value for the black surface.
Morgan (1994) has shown that the exponent 1/4 and heat transfer coefficient h in
equation (9.78) vary with the temperature rise of the surface of the cable. By performing
numerous calculations, he determined the values of constants Z, g, and E listed in Table
9.9, as well as of the exponent q — | to which the cable surface temperature rise is raised
[1/4 in (9.78)]. The values in Table 9.9 are displayed for various temperature ranges and as
a function of the number n of bundled cables.

TABLE 9.9 Values for Constants Z, E, and g for Black Surfaces of Cables
in Free Air as a Function of the Cable Surface Temperature Rise (Morgan, 1994)

il n=3

W X Y W Ya W ae ¥

Zz 0.880 0.825 0.835 0.522 0.142 0.860 0.605 0.460

g 0.350 0.300 0.263 0.447 0.521 0.335 0.349 ~=0.316
E 4.50 3.30 2.00 4.00 1.85 4.00 3.00 1.80
q-1 0.165 0.230 0.320 0.120 0.340 0.135 0.220 0.330

W:10 KSAO, < 33 K; X:33 K <A@, $66 K; ¥:66 K<A,< 100K; 0,4, = 25°C, € = 0.9

As can be seen from Table 9.9, the IEC value of g — 1 = 0.25 (for n = 1) is
approximately correct only for the case X (A@, = 33 + 66 K), which is the usual operating
range of cables.
The external thermal resistance given by equation (9.78) depends on the unknown
cable surface temperature rise A@,. This temperature rise can be obtained iteratively by
considering the energy balance equation at the surface of the cable. The energy conservation
principle states that the heat dissipated by conduction inside the cable is equal to the heat
dissipated outside the cable by convection, radiation, and conduction. Thus, the energy
balance equation for the cable surface can be written as

nm
D*h(A@,)>/4T
6. =aamb i Ad; GiAd; = (9.81)
1+A, +A,
where T is the internal thermal resistance of the cable given by equation (4.4) and A@’,is
the temperature rise caused by dielectric losses. The latter is obtained using equations (4.4)
and (4.7):

|
Ad,
=Wa[ TERS eye)
l
)n NXxT)
1+ ry}+o
(9.82)

If the dielectric losses are neglected, AQ, = 0.

Denoting

Ka
—eeD hr
(9.83)
THA HAY’
Chapter 9 m Thermal Resistances and Capacitances 243

equation (9.81) can be written as

ik NY 1/4
(9.84)
Le KA(AG)

where AO = 6. —Oampis the permissible conductor temperature rise above ambient. Equa-
tion (9.84) suggests the following iterative procedure for determination of (A@,)!/4. Set the
initial value of (A@,)'/4 = 2 and compute consecutive values from

(Aé;)
Oe
MAE Rael
KR(AG.), és
Iterateuntil(A0,)\/", —(A@,)h/*
< 0.001.
If the cables are directly exposed to solar radiation, the effect of solar heating must be
taken into account by modifying slightly equation (9.85):

(AO;)
=
ores A+A@!
[ee HK,/xh]'4
=
1+ Kx(A6,)1/4 (9.86)
In (9.86), o is the absorption coefficient for the cable surface (see Table 9.10) and H
is the intensity of solar radiation. IEC Standard 287 recommends values for H for various
countries. When the actual value of H is unknown, this value should be taken as 1000
W/m? for most latitudes.

TABLE 9.10 Absorption Coefficient of Solar

Radiation and Emissivity Values for Cable Surfaces

Material (oy ee

Bitumen/jute serving 0.8 0.95

Polychloroprene 0.8 0.9
RVG 0.6 0.9
PE 0.4 0.9
Lead 0.6 0.63

EXAMPLE 9.11
Consider cable model No. 2 installed in free air with air temperature of 25°C. The parameters of
this cable taken from Appendix A are: D, = 72.9 mm, A; = 0.0218, 7; = 0.307 K - m/W, and
T; = 0.078 K - m/W.
We will determine the external thermal resistance and cable rating for the following conditions:
(1) single cable not clipped to a vertical wall and shaded from solar radiation, (2) the same case with
the cable being exposed to solar radiation with intensity of 1000 W/m’, and (3) the cable is clipped
to a vertical wall and shaded from solar radiation.
(1) Cable shaded from solar radiation
The heat transfer coefficient is obtained from equation (9.80) with the constants taken from
Table 9.8: 2 0.21
= reper + 3.94 = 4.95 Wim? - K*/4
244 Part I] » Evaluation of Parameters

Constant K, is obtained from equation (9.83) with the required cable parameters listed above

Kt ED © geee?Te?lew+(t+
1+, +A2 0.0218).
0.078]
=0.202
1+ 0.0218 3

The cable surface temperature rise is obtained from equation (9.85) with A@, = 0:

| A6
+A@,
Seiotdonlhawgave
RL ls
(AG) =Fee
oS= | -=2.6085
(A8,)2"
=Feces
=a =aa| cca
(A9,)3""
=Feces
oeof =aa| ato,
(AB) =Feces
“Se ~=| Se
1/4
The iterations are stopped since (A@,),°° —(A6@,
aes < 0.001. The external thermal resistance is now
obtained from equation (9.78):

Tos 1
— 1 = 0.345 K- m/W
mDth(A6,)'/4 1 -0.0729-4.95-2.558
The rating of this cable is obtained from equation (4.3):

_ 90 —25 npieesiabot
~ | 0.798- 10-4(0.307+ 3(1+ 0.0218)- (0.078+ 0.345)) 2
(2) Cable exposed to solar radiation
The first iteration of equation (9.86) is

(agit =| At AG FONKasmh| 1/4 E —25+ 0.60- 1000-0.202/(x-4.95) ]'4

1+ K, (A0,)9" a 1 + 0.202-2
= 2.6834

Repeating the application of equation (9.86), convergence is achieved in the third iteration with
(Aé, i = 2.626. The external thermal resistance is obtained from equation (9.78):

Tis | 336K-m/W
it FED
0109054
05,3 AiG
The rating of this cable is obtained from equation (4.15):

|0.5
- 90 — 25 — 0.6 - 0.0729 - 1000 - 0.336
~ [0.798 - 10-4[0.307 + 3(1 + 0.0218) - (0.078 + 0.336)] =632A
In this case, solar heating of the cable surface decreases the cable rating by about 11%.
Chapter9 = ThermalResistancesand Capacitances 245

(3) Cable clipped to a vertical wall and shaded from solar radiation
The heat transfer coefficient is equal to

1.69
= 0.:072902540.63 = 3188WimeukKo

Constant K, is obtained from equation (9.83) and is equal to 0.158. The cable surface temperature
rise is obtained after three iterations of equation (9.69) and is equal to (A@,)es* — 2.605. The external
thermal resistance is obtained from equation (9.78):

l
Ts = = 0.432 K - m/W
m - 0.0729 - 3.88 - 2.605

The rating of this cable is obtained from equation (4.3):

0.5
-| 0.798 - 10-4 (0.307 + (1 + 0.0218) 90—
25 - (0.078 + |
= 660 A

The ampacity of the cable is smaller than in case 1) because heat dissipation is now more difficult
with the cable clipped to a vertical wall.

9.6.8.3 IEC Standard 287—Derating Factors for Groups of Cables. Cable instal-
lations discussed in Section 9.6.8.2 were limited to a maximum of three cables touching,
either in trefoil or in flat formation. In this section, we will consider installations of groups
of cables in air belonging to one of the configurations shown in Fig. 9.12.
The discussion in this section is limited to:

$6

(b)

Figure 9.12 Typical groups of (a) multicore cables, and (b) trefoil circuits.
246 Part II m Evaluation of Parameters

1. amaximum of nine cables in a square formation (the last arrangement in Fig. 9.12a),
2. a maximum of six circuits, each comprising three cables mounted in trefoil, with
up to three circuits placed side by side or two circuits placed one above the other
(the last arrangement in Fig. 12b),
3. cables for which dielectric losses can be neglected (usually, only lower voltage
polymeric cables are installed in groups).
When cables are installed in groups as shown in Fig. 9.12, the rating of the hottest
cable will be lower than in the case when the same cable is installed in isolation. This
reduction is caused by mutual heating. A simple method to account for this mutual heating
effect is to calculate the rating of a single cable or circuit using the method described in the
previous section and apply a reduction factor. This is defined as follows.

pedgowh (9.87)

where I, = rating of the hottest cable in the group, A

I, = rating of the same cable or circuit isolated, A

F, = group reduction factor

The reduction in current is a result of an increase in the external thermal resistance of

a cable or circuit in a group as compared to the case when the cable or circuit is installed in
isolation. A basic criterion is that the temperature rise above ambient of the conductor be
the same for the grouped and isolated cases. Since we neglect dielectric losses, equation
(4.5) yields

If (RT + R,Ta) = F271,(RT + R,Tag) (9.88)

where R, = R(+A, +A2)

Ty, = external thermal resistance of the hottest cable in the group, K - m/W

= I external thermal resistance of one cable, assumed to be isolated, used to

compute the rated current 7, K - m/W

Defining k; as the cable surface temperature rise factor of one multicore cable or one
single-core cable mounted in trefoil and assumed isolated in free air, we have

cable surface temperature rise ifZR,T4)

conductor temperature rise 1? (RT + R,T4y)

Substituting this into (9.88), we obtain

(9.89)

Also,
Chapter 9 m Thermal Resistances and Capacitances 247

Substituting this value into equation (9.89), yields

‘ I; hie Ki Tag) tap)
1
EF,= =| —____—__ (9.90

Since the cable surface temperature rise equals W, - T4;, we have

W, ‘ 141
k)1 = ————
ee (9.91)

where W; (W/m) is the power loss from one multicore cable or one single-core cable
mounted in trefoil, assumed to be isolated, when carrying the current /,.
To compute the term Ty, /74;, we use equations (9.78) and (9.87):

irra (Cale etr 1?R,T41 0.25

thy ae he 0.25
Tar tg \ Ase hg FEUER,The © Ag Tae)Tal
Thus, the term 74,/T4; can be calculated from the ratio (h;/h,) by using the iterative
relationship
ese
T4g) netS| tg
hy| ky
(Tag/
1—Tar),
ky 0.25
(9.92)
and starting with (T4./T47)) = (hy/hg).
Equation (9.92) converges quickly; one iteration with (74/741); = (hy/h g) is usually
sufficient. Alternatively, when (h;/h,) is less than 1.4, it is sufficient to substitute (h;/h,)
for (74/747) in equation (9.90). ,
Values for the ratio (h;/h,) have been obtained empirically and are given in Table
9.11 (IEC 1042, 1991).
If a clearance exceeding the appropriate value given in column 2 of Table 9.11 cannot
be maintained with confidence throughout the length of the cable, the reduction coefficient
is determined as follows.
1. For horizontal clearances, we assume that the cables are touching each other or the
vertical surface. Appropriate values are given in column 4 of Table 9.11.
2. For vertical clearances, the reduction coefficient due to grouping is derived accord-
ing to the value of the expected clearance:

a. Where the clearance is less than the appropriate value given in column 2 of Table
9.11, but can be maintained at a value equal or exceeding the minimum given
in column 3, the appropriate value of (h;/hg) is obtained from the formula in
column 4.
b. Where the clearance is less than the minimum given in column 3, we assume
that the cables are touching each other. Suitable values of (h;/h,) are provided
in column 4 of Table 9.11.

The values in column 4 are the average values for cables having diameters from 13
to 76 mm. More precise values for multicore cables may be evaluated for a specific cable
diameter, both inside and outside this range, by consulting Table 9.8.
248 Part IJ » Evaluation of Parameters

TABLE 9.11 Data for Calculating Reduction Factors for Grouped Cables

Thermal Proximity Thermal Proximity

Effect Is Negligible Effect Is Not Negligible
Arrangement if e/D, Is Greater Than If e/D, Average Value
of Cable or Equal To: Is Less Than of hylh,”
l 2 3 4

Side by side
sealer
2multicore€ 7 ae 0.5 0.5 141

3multicore
XD e) G) 0.75 0.75 1.65
2trefoils> 1.0 1.0 Le

3trefoils a 16S) 1fF 1.25

One above the other

2multicore G>| 4
2
or 0.5
in} 1.085
(e/D,)
or 1.35

3multicore oe 4
4 1.19
(e/D,)
or 1.57
or 0.5

2 trefoils o 4 il 106(e/D,) °°8

SoGoal or1.39
és a or0.5

Near to a vertical surface 0.5 0.5 2S

or to a the
below horizontal
cable surface —| e eg

* The formulas for (h,/h,) given in column 4 of this table shall not be used for values of (e/D,) less than 0.5 or greater than the
appropriate values given in column 2.
Chapter 9 » Thermal Resistances and Capacitances 249

The reduction factor given by equation (9.90) is used when the rating of a single cable
or a single circuit is known. When the ampacity of the hottest cable in a group has to be
determined from the beginning, an alternative way to do so is to use equation (9.78) to
obtain the external thermal resistance with the heat transfer coefficient h, substituted for
the coefficient h. For the group configurations covered in Table 9.11, values of the heat
transfer coefficient h, are derived from

h
—————————
* Gulhe) 9.93
bass

where the parameter h is given in equation (9.80) for one multicore cable or for a single
cable mounted in a trefoil group, assumed to be isolated, and the ratio (Ar[h a) is obtained
from Table 9.11.
When the cables are installed in more than one plane, factors for current-carrying
capacities for the hottest cable in a group are evaluated by using the appropriate value
of (h;/h,) for the vertical clearance and ensuring that the horizontal clearance between
cables, e, is not less than the value given in Table 9.11 for neglecting the side-by-side
thermal proximity effect.

EXAMPLE 9.12
We will again consider cable model No. 2 with the installation examined in part | of Example 9.11,
but this time the cable arrangement is as shown in Fig. 9.13.

: e

Figure 9.13 Installation of cables in air for Ex-

ample 9.12.

The rating of a single cable computed in part 1 of Example 9.11 was 713 A. The ratio (A; /hg)
is calculated from the following equation, obtained from column 4 of Table 9.11:

(A;
/he) é —0.128
=1.085
(<) (F)—0.128
=1,085 =/1,085 e

The cable surface attainment factor is obtained from equation (9.91) as

‘ W,-- Tar 3I7R(L+A;)-T 3+ 7137- 0.798 - 10-*(1 + 0.0218) -0.345 _

1= pig ein ee = = 0.66
0. <. amb GO.
“_Darn 90 —25
To compute the reduction factor F,, we first apply equation (9.92). Since the ratio (h;/h,) is smaller
250 Part I] » Evaluation of Parameters

than 1.4,wesubstitute(h;/h,) for (T,/ Ts). Thus,fromequation(9.90),wehave

Nie = uae = 0.973
; 1 —ky +k, (Tag/Ts1) 1 —0.66 + 0.66 - 1.085

The ampacity of the hottest cable is therefore equal to

I, = 0.973 -713 = 694A

9.6.8.4 The Effect of Wind Velocity and Mixed Convection. Equation (9.78) was
developed under the assumption that the cable is subjected to natural cooling only and
that / is constant for a fixed value of D*. Cables installed outdoors may be subjected to
wind, resulting in forced convective cooling. Morgan has presented forced convective heat
transfer data for a single-core cable (Morgan, 1982) and bundles of two, three, and four
single-core cables (Morgan, 1993). With low wind speed, natural convective cooling and
forced convective cooling may occur at the same time. We will briefly review the issues
involved in computation of the heat transfer coefficient when both modes of cooling are
present.
Neglecting conductive heat transfer, which is very small in free air, the heat transfer
coefficient is given by equation (3.10) and is equal to

hy = Aeony+ Arad. (9.94)

The radiative heat transfer coefficient h,,qgcan be computed from equation (3.9), and can
then be approximated by (Black and Rehberg, 1985) with not-too-high accuracy:

o&[( + 273)*—(Oamb
+ 273)"|
hrad= nA ~ ope (1.38- 10°+ 1.39- 10°amb) (9.95)

In the following, we concentrate on the computation of the convective heat transfer

coefficient Agony. We denote the natural convective heat transfer coefficient as h, ,. Its value
is obtained from standard correlations (Morgan, 1982; Incropera and de Witt, 1990):

hair PCG? Pr)”

Non = fps (9.96)

where —Pr = Prandtl Number (a property of air); its computation is described in

Appendix D
De gBA%|
Gr = ——— is the Grashoff Number; all of the symbols were defined
py
in Section 9.6.4.1.

The values of coefficients c and m are given in Table 10.4 (m = n) for various ranges
of the Rayleigh Number (the product of Prandtl and Grashoff Numbers). In the temperature
ranges —10°C < @amb< 40°C and 10°C < 6, < 100°C, and for 10* < (Gr- Pr) < 10’,
the following approximation may be made (Morgan, 1993):

BePr 0.25
0.48k,ir | —— & 1.23(44%) (9.97)
De
Chapter 9 = Thermal Resistances and Capacitances 251

In the case of forced convection, the convective heat transfer coefficient is correlated with
the Reynolds Number, Re, which is defined by

Re WUDY (9.98)
ns v
where U (m/s) is the wind velocity.
We denote the forced convective heat transfer coefficient as ho,¢. Its value is obtained
from the following correlations (Morgan, 1982; Incropera and de Witt, 1990):

Kairb(Re)?
ho,—.
¢ = ———— 9:99
(9.99)

Values of the coefficients b and p for various ranges of the Reynolds Number are given in
Table 10.5.
As mentioned before, both natural and forced convection are present at low wind ve-
locities. For this mixed convection, an equivalent Reynolds Number Régq, is first calculated
(Morgan, 1992):

Ree = “soeGr - Pr)”1/p (9.100)

Next, assuming that forced and natural flows are perpendicular to each other, we compute
the effective Reynolds Number from

Ree=(Re? (05)
+Rez.) (9.101)
The effective Reynolds Number is then inserted in place of Re in equation (9.99) to obtain
the mixed convection heat transfer coefficient hy »:

7 Kairb
(Reegf)
P
Noam De (9.102 )

Morgan (1992) has shown that for cables with the outside surface temperature rise
above ambient not exceeding 20 K, as is often the case in outdoor cable installations, the
natural convective cooling can be ignored when the wind velocity exceeds 1 m/s. By
substituting for Aconyeither hy» Or ho, ¢OFAo,m,as the case may be, we can solve equations
(9.73) and (9.74) to determine the cable rating in free air. For groups of cables, we can
apply derating factors as described in Section 9.6.8.2. An alternative, simpler approach to
account for the effect of wind velocity was proposed by Neher and McGrath (1957) and is
discussed in the next section.

EXAMPLE 9.13
Consider the cable installation described in part 1 of Example 9.11, and assume that the wind velocity
is 3 m/s. We will determine the external thermal resistance and cable rating. We will consider two
approaches: (1) by solving the energy conservation equations, and (2) by applying the approximations
described in this section.
The loss factors and convection coefficient are temperature dependent. Therefore, an iterative
procedure is required for the determination of the cable rating. However, for the purpose of illustrating
252 Part II » Evaluation of Parameters

the procedure, we will take the values of R and A, from Table A1, that is, R= 0.798 - 10-4 Q/m and
d, = 0.0218. Other required parameters of this cable are: D* = 0.0729 m, ¢ = 0.9, T; = 0.307 K -
m/W, and 73 = 0.078 K - m/W.
The Reynolds Number is obtained from equation (9.98) with the air viscosity computed from
equation (D1) in Appendix D. We will assume an average cable surface temperature of 55°C. The air
properties are evaluated at the average film temperature = (55 + 25)/2 = 40°C:

v=1.7-1075 m/s, kai = 0.0272 W/K - m

ee UD) mou = 1.29- 104

Me Fe Te:

The convective heat transfer coefficient is obtained from equation (9.99). From Table 10.5,
b = 0.193 and p = 0.618. Thus,

kairb(Re)’? 4)0.618
0.0272 - 0.193 (1.29 - 10*)
hinpe = 25.0 Wim’ -K
: Ds 0.0729

(1) Exact solution

Since the cables are shaded from solar radiation, there is no solar heat gain term in equation
(9.73). Also, dielectric losses for this cable are neglected. Substituting the numerical values into
equation (9.73) and (9.74), we obtain

nI?R(1 +A, +A2) + Wa +0 Dt —1 Déheony (0%—0%

e amb ) —x Deg(9:4—Oxny)
= 3- J]?.0.798- 1074- 1.0218— - 0.0729 -25(67—298)
—x 0.0729-0.9-5.667: 10-8(@2)*= (298)*]= 0
9*—6" =n(P-R-T + Aby)
0.0307
363
—0*
=3.[1°
-0.798.
1o-*( +.0218
-0.078)
Solving these equations, we obtain / = 910 A and 6* = 327 K or 54°C. Since the outside cable
temperature is very close to the assumed value, no further iterations are required, and the above values
represent the solution.
We can observe that the presence of a moderate wind increases the cable rating by almost 28%.
(2) Approximate solution
With a jacket emissivity of 0.90 (Table 9.10), the radiative heat transfer coefficient is obtained
from equation (9.95):

Arad= OREs(1.38 - 10° + 1.39 - 10°Aamb)= 5.667 - 10-8 - 0.9 - (1.38 - 108 + 1.39 - 108 - 25)

= 8.8 W/m’ - K

The total heat transfer coefficient is the sum of the convective and radiative coefficients:

>
h, = 25.0 + 8.8 = 33.8 W/m* -K

The external thermal resistance is now obtained from equation (9.76) as

Ty= ——= l
——______ =0.129K-m/w
1 7Dth, 1 -0.0729
-33.8 df
Chapter 9 m Thermal Resistances and Capacitances 253

The rating of the cable is thus

; 0.5
pi 90 — 25
0.798 - 10-4[0.307 + 3(1 + 0.0218) - (0.078 + 0.129) =930
A
The approximate method gives, in this case, a somewhat optimistic cable rating with the ampacity
less than 3% higher than that provided by the exact solution.

9.6.8.5 Neher-McGrath Approach. The approximation in equation (9.95) is rather

crude, except for A@,;= 0 K and Aé, = 100 K (Morgan, 1993). The approach proposed by
Neher and McGrath (1957) is based on the work performed by Buller and Neher (1950), and
involves approximating the convective and radiative heat losses as functions of the cable
surface temperature rise.
The approximation adopted by Neher and McGrath for the radiative heat transfer
coefficient is to select a temperature 6,, about midway between maximum and minimum
expected temperatures for the surface of the cable and use this value in the following equation
[see also equation (9.53)]:

Wrad
= 4.20 Dee, (1 + 0.01676,,) (9.103)

Substituting equation (9.103) into (9.76), we obtain

1
Tt; (9.104)
ua D* [Reon 4d £5 ni 1E 0,01678,,)]

Further, Neher and McGrath (1957) proposed the following simplified version of equation
(9.80):
1.05
hon = (9.105)
(Dz)

The constant in equation (9.105) was selected to best fit experimental data for 1.3, 3.5, and
10.8 in (33.02, 88.9, and 274.32 mm) black pipes.
The forced convection heat transfer coefficient was approximated by Neher and
McGrath (1957) by

- ia U (9.106)
ho. ¢ = 2.87./ — 9.106

EXAMPLE9.14
Consider again part 1 of Example 9.11 and Example 9.13. This time we will apply the equations
developed by Neher and McGrath and presented in this section.
Assuming a mean cable surface temperature of 50°C, the radiative heat coefficient obtained
from equation (9.103) is equal to

Nyaa= 4.2€, (1 + 0.01676,,) = 4.2 -0.9 - (1 + 0.0167 - 50) = 6.94 W/m? - K

(1) Free convection

The convection heat transfer coefficient is given by equation (9.105) and is equal to
254 Part II » Evaluation of Parameters

h wasteedeO
dpsegue
NORryegigdaar
oun(p:)"" (0.0729)!/4
The external thermal resistance and the rating are obtained from equations (9.104) and (4.3), respec-
tively, as

1 1
Ty = = = 0.487 K- m/W
1 D* [heon + Arad] 1 - 0.0729[2.02 + 6.94]

il eee 0.5 = 632A

0.798 - 10-* (0.307 + 3(1 + 0.0218) - (0.078 + 0.487))

(2) Forced convection

From equation (9.106), we have

RT U
eT 3 = :
18.4Wim? -K
ON teen
A 5: 0.0729 a

The external thermal resistance and the rating of the cable are equal to

il 1
Te 1D*(Aeon eee
+hraal 1 -0.0729-(18.4
+ 6.94) IhRe
jes ‘eh. ©. Se Sek We" 2et 0.52g874A
0.798
-10-4(0.307
+ 3(1+0.0218)
-(0.078+0.17))
When equation (9.103) is used to approximate radiation losses in Example 9.13, the following
results are obtained: 7, = 0.136 K - m/W, J = 920 A. Comparing the approximate results with the
exact solution obtained in part 1 of Example 9.13, it appears that, in this case, the best approximation
is provided by equation (9.99) for the convective heat transfer calculations and by the Neher-McGrath
approximation (equation 9.103) for the radiative heat transfer.

9.7 THERMAL CAPACITANCES

In Section 3.2.2, thermal capacitance was defined as the product of the volume of the material
and its specific heat [equation (3.13)]. In the following section, formulas for special and
concentric layers are presented. In all formulas, Q is the thermal capacitance (J/K - m) and
c is the specific heat of the material (J/K - m°).

9.7.1 Oil in the Conductor

The thermal capacitance of oil inside the conductor is obtained from

si His
Qo= pepsi kt -C (9.107)
where —_d*
= conductor internal diameter, m
S = conductor cross section, m2

F = factor representing unfilled cross section (usually 0.36).

Chapter 9 m Thermal Resistances and Capacitances 255

9.7.2 Conductor

The thermal capacitance of the conductor is given by

Oa ae (9.108)

9.7.3 Insulation

Representation of the capacitance of the insulation depends on the duration of the

transient. For transients lasting longer than about 1 h, the capacitance of the insulation is
divided between the conductor and the sheath positions according to a method given by
Van Wormer (1955) (see Section 3.3.1). This assumes logarithmic temperature distribution
across the dielectric during the transient, as in the steady state. The total thermal capacitance
is obtained from equation (3.13) as

1
QO;
= 4—(D?”—d*’)-c (9.109)
The portion pQ; is placed at the conductor and the portion (1 — p)Q; at the sheath,
where p is the Van Wormer coefficient defined in equation (3.19) as
1 1
(9.110)
Fa2InDNCie ee
(3) (F) 7
d- d.
From the thermal point of view, the thickness of the dielectric includes any nonmetallic
semiconducting layer either on the conductor or on the insulation.
For shorter durations (less than 1 h), it has been found necessary to divide the insulation
into two portions having equal thermal resistances. The thermal capacitance of each portion
of the insulation is then assumed to be located at its boundaries, using the Van Wormer
coefficient to split the capacitances as shown in Fig. 3.4. The thermal capacitance of the
first part of the insulation is given by

Ore gD ad, dr) (9.111)

This capacitance is split into two parts using the VanWormercoefficientas follows:
Qi=p*Qn, G2=UA-—p)On (9.112)

i 1 1
DYaap eye (9.113)
In— —-l
d. d,.
Thetotalthermalcapacitanceof thesecondpartis givenby
Ons 7 (DPD aye (9.114)

which leads to the third and fourth part of the thermal capacitance of the insulation, defined

as

03 = Pon Cu—U— pen (115)

256 Part II m Evaluation of Parameters

In the case of dielectric losses, the cable thermal circuit is the same as shown in Fig. 3.4,
with the Van Wormer coefficient apportioning to the conductor a fraction of the dielectric
thermal capacitance given by equation (9.110). For cables at voltages higher than 275 kV,
the Van Wormer coefficient is given by equation (3.23).

9.7.4 Metallic Sheath or Any Other Concentric Layer

The thermal capacitance of all other concentric layers of cable components such as
sheath, armor, jacket, armor bedding, or serving is computed by using equation (3.13).
However, one should remember that the thermal capacitances of nonmetallic layers have
to be divided into two parts using the Van Wormer factor given by equation (3.20). The
appropriate dimensions for the inner and outer diameters must be used in order to attain
sufficient accuracy for short-duration transients.

9.7.5 Reinforcing Tapes

The thermal capacitance of the reinforcing tapes over the sheath is

Or=n,
zene?+(ras)c (9.116
where n, = numberof tapes
w; = tape width, m
~;* = tape thickness, m

é* = length of lay, m
d; = mean diameter of tapes, m.

9.7.6 Armor
Thethermalcapacitanceof thearmoris obtainedfrom
nd‘
Oe nti ri ais be (9.117)

where —2, = number of armor wires

dy = mean diameter of armor wire, m
€* = length of lay, m
d? = mean diameter of armor, m

9.7.7 Pipe-type Cables

The thermal capacitances of cable components are computed as described above, and
the thermal capacitance of the oil in the pipe is obtained from

1
Ou = —(DF —3D") ¢ (9.118)
4
where D? = internal diameter of the pipe, m
D; = external diameter of one cable, m

The capacitance of the oil is divided into two equal parts.

Chapter 9 m Thermal Resistances and Capacitances 257

The thermal capacitance of the skid wires is generally neglected, but may be computed
using equation (9.117) if needed. The thermal capacitance of the metallic pipe and of the
external covering are computed using equation (3.13).

EXAMPLE 9.15
We will compute the insulation thermal capacitances of cable model No. 2 located in a PVC duct for
short-duration transients (capacitances for long-duration transients were computed in Example 3.8).
We recall that for short-duration transients, the mutual heating of the cores is neglected. The
time limit for the short-duration transient is equal to ©T - ©Qwhere ST and DQ refer to one core
only (see Section 3.3.1). The short-duration transients for cable No. 2 located in duct in air last less
then 30 min. The numerical values required for this case are as follows: D; = 30.1 mm, d. = 20.5
mm and.¢=2.4~ 10>° J/(m>- K).
From equations (9.111) and (9.114), the thermal resistance of the first and second part of the
insulation is equal to

a
Die (D}-d? —d”)-c= > (30.1-20.5—20.5”)- 10°-2.4- 10° = 371.0/K-m

Oy, 8 (DR Divas) = 7 (30.17—30.1-20.5)- 10°-2.4- 107°= 544.7/K-m

The Van Wormer coefficient is obtained from equation (9.113) as

Siacsacenae oe Seatebits
aeahlyean n® heScice
Pemaiteslee AD, ie 30.1ha8 he
Ipsee base, Win
oe
a ued, 20.5 20.5
Thus, the capacitances per core of the four parts of the insulation, obtained from equations (9.112)
and (9.115), are equal to

Oi = p*On = 0.468 - 371.0 = 173.6 J/K - m,

QO. = (1 — p*)On = (1 — 0.468) - 371.0 = 197.4 J/K - m,
O33 = p* 0; = 0.468 - 544.7 = 254.9 J/K - m,
Oia = (1 — p*) Op = (1 — 0.468) - 544.7 = 289.8 J/K-m

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load capability of cable systems,” AJEE Trans., vol. 76, part 3, pp. 752-772.
Poritsky, H. (1931), “The field due to two equally charged parallel conducting cylinders,”
J. Math. Phys., vol. 32, no. 11, pp. 213-217.
Raithby, G. D., and Hollands, K. G. T. (1985), “Natural convection,” in Handbook of Heat
Transfer, W. M. Rohsenow and J. P. Hartnett, Eds. New York: McGraw-Hill.
Russel, A. (1914), Theory of Alternating Currents. Cambridge: Cambridge University
Press.
Sellers, S. M., and Black, W. Z. (1995), “Refinements to the Neher-McGrath model for
calculating the ampacity of underground cables,’ IEEE PES Winter Meeting paper 95
WM 0115-8 PWRD; to be published in JEEE Trans. Power Delivery.
Simmons, D. M. (1923), “Cable geometry and the calculation of current-carrying capacity,”
Trans. AIEE, vol. 42, pp. 600-615.
Simmons, D. M. (1932), “Calculation of the electrical problems of underground cables,”
Elec. J., vol. 29, no. 9. pp. 395-426.
Symm, G. T. (1969), “External thermal resistance of buried cables and troughs,” Proc. IEE,
vol. 166, no. 10, pp. 1696-1698.
Tarasiewicz, E., El-Kady, M. A., and Anders, G. J. (Jan. 1987), “Generalized coefficients
of external thermal resistance for ampacity evaluation of underground multiple cable
systems,” JEEE Trans. Power Delivery, vol. PWRD-2, no. 1, pp. 15-20.
Van Geertruyden, A. (1992), “External thermal resistance of three buried single-core cables
in flat and in trefoil formation,” Laborelec Report No.DMO-RD - 92-003/AVG.
Van Geertruyden, A. (1993), “External thermal resistance of two buried single-core cables
in flat formation,” Laborelec Report No. DMO-RD - 93-002/AVG.
Van Geertruyden, A. (1994), “Internal thermal resistance of extruded cables,’ Laborelec
Report No. SMI-RD - 94-002/AVG.
Van Geertruyden, A. (1995), “Thermal resistance of the external serving for three single-core
touching cables in trefoil and in flat formations,” Laborelec Report No. SMI-APP-95-
007/E/AVG.
Wedmore, E. B. (1923), “Permissible current loading of British standard impregnated paper-
insulated electric cables,” J. EE, vol. 61, p. 517.
Weedy, B. M. (1988), Thermal Design of Underground Systems. Chichester, England:
Wiley.
Whitehead, S., and Hutchings, E. E. (1938), “Current ratings of cables for transmission and
distribution,” J. IEE, vol. 38, pp. 517-557.
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 10

Special Cable Installations

10.1 INTRODUCTION

The majority of power cables are installed underground or in free air, and the rating of
cables in such installations has been discussed in earlier chapters. However, there exists a
large number of installations for which the rating techniques discussed earlier do not apply
directly. Examples of such installations are: 1) cables on riser poles, 2) cables in open and
covered trays, 3) cables in tunnels and shafts, and 4) cables in shallow troughs. What these
installations have in common is the heat transfer mechanism from the cable surface to the
outside environment. In general, we can divide cable installations considered in this chapter
as either located within protective walls (e.g., protective riser, tray cover, tunnel walls) or
located in trays without cover. In the former case, we will talk about heat transfer between
the cable surface and the protective wall, as well as the heat transfer from the outside surface
of the wall. In the latter case, we will consider heat transfer between the cable surface and
the surrounding air.
For cables installed in air, the significant modes of heat transfer are as follows:

1. By natural or free convection when no longitudinal induced flow is present

2. By forced convection by air flow along the cables
3. By radiation of the heat from the cable surface to the ambient air, walls, or covers.

Since conduction in air accounts for a small fraction of heat transfer in the installations
under consideration in this chapter, we will, in agreement with common practice, ignore
this mode of heat transfer in further analysis.
Rating of cables in air is often based on the assumption that only natural convection and
radiation are present. Forced convection, if present, will result in a lower cable temperature
for the same current compared to the natural convection case only.

263
264 Part III m Advanced Topics

The convective heat transfer process is determined by the following factors:

1. The length, diameter, and thickness of the jacket and wall

2. Venting conditions at the ends
3. The heat flux generated in the cable, which depends on the electric current and
cable type
4. The environmental conditions such as temperature, humidity, solar radiation, wind
speed, and others.

Thermal radiation is an important heat transfer mode in air-filled cable-wall systems.

Thermal radiation transfers energy from the cable surface to the wall inside surface. This
is different from convective heat transfer. Thermal radiation from the cable surface ac-
counts for 40-60% of the total heat transfer (Keyhani and Kulacki, 1985). Thus, with free
convection and air flow at low velocities, the proportion of heat removed by radiation is
substantial and must be accounted for in calculations. The amount of heat transferred by
radiation depends upon a number of factors, including surface temperatures and emissivi-
ties.
To compute the rating of cables in air considered in this chapter, the temperatures at
various points of the thermal circuit are required. To obtain the required temperatures, a
set of energy conservation equations has to be solved. In all of the cases considered here,
the same system of equations can be solved, with the differences arising in the selection
of coefficients and constants. In the next section, we will develop a general set of energy
balance equations for a cable system surrounded by a wall, and the selection of appropriate
coefficients will be discussed in detail in the sections discussing specific installations. Since,
for cables installed on riser poles, in trays, in tunnels and in shafts, the same set of energy
conservation equations is solved, we will limit the number of numerical examples presented
in this chapter.

10.2 ENERGY CONSERVATION EQUATIONS

The following assumptions are introduced to simplify the thermal resistance calculation:

1. The process is steady state.

2. The length of the wall and the cable are large, so the heat transfer can be considered
as one dimensional.

3. The wall is opaque and the cable jacket material is radiatively gray and opaque; the
air inside the protective wall is radiatively transparent.
4. The physical properties of all materials in the cable system are temperature de-
pendent. The model takes into account the variation of physical properties with
temperature.

