KWAME NKRUMAH UNIVERSITY OF SCIENCE AND

TECHNOLOGY, KUMASI

W
SANE NO B A

DEPARTMENT OF ELECTRICAL & ELECTRONIC

ENGINEERING
B.Sc. Electrical and Electronic Engineering, 1

EE 162: ELECTROMECHANICAL ENERGY

CONVERSION AND TRANSFORMERS
Credit: 3.

P. Y. OKYERE
SENIOR LECTURER
KNUST
SEPTEMBER 2012 (REVISED EDITION)
Contact Address

Dr. Philip Yaw OKYERE

Department of Electrical & Electronic Engineering
College of Engineering
Kwame Nkrumah University of Science and Technology
Kumasi, Ghana

Phone: 0322063952
0208124340

E-mail: okyerepy@yahoo.com
pyokyere.soe@knust.edu.gh

ii
Course Author

Dr. Philip Yaw Okyere graduated with a B.Sc. (First Class Honours) degree in electrical
engineering in 1979. After one-year national service in the Department of Electrical & Electronic
Engineering as Teaching Assistant, he proceeded to L’Institut National Polytechnique de
Grenoble (INPG), France to pursue postgraduate studies in electrical engineering on a
scholarship from the French Government. He graduated with DEA (“Mention: Bien”) and
Docteur-Ingenieur degrees in 1982 and 1985 respectively. He returned to Ghana to join the
Department of Electrical & Electronic Engineering of KNUST as a Lecturer in 1985 and was
promoted to Senior Lecturer in 1995.

Dr. Philip Yaw Okyere has taught courses in Electrical Machines, Electric Drives, Power
Electronics and Power Systems and has rendered professional services to industry and the
University. He has produced publications in diverse areas, including power system modelling,
lightning protection, earth electrode resistance enhancement, application of artificial neural
networks to power systems, and renewable energies. His current research interest is in
integration of renewable energy sources into power systems.

iii
Course Outline

EE 162 Electromechanical Energy Conversion and Transformers (3 0 3)

Principles of Electromechanical Energy Conversion:

Force and torque as rate of change of energy with position. Basic Transducers: Single
Excitation, Alignment Force and Torque; Double Excitation, Alignment and Interaction Forces
and torque

Transformers:
Construction; Basic theory; Phasor Diagram; Equivalent Circuits; No-load and Short-Circuit
Tests; Voltage Regulation; Efficiency; Cooling methods; Polarity; Polyphase transformer
Connections; Per-Unit Calculation; Parallel Operation of Transformers; Auto transformers;
Tap-Changing transformers; Instrument Transformers.

Recommended Textbooks

1. T. Wildi (1988): Electrical Power Technology, Prentice Hall Professional Technical

Reference.

2. John Hindmarsh (1977):Electrical Machines and their Applications, Pergamon Press, Oxford

3. Robert Stein, William T. Hunt, Jr. (1979): Electric Power System Components, Transformers
and rotating Machines, Van Nostrand Reinhold Company, New York

4. Richard A. Pearman (1994): Electrical Machinery & Transformers, Saunders College

Publishing, Harcourt Brace College Publishers, New York.

5. B. A. Theraja, A. K. Theraja (2005): A Textbook of Electrical Technology, S. Chand &

Company Ltd, New Delhi

iv
Contents
Contact Address............................................................................................................................................ ii
Course Author .............................................................................................................................................. iii
Course Outline.............................................................................................................................................. iv
Recommended Textbooks ............................................................................................................................ iv
CHAPTER ONE ...............................................................................................................................................1
TRANSFORMERS............................................................................................................................................ 1
1 Introduction .......................................................................................................................................... 1
2 Principle of operation of a transformer................................................................................................ 1
3 Polarity and terminal markings of a transformer ................................................................................. 2
4 Ideal transformer .................................................................................................................................. 2
4.1 Properties of an ideal transformer .............................................................................................. 2
4.1.1 Emf equation and voltage ratio (see Fig. 3) .......................................................................... 2
4.1.2 Current ratio and power equation........................................................................................3
4.1.3 Phasor diagram of an ideal transformer (Fig. 4) ................................................................... 4
4.1.4 Impedance ratio (see Fig. 5).................................................................................................. 5
4.1.5 Equivalent circuits of an ideal transformer........................................................................... 5
5 Practical single-phase transformer ....................................................................................................... 7
5.1 Hysteresis loss .............................................................................................................................. 7
5.2 Eddy-current loss.......................................................................................................................... 7
5.3 Phasor diagram on no load .......................................................................................................... 9
5.4 Mutual and leakage fluxes in a transformer (see Fig. 9)........................................................... 10
5.5 Equivalent circuit of a practical transformer.............................................................................11
5.6 Equivalent circuit referred to the primary side (Fig. 12)........................................................... 12
5.7 Equivalent circuit referred to the secondary side (Fig. 13).......................................................13
5.8 Approximate equivalent circuits................................................................................................13
5.9 The complete phasor diagrams for loaded conditions .............................................................14
5.10 Rating of transformers ............................................................................................................... 15
5.11 The turns ratio............................................................................................................................ 15
5.12 Definition of per-unit impedances.............................................................................................17

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 5.13 Voltage regulation......................................................................................................................17
5.14 Transformer output....................................................................................................................21
5.15 Efficiency..................................................................................................................................... 21
5.16 All-day efficiency ........................................................................................................................23
5.17 Open-circuit and short-circuit tests on a transformer ..............................................................24
5.17.1 Short-circuit test ................................................................................................................. 24
5.17.2 Open-circuit test or no load test......................................................................................... 25
5.18 Construction of transformers .................................................................................................... 28
5.19 Polarity tests...............................................................................................................................28
5.19.1 Using a low-voltage ac source.............................................................................................29
5.19.2 Using dc source ...................................................................................................................29
5.20 Parallel Operation of single-phase transformers ...................................................................... 30
5.21 Load sharing of parallel-connected transformers..................................................................... 31
6 Three-phase transformers .................................................................................................................. 33
6.1 A three-phase transformer bank ...............................................................................................33
6.2 A three-phase transformer unit.................................................................................................34
6.3 Winding arrangement ................................................................................................................ 35
6.3.1 Star connection ...................................................................................................................35
6.3.2 Delta connection................................................................................................................. 35
6.3.3 Zigzag (or interconnected star) connection........................................................................35
6.4 Vector groups ............................................................................................................................. 35
6.5 Three-phase transformer connections ......................................................................................37
6.5.1 Delta-Delta connection (Fig 29.a) .......................................................................................37
6.5.2 Delta-Star connection (Fig. 29.b) ........................................................................................37
6.5.3 Star-Delta connection .........................................................................................................38
6.5.4 Star-Star connection ...........................................................................................................38
6.5.5 Star-Zigzag connection........................................................................................................38
6.6 Parallel operation of three-phase transformers (Fig. 30) ......................................................... 42
7 Cooling methods .................................................................................................................................43
7.1 Air Cooling (Dry type transformers) .......................................................................................... 44
7.1.1 AN........................................................................................................................................ 44
7.1.2 AF ........................................................................................................................................ 44

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 7.2 Oil-immersed, Oil cooling...........................................................................................................44
7.2.1 ONAN ..................................................................................................................................44
7.2.2 ONAF ...................................................................................................................................44
7.2.3 OFAF....................................................................................................................................44
7.3 Oil-immersed, Water cooling..................................................................................................... 45
7.3.1 ONWF..................................................................................................................................45
7.3.2 OFWF...................................................................................................................................45
8 Tap-changing Transformers ................................................................................................................ 45
8.1 Changing the taps of transformers ............................................................................................ 45
8.1.1 Off-load tap changing..........................................................................................................46
8.1.2 On-load tap changing..........................................................................................................46
9 Autotransformers................................................................................................................................48
9.1 Autotransformer equations ....................................................................................................... 49
9.2 Advantages and disadvantages of autotransformer over two-winding transformer.............. 49
9.3 Two-winding transformer connected as an autotransformer..................................................50
9.4 Applications of autotransformers..............................................................................................52
10 Instrument transformers ................................................................................................................ 52
10.1 The voltage or potential transformers (VTs or PTs)..................................................................53
10.2 Current transformers (CTs) ........................................................................................................53
11 Further exercises.............................................................................................................................57
CHAPTER TWO ............................................................................................................................................59
ELECTROMECHANICAL ENERGY CONVERSION ........................................................................................... 59
1 Introduction ........................................................................................................................................ 59
2 Forces and Torques Developed by Electromechanical Devices.......................................................... 59
3 Energy balance....................................................................................................................................59
4 Force in Singly Excited Electric Field System....................................................................................... 60
5 Force in Singly Excited Magnetic Systems .......................................................................................... 63
6 Useful relations for calculating forces and torques............................................................................64
7 Magnetic Field Energy and Coenergy .................................................................................................66
8 Torque in Singly-Excited Systems........................................................................................................70
9 Torques and Forces in Doubly Excited Magnetic Field Systems ......................................................... 72
10 Alignment and Interaction Forces...................................................................................................74

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 10.1 Alignment Forces........................................................................................................................74
10.2 Interaction forces .......................................................................................................................74
11 Motional emf ..................................................................................................................................75
12 Examples of Electromechanical Devices ......................................................................................... 76
12.1 Solenoid Relay (Fig. 10) ..............................................................................................................76
12.2 Plunger relay (Fig. 11)................................................................................................................. 77
12.3 Horse shoe electromagnet.........................................................................................................78
12.4 Electrostatic voltmeter............................................................................................................... 79
12.5 Electrostatic loudspeaker...........................................................................................................81
13 Further exercises.............................................................................................................................83

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 CHAPTER ONE

TRANSFORMERS

1 Introduction

The transformer transfers electrical energy from one circuit to another via the medium of a
pulsating magnetic field that links both circuits. The widespread development of ac power
systems is principally due to the transformer. It enables us to produce and transmit power at
economical voltages and to distribute it safely in factories and homes. In low-power low-current
electronic and control circuits, it is used to provide impedance matching between a source and its
load for maximum power transfer, to isolate one circuit from another, to isolate direct current
while maintaining ac continuity between two circuits and to provide reduced ac voltages and
currents for protection, metering, instrumentation and control.

2 Principle of operation of a transformer

The transformer is a straight-forward application of Faraday’s Law of Electromagnetic

Induction. Consider the general arrangement of a single-phase transformer shown in Fig. 1. An
alternating voltage applied to coil 1, causes an alternating current to flow in the coil and this
current produces an alternating flux in the iron core. A portion of the total flux links the second
coil. The alternating flux induces a voltage in the second coil. If a load should be connected to
the coil, this voltage would drive a current through it. Energy would then be transferred through
the medium of magnetic field from coil 1 to coil 2. The combination of the two coils is called a
transformer. The coil connected to the source is called the primary winding (or the primary) and
the one connected to the load is called the secondary winding (or the secondary).

Fig. 1 Elementary transformer

1
3 Polarity and terminal markings of a transformer

Voltage E1 is induced in coil 1 and voltage E2 in coil 2. These voltages are in phase. Referring to
Fig.2, suppose at any given instant when the primary terminal 1 is positive with respect to
primary terminal 2, the secondary terminal 3 is also positive with respect to secondary terminal
4. Then terminals 1 and 3 are said to have the same polarity. To indicate that their polarities are
the same, a dot is placed beside primary terminal 1 and secondary terminal 3. Alternatively,
letters of the same suffix, A1 (for the high-voltage winding) and a1 (for the low-voltage winding)
say can be used. We note that current I1 entering coil 1 through the dotted terminal 1 and current
I2 entering coil 2 through the dotted terminal 3 create fluxes in the same direction.

(a) Magnetic Circuit (b) Electric Circuit

Fig. 2 Polarity and terminal markings

4 Ideal transformer

An ideal transformer has no losses, no leakage flux and its core is infinitely permeable. An ideal
transformer is shown in Fig. 3. The mutual flux Φm is confined to the iron. The primary current I1
is zero on no load.

Fig. 3 Ideal transformer

4.1 Properties of an ideal transformer

4.1.1 Emf equation and voltage ratio (see Fig. 3)

With the primary connected to an ac source V1, an alternating flux Φm is produced in the core.
Let the flux be expressed as
 m   max sin t webers (1)

2
The induced emf e1 as indicated in Fig. 1 is given by
d d
e1   N1 m   N1  max sin t   N1 max cos t
dt dt

e1  N 1 max sin t  90 (2)

Hence
N1 max 2fN1
E1    max
2 2

E1  4.44 fN 1 max (3)

Similarly
N 2  max 2fN 2  max
E2    4.44 fN 2  max (4)
2 2

From equations (3) and (4), we obtain

E1 N 1
 (5)
E2 N 2

The ratio a = N1/N2 is called the turns ratio. A step-up transformer has a < 1 and a step-down
transformer has a > 1. In an ideal transformer, the applied voltage V1 and the induced voltage E1
must be identical. Hence we may write
V1  E1 (6.a)

and
V1  4.44 fN 1 max (6.b)

Equation (6.b) indicates that for a given frequency, number of turns and voltage, the peak flux
 max must remain constant.

4.1.2 Current ratio and power equation

On no load I1 = 0. Now if a load is connected across the secondary terminals (i.e. switch S is
closed) current I2 flows through the load. This current produces mmf N2I2 which if it acted alone
would by Lenz’s law, cause the mutual flux to reduce. Since when V1 is fixed the flux Φmax is
also fixed, the primary develops mmf N1I1 which is such that
N1 I1  N 2 I 2 (7.a)
I1  I 2 a (7.b)
In ideal transformer the secondary voltage
V 2 E 2 (8)

3
The voltage V2 remains constant since E2 is fixed when the Φmax is fixed. It can be deduced from
equations (5), (6.a), (7.a) and (8) that for an ideal transformer
V1 I 1  V2 I 2 (9)

That is there are no reactive and active losses in an ideal transformer.