10.2.1 Energy Conservation Equation for the Cable Outside Surface

Considering the outside surface of the jacket under steady-state conditions, the con-
duction heat flux from its inner surface is equal to the heat loss through free convection
and thermal radiation in the air between the cable surface and the wall. The energy balance
Chapter 10 m Special Cable Installations 265

equation (2.8) takes the form

W,= Weonv.s
a Weave (10.1)
where Weonv,s= natural convection heat transfer rate between the cable outside
surface and the air per unit length, W/m
Wrad,s—w = thermal radiation heat transfer rate between the wall inner surface
and the cable outside surface, per unit length, W/m

W, = total energy per unit length generated within the cable. Its value is
given by equation (4.6):

W,= W,+ Wa (10.2)

ThejoulelossesW,aredefined
inequation
(4.1):
W,=n-W-(1 +A, +A2)=n-I?RO +A, +A2) (10.3)
where / is the conductor current and R is the conductor ac resistance at operating temper-
ature. 4; and Az are the sheath and armor loss factors and n is the number of conductors in
the cable.
The dielectric losses are obtained form equation (6.4):

Wa= wCU;
tans (10.4)
The dielectric loss formula is developed in Chapter 6 and the variables are defined there.
Computation of all of these quantities is discussed in detail in this book.
Newton’s law of cooling [equation (2.2)] is used to determine the convection heat
transfer rate between the outside surface of the cable and the air inside the walls:

Weonv,s
= hs (8s a gas) As (10.5)

where h, = natural convection coefficient at the surface of the cable, W/K-m?

0, = average temperature of the cable outside surface, °C
Oa; = air temperature in the gap, °C
A, = area of the cable (or cable mass) outside surface, (m7), for unit length.

The net radiation heat transfer rate between the cable and the inside surface of the wall
is based on the radiative exchange between two surfaces [equation (2.6)]:

Wrads—w
= AsrFywo (0%catOe) (10.6)
where og = Boltzmann constant, equal to 5.67 x 10-8 W/m? - K*)
6* = average temperature of the wall inner surface, K
F, » = thermal radiation shape factor; its value depends on the geometry of
the system
A,, = area of the cable surface effective for heat radiation, (m), for unit length
of the cable.
266 Part II] m Advanced Topics

10.2.2 Energy Conservation Equation for the WallInside Surface

For the wall inside surface, the energy transferred by conduction through the wall
material is equal to the energy transferred through convection and radiation on the inner
surface of the wall. Thus, the energy conservation equation under steady-state conditions

1S
Weond,
w—o= YY
conv,
war Wyrad,s—w (10.7)
where Weond.w—o = heat conduction rate from the wall inner surface to its outside
surface per unit length, W/m

Weonv.w = natural convection between the wall inside surface and the air per
unit length, W/m.

Similar to the heat transfer at the cable surface, the free convection heat transfer in the
wall inner surface is given by

Weony,w
= hw(Ogas
—Oy)Aw (10.8)
where h,, is the natural convection coefficient at this surface, W/K-m? and A,, is the area
of the wall per unit length, (m7).
The conduction part of the heat transfer is simply

Owa A
Weond,w—o
> (10.9)
Ty

where 0, = average temperature at the wall outside surface, °C

T,;' = thermal resistance of the wall per unit length, K-m/W

10.2.3 Energy Conservation Equation for the WallOutside Surface

In this section, we will consider cable installations which have air as amedium outside
the wall. At the outside surface of the wall, the energy transferred through the wall material
by conduction and the energy gain due to solar radiation are balanced by the convective and
radiative energy losses to the atmosphere. Thus, the energy conservation equation is

Weond,w—o
AeWoot=aWeonv,o
“hrWrad.o—sur (10.10)
where Weonv,o
= naturalconvectionheattransferbetweenthe walloutsidesurface
and atmosphere air, per unit length, W/m
W,ad.o—sur
= thermal radiation heat transfer rate between wall surface and
surrounding objects, per unit length, W/m
Wo, = solar radiation absorbed by the wall surface, per unit length, W/m.
This quantity is only considered for installations exposed to
solar radiation.

The convection heat transfer between the wall and the surroundings is

Weonv,o= ho(% —Pamb)Ao (10.1 1)

Chapter 10 # Special Cable Installations 267

where ho = natural or forced convection coefficient at this surface, W/K-m2

Oamb= air temperature in atmosphere, °C
A, = area of the outside wall surface, per unit length, (m7).

The thermal radiation heat transfer rate between the outside surface of the wall and
the surroundings is based on the assumption that the wall is a small body enclosed by the
surroundings. The net radiative exchange between these surfaces is

Wradjo—sur
= Ao€ooR
(04 = as) m
(10.12)
where €&, = emissivity of the wall outside surface.

If the solar flux incident on one square meter of the surface is H,, then the solar radiation
absorbed by the wall is

Woot= AosoH (10.13)

where Q@_
= wall surface absorptivity to solar radiation, which is generally different
from the emissivity of the wall
Aos = equivalent area of the wall per unit length perpendicular to sun rays, m*

10.2.4 Energy Conservation Equations

Equations (10.3)—(10.13) are the basic energy conservation equations for the cable-
wall system. There are three unknown temperatures, 0,, 6,,, and 6,. These temperatures
are thus computed from the following three equations:

n- PR a Ayar A2)=I Wa= (hh —Deas

)AsSEAylonosor” = Ge
Oy—95 = hy,(Ocas
—Ow Aw+Ar Fewor(O™ —04
Bi
(Ozas ) Ae q oR( S ae) (10.14)

Ow
ar * 6 + AostoH= ho(Bo—Pamb) *
Ao+ Aveoon(O** *
—0%)
4
In the above equations, the equivalent conductor electric resistance and the convection
coefficients are temperature dependent. Before equations (10.14) can be solved, the values
of the intensity of solar radiation and the convection coefficients have to be selected. These
subjects are discussed in the following sections.

10.3 CABLES ON RISER POLES

10.3.1 Introduction

Power delivery systems frequently consist of a combination of overhead lines and

underground cables. In most cases, the underground cable system is connected to the
overhead line through a short section of cable located in a protective riser. Figure 10.1
shows a cross section of a submarine cable installed on a riser pole with a protective guard.
268 Part III m Advanced Topics

Jute serving

Steel-wire armor

Semiconducting layers

XLPE insulation

Conductor

Air under guard

Cable guard

Wood pole

Figure 10.1 Cross section of a submarine cable on riser pole (Cress and Motlis, 1991).

The protective guard is often simply referred to as a riser. The current-carrying capacity
of the composite system is limited by that segment of the system that operates at the
maximum temperature. Very often, the riser-pole portion of the cable system will be the
limiting segment.
Considering the importance of accurately rating power cable systems consisting of
cables on riser poles, Hartlein and Black (1983) introduced a mathematical model to repre-
sent such systems. The model is based on a modified thermal circuit consisting of thermal
resistances separated by local system temperatures. The analysis results in a number of
algebraic equations, similar to equations (10.14), that are simultaneously solved for the
system temperatures for a given cable ampacity. The theoretical developments were sub-
stantiated by experimental evidence for cables in protective risers located indoors without
solar radiation and wind.
The pioneering work by Hartlein and Black suffered from gaps in knowledge (no for-
mulas were given for the computation of heat transfer coefficients under certain conditions)
and, in several cases, required assumptions which are incompatible with typical cable-riser
geometry. Much new experimental work has been reported during the 12-year period since
the publication of their paper. Anders and Gu (Anders, 1996) have updated the work of
Hartlein and Black (1983) by redefining the mathematical model and supplementing infor-
mation lacking in their work. A comparison of both models was offered. The new model
was tested against Hartlein’s and Black’s experimental data, as well as the data for outdoor
tests reported by Cress and Motlis (1991). The new model is implemented in the CEA’s
Cable Ampacity Program (CAP) (Anders et al., 1990).

10.3.2 Thermal Model

The assumption used in developing the mathematical model for the cable-riser system
is that the cable and the riser are concentric bodies with their length much greater than their
diameters. Equations (10.14) can be used to determine the required temperatures. The
Chapter 10 m Special Cable Installations 269

radiation shape factor is obtained considering two long concentric cylinders. In the case of
a single cable in the riser, we have

Peeler) eerA.Gn)/ Awen) (10.15)

where _o,= reflectivity of the cable outside surface

€; = emissivity of the cable outside surface
Oy = reflectivity of the wall inner surface
Ey = emissivity of the wall inner surface
A;;= 0 D>, Ay =1D3, Ao = 1 D5.

D>, D3, and D*, (m) are the cable outside diameter and riser inside and outside
diameters, respectively. The maximum area exposed to solar radiation is LD*. When
several cables are present, the mutual radian area between them must be subtracted from
the area radiating to the riser inner surface. The most common installations have either one
or three cables inside the guard. The effective radiating area to the guard walls is obtained
as follows (Weedy, 1988).
Elastic bands are imagined stretched around two arbitrary concave surfaces (see
Fig. 10.2). The lengths of the bands stretched between the two surfaces are found as
follows. The length of the internal band is Jin, = AD + BC’C where C’ is the point of
intersection of BCwith surface C D. The length of the external bands is /.x, = AC + BD.
The mutual radiation area A,, per unit axial length is equal to half the difference in lengths;
that is,

A Tint2—Text_ ———$—__——
AD+BC'’C
Z
AC+BD
Z \ 10.16

Application of the method to three cables in touching trefoil formation yields (Weedy, 1988)

Asp = 30D? =—3.-0.618D; (10.17)

In this case, the radiation shape factor has the form

Foy = (1+ 65/€s + CAsow/AwEw)

eens col (10.18)

Figure 10.2 Calculation of mutual radiation area.

(Weedy, 1988)
270 Part II] m Advanced Topics

Also,

10.3.3 Intensity of Solar Radiation

The solar radiation absorbed by the riser depends on the solar altitude angle a, the
angle of the solar beam with respect to the axis of the conductor, the diffuse sky radiation,
and the albedo (reflectance) of the surface of the ground beneath the conductor (Morgan,
1982). Assuming that the reflected-beam component hits the guard at the same angle a
as the direct beam, and neglecting the diffuse radiation, a simplified formula for the solar
intensity is obtained as

H = Hgcosa (10.19)

where Hz is the intensity of the direct solar beam on a surface normal to the beam. Thus,
if the intensity of solar radiation is obtained from measurements, the intensity of solar
radiation incident on the guard is obtained by multiplying the measured value by cos a.
The angle a depends on the altitude angle @,the declination 5,, and the hour angle of
the sun w,, with solar noon being zero, mornings being positive, and afternoons negative
(Morgan, 1982; Cress and Motlis, 1991):

sina = sind, sing + cos 4, Cos @COSWp (10.20)

The declination angle depends on the day of the year Ny as

ds;= 23.45 sin[360(284 + Naz)/365] (10.21)

The hour angle w, increases by 15° for every hour from zero at solar noon, and is
positive before noon. The hour angle at sunrise wo is found from equation (10.20), putting
C= 0;
wo = cos !(—tan 6, tand)

hence, the hour angle at any time f, after sunrise is given by

@, = cos '(—tan6, tan b) — St, (10.22)

To obtain solar time, add 4 min/degree of longitude east of standard time, or subtract
4 min/degree west of standard time.
The conductor temperature will normally show significant dependence on the preced-
ing as well as the immediate solar radiation intensity. However, only a single value of H
is used in the ampacity computations. Cress and Motlis (1991) introduced the concept of
equivalent solar radiation intensity H defined as a constant value that, at time tn, would
bring the conductor to the same temperature as would exposure to the time-varying solar
radiation values H;. They derived the following equations to compute the value of H:
Chapter 10 # Spécial Cable Installations 271

>aw; (10.23)
Dy= ; az=e!/*, b=1-a
where T is the time constant of the cable/guard system.
Another factor to consider is the movement of the sun from east to west during the day.
This movement results in only partial exposure of the guard to the sun’s rays. Considering
an azimuth angle and a usual shape of the riser (see Fig. 10.3), Cress and Motlis (1991)
developed a computational method to determine the maximum amount of guard surface
that could be exposed to incident solar radiation at any one time.

Figure 10.3 Angular solar radiation on riser sur-

face. (Cress and Motlic, 1991)

Considering Fig. 10.3, we note that the solar power, per meter of riser, on the short
tangent of length x is
i cos b(s) =" @ cos B)
Thus, the portion of the semi-circular guard length that is exposed to H can be cal-
culated as the sum of the incremental projections similar to x cos B. Including the straight
and curved portions of the riser, the length L. exposed to H can be expressed as

180
Le =s-cos(90 —y) + Daal - tan(Ae) - cos(90 —y —€) (10.24)
e=0
where all the quantities are defined in Fig. 10.3. For a given riser geometry and azimuth
angle, the exposed length is computed from equation (10.24) with the aid of a computer
272 Part III m Advanced Topics

program. Numerical studies performed by Cress and Motlis (1991) have shown that, fora
typical aluminum guard, the maximum surface that could be exposed to solar radiation was
about 32% of the total riser surface. This occurred when y = 63°.

10.3.4 Convection Coefficients

The cable and the riser form a vertical annulus. If the temperature of the cable or riser
is different from the air temperature in atmosphere, natural convection occurs in the annulus
gap. This natural convection makes the heat transfer processes in cable-riser systems very
complicated.
Convection coefficients required in equations (10.14) are summarized in Tables 10.1—
10.3. The basis for the selection of these coefficients is discussed in Anders (1995) and
briefly described below.

TABLE 10.1 Convection Correlations for the Cable Outside Surface

Case Model Correlation Comments References

I. Closed h,=Nu-k,i,/6 1. Correlations apply 1. Keyhani and Kulacki

at top Nu = 0.797Ra077G~0.052 K0-505 for vertical annulus. (1985)
and for Ra < 363K
025G0.7 2. Air properties 2. Keyhani et al.
bottom A ae hyrk evaluated at film (1985)
Nu=0.188Ra~"G x temperature: 3. Keyhani er al.
for 363K "Gor Raa 310° O,=(0, + ,)/2 (1986)
Ra = Pr-gB(0,- 0,,) 6°/v? :

II. Open SE NWOVe ORIRS 1. C,1s applicable for 1. Raithby and Hollands
at Nu = Gr: Pr /9.07 if Gr-Pr < 10 fully developed (1985)
both mK -p.\0.25 flow. It may vary 2. El-Shaarawi et al.
ends uaa) fe in other flow (1981)
Gra $P%= Gamv)S regimes. 3. Al-Nimr (1993)
Ly? 2. C, is fitted when
C,, = 0.46K + 0.54 ED Da
4. 6 = (0, + Pamp) / 2

II. Open h,=Nu-k,;,/L 1. Applies to vertical Morgan (1982, 1992,

at the Nu = c(Gr: Pr)" cylinder, not to 1993)
top and te 39 vertical annulus.
closed GES BBCP )Laty Annular effects are

at not considered.
bottom 6, = (8, + A, + 20 amb)/ 4

10.3.4.1 Riser Outside Surface. The convection heat transfer on the outside surface
of the riser includes natural and forced convection. Normally, the forced convection is much
stronger than free convection. In the case of natural convection, the vertical cylinder free
convection correlation (Morgan, 1982) can be used. In the case of forced convection, the
heat transfer coefficient can be calculated as described in Holman (1990), Incropera and De
Witt (1990), or Burmeister (1983).
Mixed natural and forced convection may exist with very low wind speed conditions
only. The natural convective flow is vertically upwards if the cable is warmer than the air. If
the wind direction is perpendicular to that of the natural convective flow, the total convective
Chapter 10 » Special Cable Installations 273

TABLE 10.2 Convection Correlations for the Riser Inside Surface

—_—_———_:?.XX ... —ma_axe
Case Model Correlations Comments References

I. Closed Same as the cable outside surface Same as case I in 1. Keyhani et al. (1985,
at top Re LP Table 10.2. 1985, 1986)
and sitet 58
bottom

II. Open p= TESNI OS OR VSS) . C, is applicable for . Raithby and Hollands

at Nu= Gr: Pr / 9.07 if Gr: Pr < 10 fully developed (1985)
both Nu = 0.62 (Gr: Pr)? flow. It may vary . El-Shaarawi et al.
ends Pr gB (0, — Ormp)54 in other flow (1981)
Lv2 regimes. . Al-Nimr (1993)
C,=0.6/K+04 . C, is fitted when
1<Dy/D.<5
. 0, = (8, + Ceri 2

III. Open at h,, = Nu> k,j,/ (Dg/ 2) . Developed for a . Dryer (1985)
top and Nu = (Gr: Pr) / 400 vertical circular . Martin and Cohen
closed at if Gr-Pr< 200 duct closed at its (1954)
bottom Nu = 0.35(Gr° Pr)?-28 lower end, not for . Martin (1955)
Gr= gB(Oy, - 9,,)(Dj/ 2) / (Lv?) annularlayer.
. OF= 8,

TABLE10.3 ConvectionCorrelationsfor the RiserOutsideSurface

Correlationsin ProposedModel Comments References

No wind, h,,=Nu:k,i,/ L O-=(0, + Oamp)/ 2 Morgan (1982)
natural Nu = c(Gr- Pr)”
convection Gr=26(6, = 0.) Ti"
Forced hop= Nuckyip/Di O-= (05+ amp) /2 Holman(1990)
convection Nu=b(Re)? Pr!4
(Re)= Vy;D,/¥

heat transfer coefficient on the riser outside surface can be calculated by (Morgan, 1982,
1992, 1993)
2

hahaa Phe
a?
0,
(10.25)

We will assume in the proposed model that the wind direction is perpendicular to the
natural convective flow.

10.3.4.2 Convection in the Air Gap.

Case I: Riser Closed at the Top and Bottom
An increasing number of applications in microelectronic packaging, nuclear engineer-
ing, and solar systems resulted in remarkable attention being focused on natural convection
heat transfer inside confined spaces. The correlations for this case are taken from Keyhani
and Kulacki (1985). The gas temperature is equal to 6, in equation (10.5) and 6, in equa-
tion (10.8).
274 Part II] m Advanced Topics

Case II: Riser Open at the Top and Bottom

Considerable research has been done in recent years on the laminar natural convection
in open-ended vertical concentric annuli. Joshi (1987, 1988) studied the natural convection
in an isothermal vertical annular duct, and discovered that the heat transfer strongly depends
on the radius ratio K = Dy/D,. If K is less than 1.2, the solutions for annular ducts
coincide with those for parallel plates. Natural convection in the air gap formed by two
vertical parallel plates has been studied for many years, and corresponding heat transfer
correlations can be found in many publications. Al-Nimr (1993) studied free convection in
vertical concentric annuli and the dependence of heat transfer rate on the radius ratio K. If
K = 1, the Nusselt Number at the cable outside surface is equal to that on the riser inside
surface. As K grows, the Nusselt Number increases by a factor of about (0.46K + 0.54)
on the cable outside surface, and decreases by a factor of about (0.6/K + 0.4) on the
riser inside surface. In the proposed model, the convective heat transfer coefficients are
calculated based on these results and the convection correlations for vertical slot formed by
two parallel plates. The gas temperature is equal to amp in equations (10.5) and (10.8).

Case III: Riser Open at the Top and Closed at the Bottom
There are very few published data available for the heat transfer in this particular
configuration. Hartlein and Black (1983) used the natural convection correlation for vertical
plates to calculate the convection coefficient at the cable outside surface. They also applied
the glycerin free convection correlation in open thermosyphons to calculate the heat transfer
coefficient of air at the inside surface of the riser. In the model presented in Table 10.1,
the natural convection heat transfer correlation for vertical cylinders is used for the outside
surface of the cable (Morgan, 1982).
The correlation for air natural convection in open thermosyphons is used for the
inside surface of the riser. An open thermosyphon is a device to transfer heat from a
high-temperature region to a low-temperature reservoir or atmosphere. The simplest open
thermosyphon is a vertical tube sealed at its lower end. The tube is heated by the high-
temperature heat source from which energy is transferred to the outside by the natural
convection of the fluid in the tube. In the proposed model, the natural convection corre-
lation for open thermosyphons is adopted to calculate the heat transfer between the riser
inside surface and the gas (Dryer, 1978; Martin and Cohen, 1954; Martin, 1955). The gas
temperature is equal to 6, in equation (10.5) and (6, + 0, + 26amb)/4 in equation (10.8).
As an alternative to the correlations given in Table 10.3 for this case, Morgan (1995)
suggested to use the correlation for water given by Seki et al. (1980), Nu = 0.204(Gr-Pr)°25,
in the range 4- 10* < Ra < 4- 10°, where the Nusselt and Rayleigh Numbers are based on
(Dw ae D;)/2Ds.
In the above tables, 6 is the temperature of the film at which gas properties are
evaluated (see Appendix D). Also,

G = aspect ratio, G= L/6

K = diameter ratio, K = Dz/D,

The constants c, n, b, and p are given in Tables 10.4 and 10.5.

ee ee ee eee, eee eee

EXAMPLE10.1
The sample calculations presented in this example illustrate the steps required to determine the tem-
perature profile for a given cable and riser system. We will examine model cable No. 1 located
Chapter 10 » Special Cable Installations 275

TABLE 10.4 Constants for Nusselt Number in Natural

Convection (Morgan, 1982)

c n Restrictions

0.675 0.058 10-!°< GR: PR < 107

1.02 0.148 10°2< GR: PR < 10?
0.850 0.188 10?< GR: PR < 104
0.480 0.250 10*< GR: PR < 10’
0.125 0.333 10’<GR:PR< 10!”

TABLE 10.5 Constants for Nusselt Num-

ber in Forced Convection (Holman, 1990)

Re b Dp

0.4-4 0.989 0.330

4-40 0.911 0.385
40-4000 0.683 0.466
4000-40 000 0.193 0.618
40 000-400 000 0.0266 0.805

in a riser as specified in Hartlein and Black (1983). Riser dimensions are: L = 4.87 m, D} =
0.1016 m, and D* = 0.1144 m, thermal resistivity p¢ = 3.33 K-m/W, emissivity ¢, = €, = 0.7,
and solar absorption coefficient w, = 0.4. Assume that the riser is closed at the bottom and vented
at the top. The wind velocity is equal to 3 m/s and the intensity of solar radiation is 500 W/m’.
Each of the three single-core cables is located in a separate riser. The air ambient temperature is
20°C. The computed parameters listed in Appendix A are: W. = 30.85 W/m, A; = 0.214, and
T; = 0.104 K-m/W.
We will take the values of the skin and proximity effects as well as the loss factors from Table
A1. For the installation conditions examined in this example, these values will be somewhat different
from those listed in Table Al. However, the error will be very small and the computations much
shorter.
The solution of equations (10.14) is iterative in nature. To illustrate the method of calculation
without resorting to an iterative process, the correct conductor temperature is initially assumed.
Assume the conductor temperature 6, is 66°C. The conductor losses are 30.82 W/m and the total
losses are 33.38 W/m. Dielectric losses are neglected.
The temperature at the outside surface of the cable is obtained from equation (4.2) and is equal to

0, = 0 — WeT, — [WoC + Ay)I(T3) = 66 — 30.85 - 0.214 — 33.59 - 0.104 = 55.9°C

The temperature at the inside surface of the riser can be calculated from the first equation in (10.14).
The air thermophysical properties are evaluated at a film temperature 0, = (6; + Oy + 26amv)/4-
At this point, 0, is not known; however, it can be approximated for the purpose of performing the
calculations. If 6,, = 26°C, then

O¢ = (6, + Oy + 2Oamb)/4 = (55.9 + 26.0 + 2 - 20)/4 = 30.5°C

At this temperature, the air thermophysical properties are (see Appendix D)

p=3.3- 10 WK
v = 16.1 -10~° m?/s
Pe—10/71
Kair= 0.0265 W/K-m
276 Part III m Advanced Topics

Because the cable and riser are opaque,

Casale 65 —0) I One 05

The thermal radiation shape factor is obtained from equation (10.15) and is equal to

Fy = (1 +6;/€; + AsOw/AwEw)! = [1 +.0.1/0.9 + 2 - 0.0358 - 0.3/(7 - 0.1144 -0.7)]“' = 0.8

The convection coefficient is obtained from Table 10.1. For the computation of the Grashoff
Number, we assume the inside riser temperature to be 26°C:

Gr = gB(0, —0,)L3/v? = 9.81 - 3.3 - 1073(55.9 —26)4.877/(16.1 - 10-°)? = 4.3 - 10'!

Gr-Pr = 4.3- 10!' .0.71 =3.1- 10"!
Nu = ¢(Gr-Pr)” = 0.125(3.1 - 10!'')°333 = 838.6

h, = Nu- k,ir/L = 838.6 - 0.0265/4.87 = 4.56 W/m?-K

The first equation in (10.14) thus becomes

W, = hs (0, —Ogas)As+ AsrFs,won(O44—0*4)

33.38 = 4.56 - (55.9 —6,,) -0.0358
409: -0:0358.2.0.8 +5.667 110° 1(55:9: 273)° —(One 273) 7]

The solution to this equation is 07 = 299.7 K or 6,, = 26.7°C.

The value of the convective heat transfer from the surface of the cable is given by the first term
in the last equation:
Ween lee Win

Next, we will use the second equation in (10.14) to compute the riser outside temperature. The air
properties are evaluated at the riser inside surface temperature. At 27°C, the air properties are as
follows:
6=383° 104K
P= 15.8>10-° mss
lor (OETA
Kaiy= 0.0262 W/K-m
The heat convection coefficient for the inside surface of the riser is obtained from Table 10.2:

Gr = 8B(Ogas- Ow)(D*/2)*/(Lv?)

O813 Sal ( 55.9

ts+26.7
mii+e2.20 26.1)
0.05084
*, 4.87-(15.8-10-6)2 ea
Gr-Pr = 700 - 0.71 = 497
Nu = 0.35(Gr-Pr)°”® = 0.35(497)°?8 = 1.99
hy = Nu: kyi,/(Da/2) = 1.99 - 0.0262/0.0508 = 1.03 W/m2-K

The thermal resistance per unit length of the riser is equal to

i” In Daven
ays 1144
ass Dyan
= OLTO
IGE
iememe
=0,063
K.
Chapter 10 m Special Cable Installations 277

The energy balance equation (10.7) [second equation in (10.14)] becomes

Owa. Oo
= hy(gas —Ow)Aw+ AgrFs,woR
(04 —0*4)

[(55.9 + 273)* —(26.7 + 273)*]

The solution to this equation gives 6, = 25.4°C.

Finally, equation (10.10) is used to compute the ambient temperature. The air properties are
evaluated at the outside riser film temperature:

O¢ = 0.5(25.4 + 20) = 22.7°C

At this temperature, the air properties are as follows:

6 =3.38- 107 1/K

v = 15.4-10-° m2/s
Pps (OL
7/il
Kair= 0.0259 W/K:m

And the heat convection coefficient is obtained from Table 10.3:

Re = V,i,D?/v = 3.0: 0.1144/(15.4 - 10-°) = 2.2 - 104
Nu = b(Re)?Pr'/? = 0.193(2.2 - 10*)6!8 . 0.719333= 83.1
ho, ¢= Nu- kgir/ D*= 83.1 -0.0259/0.1144 = 18.8 W/m?-K

Before we apply equation (10.10), we have to determine the area of the riser exposed to solar
radiation. Considering the results obtained by Cress and Motlis (1991) discussed in Section 10.3.3,
we assume that A,, = 0.32 - 4.87 -0.1144 = 0.178 m? per unit length. The energy balance equation
(10.10) [third equation in (10.14)] becomes

hy = Ch,
T! a ANgsoalal
= Aoho(@a Oamb) ie Ao€oOR (ox*= hak)
a
PRY,
Uf PP + 0.178 - 0.4 - 500
0.063
=n -0.1144[18.8 - (25.4 + 273 —6*,,) + 0.7 - 5.667 - 10-8
CLES Tay

Again, a numerical technique was used to find @:mp= 18.5°C This value is close to the actual
ambient temperature; hence, the assumed conductor temperature is correct. If the calculated ambient
temperature is significantly different from the given value, the equations should be solved again with
different values for the initial conditions. In acomputer program, equations (10.14) should be solved
simultaneously using iterative techniques.

10.4 CABLES IN TRAYS

A typical cable tray installation which is found in the electric power generation and dis-
tribution industry can be visualized as a 3 in deep, 24 in wide metal trough containing
278 Part II] m Advanced Topics

anywhere from to 20 to 400 randomly arranged cables ranging in size from #12 AWG to
750 kemil. This array of cables is usually secured along the cable tray to prevent it from
shifting if additional cables should be pulled into the tray. In many cases, especially in
nuclear power plants, the trays are covered with fire-protection wrap around the raceway.
Because of the very strong mutual heating effects, the ampacity of cables in trays is usually
lower than computed with the formulas given in earlier chapters. In the following sections,
we will develop models for the ampacity computations of cables in open-top and enclosed
cable trays. As a special case of the enclosed tray, we will consider cables in fire-protection
wrapped cable trays.

10.4.1 Cables in Single Open-Top Cable Tray

10.4.1.1 Introduction. Since 1975, the accepted technique in North America for
calculating the ampacity of power cables in trays has been the use of tables provided by
ICEA/NEMA Standard P54-440 (ICEA, 1986). These tables are based upon a thermal
model originally proposed by Stolpe (1971) that assumes that every cable in the cable
tray carries the maximum current producing the maximum cable bundle temperature. The
ICEA Standard also assumes that the heat generated in the cables is uniformly distributed
throughout the cross section of the cable mass.
In order to remove the conservatism in the thermal model based on the above as-
sumptions, Harshe and Black (1994) proposed a simple thermal model that can be used to
determine the maximum temperature of a mass of cables in a single, open-top horizontal
cable tray. The model accounts for load diversity within the tray by providing two different
loading options. The thermal model has been verified by comparing predicted cable bundle
temperatures with temperatures measured at an actual installation. The model is a subset
of equations (10.14), and is described in the following section.

10.4.1.2 Thermal Model. The basis of the thermal model that relates the current in
the conductors and the temperature distribution within the cable bundle is the conservation
of energy; that is, all of the heat that is generated in the cable bundle must be transferred
through the cables to the surface of the cable bundle and from the surface to the air by
radiation and convection. In the following model, the cable bundle may consist of any
number of cables with arbitrary conductor sizes. Each cable in the bundle is assumed to
have a known current, and the thermal model calculates the temperature profile throughout
the cable bundle. The model accounts for nonuniform heating within the cable tray by
providing two different loading options. The first option assumes that the cables generate
heat uniformly across the cable-tray cross section; Harshe and Black called this the Uniform
Model. The second option assumes that the heavily loaded cables are concentrated at the
cable tray centerline, and that they are surrounded by more lightly loaded cables. This
model is referred to as the Hot-Spot Model, and is schematically shown in Fig. 10.4. The
mathematical formulation assumes that the cables are located indoors in still air.
The energy balance for the bundle of N cables in the cable tray is a combination of
equations (10.2) and (10.10) with the cover removed, and can be expressed as

N
Wrotal
= Ss nj Ril?
= hsAx(0;—Oamb)
+ 6A,(0%*
—0*4) m
(10.26)
i=]

where nj is the number of cables in the subgroup of cables of the same size, “total” refers to
the total value for the entire cable bundle, and the subscript s denotes the surface value. The
Chapter 10 m Special Cable Installations 279

O.7

cantr|_
Heavily
loaded
cables
O.p
pf
WwW -

Figure 10.4 Geometry of the Hot-Spot Model assumption (Harshe and Black 1994).

remaining quantities are defined in Section 10.2. The temperature at the bottom surface of
the bundle 6, will, in general, be different from the temperature 6,7 at the top of the cable
mass because the free convection coefficients are different. The convective heat transfer
coefficient for the bottom surface of the cable trays is given by (Harshe and Black, 1994)

hp=0.248kai.w ea)0.25
>[(%)(Ose (10.27)
and the convective coefficient for the top surface of the cable tray depends upon the value
for the Rayleigh Number, Ra:

hr = 0.496k,iew°° (2) Or —a) 0.25

eB
for 10°<Ra< 10’
v 0.333 (10.28)
hr = 0.13
4kir (2) (@.r =ian) for 10’<Ra< 10"°
v
The thermal properties in equations (10.27)—(10.28) are evaluated at an average film
temperature 0¢ = (6, + amb) /2 and their computation is described in Appendix D, with the
value of @equal to 6, or 6,7, depending on which surface of the cable mass is considered.
The Rayleigh Number is given by

Ra = 0.71w°* (2) (6, —Oamb) (10.29)

The relationship between the centerline, or maximum temperature within the cable
bundle, and the temperature of the surface of the bundle is obtained by applying equation
(3.6):

Omax
=O53 H+Hy
+= iz (+ +iw)+WuH.tSi) sewH
(> i | (10.30)
w
where p is the thermal resistivity of the cable bundle including conductor, insulation, and
encapsulated air, K-m/W. The Wy, Wm, and W, are the watts generated per unit length of
cable tray in the three layers.
280 Part III m Advanced Topics

If we make a simplification proposed by Harshe and Black (1994), and assume that
6,~ = Os7 = O;, then the product of the convective heat transfer coefficient and the surface
area 1S

h,As = wihg +hr) (10.31)

and the relationship between the centerline, or maximum temperature within the cable
bundle, and the temperature of the surface of the bundle is

ie Seen iy (eee H ae iw)Wi(H i i) aiwH ine| (10.32

Aw 2 5 2
If the heat is generated uniformly throughout the cable bundle, equation (10.32) reduces to

Omax—0; ==pik (=) (10335)

If the current in each cable is known, equation (10.32) can be used to calculate the cable
bundle surface temperature. The maximum cable temperature within the bundle occurs at
the centerline, and is a function of the distribution of energy generated throughout the cable
bundle. Referring to Fig. 10.4, the depths of the three cable layers are

frenet Yind?; x=H,M,L (10.34)

S|

where w = widthof the cable tray,m

H, M, L = subscripts representing the value of heavily, moderately, and lightly
loaded cables, respectively
€ = geometric parameter that accounts for the packing between the
cables in the bundle and
€ = 1.0 for cable packing used in ICEA Standard P54-440
—€
= ().786 for cable packing using circular cross-section area.

The thermal model given by equations (10.26)—(10.30) is solved iteratively because the
heat transfer coefficients and electrical resistances of the metallic parts of the cables are both
functions of the cable temperature. In the computational algorithms, the resistances of the
cables outside the centerline can conservatively be assumed at the maximum or centerline
temperature for the given cable loading. Harshe and Black (1994) report that selecting the
initial surface and centerline temperatures at 10 and 20°C, respectively, above the ambient
temperature results in convergence within five iterations. The computational procedure is
very similar to that described in Example 10.1, and therefore will not be repeated here.

10.4.2 Cables in Covered Trays

Raceway systems in electric generating stations are often enclosed. Engmann (1984)
presented a method for the calculation of ampacity of cables in a covered tray. He later
extended the model to trays with raised covers (Engmann, 1986). In 1989, Save and
Engmann further extended the above techniques to fire-protection wrapped cable trays.
Chapter 10 m Special Cable Installations 281

Fire barrier material, material support

or cable tray cover

Fire barrier material

SAAAOALOA
GOI
OTU
STI OOO STUNSTINSTITITSN

aK

Packed cable mass

SOC ESS OOOO SSDS

Fire 2b Bottom of packed

barrier material cable Mass

Figure 10.5 Cables in fire-protection wrapped cable tray (Save and Engmann, 1989).

The thermal model of the covered tray cable system is a combination of the models
describing cables in riser poles and cables in open-top trays. The geometry of a typical
thermal system is shown in Fig. 10.5.
The energy balance equations (10.14) can be used to describe the heat transfer process
from the top of the covered tray to the environment. With the absence of solar radiation,
the modified equations (10.14) are as follows:

Wiotal= hs(9s a Deas)

Asa AvrPo yORCe aeo)
Oyi A = hy (Oas—Ow)
Aw + Asr woB
Fs,woR
0%
Ti (gas JAw+ sr —OM
— Fy) (10.35)

by —Wp
—ho(8%a, Camb
Ae a AogEoOB
oe a pe) m
Ty

Subscripts s, w, and o denote the top surface of the cable bundle, the inside surface of
the cover, and the outside surface of the cover, respectively. The heat transfer coefficients
h, and h,, are defined as follows.
Coefficient h, for the outside surface of the tray cover is the same as h,,, given in Table
10.3 with the length L replaced by w/2 and constant c = 0.57. Even though turbulent flow
conditions normally do not apply in this case (10’ < Ra < 10!) ,,the following correlations
proposed by Raithby and Hollands (1985) can be used for a wide range of the Rayleigh
Number (Ra > 1):

Nu = [(Nu;)!° + (Nu,)!°]*!
(10.36)
Nu; = 1.4/In(1 + 0.602Ra°?>), Nu, = 0.14Ra®?*?

The coefficient h; is given by

282 Part III » Advanced Topics

h, = hy =Nu- kair/5
Nu=1 if Gr< 1708
Nu = 0.195Gr°2> ss if--:10* < GrPr < 4- 10° (10.37)
Nu = 0.069(Gr-Pr)°?*8 if 4-10° < GrPr < 10°
Gr = eB(Os“FW
Oy)5° /v*

where 6 (m) is the thickness of the air gap in the tray. Air thermal properties are evaluated
at the average film temperature.
Equation (10.26) can be applied for the bottom of the tray. Subscript s pertains here to
the bottom surface of the tray, and the convection coefficient is given by equation (10.27).
Non-uniformly heated bundle can also be considered in this case by applying equation
(10.30) [Harshe and Black, (1996)].

10.4.2.1 Cables In Fire-Protection Wrapped Tray. Raceway systems in nuclear

generating stations are often enclosed with protective wrap to meet regulatory or underwriter
requirements. Fire- protective wrap systems are designed to reduce the heat transfer from a
fire source to the raceway interior. The fire wrap may also impede the transfer of heat from
the cables inside the raceway to the environment.
The energy balance equations (10.35) are also applicable in this case. Let 6rg be the
temperature at the bottom of the fire-protection material. Since the heat is transferred by
conduction through the wrap, the following relationship holds at the bottom of the fire-
protection barrier:

Wiotal Pw *ZB
Org = Osn — 5 f (10.38)

At the top of the tray, the thermal resistances in equation (10.35) are modified to
represent the combined thermal resistance of the tray cover and fire barrier material:

_ Pctic + Pw2r
1,A (10.39)
Ww
where pc = thermal resistivity of the cover material, K-m/W
Pw = thermal resistivity of the wrap material, K-m/W
zp = thickness of the fire barrier material at the bottom of the tray, m
zr = thickness of the fire barrier material at the top of the tray, m
zc = thickness of the tray cover material, m
w = width of the tray, m.

The model is applicable to cables in ladders, troughs, or solid bottom trays. The cables
may be installed randomly, without maintained spacing, and no maintained segregation of
power and control cables.
The model is applicable to a fire-wrap system that is made of a single, relatively
homogeneous wrap material. The wrap material support configuration (if any) need not be
considered if the support is relatively thin or has relatively high thermal conductivity.
Chapter 10 m Special Cable Installations 283

10.5 CABLES IN TUNNELS AND SHAFTS

10.5.1 Horizontal Tunnels

Cables are sometimes installed in tunnels provided for other purposes. In generating
stations, short tunnels are often used to convey a large number of cable circuits. Long
tunnels are built or existing tunnels adapted solely for the purpose of carrying major EHV
transmission circuits which for various reasons cannot be carried overhead. River crossings
are obvious cases where tunnels would be used either for technical or environmental reasons.
The cost of such installations is very considerable, and it is desirable to optimize as far as
possible the current-carrying capacity, groupings, and number of circuits to be installed to
meet a given transmission capacity.
Detailed investigations into the rating of cables in tunnels have been given by Burrell
(1951), Giaro (1960), Germay (1963), Kitagawa (1964), and Whitehead and Hutchings
(1938). These are based on heat transfer evaluations from equations established for me-
chanical and chemical installations. Weedy and El Zayyat (1972, 1973) presented a method
more suitable for cable installations. They performed measurements to determine suitable
convection coefficients. Their results were later adopted by CIGRE and published in Electra
(CIGRE 1992a, 1992b). Only the steady-state ratings with natural cooling are considered
here. Transient analysis using numerical methods and forced cooling computations are
described in the cited Electra papers.

10.5.1.1 Thermal Model. It may be desirable to rate cable circuits in horizontal

tunnels on the basis of free convection only in view of the difficulty in assessing air velocities
along the sides and flow of a tunnel, as distinct from the bulk velocity. In this case, since
the equivalent diameter of the tunnel is assumed to be much greater than the diameter of
the cable, the heat transfer to the outside wall of the tunnel can be described by the energy
balance equations (10.1) and (10.7). The heat transfer for the tunnel outside surface is
described by a conduction equation. Thus, the energy balance equations take the form

Wiotat
= As(Os—Ogas)As
+ AsrFswOpCe a 0%")
Ow
Fr <A
yD= hy(Ogas
Op) Awio As,FywoR(0 = ae (10.40)

Co= 0, Ao= amb

si ainis
where 7,/ and T,” are the thermal resistances of the tunnel wall and the surrounding soil,
respectively. For deep tunnels, adiabatic conditions are sometimes assumed, which means
that @, = Gimp(CIGRE, 1992a),
The thermal resistance of the soil for circular tunnels is computed from equation
(9.20) with the tunnel diameter replacing cable external diameter. For a square tunnel, the
expression derived by Goldenberg for a buried square trough and reported by Symm (1969)
can be used. This expression is

(Hea 3.388)
L (10.41)
20 a
284 Part II] » Advanced Topics

where Ps = thermal resistivity of the soil, K-m/W

Ly = depth of tunnel centerline, m
a = height and width of square cable tunnel, m.

The radiation shape factor is given by

Few — ( at O5/Esar CA Ce Agen es

a av (10.42)
\ N(N — 1) As €s0w
Aw Ew
where N is the number of cables. The mutual radiation area is computed from equation
(10.16). For two cables, it can be shown to be

Ay Shy SAAD
2 ese (10.43)

where s is the spacing between the axes of the cables. For three cables in trefoil, the mutual
radiation area is A,, = 0.618D,, and for three cables spaced horizontally, it is twice the
value given by equation (10.43).
Since cables in tunnels are often installed in horizontal and vertical groups, the mutual
heating effect reduces cable ratings. Taking this into account, the convection coefficient for
groups of cables in horizontal tunnels is obtained from the correlations given in Table 10.3
with the modification shown in equation (10.44):

hs = eee =Ne* Nwh*Nwv*Nu - kair/ De

Nu = c(Gr-Pr)” (10.44)
Gr = gB(@;,—G22)Dofus

where 7c, Nwn, and Nwy are the correction factors for multicable, horizontal wall, and
vertical wall effects, respectively. These correction factors are obtained from Table 9.11
and equation (9.93).
The values of coefficients c and n obtained from experimental results by Weedy and
El Zayyat (1972a) are summarized in Table 10.6.
The heat convection coefficient for the tunnel wall /,, can be taken from the correlations
describing a small heat source near a vertical wall:

hy = Nu- kair/L
Nu = 0.59(Gr-Pr)°?5 (10.45)
Gr = 8B (Beas= Gy) Lyre

where L is the height of the tunnel. For circular tunnels, L can be replaced by the tunnel
inside diameter. If the cables are located on one side of the tunnel only, then the surface
area A,, in equations (10.40) should be equal to the area of the wall close to the cables. In
the case of circular tunnels, half of the area of the tunnel should be used.
Chapter 10 m Special Cable Installations 285

TABLE10.6 Parameters for Convection Formulas for Cables in Tunnels

(Weedy, 1988)
i

Cableof
Grouping Spacing Group c n

See Not touching ; hie pas

Middle 0.785 0.21
s/D,>2
Outer 0.815 0.21
cee Touching Middle 0.55 0.20
Outer 0.69 0.20
; Touching Middle 0.40 0.215
e Upper 0.54 0.215
Lower 0.715 0.215
Not touching Middle 0.60 0.225
° MID SP Upper 0.70 0.225
Lower 0.78 0.225
te Touching Upper 0.48 0.20
Lower 0.68 0.20

EXAMPLE 10.2
We will compute the ratings of six circuits each in trefoil formation located in a horizontal tunnel as
shown in Fig. 10.6. Cable model No. 1 will be used with the following parameters (see Table A.1):
D,. = 0.0358 mm, R = 0.0781 - 10-7Q/m, A; = 0.09, T; = 0.214 K-m/W, and T; = 0.104 K-m/W.

Figure 10.6 Installation of cables in tunnel in Example 10.6.

The tunnel has a square cross section with a height of 2.0 m, shallow installation with 0.5 m
concrete walls, and 1 m of soil above the roof. The thermal resistivities of concrete and soil are 0.6
and 1.2 K-m/W, respectively. The soil ambient temperature is 15°C.
286 Part II] = Advanced Topics

The external thermal resistances of the tunnel wall and the soil are obtained from equations (3.4)
and (10.41), respectively, and are equal to

pat 0.6-0.5
Tj1 =
ae = = 0.15 K-m/W
S a |

(ig | | 3388L | 1.2In 3.338

2.5) = 0.273K-m/W
Pas a 2 2
The heat transfer coefficient for the hottest cable is obtained from equation (10.44). The air
thermophysical properties are evaluated at the film temperature 6, = (@, + 6,)/2. At this point,
neither temperature is known; however, they can be approximated for the purpose of performing the
calculations. Let us assume that 6, = 30.5°C. Since only one iteration of the process is illustrated
in this example, if the film temperature computed at the end of the example is different from the one
assumed here, computations should be repeated from this point on. At the assumed film temperature,
the air thermophysical properties are (seeAppendix D)

poss 10 >1K
y= .16,1 105om2/s
Pr 071
Kairp
= 0.0265W/K-m
The cable surface convection coefficient is obtained from equation (10.44) assuming that con-
vection takes place on one wall only. The required constants are taken from Table 10.4.

9s + Ow
Gr = gB(O,—9g23)D;/v>
= 9.81-3.3 - 10-7(« ) 0.0358°/(16.1- 10~°)?
2

=5.7-10°
@. ne)
0, Gh
2
Gr-Pr
=5.7
-10°
(«==
0,5+4,
)-0.71
=4.0
-10°
(«pve
et“ 6.)
Nu
=c(Gr-Pr)"
=0.48
[+010° 6, 0.20
(«==") =2.52 ;3) 0.2
(«Sits a

hs= Ne}
Nok“Tov
“NW
Kag/
Do NoTov©
2.52(« 0; + Ow 0.2
) -0.0265
/0.0358
a a

(«— )0.2
phy*(phrl
lke! W/m2-K Os + Ow
2

From Table 9.11, and equation (9.93), the reduction factors for a group of cables are as follows: the
mutual heating effect n= 1/1.39 = 0.72 and the vertical wall effect vee lee oe Ors ieeinuss

h,=0.72
-0.81
-1.87 0.ao)0,2
(4=—— @+est 0.2
=1.09 W/m?-K
The heat convection coefficient for the inside surface of the tunnel is obtained from the equations
(10.45). Hence,
Chapter 10 m=Special Cable Installations 287

0, + Oy
Gr = gB(Oa;—Oy)L?/v*= 9.81 -3.3 - 1073(“S" “ a) 2.0 CoiledOnt)

240%
10°
(<*
0, Ww -@
2
Gr-Pr
=1.0- 6,:+6,=) 50,7
10”( ==
7 a0(oo ; 2)
Nu=0.59(Gr-Pr)°?>
=0.59[7.1
LOnQ. =ae)0.25
(“4% ;
=96.3
(-oe=a)0.25
a6, 0.25
hy=Nu-
kgir/L
=96.3( peat Ow -0.0265/2.0
6, ODa)Bee
=1,28
(At*wa W/m2-K
The emissivity of the cable surface is 0.9 (see Table 9.10) and of the concrete wall is 0.63. Because
the cable and the tunnel wall are opaque, we have

Gy = las, SHU SOc Om Sly SOs] O37

The thermal radiation shape factor for one circuit is obtained from equation (10.42) and is equal to

ge N felSecepaneet
atabteti 73 etree tated
: 1 NOV
= DRA;fw © 33 = 1)70.03580.9.0.37
Aw Ew 0n 0 0.63
Few = (oF 0;/Es “ir GAgo@/Agewil

= [1+ 0.1/0.9 + 6.82 - m - 0.0358 - 0.37/(2.0- 1 -0.63)]-! = 0.75

The mutual radiation area for a single circuit is equal to A,, = 3 -0.618 - 0.0358 = 0.066 m2. Hence,
As, = 3-7 - 0.0358 — 0.066 = 0.271 m” per unit length.
Assuming that the maximum allowable conductor temperature is 90°C, the total heat generated
in the cable can be obtained from equation (4.1) and the constants taken from Table A.1. We have

W, = I?R(L +A, + Aa) = 170.0781 - 107-7(1 + 0.09) = 0.085 - 10-77?

The internal thermal resistance of the cable is obtained from equation (4.4) as

T=7,+ (1 +A))T3 = 0.214 + 1.09 - 0.104 = 0.327 K-m/W

and the relationship between conductor and surface temperature is thus given by

R- I? -T =0.085 - 1073 - 1? -0.327 = 2.8- 107° «J?= 90-6,

The energy balance equations (10.40) together with the last equation can be written in the
following form:
288 Part II] m Advanced Topics

Wirotal
= hs(9;es Oza)AsWeAgrIOofOHe
(ox4=?aS)
12
18+2.8+10-7. 7*=6.41.09 (« Bs oe) x -0.0358+ 6.0271-0.75-5.667- 10-*(0"*—03")

Oyr @: * *
1" = Aw(gas—Oy)Aw+ AsrFywon(04 = pe)
T,
—* = 6- 1.28(" Bee - a) 1.25
2.0-1+6-0.271 -0.75-5.667- 10-8(@**
—6%")

Oya 0, oes - Camb

fb! Le
Owi. 0, Ea0, = NS)
OSE 0272
20°10 ° 17 = 900—Oy

The solution of these equation yields

Pe AlPAT (Gra O) Cl. Oy oe age 0

Since the film temperature is higher than assumed in the computations, the iterations should be
repeated with new air properties until convergence is achieved.

10.5.2 Vertical Shafts

We will, again, consider installations with natural convection only. There is very little
experimental evidence available to determine heat convection coefficients for this type of
installation. As a first approximation, cables in vertical shafts can be studied as the cables
on riser poles with the open top and closed bottom and adiabatic conditions at the shaft
external surface. Hence, the energy balance equation will take the form

2 0. + Oamb
n{l RG +4; +49) + Wal=h; 2 ame a As
(10.46)
LsAsrFywOpB
GeeFFee) m

The heat convection coefficient is obtained from correlations given at the bottom of
Table 10.1.
An alternative equation is given by Endacott et al. (1970). Assuming an effective
emissivity of 0.9 and natural convection, the heat dissipated from the cable/pipe in a shaft
is given by the following empirical formula (Weedy, 1988):

D* D*
W;= 3.23. 1o-Nm OS —0") 41.21 pep Ubi
Dea A (10.47)
Ww

where m is the proportion of the cable/pipe surface available for radiation (affected by
obstructions and the proximity of other cables).

10.6 CABLES IN BURIED TROUGHS

Surface trough types of installation has been widely used in North America and Europe
for cable routes running alongside railways, canals, and in substations. The cables are
Chapter 10 m Special Cable Installations 289

normally placed with a small separation between phases at a depth of approximately 0.3 m
in a concrete trough provided with a reinforced concrete lid to afford mechanical protection.
The important thermal resistances external to the cable are those associated with the trough
fill and between the trough surface and the air.

10.6.1 Buried Troughs Filled with Sand

When cables are installed in a sand-filled trough, either completely buried or with the
cover flush with the ground surface, there is a danger that the sand will dry out and remain
dry for long periods. The cable external thermal resistance may then be very high, and the
cable may reach undesirably high temperatures.
For cables laid at depths of less than 0.6 m, cable temperatures may show significant
daily or even hourly variations due to changes in solar radiation, wind velocity, and air
temperature. The maximum cable surface temperature in a surface trough system derived
from analog investigations can be written as (Endacott et al., 1970)
0.33 W, 0.29H
(@.—W,T4+ Sa ee CO ——
0.747*0.2" 4,0.89 + Oamb
7+0.07 (10.48)

where @, = maximum cable surface temperature, °C

W, = total cable heat dissipation, W/m
T4 = thermal resistance from the hottest cable computed in accordance with
equation (9.20), K-m/W
L* = vertical depth below ground surface of the cable axis, m
v = wind velocity at a height of 10 m, m/s
Oamb= average ambient temperature, °C
H = steady net solar radiation to the ground surface, W/m’.

From equation (10.48), the effective external thermal resistance is equal to

Ses Wi9S2-19.28 0.33 0.29H
Tg ee Til
W, porapo2
+Loa,007
w, (10.49)
This value of the external thermal resistance is substituted in equation (4.3) for 74. Since
EE depends on W,, equation (4.3) has to be solved iteratively in this case. Numerical
studies performed by the author have shown that on some occasions, equation (10.49) may
give very high external thermal resistance. Since the upper limit of the external thermal
resistance is obtained when the sand is assumed completely dry, this value should be used
when the iterative procedure results in ie reaching values which are too high.
A more approximate method, which is used in Great Britain, consists of calculating
the temperature rise of the cable assuming the ground surface to be isothermal with an
effective ground ambient temperature which takes account of wind velocity, solar radiation,
and ambient air temperature effects. These temperatures, for different weather condi-
tions considered for overhead lines specified in column 2, are given in column 4 of Table
10.7. Corresponding values used for directly buried systems at normal laying depths (1.e.,
> 0.6 m) are given for comparison in column 3 (Endacott er al., 1970).
An alternative simpler approach is suggested in IEC 287 (1982). The standard rec-
ommends that the cable rating be computed in the case of surface troughs filled with sand
290 Part II] m Advanced Topics

TABLE 10.7 Equivalent Ground Temperatures for Shallow Filled Troughs

i

Overhead Line Ground Temperature

Rating Condition Air Temperature (°C) At Depth > 0.6 m At Depth 0.3 m

Cold weather <4.5 10 10

Normal weather 4.5-18 15 30
Hot weather >18 15 40

using a value of 2.5 K-m/W for the thermal resistivity of the sand filling unless a specially
selected filling has been used for which the dry resistivity is known.
Thethreeapproaches
discussedabovecanleadto radicallydifferentcableratings,as
illustratedinExample10.3.Themostpessimisticratingis usuallyobtainedbyapplying
theIECmethod.
EXAMPLE 10.3
Consider a three-phase cable system composed of model cable No. | laid in a surface trough filled with
sand of thermal resistivity 0.6 K-m/W when moist and 2.5 K-m/W when dry. The laying information
is shown in Fig. 10.7. We will assume wind velocity of 3 m/s and solar radiation intensity of
200 W/m?. We will determine the rating of this cable system.

0.36m
po aa

p=1K-m/W G.mb= == 1) ° G

Figure 10.7 Three cables in a buried trough.

The thermal resistances of the internal parts of the cable are given in Table Al. The external
thermal resistance of the middle cable is computed in two stages. From equation (9.32), we have

4-03
T;(wet) =p,[0.475 In(2u) —0.346] =0.6 (047s In0aa =0.346) =0.845 K-m/W
T,(dry) = p;(0.475 In(2u) —0.346] = 2.5 { 0.475 In 4-0.36
00358 ~ 0.346) = 3.52 K-m/W

The equivalent radius of the trough (ignoring the concrete walls) is obtained from equation (9.61) as

Wee
/e4ae
ar2 5=(-
ee toc
=) y?eine
( =)
a)inlees) xEWO
Np S seZpath
(-eee0S!
ra)in(11.02
+SS) 0.3
tineas
baile
togeg
baas
Hence, rp = 0.22 m and u = 0.2/0.22 = 0.91.
Chapter 10 Special Cable Installations 291

Since u < 1, the correction factor in equation (9.60) is obtained from the table of extended
values. For wet sand it is equal to 0.0464, and for dry sand is equal to —0.174. Thus, T;(wet) =
0.845 + 0.0464 = 0.891 K-m/W and 7,(dry) = 3.52 — 0.174 = 3.35 K-m/W. In order to compute
T;", we need the value of W,. We will start the iterative process by neglecting first the effect of wind
and solar radiation. The ac resistance of the conductor is in this case equal to 0.791 Q/km, and the
concentric neutral loss factor for the hottest cable is equal to 0.101. Therefore, from equation (4.3),
we have
ove es vl
I= -, = 851A
0.0791 - 10-3 - [0.214 + 1.101 - (0.104 +0.891)]
The total losses are equal to 1.101 - 0.0000791 - 8517 = 63.1 W/m. The effective external thermal
resistance is now obtained from equation (10.49) including the effect of wind and solar radiation:

0.33 OZone
De = List 0.7470.24 p89 7p 0.883 0.33 0.29-200 = 1.44K-m/W
0.07
Wy “a 30.749)
360.2a 30.8909
360.07
. 63.|
The new ampacity is obtained again from equation (4.3), and is equal to 706 A. Corresponding
losses are 43.4 W/m, and the new value of 7" is equal to 1.6 K-m/W. Continuing the same way, the
process ends after four iterations with the ampacity of 673 A and 7" = 1.7 K-m/W. Since the value
is smaller than the thermal resistance with completely dry sand, the calculations are completed.
For comparison, we observe that the ampacity of this system using the IEC method is 484 A,
and using the equivalent soil ambient temperature of 30°C, taken from Table 10.6, the current rating
is equal to 764 A.

10.6.2 Unfilled Troughs of Any Type, with the TopFlush with the Soil Surface
and Exposed to Free Air

IEC 287 (1982) suggests using an empirical formula which gives the temperature rise
of the air in the trough above the air ambient temperature as

Sa, = (10.50)

where p (m) is that part of the trough perimeter which is effective for heat dissipation. Any
portion of the perimeter which is exposed to sunlight is therefore not included in the value
of p. The rating of a particular cable in the trough is then calculated as for a cable in free
air, but the ambient temperature shall be increased by A@,,.

EXAMPLE 10.4
We will compute the rating of the cable system analyzed in Example 10.3 assuming now that the
trough is unfilled. Since the trough is exposed to solar radiation, we will assume that the value of p
is equal to 1.2 m.
The heat dissipation coefficient is obtained from equation (9.80) with the required constants
givenin Table9.8.
Hd,
es (De 0.62
h | ~~= 0.035802 + 1.95 = 3.38 Wim?-KO"

To compute cable surface temperature rise, we use equations (9.83) and (9.86):
a Da m - 0.0358 - 3.38
A [0.214 + 1.09 - 0.104] = 0.114
loa 1.09

oHK, 0.6-200-0.114
Sally
nae ene a8
292 Part III m Advanced Topics

We will start an iterative process by setting the value of (A@,) '/4 in the denominator of equation
(9.86) to 2. Thus, we have

(A6;)0 ys=[Seta te Ealat)

= 1/4
| 79+1,2999°C
1+ Ky(A0,)4 i Ond4 «2

In the second iteration, the value of (A6,)3" is equal to 2.76°C. Convergence is achieved at this
iteration. The external thermal resistance is obtained from equation (9.78) and is equal to

l 1
tT, = —————_,, = = 0.95 K-m/W
maD*h(A@,)'/4 =x- 0.0358 - 3.38 - 2.76

The cable rating is given by equation (4.3) as

z 75 — 0.6 - 0.0358 - 500 - 0.95 0.5 entn

0.0781 - 10-3(0.214 + 1.09(0.104 + 0.95)

The corresponding total losses are equal to 51.8 W/m. Thus, the ambient temperature reduction
obtained from equation (10.50) is equal to

Ab, =
Weeeo
— =
1es = 43.2°C
Dyin eaee)

The new value of (A@,)'/4 is 2.3°C, and the corresponding value of JT, = 1.15 K-m/W. The
permissible current rating is equal to 397 A, and the corresponding losses are 10.3 W/m per cable.
The ambient temperature reduction is equal in this case to 10.3°C. Continuing the iterative process,
the current rating is obtained at 24th iteration and is equal to 590 A.

REFERENCES

Al-Nimr, M. A. (1993), “Analytical solution for transient laminar fully developed free con-
vection in vertical concentric annuli,” /nt. J. Heat and Mass Transfer, vol. 36, pp. 2385—
2395.
Anders, G. J., Roiz, J., and Moshref, A. (July 1990), “Advanced computer programs for
power cable ampacity calculations,” JEEE Comput. Appl. Power, vol. 3, no. 3, pp. 42-45
(1996 revision).
Anders, G. J. (Jan. 1996), “A unified approach to rating of cables on riser poles, in trays, in
tunnels and in shafts—A review,” [EEE Trans. Power Delivery, vol. 11, no. 1, pp. 3-11.
Anders, G. J. (1995), “Rating of cables on riser poles,” in Proc. Jicable ‘95 Conf., Versailles,
France, June 1995.
Burmeister, L. C. (1983), Convective Heat Transfer. New York: Wiley.
Burrell, R. W. et al. (1951), “Forced air cooling for station cables.’ AJEE Trans., vol. Il
(70), pp. 1798-1803.
CIGRE (1992a), “Calculation of temperatures in ventilated cable tunnels—Part 1,” Electra,
no. 143, pp. 39-59.
CIGRE (1992b), “Calculation of temperatures in ventilated cable tunnels—Part 2.” Electra,
no. 144, pp. 97-105.
Chapter 10 m Special Cable Installations 293

Cress, S. L., and Motlis, J. (Jan. 1991), “Temperature rise of submarine cable on riser poles,”
IEEE Trans. Power Delivery, vol. 6, no. 1, pp. 25-33.
Dryer, J. (Oct. 1978), “Natural convective flow through a vertical duct with restricted entry,”
Int. J. Heat and Mass Transfer, vol. 21, pp. 1344-1354.
El-Shaarawi, M. A., and Sarhan, A. A. (Nov. 1981), “Developing laminar free convection in
a heated vertical open-ended concentric annulus,” Indust. & Eng. Chem., Fundamentals,
vol. 20, no. 4, pp. 388-394.
Endacott, J. D., Flack, H. W., Morgan, A. M., Holdup, H. W., Miranda, F. J., Skipper,
D. J., and Thelwell, M. J. (1970), “Thermal design parameters used for high capacity
EHV cable circuits,’ CIGRE, Report 21-03.
Engmann, G. (1984), “Ampacity of cable in covered tray,’ JEEE Trans. Power App. Syst.,
vol. PAS-103, pp. 345-350.
Engmann, G. (1986), “Cable ampacity in tray with raised cover,’ JEEE Trans. Energy
Conversion, vol. EC-1, pp. 113-119.
Germay, N. (1963), “Calculation of the temperature rise of cable in a gallery with forced
ventilation,” Rev. Elec., Suppl. to Bull. Soc., Roy. Belge. Elec., vol. 4, no. 1, pp. 3-13.
Giaro, J. A. (1960), “Temperature rise of power cables in a gallery with forced ventilation,”
CIGRE, Report N1 213.
Harshe, B. L., and Black, W. Z. (1994), “Ampacity of cables in single open-top cable trays,”
IEEE Trans. Power Delivery, vol. PWRD-9, no. 4, pp. 1733-1740.
Harshe, B. L., and Black, W. Z. (1996), “Ampacity of cables in single covered trays,’ Paper
96 WM 209-7 presented at the Winter Meeting of PES in Baltimore, MD.
Hartlein, R. A., and Black, W. Z. (June 1983), “Ampacity of electric power cables in vertical
protective risers,’ JEEE Trans. Power App. Syst., vol. PAS-102, no. 6, pp. 1678-1686.
Holman, J. P. (1990), Heat Transfer. New York: McGraw-Hill.
ICEA/NEMA (1986), “Ampacities of cables in open-top cable trays,” ICEA Publication
No. P54-440, NEMA Publication No. WC51, Washington, DC.
IEC (1982), “Calculation of the continuous current rating of cables (100% load factor),”
IEC Standard Publication 287, 2nd ed.
Incropera, F. P., and De Witt, D. P. (1990), Introduction to Heat Transfer. New York:
Wiley.
Joshi, H. M. (Nov.—Dec. 1987), “Fully developed natural convection in an isothermal vertical
annular duct,’ Int. Commun. in Heat and Mass Transfer, vol. 14, no. 6, pp. 657-664.
Joshi, H. M. (1988), “Numerical solutions for developing laminar free convection in vertical
annular ducts open at both ends,” Numer. Heat Transfer, vol. 13, no. 3, pp. 393-403.
Keyhani, M., and Kulacki, F.A. (1985), “Natural convection in enclosures containing tube
bundles,” in Natural Convection, S. KaKac, W. Aung, and R. Viskanta, Eds. New York:
Hemisphere.
Keyhani, M., Kulacki, F. A., and Christensen, R. N. (Aug. 1985), “Experimental inves-
tigation of free convection in a vertical rod bundle—A general correlation for Nusselt
numbers,” J. Heat Transfer, Trans. ASME, vol. 107, no. 3, pp. 611-623.
Keyhani, M., Prasad, V., and Kulacki, F. A. (1986), “An approximate analysis for thermal
convection with application to vertical annulus,” Chem. Eng. Commun., vol. 42, no. 4-6,
pp. 281-289.
294 Part II] m Advanced Topics

Kitagawa, K. (1964), “Forced cooling of power cables in Japan,’ CIGRE Paper 213.
Martin, B., and Cohen, M. (1954), “Heat transfer by free convection in an open ther-
mosyphon tube,” Brit. J. Appl. Phys., vol. 5, pp. 91-95.
Martin, B. (1955), “Free convection in an open thermosyphon with special reference to
turbulent flow,” Proc. Roy. Soc., vol. 230(ser. A), p. 502.
Morgan, V. T. (1982), “The thermal rating of overhead-line conductors, Part I. The steady-
state thermal model,” Elec. Power Syst. Res., vol. 5, pp. 119-139.
Morgan, V. T. (Mar. 1993), “External thermal resistance of aerial bundled cables,” Proc.
IEE, vol. 140, part C, no. 2, pp. 65-62.
Morgan, V. T. (Mar. 1992), “Effect of mixed convection on the external resistance of single-
core and multicore bundled cables in air,” Proc. IEE, vol. 139, part C, no. 2, pp. 109-116.
Morgan, V. T. (1995), Discussion of Anders (1996); IEEE Trans. Power Delivery, vol. 11,
no. 1, pp. 3-11.
Neher, J. H., and McGrath, M. H. (Oct. 1957), “The calculation of the temperature rise and
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Raithby, G. D., and Hollands, K. G. T. (1985), “Natural convection,” in Handbook of Heat
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Weedy, B. M., and El Zayyat, H. M. (1973), “The current capacity of power cables in
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 11

Ampacity Computations
Using Numerical Methods

11.1 INTRODUCTION

We have shown in Chapter 2 that if the thermal resistance is constant, the heat conduction
equation (2.12) can be written as

070)| 070 1 00
Gun)
aa
Fedeled
sf)Une Ta KOe
where 5 = 1/pc is the thermal diffusivity of the medium (m/s) and the remaining symbols
are defined in the list of symbols.
The boundary conditions associated with (11.1) can be expressed in two different
forms. If the temperature is known along a portion of the boundary, then

= 03(s) C12)

where 63 is the known boundary temperature that may be a function of the surface length
s. If heat is gained or lost at the boundary due to convection h(@ — Oamp)or a heat flux q,
then

1 00
ee ee Ceo) —U0 (11.3)
p on

where n is the direction of the normal to the boundary surface, h is a convection coefficient,
and @ is an unknown boundary temperature. The solution of these equations yields the
temperatures at all points of the region, including the cable conductor.

295
296 Part III m Advanced Topics

As discussed in Chapter 2, if the medium surrounding the cables is not uniform, and
the earth surface boundary is not isothermal, an analytical solution is generally not feasible
and a numerical approach is indicated.
The usual ampacity problem is to compute the permissible conductor current so that
the maximum conductor temperature does not exceed a specified value. When numerical
methods are used to determine cable rating, an iterative approach has to be used for the
purpose. This is accomplished by specifying a certain conductor current and computing
the corresponding conductor temperature. Then, the current is adjusted until the specified
temperature is found to converge within a specified tolerance.
The limitations of the classical methods will be apparent from a few examples. In the
transient analysis discussed in Chapter 5, separate computations were performed for the
internal and external parts of the cable. Coupling between internal and external circuits
was achieved by assuming that the heat flow into the soil is proportional to the attainment
factor of the transient between the conductor and the outer surface of the cable. The validity
of the methods did not rest on an analytical proof, but on an empirical agreement of the
responses given by the recommended circuits and the temperature transients calculated by
more lengthy but more accurate computer-based methods.
In the analytical methods derived in Chapters 4 and 5, the case of a group of cables is
dealt with on the basis of the restricted application of superposition. To apply this principle,
it must be assumed that the presence of another cable, even if it is not loaded, does not
disturb the heat flux path from the first cable, nor the generation of heat within it. This
allows separate computations to be performed for each cable, with the final temperature rise
being an algebraic sum of the temperature rises due to the cable itself and the rise caused by
the other cables. Such a procedure is not theoretically correct, and for better precision, the
temperature rise caused by simultaneous operation of all cables should be considered. A
direct solution of the heat conduction equation employing numerical methods offers such
a possibility.
Numerical methods allow not only better representation of the mutual heating effects,
but also permit more accurate modeling of the region’s boundaries (e.g., a convective bound-
ary at the earth surface, constant heat flux circular boundaries for heat or water pipes in
the vicinity of the cables, or an isothermal boundary at the water level at the bottom of a
trench).
In what follows, we will develop a solution to equations (11.1)—(11.3) using finite-
element and finite-difference methods. Even though both methods have been applied to
solve the heat conduction problem around loaded power cables, our view is that the finite-
element method is better suited for this application. Comparison of the two methods is
given in Section 11.5.

11.2 GENERAL CHARACTERISTICS OF NUMERICAL METHODS

Both the finite-element and finite-difference methods discussed in this chapter require dis-
cretization of the partial differential equations in space and time domains. The finer this
discretization is, the more accurate are the results, yet the heavier is the computational
burden. The size of the region around power cables to be discretized is also of importance
for both the accuracy and efficiency of the computations. The following sections give some
guidance on these topics.
Chapter 11 Ampacity Computations Using Numerical Methods 297

11.2.1 Selection of the Region to be Discretized

The location of boundaries is an important consideration in numerical studies. The

objective is to select a large enough region so that the calculated values along the boundaries
agree with those that exist in the physical problem. For the cable rating problem, the earth’s
surface is an obvious boundary and the side and bottom boundaries must be selected in
such a way that the nodal temperatures at those boundaries all have the same value and the
temperature gradient across the boundary is equal to zero.
Experience plus a study of how others modeled similar infinite regions is probably the
best guide. In the experience of the author, a rectangular 10 m wide and 5 m deep field, with
the cables located in the center, gives satisfactory results in the majority of practical cases.
For transient analysis, the radius of the soil, out to which heat disperses, will increase
with time, and for practical purposes, it is sufficient to consider only that radius within
which a sensible temperature rise occurs. This radius can be estimated from a simplified
form of equation (5.14) assuming that each cable is a line source of heat:

ie W1
AaPs (=—r?) ( )
6 —Ei| — 11.4

where 6,, is the threshold temperature value at the distance r from the cable axis and the
remaining quantities are defined in Chapter 5 and also in the list of symbols. This value
can be taken as 0.1 K when the number of cables is not greater than 3 and suitably smaller
for a large number of cables. Equation (11.4) is applied for each cable. The region to be
discretized will be an envelope around all the circles representing the individual cables.

11.2.2 Representation of Cable Losses

Conductor, sheath, and dielectric losses are represented as heat sources, and provision
should be made to vary these as required with time and/or temperature. Values of these
losses are recalculated at each time step, using methods given in Chapters 4, 6, 7, and 8.
In finite-difference and finite-element methods, each node contains an initial temper-
ature rise at t = O. In the general case where the transient due to the dielectric loss may not
have reached its steady state, these initial temperature rises (relative to the ambient tem-
perature outside the mesh) must be obtained from a prior calculation. From the beginning
of the transient, the computation must in all cases take account of both the joule and the
dielectric losses. The two usual situations are

1. The transient due to dielectric losses has reached a steady state (the voltage has
been applied for a very long time). The initial temperature rise at each node 1s put
equal to the steady state temperature rise at that point caused by the dielectric losses
only.
2. The voltage is applied to the cable at the same time as the load current. In this case,
the initial temperature rises are zero, and the dielectric loss generators should be
allocated their proper values from time zero.

11.2.3 Selection of a TimeStep

Since, in general, the computations involve evaluation of temperatures in increments
of time, care must be taken in the selection of the time step. In principle, one should select
298 Part III m Advanced Topics

as large a time interval as possible to reduce the amount of computation. Unfortunately, too
large a time step can compromise the accuracy of the computations. CIGRE (1983) and
Libondi (1975) provide some guidance on the selection of suitable values. The duration of
the time step At will depend on: (1) the time constant U7- &Q of the network [defined as
the product of its total thermal resistance (between conductor and outer surface) and its total
thermal capacitance (whole cable)], (2) time elapsed from the beginning of the transient T,
and (3) the location of the time t with relation to the shape of the load curve being applied.
Requirement (3) can be illustrated as shown in Fig. 11.1 where the value of At is selected
to coincide with the change of the shape of the load curve.

Figure 11.1 Relationship among the time step,

t At the load curve, and the time elapsed from the
beginning of the transient.

The following conditions are suggested for the selection of the time step At (CIGRE,
1983):

10gfea eselope ——
aa 8 OREN, 0
Sm we he 3 ae
et Sil eeOn mel eeeeS
Sie Oe TO 3
Adjusting the time step automatically during the computations is the preferable approach.