4.1.3 Phasor diagram of an ideal transformer (Fig. 4)

(a) No load (b) Load is resistive inductive

Fig. 4 Phasor diagram of ideal transformer

Example 1
An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is
connected to a 200-V, 50-Hz source. The load across the secondary draws a current of 2 A at a
power factor of 0.80 lagging. Calculate (a) the rms value of the primary current (b) the flux
linked by the secondary winding (c) Draw the phasor diagram.

Solution
(a) N 1 I 1  N 2 I 2  90 I 1  2250  2  I 1  50 A
V1 200
(b)  max    0.01 Wb
4.44 fN 1 4.44  50  90
N 
(c) E 2   2  E1  5000 V
 N1 
The phase angle between V2 and I 2 is cos -1 (0.8)  36.9

4
Example 2
A 200-kVA, 6600-V / 400-V, 50-Hz 1-ph transformer has 80 turns on the secondary. Calculate
(a) the approximate values of the primary and secondary full-load currents (b) the approximate
number of primary turns and (c) the maximum value of the flux.

Solution
(a) Full - load primary current  (200  1000) 6600  30.3 A
and full - load secondary current  (200  1000) 400  500 A
(b) N 1  (80  6600) 400  1320
V2 400
(c)  max    0.0225 Wb
4.44 fN 2 4.44  50  80

4.1.4 Impedance ratio (see Fig. 5)

The impedance seen by the source

Z e  V1 I 1  E1 I 1  aE 2 I 2 a   a 2 E 2 I 2  a 2 Z

Hence
Ze  a2Z (10)

Fig. 5 Impedance ratio

4.1.5 Equivalent circuits of an ideal transformer

From equation (10), we can represent the transformer in Fig. 5 by equivalent circuit shown in
Fig. 6.a. We may also write
I 2 Z  E 2  E1 a  V1 a

V1
 I2Z (11)
a

and then represent the transformer by an equivalent circuit shown in Fig. 6.b

5
 (a) Circuit referred to the primary side (b) Circuit referred to the secondary side

Fig. 6 Equivalent circuits of ideal transformer

Example 3
Calculate the voltage V and current I in the circuit of Fig. 7, knowing that the ideal transformer
has a primary to secondary turns of 1:100 (i.e. a = 1/100).

Fig. 7 See Example 3

Solution
We shall shift all impedances to the primary side to obtain the equivalent circuit:

V   aV

Total impedance Z e  R 2   X L  X C   4 2  32  5 
2

V1 10
Total current I    2A
Ze 5
Voltage across the 4-ohm resistance V   IR  2  4  8 V.
V
The actual voltage V   100  8  800 V
a

6
5 Practical single-phase transformer

The windings of a practical transformer have both resistance and leakage inductance. The core is
also imperfect: it has a core loss and finite permeability. The core loss has two components:
hysteresis loss and eddy current loss.

5.1 Hysteresis loss

When ferromagnetic material is subjected to alternating magnetization, the energy put into the
magnetic field when the flux is increasing is not completely given back when the flux dies away
but a certain portion is wasted. The area of the hysteresis loop gives the value of the energy loss
per unit volume (WH) taking place for each complete cycle of magnetization. The loop area is
x
found experimentally to vary as Bmax up to moderate values of flux density (1.0 – 2.0 Wb/m2).

The index x is named after Steinmetz and is about 1.6 though it may be higher. In practice the
2
hysteresis loss, for simplicity, is often taken as proportional to Bmax

If f = frequency of magnetization, the power wasted in magnetic hysteresis

Ph  fW H W/m 3
Ph  fBmax
1.6
 fBmax
2
W/m 3 (12)

The hysteresis constant η depends upon the magnetic material.

5.2 Eddy-current loss

When a changing magnetic flux permeates any mass of metal, eddy currents are induced. The
eddy currents cause the metal to heat. Eddy currents are also induced in a core revolving in a
stationary or constant magnetic field. If iron or metal subjected to alternating magnetization is
built up of laminations (or thin sheets) insulated from each other, the eddy current loss is
reduced.

The eddy current loss in a laminated core is given by

Pe  (kf 2 t 2 Bmax
2
) /  W/m 3 (13.a)

where
t = thickness of the laminations
ρ = resistivity of the material
k = a constant which depends on the waveform of the alternating flux

At high frequencies where it is not practicable to make very thin laminations, core losses may be
reduced by using ferrite cores or dust cores. Ferrite is a ceramic material having magnetic
properties similar to silicon steel, and dust cores consist of fine particles of carbonyl iron or

7
permalloy (i.e. nickel and iron), each particle of which is insulated from its neighbour by a
binding material. Such materials have a very high value of resistivity.

For a given thickness and waveform, (13.a) reduces to

Pe  k f 2 Bmax
2
W/m 3 (13.b)

Example 4
The core of a transformer operating at 50 Hz has an eddy current loss of 100 W/m3 and the core
laminations have a thickness of 0.50 mm. The core is redesigned so as to operate with the same
eddy current loss but at a different voltage and at a frequency of 250 Hz. Assuming that at the
new voltage the maximum flux density is one-third of its original value and the resistivity of the
core remains unaltered, determine the necessary new thickness of the laminations.

Solution
From equation (13.a), we can write
2 2 2
Pe 2  f 2   t2   Bmax 2   1 
       
Pe1  f1   t1   Bmax 1   2 

B  1
Since eddy current loss and resistivity remain the same, and  max 2   , we have
 Bmax 1  3
2 2 2
 250   t 2   1 
1      
 50   0.50   3 
 50   3 
Therefore the new thickness t 2  0.50        0.30 mm
 250   1 

We note that if the new voltage had been given rather than Bmax2, then we would have used
equation (6.b) to find Bmax2. From equation (6.b), we can write
V1, 2  f 2  N 1, 2   max 2   f 2  N 1, 2  Bmax 2  A2 
          
V1,1  f 1  N 1,1   max 1   f 1  N 1,1  Bmax 1  A1 

If the number of turns and the cross section remain the same, then the equation reduces to
V1, 2  f 2  Bmax 2 
   
V1,1  f 1  Bmax 1 

We can use the above equation to find the new voltage (V1,2) in example 4 if it is required.

Example 5
In a transformer core of volume 0.16 m3 the total iron loss was found to be 2170 W at 50 Hz.
The hysteresis loop of the core material, taken to the same maximum flux density, had an area of
9.0 cm2 when drawn to scales of 1 cm = 0.1 Wb/m2 and 1 cm = 250 AT/m. Calculate the total
iron loss in the transformer core if it is energized to the same maximum flux density, but at a
frequency of 60 Hz.
8
Solution
Hysteresis loss WH = area x scale factors = 9 x 0.1 x 250 = 225 J/m3
At 50 Hz, hysteresis loss Ph  fWH  volume of core = 225 x 50 x 0.16 = 1800 W
Therefore eddy-current loss = 2170 – 1800 = 370 W
From equation (12), at 60 Hz, hysteresis loss = 1800 x (60/50) = 2160 W
and from equation (13.b), eddy-current loss = 370 x (60/50)2 = 533 W
Therefore total iron loss = 2160 + 533 = 2693 W

Example 6
For the same maximum flux density, the total core loss in a core is 500 W at 25 Hz and 1400 W
at 50 Hz. Find the hysteresis and eddy-current losses for both frequencies.

Solution
Since Bmax is constant, the losses can have the following forms:
Ph  Af , Pe  Bf 2 and Pc  Ph  Pe  Af  Bf 2
For a frequency of 25 Hz, the core loss = 500 = A(25) + B(25)2
For a frequency of 50 Hz, the core loss = 1400 = A(50) + B(50)2
Solving the two equations, we obtain A = 12, B = 0.32 and the individual losses
Ph = 300 W, Pe = 200 W at 25 Hz and Ph = 600 W, Pe = 800 W at 50 Hz

5.3 Phasor diagram on no load

To furnish the power loss in the core, i.e. the core loss, a small current must be drawn from the
source. This current Ip must be in phase with induced voltage E1. Also to create the mutual flux
Φm, a magnetizing current Im in phase with Φm and lagging 90° behind E1 must be drawn to
produce the required mmf.

Fig. 8 Phasor diagram for practical transformer on no load

We note that
(i) the no-load current Io taken by the primary is the phasor sum of Ip and Im
(ii) the difference between the value of the applied voltage V1 and that of the induced emf E1 is
only about 0.05% when the transformer is on no load so the two can be considered to be
equal
(iii) Ip is very small compared with Im. Therefore the no-load power factor is very low.

9
Example 7
A 1-ph transformer has 480 turns on the primary and 90 turns on the secondary. The mean length
of the flux path in the iron core lc is 1.8 m and the joints are equivalent to an air gap lg of 0.1 mm.
If the peak value of the flux density is to be 1.1 T when a voltage of 2200 V at 50 Hz is applied
to the primary, find (a) the cross-sectional area of the core (b) the secondary voltage on no load
(c) the primary current and power factor on no load.

Assume the value of the magnetic field strength Hc for 1.1 T in iron to be 400 A/m, the
corresponding iron loss to be 1.7 W/kg at 50 Hz and the density of the iron to be 7800 kg/m3.

Solution
The magnetic circuit of the transformer consists of iron core in series with the air gap.
V1 2200
(a)  max    0.0206 Wb
4.44 fN 1 4.44  50  480
 0.0206
Cross - sectional area of core Ac  max   0.0187m 2
Bmax 1.1
N 90
(b) Secondary voltage on no load V2  V1  2  2200   412.5 V
N1 480
(c) Peak value of mmf required for the iron core Fg max  H c  l c  400  1.8  720 A
Bg 1.1
Peak value of mmf for the air gap Fg max  H g l g  lg   0.0001  87.5 A
o 4  10 7
Peak value of total mmf required to produce  max  Fc max  Fg max  720  87.5  807.5 A
Peak value of total mmf 807.5
Peak value of magnetizing current    1.682 A
N1 480
1.682
Its rms value assuming it to be sinusoidal I m   1.19 A
2
Volume of iron  l c  Ac  1.8  0.0187  0.0337 m 3
Mass of iron  volume  density  0.0337  7800  263 kg
And iron loss  263  1.7  447 W
I p  Iron loss V1  447 2200  0.203 A
No load current  I o  1.19 2  0.203 2  1.21 A
No load power factor  I p I o  0.203 1.21  0.168 lagging

5.4 Mutual and leakage fluxes in a transformer (see Fig. 9)

The actual flux linking a coil can be considered to have two components: the mutual flux Φm
linking both coils and the leakage flux Φl. The reluctance of the paths of the leakage flux is
almost entirely due to the long air paths and is therefore practically constant. Consequently, the
value of the leakage flux is proportional to the current in the coil.

10
 Fig. 9 Transformer possessing two leakage fluxes and a mutual flux

The secondary induced voltage Es is composed of two parts:

(a) a voltage El2 induced by leakage flux Φl2 given by El2 = 4.44fN2Φl2,max
(b) a voltage E2 induced by mutual flux Φm given by E2 = 4.44fN2Φmax

Similarly, the primary induced voltage is composed of El1 =4.44fN1Φl1,max and

El1 = 4.44fN1Φmax

We can separate the four voltages E1, E2, El1 and El2 by rearranging the transformer circuit as
shown in Fig. 10. The rearrangement of the transformer circuit makes it clear that El1 and El2 are
voltage drops across reactances. These reactances called leakage reactances are given
by X 1  El1 I 1 and X 2  El 2 I 2

Fig. 10 Separation of induced voltages

5.5 Equivalent circuit of a practical transformer

The behaviour of a practical transformer may be conveniently considered by assuming it to be
equivalent to an ideal transformer and then allowing for the imperfections of the actual
transformer by means of additional circuits or impedances inserted between the supply and the
primary winding and between the secondary winding and the load. The complete equivalent
circuit of this transformer is shown in Fig. 11. R1 and R2 are resistances of the primary and
secondary windings. X1 and X2 are the leakage reactances. The reactance Xm is such that it takes a
reactive current Im (i.e. the magnetizing current) of the actual transformer. The core loss is
accounted for by the resistance Rm which takes the component Ip of the primary current.
11
 Fig. 11 Complete equivalent circuit of a practical transformer

Ideal transformer equations still apply: I 2 N 1  N 2 I 2 and E1 E 2  N 1 N 2

5.6 Equivalent circuit referred to the primary side (Fig. 12)

Fig. 12.Exact equivalent circuit referred to the primary side

Let a  N 1 N 2 (14.a)

Then
X 2  ( N 1 N 2 ) 2 X 2  a 2 X 2 , R2  ( N 1 N 2 ) 2 R2  a 2 R2 , Z   ( N 1 N 2 ) 2 Z  a 2 Z (14.b)
 
V2  ( N 1 N 2 ) V2  aV2 , I 2  I 2 a (14.c)

 
It is worth noting that for a practical transformer, R2  R1 and X 2  X1

12
5.7 Equivalent circuit referred to the secondary side (Fig. 13)

Fig.13 Exact equivalent circuit referred to the secondary side


R1  ( N 2 N 1 ) 2 R1  R1 a 2 , etc (15.a)
 
N 2 I p  N1 I p or I p  aI p etc (15.b)

5.8 Approximate equivalent circuits

The exact equivalent circuit of the transformer is too exact for most practical engineering
applications. Consequently, we can simplify it to make calculations easier. Approximate
equivalent circuits commonly used for power transformer calculations are given in Fig. 13.

(a)

(b)

13
 (c)

Fig.14 Three approximate equivalent circuits

The exciting or no load current Io causes voltage drop in the primary leakage impedance. In Fig.
14.a, this voltage drop is neglected.

In Fig. 14.b, the exciting current is neglected entirely. Using this circuit may not cause serious
errors in some applications such as voltage regulation calculations.

In Fig. 14.c, Re is ignored. For transformers above 500 kVA where Xe is at least 5 times greater
than Re, this circuit can be used to calculate say, voltage regulation.

5.9 The complete phasor diagrams for loaded conditions

They are shown in Fig. 15

Fig. 15 Complete phasor diagrams for loaded conditions (drawn in two parts)

14
5.10 Rating of transformers
To keep the transformer temperature at an acceptable level, limits are set to both the applied
voltage (this determines the iron loss at a given frequency) and the current drawn by the load
(this determines copper loss in the windings). The limits determine the rated voltage and rated
current of transformers.

The power rating of transformer Srated = rated voltage x rated current can be expressed in VA,
kVA or MVA depending on the size of the transformer. The rated kVA (i.e. the rated power),
frequency and voltage are always shown on the name plate. In large transformers, the
corresponding currents are also shown.

We note that
Rated kVA = V1r x I1fl x10-3 = V2r x I2fl x 10-3 = V2n x I2fl x10-3 = E2n x I2fl x10-3

where
V1r = rated primary voltage
I1fl = rated primary current = primary full load current
V2r = rated secondary voltage = V2n (no load secondary voltage corresponding to the rated
primary voltage) = E2n (no load induced secondary voltage corresponding to primary rated
voltage)
I2fl = rated secondary current = secondary full load current

5.11 The turns ratio

It is given by
a  E1 E 2

E1  V1r on no load because I o Z1 is very small. Since the induced voltage in the secondary is
equal to the secondary terminal voltage on no load
a  V1r V 2 r (16.a)

and
I 1 fl  I 2 fl a (16.b)

Example 8
A transformer is rated 10 kVA, 2400 / 240 V, 60 Hz. The parameters for the approximate
equivalent circuit of Fig. 14.a are Rm = 80 kΩ, Xm = 35 kΩ, Re1 = 8.4 Ω and Xe1 = 13.7 Ω.
Determine the voltage to be applied to the primary to obtain the rated current in the secondary
when the secondary terminal voltage is 240 V. What is the input power factor? The load power
factor is 0.8 lagging.

15
Solution

The secondary current when referred to the primary I 2  I 2 fl a  I 1 fl  10000 2400  4.17 A
The load power factor angle   cos 1 0.8  36.9
If we choose the load current I2 as the reference phasor, then V 2  24036.9
The load voltage referred to the primary V   aV  240036.9  1920  j1440 volts
2 2

The voltage across the equivalent leakage impedance is

Z I   8.4  j13.7 4.17  j 0  35  j 57 volts
e1 2

The primary voltage required

 
V1  V 2  Z e1 I 2  1920  j1440  35  j 57   1955  j1497  246237.4 volts
V1 1955  j1497
Ip    24.4375  j18.7125 mA
Rm 80 k
V 1955  j1497
Im  1   42.7714  j 55.8571 mA
jX m j 35 k

I 1  I 2  I p  I m  4.17  j 0   0.0244  j 0.0187   0.0427  j 0.0558
 4.2371  j 0.0371  4.237  0.50 A

Phase angle between V1 and I 1  37.4   0.50  37.9

Input power factor  cos37.9  0.79 lagging

Example 9
A 1-ph transformer operates from a 230-V supply. It has an equivalent resistance of 0.1 Ω and an
equivalent leakage reactance of 0.5 Ω referred to the primary. The secondary is connected to a
coil having a resistance of 200 Ω and a reactance of 100 Ω. Calculate the secondary terminal
voltage. The secondary winding has four times as many turns as the primary.

Solution
Refer to approximate equivalent circuit of Fig. 14.b
N 1
a  1  , Z  200  j100  , Z   a 2 200  j100  12.5  j 6.25 
N2 4
Total impedance referred to the primary  Z e1  Z   0.1  j 0.5  12.5  j 6.25  12.6  j 6.75
 2300  2300
I2  and V 2  12.5  j 6.25
12.6  j 6.75 12.6  j 6.75
 230  12.5  j 6.25 230  13.9754
The magnitude of the voltage V 2    224.8719 volts
12.6  j 6.75 14.2941

V
The actual secondary load voltage V2  2  4  224.8719  899 volts
a

16
5.12 Definition of per-unit impedances
The leakage impedances Z1 and Z2 on the primary and secondary side are expressed in per unit as
follows:
Z1
Zˆ1  (17.a)
Z 1base
Z2
Zˆ 2  (17.b)
Z 2 base

where
V1r V2 V2 r V2
Z 1base   1r and Z 2 base   2r (17.c)
I 1 fl S rated I 2 fl S rated

The impedances are said to be expressed in per unit with reference to a base voltage and a base
power which have been chosen in the case of transformers to be V1r, Srated for the primary circuit
and V2r, Srated for the secondary circuit. The total impedance of the transformer in per unit
 Zˆ1  Zˆ 2 .

Example 10
A single-phase transformer that is rated 3000 kVA, 69 kV / 4.16 kV, 60 Hz has an impedance of
8 per cent. Calculate the total impedance of the transformer referred to (a) the primary side (b)
the secondary side

Solution
Vr2 69 2  10 6
(a ) Z1base    1587 
S rated 3000  10 3
Z e1  Zˆ e1  Z1base  0.08 1587  127 
V22r 4.16 2  10 6
(b) Z 2base    5.7685
S rated 3000  10 3
Z e 2  Zˆ e 2  Z 2 base  0.08  5.7685  0.46 
2 2
N   4.16 
Alternatively Z e 2   2  Z e1     127  0.46 
 N1   69 

5.13 Voltage regulation

With the primary voltage maintained constant, the secondary terminal voltage at no load differs
from the secondary voltage under load. The voltage regulation or simply regulation of a
transformer is the change in secondary voltage which occurs when the rated kVA output at a
specified power factor is reduced to zero, with the primary voltage maintained constant. It is

17
usually expressed as a percentage (called percentage regulation), or a fraction of the rated or no-
load terminal voltage (in per unit).

The equivalent circuit given in Fig. 14.b is used to calculate voltage regulation. The circuit may
be either referred to the primary or the secondary side. Therefore the circuit is represented in
general form as shown in Fig. 16.

If the circuit is referred to the primary side, then


E  V1 , V  V2 , Re  Re1 and X e  X e1

Re Xe

E Load V

Fig. 16 Circuit for computing voltage regulation


E  V V1  V2
Voltage regulation    pu (18.a)
E V1

If the circuit is referred to the secondary side, then

V
E  V2 n  1 , V  V2 , Re  Re 2 and X e  X e 2
a

V2 n  V 2
Voltage regulation   pu (18.b)
V2 n

The voltage regulation given by equations (18.a) and (18.b) are equal as shown below:
 
V  V2 V a  V2 a V2 n  V 2
 1  1  (19)
V1 V1 a V2 n

When voltage regulation is known the terminal voltage can be calculated as follows:
V2  1   V2 n (20)

In general, let the load current be I lagging behind the load voltage V by  . Then taking the load
voltage as the reference phasor, we can write
E  V 1  j 0  Re  jX e I cos   jI sin  
 V  IRe cos   IX e sin    j IX e cos   IRe sin  

18
and hence
E E  V  IRe cos   IX e sin  2  IX e cos   IRe sin  2
The second term under the root is usually negligible except at low leading power factors.
Considering the first term only gives
E  V  IRe cos   IX e sin 
E  V  IRe cos   IX e sin  and (21.a)
E  V IRe IX
  cos   e sin  (21.b)
E E E

The angle  is negative when current is leading and positive when current is lagging.

The voltage regulation is maximum for a given current or power when  =  where
X R X
tan   e or cos   e or sin   e (22.a)
Re Ze Ze

Proof:
d X
From E  V   IRe  sin    IX e cos   0 , we have tan   e  tan 
d Re

The maximum voltage regulation is given by

R  X   R 2  X e2 
E  V max  IRe cos   IX e sin   IRe  e   IX e  e   I  e

  I Ze

(22.b)
 Ze   Ze   Z e 

where
2 2
Z e  Re  X e (22.c)

From equation (21.b), if E is the rated voltage, then

E  V  I fl  IR   I  IX 
  rated    cos    fl   sin 
E rated I  E I  E
  fl  rated   fl  rated 

E rated  V  I  I fl R   I  I fl X 
    cos      sin 
E rated  I  E I  E
 fl  rated   fl  rated 

E rated  V
  I p.u . R p.u. cos   I p.u . X p.u. sin  (23)
E rated

where
I p.u.  I I fl  current in per unit R p.u.  I fl R E rated  resistance in per unit , etc

19
We note the following:
(i) Usually the quantities will be referred to the secondary side.
(ii) Where E is the rated voltage and transformers parameters are given in per unit, equation (23)
may be used. We should remember that with impedances in per unit, primary and secondary
impedances can be added directly.
(iii)The voltage regulation equations (21.a), (21.b), (22.b) and (23) are correct at any current or
kVA.
(iv)At rated current or kVA, Ip.u.=1.

Example 11
A 100-kVA 1-ph transformer has 400 turns on the primary and 80 turns on the secondary. The
primary and secondary resistances are 0.3 Ω and 0.01 Ω respectively, and the corresponding
leakage reactances are 1.1 Ω and 0.035 Ω respectively. The supply voltage is 2200 V. Calculate
(a) the equivalent impedance referred to the primary circuit and (b) the voltage regulation and the
secondary terminal voltage for full load having a power factor of (i) 0.8 lagging and (ii) 0.8
leading (c) the maximum voltage regulation

Solution
(a) Re1  R1  ( N1 N 2 ) 2 R2  0.3  0.01  (400 80) 2  0.55 
X e1  1.1  0.035  (400 80) 2  1.975 
 1
Z e1  0.552  1.9752 2  2.05 
(b) (i ) cos   0.8  sin   0.6
I 1 fl  100  10 3 2200  45.45 A
 R cos   X e1 sin   0.55  0.8  1.975  0.6 
  I 1 fl  e1   45.45   0.0336 pu
 V1   2200 
N  80
The secondary no load voltage V2 n  V1  2   2200   440 V
 N1  400
The secondary terminal voltage V2  V2 n 1     4401  0.0336   425.2 V
 R cos   X e1 sin   0.55  0.8  1.975  0.6 
(ii )   I 1 fl  e1   45.45   0.0154 pu
 V1   2200 
The secondary terminal voltage V2  4401   0.0154  446.8 V
I 1 fl Z e1 45.45  2.05
(c) Maximum voltage regulation  max    0.0424 pu.
V1 2200

Example 12
The primary and secondary windings of a 30-kVA, 6000-V / 230-V transformer have resistances
of 10 Ω and 0.016 Ω respectively. The total reactance of the transformer referred to the primary
is 23 Ω. Calculate the percentage regulation of the transformer when supplying full load current
at a power factor of 0.8 lagging.

20
Solution
Re1  R1  ( N1 N 2 ) 2 R2  10  0.016  (6000 230) 2  20.89 
X e1  23 
I1 fl  30  10 3 6000  5 A
 R cos   X e1 sin   20.89  0.8  23  0.6 
  I 1 fl  e1   5   0.0254 pu
 V1   6000 
Percent regulation  2.54%

5.14 Transformer output

The transformer supply voltage and frequency are substantially constant. The heating therefore
depends on the current taken by the load. Since the secondary voltage of the transformer is also
substantially constant it means that the heating also depends on the load kVA. The transformer
output is therefore usually quoted in kVA. The transformer load in kVA is given by
S  V2 I 2  10 3 kVA (24)

where
V2 = actual load voltage and
I2 = actual load current.

When S is given, V2 is unknown and the load current is required, it can be estimated using the
approximate equation
S  V2 n I 2  10 3 kVA (25)

5.15 Efficiency
The losses which occur in a transformer on load are composed of
2 2
(a) Copper losses in primary and secondary windings, namely I12 R1  I 22 R2  I 1 Re1  I 2 Re 2
(b) Iron losses in the core due to hysteresis and eddy currents. The iron losses depend on the
peak value of the mutual flux Φm and frequency. It is therefore independent of load current if
voltage and frequency are constant.

Let
Pi = the iron losses (fixed loss) in kW and
Pc = the copper loss with full-load Srated kVA in kW

Then the total loss at any load S = xSrated kVA at power factor cos  is Pi + x2 Pc and the
efficiency is
output xS rated cos  S rated cos 
  
input xS rated cos   Pi  x Pc
2
P 
S rated cos    i  xPc 
 x 

21
For a given power factor, the efficiency is maximum when the expression in brackets is a
minimum. Hence for a maximum efficiency, we have
Pi
x (26)
Pc

Proof:
d  Pi  Pi
  xPc   0 , we obtain  2  Pc  0 or Pi  x Pc
2
From
dx  x  x

According to equation (26), efficiency is maximum when the copper loss, x 2 Pc is equal to the
fixed loss or iron losses Pc, or efficiency is maximum when the load is such that the resulting
copper loss is equal to the fixed loss or iron losses.

The efficiency of a transformer is calculated using this form of efficiency equation:

losses Pi  x 2 Pc
  1  1 (27)
losses  output xS rated cos   Pi  x 2 Pc

Example 13
The primary and secondary windings of a 500-kVA transformer have resistances of 0.42 Ω and
0.0011 Ω respectively. The primary and secondary voltages are 6600 V and 400 V respectively
and the iron loss is 2.9 kW. Calculate the efficiency on full load at a power factor of 0.8.

Solution
500  1000
Full - load secondary current I 2 fl   1250 A
400
1000
Full - load primary current I 1 fl  500   75.8 A
6600
Secondary copper loss on full load  1250 2  0.0011  1720 W
Primary copper loss on full load  75.8 2  0.42  2415 W
Total copper loss on full load , Pc  1720  2415  4135 W  4.135 kW
Total loss on full load  Pc  Pi  4.135  2.9  7.035 kW
Output power on full load at 0.8 pf  S rated cos   500  0.8  400 kW
Pi  Pc 7.035
Efficiency on full load   1   1  0.9827  98.27 %
S rated cos   Pi  Pc 400  7.035

Example 14
Find the output, at which the efficiency of the transformer of example 11 is maximum and
calculate its value assuming the power factor of the load to be 0.8.