11.3 THE FINITE-ELEMENT METHOD

11.3.1 Overview

The finite-element method is a numerical technique for solving partial differential

equations. Among the many physical phenomena described by such equations, the heat
conduction problem and heat and mass transfer in the vicinity of power cables have been
often addressed in the literature (Abdel-Hadi, 1978; Anders et al., 1987; Anders and Rad-
hakrishna, 1988a, 1988b; Anders et al., 1988; El-Kady, 1985; El-Kady and Horrocks, 1985;
Flatabo, 1973; Labridis and Dokopoulos, 1988; Konrad, 1982; Mitchell and Abdel-Hadi,
1979; Mushamalirwa et al., 1988; Tarasiewicz et al., 1987; Thomas et al., 1980). The
fundamental concept of the finite-element method is that temperature can be approximated
by a discrete model composed of a set of continuous functions defined over a finite num-
ber of subdomains. The piecewise continuous functions are defined using the values of
temperature at a finite number of points in the region of interest.
Chapter 11 m Ampacity Computations Using Numerical Methods 299

The discrete solution is constructed as follows:

1 A finite number of points in the solution region is identified. These points are called
nodal points or nodes.
a The value of the temperature at each node is denoted as a variable which is to be
determined.

The region of interest is divided into a finite number of subregions called elements.
These elements are connected at common nodal points, and collectively approxi-
mate the shape of the region.
. Temperature is approximated over each element by a polynomial that is defined
using nodal values of the temperature. A different polynomial is defined for each
element, but the element polynomials are selected in such a way that continuity
is maintained along the element boundaries. The nodal values are computed so
that they provide the “best” approximation possible to the true temperature distri-
bution. This selection is accomplished by minimizing some quantity associated
with the physical problem (this is the so-called Rayleigh—Ritz method), or by using
Galerkin’s method (Zienkiewicz, 1971) which deals with the differential equations
directly. Either approach results in a matrix equation whose solution vector con-
tains coefficients of the approximating polynomials. The solution vector of the
algebraic equations gives the required nodal temperatures. The answer is then
known throughout the solution region.

In the following section, we will develop the algebraic equations from which nodal
temperatures are obtained.
In cable rating applications, two-dimensional elements are most commonly used. The
elements in the two-dimensional domain are functions of x and y and are generally ei-
ther triangular or quadrilateral in shape. The element function becomes a plane (Fig. 11.2).

Figure 11.2 Modeling of atwo-dimensional scalar function using triangular or quadrilateral

elements (Segerlind, 1984).
300 Part II] m Advanced Topics

The plane is associated with the minimum number of element nodes, which is three for the
triangle and four for the quadrilateral.
The element function can be a curved surface when more than the minimum number
of nodes are used. An excess number of nodes also allows the elements to have curved
boundaries (Fig. 11.3). For the purpose of introducing the method and explaining how it
is used in cable rating computations, we will use the simplest and the most common shape
for two-dimensional elements, the triangle. In this chapter, the words “triangle,” “element,”
and “finite element” will be used interchangeably.

Figure 11.3 Modelling of a two-dimensional scalar function using a quadratic triangular

element (Segerlind, 1984).

11.3.2 Approximating Polynomials

Consider a simple triangular element shown in Fig. 11.4. For this element, the tem-
perature 6 at any point inside can be uniquely specified as (Flatabo, 1972)

6 = Aa; + Bo; + Con (11.6)

where @;, @;, and @,, are the area coordinates defined as in Fig. 11.4. These area coordinates
uniquely define the position of any point P inside the triangle ijm. To determine the constant
A, the temperature at node i is written as [equation (11.6)]:

6=1xA+0xB+0xC

This gives A = 6;. Similarly, applying (11.6) for nodes j and m, we obtain B = 6; and
@— 0, lneretores

6;
d= 5 OjaDOr + OmOm= [w;, Wj,Om| 0; = N*.-0° (ilI$7)
Om
where N°p = [@;, 0), @m], Of p = [0;. 0}, m] t , and the superscript
. ¢ denotes transposition.
ne
Assuming that the time derivatives are prescribed functions of the space coordinates
at any particular instant of time, we can write the time derivative for the temperature within
Chapter 11 = Ampacity Computations Using Numerical Methods 301

Figure 11.4 Area coordinates.

each element as

00 00; 00; A OOmn

— Oma eae i l> j> f(a) ee, = y
OOn
ot

because N° is a function of the coordinate system, but not of the time.

The relationship between area coordinates and Cartesian coordinates is

36 Wj
Yl =|) Le wip ae Wj
| We Ghyll Om

The inverse relationship yields the coefficients of vector N°:

Wj 1 (yj = Ym) (ae —x;) (XjYm—Xmyj) x

Wj = 7A (OFBeyi) Cyeer) (NaVii y C19)
Om (vi-yj) (Gi-%) (xi¥y—xy) 1
where A is the area of the triangle.
We can observe from equations (11.7) and (11.9) that the temperature is a linear
function in x and y. This means that the gradients in either x or y directions are constant.
A constant gradient within any element means that many small elements have to be used to
approximate a rapid change in the value of @.

EXAMPLE 11.1
Consider a triangular element shown in Fig. 11.5. We will evaluate the element equation and calculate
the value of the temperature at point P for the following nodal temperature values: 6; = 40°C,
}; = 34°C, and 0, = 46°C. P is located at (2.0, 1.5). y OF
at Gah, ®)
(2)
302 Part II] m Advanced Topics

yA m (2,5)

j (4, 0.5)
—
i (0, 0) x Figure 11.5 Illustrationfor Example 11.1.

The temperature 6 is given by equation (11.7) with the shape function described by equation
(11.9). First, we need to compute the area of the triangle. This is obtained from

ee
any: i @ @
DAPail Kael ey A ee OSa= 19
Lens I" wPh
ge3B
Fromequation(11.9),
Wj (9; =In) Gm—7). (23m —FmYj) x
DME | Cine Ne tedXm) XmYi—XiYm) y
Om (vi-yj)) Gs -*) (ix) —35¥) I

OS—5)5@=4)) 94-5 —2530/5) 2 0.368

=5 @O=O) (WS), oO] Weay) her |[ || stots
(0O—0.5) (4-0) (0-05 —4-0) 1 0.264
Thetemperatureat pointP is obtainedfromequation(11.7):
6j
0 =[a;,0j,@m]-| 0; | = 0.368-40 + 0.358-34 + 0.264-46 = 39.4°C
Om

In triangular elements, the temperature varies linearly between any two nodes. Any
line of constant temperature is a straight line and intersects two sides of the element. The
only exceptions are when all nodes have the same value. These two properties make it easy
to locate isothermal contour lines.

EXAMPLE 11.2
We will determine the 41°C contour line for the triangular element used in Example 11.1.
The temperature isotherm for 41°C intersects sides im and mj. The coordinates at which this
isotherm intersects the sides of the triangle are obtained from the following simple ratios:

46 — 41 Pe
ines = Rg Obmexe
Pees
Chapter 11 m Ampacity Computations Using Numerical Methods 303

and
200 Leenaaay 3.19
Te Gate ama

The contour is shown in Fig. 11.6.

64

8; = 40°C

Om = 46°C

41°C isotherm

Figure 11.6 41°C isothermal contour.

11.3.3 Finite-element Equations

In the previous section, we have discussed how to compute the temperature at any
point inside an element if the temperature values at the nodes are known. To obtain node
temperatures, we use a property, known in the variational calculus, which states that the
minimization of the functional

f=[ p1| voy'v6

+(Win
=f)do| dS+i!a0Siz
ae
1 (Ch
Bans)
dC
(oh (11.10
over the area S bounded by the closed curve C requires that the differential equation (11.1)
with the boundary conditions (11.2) and (11.3) be satisfied. The temperature gradient is
defined as
00
ax
Vor
00
dy
Clearly, any temperature distribution that makes x a minimum also satisfies the governing
differential equations, and is therefore a solution to the problem being studied.
Equation (11.10) is a starting point for determining the temperature at each node. We
minimize (11.10) by using our set of element functions, each defined over a single element
and written in terms of the nodal values. The nodal values 6, are the unknown values in
304 Part III m Advanced Topics

our formulation. These values are obtained by taking derivatives of x with respect to each
9, and equating them to zero.
Recalling that the functions @are defined over each individual element, the integrals
in (11.10) must be separated into integrals over the individual elements and the derivatives
computed for each element; that is,

ese (11.11)
where x° is the functional defined for element e and E is the total number of elements.
Let us consider a single element first. As any element contributes to only three of the
differentials associated with its nodes, these contributions can be listed as
ax*
06;
( axsi ) \*wees axe (11.12)
06;
Oe
dO
The derivatives in equation (11.12) cannot be evaluated until the integrals in (11.10)
have been written in terms of the nodal values ©°. This is done by first computing the
derivatives of 6 with respect to x and y. Because the sum of the area coordinates equals
one, only two of them are independent. Assuming that these are w; and w;, we have

a0 00
Ox Toy
Vo = = J
RY2) 00
dy J;
00
1 gle
1 | tof
Yj —Ymmn)(Ym—Yi) dw; (11.13)
2A (Xp = x;) (Xi—Xm) 06
J;
00
ug bb; 0;
en Weaning 06
J;
where the Jacobian J is obtained by differentiating equation (11.9). Further, from equation
(11.7) and the fact that @; + ©; + @», = 1, we obtain

a0
dw; (tage vs
+ peo bad bade bee 6; |=Ve (11.14)

J;
Chapter 11 m Ampacity Computations Using Numerical Methods 305

Thus, for a single element, we have

Ve = J Vie0° (11515)

Substituting (11.15) into (11.10), with S and C corresponding to a single element, and
differentiating with respect to O°, after some routine but tedious computations, equation
(11.12) can be written as (Flatabo, 1972)

dx \°= h°O°“ehé@
(5%) I@*
apes (11.16)
». (2 .
Denoting by dj;, djm, and d,,; the distance between nodes ij, jm, and mi, respectively,
the element conductivity matrix h* is equal to

1 OF aja; Aj;am be bib; bibm

he = AAp aja; a’ ajam =e bib; by bjBm
Qjdm jan a, DiDm,bj Dy Do

hd eat hd es ie hd sae & CUEN

Pept et oe?ae gs| a a?ial a ee | a Oe ate
6 6 6
OY© @ 0 i 2 i @ 2

Qi=Xm —Xj Aj = Xi-Xm An = Xj—-Xi

bi = Yj —Yn D =n = i bm = Yi —Yj

If there is no convective boundary along any segment of the element, the relevant term
in equation (11.17) is omitted (see Example 11.3).
The element capacity matrix is given by

; sas |
Cc
c= ——
Var 1 1a. 11.18
(11.18)
8 ey)
and the element heat generation vector is equal to

1
res THAD 1
i UEC PCSpePM a OS
3 | D 0 5
(11.19)
=(h@amb 1
: q)ani 0
1
Here, again, the last three terms apply only if the appropriate boundary exists along the
element edge. Factor Win:A represents the total heat in W/m generated in the element.
306 Part III m Advanced Topics

Performing computations given by equations (11.16)-(11.19) for each element, we

finally obtain the following set of linear algebraic equations for the whole region:

Date
dO
pee
gee KOAE
Selaen) EI)
AOS,
= HO+Q0——-K=0 (11.20
In this equation, H is the heat conductivity matrix, Q is the heat capacity matrix, © and
9@/dt are vectors containing the nodal temperatures and their derivatives, and K is a
vector which expresses the distribution of heat sources and heat sinks over the region under
consideration.
In a steady-state analysis, equation (11.20) simplifies to

HO -K=0 (11.21)

EXAMPLE 11.3
Consider the element examined in Example 11.1. Assume that this element experiences convection
on surface ij and a constant heat flux on surface mi. We will calculate element matrices given the
numerical dimensions and properties shown in Fig. 11.7.

Ya

8 amb = 20

x7
Figure 11.7 Illustration for Example 11.3.

The lengths of the boundary sides are

dij = JO —4)? + 0 —0.5) = 4.03, dn; = (2-02 + 6 —0)? = 5.39

Substituting the numerical values in equation (11.17), we obtain the following conductivity matrix:

Aj =Xm —Xj =2—4=-2 aj =X; —Xm =0-2=-2 Am =x; —x; =4-0=4

bi = LAB 1 Pig05-5 =—4.5 b; = Ym —-Y=S5S-O=5 dn = Vi yp 0—0.5 = —0.5

Chapter 11 m Ampacity Computations Using Numerical Methods 307

i ae aja; diam b; bib; bibm

hé = aAp aja; a; aj am AF bib; bj bjbm

hdj;
ies savin hdjm
Ga Gale Rasea: eae hd
waa ce ean lad Poe
0 0 0 eat [ee ae
(—2)2 (—2)(—2) (—2)(4)
PT Sree Ne ee
(—2)(4) (—2)(4) 4
(4552 (—4:5)6) (©€4.5)G0:5)
+ (—4.5)(5) (5)? (5)(—0.5)
(—4.5)(—0.5) (5)(—0.5) (05
5-403 5-4.03
3 6 6.85 520m 0103

5 5-403 5-4.03 che 3226 6.87 —0.06

6 3 =08 OG O09

0 0 0

The heat generation vector is obtained from equation (11.19):

xe = Win
Wid ]
|p |, FOamy
+9)4j | 1 |, Pm +9)dim 0
|, |, Oem+4)dui |g
A Gar dij am jm am mi
3 oD; 2, 2
1 0 1
1 1 276
x ils).
o)5) \ belles se10-5.39 749
aulde 2 2
1 0 ] 74.5
EXAMPLE 11.4
We will consider now a domain composed of three elements, one being the same as examined in Exam-
ple 11.3 and two adjacent elements as shown in Fig. 11.8. We will determine nodal temperatures in the
steady state for this system, assuming that the other boundary surfaces have zero temperature gradient.
The matrix for element | was obtained in Example 11.3, and is equal to
6.85 3.26 —0.03
he B26 6.87 —0.06
—0.03 —0.06 0.09

The element matrices for elements 2 (nodes j, n, m) and 3 (nodes j, k, n) are obtained from equation
CIM
GIDE
OBS eue—0) 225 Osi 0.5 —0.5 0
hits —0.22 20508 P=0:68 and@ho=1/ 8 =0:5.9 90-68: 0713
—0.11 —0.68 0.79 0 —0.13 0.13
308 Part III m Advanced Topics

Figure 11.8 Illustration for Example 11.4.

Since there are five nodes in this system, the matrix H will have five rows and five columns:

i J m ies

6.85 3.26 —0.03 0 O 0 0 0 0 0

3.26 6.87 —0.06 0 O 0 033 —0.11 -—0.22 0

H=| -0.03 -0.06 009 0 0j{+] 0 -011 0.79 -—0.68 0

0 0 0 0 O 0 -—0.22 —0.68 0.90 0

0 0 0 0 O 0 0 0) 0 0

0 0 0 0) 0 6.85 3.26 —0.03 0 0

a) OS. © 0 —0.5 5226 tell —0.17 -—0.22 —0.5

+ 0 0 0 0 0 =| -0.03 -0.17 0.88 —0.68 0

0 0 OP SONS Ors 0 =—0:22 —0.68 1.03 —0.13

0.57 ORR 0213 2 0163 0 —0.5 0 —0.13 0.63

Since elements 2 and 3 do not generate any heat and have zero temperature gradient, vector K is the
same as obtained in Example 11.4 with the components corresponding to nodes n and k equal to zero;
that is,

With the conductance matrix and heat generation vector given above, the following nodal temperatures
are obtained by solving equations (11.21):

O = '24.6°C.9 9.0)= 34.9°C, Ot 2119°C a 6) = 154.9). 6, = 597°C

The set of ordinary differential equations (11.20) which define the discretized problem
can be solved using one of the many recursion schemes. There are two popular procedures
Chapter 11 m Ampacity Computations Using Numerical Methods 309

for solving these equations to obtain the values of © at each point in time. The first is to
approximate the time derivative using a finite-difference scheme. The alternate procedure
is to use finite elements defined in the time domain. Flatabo (1972) used the midinterval
Crank—Nicolson finite-difference algorithm for the solution of this equation. This method
requires an iteration within each time step. Here, we propose to use Lees’ (1966) three-
level, time-stepping scheme in which the discretized equation is replaced by the recurrence
relationship

5 H"
ott=| 5 n
ST
2ATt =| H’@"
3 4H"
Qe"!n
2) Ee"!
siya
Pa -K] (11.22)
where the superscript n refers to the time level and Art is the time step. The procedure is
unconditionally stable, and has the advantage of producing the solution at time level n + 1
without the need for any iteration as the coefficient matrices are evaluated at level n. The
initial conditions have to be specified, and the first time step iteration is performed by a
modified version of equation (11.22) requiring only one previous time step solution.

11.3.4 Some Comments on Computer Implementation

The implementation of the finite-element method proceeds in two stages. In the first
stage, the region under consideration is discretized into a finite number of elements and
relevant matrices are formed. In the second stage, the resulting set of linear equations is
solved. The efficiency of the computations depends to a large extent on the discretization
process. This process can be divided into two general parts: the division of the region into
elements, and the labeling of the elements and the nodes. The latter sounds quite simple,
but is complicated by the desire to increase the computational efficiency. We will briefly
discuss this process using the example of triangular elements.
The division of any two-dimensional domain into elements should start with the divi-
sion of the body into quadrilateral and triangular regions. These regions are then subdivided
into triangles. The subdivision between regions should be located where there is a change
in geometry, or material properties, or both.
In most cable rating computations, the region which is divided into elements is usually
a rectangle. This is simply accomplished as illustrated in Example 11.5 below. When
a rectangle is divided into two triangles, division using the shortest diagonal is, prefer-
able because elements close to an equilateral shape produce more accurate results than
long narrow triangles. The spacing between boundary nodes can be varied to obtain de-
sired element sizes (the elements should be smallest closer to the cables). There will be
2(n — 1)(m — 1) elements in a quadrilateral where n and m are the number of nodes
on adjacent sides. The ability to vary the element size is an important advantage of the
finite-element method.
The labeling of nodes (assigning numbers) influences computational efficiency. The
conductance matrix in equation (11.20) is usually very sparse. Good general advice is to
number the nodes in such a way that the largest difference between the node numbers in
a single element is as small as possible. The numbering of elements does not affect the
computational aspects of the problem.
There are commercially available computer programs which automatically generate
finite-element meshes for thermal problems. However, we would recommend that a person
310 Part III] m Advanced Topics

wishing to use the finite-element method read some specialized books on the subject and
do some experimentation on his/her own.
We will illustrate an application of the finite-element method to determine the geo-
metric factor of a backfill or duct bank.

EXAMPLE 11.5
The following procedure to obtain the geometric factor for cables located in duct banks and backfills
was proposed by El-Kady and Horrocks (1985).
Consider the thermal circuit configuration given in Fig. 11.9 where the cable bank is represented
by a rectangular cross-sectional surface C of height A and width w. For this configuration, the total
thermal resistance between the duct bank surface and the ground ambient is given by Goldenberg
(1969):
AsPs (6. = Oamb)
—_ (U23)
ii —ds
00
Ccon
where p, is the thermal resistivity of the soil, C represents the duct bank surface, and 0@/dn denotes
differentiation in the direction perpendicular to C.

Isothermal surface S, O,

ny

Isothermal surface C. © Figure 11.9 Thermal circuit configuration in

"CC Example 11.5.

In the finite-element solution, the medium surrounding the surface C is partitioned into small
triangles constituting a finite-element grid such that the first grid layer, enclosing the bank surface C,
is carefully structured, as shown in Fig. 11.10, to attain an efficient subsequent evaluation of equation
(lal289).
The surface C is partitioned into K small segments, as shown in Fig. 11.10, where the tempera-
tures 0), #2, --+ of the middle points ofthe first grid layer (which constitute nodes of the finite-element
grid) are evaluated. The accuracy ofthe solution can be controlled by adjusting the size of the elements
of the grid. Equation (11.23) can now be written in the discretized form
T= p
AG, 1S, (11.24)

fo, ANi Oci —Pamb

where, as shown in Fig. 11.10, 6; is the temperature of segment i along the first finite-element grid
layer surrounding the duct bank surface, 6,; is the temperature at the duct bank surface C of segment
Chapter 11 m Ampacity Computations Using Numerical Methods 311

Surface S, O = 0°C

First Grid
Layer

pte7;
Surface C
Oc = IG;

Figure 11.10 Finite-element grid structure for an outer layer of a duct bank.

1, and J, is the index set of segments along C. By choosing AS;/An; = 1 for all i, equation (11.24)
reduces to (see Section 9.6.5.2 for the discussion of a geometric factor)
p p
ge eS eae a CU2S)
i€leBei—amb
Hence,
Gee (11.26)
Aci—9
ie€l, Oi—Gam
Equation (11.26) provides the value of the geometric factor in terms of the temperature results
from the finite-element analysis. We note that the surface C is not assumed to be isothermal. If, in
fact, the duct bank surface is an isotherm, then @,; = 9, for all 7 in equation (11.26), leading to
= 20 (8. = amb)
22 seep Cul 2”
Y> @ — 4)
i€le
If we set 0. = 1 and O2mp= 0, equation (11.27) further simplifies to

C2 Mise2 (11.28)

Equation (11.28) was used by El-Kady and Horrocks (1985) to obtain the extended values of
the geometric factor for duct banks and backfills reported in Table 9.7.
312 Part II] m Advanced Topics

11.4 THE FINITE-DIFFERENCE METHOD

11.4.1 Overview

Historically, numerical methods employing finite differences have been used more
frequently than those based on finite elements to solve partial differential equations (Black
and Park, 1983; Groenveld et al., 1984; Hanna et al., 1993; Hartley and Black, 1981;
Hiramandani, 1991; Kellow, 1981; KEMA, 1981; Radhakrishna et al., 1984). However,
the reader should be aware of the manner in which the approximations used in the finite-
difference methods are used. Finite-difference methods are approximate in the sense that
derivatives at a point are approximated by difference quotients over a small interval. This
is in contrast to the finite-element methods where the temperature as a function of space
and/or time is approximated by polynomials.
In finite-difference methods (Fig. 11.11), the area of integration, that is, the area S
bounded by the closed curve C, is overlaid by a system of rectangular meshes formed by
two sets of equally spaced lines. An approximate solution to the differential equation (11.1)
is found at the points of intersection of the parallel lines, the so-called nodes or mesh points.
This solution is obtained by approximating the partial differential equation over the area S
by n algebraic equations involving the values of 6 at the n mesh points internal to C. The
approximation consists of replacing each derivative of the partial differential equation at
the point Pj; (say) by a finite-difference approximation in terms of the values of 6 at P;,;
and at neighboring mesh points and boundary points, and of writing down for each of the
n internal mesh points the algebraic equation approximating the differential equation.

Y,

P09,
|[ey
| e fe
Figure 11.11 Discretization of region S by a rectangular mesh.

This process, illustrated in the next section, clearly gives n algebraic equations for the
nunknowns 6,1, 6),2, +++, j,;, >>. Accuracy can usually be improved either by increasing
the number of mesh points or by including “correction terms” in the approximations for the
derivatives.

11.4.2 Finite-difference Approximations to Derivatives

The temperature 6 is a function of space and time coordinates. The partial derivative
with respect to time can be approximated as follows. Let At be a small time interval.
Chapter 11 m Ampacity Computations Using Numerical Methods 313

Expanding the value of 6 in a Taylor series, we have

6¢ go 1 dOaan t 0°60
( + At)) = 0(t)
(t) + Ar— + —(Ar)?
Lire+ 5| T) a2 any= Sa)Gels
ap = Oe
ai @ , : (11029)
6@
( Ax)ere = 0) = At—+
Ao =(Ary
5 “Sebool
— me
eo 0-6
= =Gap 2 Tt-

Addinguptheseexpressions,
, 076 4
A(t+ At) +00 —At) = 26(1)+ (Ary =>+ 0 [(At)'] (11.30)

where O [(Ar)*] denotes terms containing fourth and higher powers of Ar. Assuming
that these are negligible in comparison with lower powers of At, we obtain from equation
(11.30)

d°O(t)A(t +Ar)—20(t)+O(t—At) (11.31)

ar2 (At)2
Substituting (11.31) into (11.29) and neglecting the terms of order (At)* and higher leads
to
d0(t) O(t+At)—O(t —At)
—Ot = QING ( 1132 )

Equation (11.32) is called the central-difference approximation of the time derivative. Two
other approximations often used in practice are
The forward-difference formula:

d0(t) O(@¢+ At) —O(t)

——S-
ot SS AT ((383)
andthebackward-difference
formula:
() _ (1) —Ot
- At)
d0(t 6(t) —-O(t—A er
ot AT

Similar expansion into a Taylor’s series can be performed for space coordinates. Let
us assume, for the purpose of illustrating the concepts, that the region S is a rectangle, and
we divide it as shown in Fig. 11.12.
Consider a representative mesh point P in Fig. 11.12. Denoting the value of @at P by

Op = &,j

the partial derivatives appearing in equations (11.1)—(11.3) can be approximated as follows:

026 ie 0°76 va Oiaaiance

Oy rarOS)5
Ox amor) he aoe
(11.35)
VO\ —(8°O\Gin —261.5
+Oi,5-1
DE/ie RAGED
es Ay?
314
Part II] m= Advanced Topics

Figure 11.12 Rectangular finite-difference mesh.

We can combine equations (11.33) and (11.35) by introducing the notation

X=iAx,y = jAy,t =nArt, andeOGAr Ay AAT! = Fin

With this notation, equation (11.1) can be written as

A Gijnt1 = 9,jn _ Gittjn —26ijn+ 9-1 in

dij AT Ax?
(11.36)
zuOj,41,0=e26%,
junair6;j-1,n+ Winti,jn Pi,j
Ay*
The step sizes Ax and Ay could be varied along the x and y axes. Also, in equation (11.36),
the material properties can change in space and the heat generated can vary in space and
time. Since the problem becomes computationally complex even for a small number of
nodes, we will illustrate the above concepts for a steady-state heat transfer problem.

EXAMPLE 11.6
We will consider a single cable located | m below the ground and generating 12 W/m of heat. The
earth surface is a convective boundary with convection coefficient | W/m? - K and the ambient air
temperature is 0°C. We will also assume that the soil ambient temperature at the outer border of the
mesh is also equal to 0°C.
Since the computations are performed without the aid of a computer, we will consider a small
(2 m x 2 m) mesh with the cable located in the center. Placing the origin of the coordinates at the
cable center, the region is bounded by straight lines x = +1, y= +1. We will use the division
Chapter 11 m Ampacity Computations Using Numerical Methods 315

No at SY ; and label the mesh points as indicated in Fig. 11.13, equal numbers denoting equal
values of 0.

Figure 11.13 Finite-difference mesh for Example 11.6.

For simplicity, we will allocate the whole heat source at node (0, 0). Soil thermal resistance in
each square is 1 K - m/W. The boundary conditions can be written as

00
@= 0/onx == sandy = ——
I; Ae aa ae
n
We will denote the mirror image of the point 2 (say) in y = 1 by —2, and the value of @at point 2 by
9, and so on. One finite-difference representation of equation (11.1) is
1
052 ith 20; 5eyS45) A 0 52(Gi;
jams 20; jekG521)=
O; 8
or
OG FO EG +t8jy = 4075 = 2/4
where g = —12 for node 7 and zero otherwise. At the boundary point 1, we imagine the heat
conducting area extended to the first row of external mesh points, and therefore we have

0, + 64 + 02 + 04 —40, = 0

and from the boundary condition, using central differences,

(0-4 — 04)/2Ay = (0-4 — 4) = —O

 Part III m Advanced Topics

Elimination of 6_4 gives the equation

—50, + 20, + 2064= 0

In the same way, remembering that 0; = 0, point 2 yields the equation

6, —50, + 20; = 0

Similarly, points 4, 5, 7, 8, 10, and 11 yield the equations

0, —404 + 205 + 67 = 0

0 + 64 —405 + 63 = 0

04 — 40; + 203 + O19= —3

0; + 07 —40g + 61; = 0

6, —4610 + 201; = 0

03 + O19 —46;; = 0

Written in a matrix form, these equations become

-5 2 2 0, 0

lee) een: 2 0, 0

| 0 -4 2 l 04 0

1 1 -4 0 1 65 0
= (11.36a)
1 0 -4 2 1 0, —3

1 1 -4 0 l Ag 0

| 0 -4 2 Ai 0

1 1 —4 A 0

where the empty area in the matrix is filled with zeros. We can observe that the coefficient matrix is
of the form
(B-—I) 2I

I B I 4y 2
where B =
I B I 1 —4

I B

The solution of (11.36a) is

Cie— 10.20 Cree

— 0.6 Caoge—
10.40" Geman
— (7 Ge
0; => IPAbeeCee Ag = Oa eee Ai0 = 0.40°C, 61, = 020°C

In Example 11.6, uniform soil was assumed. When different soil layers are present, the
grid lines should coincide with the boundaries between soils with different characteristics.
In this case, it may be more convenient to write equation (11.36) in the form
Ohjnel —Orjn Kitt s@reign — 2k ji jot Kinl jedi
Qi;
At Ax?
kit Gnjain 2k enya kj len join
-f + Winti,j,n Chis7)
Ay?
Chapter 11 m= Ampacity Computations Using Numerical Methods 317

where k; ; and Q; ; denote the thermal conductance and thermal capacity of the material in
which node ij is located. We will demonstrate how the node conductivities are computed
by again considering Example 11.6.

EXAMPLE 11.7
Assume that cable in Example 11.6 is located in a thermal backfill with thermal resistance of
0.6 K-m/W, which encompasses six squares bounded by corner nodes 2, 2, 11, and 11 in Fig. 11.13.
We will reformulate the finite-difference equations to reflect varying thermal resistivities.
Equation (3.37) will now take the form

Kesh cient ki,pny vip kip Gian

jini Rigaigein —4679 jm= 9/4
where the conductivity values are computed as follows.
The thermal conductivity associated with a node is an arithmetic average of the conductivities
of the rectangles to which the node belongs. For the nodes inside the backfill (nodes 1, 4, and 7),
k;,; = 1/0.6 = 1.67. For the nodes located inside the native soil, kj; = 1/1 = 1. For nodes 2,
5, 8, and 10, on the perimeter of the backfill, k;; = (1/0.6 + 1/1)/2 = 1.33, and for node 11,
y= (1/0.6+3 1/1) /4=17:
At the boundary point 1, we have

1.336) + 1.670_4 + 1.336) + 1.676, —4- 1.676, = 0

and from the boundary condition, using central differences,

1.67(6_4 —0) = —0;

Elimination of 6_4 gives the equation

=2.30; + 0. +0, =0

Similarly, point 2 yields the equation

1.670, — 6.320) + 2.660; = 0

Likewise, points 4, 5, 7, 8, 10, and 11 yield the equations

0, —404, + 1.605 + 6, =0

62 + 1.2564 — 405 + 63 = 0

1.2504 — 507 + 203 + O19= —3

1.3305 + 1.670, — 5.320, + 1.170;; = 0

1.6707 — 5.32619 + 2.346), = 0

03 + O10— 3.501; = 0

Their solution is

6, =0.27°C, 0 =0.19°C, 6,=0.43°C, 05 = 0.29°C,

0; = 0.96°C, Ag=> 0.42°C, O10= 0.40°C, O11= 0.24°C

a ee
318 Part III m Advanced Topics

Equation (11.36) is called the explicit finite-difference representation of equation

(11.1). This is because the values of 6 can be computed explicitly from this equation
given the initial and boundary conditions. Although the explicit method is computationally
simple, it has one serious drawback. The time step At is necessarily very small because
the process is valid only for

platSelsey
za]
1 finde
i
oats ae)
alWoy
reBe1 (11.38
and Ax and Ay must be kept small in order to attain reasonable accuracy. Crank and
Nicolson (Smith, 1965) proposed a method that reduces the total volume of calculation and
is stable. In this approach equation (11.36) is replaced by

Span
é WACie
AT A 2(=elsea dy?oly ee
ax2° dy?Pk +Wino(11.39
im? ( )
which yields somewhat more complex finite-difference equations.
When the cable is not a line heat source but has a specified diameter, the cable outer
surface cuts the vertical and horizontal lines of the grid as shown in Fig. 11.14, producing
an irregular boundary inside the trench. The finite-difference equation (11.37) has to be
modified to reflect the new boundary points. For example, in the case shown in Fig. 11.14,
we have

(33)=($5)hte| 26341,musi
eaeat
Ox)
p One};“AX?PGC
ates)eae
peClee) (11.40)

(SS)=(5) Ojai—20;git
dy?
Pha)
dy?
ge Ay?
The finite-difference equations become even more complex if the internal structure of
the cable is to be represented properly or when normal derivatives at the cable surface are to
be approximated by finite differences. When a circular symmetry with respect to the cable

Figure 11.14 Cable outer diameter cuts the grid

lines.
Chapter 11 m Ampacity Computations Using Numerical Methods 319

center can be assumed, a radial form of the finite-difference equations could be applied.
In practice, only under certain restrictive assumptions can such a symmetry be assumed.
One such case is a single-core cable located in a uniform soil with an isothermal boundary
condition at the earth surface. This has been studied by CIGRE WG21.02 (CIGRE, 1983).
We will discuss this case next.

11.4.3 Application of the Finite-Difference Method for the Calculation

of the Response of Single-Core Cables to a Step Function Thermal Transient!

11.4.3.1 General Characteristics of the Method. —The finite-difference method de-

scribed in the previous section can be viewed as being equivalent to a representation of
the cable and its surrounding by a ladder network of thermal resistances and capacitances,
in the same general way as was described in Chapter 5, but with many more subdivi-
sions.
To show this, we will assume that the heat flow is radial both inside and outside the
cable so that it is possible to divide the cable and the soil into a number of concentric layers
corresponding to elements in the ladder network. The outer radius of the soil where the
heat disperses will increase with time, and for practical purposes, it is sufficient to consider
only that radius within which a sensible temperature rise occurs. In this case, this radius
can be estimated from equation (11.4).
For long-duration transients, this cylinder of soil may include adjacent cables. In the
case of isothermal earth surface boundary, the radius will also include image sources to
both the cable under consideration and adjacent cables (Fig. 11.15), dissipating (at each
time step) losses of the same value but opposite sign. The time steps can be varied during
the computation, and their durations should be governed by equations (11.5).
When calculating the thermal field of a cable by the method presented in CIGRE
(1983), the space occupied by other cables is assumed to have the same thermal properties
as the soil (restricted application of superposition).

11.4.3.2 Description of the Method. The insulation (including the semiconducting

screen) should be divided into ten elementary layers of equal thickness. The semicon-
ducting screens are part of the thermal circuit, and are assumed to have the same thermal
characteristics as the insulation. The dielectric loss is calculated disregarding the screens
but, for simplicity, it is approportioned over the total thickness (i.e., insulation plus screen).
A single layer is used for cable covering.
The number of elementary layers in the soil n; must fulfill the following conditions:

2r — D.
< 0.16 (11.41)
Qn, D>

where r is given by equation (11.4). The division of the cable and the soil into layers is
illustrated in Fig. 11.16.

1 Some of the material in this section is based on an article published in Electra (CIGRE, 1983).
2 An alternative method, which can offer advantages in computation, is to divide the thickness of each
material so as to obtain equal ratios of the radii, that is, each layer in a particular material would then have the
same thermal resistance.
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Chapter 11 m= Ampacity Computations Using Numerical Methods
321

Sheath TextileTape
Screen Reinforcement Soil
Screen Insulation Serving
Conductor
Spiral He
OilDuct
pee \
oe
a.
F
-<=:L-a 3
' '
Sis
I

“Hl
=

Wii W2i Woi Wioi

Figure 11.16 Division of the cable and the soil into layers (CIGRE, 1983).

Each layer has a thermal resistance obtained from applying equation (3.6):

fp RS fry Ot 11.42
: 21 rjis ( )

where r; and r;_,; are the external and the internal radii of the layer, respectively. The
thermal capacitance of each layer is divided at the midlayer radius, and allocated to the
nodes at the inner and outer radii of the layer to give a a scheme for the equivalent circuit
of each layer. The capacitance at each node is obtained from equation (3.14):

Q;=xc(ae) 2Pf(24) 2 (11.43

In developing equations for temperature variations with time, we will consider first
durations so short that no adjacent cables or image sources are involved. Next, we will
discuss the case of longer duration transients.