22
Solution
Let the load S be a fraction x of the rated power, i.e. S = xSrated. Then
Pi 2.9
x   0.837
Pc 4.135

Hence output at maximum efficiency  xS rated  0.837  500  418.5 kVA

At maximum efficiency, total loss  Pi  x 2 Pc  Pi  Pi  2 Pi  2  2.9  5.8 kW
and output power at 0.8 pf  xS rated cos   418.5  0.8  334.8 kW
5.8
Therefore, maximum efficiency  max  1   0.983  98.30 %
5.8  334.8

Example 15
A 400-kVA transformer has an iron loss of 2 kW and the maximum efficiency at 0.8 pf occurs
when the load is 240 kW. Calculate (a) the maximum efficiency at unity power factor and (b) the
efficiency on full load at 0.71 power factor

Solution
(a) Total loss at maximum efficiency  2 Pi  2  2  4 kW
240
Output kVA at maximum efficiency   300 kVA
0 .8
Output power at maximum efficiency at unity pf  300  1  300 kW
4
Maximum efficiency at unity pf  max  1   0.9868  98.68 %
4  300
300
(b) The fraction x of full load kVA at which efficiency is maximum   0.75
400
P 2
Hence full load copper loss Pc  2i   3.56 kW
x 0.75 2

Pi  Pc 2  3.56
Full load efficiency at 0.71 pf  1   1  98.08 %
S rated cos   Pi  Pc 2  3.56  400  0.71

5.16 All-day efficiency

Power transformers used in generating stations and transmission substations operate at
substantially constant load. They are designed with minimal copper losses. Their maximum
efficiency occurs at full load.

Distribution transformers used in distribution networks on the other hand supplies load which
varies widely over a 24-h period. They are always connected in circuit and operate less than full
load for most of the time. Hence they are designed with minimal core losses. Their maximum
efficiency occurs at 60 to 70 % of full load. The efficiency of these transformers is better

23
assessed on energy basis. The output and losses are calculated in kW hours over a 24-hour day.
The all-day efficiency is defined as
losses in kWh
η allday  1  (28)
losses in kWh  output in kWh

Example 16
A 200-kVA 1-ph transformer has full load copper loss of 3.02 kW and iron loss of 1.6 kW. The
transformer is in circuit continuously. For a total of 8 hours, it delivers a load of 160 kW at 0.8
pf. For a total of 6 hours, it delivers a load of 80 kW at unity power factor. For the remainder of
the 24-h cycle, it is on no load. What is the all-day efficiency?

Solution
At 160 kW, 0.8 pf, kVA  200 kVA (full load), copper loss  3.02 kW
iron loss  1.60 kW
total loss  4.62 kW
2
 80 
At 80 kW, upf, kVA  80 kVA , copper loss     3.02  0.48 kW
 200 
iron loss  1.60 kW
total loss  2.08 kW
On no load (there is no copper loss on no load), Total loss  1.6 kW

For 8 h output  160  8  1280 kWh loss  4.62  8  37 kWh

For 6 h output  80  6  480 kWh loss  2.08  6  12.5 kWh
For 10 h output  0  10  0 kWh loss  1.6  10  16 kWh
In 24 h, total output  1760 kWh total loss  65.5 kWh
65.5
All day efficiency  1   0.9641  96.41 %
65.5  1760

5.17 Open-circuit and short-circuit tests on a transformer

These two tests enable the efficiency and voltage regulation to be calculated without actually
loading the transformer.

5.17.1 Short-circuit test

This test is used to determine the leakage impedance. During this test, one winding is short-
circuited and a reduced voltage Vsc applied to the other to cause rated current to flow. The test
circuit and the equivalent circuit are shown in Fig. 18.a and Fig. 18.b respectively. The
magnetizing branch is neglected because its current under this condition is less than 1 % of the
total.

24
Voltage Vsc, current Isc and power Psc measured by the instruments are used to make the
following calculations:
V P
(a ) Z e1  sc (b) Re1  sc2 (c) X e1  Z e21  Re21 (29)
I sc I sc

We note that the following:

(i) Isc need not be the rated current since the equivalent circuit is linear. However, it is desirable
that it should be near to the rated value so that stray losses (they are due to eddy currents set
up in large section conductors, tank and metallic supports by leakage fluxes) are normal.
(ii) The supply could be fed to either winding. It is often convenient on the high voltage
transformers to supply the high-voltage winding, thus using a smaller current. Vsc which will
be about 3-15 % of the rated value may also be more suitable for test facilities.
(iii)In laboratory experiments using small transformers, the instruments positions shown
minimize measurement errors.
(iv)The circuit parameters obtained with equations (29.a) – (29.c) are referred to the side which
the test voltage is applied.

Fig. 18 Short-circuit test

5.17.2 Open-circuit test or no load test

During this test, one winding is open-circuited and rated voltage at rated frequency is applied to
the other. Quite often the low-voltage side is supplied to reduce the test voltage required for
safety reasons. As with the short-circuit test, the equivalent circuit parameters will be referred to
the side to which the test voltage is applied. The test circuit and equivalent circuit are shown in
Fig. 19.a and b.

The following calculations can be made:

P 2 2 V V1 V1
(a ) I p  o ( b) I m  I o  I p (c ) R m  1 (d ) X m  ( e) a  (30)
V1 Ip Im V2

Psc and Po represent the full load copper loss and the core loss (or iron losses) respectively. They
can be used directly to calculate efficiency.
.

25
 Fig. 19 Open-circuit or no-load test

Example 17
The circuit shown below was used in a test on a 3-kVA transformer. A variable voltage supply of
fixed frequency was connected to terminals A and B and two tests were performed:
(a) The voltage was raised to normal rated voltage and the meter reading were then 200 V, 24
W, 1.2 A
(b) The terminals C and D were short-circuited and the voltage was raised until the transformer
full-load current was flowing. The meter readings were then 6.4 V, 28 W, 15 A

From the results of the tests, obtain:

(i) the no-load current and its power factor
(ii) the iron losses of the transformer at normal frequency and voltage
(iii)the full-load copper loss
(iv)the transformer resistances, Re1 and Rm and reactances Xe1 and Xm
(v) the efficiency of the transformer at full load at a power factor of 0.8
(vi)the efficiency of the transformer at half full load and 0.8 power factor.

C
A A

B D
Circuit diagram for Example 13

Solution
(i) No - load current I o  1.2 A
input power 24
No - load power factor    0.1 lag
input apparent power 200  1.2
(ii) Iron losses Pi  Po  24 W

26
(iii) Full load copper loss Pc  Psc  28 W
(iv) Open - circuit test calculations :
P 24 2 2
Ip  o   0.12 A I m  I o  I p  1.2 2  0.12 2  1.19 A
V1 200
V 200 V 200
Rm  1   1.67 kΩ Xm  1   168 
I p 0.12 I m 1.19
Short - circuit test calculations :
V 6.4 P 28
Z e1  sc   0.43  Re1  sc2
 2  0.12 
I sc 15 I sc 15
X e1  Z e21  Re21  0.432  0.12 2  0.41

Pi  x 2 Pc 24  28
(v)  fl  1   1  0.9788 pu  97.88 %
xS rated cos   Pi  x Pc
2
(3000  0.8)  24  28

Pi  x 2 Pc 24  28( 12 ) 2
(vi)  1 fl  1   1   97.48 %
2 xS rated cos   Pi  x 2 Pc  12 (3000  0.8)  24  28( 12 ) 2
Example 18
A 10-kVA 1-ph transformer has a voltage ratio 1100 / 250 V. On no load and at normal voltage
(1100 V) and frequency the input current is 0.75 A at a pf of 0.2 lagging. With the secondary
short-circuited, full-load currents flow when the primary applied voltage is 77 V, the power input
being 240 W. Calculate
(a) the transformer equivalent resistance and reactance referred to the secondary side
(b) the maximum value of the voltage regulation at full load and the load power factor at this
regulation
(c) the percentage of full-load current at which the transformer has maximum efficiency

Solution
(a) Referring the circuit to the secondary side,
V 250
Vsc 2  sc1  77   17.5 V
a 1100
S 10 000
I sc 2  I 2 fl  rated   40 A
V2 n 250
V 17.5 P 240
Z e 2  sc 2   0.438  , Re 2  2sc  2  0.15  and X e 2  0.4382  0.152  0.41
I sc 2 40 I sc 2 40
(b) The maximum voltage drop at full load = I 2 fl Z e 2  17.5 V
The maximum regulation occurs when    or when
R 0.15
cos   cos   e 2   0.342 lag
Z e 2 0.438

27
(c) Maximum efficiency occurs when x 2 Pc  Po . Therefore
Po Vo I o cos  o 1100  0.75  0.2
x    0.829 or 82.9 %
Pc Pc 240

5.18 Construction of transformers

Power transformers are designed so that their characteristics approach those of an ideal
transformer
(a) To attain high permeability and consequently a small magnetizing current, the core is made
of iron and the core forms a closed magnetic circuit
(b) To keep the hysteresis loss down, high-grade grain-oriented steel having a narrow hysteresis
loop is used
(c) To keep the eddy current loss down, the core is laminated and 3 to 4% silicon is added to
increase the resistivity of the steel. The laminations (about 0.4 mm thick) are insulated from
each other by a thin layer of insulation, thus overall cross-sectional area is slightly greater
than the actual cross-sectional area of the iron.
(d) To keep the leakage fluxes low, the windings are arranged as shown in Fig. 20 in the case of
1-ph transformers. Two main forms of magnetic circuits are used: core- and shell-type
arrangements. The core-type construction has the primary and secondary windings
distributed over both core legs in order to reduce the amount of copper (i.e. length of turns is
reduced).

The transformer windings are carefully insulated from each other and from the core. Winding
resistances R1 and R2 are also minimized to minimize copper losses.

(a) Core - type constructi on with (b) Shell - type constructi on with
concentric windings sandwiched windings

Fig. 20 Common Single-phase transformer construction

5.19 Polarity tests

The four terminals of a single-phase transformer may be mounted so that it has either additive or
subtractive polarity as shown in Fig. 21. If it becomes necessary to determine whether a
transformer has additive or subtractive polarity then either of the following polarity tests may be
used:

28
 (a) additive polarity (b) subtractive polarity
Fig. 21 Two standard ways of mounting transformer terminals

5.19.1 Using a low-voltage ac source

Referring to Fig. 22,

(i) Connect the HV winding to ac source
(ii) Connect a jumper J between any two adjacent HV and LV terminals (1, 3)
(iii)Connect a voltmeter (V2) between the other two adjacent HV and LV terminals (2, 4)
(iv)Connect a voltmeter (V1) across the HV winding.

Reading of voltmeter V2 = E1 – E2 = V1 – E2 if 1 and 3 possess the same polarity

= E1 + E2 = V1 + E2 if they do not
Therefore if V2 is lower than V1 then the polarity is subtractive and if V2 is higher than V1 then
the polarity is additive

Fig. 22 Polarity test using ac source

5.19.2 Using dc source

Referring to Fig. 23,

(i) Connect a dc source to the LV winding
(ii) Mark the terminal connected to the positive side of the source a2
(iii)Connect a dc voltmeter across the HV winding
(iv)Close the switch and observe the movement of the pointer of the voltmeter
(v) If the pointer moves upscale, the transformer terminal connected to the (+) terminal of the
voltmeter is marked A2 and the other marked A1.

Fig. 23 Polarity test using dc source

29
5.20 Parallel Operation of single-phase transformers
Parallel connection of several transformers is widely used in electrical systems for the following
reasons:
(a) In many cases, the amount of power to be transformed is greater than that which can be built
into one transformer
(b) Frequently, the growth of load requires that the installed transformers supply an output
greater than their total kVA capacity. Additional transformers are then installed to run in
parallel with the existing transformer.
(c) It is sometimes found desirable to supply a load through two or more units in order to reduce
the cost of the spare unit required to ensure continuity of service in case of damage.

The following conditions must be fulfilled when operating two or more single-phase
transformers in parallel:
(a) The polarity should be the same. The polarity can be either right or wrong. A wrong polarity
results in a severe short circuit. Terminals of the same markings are connected together to
ensure correct polarity. See Fig. 24. If the polarity markings are either incorrect or not
present, the polarity of the incoming transformer can be checked by connecting a voltmeter
across the paralleling switch
(b) The voltage ratio should be the same. This is to avoid no-load circulating current and also
over-loading on one transformer when the paralleled transformers are loaded.
(c) The per-unit impedances should be equal in magnitude and have the same angle. When they
are equal in magnitude, the transformers share kVA loads in proportion to their respective
ratings. If both their magnitudes and angles are the same they will not only share kVA loads
in proportion to their respective ratings but also the combined load kVA will be the algebraic
sum of the kVA carried by each transformer. If the angles are different, the resultant kVA
capacity of the paralleled group will be slightly smaller than the sum of their individual
ratings if none should be overloaded. It is not very necessary that the angles should be the
same

Fig. 24 Connection ensuring correct polarity

30
5.21 Load sharing of parallel-connected transformers
The equivalent circuit of two transformers in parallel feeding a common load ZL is shown in Fig.
25. The voltage ratios are supposed to be equal and the magnetizing branch is neglected. The
circuit is referred to the secondary side but it may also be referred to the primary side.

I1 Z1
I
I2
Z2
V1
V ZL
a

Fig. 25 Equivalent circuit of two transformers in parallel

Z1 is the impedance of transformer 1 referred to the secondary side and Z2 is the impedance of
transformer 2 referred to the secondary side.

Using the current divider rule,

Z2
I1  I
Z1  Z 2
Z1
I2  I
Z1  Z 2

Multiplying by the terminal voltage V gives

Z2
S1  S (31.a)
Z1  Z 2
Z1
S2  S (31.b)
Z1  Z 2

where
S1  VI1  apparent power or load carried by transformer 1 (32.a)
S 2  VI 2  apparent power or load carried by transformer 2 (32.b)
S  VI  combined apparent power or load (32.c)

The apparent powers S1, S2 and S are in complex notation. With the load voltage V as reference
and the load current I lagging behind the voltage V by  , the combined load can be expressed as
S  VI    VI  cos 1 (load pf ) and a transformer 1 load power factor say, obtained as cos 1
where 1 is the phase angle of S1. V  V  magnitude of the load voltage and I  I 
magnitude of the load current.

31
Equations (31.a) and (31.b) hold for per-unit impedances provided that all are expressed with
reference to a common base power. The following equation can be used to obtain a new per unit
value with reference to a new base power.
S base,new
Ẑ new  Ẑ old  (33)
S base,old

Example 19
A 500-kVA transformer (Transformer 1) is connected in parallel with a 250-kVA transformer
(Transformer 2). The secondary voltage of each is 400 V on no load. Find how they share a load
of 750 kVA at power factor of 0.8 lagging if
(a ) Z 1  0.01  j 0.05 pu  0.0509978.69 pu and
Z 2  0.015  j 0.04 pu  0.0427269.44 pu
(b) Z 1  0.01  j 0.05 pu  0.0509978.69 pu and
Z 2  0.01  j 0.05 pu  0.0509978.69 pu
(c) Z 1  0.01  j 0.05 pu  0.0509978.69 pu and
Z 2  0.025  j 0.0444 pu  0.0509560.62 pu

Solution
If all the impedances are referred to a base power of 500 kVA, then only the impedances of
transformer 2 will change.
S base,new 500
Ẑ new  Ẑ old   Ẑ old   2 Ẑ old
S base,old 250
Case (a)
Z 1  0.01  j 0.05  0.05099 pu78.69 and
Z 2  20.015  j 0.04  0.03  j 0.08  0.08544 pu69.44

Further
Z 1  Z 2  0.01  j 0.05  0.03  j 0.08  0.04  j 0.13  0.1360 pu72.90
Total kVA load S  750  cos 1 0.8  750 kVA  36.9
Z2 0.0854469.44
S1  S   750  36.9   471 kVA  40.36
Z1  Z 2 0.13672.9
 471 kVA at power factor of cos 40.36  0.762 lagging

Similarly
0.0509978.69
S 2  750  36.9   281 kVA  31.11
0.13672.9
 281 kVA at power factor of 0.856 lagging

Remark:

32
Transformer 1 with larger per-unit impedance is under-loaded whereas transformer 2 with lower
per-unit impedance is overloaded.

Case (b)
Z 2  20.01  j 0.05  0.02  j 0.1  0.10198 pu78.69
Z 1  Z 2  0.03  j 0.15  0.15297 pu78.69
0.1019878.69
S 1  750  36.9   500 kVA  36.9
0.1529778.69
 500 kVA at power factor of 0.8 lagging
0.0509978.69
S 2  750  36.9   250 kVA  36.9
0.1529778.69
 250 kVA at power factor of 0.8 laggging

Remark:
Load shared in proportion to transformer ratings. Arithmetic sum of loadings is equal to the
combined load. A shorter approach can be used on recognizing that Z 1 pu  Z 2 pu

Case (c)
Z 2  20.025  j 0.0444  0.05  j 0.0888  0.1019 pu60.62
Z 1  Z 2  0.06  j 0.1388  0.1512 pu66.62
0.101960.62
S 1  750  36.9   505 kVA  42.9
0.151266.62
 505 kVA at power factor of 0.732 lagging
0.0509978.69
S 2  750  36.9   253 kVA  24.82
0.151266.62
 253 kVA at power factor of 0.91 lagging

Remark:
Transformers are slightly overloaded when the combined load is equal to the sum of individual
kVAs

6 Three-phase transformers

These are required to transform 3-phase power. The three-phase transformer may be either of the
following:

6.1 A three-phase transformer bank

This consists of three identical single-phase transformers having their windings externally
connected for three-phase working. The single-phase transformers retain all their basic single-
phase properties such as current ratio, voltage ratio and the flux in the core. The kVA capacity of
the bank is the sum of their individual ratings. See Fig.26.

33
6.2 A three-phase transformer unit
This is a single unit of special construction for three-phase working. Modern large transformers
are usually of the three-phase three-legged core type shown in Fig.27. A leg carries the primary
and secondary windings of a phase. The windings are internally connected. For a given total
capacity, 3-phase units are much cheaper in capital cost, lighter, smaller and more efficient.

Fig. 26 Delta-Star connection of single-phase transformers

Fig. 27 Delta-Star connection of three-phase, three-legged core-type transformer

34
6.3 Winding arrangement
The three windings, primary or secondary, can be connected in three different ways:

6.3.1 Star connection

For this connection, phase voltage = line voltage/√3 which is about 58 % of the line voltage. This
enables the insulation of the winding to be reduced to a minimum for a given supply voltage. For
star connection line current = phase current. It is the most economical connection for a high-
voltage winding.

6.3.2 Delta connection

For this connection, phase voltage = line voltage. Therefore winding must be insulated for the
full Line voltage. More turns are also required. With very high voltages a saving of 10 % may
be achieved by using star-connection rather than delta connection on account of insulation. The
saving is small, however, at voltages below 11 kV. For delta connection, phase current = line
current/√3 so the winding cross-sectional area is 58 % of that required for the star connection. It
it is the most economical for low-voltage winding.

6.3.3 Zigzag (or interconnected star) connection

It is a modification of the star connection. Each phase winding is divided into two sections and
placed on two different legs. The two sections are then connected in phase opposition. 15 %
more turns are required for a given phase terminal voltage compared with a normal star.

The three different winding arrangements give rise to several possible connection combinations:
star-star, star-delta, star-zigzag, delta-star, delta-delta, etc

6.4 Vector groups

The voltage induced in a primary phase winding is in phase with its corresponding secondary
phase winding. However, the phase angles of the primary and secondary line voltages may
differ depending on the type of winding connection. Three-phase transformers are given symbol
or code which indicates the type of connection used for the high-voltage and the low-voltage
winding and the phase displacement between the high-voltage side line to line voltage and its
corresponding low-voltage side line to line voltage. The symbol is in the form Xxn where X (the
capital letter) and x (the small letter) indicate the type of connection for the high-voltage and
low-voltage windings respectively and n is a clock hour number which indicates the phase
displacement. In this method of indicating the phase shift, the high-voltage side line to neutral
voltage is represented by the minute hand of a clock always set at 12 and the corresponding low-
voltage side line to neutral voltage represented by the hour hand. Thus when n = 11 it means the
clock reads 11 O’clock and the low-voltage side line to line voltage or line to neutral voltage
leads its corresponding voltage by 30o. In delta connection, the neutral point does not physically

35
Fig. 28 Vector symbols, diagrams and groups, winding connections,
and phase displacement of 3-phase transformers

36
exist and it has to be found geometrically. Two-winding three-phase transformers are classified
into four vector groups depending on the phase displacement as shown in the table below:

Vector Group Phase Displacement Vector Symbol or Code

1 zero Yyo Ddo Dzo
2 180o Yy6 Dd6 Dz6
3 30o lag Dy1 Yd1 Yz1
4 30o lead Dy11 Yd11 Yz11

Fig. 28 above shows in addition vector diagrams and winding connections. Before connecting
two or more transformers in parallel, their vector groups will have to be determined.

6.5 Three-phase transformer connections

More common transformer connections are

6.5.1 Delta-Delta connection (Fig 29.a)

This connection is economical for large low voltage transformer. Because there is no neutral
point, a four-wire supply cannot be given. It is used in a 3-phase transformer bank but rarely in
3-phase transformer unit. It is possible to use this arrangement to provide 3-phase power with
one transformer removed. This connection, known as open-delta or vee connection, can supply
up to 57.7 % of the load capacity of the delta-delta connection.

IL
L1 
L3
A1 I ph A2 a1 a2 
L3 L1
C2 B1 c2 b1

VL V ph

C1 B2 c1 b2

L2 L2
VL  V ph
I L  3I ph
Fig. 29.a Delta-delta connection

6.5.2 Delta-Star connection (Fig. 29.b)

It is commonly used to step up alternator voltage to transmission line voltage. Another common
application is in distribution service where as a step-down transformer, the windings are not the
most economical. The secondary star point can be earthed and a four-wire supply given.

37
6.5.3 Star-Delta connection

There is no secondary neutral and four-wire supplies cannot be given. The main use is as a step-
down transformer at the load end of transmission line.

Fig. 29.b Delta-star connection

6.5.4 Star-Star connection

This is economical for high-voltage transformer. However, if the secondary load is unbalanced,
the neutral point will be displaced and the line-to-neutral voltages will become unequal, a
situation referred to as floating neutral. If the neutral of the primary and the neutral of the source
are connected together usually by way of ground, the floating neutral problem can be eliminated.
Again if there is no primary ground, the primary and secondary phase voltages will contain
pronounced third harmonic voltages, which will cause the neutral point to oscillate above and
below the earth at a voltage equal in magnitude to the third harmonic voltage. To eliminate the
oscillation problem, a third delta-connected winding called tertiary winding is provided.
Although it can be used to supply additional power, the tertiary winding generally has no
external connection.

6.5.5 Star-Zigzag connection

The star-zigzag arrangement reduces third harmonic voltages and its neutral point is not affected
by unbalanced loading. The zigzag is restricted to comparatively low-voltage windings.

Example 20
Three single-phase step-up transformers rated 40 MVA, 13.2 kV/80 kV are connected in delta-
star on a 13.2 kV transmission line. If they feed a 90 MVA load, calculate the following:
(a) the secondary line voltage
(b) the currents in the transformer windings
(c) the incoming and outgoing transmission line currents.
Assume transformers are ideal.

38
Solution
(a) Let the voltage applied to the primary of a transformer be Vph1 and the secondary terminal
V ph 2 N 2 80
voltage be Vph2. Then considering the transformer to be ideal  
V ph1 N 1 13.2
Since the primary connection is delta, the voltage applied to the primary of each single-phase
transformer is equal to the transmission line voltage. Therefore
N 80
V ph 2  2  V ph1   13.2  80 kV .
N1 13.2
Since the secondary side is connected in star, the line voltage on the secondary side
 80 3  138 kV

V ph 2

V ph 1
13.2 kV
I ph 1

I ph 2

90
(b) The load carried by each transformer   30 MVA
3
30 MVA
From S  V ph1  I ph1 , current in the primary winding I ph1   2272 A
13.2 kV
30 MVA
From S  V ph 2  I ph 2 , current in the secondary winding I ph2   375 A
80 kV
(c) Current in incoming line  3I ph1  2272 3  3932 A
Current in outgoing line  I ph 2  I L  375 A

Alternatively, we can first calculate the line currents and use them to calculate the phase currents
or currents in the windings. To calculate a line current we use the formula, S  3V L I L .

Example 21
Three single-phase transformers have their primaries joined in delta to a 6600 V, three-phase,
three-wire supply. Their secondaries are connected to give a three-phase, four-wire output at 415
V across lines. The total load on the transformers is a balanced load of 150 kW at 0.8 pf lag. If
the voltage per turn on the primaries is 4, find
(a) the number of turns on the primary winding and the secondary winding
(b) the currents and voltages in all windings and lines, including the neutral wire on the
secondary side

39
(c) kVA load on each transformer

Assume transformers are ideal

Solution
V ph1 V ph 2
(a) Voltage per turn   4
N1 N2
For delta connection V ph1  V L1  6600 V
V ph16600
Therefore N 1    1650 turns
4 4
V 415
For star connection V ph 2  L 2 
3 3
V ph 2 415
Therefore N 2    60 turns
4 4 3

I L2
I L1
I ph1
415V 150 kW
V ph 2 Load 0.8 pf lagging
6.6 kV V ph1

Power 150  10 3
(b) Secondary I L 2  I ph 2    261 A
3VL cos  3  415  0.8
IN  0 A because load is balanced
 N 2 I ph 2  60  261
Primary phase current I ph1      9.5 A
 N1  1650
Primary line current I L1  9.5 3  16.4 A .

Example 22
A three-phase 415 V load takes a line current of 800 A from a 3300 / 415 V delta/star
transformer. The 3300 V system is supplied from an 11000/3300 star/star transformer. Draw the
circuit diagram and assuming no losses, find both line and phase values of voltages and currents
in each part of the circuit. What will be the turns ratios of both transformers?

40
Solution
Transforme r B Transformer A
I1 I 2  I3 I 4  800 A

V ph1 I ph 1 V ph 2 I ph 4 
I ph2 I ph 3 V ph 4 V 4
V1  11000 V2 V3 V ph 3 415 V

Voltages:
V4 415
V4  415 V; V ph 4   240 V
3 3
V3  3300 V; V ph 3  V3  3300 V
V2 3300
V2  V3  3300 V; V ph 2  
 1905 V
3 3
V 11000
V1  11000 V ; V ph1  1   6350 V
3 3

Turn ratios:
V ph3 3300
Transformer A    13.75
V ph 4 240
V ph1 6350
Transformer B    3.33
V ph 2 1905
Currents:
I 4  800 A; I ph 4  I 4  800 A
I ph 4 800
I ph 3    58.2 A
turns ratio A 13.75
I 2  I 3  3 I ph 3  3  58.2  100.8 A
I ph 2  I 2  100.8 A
I ph 2 100.8
I ph1    30.2 A
turns ratio B 3.33
I 1  I ph1  30.2 A

Check:
Input kVA  3 V1 I 1  3  11000  30.2  575000 VA  575 kVA
Output kVA  3 V4 I 4  3  415  800  575000 VA  575 kVA

41
6.6 Parallel operation of three-phase transformers (Fig. 30)
Three-phase transformers operating in parallel should have
(a) the same line voltage ratios
(b) the same per-unit impedances, i.e., they are equal in magnitude and in phase
(c) the same phase displacement between primary and secondary line voltages
(d) the same phase sequence

The last two conditions which are absolutely essential ensure that the secondary line voltages of
the transformers are in phase. When these conditions are not fulfilled a potential difference
appears across the paralleling switches S1 and S2.

From the view point of phase sequence and phase displacement, three-phase transformers which
can operate in parallel are:
(a) transformers of the same vector group. In this case the terminals with the same letter must be
connected to the same line as shown in Fig. 30.
(b) transformers having -30o phase displacement (vector group 3 transformers) and those having
+30o phase displacement (vector group 4 transformers). In this case two of the high voltage
connections and the corresponding low voltage connections are interchanged as shown in
Fig. 31.