11.4.3.3 Computation of Temperature Transient for a Single Cable. We will

develop the finite-difference equations for the ladder network in Fig. 11.16 starting from
the heat balance constraint at each node (Hanna et al., 1993; Selsing, 1985; Weedy, 1988).
322 Part II] m Advanced Topics

The heat balance equation, when differentiated, leads to the partial differential equation
(11.1) (see Section 2.3.1). The heat input to the node j is equal to

kj (Oj-1,n S Ojn) ar kj+1(@j4i.9ty Oj.) oFqj d 1.44)

where k; is the thermal conductance from node j — | to j, n is the index of elapsed time,
and q; is the heat generated in the volume represented by node j from the joule or dielectric
loss. Over a finite time interval At, the average rate of flow to the region is given by

Git At) = OF) Ojnti —Oj.n

2; > At = F729
2; At ( 11.45

where Q is the thermal capacitance of an elementary layer between nodes j — | and j.

Equating now (11.44) and (11.45), we obtain

Arq;
Ojn+1= ee a, neNee Oj4in a te a (kj mrka] Oj.n= (11.46)
Q; Qj; Qj; J

beginning at j = N — | and ending at j = 0 where 6,,, is the temperature rise at node j

at time n and 6y., is always zero. N is the number of nodes, and we start with known or
assumed initial temperatures.
Equation (11.41) can be applied to each core of a pipe-type cable. As Weedy (1988)
pointed out, an accurate modeling of the oil in the pipe is much more difficult, and usually
the oil is represented by a single lumped thermal capacity.
To assure numerical stability of equation (11.41), the coefficient of 6; , must be positive.
This yields the following requirement for the time step at each node:

Q;
AT ~—<,
a (11.47)

The limitations of the time step in the explicit method may be overcome by the use
of implicit equations such as given by Crank and Nicolson [equation (11.39)] by Lees’
approach (1966), or by a Hopscotch method (Gourley, 1970).
CIGRE (1983) proposed the following algorithm equivalent to equation (11.46).

1. For each node, we compute

Js
Ait j-1,))2 ( Ti
[ 1—Aj_ et)+1 4 ( QjTj+1
1¢ Atnea 7 | ¥ ( 11.48)

beginningat 7= 0 andendingat j = N —1,withA_,,, = 1. Notethat An

does not change during the computation unless the time step At; is changed after
a given time t during the computation.
2. Compute initial values for each node:

ii
B;, =A iL ma djn+ Oj, (2 Q;
Waj.n Gane pani sr T;
Bj-\
ena (11.49)

beginning at j = 0 and ending at 7= N — 1, with B_,,, = 0.

Chapter 11 m Ampacity Computations Using Numerical Methods 323

3. Compute the values of temperature rise from

Oj.n= (Aj t Oj41.n)+ Bin (11.50)

beginning at j = N — 1 and ending at j = 0, where 6;,,, is the temperature rise at
node j at time n and Oy, is always zero.
4. Repeat steps 2 and 3 for as many times as is necessary to complete the transient

At = Tj — T-1

The value of At; may be changed during the computations of the transient provided
it remains with the limits specified by equation (11.5). Coefficient A;,, must then
be recalculated.
11.4.3.4 Mutual Heating from Other Cables. | When there is more than one cable,
each will have its own thermal diagram and equivalent network from which the transient
temperature rises caused by its own losses are calculated. If the duration of the transient is
sufficiently long for mutual heating from other cables and image sources to be significant,
this must be taken into account in the calculation of the true temperatures used to update
cable losses at each time step. Since the presence of other cables violates cylindrical sym-
metry, a restricted application of superposition is applied.
The true temperature for each cable is obtained at each time step by adding to the
cable’s own temperature the temperature rise at the same location from every other cable
and image source (see Fig. 11.15). These thermal fields are computed from the equations
corresponding to the equivalent networks centered on each cable or image source. This
process is carried out for all cables and image sources in the group. Note that the thermal
diagram and network of an image source are identical to those of its corresponding cable
except for the negative sign of its heat flux and temperature rise; separate networks for
image sources are therefore not needed.

EXAMPLE 11.9 (CIGRE, 1983)

We will consider one cable in a circuit of three single-core cables with the duration of a transient
lasting 4 h. This time is short enough for neglecting the effect of the two adjacent cables. The
computations deal only with the self-heating of the center cable. The cables are 400 kV, 2000 mm?
with paper/fluid insulation laid directly in the ground, | m deep, in flat configuration, spaced 70 cm
apart, and with the sheaths cross bonded.
The additional information is as follows:

Conductor de resistance at 20°C = 8.95 wQ/m

Diameter of the conductor = 57.5 mm
Diameter over conductor screen (semiconducting paper tapes) = 59.0 mm
Diameter over paper insulation = 105.0 mm
Relative permitivity of the insulation = 3.3
tan 6 = 0.0025
Diameter over semiconducting screen = 106.0 mm
Diameter over lead sheath = 114.0 mm
DC resistance of sheath at 20°C = 154.8 4&2/m
Diameter over polyethylene jacket = 122.0 mm.
324 Part I] m Advanced Topics

Thermal resistivities of the metallic parts are assumed to be zero. All remaining thermal resis-
tivities and heat capacities are as specified in Table 9.1.
The steady-state ampacity of this cable is equal to 1800 A. We will investigate variation of
conductor temperature with time when an emergency current of 3600 A is applied to this cable for
4h. Prior to the emergency condition, the cable was energized long enough for the temperature rise
due to dielectric losses to reach its steady-state value.
Figure 11.17 shows the results of the computations.

120
Conducto

100

(°C)
Temperature
—
fo)
=) 40

20
Sheath

Cable Surface

0
0 40 80 120 160 200 240

Time, minutes —

Figure 11.17 Calculated transient temperatures, 3600 A for 4 h.

We will close this chapter with a brief discussion of modeling and computational issues
involved in the application of numerical methods.

11.5 MODELING AND COMPUTATIONAL ISSUES

Finite-difference and finite-element methods for solving the heat conduction differential
equation often lead to a very large number of algebraic equations, and their solution is a
problem in itself. Large sets would generally be solved iteratively and small sets by direct
elimination methods. Iterative methods are more efficient then direct methods in that they
take advantage of the large number of zero coefficients in the matrices. There is a vast
literature on the subject dealing with sets of linear equations and the economics that can
be achieved when the matrices are sparse, and the interested reader is referred to the many
books on numerical methods.
The basic difference between finite-element and finite-difference methods lies in the
manner in which approximations are performed. As stated earlier, in the finite-element
method, the temperature is approximated by a discrete model composed of a set of con-
tinuous functions defined over a finite number of subdomains. The piecewise continuous
functions are defined using the values of temperature at a finite number of points in the
region of interest. In finite-difference methods, the derivatives at a point are approximated
Chapter 11 m Ampacity Computations Using Numerical Methods 325

by difference quotients over a small interval. Both approximations can be quite accurate if
the meshes are made suitably small.
Both methods allow a choice for the multinodal elementary subdomains. However,
because of the nature of the approximations involved, a rectangular mesh with division
lines parallel to the coordinate axes is usually chosen for the finite-difference method. This
may create difficulty in representing circular boundaries which are quite common in cable
representation. This difficulty does not exist in the finite-element method because curved
boundaries can be well approximated by a suitable selection of elements. In addition, finite-
difference equations may become very complex when nonuniform spacing is used along
any of the axes, whereas the finite-element methods handle differing element sizes quite
naturally.
Because of their flexibility in representing the region around the cables and the easy
modeling of boundary conditions, finite-element methods appear to be more suitable for
the numerical solution of the heat conduction equations considered in this chapter.

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Lees, M. (1966), “A linear three-level difference scheme for quasilinear equations,” Math.
Comp., vol. 20, pp. 516-622.
Libondi, L. (Feb. 1975), “Calcolo numerico di transitori termici in cavi unipolari,’ Elet-
trotecnica, vol. LXII, no. 2.
Mitchell, J. K., and Abdel-Hadi, O. N. (1979), “Temperature distributions around buried
cables,” JEEE Trans. Power App. Syst., vol. PAS-98, no. 4, pp. 1158-1166.
Mushamalirwa, D., Germay, N., and Steffens, J. C. (1988), “A 2-D finite element mesh gen-
erator for thermal analysis of underground power cables,” JEEE Trans. Power Delivery,
vol. 3, no. 1, pp. 62-68.
Radhakrishna, H. S., Lau, K. C., and Crawford, A. M. (1984), “Coupled heat and moisture
flow through soils,” J. Geotech. Eng., vol. 110, no. 12, pp. 1766-1784.
Segerlind, L. J. (1984), Applied Finite Element Analysis, 2nd ed. New York: Wiley.
Selsing, J. (1985),“A versatile computer method for computation of conductor temperatures
in cable terminations and pipe cable systems,” JEEE Trans. Power App. Syst., vol. PAS-
104, no. 4, pp. 768-774.
Smith, G. D. (1965), Numerical Solution ofPartial Differential Equations. London: Ox-
ford University Press.
Chapter 11 m Ampacity Computations Using Numerical Methods 327

Tarasiewicz, E., El-Kady, M. A., and Anders, G. J. (Jan. 1987), “Generalized coefficients
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systems,” JEEE Trans. Power Delivery, vol. PWRD-2, no. 1, pp. 15-20.
Thomas, H. R., Morgan, K., and Lewis, R. W. (1980), “A fully nonlinear analysis of heat
and mass transfer problems in porous bodies,” Int. J. Numer. Methods in Eng., vol. 15,
pp. 1381-1393.
Weedy, B. M. (1988), Thermal Design of Underground Systems. Chichester, England:
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Zienkiewicz, O.C. (1971), The Finite Element Method in Engineering Science. New York:
McGraw-Hill.
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 12
Economic Selection
of Conductor Cross Section

12.1 INTRODUCTION

Selection of cable sizes is currently based on ampacity considerations; that is, a cable with
minimum admissible cross-sectional area is usually selected without consideration of the
cost of the losses that will occur during the life of the cable. Choosing a minimal cross
section results in a minimal initial cost investment. However, the cost of losses over the
lifetime of the cable may be quite substantial. Selection of a larger cable size than required
for ampacity consideration will often result in a smaller value of losses, and hence may lead
to a lower overall cost.
The importance of the cost of losses in selecting the most economical cable size has
been discussed by Parr (1989), and an international standard sponsored by the IEC has been
issued (IEC 1059, 1991).
In this chapter, we will introduce a mathematical model for the selection of the most
economical cable size for a particular application. The model includes the estimation of the
energy losses over the life of a cable, making allowance for growth in load and usage, and
calculates the present value of these losses using projected interest rates. This present value
of the cost of losses is added to the primary investment to give the total cost. To obtain the
most economic conductor size, the total cost is minimized.
We will consider the representation of changes in the load pattern during the life span
of the cable, as well as the effect of charging current and dielectric losses. A full coupling
with the ampacity computation techniques described in earlier chapters is included in the
optimization process.
To calculate the present value of the costs of the losses, the following parameters are
required:

329
330 Part II] m Advanced Topics

® Basic economic data: cost of installation, unit cost of losses, forecast of discount
rate, and unit demand charges
e Basic construction of the cable: number of conductors, insulation type, outer pro-
tection type
e Basic installation information: depth of burial and spacing
e Basic cable operational data: voltage level, estimated economic life, current in the
first year of operation, annual load growth, load curve characteristics, operating and
ambient temperatures, soil thermal resistivity.
The mathematical models described below are partially based on the IEC Standard
1059 (1991). The optimum cross-sectional area of the conductor for the required load is
first computed, and then the closest standard conductor size is selected.

12.2 COST OF JOULE LOSSES

The optimization model proposed below considers the cost of purchase and installation of
the cable and the cost of losses over its economic life. Both costs are expressed in “present
values.” The future costs of the energy losses are converted to their present values by
utilization of a discount rate which is linked to the cost of borrowing money.
In the procedure given below, inflation has been omitted on the grounds that it will
affect both the cost of borrowing money and the cost of energy. If these items are considered
over the same period of time, and the effect of inflation is approximately the same for both,
the choice of an economic size can be made satisfactorily without introducing the added
complication of inflation.
The total cost of installing and operating a cable during its economic life, expressed
in present values, is calculated as follows.
The total cost

ChaOl-- eg Cl2.))
where C/ =cost of the installed length of cable, $
CL = equivalent cost, at the date the installation was purchased, of the losses
during economic life of N years, $.

Theoverallcostofthefirstyear’sjoulelossesis
I, -R-L-N,-N.-(T-P+D)
If these costs are paid at the end of the year, then at the date of the purchase of the
installation, their present value is

12ORME
CNOUNDER
OPED)
Wet
where D = demand charge per year, $/W-year
/y = maximum load on the cable during the first year, A
i = discount rate, not including the effect of inflation
Chapter 12 m Economic Selection of Conductor Cross Section 331

L = length of cable, m
N, = number of phase conductors per circuit
N. = number of circuits carrying the same value and type of load
P =cost of | watt hour of energy at the relevant voltage level, $/Wh
R = apparent ac resistance of a conductor per unit length, taking into account
both skin and proximity effects and losses in metal screens and armor;
since the resistance is dependent on the cable operating temperature, its
value should be computed iteratively or an approximate mean value
should be used as discussed in Section 12.2.2

T = operating time at maximum joule loss, hours/year

= number of hours per year that the maximum current Jp would need to
flow in order to produce the same total yearly energy losses as the actual,
variable, load current:

(- I(t) dt
PS 5
0 fy

where / (t) is a load current as a function of time (A).

12.2.1 Constant Load Factor

If the load loss factor (j) is known and can be assumed to be constant during the
economic life, then

T, = j= 8760 (12.2)

The load loss factor jz can be approximated using the Neher/McGrath (1957) approach
as

pao peLe + U—p)-Le (12.3)

where LF is the load factor and the value of constant p can be taken as equal to 0.3 for
transmission circuits and equal to 0.2 for distribution networks (IEEE, 1990).
Similarly, the present value of the cost of losses during N years of operation discounted
to the date of purchase is

io RoLAN, Newt te =O pUNy DO7 (N))

he
where Qp(N) and Qp(N) are coefficients taking into account the increase in load, the
increase in cost of energy over the N years, and the discount rate.

% N Wen;
famiyegs
E N
Oalips gob 2
(1=ha)*Cick
Sakb) 12.4
Op(N) oe omaey es (12.4)

N w l—r N (+a) 2CU+c)

Op(N) a y rp(n=1) 2 Tae D rp ee oeas See 2-5)
(12.5)
332 Part II] m Advanced Topics

a = increase in load per year

b = increase in cost of energy per year, not including the effect of inflation

c = demand charge escalation factor.

Where a number of calculations involving different sizes of conductor is required, it is

advantageous to express all the parameters except conductor current, resistance, and length
in two coefficients F; and F>, where

pi OW
eaACEcm
a NeLN N aap. D-N,-N--:
ame (N) rigs
1+i 1+i
ThepresentvalueCL of the costofjoule lossesis thenobtainedfrom
CL=1;-R-L-[F\(N) + FA(N)] (12.7)
Denoting F(N) = F\(N) + F>(N), the total cost of investment and joule losses is obtained
from equation (12.7):

CT = Cl fig-R-L- FW) (12.8)

In this model, the installed cost CJ is assumed to be paid fully at the time when the
cable is energized for the first time. In reality, utilities have to borrow money to pay for cable
purchase and installation, and this money is recovered through the rate structure. Since the
cable installed costs are usually small in comparison with other capital costs (notably the
costs of new generating facilities), utilities often adopt the assumption used in this model.

12.2.2 Mean Conductor Temperature and Resistance

Itis convenient, and usually sufficiently accurate, to assume that conductor resistance is
constant during the life of the cable. A simple formula for making an estimate of conductor
operating temperature, and hence its resistance, is given in IEC 1059 (1991). This is based
on observations of typical calculations that the average operating temperature rise of an
economic size of conductor, taken over economic life, is in the region of one third of the
rise occurring with its maximum permissible thermal rating. Thus, the average conductor
temperature can be taken as

On
=Oamb
Sa— (12.9)
Here, 6 is the maximum rated conductor temperature for the type of cable concerned
and mp 1s the ambient average temperature.
In general, a more precise value of conductor resistance will affect the selection of
an economic size only in very marginal cases. If greater accuracy is desired for particular
cases, refined values for conductor temperature and resistance can be made. The following
derivation gives the formulas for one such estimate.
Conductor temperature is computed as a mean of the values and 0+ during the first
and the last years of an economic period. Assuming that the temperature rise is proportional
to the conductor losses, we can write

TsNak
0 = (Oeamb)(2) sit + Oamb 2a)
are LY
Chapter 12 m Economic Selection of Conductor Cross Section 333

where /, is the current-carrying capacity, for a maximum permitted temperature rise of

0 —Oamp.If the average load factor during the economic life of the cable is different from
unity, the value of [, is computed by either using the Neher/McGrath modification of the
external thermal resistance as discussed in Section (9.6.7) or by multiplying the current
corresponding to 100% load factor by the cyclic rating factor M as discussed in Section
5.0;
Also, we have

acciaithy
6 R,= opuaeeee 0 (12411)
Ro = Rx
B +20 B +420
where £ is the reciprocal of the temperature coefficient of resistance of the conductor
material in degrees Kelvin. For aluminum, 6 = 228; for copper, B= 234.4.
Substituting (12.11) into (12.10), we obtain

6 In\* Gear
(6 owns)(2) (B+ 9% b ( )
—(9 — = hes AD

Denoting

e)ey(ace
! ={—a B+0) oe
WAN
equation (12.12) can be written as

6 = Pam
LE+ (12.14)
a,
Theequationsaresimilarforthefinalyeartemperature:
0¢ = (0 —Oamb) ( IVa
i. ) Ry
R. + Oamb
= =—¢ at aa Oam
I; = 1b
where
ga (pa) ee (12.15)
Thus,
0; = YB8+ Pamd (12.16)
cay
From equations(12.14)and (12.16),we obtainthe meanconductortemperatureas
Os ep eS (= —2y? eB—yg0amv+ ¥8B —¥Pamb+ ve)
Am (2:17)
iene ema (d—y) gy)
Adding f inside the brackets and subtracting it outside the brackets, equation (12.17),
after simplification, can be written as

iy=PP ( eee, )-e (12.18

D l-y l—gy
334 Part III m»Advanced Topics

The mean conductor resistance is obtained as a mean of the first and last year resistances;

that is,

tea sa Af (12.19)
2
Combining equations (12.11) and (12.14), we obtain

R29 | B + | aF am
l—y Ro(B + Oamb) |
Ro = = —— (12.20)
B+ 20 B+ 20 l-y

Similarly,

= R20(B + Gamb) l (12.21)

I B+ 20 l—yeg

Substituting (12.20) and (12.21) into (12.19), we obtain

Ringe cee (+ ) (12.22)

al B +20 |e et

In order to take into account the skin and proximity effects, as well as the losses in sheath
and armor, we multiply the value of R,, obtained in equation (12.22) by a factor B given by

B=(1+ ys + yp)GU+A; + Aa) (i224)

Then, the value of R,, can be substituted directly in equations (12.7) and (12.8).
Similarly, the following equation can be used to obtain a value of p,, which can be
substituted for p21 + @29(A. —20)]:

A re
(20 B+ CambaN” 1 l
ae = aoa ag ae 12.24

where 99 = resistivity of the conductor at 20°C, Q - m.

The economic conductor size is unlikely to be identical to a standard size, and so it is

necessary to provide a continuous relationship between resistance and size. This is done by
assuming a value of resistivity for each conductor material. The values recommended in
IEC 1059 (1991) for p29 are 18.35 - 10~? for copper and 30.3 - 10~° for aluminum. These
values are not the actual values for the materials, but are compromise values chosen so that
conductor resistances can be calculated directly from nominal conductor sizes, rather than
from the actual effective cross-sectional area.

EXAMPLE 12.1
Consider cable model No. | with installation conditions as specified in Appendix A. The ampacity of
this cable with unity load factor is equal to 629 A. The following are the relevant parameters of this
cable taken from Table Al: Rx = 0.0601 - 10-3 Q/m, y, = 0.014, Yp = 0.0047, and A, = 0.09. We
will determine the present value of the cost ofjoule losses for this cable with the following operational
and economic parameters:
Chapter 12 s Economic Selection of Conductor Cross Section 335

Ns Il
N,=3
Io = 160 A
N = 30 years

ibJe = OI

pSOZ

P = 0.05 $/kWh
D = 0.003 $/W-year
To
(3%

@=

(= 3%

i — 00K

First, we will compute the average conductor temperature and resistance during the economic
life. This resistance will depend on the cable rating, including the effect of a nonunity load factor.
With a load factor of 75%, the cable ampacity is increased to 740 A (see computation of the external
thermal resistance in Section 9.6.7). Also, from equations (12.3) and (12.2), we have

p=p-LF+(0—p)-LF? =02-0.75 + (1 —0.2)- 0.75?= 0.6

T = pw-8760= 0.6- 8760= 5256h
The mean value of the temperature is obtained from equations (12.13), (12.15), and (12.18):

nya ieHOwa
el ea 160\* / 90-15 = 0.0108,
ie B+0 740 234.4 + 90

g = (1 +a") = (1 + 0.02)?6°-) = 3.15

BateCarne 1 1 )
he = 2 1-y =F 1l-gy =P

_234.4415
a 2) ( I
1 —0.0108 :
1 —3.15- 0.0108)~234.4
=20.75°C
The mean conductor resistance is computed from equation (12.22), which yields

R, = *R a) Bam(+ I 1 )
2 B+20 l-y Il-gy
_ 0.0601 =3
-107?(234.4
+ 15 I ) aN ice
2 234.4+20) \1—0.0108
©1—3.15-
0.0108
To include the effect of joule losses in the screen wires, we multiply this resistance by the factor
B given by equation (12.23). For this example, we will use the value of the screen loss factor given
in Table Al. In a computer program, the screen and armor loss factors should be computed for the
average temperature obtained above. We have

R,, = 0.0603: (1+ yp +ys)(1 +A; +A2) = 0.0603 - (1 + 0.014 + 0.0047) (1 + 0.09) = 0.0669 2/km
336 Part II] m Advanced Topics

Next, we will compute the auxiliary quantities required in equation (12.6). Since, in this example,
b =, we have from equations (12.4) and (12.5)

ee a 27021
ae fay 1 +0.05
Ege eae
Se ee a OM
QO
p(30)= Qp(30)= 12 7p 1 0!

Factors F are obtained from equation (12.6):

F (30) =
TP <=N,-Ne:
Op(N) 5256:0.0510m galoil = SY
feed i? 1.05
D-N,-Ne-Qp(N) _ 0.003-3- 1-41.21
F,(30) = ray a 1.05 = (0)35

Finally, the cost of joule losses is obtained from equation (12.7) as

CL=12-R-L-[F\(N) + Fy(N)] = 160? - 0.0669 - 10~* - 500 - (30.94 + 0.35) = $26 794

12.2.3, Growing Load Factor. In many systems, the load factor will grow with
time due to many reasons such as increases in load diversity with load growth, increases in
energy consumption per kW of connected load with time, measures taken by the utilities
to flatten the load curves for improving system efficiency and to curb the growth of peak
demand, and so on. The load growth on a particular cable circuit is limited by the circuit
ampacity, and when this value is reached, new transmission facilities have to be installed.
On some occasions, the load growth may stop even before circuit ampacity is reached;
this may be the case, for example, when the cables are installed to transport power from a
generating station whose output will grow in the future until the full station is completed.
These considerations may be taken into account in the above model (Anders et al., 1991).
Adopting an assumption made by Sheer (1966), that the system load factor grows,
reducing the difference between an ultimate load factor L F,, and the present load factor
LF, by one half over a period of 16 years, the load factor at any year n is given by

GF LF, — (0.5)" oe(Liye — Le)

(12.23)
= LF, —(0.9576)"=(LF, —LF,)

Assuming that the annual growth in the load factor is achieved during the last day of the
year, the value of 7 becomes

T, = 8760 - {p[LF, —(0.9576)""! - ALF]

+(1 —p)[LF, —(0.9576)""! - ALF}?}
II c —0.9576""' -d +0.9170"! -e (12.26)

where
c = 8760-[pLF, + (1 —p)LF?]
d = 8760-[pALF + (1 —p)2LF, - ALF] (12.27)
e = 8760-(1— p)- ALF?
Chapter 12 # Economic Selection of Conductor Cross Section 337

with
ALF LE, aLEs.
The factor F;(N) then becomes

P-Np- Ne: Qp(N)

EXCN)
i(N) = pati (12.28)

where
QO’n(N)= cOp(N) —dQp\(N) + eQp2(N) (12.29)

and
Sir
Op,(N) = ———., Kawlee, (12.30)
b= Tp,
Factors rp, are given by
0.9576(1 + a)*(1 +b)
Ut i Sa
(12.31)
0.9170(1 + a)*(1 +b)
BO a a
1+1
EXAMPLE 12.2
Let us assume that the cable system studied in Example 12.1 experiences a growing load factor with
an initial value of 75% and a final value of 95%. We will determine the cost of joule losses in this
case.
The increase in the load factor is equal to

ISAT =) 0 SUP

The constants c, d, and e are computed from equation (12.27):

c = 8760: [pLF, + (1 — p)LF?] = 8760(0.2 - 0.95 + (1 — 0.2)0.95*] = 7989.12

d= 8760: [pALF + (i — p)2LF, - ALF] = 8760[0.2 - 0.2 + —0.2)2 - 0.95 - 0.2] = 3013.44

e = 8760- (1 — p)- ALF? = 8760- (1 — 0.2)0.2? = 280.32

From equations (12.31), (12.30), and (12.29), we have

0.9576(1 + a)?(1 +b) 1 0.25760 0.02)?(1 + 0.03)

f = gees 1+0.05 = 0.9773

0.917011 +.4)?(1+b) 0.9170(1 + 0.02)7(1 + 0.03)

rp. = = = 0.9359
L+i 1+ 0:05
Ome ep = pu 0 Oo.
La? = 21,93
aye = 0.9773
fr e0,0359"
NY lea
a Tio a tel 0/9359 St ierG

Q', (30) = cOp(N) —dQp\(N) + eQp2(N)

= 7989.12 - 41.21 — 3013.44 - 21.93 + 280.32 - 13.46 = 266 920.
338 Part III m Advanced Topics

The new factor F; is computed from equation (12.28):

P-N,-Nce-Qp(N) _ 0.05-10~* -3-1- 266920
F,G0) = = 38.13
1+i a 1.05

The cost ofjoule losses in this case is obtained from equation 1257):
CL=I2-R-L-[F\(N) + F(N)] = 160? - 0.0699 - 10~* - 500(38.13 + 0.35) = $34 429

This value is 28% higher than the cost computed in Example 12.1.

If we assume that the load growth on the circuit of interest stops after No years (No <
N), then the growth of the L F on this circuit is limited to No years; the increase in L F due
to other factors beyond No years is ignored. In this case,

ee a a ot I Sty

where
O>(N)= Q'p(No)
+ fLQD(N)
—QO(No)] BER
Q7,(N)
= Qp(No)
+ (1+a)? PLO(N)— O5,(No)]
with
f =8760(1+4)? ny, (12.34)
Hoya a, No ya ttt
extn ie (12.35)
OES cymes: Ser bag
where jy, 1S the ultimate load loss factor.

EXAMPLE 12.3
Suppose that in Example 12.2 the load factor growth is achieved in a period of ten years. We will
compute the cost of joule losses in this case.
If the load stops growing after ten years, the final load factor is obtained from equation (12.25):

LF = LF, — (0.9576)" - (LF, — LF,) = 0.95 —0.9576"° - 0.2 = 0.82

and the ultimate load loss factor is computed from equation (12.3):

Pio = p> LF +(1 — p)- LF? =0.2-0.82 + (1 —0.2) - 0.82? = 0.70

The auxiliary quantities are obtained from equations (12.35) and (12.34):
l+b 1+0.03
,=r,=
POET a | OS = 0.98,

f = 8760(1 + a)? wy, = 8760(1 + 0.02)2°- . 0.70 = 8758.0

L—r,° 1 —0.98"
>= OCU) = — = 9.15;
mid ae Bek= 1 —0.98

05630).
= 07,30)
= os er
Chapter 12 m Economic Selection of Conductor Cross Section 339

Also, from equations (12.4), (12.30), and (12.29), we have

1—r | dts Lozi

Qp(10) = Qp(10) = Sle)
Nere «6h
— 1.021

CGD 10) =
1—r},
Gaeta
mit —0.9773'°
EM an gyyaN
= —_____
— = 9 (04

1—r%, 1-—0.9359!0
o(10)r
QPp2(10) Dre Po 1 20.9359 = 7.56

Q',(10) = cOp(N) — dQp,(N) + eOpo(N) = 7989.12 - 11.0 — 3013.44 - 9.04 + 280.32 - 7.56

= 62758.

The new values of constants F, and F> are obtained from equations (12.33) and (12.32):

Q>(N) = Q(No) + fLOS(N) — Q',(No)]= 62 758 + 8758 - (22.73 —9.15) = 181692

Q7(N) = Qn(No) + (1 +4) O7,(N) — Q,(No)]= 11 + (1 + 0.02)? (22.73 —9.15)

= 30.4
Pi NoOIN CMOS
(NY©2 005910 43'-181811692
BE0)
se) a 1+i ee 1.05 ee DG
260
DUNSNetoPtNy 17000373"1304.
iG).
230) ee iheice 1.05 = 0.26.
Assuming that the average resistance is the same as computed in Example 12.1, the present value of
the cost of joule losses is equal to

CL=1,-R-L-[F\(N) + Fx(N)] = 160° - 0.0669 - 10-7 - 500(26.0 + 0.26) = $22 487.

This cost is substantially lower than the value obtained in Example 12.2.

EXAMPLE 12.4
We will revisit Example 12.3, but we will change the assumption on the load growth. Let us assume
that the load factor is constant and equal to 0.75 during the entire 30 year period, but that the load
stops growing after the first ten years and then remains constant for the remaining 20 years. We will
compute the present value of the cost of losses.
From Example 12.3, we have

0',(10) = Q%,(10)= 9.15, 0’, (30) = Q%,(30)= 22.73

Factor f will now be equal to

f = 87600 +4)? wy,= 8760(1 + 0.02)7"°” -0.6 = 7506.9
Since, in this example, AL F = 0, we have d = 0 and e = 0. From Example 12.3, Qp(10) = 11,
and from equation (12.29), we have

Q',(10) = COp(N) —dQp,(N) + eQp2(N) = 8760 -0.6 - 11 = 57816

From equation (12.33), we obtain

Q*,(30) = O',(No) + fLOR(N) — O(No)] = 57 816 + 7506.9 - (22.73 —9.15) = 159760
Q*,(30) = Op(No) + (1 +.) [QR(N) — QF(No)] = 11+ (1 + 0.02)? (22.73 —9.15)

= 30.4
340 Part III m Advanced Topics

From equation (12.32), we have

P.-Np~Ne+
Qp(N)
2 | 0.05:
10-3
+ 3
1-159760
are ab1. _5,
Hee l+i 1.05
030)=DN:Ne:Qb(N)
_0.003-3-1-304
_94.
As Li Pa 1.05 =
Assuming again that the average resistance is the same as computed in Example 12.1, the present
value of the cost of joule losses is equal to

CL=I2-R-L-[F,(N) + Fx(N)] = 160° - 0.0669 - 10~? - 500(22.82 + 0.26) = $19 764

12.3 EFFECT OF CHARGING CURRENT AND DIELECTRIC LOSSES

Dielectric losses and the losses due to charging current are always present when the cable
is energized, and therefore operate at 100% load factor. Both types of losses are significant
only at high-voltage levels and are dependent on cable capacitance. Evaluation of transmis-
sion cable systems often assumes the placement of shunt reactors at the ends of the cable
system to supply the reactive vars required by the cable. The reactors have losses equal to
about 0.8% of power rating. Those losses should be considered in the evaluation of cable
system losses, and the cost of the reactors should be added to the cable purchase cost.
The dielectric and charging current losses are sometimes referred to as voltage-dependent
losses, in contrast to the joule losses which are referred to as current-dependent losses.
Cable capacitance C is given by equation (6.2):

i aa lee ga (12.36)
18
In(3)
Dj
de
where ¢ is the relative permitivity of insulation, d. (mm) is the diameter of the conductor
including the screen, and D; (mm) is the diameter over insulation.
Charging current is calculated from

les 217 GUp C1237)

where f is the system frequency and Up (V) is phase-to-ground voltage.

Charging current is not uniform along the cable. In a cable with all charging current
flowing from one end, the charging current losses are computed from (IEEE, 1990)

Weaieets
[oiLae € (12.38)
If the system has equal charging current flowing from each end, either due to natural
system conditions or to the addition of reactors to force the equal flow, the losses per phase

Wa — 2 3 I ec eed
) ; R d Dbs39 )
Chapter 12 m Economic Selection of Conductor Cross Section 341

Thus, in general, the charging current will be obtained from

Wenge
ntl otha eR (12.40)

where g = 1/3 or 1/12, depending on whether equation (12.38) or (12.39) applies.

The dielectric losses are proportional to the square of the voltage [equation (6.4)]:

Wa=2n7-L:C-U- -tané (12.41)

where tan 6 is the loss factor of the insulation.
The total cost, including the effect of charging current and dielectric losses, can be
represented by extending equation (12.8) to

CT =CI+15-R-L- FIN) +(g-17-R-L? + Wa): F3(N) (12.42)

with
Np: N. (8760: P- O(N) + D- O,(N)]
F3(N) = (12.43)
iad
where Q',(N) and Q/,(N) are definedin equation (12.26).

EXAMPLE 12.5
We will consider a 5.8 km long cable (model No. 3) with laying conditions as described in Appendix
A. The parameters ofthis cable are: Roy = 0.1817: 10~* Qhm, y, = 0.05, yp = 0.054, 4, = 0.01, and
Az = 0.311. The load is assumed to grow from 250 to 795 A during the 40 year economic life. The
charging current is assumed to flow from one end only (the worst case scenario). We will compute
the cost of current- and voltage-dependent losses for the following nominal conditions:

N=
Nea
Lk ODDIN
N = 40 years

JEIE3 O95)
B= 03

P = 0.05 $/kWh
D = 1.0 $/W-year

C= 36
bese
C= Use

Dio

L = 5800 m.

Since we assume that the load grows steadily over the 40 year economic life period, the load
growth factor is obtained by solving the following equation:

795 = 950° (1 a)”

which yields a = 3.0%.

342 Part III = Advanced Topics

First, we will compute the average conductor temperature and resistance during the economic
life. This resistance will depend on the cable rating, including the effect of a nonunity load factor.
With a load factor of 85%, the cable ampacity is equal to 902 A. Also, from equations (12.3) and
(12.2), we have

w= p-LF +(1— p)- LF? =0.3-0.85 + (1 —0.3) - 0.85?= 0.761

T = 8760: « = 8760- 0.761= 6664h
The mean value of the temperature is obtained from equations (12.13), (12.15), and (12.18):

y= lo 2 @fs
=amb= 250 2(ser —2
40)= 0.0114,
1. B+O 902) \234.4+70
g=(1+ay- = (1 +.0.03)?-» = 10.03
B + Oamb
( 1 1 )
An— 2 =1.29 etFe tere ae P

234.44 25 1 r 1 ) —234.4 = 43.2°C

‘ 2 1—0.0114 1 —10.03-0.0114

The mean conductor resistance is computed from equation (12.22), which yields

Rites Rr B algOamb 1 i |
2) B+ 20 l-y 1l-gy

1835 10°? (234.4 + 25 1 _ 1

~ 2-0.00101\ 234.4+20 1—0.0114 1—10.03-0.0114 )=0.0198
Q/km
To include the effect of joule losses in the screen wires, we multiply this resistance by the factor B
given by equation (12.23). The values of y,, y,, Ay, and A are given in Table Al. Thus, we have

Ry = 0.0198-(1+y,+yp)(1+A)+A2) = 0.0198-(1+0.05+0.054)(1+0.01+0.311) = 0.0289 Q/km

Next, we will compute the auxiliary quantities required in equation (12.6). Since, in this example,
b =, we have from equations (12.4) and (12.5)

_—(+ay(1+b) _ (1+0.03)?(1 + 0.078)

in? = {eu = fe Oot ==wROS
q

ih eM O32
Qp(40)
40) == Qp(40)
40) = [aps r= [Sie = 75.4

Factors F are obtained from equation (12.6):

Pt N en eee (N) 66642 01051 0nses lees

F, (40) = ! as = 67.9
I+i ie!
D-Ny:Ne-Qn(N) _ 1.0-3-1-75.4
F,(40) = = 203.8
1+i te 1.11

The cost of joule losses is obtained from equation (12.7) as

CL =1)-R-L-[F\(N) + F(N)] = 250? - 0.0289 - 10-3 - 5800 - (67.9 + 203.8) = $2 846 397.
Chapter 12 m Economic Selection of Conductor Cross Section 343

The dielectric capacitance of the insulation is computed from equation (12.36) as

i € LO et ea 3:5) oe O04 Oem

18In(— Sin
d. 41.45
The charging current and charging current losses are obtained from equations (12.37) and (12.38),
respectively:

I, = 2 f CUp = 2m - 60 - 0.404 - 10-° - 138 000//3 = 0.012 A

Wen = 512-13. R = + 0.012? - 5800° - 0.0298 - 10-? = 271 W

The dielectric losses are given by equation (12.41):

W, = 2nf -L-C-U, - tand = 27 - 60 - 5800 - 0.404 - 10-° - (138. 000/73)? - 0.005 = 28 038 W.