Faulty internal connections in the transformer tank can cause the phase sequence of a transformer
to be reversed. Voltage across paralleling switches should therefore be monitored before the
switches are closed.
C B A a b c
Transformer 1

A2 a2

B2 b2

C2 c2

Transformer 2

A2 a2
S1
B2 b2
S2
C2 c2

Fig. 30 Parallel operation of 3-phase transformers

42
 a b c
C B A
Transformer 1

A2 a2

B2 b2

C2 c2

Transformer 2

A2 a2
S1
B2 b2
S2
C2 c2

Fig. 31 Parallel operation of group 3 & 4 transformers

7 Cooling methods

Cooling of transformer windings and core is provided to prevent rapid deterioration of the
insulating materials. There are several methods of transformer cooling. Each method is described
by a standard designation (or nomenclature) consisting of letter symbols. They are
(a) Letters for medium: air A, gas G, synthetic oil L, mineral oil O, solid insulation S, water W
(b) Letters for circulation: natural N, forced F

Up to four letter symbols are used for each method for which the transformer is assigned a
rating: some big transformers are designed to have a variable rating, depending on the method of
cooling used. The order of the symbols is
(i) the medium and
(ii) the circulation of the coolant in contact with the windings; and
(iii)the medium and
(iv)the circulation of the coolant in an external heat exchanger system.

43
Common methods of cooling transformers are:

7.1 Air Cooling (Dry type transformers)

7.1.1 AN

They use the ambient air as coolant and natural circulation by convection. The metallic housing
is fitted with ventilating louvers. They have ratings up to 50 kVA for low-power transformers.
Using high-temperature insulating materials, glass and silicone resins, make ratings up to 1.5
MVA possible. The low-power transformers are used inside buildings where the air is clean and
high-power ones are for special conditions such as those in mines.

7.1.2 AF

Forced air circulation is used to raise transformer loadings

7.2 Oil-immersed, Oil cooling

Oil is a much better insulator than air. Consequently, it is invariably used on high-voltage
transformers.

7.2.1 ONAN

Natural oil circulation and natural air flow over the tank are used. It is very common for
transformers rated up to 5 MVA. With radiators, it is possible to build units up to 40 MVA. We
note that limit of output is determined by tank size and cost.

7.2.2 ONAF

Cooling fans blow air over the radiators to enable a much bigger output from a transformer of a
given size. With this method of cooling ratings up to about 75 MVA can be built.

7.2.3 OFAF

Pumps are used to circulate the oil and cooling fans to blow air over radiators. This is the usual
method for transformers of 30 MVA and upward. Both OFAF and ONAN may be used on a unit
with ONAN up to 0.5 per unit rating. Change over is initiated automatically by temperature-
sensing elements. Three-phase type OFAF step-up transformer rated 1300 MVA installed at a
nuclear power generating station is one of the largest units ever built.

44
7.3 Oil-immersed, Water cooling
7.3.1 ONWF

Copper cooling coils are mounted in the tank above the level of the transformer core, but below
the oil surface.

7.3.2 OFWF

Oil is circulated by pump from the top of the transformer tank to an external oil/water heat
exchanger. Oil returns when cold to the bottom of the tank. Its advantages over ONWF include
(a) The transformer is smaller and the tank does not have to contain the cooling coils.
(b) Leakage of water into oil is improbable if oil pressure is greater than that of water

This method is used for large installations. It is commonly used in generating stations,
particularly hydro stations where ample supply of water is available.

8 Tap-changing Transformers

Most power transformers have tappings on coils brought out to terminals so that the number of
turns on one winding can be changed. The turns-ratios are changed in order
(a) to maintain the secondary voltage at their rated value under the varying conditions of load
and power factor. The secondary terminal voltage may vary with changes in load over an
undesirably large range, because of changes in the impedance drop in the transmission lines
and transformers
(b) to control the flow of reactive power between two interconnected power systems or between
component parts of the same system, at the same time permitting the voltages at specified
points to be maintained at desired values.

Tappings on power transformers permit voltage adjustment within  5% . Low- and medium-
power transformers usually have three taps per phase: +5%, 0, and -5% variations in the turns
ratio. Higher power ratings usually have five: +5%, +2.5%, 0, -2.5% and -5% variations in the
turns-ratio. The principal tapping “0” is that to which the rating of the winding is related. A
positive tapping includes more, and a negative less turns than those of the principal tapping.

Tappings are usually fitted on the higher voltage winding to obtain tappings within fine limits.
Consider say, 11kV / 433 V, 600 kVA delta-star distribution transformer having volts / turn = 10.
On the low-voltage side, N  433 10 3  25 . Adjustment can then be in steps of 4%. If 5 %
and 10 % are required, we shall use 4 % and 8 % or 12 %.
On the high-voltage side N  11000 10  1100 and it is possible to make adjustment in steps of
0.09 %.

8.1 Changing the taps of transformers

Tap-changing may be either on-load or off-load tap-changing:

45
8.1.1 Off-load tap changing

The changes are made when the transformer is disconnected from the primary circuit. The most
common off-load tap changing transformer has tappings inside the tank and connected to an
internal switch which is operated by an external switch handle (usually by rotary movement of a
hand wheel). Fig. 32 shows a three-phase star connected winding with taps made at the neutral.
Off-load tap changing is simple and inexpensive and it is commonly used with distribution
transformers where occasional adjustments are required.

Fig. 32 Three-phase star connected winding with taps at the neutral point

8.1.2 On-load tap changing

Daily and short-time adjustment is generally by means of on-load tap-changing gear. Tap
changing is done without breaking the circuit. Momentary connection must be made
simultaneously to two adjacent taps during the transition, and the short-circuit current between
them must be limited by some form of impedance known as transition impedance. Centre-tapped
iron-core inductors (the reactor method) or resistors (the resistor method) are used for this
purpose.

Reactor method
This method has now almost entirely been superseded by the resistor method. It is manufactured
and used only in the USA.

Resistor method
In modern designs the transition impedance is almost invariably obtained by means of a pair of
resistors. An arrangement of such a tap changer for one phase is shown in Fig. 33. In this figure,

46
the diverter switch and the even tap selector are shown in the position when the tap T2 is brought
in circuit.

To move to the next tap, T3, the odd tap selector should first be moved to that tap (see the dashed
lines in Fig. 33.a), and the diverter switch may then be rotated clockwise. The ensuing sequence
of events is as follows:

T8
T7
T6
T5
T4 even tap selector
odd tap selector T3
T2
TS 2
T1
TS1

R1 R2
I
2 3
1
4
I

Fig. 33.a. Resistor tap changer

- contacts 3 and 4 break, contacts 1 and 4 make, Fig. 33.b

R1 R2
2 3 1
1
I-i
1
i Ii
2 2
4
I

Fig 33.b

47
- contacts 1 and 4 break, and contacts 1 and 2 make, Fig. 33.c

Fig 33.c

The tap selectors may be moved from tap to tap only when their circuits are de-energized. The
resistors are short-time rated so their time of duty must be kept low. For the same reason, means
must be adopted to ensure that it cannot be inadvertently left in the bridging position.

On-load tap changer control gear can vary from simple push-button initiation to a complex
automatic control of as many as four transformers in parallel.

9 Autotransformers

An autotransformer has a single tapped winding which serves both primary and secondary
functions as shown in Fig. 34. The circuit diagrams are shown in Fig. 35.

I1 A
I2 C

C I1 A
I2
N1
V1 N2
N2 V1 N1 V2 Load
B
V2 Load
B

B B
Fig. 34.a Step - down autotransf ormer Fig. 34.b Step - up autotransformer

48
9.1 Autotransformer equations
If we neglect losses, leakage flux and magnetizing current then the turns ratio
N V I
a 1  1  2
N 2 V2 I 1

9.2 Advantages and disadvantages of autotransformer over two-winding

transformer
The main advantage gained in the use of autotransformer is the saving of copper. For a two-
winding transformer and an autotransformer which can perform the same duty (they
should have the same voltage per turn and therefore the same flux. We can also assume the same
mean length per turn)

Volume of copper in autotransformer V

 1 L
Volume of copper in two - winding transformer VH

Or
 saving of copper effected by   VL 
      Volume of copper in the two winding transformer 
 using an autotransformer   VH 

where
VL  voltage on the low voltage side
VH  voltage on high voltage side

In practice, voltage ratios VL V H less than about 1 3 show little economic benefit over two-
winding transformer because of other factors such as cost of insulation.

The main disadvantage is that the primary and secondary circuits are not isolated from each
other.

49
Example 23
An autotransformer is required to step up a voltage from 220 to 250 V. The total number of turns
is 2000. Determine (a) the position of the tapping point (b) the approximate value of the current
in each part of the winding when the output is 10 kVA and (c) the economy in copper over the
two winding transformer having the same peak flux and the same mean length per turn.

Solution
I2

I1
N2 V2

V1 N1

N 1 V1 220 220 220

(a )   or N 1   N2   2000  1760
N 2 V2 250 250 250
Position is 240 turns from one end.

10  10 3
(b) V2 I 2  10  10 3  I 2   40 A
250
10  10 3
V1 I 1  10  10 3  I 1   45.45 A
220

Current in series winding is 40 A and current in common winding  45.45  40  5.45 A

V 220
(c) Saving in copper  L   0.88 pu or 88 per cent of what is used in the two-winding
VH 250
transformer.

9.3 Two-winding transformer connected as an autotransformer

A two-winding transformer can be changed into an autotransformer by connecting the primary
and secondary windings in series. The following rules apply whenever a two-winding
transformer is connected as autotransformer:
(a) the current in any winding should not exceed its current rating
(b) the voltage across any winding should not exceed its voltage rating
(c) rated current in one winding gives rise to rated current in the other
(d) rated voltage across one winding gives rise to rated voltage across the other
(e) if current in one winding flows from say A2 to A1, then current in the other winding must
flow from a1 to a2 and vice versa

50
(f) the voltages add when terminals of opposite polarity (A1 and a2 or A2 and a1) are connected
together by a jumper. The voltages subtract when terminals of the same polarity (A1 and a1 or
A2 and a2) are connected together.

Example 24
A two-winding single-phase transformer rated 15 kVA, 600 V / 120 V, 60 Hz. We wish to
reconnect it as an autotransformer in three different ways to obtain three different voltage ratios:

(a) 600 V primary to 480 V secondary

(b) 600 V primary to 720 V secondary
(c) 120 V primary to 480 V secondary

Calculate the maximum load the transformer can carry in each case.

Solution
Let the voltage on the high voltage side of the two-winding transformer be VH and voltage on the
low voltage side be VL. Then one of the rated voltages of the autotransformer is always given by
VR = VH or VR = VL. The other must be derived from the two windings in series as VD = VH +VL
by connecting terminals of opposite polarity (A1 and a2 or A2 and a1) together by a jumper or as
VD = VH – VL by connecting terminals of the same polarity (A1 and a1 or A2 and a2) together by a
jumper.

(a) Here VR = 600 V and VD = VH – VL = 600 – 120 = 480 V. Therefore any two terminals of the
same polarity can be connected in series to derive the 480 V. The 480 V, being the secondary
voltage of the autotransformer, goes to load.

To obtain the maximum load the autotransformer can carry, we note that the current in the
120-V winding, connected in series with the load in the autotransformer connection, is not
supposed to exceed its current rating given by I 2 fl  S rated 120  15  10 3 120  125 A . This
then limit the load current to 125 A. And at a load voltage of 480 V, the maximum power
becomes 125  480  60 kVA .

51
(b) The secondary voltage 120 V must be added to the 600 V to obtain 720 V

(c) The 120 V becomes the primary of the autotransformer and the 120 V is subtracted from the
600 V to obtain its secondary

9.4 Applications of autotransformers

They are mainly used for
(a) variac
(b) interconnecting power systems that are operating at roughly the same voltage (e.g. 132 kV,
275 kV, 400 kV) and
(c) starting squirrel-cage induction motors.

10 Instrument transformers

They are used in ac circuits to serve these purposes:

(a) to make possible the measurement of high voltages with low-voltage instruments or large
currents with low current ammeters
(b) to insulate high voltage circuits being monitored from measuring circuit in order to protect
the measuring apparatus and operator
(c) to energize relays for the operation of protective and automatic control devices

52
The load on the secondary of an instrument transformer is called its burden and is expressed in
volt-amperes (VA). There are two types of instrument transformers: the voltage (or potential)
transformer and the current transformer

10.1 The voltage or potential transformers (VTs or PTs)

The construction is similar to a power transformer. The primary is connected directly to the
power circuit either between two phases or between a phase and ground and the secondary is
connected to instruments and coils of relays. Sufficient insulation is provided between the
primary and the secondary to withstand the full line voltage as well as the very high impulse
voltage. Voltage transformers are designed to step down the primary voltage to a nominal or
rated voltage of 110 V so that standard instruments and relays can be used. They introduce
errors of two kinds into measurement being made: the ratio errors (the ratio between input and
output voltages is not constant under all conditions of load) and the phase angle errors (the phase
shift between input and output voltages is not zero). These errors are due to the exciting current
and the equivalent series impedance of the transformer and they are kept low by using high
quality iron (high permeability and low loss) and operating it at low flux densities so that the
exciting current is very small. The resistance and reactance of the windings are also made very
low.

Fig. 36 shows the circuit for a potential transformer. One terminal of the secondary winding is
always earthed. The windings though insulated from each other, are connected invisibly together
by distributed capacitance between them. By earthing one of the secondary terminals, the
highest voltage between the secondary lines and earth can never rise above that of the secondary
voltage.

H.V. ac circuit

PT

110 V

V
0  150 V
Fig. 36 Potential transformer installed on H.V. circuit

10.2 Current transformers (CTs)

Their primary consists of small number of turns connected in series with the power circuit load.
The secondary consists of a larger number of turns and it is connected to ammeter, current coils
of other instruments or current coils of relays. Current transformers have the ratio of primary to

53
secondary current approximately constant. The nominal or rated secondary currents are usually
5 A or 1 A, irrespective of the primary current rating. The transformer ratio is usually stated to
include the secondary current rating. Current transformers also introduce two errors in
measurement: the ratio error (the ratio between primary and secondary currents is not constant)
and phase angle error (the phase angle between the primary and secondary currents is not zero).
The basic cause of ratio and phase angle errors is the exciting current. To keep the exciting
current small, a high quality iron operating at very low flux densities is used as in PTs. In CTs
the secondary leakage impedance and impedances of the secondary leads and instruments should
also be very low; for any increase in these impedances increases the core flux and therefore the
exciting current. The transformer is connected in the power circuit as shown in Fig. 37. As in the
case of PT (and for the same reasons) one of the secondary terminals is always earthed

Current transformer secondary circuit must not be opened while current is flowing in the
primary. Without opposing ampere-turns the line current, which may be 100 to 200 times the
normal exciting current, becomes the exciting current. The iron core becomes saturated and very
high voltage spikes (several thousand volts) are induced across the open-circuited secondary.
These voltages are dangerous to life and to the transformer insulation. The core when it becomes
saturated can also cause excessive heating of the core and windings. Therefore when it is desired
to remove a load from the secondary circuit, the secondary winding must first be short circuited.

When the line current exceeds 100 A we can sometimes use a toroidal or bar-primary (N1=1)
transformer shown in Fig. 38. It consists of a laminated ring-shaped core which carries the
secondary winding. The primary is composed of a single conductor that simply passes through
the centre of the ring as shown in the figure. Toroidal CTs are simple and inexpensive and are
widely used in HV and MV indoor installations.

Load current

ac supply Load

CT

1 A or 5 A

A
1 A or 5 A
ammeter

Fig. 37 Current transformer installed in H.V. circuit

54
 I 1  600 A

Bar  primary
N1  1

I2  3 A
A

Fig. 38 1000A / 5A bar-primary transformer measuring current in a line

Current transformers are also commonly used for the measurements of large currents even when
the circuit voltage is not dangerously high. This avoids bringing heavy leads to the instrument
panels. Whereas instrument CTs have to remain accurate up to 120 % rated current, protection
CTs must retain proportionality up to 20 times normal full load

Example 25
A potential transformer rated 14400 V / 115 V and a current transformer rated 75 A / 5 A are
used to measure the voltage and current in a transmission line. If the voltmeter indicates 111 V
and the ammeter reads 3 A, calculate the voltage and current in the line.

Solution
V1 14400
Treating the PT as an ideal transformer, we have  .
V2 115
14400 14400
Therefore the transmission line voltage V1   V2   111  13900 V
115 115
I 75
Treating the CT also as ideal transformer, we have 1 
I2 5
75 75
Therefore the transmission line current I 1   I2   3  45 A
5 5

Example 26
The toroidal current transformer of Fig. 38 has a ratio of 1000 A / 5 A. The line conductor carries
a current of 600 A.
(1) Find the voltage across the secondary winding if the ammeter has an impedance of 0.15 Ω
(2) Find the voltage drop the transformer produces on the line conductor
(3) If the primary conductor is looped four times through the toroidal opening, find the new
current ratio

55
Solution
5
(a) Current in the secondary I 2   600  3 A
1000
Voltage drop across the burden  3  0.15  0.45 V
(b) Voltage drop on line conductor = primary voltage V1
I 3
V1 I 1  V2 I 2 . Therefore V1  2  V2   0.45  2.25 mV
I1 600
(c) For a bar primary transformer I 1rated x1  I 2 rated N 2 .
If the primary conductor is looped 4 times, the number of turns on the primary side becomes
4 and we should have I 1rated , new x 4  I 2 rated N 2 .
I 1rated 1000
From the two equations, we obtain I 1rated , new    250 A
4 4
Therefore the new current ratio =250 A/5 A

56
11 Further exercises

(1) A single-phase transformer has a primary winding with 1,500 turns a nd a secondary winding
with 80 turns. If the primary winding is connected to a 2300-V, 50-Hz supply, calculate (a)
the secondary voltage (b) the maximum value of the core flux. Neglect the primary
impedance

(2) A single-phase 2,300/230-V, 500-kVA, 50-Hz transformer is tested with the secondary open-
circuited. The following test results were obtained: V1 = 2,300 V, Io = 10.5 A, and Po = 2,300
W. Calculate (a) the power factor (b) the core-loss current Ip (c) the magnetizing current Im.

(3) A 150-kVA, 2,400/240-V, 50-Hz single-phase transformer has the following resistances and
reactances: R1 =0.225 Ω, Xl1 = 0.525 Ω, R2 = 0.00220 Ω and Xl2 = 0.0445 Ω. Calculate the
transformer equivalent values (a) referred to the primary (b) referred to the secondary.

(4) A 150-kVA, 2,400/240-V, 50-Hz single-phase transformer has the following resistances and
reactances: R1 =0.225 Ω Xl1 = 0.525 Ω, R2 = 0.00220 Ω, Xl2 = 0.00445 Ω, Rm = 10 k Ω and
Xm = 1.5 k Ω. The transformer is supplying full-rated load at 0.85 lagging power factor and
rated secondary terminal voltage. Calculate (a) I2 (b) Ip (c) Im (d) Io (e) I1 (f) V1. Use the
exact equivalent circuit referred to the primary side.

(5) The results of open- and short-circuit tests carried out on a 230/115-V, 60-Hz single –phase
transformer are

Test type Primary Secondary

Open circuit Open 115 V, 6.5 A, 192 W
Short circuit 17.5 V, 43.5 A, 234 W Short-circuited

Calculate the parameter of the approximate equivalent circuit of Fig 13.a

(6) The results of open- and short-circuit tests on a 100 kVA, 11,000/2,200-V, and 50-Hz single–
phase transformer are

Test type Primary Secondary

Open circuit Open 2,200 V, 1.5 A, 800 W
Short circuit 600 V, 10.0 A, 1,000 W Short-circuited

Determine (a) Re1 and Xe1 (b) Re2 and Xe2 (c) the percentage regulation at 0.75 power factor
leading, unity power factor and 0.85 power factor lagging

(7) From the data of Exercise 6 calculate the transformer efficiency at 0.8 power factor lagging
for (a) 50 % (b) 100 % of rated full load.

57
(8) A 100-kVA, 11,000/220-V, 50-Hz transformer has a core loss Po of 800 W and Re2 of 0.48
Ω. Calculate the secondary current for maximum efficiency.

(9) A 10-kVa, 2,400/240-V, 60-Hz single-phase distribution transformer has a full-load copper
loss of 125 W and a core loss of 63 W. It supplies the following loads over a 24-h period:

No load for 2.5 h

15 % rated load at 0.7 power factor lagging for 3 h
40 % rated load at 0.75 power factor lagging for 4.5 h
75 % rated load at 0.8 power factor lagging for 5 h
100 % rated load at 0.85 power factor lagging for 6 h
110 % rated load at unity power factor for 3 h

Calculate (a) the total energy loss over the 24-h period (b) the total energy output over the
24-h period (c) the all-day efficiency.

(10) A 10-kVA 2,400/240-V 50-Hz single-phase transformer is reconnected to step down a

voltage from 2,640 to 2,400 V. Calculate (a) the kVA rating as an autotransformer

(11) A 75-kVA 230-V three-phase load is supplied from a 6,600-V three-phase supply using a
star-star-connected transformer bank. What are the voltage, current and kVA ratings of the
single-phase transformers?

(12) Three single-phase transformers are connected in delta-delta, and are used to step down a
line voltage of 110 kV to 66 kV to supply an industrial plant drawing 50 MW at a 0.80
power factor. Calculate 9a) the high-voltage-side line current (b) the low-voltage-side line
current (c) the primary phase currents (d) the secondary phase currents

58
 CHAPTER TWO

ELECTROMECHANICAL ENERGY CONVERSION

1 Introduction

In electromechanical energy conversion, energy is converted between electrical and mechanical

systems. The energy conversion process takes place through the medium of electric or magnetic
field of the conversion device which links the two systems. In the conversion process, the device
releases energy from what is stored in its field to its output and takes a new supply from its input
by producing a reaction in the system at its input. Although the various conversion devices
operate on similar physical principles, the structures of the devices depend on their function. The
conversion devices can be placed in three groups:

(a) Signal handling devices: They are used to generate output signals which have linear
relationship with input signals. They generally operate with small signals. Examples are
pickups, loudspeakers, microphones and tachometers.

(b) Force-producing devices: They are used to produce bulk and non-controlled forces or
torques. Examples are lifting magnets, iron-core, solenoid-actuators, relays, contactors and
iron-vane instrument.

(c) Continuous energy-conversion devices: These devices deal with large amounts of energy.
Examples are motors and generators.

A transducer is a device which takes energy from one system and supplies it to another. This
term, however, is usually used for the devices for obtaining signals for measurement and control
which operate at very low power levels. Transducers are signal handling devices.

2 Forces and Torques Developed by Electromechanical Devices

A method which is simple and is applicable to all the conversion devices will be used to develop
expressions for the forces and torques. The method is based on the law of conservation of
energy.

3 Energy balance

Suppose that the energy conversion device is connected at its input to an electrical source and at
its output to a mechanical system. Then calculating from an initial zero-energy condition, the
energy balance is

59
(electrical energy input, welec ) = (stored field energy, w fld ) + (mechanical energy output, wmech ) +
(energy lost, wloss ) (1)

This relation equally applies to energy changes (i.e. changes after an established condition or
state) and to energy rates, i.e. power:
dwelec  dw fld  dwmech  dwloss (Energy balance) (2)
dw fld dwloss
Pe   Pm  (Power balance) (3)
dt dt

The energy loss generally arises from these causes:

(a) part of the electrical energy converted into copper loss in the electric circuit
(b) part of the mechanical energy converted into friction and windage loss dissipated away as
heat
(c) part of the energy absorbed by the coupling field converted into iron loss (for magnetic field
coupling) or dielectric loss (for electric field coupling) dissipated away as heat.

The copper, friction and windage losses do not play basic part in the energy conversion process.
They can therefore be considered as losses in the electrical and mechanical systems on the two
sides of the conversion device. With the losses considered this way, welec in equations (1) and
(2) becomes the electrical energy supplied to the coupling field, wmech becomes the gross
mechanical output from the coupling field and wloss becomes iron or dielectric loss only. The
schematic diagram of such a device is shown in Fig. 1

Fig. 1 General representation of electromechanical device

4 Force in Singly Excited Electric Field System

These devices are built with only one electric circuit. A model of such a system is shown in Fig.
2. The conversion device consists of two parallel plates separated by air. One of the plates is

60
fixed and the other is movable. The movable part is designed to have one degree of freedom. In
the equilibrium
f mech  f fld (4)

where
fmech = the gross mechanical force exerted by the mechanical system and
ffld = the force produced by the field called the field force.

Fig. 2 Singly excited electric field system

We are interested in the expression for the field force. To derive it, we shall apply the principle
of virtual work: we assume that a small displacement takes place, calculate the work that would
be done in such a virtual displacement, and then use the work done to determine the actual force.

Suppose an incremental displacement dx of the movable plate is made in a time dt . During this
displacement
f fld dx  dwmech
f fld dx  dwelec  dw fld  dwloss (5)

The electrical energy input to the device is

dwelec  eidt
dwelec  edq (6)

The dielectric loss is negligible so

dwloss  0 (7)

The energy supplied to the coupling field during the small displacement is
1 
dw fld  d  qe 
2 
1 1
dw fld  qde  edq (8)
2 2

61
Substituting equations (6), (7) and (8) into equation (5) gives
1 1 1 1
f fld dx  edq  qde  edq  edq  qde
2 2 2 2

From which
1 dq 1 de
f fld  e  q (9)
2 dx 2 dx

Now
q
c (10.a)
e

Therefore
dc 1  dq de 
 2 e -q (10.b)
dx e  dx dx 

Substituting equation (10.b) into equation (9) yields

1 dc
f fld  e 2 (11)
2 dx

The expression for the field force derived is general and is applicable to electromechanical
devices of similar method of excitation. To apply the force expression, it will be necessary to
evaluate capacitances of the systems as functions of position.

Example 1
Two parallel plates as shown in Fig. 2 are maintained at a potential difference e of 104 V. Each
plate has an area of 0.02 m2
(a) Find the force between the plates as a function of their spacing x
(b) Find the energy converted to mechanical form as the plate spacing is reduced from 1 cm to
0.5 cm

Solution
 o r A
(a) For the two parallel plate the capacitance as a function of the spacing x is C( x) 
x
1 2 dc   Ae 2
Therefore f fld ( x)  e  o r 2
2 dx 2x
The negative sign implies that the direction of the force is the reverse of that shown, i.e.
f fld will rather act to reduce the plate spacing.
1 2 dc 1
(b) For a small change dx , the work done dwmech  f fld ( x)dx  e dx  e 2 dc
2 dx 2
1 C2 1 1 1 1
Therefore Wmech  e 2  dc  e 2 (C 2  C1 )  e 2 o  r A  
2 C1 2 2  x 2 x1 

62
 With e  10 4 V,  o  8.85  10 12 ,  r  1, A  0.02 m 2 , x 2  0.005 m and x1  0.01 m
Wmech  8.85  10 4 Joules

5 Force in Singly Excited Magnetic Systems

These devices are also built with only one electric circuit. They have no permanent magnet yet
they are capable of producing motor and generator action. They are used for instruments,
microphones and relays. Fig. 3 shows a model of the system. The magnetic circuit of the device
has air gap of length x between fixed and movable members.

Movable
Fixed member member
x
ffld
i
+ fmech

+ e
V(t)
- -

Mechanical
system

Electrical
system

Fig. 3 Singly Excited Magnetic field system

Applying here also the principle of virtual work, we obtain

f fld dx  dwmech
f fld dx  dwelec  dw fld  dwloss (12)

The electrical energy input in the time dt

dwelec  ei dt (13)

The induced voltage e arises only from the rate of change of flux linking the coil. From
Faraday’s law
d
e (14)
dt

The reluctance of the iron part in the magnetic circuit is assumed to be linear. In this case the
induced voltage can be expressed as
d di dL
e  ( Li )  L  i (15)
dt dt dt

63
Substituting equation (15) into equation (13) yields
dwelec  Li di  i 2 dL (16)

The iron loss is assumed to be negligible. Therefore

dwloss  0 (17)

The energy supplied to the coupling field

1  1
dw fld  d  Li 2   i 2 dL  Lidi (18)
2  2

Substituting equations (16), (17) and (18) into equation (12) gives
1 1
dwmech  Li di  i 2 dL  i 2 dL  Li di  i 2 dL
2 2

The above expression for the work done can be used in equation (12) to obtain
1 dL
f fld  i 2 (19)
2 dx

To apply this force expression, the inductance of the system must be evaluated as function of
position.