From equation (12.35), we have

1+b
regs 1+0.078
Saida Sa,67
ie
EES:
dee melee0.0770
(40)
sme = sitaan
07,(40)
= lr), = ——__
1—0.97 =73.48
Factor F3(N) is obtained from equation (12.43):

N, +Ne - [8760- P- O%(N) + D- ON(N)]

F (40)
1%
3 - 1 - (8760 - 0.05 - 1073 23.48 + 1-23.48) _
91.3
eal a

Finally, the cost of charging current and dielectric losses is computed from equation (12.42) and is
equal to

CLensa = (g 12 RL? + Wa) F(N) = (271 + 28038) - 91.3 = $2584612

In this example, the cost of voltage-dependent losses is comparable to the cost of joule losses, mainly
because of the very high dielectric losses.

12.4 SELECTION OF THE ECONOMIC CONDUCTOR SIZE

The economic conductor size S,. is determined in two stages. First, we neglect voltage-
dependent losses and find the cross section that minimizes the cost function:

[er = CI(S)+ 12- R(S): F(N) +L (12.44)

where C/(S) and R(S) are expressed as functions of the conductor cross section S.
The equation for the relationship between C/(S) and conductor size can be derived
from the known costs of standard cable sizes. In general, if a reasonably linear relationship
can be fitted to the costs, possibly over a restricted range of conductor sizes, it should be
344 Part II] m Advanced Topics

used. This will cause little error in the results in view of the possible uncertainties in the
assumed financial parameters for the economic life period chosen. If a linear model can be
fitted to the values of initial cost for the type of cable and installation under consideration,
then

CICS) SLA} 5.4G) (12.45)

where A = variable component of cost related to conductor size, $/mm?/m
G = constant component of cost unaffected by size of cable, $/m.

Then, the optimum size S,. (mm7) can be obtained by equating to zero the derivative
of CT with respect to S, giving for the case when voltage-dependent losses are neglected

Ij - FCN) - p20 - BU + @20(8ay —20)]

Se 000 i ( 12.46 )

B is an auxiliary value which takes into account skin and proximity effects as well
as sheath and armor losses. As the economic size is unknown, it is necessary to make
assumptions as to the probable cable size in order that reasonable values of yp, ys, A1,
and A» can be calculated. Recalculation may be necessary if the economic size is too
different.
We can observe that S,. does not depend on the value of the constant component G of
the cost which is unaffected by the size of the cable. Therefore, in performing a comparative
analysis for the selection of an optimal conductor size for a specific installation, we can
compare only the size-dependent component CT— C/ of the total cost. This approach is
used in the numerical example discussed below. However, in the computer program, the
user should be able to enter the value of G so that the true total cost can be computed.
Sec is unlikely to be exactly equal to a standard size, and so the cost for the adjacent
larger and smaller standard sizes must be calculated and the most economical one chosen.

EXAMPLE 12.6
We will compute the optimal conductor size for the cable system studied in Examples 12.1—12.3.
We will assume that the coefficient of that part of the installation cost which depends on conductor
size is equal to 0.1133 $/m/mm?. We will also assume that the installation cost is independent of the
conductor size. This is normally the case when we consider the narrow range of cable sizes.
First, we have to assume an initial value of the conductor cross section. For all the cases
studied, we will start with § = 300 mm?. Table 12.1 summarizes cable information for three standard
conductor sizes which will be considered in the example.

TABLE 12.1 Cable Data

a a
ConductorSize Ampacity Resistanceat 90°C
(mm?) (A) (Q/km) Vs Yp A,

300 629 0.0781 0.014 0.005 0.090

400 862 0.0599 0.023 0.006 0.098
500 971 0.467 0.037 0.011 0.119
eee
Chapter 12 m Economic Selection of Conductor Cross Section 345

(1) Constant load factor

From equation (12.24), the mean electrical resistivity is equal to

pn=Sn(FeSeealec
ZeXepr20 l—y~1-—gy
SRELOS: ( =)( 1 in 1 )=18.4-10°?Q-m
2 234.4
We will
+20 1—0.0108
take
1—3.15-0.0108
factor B as given in Table 12.1 with the loss factor values corresponding to 90°C.
A small error introduced by this assumption will have no effect on the selection of the economic
conductor size. To be consistent, in the remainder of this example, the values of B will be computed
for the conductor temperature of 90°C.

B=(1+y,; + yp) + A; + Az) = 2 +.0.014 + 0.005) (1 + 0.090) = 1.11.

The optimal conductor cross section is obtained from equation (12.46), with the factor F obtained
in Example 12.1, and is equal to

Th a
I? - F(N)- - BUI i
hy = AD

= 1000,/160%. (30.94
602- (30.94+ 0.35).
184-10
+0.35)- 18.4-10°« ill = 380.0 mm?
0.0133

The closest standard conductor sizes are 300 and 400 mm*. The factor B for the 400 mm?
conductor is recomputed using equation (8.62) for the middle cable and applying equation (12.23):

B=(1+y,+yp)(1 +A; +2) = (1 + 0.023 + 0.006)(1 + 0.098) = 1.13

The auxiliary variables in equations (12.13) and (12.15) are equal to

In\? (0 —Gam 160\* ( 90-15

y= a Say Weepeta ——_—— ] = 0.008,
if pre? 862 234.4 + 90
g = (1+4)-) = (1 +.0.02)7O°-) = 3.15

From equation (12.24), the mean electrical resistivity is equal to

p20( B| ( 1 1 )
a= a5
2 B+ 20 l-y l-gy
18.35-10-? (234.4+ 15 | re 1
D, 234.4+ 20 1—0.008 1-—3.15-0.008 )=18.3-10°
Q.m
The revised optimal conductor cross section is obtained again from equation (12.46):

PURtr pac
s, =100]! CNaad BILoa0ne. 20)

Ny
1000-1aees ae Pe LORE
; On oe an?
ral, 5
This is also within the 300-400 mm” range.
346 Part III m Advanced Topics

The total cost of cable and joule losses for each of the possible conductor sizes is now calculated
with the aid of equations (12.44) and (12.45). For the 300 mm?’ cable, the joule losses were calculated
in Example 12.1. Thus,

CT300 = A+ L + S + CT390joute= 0.1133 - 500 - 300 + 26794 = $43 789

CTs = CI(S) + Ig - R(S)- F(N)-L

= 0.1133 - 500 - 400 + 160? - 18.3 - 10~° - (30.94 + 0.35) - 500/0.0004 = $40 983

Therefore, the 400 mm? conductor is the more economical size.

(2) Growing load factor
The factor F computed in Example 12.2 is equal in this case to 38.48. Hence, the optimal
conductor size is

nO To
eaO RUNY
to Bion ne
ea yee ae 20y)
160 - 38.48 - 18.4- 107° - 1.11
= 1000- = 421.4 mm
0.1133

The two neigboring standard conductor sizes are 400 and 500 mm?. For a conductor size of 500
mm, the revised value of factor B is 1.17 and the mean resistivity is equal to 18.2 - 10-°Q-m. This
increases S,. to 430.6 mm?. The present values of the cost of cable and joule losses of the two cables
are equal to

CTy9 = C1(S) + If - R(S)- F(N)-L

= 0.1133 - 500 - 400 + 1607 - 18.3 - 10-° - 500 - 38.48/0.0004 = $45 194

CTs9 = CI(S) + 1G - R(S)- F(N)-L

= 0.1133 - 500 - 500 + 160° - 18.2 - 10-° - 500 - 38.48/0.0005 = $46 254

Since the installation costs were assumed to be the same for all standard conductor sizes, we
select the optimal conductor size equal to 400 mm’.
(3) Load growth stops after ten years
In this case, F = 26.26 and

ie F(N) 20a Bil + 29 (Bayis 20)]

S000
A
= 1 po 1602
cae- 26.26- 18.4. 10-9- 1.11 = 348. 2
0.1133 48.1 mm

The closest standard conductor sizes are 300 and 400 mm? with the total costs equal to

CT399 = C1(S) + If - R(S)- F(N)-L

= 0.1133 - 500 - 300 + 160? - 18.4. 10-° - 500- 26.26/0.0003 = $37 611
CTso9 = C1(S) + If - R(S)- F(N)-L

= 0.1133 - 500 - 400 + 160? - 18.3 - 10-° - 500 - 26.26/0.0004 = $38 038

Thus, in this case, the optimal conductor size is 300 mm*?.

Chapter 12 m Economic Selection of Conductor Cross Section 347

EXAMPLE 12.7!
A 10 kV cable circuit has to be sized to supply ten 10 kV/0.4 kV substations equally spaced along
a route from a 150 kV/10 kV station (see Fig. 12.1). Our aim is to select a conductor size for each
section based on the following considerations: 1) different economic conductor size for each section,
2) conductor size based on thermal current-carrying capability, and 3) the same most economical
conductor size for all sections.

Main Station

Cable 1 Cable 2 Cable 9 Cable 10

(160 A) (144 A)

(16 A)

Substation Substation Substation Substation Substation

1 2 8 9 10

Figure 12.1 Distribution feeder supplying ten stations. (IEC Standard 1059, 1991)

There is only one three-phase circuit, so N. = 1 and N, = 3. The cable length between
substations is 500 m.
The highest hourly mean value of current / in the first year in the first section of the route is
160 A. In subsequent sections, the current is reduced by 16 A at each substation; thus, in the second
section, the current is equal to 144 A and in the final section 16 A. The cyclic rating factor for all
loads is 1.11. It is assumed that this factor remains constant during the economic life of the cable.
The financial data are as follows:

N = 30

P = 0.0609 $/kWh
D = 0.003 $/W-year
A = 0.1133 $/m-mm?
a=0.5%

b=2.0%

C= 20%

70

For the purpose of this example, a fictional three-core 6/10 kV type of cable has been assumed.
The ac resistances of the conductors at 40 and 80°C are given in columns (2) and (3) of Table 12.2, and
the financial details are given in columns (4) and (5). The cable has a permissible maximum conductor
temperature of 80°C, and when laid in the ground, the steady-state ratings at this temperature, for a
20°C ground ambient temperature, are given in Table 12.3.

! This example is extracted from IEC Standard 1059 (1991).

348 Part III m Advanced Topics

TABLE12.2 CableDetails
ee ee eee
Resistance
byPhase Primary
Cost
Cable
Size 40°C 80°C
(mm?) (Q/km) (Q/km) Cable
($/m) Laying
($/m)—Sum($/m)
(1) (2) (3) (4) (5) (6)

25 1.298 1.491 10.62 17.23 27.85

35 0.939 1.079 11.65 17.33 28.98
50 0.694 0.798 13.19 17.49 30.68

70 0.481 0.553 15.24 17.71 32.95

95 0.348 0.400 17.81 17.97 35.78
120 0.277 0.318 20.37 18.24 38.61
150 0.226 0.259 23.45 18.55 42.00
185 0.181 0.208 27.04 18.92 45.96
240 0.140 0.161 32.69 19.51 52.20

300 0.114 0.131 38.85 20.14 58.99

400 0.091 0.104 49.11 21.20 70.31

TABLE 12.3 Cable Ratings

Nominal Size
(mm?) 25 35 50 70 95 120 150 185 240 300 400

Current- 103 125 147 181 PUD 255 281 328 382 429 482
Carrying
Capacity(A)

(1) Economic conductor size for each section

Since, in this example, b = c, we have from equations (12.4) and (12.5)

> 9d+e70 Fb) (140.057.4002)

DP Lippy ae = 1+0.05 —(UR
Xe5H
N7)
l —rp _ 1—0.9812%
0p(30) = Qp(30) = i = 2713}Oeil
=p 1—0:98i2
Factors F are obtained from equation (12.6):

T-P-N,-Nce:Qp(N) _ 2250- 0.0609 - 10-3 -3- 1 - 23.081

F\ (30) = = 90363
1+i ms 1.05

D-Ny-Ne:Qv(N) _ 0.003 -3-1- 23.081

F,(30) = = 0.1978
1+i a 1.05

Assuming initially that aconductor size of 185 mm? could be the economic optimum, the factor
B is equal to 1.023. The average conductor temperature is computed from equation (12.9) as

6 ae Com 80 = 20
On = amb aig —— = 20+ 3 = 40°C
Chapter 12 = Economic Selection of Conductor Cross Section 349

From equation (12.46), we have

s, =100]‘Kg
0°FIN):
p20
Ee ae etgty20)]
1000 - [1602 - 9.2341 - 30.3 - 10-°01133
- 1.023[1Ua + 0.0039(40 — 20 J = 264 mm?”
II

Thus, either a 240 mm? or a 300 mm? conductor size could be chosen.
The initial choice of a 185 mm? conductor for the estimation of B can now be improved.
Recalculating with the value of B= 1.057 for a 300 mm2 conductor gives a value of S,. of 269 mm”,
which is also within the 240-300 mm? range.
The total cost for each of the possible conductor sizes is now calculated with the aid of equation
(12.8):

CTn49 = CI + I? -R-L- F(N) = [52.2 - 500] + [1607 - (0.140/1000) - 500 - 9.2341 = $42 648

CT399 = CI + Te -R-L- F(N) = [58.99 - 500] + [1607 - (0.114/1000) - 500 - 9.2341 = $42 969

The 240 mm? conductor is therefore the more economical size.

Sizes and costs for the other sections have been calculated in a similar manner. The values of
optimal sizes and corresponding costs are given in Table 12.4.

TABLE 12.4 Economic Loading

Section Number 1 2 3 4 5 6 I 8 9 10

Load
Ip (A) 160 144 128 112 96 80 64 48 BY) 16

Cable
Size (mm?) 240 240 185 185 150 120 95 710 50 25
Capacity (A) 382 382 328 328 281 255 221 181 147 103

Cost per Section

and total
Cable ($) GIS45 ee GIS 4 5 3152.0 Si 52 OMe 2 SeeON
Ss) 8905 7620 6595 5310 110 070
Laying ($) S755 9755 9460 9460 9275 9120 8985 8855 8745 8615 92 025
CI ($) 26100 26100 22980 22980 21000 19305 17890 16475 15340 13925 202 095
CL ($) 16548 13403 13692 10483 9616 8185 6581 Sly, 3281 1534 88 440

CT ($) 42648 39503 36672 33463 30616 27490 24471 21592 18621 15 459 290 535

(2) Conductor size based on maximum load—choice based on thermal ratings

We will now select a cable size for each section so it can carry the anticipated maximum load for
the last year of the economic life and not exceed the maximum permissible conductor temperature.
For section 1:

Io (first year) = 160 A

Maximum current in last year = 160 - (1 + 0.005)*°"! = 185A

The required current-carrying capacity (100% load factor) / for the final year shall be not less
than
LSS/ ee eG eA
350 Part II] » Advanced Topics

where the number 1.11 is the cyclic rating factor assumed above. From Table 12.3, the required
conductor size is 70 mm2. In order to make a fair comparison with the losses and financial figures
calculated for the economic choice of conductor size, we have to assume an appropriate conductor
temperature at which to calculate the losses. For the economic choice, we assumed that the temperature
of the conductor would be about 40°C. We propose that a comparable assumption for the temperature
of conductors chosen on the basis of thermal ratings would be the maximum permissible value of
80°C. The conductor resistances at a temperature of 80°C are given in Table 12.2.
The total cost of section | during the 30 year period is obtained from equation (12.44):

CT =CI(S) +15 - R(S)- F(N)- L = (32.95 - 500) + (160° - 0.000553 - 500 - 9.2336) = $81 834.

Comparison with the cost for this section when using the economic size of conductor shows that the
saving in cost for this section is (81 834 — 42 648) - 100/82 834 = 48%. Similar calculations using
sizes based on maximum thermal current-carrying capacity have been made for all the sections and are
given in Table 12.5. The total saving for the ten sections is (547 864 —290 535) - 100/547 864 = 47%.

TABLE 12.5 Current-Carrying Capacity Criteria

Section Number | 2 3 4 5 6 i 8 9 10 Sum

Load
1) (A) 160 144 128 112 96 80 64 48 32 16
Tena (A) 185 166 148 129 111 92 74 aS ahi) 18
Teng/ 11(A) 167 150 133 117 100 83 67 50 53 17

Cable
Size (mm?) 70 70 50 35 25 25 25 25 25 25
Capacity (A) 181 181] 147 125 103 103 103 103 103 103

Cost per Section

and total
Cable ($) 7620 7620 6595 5852 5310 5310 5310 5310 5310 5310 59:52
Laying ($) 8855 8855 8745 8665 8615 8615 8615 8615 8615 8615 86 81
CI($) 16475 16475 15340°- 14490: 13925 13925 13925 13925) _ 13925 13925 146 33
CL ($) 65363 52944 60365 62492 63443 44058 28197 15861 7049 1762 401 53

CT ($) 81834 69419 75705 76982 77368 57983 42122 29786 20974 15687 547 86
LL

(3) Calculations based on the use of one standard conductor size for all sections
It is first necessary to assume a probable conductor size, and the total cost is calculated with
equation (12.44) using this size for all sections. Then, costs assuming the use of the next smaller and
larger size of conductor are calculated in order to confirm that the assumed size is indeed the most
economical. For the purpose of this example, we assume that a 185 mm? conductor would be the best
choice.
Although only one conductor size is used, the current is different for each cable section so that
the average losses must be computed (all sections are assumed to operate at the same temperature and
hence the same conductor resistance).

averagelosses S00- 160°+ 500- 1447+ --- + 500- 167

0.385
maximumlosses 10-500- 1602 ct

From equation (12.46), using B for a 185 mm? conductor, we obtain

Chapter 12 m Economic Selection of Conductor Cross Section 351

S,. = 1000 [15 - FCN) - pr « = + 020(Gay— 20)]

= 1000. fo - 1.023 - 30.3 - 10-9 - 9.2341 - 0.385 - [1 + 0.00403(40 — 20)]

= 164 mm?
0.1133
so that either 150 or 185 mm? conductors could prove to be the most economic.
The total costs for each of these conductors are

CT\s0 = 42.00 - 500 - 10 + 160? - (0.226/1000) - 500 - 10 - 9.2341 - 0.385 = $312 843
CTig5 = 45.96 - 500 - 10 + 160? - (0.181/1000) - 500 - 10 - 9.2341 - 0.385 = $312 165

Thus, the 185 mm? conductor is confirmed as the most economic size to use if only one conductor
size is to be used throughout the route.
It is clear, by comparison with the sizes chosen in Table 12.4, that a 185 mm? conductor is
thermally adequate to carry the maximum load at the end of the 30 year period.
Summary of results
The summary of the results of the calculations for the cable and conditions studied in this
example is given in Table 12.6.

TABLE 12.6 Summary of Costs

Cl Gi Total

Basis of Costing $ $ $ %

Thermal current-carrying capacity 146 330 401 534 547 864 100
for each section
Economic size for each section 202 095 88 440 290 535 53
Economic size using one standard
size of 185 mm? throughout 229 800 82 365 312 165 S7

Equation (12.44) does not reflect the effect of voltage-dependent losses. The capaci-
tance of the cable depends on the ratio of insulation thickness ¢; and conductor diameter d..
For a given voltage level, an increase of conductor diameter results in an increase of cable
capacitance (the insulation thickness usually remains unchanged or may even decrease),
and hence an increase in voltage-dependent losses. Because of this, the optimization proce-
dure will tend to decrease conductor diameter as opposed to the effect of current-dependent
losses. Since the dielectric losses in high-voltage cables can sometimes be higher than the
conductor losses, this effect may be quite significant. The losses caused by charging current
I, are proportional to 17, where R is the same as that used to calculate the losses due to the
load current, and this will tend to increase conductor cross section. Since, on the other hand,
I, depends on cable capacitance, the increase in the conductor size will tend to increase
the value of the charging current, and this will increase the charging current losses, forcing
a lower optimal conductor cross section. The influence of dielectric and charging current
losses is assessed through an iterative procedure in the second stage of the optimization
process.
352 Part III m Advanced Topics

aES
EXAMPLE 12.8
We will reconsider Example 12.5, and we will include the effect of dielectric and charging current
losses. The cost coefficient A is equal to 0.22 $/m/mm’.
The smallest standard conductor size which satisfies ampacity requirements for this system is
887 mm? (1750 kcmil). The economic conductor size is first computed from equation (12.46) with
the required parameters computed in Example 12.5:

Pm(20
(f“tee
) (
= 2 B+20 SSF l 1 )
l-y 1l-gy
Y ~9
Pt18.35
-10(aa =)(
234.4+ 20 1 1
1—0.0114as 1—10.03-0.0114)250.0.
10°O m
Ip - F(N) - p29 +BEL + @29(Oa,— 20)]
x 1000
i A

— 1000. 2502-
2, (67.9 + cane 20.0 -- 1010-2
- 1.46
1.4: ier. :

This is very close to the standard conductor size of 1520 mm? (3000 kemil). The ampacity of
a 1520 mm” cable with 85% load factor is equal to 943 A. We will compute the mean resistance and
the factor B for this cable. From equations (12.13), (12.22), and (12.24), we have

RY? B.
ef (° Cunby 2
a (67ers a ) = 0.0104
< II
L. B+0 943) \ 234.4470
Ry= R
5 (Foe ee fel
2 B+20 1—y Il-—gy
we1S.35-10.( Ss) ( 1 1 )=0.0129
~ 2.0.00152 \234.4+20/) \1—0.01041 —10.03-0.0104 Q/km
(20(ee
Pm= — | ———] (|——+ 1 I
2 B+20 l-y I-gy
_ 18.35-10-° (234.4+ 25 1 Ps | RE Oneee
= 2) Dads elon 0) eee10,05n0.01
04 jeeiancme
« in
Factor B is equal to 1.78. This gives S,. = 1641 mm*. This is very close to a standard conductor
size of 1647 mm? (3250 kemil). Neglecting the effect of charging current and dielectric losses, we
would select the 3250 kemil conductor as the most economical one.
To account for the effect of charging current and dielectric losses, we should compute the values
of these losses for several conductor sizes in the neighborhood of the selected optimal cross section.
We will start with the 1520 mm? conductor cable.
The dielectric capacitance of the insulation is equal to

rT
Cok ELe -10-”=——7,
ore She) : 10—-9
__ -—9
= 0.499-10°?Fim
18In| — isin <=
ae 52.22
Charging current and dielectric losses are obtained from equations (12.38) and (12.41), respectively:

I, = 2nf CU = 27760 - 0.499 - 10-° - 138 000/./3 = 0.015 A

Wen= 312 -L?- R = 3 - 0.015? - 58008 - (0.0129 - 10-3 - 1.78) = 336 W

Wa = 2nf -L-C-U?-tand = 0.015 - 5800 - 0.005 - 138 000//3 = 34658 W

Chapter 12 m Economic Selection of Conductor Cross Section 353

The costs of cable and losses are computed from equations (12.45) and (12.42):

Cl=AVES = 0.22 5800- 1520 ='$1 939 520

Cine 1g ROS) LF (N)- = 250? 20.023 10 -5800-271.7 = $2265 299

CLenta = (336 + 34658) - 91.3 = $3 194.952

Thus, the total cost is equal to

CT = 1939520 + 2265 299 + 3 194952 = $7399771

We observe that this cost is greater than the total cost of the cable with a 1010 mm? conductor.
Indeed, the cost of losses for this cable was computed in Example 12.5; thus, the total cost for the
1010 mm* cable is equal to

CTio10 = 0.22 - 5800 - 1010 + 2 846 397 + 2584612 = $6719 769

Hence, in spite of the fact that the cost of joule losses is higher for the smaller conductor, the
cost of dielectric losses forces the selection of a smaller conductor size. Figure 12.2 shows the total
cost for standard conductor sizes. In this case, the computations start at the 887 mm? (1750 kemil)
conductor size since this is the smallest cross section to satisfy ampacity requirements.

x10

($)
Cost
7 as eel
6

887 1010 1140 1267 1393 1520 1647

Figure
fd 12.2 Total cost as a function of conductor Conductor Cross-Section ; (mm?)
cross section.

Thus, in this case, the optimal conductor size is 1010 mm’.

The voltage-dependent losses can be neglected for distribution voltages, and the limit-
ing voltage levels for each insulation type can be those given in IEC Publication 287 (1982)
and listed in Table 6.2.

12.5 PARAMETERS AFFECTING ECONOMIC SELECTION OF CABLE SIZES

Sensitivity analysis forms an important part of engineering studies. In the actual cable
system design process, several parameters are either unknown or can be predicted with
limited accuracy. Some of these parameters can have a profound influence on the selection
of the optimal cable size and the associated cost of losses, whereas for others, approximate
354 Part III m Advanced Topics

values can be safely entered since they will have little influence on the final results. The
following study examines the effect of two of the parameters discussed in the previous
section. A more complete study is given in Anders er al. (1991).
proses fee AASSJee 8 ee eee ee eee
EXAMPLE12.97
We will consider three self-contained, insulating liquid-filled cables with paper insulation and lead
sheath with copper reinforcing tape. The system is operated at 230 kV. Cables are located 1 m
below the ground with soil thermal resistivity of | K-m/W. Ambient soil temperature is 20°C. All the
economic and laying parameters are the same as studied in Examples 12.5 and 12.8. The minimal
standard conductor size to meet ampacity requirements is equal to 460 mm? (900 kemil).
For the nominal parameters specified in Examples 12.5 and 12.8, the most economic conductor
size is 630 mm as shown in Fig. 12.3, and the optimal cable dimensions are shown in Fig. 12.4. The
optimal cable has an ampacity of 982 A with losses split as follows: conductor joule losses = 34.41
W/m, sheath joule losses = 3.0 W/m, dielectric losses = 7.52 W/m.

Total Cable Cost

18765

18159

17553

$)
(1000
Cost
Total 16947

16341

15735

1ol29

14523

13916

oo 460 510 630 760 1010 1270 1520 1770 2030

Conductor
Size(mm?)
Figure 12.3 Total cost as a function of cable cross section.

In the studies presented below, we will vary one parameter at a time, and the influence of this
parameter on the cost of losses and the selection of the optimal conductor size will be investigated.
Even though the computer program? used for this study computes the theoretically optimum economic
cross section of the conductor, results giving standard sizes only are presented below.

Case 1: Variation of the Load Growth Factor

The load growth factor a was varied between 1 and 5%. Factors F defined in equations (12.6)
and the optimal cross section in equation (12.42) were recomputed for each value of a, and the results
plotted in Fig. 12.5. Intuitively, the rate that the load grows over the economic life of the cable should
have a significant effect on the selection of the economic size of the conductor cross section. The

2 This example is extracted from Anders ef al. (1993).

3 The ECOPT computer program used in this study is a part of the Cable Ampacity Program (CAP) developed
by Ontario Hydro for Canadian Electrical Association (Anders et al., 1990, 1991).
Chapter 12 » Economic Selection of Conductor Cross Section 355

Cu, dried
S = 630 mm“ D= 36mm Dgig = 37.52 mm

Paper insulation
D=76.12mm Th=19.30 mm

Semiconducting shield
D=80.44mm Th=2.16mm

Corrugated aluminum sheath

D = 84.23mm Th=1.89 mm

Polyethylene jacket
D=89.23mm Th=2.50 mm

Figure 12.4 Optimal cable dimensions for nominal conditions.

larger the load growth factor, the higher are the conductor losses. This, in turn, will favor larger
conductor sizes for the most economic cable selection. The effect is very dramatic, as illustrated in
Fig. 12.5. The step function is a result of the fact that only standard cross sections are shown.
Since at a low load growth rate (1-2%) the current carried through the cable at the end of the
economic life will be smaller than 550 A, the optimal conductor size to meet ampacity requirements
is smaller than 460 mm’. The analysis indicated that for the voltage level of 230 kV, the dielectric
losses can become significant as shown in the example, and their influence on the total cost of losses is
considerable. In this case, the voltage-dependent losses result in the selection of a smaller conductor
size for some values of the factor a in comparison with the case when the dielectric losses are neglected.

Case 2: The Effect of Changing Financial Factors

The rate at which future energy costs will increase may have a significant effect on the future
cost of losses. If the discount rate grows at the same time, the effect of the growth in factor b may be
somewhat mitigated. In this study, the value of factor b is varied between 1 and 10% and the discount
rate is kept constant at 11%. The economic life is equal to 40 years, and so is the period over which
the load growth occurs. The results are shown in Fig. 12.6.
As expected, the increase in the cost of energy has quite a significant effect on the selection
of the conductor optimal cross section and on the cost of losses over the economic life of the cable.
The costs increase from about $8 500 000 when b is equal to 1% to over $17 000 000 with b = 10%.
Since the cost of losses is affected significantly when the value of b is large, the standard optimal
cable cross section changes by several sizes from 460 to 630 mm?” in the present case. This effect is
particularly evident when b exceeds 5%. Note the effect of dielectric losses on the optimal conductor
size shown in Fig. 12.6. For values of b between 3.5 and 10%, the cost of dielectric losses becomes
so significant that it forces the selection of the conductor cross section one size lower in comparison
with the case when the dielectric losses are neglected. The presence of dielectric losses has the effect
of shifting down the selection of a larger conductor cross section for larger values of b. This effect of
dielectric losses on the selection of the optimal conductor size was explained in Section 12.4.
An increase in the discount rate has the opposite effect to an increase in energy cost. Since
discounting has the effect of reducing future costs, the higher the discount rate, the lower is the cost
of losses. Therefore, increasing the discount rate will result in the selection of a smaller conductor
size. This effect is illustrated in Fig. 12.7.
356
Part II] = Advanced Topics

1900

(mm*)
Cable
section
cross
1500

With dielectric losses a

1100

Without
dielectric
lossesie
700

1 2 3 4 5
Annual load increase (%)

(a)

$)
(min
Total
cost
With
dielectric
lossesFe

Without
dielectric
lossesNs

1 2 3 4 5
Annual load increase (%)

(b)

Figure 12.5 The effect of varying the load growth on the optimal cable cross section
and costs.

We can observe from Fig. 12.7 that the effect is quite dramatic. Here, again, inclusion of the
dielectric losses has a significant effect on the selection of the most economic cable cross section.

Some of the important conclusions drawn from the studies reported in Anders et al.
(1991) can be summarized as follows. The optimal conductor size for a high-voltage cable
is, under almost all conditions, greater than the minimal conductor cross section needed to
Chapter 12 m Economic Selection of Conductor Cross Section 357

800

Se 700
(es
ie) . :
Without : losses
dielectric
s mane
D
Ss 600
0p)
fe)
rs)
®
8 500
With dielectric losses

400
1 2 3 4 5 6 th 8 9 10
Rate of increase of energy cost (%)

(a)

A
=
£
B Withdielectric
lossesmS
oO
i
ie)
ke

jae
Without dielectric losses

1 2 3 4 5 6 7 8 Ome 10
Rate of increase of energy cost (%)

(b)
Figure 12.6 The effect of varying the increase in the cost of energy on the optimal cable
cross section and cost.

meet ampacity requirements only. This confirms the conclusion reached in studies reported
by Parr (1989) for a low-voltage cable. In most of the cases studied, an increase in the
economic conductor size caused by including the influence of joule losses was tempered by
the associated cost of dielectric losses which may force the selection of a smaller conductor
size in comparison with the case when these losses are neglected. The effects of various
parameters on the cost of losses and the selection of the optimal conductor size are as
follows.
358 Part II] m Advanced Topics

1900

(mm2)
section
Cable
cross1500

1100 Without dielectric losses

700

0 4 8 12 16 20
Discount rate i (%)

(a)

$)
(min
Cost
Total

a With dielectric losses

0 4 8 12 16 20
Discount Rate i (%)

(b)

Figure 12.7 The effect of varying discount rate on the optimal cable cross section and cost.

1. The rate at which the load grows has a significant effect on the selection of the most
economic conductor cross section. In the example studied, when the load growth
changes in a fairly small range between | and 5% over a period of 40 years, the
optimal conductor cross section increases by 11 standard sizes from 350 to 2030
mm.
2. The length of the economic life of the cable is another factor which can have a
significant effect on the selection of the conductor size. In some countries, this
value is set to as low as ten years, and this favors selection of smaller conductors.
Chapter 12 m Economic Selection of Conductor Cross Section 359

In most cases, the longer the economic life, the larger the size of the most economic
conductor. We should perhaps point out that the length of economic life may be
selected quite independently of the actual service life of the cable, and this selection
may be based on financial or regulatory considerations and not on operational
expectations.
3. Financial factors can have a significant effect on the final conductor selection.
An increase in the discount rate favors smaller conductor sizes as the initial cost
becomes more significant. A faster increase in the energy cost, on the other hand,
will favor larger conductor sizes. These two effects may almost cancel each other
if the growth in the discount rate is approximately equal to the growth in the energy
costs.
4. When the full economic impact of cable losses on the selection of the optimal cable
cross section is studied, the importance of proper representation of the load growth
increases. If the load grows initially very fast and then levels off, a much larger
conductor size may be required as compared to the case when the load growth rate is
moderate but extends over a longer period of time. The change in the ultimate load
factor, on the other hand, has little influence on the selection of optimal conductor
size, but it should be modeled if the cost of losses is to be evaluated correctly.
The same conclusion can be drawn with regard to the representation of charging
current losses. Only for long distances and high voltages can an uncompensated
cable system have significant losses caused by charging current, and in that case,
the cost of these losses should be evaluated.

REFERENCES

Anders, G. J., Moshref, A., and Roiz, J. (July 1990), “Advanced computer programs
for power cable ampacity calculations,’ JEEE Comput. Appl. in Power, vol. 3, no. 3,
pp. 42-45.
Anders, G. J., Vainberg, M., Horrocks, D. J., Foty, M., and Motlis, J. (1991), “A user-
friendly computer program for economic selection of cable sizes,” in Proc. 3rd Int. Conf.
on Extruded Dielectric Power Cables—Jicable-91, Versaille, France, June 1991.
Anders, G. J., Vainberg, M., Horrocks, D. J., Motlis, J., Foty, M., and Jarnicki, J. (1993),
“Parameters affecting economic selection of cable sizes,” IEEE Trans. Power Delivery,
vol. 8, no. 4, pp. 1661-1667.
IEC 287 (1982), “Calculation of the continuous current rating of cables (100% load factor),”
IEC Publication 287, 2nd ed.
IEC 1059 (1991), “Economic optimization of power cable size,” IEC Publication 1059.
IEEE (Oct. 1990), “Loss evaluation for underground transmission and distribution cable
systems,” Insulated Conductors Committee—Task Group 7-39; IEEE Trans. Power De-
livery, vol. 5, no. 4, pp. 1652-1659.
Neher, J. H., and McGrath, M. H. (Oct. 1957), “The calculation of the temperature rise and
load capability of cable systems,” AIEE Trans., vol. 76, part 3, pp. 752-772.
Parr, R. G. (1989), “The economic choice of conductor size,’ Revue General de l’Electricite,
no. 10, Paris, pp. 45-50.
Scheer, C. B. (Feb. 1966), “Future power prediction,” Energy Int., pp. 14-16.
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PART IV APPENDIXES
e3xXigvs9s4 ViTAAY
>
7
«=
 A

Model Cables

Five model cables are described in this Appendix. The cables are used throughout the book
to illustrate various concepts as they are developed. Design and computed parameters for
the model cables are summarized in Table A.1.
The installation conditions of the model cables are also described below. However,
the installation information may vary in the examples to show the sensitivity of cable rating
to variations in various laying parameters.

A.1 MODEL CABLE NO. 1

This is a 10 kV single conductor XLPE cable. Conductor resistance at 20°C is taken from
IEC 228 (1978). The cable has copper screen wires with a given initial electrical resistance
(at 20°C) and a PVC jacket. All thermal and electrical parameters are as specified in IEC
287 (1982), and these values are given in the various tables in this book. The cable cross
section is shown in Fig. A.1.
The laying conditions are assumed as follows: cables are located 1 m below the
ground in a flat configuration. Uniform soil properties are assumed throughout. Spacing
between cables is equal to one cable diameter (spacing between centers equals to two cable
diameters). Ambient soil temperature is 15°C. The thermal resistivity of the soil is equal to
1.0 K - m/W. The cables are solidly bonded and not transposed.

A.2 MODEL CABLE NO. 2

This is a 10 kV three-core XLPE cable. The conductor resistance at 20°C is taken from
IEC 228 (1978). The cable has copper tape screen around each core with a given initial

363
364 Part IV m Appendixes

Conductor, copper, stranded

do = 20.50 mm

Conductor shield
Th = 0.60, des = 21.70 mm

Insulation, XLPE (unfilled)

Th = 3.40, D; = 28.50 mm

Insu. screen, semiconducting

Th = 0.80, Dj, = 30.10 mm

Concentric wires, copper

Th = 0.70, Dg = 31.20 mm,
Wires = 66

Jacket, custom
Th = 2.20, De = 35.80 mm

Voltage : 10.0 kV
Conductorarea: 300.0 mm

Figure A.1 Cross section of the model cable No. 1.

electrical resistance (at 20°C) and a PVC jacket. All thermal and electrical parameters are
as specified in IEC 287 (1982). The cable cross section is shown in Fig. A.2.
The installation conditions are assumed as follows: the cable is located in free air (not
clipped to a vertical wall). The ambient air temperature is 25°C, and the cable is shaded
from solar radiation.

A.3 MODEL CABLE NO. 3

This is a 138 kV high-pressure liquid-filled (HPLF) cable. All parameters are the same
as in the Neher/McGrath (1957) paper (see Table A.1). The cable shield consists of an
intercalated 7/8(0.003) inches bronze tape—1 in lay, and a single 0.1 (0.2) in D-shaped
bronze skid wire—1.5 in lay. The cables lie in a cradle configuration and operate at 85%
load-loss factor. Several parameters are different from those used in IEC 287 (1982). The
ones which are different are given as follows:

thermal resistivity of the insulation = 5.5 K - m/W

thermal resistivity of pipe coating = 1.0 K - m/W

dielectric constant = 3.5

tan d = 0.005

The cable cross section is shown in Fig. A.3.

Appendix A # Model Cables 365

Conductor, copper, stranded

d, = 20.50 mm

Conductor shield
Th = 0.60, des = 21.70 mm

ii Insulation, XLPE (unfilled)

Th = 5.00 mm

r Insulation screen, copper

Th = 0.20 mm

Jacket, PVC
in=13:50) D, = 72.90 mm

Voltage: 10.0 kV
Conductorarea: 300.0 mm

Figure A.2 Cross section of the model cable No. 2.

The laying conditions are assumed as follows. The cables are located in a steel pipe of
8.625 in outside diameter. The pipe is covered with an asphalt mastic covering 0.5 in thick.
The center of the pipe is located 3 ft below the ground. The soil is uniform throughout.
The ambient soil temperature is 25°C. The thermal resistivity of the soil is equal to 0.8 K -
m/W.

A.4 MODEL CABLE NO. 4

This is a 230 kV low-pressure, oil-filled cable with 1250 kcmil (633 mm?) copper hollow
core conductor. The cable has paper insulation with an insulation screen composed of four
layers of carbon tapes. On top of insulation screen is a lead sheath reinforced with three
layers of copper tapes, 50% overlap. The tapes are 25 mm wide and 1.3 mm thick. The lay
of tapes is equal to 115.2mm. Armor is composed of 51, #4 AWG copper wires with a lay
of length of 121.8 mm. The armor bedding is a saturated jute and a layer of polyethylene
with the equivalent thermal resistivity 4.27 K - m/W. The armor serving is a saturated jute.
All remaining thermal and electrical parameters are as specified in IEC 287 (1982).
The cable cross section is shown in Fig. A.4.
Three cables are laid in a thermal backfill as shown in Fig. A.5.
Soil ambient temperature is 20°C. Thermal resistivities of the soil and backfill are
shown in Fig. A.5. Cable sheaths are two-point bonded.
366 Part [V » Appendixes

Conductor, copper, 4 segments

dq. = 41.45 mm

Conductor shield
Th = 0.00, dog = 41.45 mm

Insulation, paper
Th = 12.83, Dj = 67.11 mm

Tape over insu. brass/bronze

Th = 0.08, Dy = 67.26 mm, Tapes = 1

—___— Skid wires, brass/bronze

Thi= 5:08: Dg = 67.59 mm, Wires = 1

Voltage: 138.0kV
Cond. area: 1010.0mm

Figure A.3 Cross section of the model cable No. 3.

A.5 MODEL CABLE NO. 5

This is a 400 kV paper—polypropylene—paper cable with 2000 mm? copper segmental con-
ductor and aluminum corrugated sheath. The outer covering is aPVC jacket.
The cable cross section is shown in Fig. A.6.
The cables are laid in a flat formation without transposition, directly in the soil with
thermal resistivity of 1 K - m/W and ambient temperature equal to 25°C. The sheaths are
cross bonded with unknown minor section lengths. The centers of the cables are 1.8 m
below the ground and phases are 0.5 m apart.
Appendix A m Model Cables 367

Conductor, copper, hollowcore

qd= 7/s0) amand. = 33.80 mm
Conductor shield
Win = O45, dos = 34.70 mm

Insulation, LPOF
Tilo)= 1s, D; = 67.80 mm

Insulation screen, semiconducting

Th = 0.60, Dj, = 69.00 mm

Sheath, lead with reinf. tape

in = G20. Ds = 75.40 mm

Sheath reinforcing material, copper

Th = 1.30, Dy = 78.00 mm, Tapes = 3

Armor bedding, custom

Th = 5.00, Dj),= 88.00 mm

Armor, copper wires

Une. 8), D's = 98.40 mm, Wires = 51

Armor serving, compound jute

Th = 3.30, D, = 105.00 mm

Voltage: 230.0 kV
2
Conductor area: 633.0 mm

Figure A.4 Cross section of the model cable No. 4.

368 Part IV m Appendixes

[\

Soil p = 1 Km/W

A
Backfill p = 0.6
® @ ie

Sa (OS)GH)
= 1.5m

Figure A.5 Laying conditions for model cable No. 4.

Conductor, copper, 6 segments

d. = 58.00 mm

Conductor shield
Thie20.30: dos = 58.60 mm

Insulation, Paper
Th = 18.00, D; = 94.60 mm

Insulation screen, semiconducting

Th = 0.30, Dj, = 95.20 mm

Corrugated sheath, aluminum

Th = 6.00, d, =113.5 mm

Jacket, polyethylene
Th = 6.00, D, =125.00 mm

Voltage: 400.0 kV
Conductorarea: 2000.0 mm*

Figure A.6 Cross section of the model cable No. 5.

Appendix A m Model Cables 369

TABLE A.1_ Parameters of Model Cables

rr

Cable Cable Cable Cable Cable

No. | No.2 No. 3 No.4 No.5

Construction

Cu
Cu Cu Cu hollow Cu
Conductor stranded stranded segmental core segmental
S (mm?) 300 300 1010 633 2000
d..(mm) 20.5 20.5 41.45 33.8 58.0
d.; (mm) PA 21e7, 41.45 34.7 58.6
d; (mm) LES
c (mm) 17.6
s (mm) 30.5 67.59

Insulation XILPE XLPE Paper Paper PPLP

D; (mm) 28.5 PRY 67.11 67.8 94.6

Dj, (mm) 30.1 30.1 67.26 69.0 95.2

t (mm) 9.6
t, (mm) 4.8!

Brass/bronze Leadwith Corrug’d.

Screen/sheath Cu wires Cu tape Tapeand 3 Cu tapes aluminum
1skidwire
D, (mm) 31.4 67.59 75.4 102
Do.(mm) 113
D;,(mm)
d (mm) 73.8
d5(mm) 30.5 76.0/
(78overtape)
5, (mm) 0.2
t, (mm) Bi 6.0

Armor Copper

D, (mm) 98.4
5 (mm) 5.189

Somastic Jute/poly-
pipe ethylene PVC
Jacket/bedding PVE PVG coating jute

ty(mm) 5.0
t; (mm) 3.3
t; (mm) QE 3h) 6.0
D, (mm) 35.8 WBE 244.48 105 DS)2
a a et lg a re a
Ductor pipe Steel
ERE at ee ee Fee ee
D, (mm) 206.38
Do(mm) 219.08
ne EEE

| Insulation thickness includes the thickness of semiconducting screens at the conductor and above

the insulation.
370 Part IV m Appendixes

TABLEA.1. Continued
Sas gh SeNeS98 SR ea re
GivenCableParameters
ee a eae
8 (CC) 90 90 70 85 85
f (Hz) 50 50 60 50 60
€ Sim) 2.8
tan 6 0.005 0.001
Rp (Q /km) 0.0601 0.0601
Ryo(Q /km) 0.759 0.906

Computed Parameters (Hottest Cable)

I(A) 629 713 902 771 1365

R(Q/km) 0.0781 0.0798 0.0245 0.0356 0.0126
Ys 0.014 0.014 0.050 0.021 0.132
Yp 0.0047 0.027 0.054 0 0.005
X(Q/km) 0.096 0.044 0.052 0.154 0.168
X, (Q /km) 0.102 0.170 0.186
X,, (Q/km) 0.044 0.044 0.052
1 0.090 0.0218 0.010 0.325 0.140
re 0.311 0.955
T, (K: m/W) 0.214 0.3077 0.422 0.568 0.579
T>(K- m/W) 0.082 0.082
T; (K- m/W) 0.104 0.078 0.017 0.066 0.056
T,(K: m/W) 1.933 0.345 0.343 0.814 1.276
(0.289)
Q.(3/K- m) 1035 10354 3484.5 2183.9 6900
Q; (J/K- m) 915.6 915.6 1458.4 5684.0 8952
Q,,(3/K m) 38570 1995.1>
Q, (J/K- m) 4.0 4.0 i 2133,2° 3916.3
Q, (J/K- m) 613.3
Q; (J/K: m) 394.8 432.4 318477 2108.7 5392.5
LT-ZO (h) 1.47 1.04 18.1 6.26 13.4
W..(W/m) 30.85 40.58 19.93 21.16 23.40
W,,(W/m) 4.83 6.62 6.53
W, (W/m) 33.59 41.45 26.32 48.24 26.92
W,(W/m) 33.59 41.45 31.15 54.86 33.45

* For cable No. 2, all computed values except T| are given on a per-core basis. T; for a single
core is one third of the value in Table A.1.
3 The second 7, value corresponds to the Neher/McGrath method with load-loss factor = 0.85.

* The conductor and insulation capacitances are given on a per-core basis. See Example 3.8 for
representation of this cable by an equivalent single-core cable.
> This is oil in the conductor.
®Capacitance of sheath and reinforcement.

7This value includes capacitance of the pipe and pipe covering.

 B
An Algorithm to Calculate
the Coefficients of the
Transfer Function Equation

The Laplace transform of a ladder network transfer function is given by a ratio [see equation
(al) i:
H(s)
we)
~ Os)

P(s) and Q(s) are polynomials, their forms depending on the number of loops in the
network. An algorithm for the computation of the coefficients of these polynomials is
described below.

B.1 DENOMINATOR EQUATION

The degree of the denominator equation will be the same as the number of loops of the
equivalent network. This equation is common for all the nodes of the circuit and has the
form

O(s) = 8 (bys + by_18"~! + bis + bo) (B.1)

where n is the number of loops of equivalent network.

We define

Pet 12 eee} set Ol indices

I, = {i,, in, -++, ix} any combination of k elements of /
Ty= 1 — Ip = {ixsi, +++in} complement of J; in I

S7i
372 Part 1V » Appendixes

By,= {Bi,-++ »Bn—k}Set of sequences in which B are

equal to zero or one
I = {iges —Bi, ikt2 —Bos -++ +in —Bn} Sequences of B, over J; where all
elements of /; are different.

The
coefficients
binequation
(B.1)
areobtained
from
k
be=> [1Ou| [] Gn+DUTT: (B.2)
i j=l mel, Bylel’
where Q, = equivalent capacitances, J/K - m
G, = 1/T, = equivalent conductances, W/K - m.

The first summation in (B.2) includes all the combinations of k elements out of n. There are
(7) terms. The second summation includes all the sequences B,, except B, = {0,0,--- , O}.
There are 2""-) — | terms.

B.2 NUMERATOR EQUATION

For each node i, the numerator equation P(s) is different. The degree of P(s) is equal to
the number of loops minus the node index
m=n-i1 (B.3)

and
P(s) — Ledes ae Cais qr 0°SF ays smaj (B.4)

We define

I = {1,2,--- ,n} set of indices

I, = {i,, 12, +++, ix} any combination of k elements of
I withi, > i for p= 1,---.k.
I, = 1 —I = {igsi,+++ , in} complement of J, in J but index i
excluded, that is, i, 4 i for
piss Kk+ bearermn;
By = {B, +++, Bn—«}set of sequences in which B are
equal to zero or one excluding
Bo ==(0, O7<s= 0}
I = {inst —Bi, theo —Bo, +++» in —Bn—x~}
Sequences of B, over I, where all
elements of /;’ are different.

Similarly to the denominator, the coefficients a for the numerator equation (B.4) are given
by

k
aki= SYI] Oi; I] Gm+DaI] G; (B.5)
Kh j=l mel, By lel;
 C
Digital Calculation
of Quantities Given
Graphically in Figs. 9.1—9.7

This Appendix gives formulas and methods suitable for digital calculations for several quan-
tities given graphically in Chapter 9 (Figs. 9.1—9.7). The method used is the approximation
by algebraic equations, followed by quadratic or linear interpolation where necessary. The
maximum percentage error prior to interpolation is given in each case. The formulas were
first published in IEC Publication 287.

C.1 GEOMETRICFACTORG FORTWO-COREBELTEDCABLES

WITHCIRCULARCONDUCTORS(FIG. 9.1)

Denote
9 lwhii Yee(217/1)
1 (a)
Then
G=MG, (C22)
where
M= Mie =In; BthalSe 2 : : ae 1/2
formula : (C3)

useee
«= foitt ; (CA)
+a] pope

373
374 Part IV m Appendixes

G, = G,(X, Y); that is, G, is a function of both X and Y. To obtain this function, we first
compute

G,(X, 0) = 1.06719 — 0.0671778X + 0.0179521 X?

G,(X, 0.5) = 1.06798 — 0.0661648X + 0.0158125X? (G5)
G,(X, 1) = 1.06700 — 0.0557156X + 0.0123212X?

Then G,(X, Y) is obtained by quadratic interpolation using the following formula:

(C.6)
442(2G,(X,,0) —4G,(X, 0.5) + 2G, (X, 1)]
The maximum percentage error in the calculation of G;(X, 0), G;(X, 0.5), and G;(X, 1)
is less than 0.5% compared with corresponding graphical values.

C.2 GEOMETRICFACTORG FORTHREE-COREBELTED

CABLESWITHCIRCULARCONDUCTORS(FIG. 9.2)

Equations (C.1)-—(C.3) are also applicable in this case with the following parameters:

(1 x) ;
cease 2X
colina
eee = (1
Eamesa)+ |
3 1+Y

As before, G, is a function of both X and Y. To compute this function, we first compute

G,(X, 0) = 1.09414 —0.0944045X + 0.0234464X?

G;(X, 0.5) = 1.09605 —0.0801857X + 0.0176917X? (C.8)
G,(X, 1) = 1.09831 —0.0720631.X + 0.0145909 X?

Then G,(X, Y) is obtained by quadratic interpolation using equation (C.6). The maximum
percentage error in the calculation of G,(X, 0), G;(X, 0.5), and G,(X, 1) is less than 0.5%
compared with corresponding graphical values.

C.3 THERMAL RESISTANCE OF THREE-CORE SCREENED CABLES

WITH CIRCULAR CONDUCTORS COMPARED TO THAT OF A
CORRESPONDING UNSCREENED CABLE (FIG. 9.4)

Denote

Aiea (6, Pr /dePm) ’ = ti /d-. (C9)

The screening factor K is a function of both X and Y. To calculate it, we first compute
K(X,0.2), K(X, 0.6), and K(X, 1) from the following formulas according to whether
Oe X= 6:0r Grae 25)
Appendix C m Digital Calculation of Quantities Given Graphically in Figs. 9.1—9.7 375

Oi X= 6 K(X, 0.2) = 0.998095 — 0.123369X

+ 0.0202620X? —0.00141667X?
K(X, 0.6) = 0.999452 — 0.0896589X
+ 0.0120239X? —0.000722228 x?

K(X, 1) = 0.997976 — 0.0528571X

+ 0.00345238 X
‘i
Oieek Oran) K(X, 0.2) = 0.824160 —0.0288721X can?
+ 0.000928511X? —0.0000137121X?
K(X, 0.6) = 0.853348 —0.0246874X
+ 0.000966967 X* —0.0000159967X?

K(X, 1) = 0.883287 —0.0153782X

+ 0.000260292 xX?

Then, K(X, Y) is obtained by quadratic interpolation using the following formula:

K(X, Y) = K(X, 0.2) + Z[—3K (XX,0.2) 4+4KiCX,,0:6) 2A 1] (C.11)

aOR KAO.
2 u AK X OO) KX al)|

where Z =1.25Y —,0.25.,

The maximum percentage error in the calculation of the reduction factor is less than
0.5% compared with the corresponding graphical values.

C.4 THERMAL RESISTANCE OF THREE-CORE SCREENED CABLES

WITH SECTOR-SHAPED CONDUCTORS COMPARED TO THAT
OF A CORRESPONDING UNSCREENED CABLE (FIG. 9.5)

Denote

X= (O1Pp/ On0m)s Y =t,/d, (C.12)

The screening factor K is a function of both X and Y. To calculate it, we first compute
K(X, 0.2), K(X, 0.6), and K(X, 1) from the following formulas according to whether
OpeeXe= 3757 Xx= 6, OFOe Xg= 25)

Oe Osa) K(X, 0.2) = 1.00169 — 0.0945X + 0.00752381X?

K(X, 0.6) = 1.00171 — 0.0769286X + 0.00535714X? (O3r3)
K(X) 1) = K O06)

Be KMSE K(X, 0.2) and K(X, 0.6) are given by the same formula as for 0 < X <3.
K(X,.1) = 1.00117 = 0.0752143X = 0.00533334X~
376 Part IV m Appendixes

6 = Xp5 K(X, 0.2) = 0.811646 —0.0238413X

+ 0.000994933X2 —0.0000155152X°
K(X, 0.6) = 0.833598 —0.0223155X
+ 0.000978956X2 —0.0000158311X?
K(X, 1) = 0.842875 —0.0227255X
+ 0.00105825X? —0.0000177427 X?

For 0 < X < 3 and 0.2 < Y < 0.6, K(X, Y) is obtained by linear interpolation
between K(X, 0.2) and K(X, 0.6) as

K(X, Y) = K(X, 0.2),+.2.5(Y —.0.2)[K CX,0.6) — K(X, 0.2)] (C.14)

For 3 < X < 25, K(X, Y) is obtained by quadratic interpolation between the three
calculated values from equation (C11). The maximum percentage error in the calculation
of the sector correction factor is less than 1% compared with graphical values.

C.5 CURVE G FOR OBTAINING THE THERMAL RESISTANCE

OF THE FILLING MATERIAL BETWEEN THE SHEATHS AND ARMOR
OF SL TYPE CABLES (FIG. 9.7)

We will denote by X the thickness of material between sheaths and armor expressed as a
function of outer diameter of the sheath. The lower curve is given by

OneaxX=.0.03 G = 27 (0.000202380 + 2.03214X —21.6667X7)

0.03 < X < 0.15 G = 2n(0.0126529 + 1.101X —4.56104X? + 11.5093X°3)
(ea)

The maximum percentage error in the calculation of G is less than 1%.

The upper curve is given by

Oke 0.(3 G = 27 (0.00022619 + 2.11429X —20.4762X7)

0.03 < X < 0.15 G = 27 (0.0142108 + 1.17533X —4.49737X2 + 10.6352X°)
(C.16)

The maximum percentage error in the calculation of G is again less than 1%.
 D

Properties of Air
at Atmospheric Pressure

The air thermodynamical and transport properties used in this book are from the U.S.
National Bureau of Standards, cited in Holman (1990). All of the properties are temperature
and pressure dependent. For the convenience of computer programming, formulas are
generated by curve fitting to relate the air density (~), thermal conductivity (k), viscosity
(v), and Prandtl Number (Pr) for temperature (9*) in the range of 250-450 K:
p = 352.64/0, kg/m?
kair= 10-8(—27997.7 + 989.9986* —3.542830*7), W/K-m ‘on
v = 107!!(—376936 + 3780.056* + 9.114220*), m?/s
Pr = 0.833209 — 0.5823496* - 10-7 + 0.5523360*? - 10-°

where the temperature is in Kelvin. Similar formulas are widely used in thermal engineering.
Morgan (1982) gives the following simpler relationships:
k= 240 AO 272. 1006" — 273)
== 132105 + 95-1072 (6* = 273) (D.2)
Pr = 0.715 — 0.00025(6* — 273)

The accuracy of Morgan’s formulas is compared with the equations (D.1) in Anders
and Gu (1995).

REFERENCES

Anders, G. J., and Gu, N. (1995), “Energy conservation equations for cables in air,’ Canadian
Electrical Association Report 138 D 375E.

Si)
378 Part IV m Appendixes

Holman, J. P. (1990), Heat Transfer. New York: McGraw-Hill.

Morgan, V. T. (1982), “The thermal rating of overhead-line conductors, Part I. The steady-
state thermal model,” Elec. Power Sys. Res., vol. 5, pp. 119-139.
 =

Calculation Sheets for

Steady-State Cable Ratings

The calculation procedures set up in this Appendix are applicable to single- and three-
conductor cables. They are based on the material presented in this book, and include the
basic computations covered in IEC Standard 287 and the Neher—McGrath paper.

E.1 GENERAL DATA

FOZ) = system frequency

UV) = cable operating voltage (phase-to-phase)
Le daily load factor
6= conductor temperature!
Oh = ambient temperature

E.2 CABLE PARAMETERS

E.2.1 Conductor

S (mm?) = cross-sectional area of conductor

D, (mm) = external diameter of cable
D> (im) = external diameter of cable, or equivalent diameter
of cable (cables in air)

u Subscripts tf, s, and a will be used to denote tape, sheath, and armor, respectively.

379
380 Part IV m Appendixes

d-((mm) = external diameter of conductor

d’ (mm) = conductor diameter of equivalent solid conductor
having the same central oil duct
d; (mm) = conductor inside diameter
if ee number of conductors in a cable

Three-conductor Cables.
d; (mm) = diameter of an equivalent circular conductor having
the same cross-sectional area and degree of
compactness as the shaped one
c (mm) = distance between the axes of conductors and the axis
of the cable for three-core cables (= 0.55r; + 0.291
for sector-shaped conductors)
r; (mm) = circumscribing radius of three-sector shaped
conductors in three-conductor cable

E.2.2 Insulation

D; (mm) = diameter over insulation

t; (mm) = insulation thickness between conductors and sheath
p(K- m/W) = thermal resistivity of the material?

Three-conductor Cables.
t (mm) = insulation thickness between conductors

t; (mm) = thickness of core insulation, including screening

tapes plus half the thickness of any
nonmetallic tapes over the laid up cores

E.2.3 Sheath

(Bd.(anaven)) sheath diameter

d (mm) = sheath mean diameter
in (CaaEN)== sheath thickness

P2 ; :
= ratios of minor section lengths, where minor section
ch lengths are a, p2a, gra and a is the shortest
section
¢ ($22 mj electrical resistivity of sheath material at operating
temperature

2 ‘ : ie. :
“ The same symbol is used for thermal resistivity of various materials. The appropriate numerical value
taken from the table on page 400 will correspond to the material considered.
Appendix E m Calculation Sheets for Steady-State Cable Ratings 381

E.2.4 Armor or Reinforcement

A (mm?) = cross-sectional area of the armor

Dz (mm) = external diameter of armor

dq (mm) = mean diameter of armor

dy (mm) = diameter of armor wires

dy (mm) = mean diameter of reinforcement

number of armor wires

number of tapes

length of lay of a steel wire along a cable

length of lay of a tape
thickness of tape
width of tape
angle between axis of armor wire and axis of cable

E.2.5 Jacket/serving
ty (mm) = thickness of the jacket
tz (mm) = thickness of the serving

E.3 CABLE PARAMETERS—PIPE-TYPE CABLES

i)5 (mim) == external diameter of the pipe

(dD),(Cavite)= external diameter of the pipe coating
Dena )= moisture barrier mean diameter
[Dyry(Canad)= diameter of skid wire

D, (mm) = diameter of moisture barrier assembly

Dg (mm) = internal diameter of pipe
number of skid wires
Nsw =
ee number of moisture barrier metallic tapes
Cee (min) = lay of length of skid wires
£7 (mm) = lay of moisture barrier metallic tapes
t3 (mm) = thickness of pipe coating
t= thickness of moisture barrier metallic tape in
pipe-type cable

w; (mm) = width of moisture barrier metallic tape

E.4 CABLE PARAMETERS—INSTALLATION CONDITIONS

E.4.1 Cables in Air

H (W/m?) = solar radiation

382 Part IV m Appendixes

F.4.2 Buried Cables

L (mm) = depth of burial of cables

/0)-.(Cinen) = fictitious diameter at which effect of loss factor
commences
ps (K-m/W) = thermal resistivity of soil
s (mm) = spacing between conductors of the same circuit
So (mm) = axial separation of cables; for cables in flat
formation, 52 is the geometric mean of the three
spacings

E.4.3 Duct Bank/Thermal Backfill

Lg (mm) = distance from the soil surface to the center of a

ductbank
x, y (mm) = sides of duct bank/backfill (y > x)
oo number of loaded cables in a duct bank/backfill
p-(K - m/W) = thermal resistivity of concrete used for a duct bank
or of the backfill
Pe (K - m/W) = thermal resistivity of earth surrounding a duct
bank/backfill

E.4.4 Cables in Ducts

Dg (mm) = internal diameter of the duct

(DY,(Gnnvnny)
= external diameter of the duct
Oy = mean temperature of duct filling medium

E.5 CONDUCTOR AC RESISTANCE

Resistivity (9): 10° Temperature Coefficient (9): 107

Material Q-m at 20°C per K at 20°C

Copper 1.7241 3.93

Aluminum 2.8264 4.03
rrr ————————

pre 1.02
- 10°p>
P20
5 [1 + 29 (0 — 20)]
Appendix E @ Calculation Sheets for Steady-State Cable Ratings 383

Whether
Dried and

Impregnated
Type of Conductor or Not

Copper
Round, stranded
Round, stranded
Round, compact
Round, compact
Round, segmental
Hollow, helical
stranded
Sector-shaped
Sector-shaped
Aluminum
Round, stranded Either 1
Round, 4 segment Either 0.28
Round, 5 segment Either 0.19
Round, 6 segment Either 0.12
Segmental with Either eq. (7.17)
peripheral strands

“Since there are no accepted experimental results dealing specifically with

aluminum stranded conductors, IEC 287 recommends that the values of k, given
in the above table for copper conductors also be applied to aluminum stranded
conductor of similar design as the copper ones.

2
Xs

If x, < 2.8 (a majority of the cases), then the following equations apply. Otherwise, use
equations (7.10) or (7.11).

4
x;
We=
192+ 0.8x4

Ws
384 Part IV m Appendixes

If x, < 2.8 (a majority of the cases), then the following equations apply. Otherwise, use
equations (7.26) or (7.27).

4
7)
F, = ——*__
A 192 0.8x4

s=d,+t
d, = d,
Yp = 2yp/3
For oval conductors:

d. = de minor*dc major

R= R(1 se Vets Yp)

Keo Q2/m

For cables in magnetic pipes and conduits:

LR
=R'[1+
1.5(y;
+yp)]
he Q/m
Appendix E m Calculation Sheets for Steady-State Cable Ratings 385

E.6 DIELECTRIC LOSSES

Type of Cable € tan d

Cables insulated with impregnated paper

Solid type, fully impregnated, preimpregnated 4 0.01
or mass-impregnated nondraining
Oil-filled, low-pressure
up to Up = 36 kV 3.6 0.0035
up to Up = 87 kV 3.6 0.0033
up to Up = 160 kV 3h) 0.0030
up to Up = 220 kV 3#5) 0.0028
Oil-pressure, pipe-type Sei) 0.0045
Internal gas-pressure 3.4 0.0045
External gas-pressure 3.6 0.0040
Cables with other kinds of insulation
Buty! rubber + 0.050
EPR—up to 18/30 kV 3 0.020
EPR—above 18/30 kV 3} 0.005
PME 8 0.1
PE (HD and LD) 23 0.001
XLPE up to and including 18/30 (36) kV—aunfilled DS) 0.004
XLPE above 18/30 (36) kV—unfilled DS) 0.001
XLPE above 18/30 (36) kV—filled 3 0.005
Paper—polypropylene—paper (PPL)" 2.8 0.001

*The dielectric constant and the loss factor of PPL insulation has not been
standardized yet.

tand =

: ee [=
reine a
nNieee
d,

yy U
aS

Up= V

Wy= 2nf -C-U;-tand

Wi= W/m
386 Part IV m Appendixes

E.7 SHEATH LOSS FACTOR

E.7.1 Sheath Resistance

Resistivity (P9)- 10 Temperature Coefficient (59): 10°

Material Q-m at 20°C per K at 20°C

Lead or lead alloy 21.4 4.0

Steel 13.8 4.5
Bronze Sh) 3.0
Stainless steel 70 Negligible
Aluminum 2.84 4.03

_ P20" 10°
R [1 + a9 (A; — 20)]
ied er:

ds
If ie > 0.44, R,; = 2R, computed above.
T
To calculate sheath losses, use the combined resistance of sheath and reinforcement.

Ry => Q/m
Substitute R,, for R, in what follows.

E.7.2 Sheath Reactances

For single-conductor and pipe-type cables:

X=4nf1077 2
-In>
As Q/m
Appendix E m Calculation Sheets for Steady-State Cable Ratings 387

For single-conductor cables in flat formation, regularly transposed, sheaths bonded at both
ends:

X}=4xf- 1077
-n|2-72(=)|
Xi= Q/m

For single-conductor cables in flat configuration with sheaths solidly bonded at both ends,
the sheath loss factor depends on the spacing. If it is not possible to maintain the same
spacing in the electrical section (i.e., between points at which the sheaths of all cables are
bonded), the following allowances should be made:
(1) If the spacings are known, the value of X is computed from

_ aXatleXpt---+dnXn
Xo
botelpak is ly

where Ia, lp,-++ ,l, are lengths with different spacing along an electrical section
Xq,Xp,°** , Xn are the reactances per unit length of cable, given by equations
for X or X; above

where appropriate values of spacing sa, 5p, --* , Sn are used.

(2) If the spacings are not known, the value of A, calculated below should be increased
by 25%.

Korase7te 10 (=f

Xm = Q/m

E.7.3 Single-conductor Cables

(1) Sheath bonded both ends—triangular configuration:
388 Part IV m Appendixes

(2) Sheath bonded both ends—fiat configuration, regular transposition:

N=
i =0
(3) Sheath bonded both ends—flat configuration, no transposition. Center cable
equidistant from other cables:

P=Xy+X

P= Q/m

Q= Q/m

RE Rees

RP=
| RO
=R,+Q°
RQ=
Appendix E # Calculation Sheets for Steady-State Cable Ratings 389

DR P.O:
SER PRO

C=

Rs
Re =o a
R

Re =

[jy=Ra(A+
B+)|
Outer cable carrying lagging phase.

Ai/ —_—
ae

Center cable.
pVia

Ay=Res(A+ B—C)
Outer cable carrying leading phase.

/ —
M2 =

M=
iM=0
Ratings for cables in air should be calculated using 4/,,; 4; is equal to A}, or A}, or A},
depending on which cable is the hottest.

E.7.4 Large Segmental Conductors

When conductor proximity effect is reduced, for example, by large conductors having
insulated segments, 4, cannot be ignored and is calculated as follows:

M=N=—
X
390 Part IV m Appendixes

Cables in trefoil formation.

M=N=

and

M Rs
x
N= Rs
eo BE
}Epa
3

Cables in flat formation with equidistant spacing.

Me

Ne

Eh4M?N*+(M+N)*
~ 4(M?2+ 1)(N2 +1)

jo =

i) is calculated by multiplying the value of the eddy current sheath loss factor calculated
below by F.

E.7.5 Sheaths Single-point Bonded or Cross Bonded

Lead-sheathed cables.

For corrugated sheaths, the mean outside diameter shall be used.

Appendix E m Calculation Sheets for Steady-State Cable Ratings 391

w=1+(#) t 1.74
(B,D,- 107°—1.6)

LO A= Oko 0;
E.7.6 Three Single-conductor Cables in TriangularConfiguration

A, =
= (1.14m* +
2.45 0.33)(=
Aas es
A,=

E.7.7 Three Single-conductor Cables in Flat Configuration

(1) Center cable:

392 Part IV m Appendixes

AyE50,
868 es 1.4m+0.7
as

do =

A, = 4.7m? (=)d 0.16m+2

Ss

Ai=

. De<is(n
aee

Ao =

0.74(m
1 SS +2)m°>
(Z \
I Ge OA)? \O5

Ai =
Appendix E m Calculation Sheets for Steady-State Cable Ratings 393

Ay
i =RB
Rsa (1+Ay
ain
A>)+aeteé:0
MN

E.7.8 Sheaths Cross Bonded

The ideal cross-bonded system will have equal lengths and spacing in each of the three
sections. If the section lengths are different, the induced voltages will not sum to zero
and circulating currents will be present. These circulating currents are taken account of by
calculating the circulating current loss factor A; assuming the cables were not cross bonded,
and multiplying this value by a factor to take account of the length variations. This factor
F.. is given by

pie patent)
: AIG? cia

where p2a = length of the longest section

qoa = length of the second longest section
a = length of the shortest section.

This formula deals only with differences in the length of minor sections. Any deviations in
spacing must also be taken into account.
Where lengths of the minor sections are not known, IEC 287-2-1 (1994) recommends
that the value for 4, based on experience with carefully installed circuits be

i, = 0.03 for cables laid directly in the ground

i, = 0.05 for cables installed in ducts.

E.7.9 Three-conductor Cables

=O

uN,=0
1) Round or oval conductors in common sheath, no armor:
R, < 100 wQ/m
394 Part IV m Appendixes

R, > 100 12/m

Mi

E.7.10 Three-conductor Cables with Steel TapeArmor

The value for 4/ calculated above should be multiplied by the factor F;.

is usually taken as 300.

Appendix E m Calculation Sheets for Steady-State Cable Ratings 395

E.7.11 Cables with Separate Lead Sheath (SL Type)withArmor

E.8 SHIELDING/SKID WIRE LOSS FACTOR

E.8.1 Pipe-type Cables

i 02/m

D 2 1/2
TTLUsm
t ) [1 + 20 (sw a 20)]

E.8.2 Total Sheath Loss Factor

396 Part 1V » Appendixes

E.9 ARMOR LOSS FACTOR

E.9.1 Single-conductor Cables

(1) With nonmagnetic wire armor:

a=Seat (F)| 1/2

4p-10° Di\4[1+a9(64
d + Ng: —20)]
dy ie

Oh = Q/m

rig R,Ra
Ao (R; + Ra)?

X=

Ra
era

A=

h= Rs
2= R x

AQ=
(2) With magnetic wire armor:
The following applies to cables spaced at least 10 m apart. The ac resistance of the
armor wires will vary between about 1.2 and 1.4 times its de resistance, depending on the
wire diameter, but this variation is not critical because the sheath resistance is generally
considerably lower than that of the armor wires.

uel Rs Ra
é
" Rs +R

Ka 02/m

Average values for magnetic properties of armor wires with diameters in the range of 4-6
mm and tensile strengths on the order of 40 MPa:

[ke = 400

/4; = 10 for armor wires in contact

4; = | for armor wires which are spaced

Vestas
Appendix E # Calculation Sheets for Steady-State Cable Ratings 397

If amore precise calculation is required and the wire properties are known, then it is initially
necessary to know an approximate value for the magnetizing force H in order to find the
appropriate magnetic properties.

he ik
md,
where J and I, are the vectorial values of the conductor and sheath currents, respectively.
For the initial choice of magnetic properties, it is usually satisfactory to assume that 7+/, =
0.8/7, and to repeat the calculations if it is subsequently established that the calculated value
is significantly different.

He
=26109
In(>)
2
Hie H/m

nad5 i)
Hy = Wie ei 10°“sin Bcos y

Hy,=> H/m

nad; meee
Hy =e —]10°' sinB siny
baile
Eh= H/m
H;=0.4
(u;cos”
B— 1)@ LOme
H3= H
Bi=@OUHet
Hy
+Hs)
|
(t= Q2/m

B, I 2)Ss5
398 Part IV m Appendixes

/ % Re (ee)
Ve PORN IR By) ee

N=

Ao=
E.9.2 Three Conductor Cables—Steel Wire Armor

2.77R410°\?
| i
wW

(2) SL type cables:

2 calculated above should be multiplied by (1 — A{), where A‘ is calculated in the
section on the sheath loss factor for SL type cables.
(3) Sector-shaped conductors:

R(+)2
We0358
R (sae
Ae,
W eg
A i)

E.9.3 Three Conductor Cables—Steel Tape Armor or Reinforcement

Hy = mm
Appendix E m Calculation Sheets for Steady-State Cable Ratings 399

it is usually taken as 300.

ee Oe
S| Rd

Ag = Mt

E.9.4 Pipe-type Cables in Steel Pipe—Pipe Loss Factor

Configuration A B

Cradled 0.00438 0.00226

Triangular bottom of pipe 0.0115 0.001485

Mean between trefoil and cradled 0.00794 0.00039

Three-core cable 0.0199 —0.001485

A=

B=

(2 f Atogeb:
Bs Dy
R
400 Part IV m Appendixes

E.10 THERMAL RESISTANCES

Thermal Thermal
Resistivity Capacity
(p) (c- 10)
Material (tK-m/W) [J/(m?- K)]

Insulating materials*
Paper insulation in solid type cables 6.0 AAD)
Paper insulation in oil-filled cables 5.0 2.0
Paper insulation in cables with external gas pressure She) 2.0
Paper insulation in cables with internal gas pressure
preimpregnated 6.5 20
mass-impregnated 6.0 2.0
PE BS 2.4
XLPE 35 2.4
Polyvinyl chloride
up to and including 3 kV cables 5.0 ey)
greater than 3 kV cables 6.0 1.7
EPR
up to and including 3 kV cables ahs 2.0
greater than 3 kV cables 5.0 2.0
Butyl rubber 5.0 2.0
Rubber 5.0 2.0
Paper—polypropylene—paper (PPL) 6.5 2.0
Protective coverings
Compounded jute and fibrous materials 6.0 2.0
Rubber sandwich protection 6.0 2.0
Polychloroprene SS, 2.0
PVG
up to and including 35 kV cables 5.0 Ly
greater than 35 kV cables 6.0 UTE
PVC /bitumen on corrugated aluminum sheaths 6.0 Ls.
PE 3.5 2.4
Materials for duct installations
Concrete 1.0 ae
Fiber 4.8 2.0
Asbestos 2.0 2.0
Earthenware ib 1.8
PVC 6.0 1.7
PE 3:5 2.4

“For the purpose of current rating computations, the semiconducting screening materials are
assumed to have the same thermal properties as the adjacent dielectric materials

E.11 INSULATION THERMAL RESISTANCE

For oval-shaped conductors, the diameter over the insulation is the geometric mean of the
minor and major diameters over the insulation.
Appendix E # Calculation Sheets for Steady-State Cable Ratings 401

E.11.1 Single-conductor Cables

E.11.2 Three-conductor Belted Cables

(1) With round or oval conductors:

G is obtained from the figure below.

G
3.0

ZO

2.0

15 t = thickness of insulation
between conductors

t1 = thickness of insulation
between conductor
and sheath

dc = diameter of conductor
(circular)

0.5

T=
K+ m/W
402 Part IV m Appendixes

(2) With sector-shaped conductors:

Fyj=3 Pe ot
2n(d,+t)—t

d, = external diameter of the belt insulation, mm

i= K -m/W

E.11.3 Three-conductor Shielded Cables

(1) With round or oval conductors:

The geometric factor is read from the preceding figure. The screening factor is read from
the figure below.
Appendix E @ Calculation Sheets for Steady-State Cable Ratings 403

Screening
factor K
1.0 5, = thickness of metallic screen
on core

pz = thermal resistivity of insulation

d, = diameter of conductor(circular)
0.9 t, = thickness of insulation between
conductor and screen

Pm = thermal resistivity of screening

material:
27 x 10° K.m/W for copper
48 x 104 K.m/W for aluminum
0.8

0.7

0.6

Cs

ie

0
T,;=K—G
Qn
N= _K-mw
(2) With sector-shaped conductors:

F 3+ Zt
ee (de 1)—it

Fi
404 Part IV m Appendixes

Ge

The screening factor is read from the figure below.

Screening
factor K
1.0
5, = thicknessof screen
Pz = thermal resistivity of insulation
d, = diameter of circular conductor
having the same section and
0.9 compaction
t, = thickness of insulation between
conductor and screen

Pim = thermal resistivity of screening

material:
27 x 10° K-m/W for copper
0.8
48 x 10° K-m/W for aluminum

0.7

0.6

Lies K - m/W
Appendix E » Caiculation Sheets for Steady-State Cable Ratings 405

E.11.4 Ojil-filled Cables with Round Conductors and Round Oil Ducts Between Cores

(1) Metallized paper insulation shield:

THA»()
0358 2t;
1h = K - m/W
(2) Metal tape insulation shield:

T, = 0.359| 0.923 — lj
d,.ar 2tj

= K - m/W
F.11.5 SL Type Cables

In SL type cables, the lead sheath around each core may be assumed isothermal. The
thermal resistance 7; is calculated in the same way as for single-core cables.

E.12 THERMAL RESISTANCE OF PIPE-FILLING MEDIUM

Installation Condition U V Ve

Gas pressure cable in pipe 0.95 0.46 0.0021

Oil pressure pipe-type cable 0.26 0.0 0.0026

For three cables in a pipe, D, = 2.15 x single-cable external diameter.

U
(fae
P0102 Ye,,)D;

Th = K-m/W

E.13 JACKET THERMAL RESISTANCE

E.13.1 SL TypeCables
G is the geometric factor and is obtained from the figure below.
406 Part IV » Appendixes

0.7
Sheaths touching

0.6

0.5

0.4

0.3

Equal thicknesses of material

between sheaths and between
0.2 sheaths and armor

0.1

0
0 0.05 0.10 0.15

Thicknesses of material between sheaths and

armor expressed as a fraction of the outer
diameter of the sheath

E.14 SERVING THERMAL RESISTANCE

Appendix E # Calculation Sheets for Steady-State Cable Ratings 407

For pipe-type cables:

bv K - m/W

E.15 EXTERNAL THERMAL RESISTANCE OF BURIED CABLES

For buried cables, two values of the external thermal resistance are calculated: 74 corre-
sponding to dielectric losses (100% load factor), and 74,,—the thermal resistance corre-
sponding to the joule losses, where allowance is made for the daily load factor (LF) and
the corresponding loss factor ju.

Oe (Mc OgerOL.
F)e

w=

The effect of the loss factor is considered to start outside a diameter D, defined as D, =
61 200,/65 (length of cycle in hours) where 6 is soil diffusivity (m*/h). For a daily load
cycle and typical value of soil diffusivity of 0.5 - 10~° m?/s, D, is equal to 211 mm (or 8.3
in). The value of D, is valid even when the diameter of the cable or pipe is greater than D,.

F.15.1 Mutual Heating Effect

A factor F accounts for the mutual heating effect of the other cables or cable pipes in a
system of equally loaded, identical cables or cable pipes. The distances needed to compute
factor F are defined in the diagram below. These are center-to-center distances.

1'-W

Air
Ground

Cable No. 1
408 Part 1V m Appendixes

For cable p:

ji =
There are (gq— 1) terms, with the term dy/ Appexcluded. The rating of the cable system is
determined by the rating of the hottest cable or cable pipe, usually the cable with the largest
ratio L/D,. For a single isolated cable or cable pipe, F = 1.

E.15.2 Single-conductor Cables

When the losses in the sheaths of single-core cables laid in a horizontal plane are
appreciable, and the sheaths are laid without transposition and/or the sheaths are bonded at
all joints, the inequality of losses affects the external thermal resistance of the cables. In
such cases, the value of the F factor used to calculate 74,, is modified by first computing
the sheath factor (SHF):

1 +0.5(A), +A49)
(eReWeg
Ya
lib At m

(SHE) =
and then calculating

yeu= FOSHF)

nS

(1) Equally loaded similar cables:

Directly buried cables or cable pipes.
For cables in ducts, use D, in place of D, in the following formulas.

Tees Ee 4L-F
20 Ds

co K-m/W

Tay a (inD. In D
Appendix E m Calculation Sheets for Steady-State Cable Ratings 409

(2) Cables or cable pipes buried in thermal backfill:?

ie K - m/W

Le De LWbed N
Ty,
LL = —
on (In—-+
( D. pln D, + LLuU—(pPe
on Pe — Pe) P-)G
b

T4, = K-m/W

(3) Cables in ducts:

Installation Condition U V ve

In metallic conduit SE? 1.4 0.011

In fiber duct in air S19 0.83 0.006
In fiber duct in concrete Sy 0.91 0.010
In asbestos cement
duct in air 5 iL? 0.006
duct in concrete Sy? itil 0.011
Earthenware ducts 1.87 0.28 0.0036

3 Valid for y/x <3.

410 Part IV m Appendixes

U
if ee ee
AS ili. (Ii Varia
YOn) De

Le K- m/W

i Ud pee
p eee
/D),
Sec yay
p is the thermal resistivity of duct material.
For metal ducts, 7,’ = 0.

is) mm

my Pc 4L =F N
GS er Or,
IU U

Ty"= K - m/W

ow 7 N és
D} (i D Lae (2 b

TBE— K ,m/W
Appendix E m Calculation Sheets for Steady-State Cable Ratings 411

TTA eT!

in K - m/W

Tay= TTY + Teh

1, = K - m/W

E.15.3 UnequallyLoaded or Dissimilar Cables

In this case, heating between cables is accounted for by calculating the temperature
rise at the surface of the cable being studied caused by other cables in the same group. To
do this, the losses in each cable must first be estimated; these estimates are amended as
necessary by the results of the calculation.
For cable j, the losses are

W;=n[I?Ri(1 +Ay+A2)uj + Was]

W; = W/m
The thermal resistance between cable j and cable i, the cable being studied, is
for directly buried cables

Tj = K-m/w

aed eee
Ty=>_In=—+ (Ps—pe)Go
Tj = K-m/W

The temperature rise at the surface of cable i due to the losses in cable j:

8) =W;
Ty|
Ab;; = ce
412 Part 1V m Appendixes

The temperature rise at the surface of cable i due to all other cables in the group:

N
Aéin= ye AGij
oa
fs

AGin = cS

E.16 EXTERNAL THERMAL RESISTANCE OF CABLES IN AIR

Cables with jackets or other nonmetallic surfaces should be considered to have a black
surface. Plain lead or unarmored cables should be assigned a value of h equal to 80% of
the value for a cable with a black surface.

Material oO

Bitumen/jute serving 0.8

Polychloroprene 0.8
PVC 0.6
PH 0.4
Lead 0.6

E.16.1 Heat Transfer Coefficient

t= W/m?-°C>/4
Appendix E # Calculation Sheets for Steady-State Cable Ratings 413

Installation Vb, E g Mode

1 Singlecable* 0.21 3.94 0.60 =P 203De

2 Twocablestouching,horizontal 0.21 2.35 0.50 =P rt 2 0.5 D,

2()5) 10);

3 Threecablesin trefoil 0.96 1.25 0.20 —> \t-20.5DF

20.5 D;

4 Threecablestouching,horizontal 0.62 95 0.25, —> +2 0.5 D*

20.5 D;

5 Twocablestouching,vertical 1.42 0.86 0.25 ]P i+ -205 Ds

6 TwocablesspacedD&,vertical 0.75 2.80 0.30 Pt 20 5D)

| One,

YO A

Wf Threecablestouching,vertical 1.61 0.43 0.20 —>| |< SOD e

21.0D;
|

8 ThreecablesspacedD;, vertical 1.31 2.00 0.20 —P }t-21.0D7

9 Singlecable 1.69 0.63 0.25

10 ~~‘Threecablesin trefoil 0.94 0.79 0.20 2

*Values for a “single cable” also apply to each cable of a group when they are spaced horizontally
with a clearance between cables of at least 0.75 times the cable overall diameter.
414 Part IV m Appendixes

a D*h T|
K,A = ———
eels | — + ( (14+A)
ia Bn 4+0+a rl+22F|
2)T3

ei
(1) Shielded from solar radiation:

Ad W 1 l\ 7 nNarT>
OE | TSCA Rey"TEX
AO te

1/4 Ad==A@q ue
CAGE) gta reer rere
1+ Ka (A6,),/

(Ag 1/4 = oAl/4

cu

Settheinitialvalueof (A6,)'/+= 2. Reiterateuntil(AQ), _ (A@,)

1/4< 0.001

~ 7 D*h(Ad,)'/4

T= K - m/W

(2) Exposed to solar radiation:

(AG,)/4
=[Pastel talk)
S/n+]1+K,(A6,)}/"
Aen=
1/4 “Cus 1/4
Set the initial value of (A6,)'/* = 2. Reiterate until (A@,))/,, —(A@,)14 < 0.001

3 D*h (A6,)'4

(i K - m/W

Ambient temperature should be increased by A@,,:

AGs,= o DHT,

AG, = Aw:
Appendix E m Calculation Sheets for Steady-State Cable Ratings 415

E.17 AMPACITY

E.17.1 Buried Cables

ne A@ — Wa [0.57 +n(Tr + T3+Ts)]— A%m 7°

RT, +nR(1 +A,)To +nRC + Ay + A2)(T3 + Tay)

E.17.2 Cables in Air

a KO WAST nCleiels aly eAOF Is

RT, +AR(1 + A172 HARI Ay Ag) 3 + 1)

l= A

E.17.3 Temperature Rise of Cable Components

buried cable

| AO,
=AGint
+72
{[WeC
+Aj+A2)
(3+Ta)+Wal3
+ro}|
AO, = te

cable in air

| AO,
=Abs,
+m{{Wel
+At+A2)
+Wal
(3+rT»)
|
Nove 1c

| Aé,
=AOy
+n[WeU
+A1)
+Wal
Tr,
|
AO, = =C

| ROR
AOeA
Wetct
0sWa)T
|
= eC
 i - * 06)! a
‘.& ae ‘ dé pee ar!
Q nen ey a

had
| ake
eh—
Tt&
sasinn
cloclip
@etnO
GRA‘AWw
as
ry —_——

Acitr=@ perapimeangre
%
 =
Differences between the
Neher/McGrath and
IEC 287 Methods

The two methods are, in principle, the same, with the IEC method incorporating several
new developments which took place after the publication of the Neher/McGrath (NM)
paper. Similarities in the approaches are not surprising since, during the preparation of
the standard, Mr. McGrath was in touch with the Chairman of Working Group 10 of IEC
Subcommittee 20A (responsible for the preparation of ampacity calculation standards).
The major difference between the two approaches is the use of metric units in IEC 287
and imperial units in the NM paper (the same equations look completely different because
of this). Even though the methods are similar in principle, the IEC document is more
comprehensive than the NM paper. IEC 287 not only contains all the formulas (with minor
exceptions listed below) of the NM paper, but in several cases, it makes a distinction between
different cable types and installation conditions where the NM paper would not make such
a distinction. Also, the constants used in the IEC document are more up to date.
The following is a list of the most important differences between the two approaches:

Load Factor

1. The Neher/McGrath paper considers a nonunity load factor (see Section 9.6.7),
whereas the IEC 287 document assumes a unity load factor. Another IEC docu-
ment (IEC 853, 1985, 1989) deals with cyclic and emergency ratings (see Chap-
(iB?Dy)
Circulating and Eddy Current Losses
2. Equation 30 in the Neher/McGrath paper for the eddy current effect of single-
conductor cables with single-point bonding applies only when the cables are ar-
ranged in an equilateral configuration. The IEC document, in addition to the

417
418 Part IV m Appendixes

equilateral configuration, provides formulas for calculating the eddy current ef-
fect in the more usual flat configuration. In addition, the IEC document considers
separately two- and three-core cables with steel tape armor which are not discussed
in the NM paper.
. Equation 27 in the Neher/McGrath paper for the circulating current effect of single-
conductor cables with two-point bonding applies only when the cables are arranged
in an equilateral configuration. The IEC document, in addition to this configura-
tion, provides formulas for calculating the circulating current effect in the more
usual flat configuration with and without transposition. In addition, the IEC docu-
ment accounts for the effect of variation of spacing of single-core cables between
sheath bonding points. The NM paper refers the reader to the Simmons (1932)!
paper for computation of circulating current losses for cables in ducts.
. For cables with large segmental-type conductors and sheaths bonded at both ends,
IEC 287 provides an expression for eddy current computations. In the NM method,
this contribution is ignored.
. Calculation of losses in magnetic armor is treated only qualitatively in the NM pa-
per with references to the literature for complex computational methods. Relevant
approximations are proposed in IEC 287.
Calculation of Thermal Resistances

6. IEC 287 gives analytical expressions for the computation of the geometric factor of
three-core cable insulation, whereas the NM paper makes a reference to the paper
by Simmons (1932). In addition, the IEC document differentiates between various
cable constructions, e.g., belted versus screened three-core cables, oil-filled, SL
type, and so on. The NM paper does not provide this information.
. The values of thermal resistivities specified in the NM paper are outdated (in view
of the research which was carried out after the publication of the paper). Also,
several new insulating materials are not listed.
. The external thermal resistance of cables in air is somewhat more accurately
computed in the IEC method (both methods are similar, the major difference being
the formula for computation of the cable surface temperature). In the NM paper,
the approximation is used that the heat transfer coefficient due to convection is
independent of the cable/duct surface temperature (see Section 9.6.8.5), whereas
the IEC document provides an iterative method for evaluating this coefficient as
a function of the cable surface temperature rise (a more accurate assumption).
Also, the IEC method distinguishes between various arrangements for cables in
air, whereas the NM paper does not.
. The NM paper considers the effect of wind on cable ampacity (see Section 9.6.8.5),
whereas the IEC document assumes the worst case scenario with no wind. In this
book, a general method of dealing with the effect of the wind in discussed in
Chapter 10.

. The IEC document distinguishes between trefoil and flat configurations for the
computation of the external thermal resistances. The NM paper uses one formula
only, which in reality is the same as the flat configuration formula in the IEC

| See references at the end of Chapter 9.

Appendix F » Differences between the Neher/McGrath and IEC 287 Methods 419

document. Formulas for the calculation of the external thermal resistance of

cables in touching configurations are provided in the IEC document, and not in
the NM paper.

11. The treatment of different cables types or unequally loaded cables in one installa-
tion is discussed in detail in the IEC document, and only qualitatively in the NM
paper.
12. Consideration of the drying up of soil in the vicinity of loaded power cables is
included in the IEC document, but not considered in the NM paper.

Emergency Ratings
13. The NM paper provides a formula for this rating. A corresponding formula is
given in IEC 853-2 (1989).
In this book, the IEC 287 and 853 approaches are used as a basis for the presentation.
The material which is included in the NM paper and not covered by the IEC documents is
also discussed (see, for example, several sections in Chapter 9).
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 Index

A magnetic, 175
magnetic field, 145
Absorption coefficient of solar radiation, 64,
materials, 172
243, 239
nonmagnetic, 172
Absorptivity, 25, 267
thermal capacitance, 256
AC resistance of conductors, 115, 118
Armor losses, 15
effect of harmonics, 129
Attainment factor for the cable outer surface,
current derating factor, 130
70, 78, 91, 92
maximum values (AEIC), 127
maximum values (IEC 228), 126
proximity effect. See Proximity effect
skin effect. See Skin effect Belted cables, 200, 205
AC/DC resistance ratio, 130 Bonding arrangements, 12, 138
Air properties, 225, 377 cross-bonded systems, 13, 139
Ampacity calculations single point bonded systems, 13, 138
history, 16 two-point bonded systems, 13, 138
Angular frequency, 150 Bonding arrangements. See also Sheath:
Angular time delay, 150 bonding arrangements
Armor, 8 Boundary conditions, 295
ac resistance, 175
C
bedding thermal resistivity, 209
circulating current loss factor, 175, 177 Cable bundle, 278
circulating current losses, 8, 13, 152 Cable capacitance, 340
eddy current loss factor, 177 Cable components, 4
equivalent thickness, 189 Cable conductors: economic sizing, 329
hysteresis losses, 8, 152, 176 Cable impedances, 150
length of lay, 173 average reactance, 161
loss factor definition, 58 armor-armor, 151

421
 Index
422

Cable impedances (Continued ) resistance, 117, 118, 334

conductor-armor, 151 skin effect, 119, 334

conductor-conductor, 150 thermal capacitance, 255
conductor-sheath, 151, 157 Conductor temperature rise, 58

sheath-armor, 151 Conformal transformation, 234

sheath-sheath, 143, 149, 151, 158 Convective heat transfer coefficient, 25, 239,
Cable inductances, 141 BNO)OD, AE
armor, 149 Cost of losses, 330
armor self, 143 Cradle configuration of cables in pipe, 178, 364
conductor self-inductance, 141 Critical degree of saturation, 62
conductor-conductor, 142, 147 Critical moisture content, 62
conductor-sheath, 143, 148 Critical temperature rise, 60, 62, 97
nonmagnetic armor, 149 Current distortion limits, 131
Cable installations, 9 Current rating equation
cables in air, 64
covered trays, 280
duct banks, 10, 229 with moisture migration, 62
fire protected trays, 282 without moisture migration, 59
flat configuration, 10 Cyclic loading, 237
horizontal tunnels, 283 Cyclic rating, 68, 87
in air, 11 McGrath’s approach, 88
riser poles, 267 Cyclic rating factor, 88
trays, 11, 277 groups of identical cables, 93
trefoil configuration. See triangular with moisture migration, 99
configuration without moisture migration, 92, 95
triangular configuration, 10
D
troughs
filled, 289 DC resistance of conductors, 116
unfilled, 291 stranded conductors, 117
tunnels and shafts, 12, 283 Dielectric constant. See Relative permittivity
underground cables, 9 Dielectric loss factor, 110
vertical shafts, 288 Dielectric losses, 7, 16, 109, 265, 341
Capacitance of the insulation, 110 cables in air, 75
Capacitive (charging) current, 109 Diffusivity of soil, 78
Charging current, 7, 109, 342 Distorted resistance ratio, 130
Circulating current Dry zone, 60
in armor, 152, 172
E
in parallel cables, 163
loss factor, 140, 141, 156, 157, 159, 175 Economic conductor size. See Conductors
SL-type cables, 162, 177 Eddy current
pipe-type cables, 162, 178 loss factor, 140, 177, 179, 182, 185
losses, 137, 139, 140 3-core armored cables, 188
effect of spacing, 161 3-core unarmored cables, 188
Concentric neutral wires, 7 losses, 137
Conductor losses, 15 Effective thermal conductivity, 223
Conductors, 5 Electrical resistivity, 116
conductor constructions, 5 Emissivity, 25, 239, 243, 269
economic size, 334 Emergency rating current
geometric mean radius, 164 groups of circuits, 103
large segmental eddy current, 189 single circuit, 102
mean temperature, 332 Energy balance equations, 26, 28, 239, 265
proximity effect, 123, 334 cables in air, 30, 283
Index 423

Energy conservation principle, 242, 264, 267 Heat transfer

Exponential integral, 73 conduction, 23, 224
External covering, 8 convection, 24, 266, 295
External thermal resistance. See Thermal forced, 24, 251
resistance in the air gap, 273
in the riser pole, 273
F mixed, 272
natural, 24, 223, 250
Fictitious diameter, 238
radiation, 25, 225, 267
Filament heats source simulation (f.h.s.s.)
rate equations, 25
method, 190, 200
Heat transfer coefficient, 36, 240
Finite-difference method, 312
Heat transfer equations
approximation of differential equations, 312 cylindrical form, 29
backward-difference approximation, 313
rectangular form, 28
central difference approximation, 313 solution methods, 31
forward-difference approximation, 313 Heat transfer rate, 25, 265
Finite-element method, 200, 236, 298 High-pressure fluid-filled cable. See Insulation
approximating polynomials, 300
types
area coordinates, 300
Hysteresis loss, 150
computer implementation, 309
element capacity matrix, 305
element conductivity matrix, 305
element heat generation vector, 305 Image cable, 74, 212
elements, 299 Impedance. See Cable impedances
equations, 303 Inductance. See Cable inductances
Forced convection heat transfer coefficient, 253 Insulation, 6, 109
Free convection. See Heat transfer: natural insulation types, 7
convection high-pressure fluid-filled cables, 7, 364
Forced cooling of cable circuits, 13 low-pressure fluid-filled cables, 7, 365
Fourier’s law, 24, 27, 30, 34, 39 paper-polypropylene-paper, 7, 366
Fundamental frequency, 130 losses in, 16, 109
thermal capacitance, 255
G Integral-equation method, 200, 220
Intensity of solar radiation, 64, 239
Geometric factor, 199 Internal thermal resistance, 240
extended values, 232, 312
transient analysis, 232
J
Grashoff Number, 250
Joule losses, 265
H conductor, 115
screens, sheaths and armor, 137
Harmonics. See AC resistance of conductors:
effect of harmonics
K
Heat flux, 24
Heat sources in power cables, 13 Kennelly hypothesis, 74, 212, 213
armor and pipe losses, 15 Kinematic viscosity, 225
current-dependent losses, 14
conductor losses, 15 L
sheath losses, 15
voltage-dependent losses, 16 Ladder network, 40, 57, 371
charging current losses, 16 two-loop circuit, 46
dielectric losses, 16 Lagging phase, 159, 186
 Index
424

Law of conservation of energy, 26 P

Lay-length factor, 117
Paper-polypropylene-paper cable. See
Length of lay, 117
Insulation types
armor, 144
Parallel cables, 163
Load factor, 68, 331, 336
Peripheral strands, 122
Longitudinal voltage drop, 164
Pipe-type cables, 10, 162, 211, 364
Loss factors, 151, 331
thermal capacitances, 256
armor circulating loss factor, 152, 171, 175,
Poynting’s Theorem, 144
Phos MGT
Poles and zeros of a transfer function, 71
armor hysteresis loss factor, 152
Prandtl number, 224, 225, 250, 377
sheath and armor, 193
Preloading conditions, 68
sheath circulating loss factor, 152, 157, 159,
Present value, 332
162, 163, 166, 171, 175
Principle of superposition, 78
Low-pressure fluid-filled cables. See Insulation
Properties of air, 377
types Proximity effect, 123, 334
Lumped capacitance method, 39
pipe-type cables, 124
proximity effect factor, 119

Magnetic armor. See Armor: magnetic

Magnetic permeability of free space, 147 Radiation heat transfer rate, 36
Magnetic permeability. See Relative Radiation shape factor, 269, 284
permeability Radiative heat transfer coefficient, 250
Mixed convection, 272 Rating equations. See Current rating equation
Model cables, 361 Rayleigh number, 224, 250, 279
Moisture content, 78 Reflectivity, 269
Moisture migration, 60, 61 Relative permeability
Multiple cables per phase. See Parallel cables longitudinal magnetic, 146, 150
Mutual heating effect, 246 transverse magnetic, 146
Mutual radiation area, 269, 284 Relative permittivity, 110
Representation of capacitances of the
dielectric, 41
Representation of the dielectric for
Natural convection coefficient, 250, 265
three-conductor cables, 43
Neher—McGrath method, 226, 230, 232, 237,
Residual voltage, 158
253, 417
Resistive (leakage) current, 109
Newton’s law of cooling, 25, 35, 265
Response to a step function, 69
Nonmagnetic armor. See Armor: nonmagnetic
Reynolds number, 251
Numerical methods, 296. See also finite-
Riser pole. See Cable installations
element and finite-difference method
RMS diameter, 162
modeling and computational issues, 324
selection of the time step, 297 S
size of the region, 297
Nusselt number, 274 Screen resistance, 163
Screening factor, 206
0 ~ Section length, 161
Sheath, 7
Ohm’s law bonding arrangements, 138
thermal analogy, 40 single cross bonded systems, 138
thermal equivalent of, 35 single point bonded systems, 138
Index 425

circulating current loss factor, 152, 152, 157, Temperature rise, 214
1ISORLG2Z IGS a66mliie7 5, 179 Thermal backfill, 10, 230
multi-core armored cables, 163 equivalent radius, 230
parallel cables, 166 Thermal capacitance, 254
pipe type cables, 162 armor, 256
corrugated, 199, 211 concentric layers, 256
current, 156, 158 conductor, 255
eddy current, 180 definition, 38
eddy current loss factor, 182, 185 for a coaxial configuration, 40
3-core armored cables, 188 insulation, 255
3-core unarmored cables, 188 oil in the conductor, 254
loss factor definition, 58 pipe type cables, 256
reactance. See Cable impedances reinforcing tapes, 256
reinforcement, 162 sheath, 256
residual voltage, 158 Thermal capacity, 28
thermal capacitance, 256 Thermal coefficient of expansion, 225
voltage, 157 Thermal diffusivity, 297
Single solidly bonded systems, 138 Thermal resistance, 59
Sheath currents, 151 between sheath and armor, 209
Sheath losses, 15 SL type cables, 210
eddy current losses, 15 cables in ducts and pipes, 222
sheath circulating losses, 15 cables with shaped conductors, 205, 207
Skin effect, 119, 336 definition, 34
circular conductors, 120 for conduction, 35
large aluminum conductors, 122 for convection, 35
large segmental conductors, 120 for radiation, 35
oval shaped conductors, 120 derating factors, 220
pipe-type cables, 124 equivalent cable, 59
skin effect factor, 119 external, 211, 289
tubular conductors, 120 cables in air, 239, 240
SL type cables, 162, 177, 189, 209, 210 derating factors, 245
Soil diffusivity, 238 cables in duct banks/backfills, 229
Soil thermal conductivity, 60 cables in ducts and pipes, 222
Solar absorption coefficient, 239 conformal transformation, 234
Solar declination angle, 270 cyclic loading, 237
Solar radiation, 243, 267 effect of wind velocity, 250
intensity of, 270 extended values of geometric factor, 232
Solar time, 270 geometric factor for transient analysis,
Solenoidal field, 144 232
Specific heat of materials, 198 groups of buried cables
Steel pipes not touching, 214, 215
losses in, 177 touching, 217, 218, 219
Stefan—Boltzmann constant, 25, 239 multiple soil layers, 234
Stefan—Boltzmann law, 25 pipe/duct, 228
Submarine cables, 8, 138 single buried cable, 212, 213
Superposition principle, 33 three single core cables
touching, 218, 219
T insulation
Tan delta. See Dielectric loss factor extruded cables, 202
Temperature coefficient of resistance, 116 geometric factor, 199
426 Index

Thermal resistance (Continued ) Transient temperature rise

single core cables, 199 variable loading, 78
SL-type cables, 209 Transient rating
three-core cables, 199, 206 groups of cables, 76
three-core screened cables, 206 Tunnels. See Cable installations
internal, 240 Two-zone model of the soil, 60, 97
oil-filled cables, 207
pipe/duct, 228 V
pipe/duct filling medium, 222, 227
Voltage dependent losses, 16
pipe-type cables, 211
Van Wormer coefficient, 41, 255
serving, 210
dielectric loss, 46
Thermal resistivities of materials, 198
long-duration transients, 43
Thermal resistivity, 24
short-duration transients, 44
of filler, 202
Variable cable spacing, 16
Thermal time constant, 39
Viscosity, 224
Thermal stability of the soil, 64
Volumetric thermal expansion coefficient, 224
Thermosyphon, 274
Total harmonic distortion, 130, 133
W
Total heat transfer coefficient, 36
Total transient temperature rise, 77 Wind velocity
Transfer function, 70, 72, 371 effect of. See Thermal resistance: external
Transient ratings, 68
 About the Author

Employed by Ontario Hydro since 1975, George J. Anders is presently a Principal Engi-
neer/Scientist in the Electrical Systems Technology Unit of Ontario Hydro Technologies.
For many years, Dr. Anders has been responsible for Ontario Hydro’s development of power
cable calculation methods and tools.
Throughout his 22 years with Ontario Hydro, Dr. Anders has been involved in several
aspects of power system analysis and design. His principal activities have been concentrated
in three areas: (1) ampacity computations of electric power cables, (2) the application of
probability methods in power system analysis and design, and (3) the application of novel
techniques in electric power utility practice. He is the author of a book and has written over
70 papers published in several international journals. He has been conducting seminars
on power cable ampacity issues in Canada and the United States as well as in Rio de
Janeiro, Brazil; Warsaw, Poland; Bogota and Cali, Colombia; Porto, Portugal; Sydney and
Melbourne, Australia; Santiago de Chile; and Hong Kong.
In the field of thermal analysis of electric power cables, Dr. Anders has made major
contributions in three areas: (1) the development of computational techniques using the
finite-element method to evaluate heat and moisture transfer in the vicinity of loaded power
cables, (2) the development of optimization techniques for selection of the most economic
conductor sizes, and (3) the development of new algorithms for transient ratings of buried
power cables. He has published over 20 papers on the subjects dealing with thermal analysis
of underground systems. As recognition of his work in this field, he received a New
Technology Award from Ontario Hydro in 1990.
Dr. Anders is a Canadian representative in Working Group 10 (ampacity computations
of power cables) and WG15 (short circuit temperatures of power cables) of the Interna-
tional Electrotechnical Commission. These working groups develop new computational
techniques and new standards for power cable ampacity computations. He has also been
a project leader on a number of projects dealing with ampacity computations sponsored
by the Canadian Electrical Association. In the course of these projects, a series of highly

427
428 About the Author

successful computer programs were developed for CEA. These programs are in use by over
200 institutions in 33 countries on 5 continents.
In the field of application of probability methods in power system engineering, Dr.
Anders has been involved in developing new methods and applications of probabilistic
techniques to power system problems since 1975. He has published over 50 papers dealing
with various topics on probability and optimization applications. Some of the problems
he has been working on involved the development of mathematical techniques to model
operator’s action in probabilistic load flow analysis; anew technique to assess the importance
of measures in power system reliability studies; anovel approach to frequency and duration
analysis and uncertainty considerations for radial and two interconnected systems; a new
method to evaluate the frequency of severe power system faults; and models to represent
human failure in reliability analysis. He recently has been a project leader of several large
projects on the probabilistic estimation of the remaining life of electrical equipment.
For several years, Dr. Anders has been teaching a course in the Faculty of Applied
Science and Engineering at the University of Toronto on the application of probability meth-
ods in engineering. His book, Probability Concepts in Electric Power Systems (Wiley, New
York, 1990), is well recognized around the world as a unique reference on the application
of probability methods in power system planning, design, and operation. Dr. Anders is a
Canadian representative in CIGRE WG 37.06.11 whose task is to develop methods for the
reliability assessment of interconnected power systems. He is also a member of the IEEE
Task Force on the Impact of Maintenance Strategies on Power System Reliability.
Dr. Anders also has been involved in developing new applications of optimization
methods and novel techniques in power system problems. He started by developing an in-
terval programming technique for application in reliability studies. Later, he was involved
in developing models for optimal economic power transfer. He was also involved in pub-
lishing new techniques for the selection of the most economic cable sizes. Recently, he
developed a procedure for the optimal construction of rigid-bus stations. He is also working
on the application of multiobjective decision models for selecting an optimal maintenance
strategy for power equipment. He has been a project leader of a large project undertaken
by Ontario Hydro Technologies to develop procedures, methods, and computer tools for
decision support and risk assessment.
Dr. Anders received the Master’s degree in electrical engineering from the Technical
University of Lodz, Poland in 1973, and the M.Sc. degree in mathematics and Ph.D. degree
in power system reliability from the University of Toronto in 1977 and 1980, respectively.
He is a Registered Professional Engineer in the Province of Ontario and a Senior Member
of IEEE.
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RATINGOF ELE
Ampacity
Computatio
andIndustrial
Application.
IEEEPressPower EngineeringSeries
Dr.Paul M.Anderson,SeriesEditor
RATINGOF ELECTRIC POWERCABLESis the firstand only book d
theory and practice of computing the maximum current a power
overheating. This comprehensive reference clearly describes
of cable ratings, extensively illustrating their application in f |
installations. Based on the full range of cable designs in use today, ea
with numerical examples that explain the various concepts discussed. To
* How to select cable sizes that require minimuminvestment and op:
* General theory of heat transfer 2
* Computation of the parameters required in rating equations
* Specialized applications and advanced computational procedures
Complete with calculation sheets that can be used as templates for rating power cables in the
most common installations, this book will be an invaluable tool for electrical engineers, students,
and researchers.

ABOUT THE AUTHOR

George J. Anders has been employed by Ontario Hydro since 1975. He is presently a Principal
Engineer/Scientist in the Electrical Systems Technology Unit of Ontario Hydro Technologies. For
many years, Dr. Anders has been responsible for Ontario Hydro’s development of power cable
calculation methods and tools.
Dr. Anders has been involved in several aspects of power system analysis and design, including
ampacity computations of electric power cables, application of probability methods in power -
system analysis and design, and the application of novel techniques in electric utility practice.
Iie is the author of more than 70 papers published in international journals. Dr. Anders has been
conducting seminars on power cable ampacity issues in Canada and in the United States, as well
as in Brazil, Poland, Colombia, Portugal, Australia, Chile, and Hong Kong.

ABOUT THE IEEE PRESS POWER ENGINEERING SERIES

The IEEE Press Power Engineering Series is devoted to providing comprehensive coverage of the
field, including the design, operation, and analysis of power systems. Created expressly for use
by power engineers and engineering students, this series offers extensive complementary coverage
of both theory and practical applications.

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