6 Useful relations for calculating forces and torques

N2
(a) Inductance of the coil L  where N = number of turns of the coil and R is the total
R
reluctance of the magnetic circuit.
l
(b) Reluctance R  where A is the cross sectional area and l is the length of the
o r A
magnetic circuit.
q
(c) Capacitance of a system C  where q is the charge stored in the system and e is the
e
potential across the system.
 A
(d) Capacitance of two-parallel-plate capacitor system C  o r where A is the cross
d
sectional area of the plate and d is the spacing between the plates.

64
Example 2
A steel electromagnet is used to support a solid block of steel having a mass of 908 kg as shown
in Fig. 4. A force of 8900 N is required to support this weight. The cross-sectional area of the
magnet (part 1) is 0.01 m2. The coil has 700 turns. Assume both air gaps are 0.0015 m long.
Neglect the reluctance of the steel parts. Neglect fringing in the air gaps. Find the minimum
current that can keep the weight from falling.

i
Part 1
(fixed)

x x

Part 2
(movable)
Fig. 4 See Example 2

Solution
2x
The total reluctance, R( x)  , where A is the cross sectional area of the air gaps.
0 A
N 2 o A
Therefore the inductance of the coil L( x) 
2x
1 2 dL 1 2   N  o A 
2
N 2  o AI 2
The force f fld ( x)  I  I   
2 dx 2  2 x 2  4x 2

The cross sectional area of the air gaps is equal to the cross sectional area of magnet (part 1) if
fringing is neglected.
7 2  10 4  4  10 7  0.01i 2 1960 I 2
Therefore when x  0.0015 m , f fld   
4  15 2  10 8 9

1960 I 2 8900  9
Now  8900 . Therefore I 2   13 and I  3.6 A
9 1960

Example 3
Show that the force equation (19) can be restated as
1 dR
f fld    2
2 dx

where Φ is the total flux linking the coil and R is the total reluctance of the magnetic circuit.

65
Solution
N 2 dL N 2 dR
From L   ,  2
R dx R dx

Substituting this in equation (19) yields

1 N 2 i 2 dR 1 dR
f fld   2
  2 (20)
2 R dx 2 dx

7 Magnetic Field Energy and Coenergy

The total energy stored in the magnetic field of the magnetic circuit given in Fig. 3 can be
expressed in magnetic circuit terms as
1
W fld   Fd (21)
0

To evaluate the integral, the mmf F of the coil must be expressed as a function of the total flux  .
The relation between the two quantities is described by a saturation curve. For a linear magnetic
circuit, the saturation curve is a straight line passing through the origin as shown in Fig. 5.

The total stored energy is given by the area oabo between the saturation curve and the vertical or
flux axis.

Fig. 5 Saturation curve of linear magnetic circuit

Referring to Fig.5, the area obco is given by

F1
   dF
W fld (22)
0

 given by this area has no physical meaning, but it is a useful function for
The quantity W fld
evaluating the field force. It is called the coenergy. For a given saturation curve, linear or
nonlinear
  F
W fld  W fld (23)

If the magnetic circuit is linear, energy and coenergy are equal and they are given by
  12 F
W fld  W fld (24)

66
If motion is made to occur at a constant flux or so fast that the flux remains constant, it can be
shown that the input electrical energy is zero and mechanical work is done solely at the expense
of stored magnetic energy, i.e.
0  dW fld  F fld dx (25)

The device in Fig. 2 has several saturation curves, one for every value of the gap length x . Fig. 6
shows two of such curves for x  x1 and x  x 2 where x1 is larger than x 2 . Referring to this figure,
the magnetic field energy reduces during motion at a constant flux 1 from the area oaco to the
smaller area oabo. The difference in area given by area obco represents the mechanical work
done.

Fig. 6 Motion at constant flux

If motion is made to occur at a constant current (or mmf) or so slowly that the current remains
constant, the mechanical work done is equal to increase in coenergy, i.e.
  F fld dx
0   dW fld (26)

Referring to Fig. 7, the coenergy increases during motion at a constant flux F1 from the area oabo
to the larger area oaco. The difference in area given by area obco represents the mechanical work
done.

Fig. 7 Motion at constant mmf

The equations (25) and (26) can be solved for the field force which comes out as partial
derivatives as follows:

67
  W fld 
f fld    (27)
 x   cons tan t


 W fld 
f fld    (28)
 x  i cons tan t

To evaluate equations (27) and (28) for the field force, W fld is expressed as function of  and x ,
 as function of F or i and x .
and W fld

Example 4
Use equations (27) and (28) to derive force expressions (19) and (20).

Solution
For linear magnetic circuit
  12 F
W fld  W fld

In terms of the current i and the gap length x or in terms of the current i and coil self inductance
L, L being a function of x , the coenergy is given by
1  N2  1 2
  F  Ni   i 2 
1 1 Ni
W fld  i L
2 2  R  2  R  2

Substituting it into the force equation (28) yields

1 dL
f fld  i 2
2 dx

In terms of flux  and the gap length x or in terms of flux  and the total reluctance R, R being a
function of x , the magnetic field energy is given by
1 1 1
W fld  F  R    2 R
2 2 2

Substituting it into the force equation (27) gives

1 dR
f fld    2
2 dx

Example 5
A relay mechanism has a saturation curve which can be approximated by   M F x ,
where M  9  10 8 .
(a) Find the stored energy as a function of  and x
(b) Find the coenergy as a function of F and x
(c) Find the derivatives of energy and coenergy with respect to the gap length x
(d) Find the mechanical force for x  0.01 m and   0.0006 webers

68
Solution
(a) Since the saturation curve is nonlinear, to find the field energy we use equation (21):
2 2 2
 x  
3
1 1  x  2  x  1
W fld   Fd      d      d   
2
0 0
M  M  0
M  3

(b) To find the coenergy we use equation (22)

F1  M  F1 2M 
   dF     F 1 2 dF    F 3 2
W fld
0
 x  0 3 x 
Alternatively,
2 3
M  12 1 x   M  32  M  32 1 M  32 2M  32
  F  W fld  F 
W fld F      F   F   F   F
 x  3 M   x   x  3 x  3 x 
W fld  2 x   
3
(c)   2  
x  M  3 

W fld 2
 M  2
     F 3 2
x  x  3

(d) The given values are substituted into equation (27) to obtain the force:
 W fld 
f fld     2 x   
3
  2    
 2  0.01  6  10 4 3 
  
  178 N
8 2 
 x  cons tan t  M  3  
 9  
10  
3 

Example 6
A relay mechanism has reluctance in terms of the gap length x as R  9  10 8 0.003  x  AT/Wb .
The relay coil has 1620 turns and 55 Ω resistance. The external voltage source is 110 V dc.
(a) Find the energy stored in the magnetic field when the relay is open ( x  0.006 m )
(b) Find the energy stored in the magnetic field when the relay is closed ( x  0.001 m )
(c) Find the work done if the relay is allowed to close slowly from x  0.006 m to x  0.001 m
(d) Find the work done if the relay is allowed to close fast from x  0.006 m to x  0.001 m

Solution
(a) The current in the coil is I 1  V R  110 55  2 A
The mmf is F1  NI 1  1620  2  3240AT
For x  0.006 m , the reluctance of the magnetic circuit is
R1  9  10 8 0.003  0.006  8.1  10 6 AT/Wb

The flux across the air gap is 1  F1 R 1  3240 8.1  10 6  0.0004 webers 
For a fixed value of x , the relation between  and F is a straight line. Therefore from equation
(24), the energy stored in the magnetic field is obtained as
W fld 1  12 F11  12  3240  0.0004  0.648 J

69
(b) In the steady state, the current and mmf have the same values as in (a).
For x  0.001 m , the reluctance of the magnetic circuit is
R2  9  10 8 0.003  0.001  3.6  10 6

The flux across the air gap is  2  F1 R 2  3240 3.6  10 6  0.0009 webers 
W fld 2  12 F1 2  12  3240  0.0009  1.458 J
(c) For slow closure or closure at constant current,
x2 x2 L2 2
1 dL 1 1 2
Wmech   f fld dx   i 2 dx   2i
2
dL  i L2  L1 
x1 x1
2 dx L11
2
1 2N2 N2  1  NI 1 NI 1  1
 i     NI 1     F1  2  1   0.810 J
2  R2 R1  2  R2 R1  2

Alternatively, we can plot the saturated curves for the two gap lengths as shown in Fig. 7 and
use the shaded area to find the work done.

(d) For fast closure or closure at a constant flux,

x2 x2 R2 2
1 2 dR 1 1
Wmech   f fld dx     dx     2 dR   2 R1  R2 
x1 x1
2 dx R11
2 2
1 1 1
 12 R1  R2   1 1 R1  1 R2   1 F1  F2 
2 2 2
1
F2  1 R2  0.0004  3.6  10 6  1440 AT . Therefore Wmech  1 F1  F2   0.360 J
2
Alternatively, we can plot the saturated curves for the two gap lengths as shown in Fig. 6 and
use the shaded area to find the work done.

8 Torque in Singly-Excited Systems

Electromechanical devices can be designed in such a way that the permitted motion is rotational
rather than translational. Examples are shown in Fig. 8. In this case, the mechanical work is
expressed in terms of torque Tfld and the angle θ of rotation. The torque is given by
dwmech
T fld  Nm (29)
d

By analogy with equations (11) and (19), the torque may be expressed as
1 dC
T fld  e 2 for electric coupling and (30)
2 d

1 dL
T fld  i 2 for magnetic field coupling (31)
2 d

The angular position θ is measured from arbitrarily chosen reference position for which θ is zero.

70
Fig. 8.a Simple rotating electrostatic machine

Fig. 8.b Reluctance motor

71
9 Torques and Forces in Doubly Excited Magnetic Field Systems

Multiple excited magnetic field devices are used to obtain forces and torques proportional to
electrical signals and electrical signals proportional to forces and velocities. Permanent magnets
are frequently used as one of the excitation paths. Models of doubly excited magnetic field
systems are shown in Fig. 9

Let us consider the system in Fig. 9.a which has a permanent magnet as one of the excitation
paths. The force developed by the device is given by the well known expression
f fld  Bli (32)

where B is the flux density in the air gap and l is the total active length of the coil. This
expression for the force can also be derived using the foregoing method. Let the coil be displaced
through dx in a time dt . Then the electrical energy input
dx
dwelec  eidt  B l i dt  Bl i dx
dt

and the mechanical energy output

dwmech  f fld dx

ffld
fmech
N
x

Mechanical
system
S
i

Permanent + -
magnet V(t)

Fig. 9.a Doubly-excited magnetic system with permanent magnet as one of excitation paths

72
 θ

i1
Tfld
-
e1 V2(t)
+ +
V1(t) e2
-
i2

Fig. 9.b Doubly-excited magnetic field system with two sets of electrical terminals

In this device the field energy remains constant during the conversion process so that
dw fld  0

Equating therefore the mechanical and electrical expressions, the force is found to be
f fld  dwmech dx  Bl i
Let us now consider the system in Fig. 9.b. The electrical energy input
dwelec  e1i1 dt  e2 i2 dt

From Faraday’s Law

d d di dL di dM
e1  1  L1i1  Mi2   L1 1  i1 1  M 2  i2 (33)
dt dt dt dt dt dt

Similarly
di dL di dM
e2  L2 2  i 2 2  M 1  i1 (34)
dt dt dt dt

Substituting these into the electrical energy input expression gives

dwelec  L1i1di1  i12 dL1  Mi1di2  i 2 i1dM  L2 i2 di2  i22 dL2  Mi2 di1  i2 i1dM (35)

Energy supplied to the coupling field

1 1 
dw fld  d  L1i12  L2 i22  Mi1i2 
2 2 
1 1
dw fld  i12 dL1  L1i1di1  i22 dL2  L2 i2 di2  i1i2 dM  Mi2 di1  Mi1di2 (36)
2 2

73
From the relation
dw dwelec  dw fld
T fld  mech 
d d

we obtain for the torque, the expression

1 dL 1 dL dm
T fld  i12 1  i22 2  i1i2 (37)
2 d 2 d d

Example 7
For the doubly excited system in Fig. 9.b, the inductances are approximated as follows: L1 = 11
+ 3cos2θ, L2 = 7 + 2cos2θ, M = 11cosθ. The coils are energized with dc currents. I1 = 0.7 A, I2
= 0.8 A. Find the torque for θ = +50°. What is its direction?

Solution
dL1 dL2 dM
 6 sin 2θ ;  4 sin 2θ ;  11sin θ
d d d
1 dL 1 dL dM
T fld  i12 1  i 22 2  i1i2
2 d 2 d d
1 1
 0.7   6 sin 2θ   0.8  4 sin 2θ   0.7 0.8 11sin θ 
2 2

2 2
 2.75 sin 2θ  6.16 sin θ Nm

For   50, T fld  7.43 Nm . The torque acts anticlockwise on the rotor.

10 Alignment and Interaction Forces

Electromechanical devices develop forces either by alignment or interaction or both.

10.1 Alignment Forces

The forces developed in singly excited systems are all by alignment. Alignment force acts in a
direction that tends to increase the field energy in the device. To achieve maximum field energy,
the force will try to bring the fixed and movable members together so as to increase the
capacitance of the plates in the case of the electric field system and the inductance of the coil in
the case of the magnetic field system. The torque components of equation (37) in the form
(i 2 2).(dL d ) are also due to alignment forces.

10.2 Interaction forces

They are produced by the interaction of magnetic field and current-carrying conductors. The
force is set up when a conductor properly placed in a magnetic field is made to carry current. The
force is also set up between two or more current-carrying coils. The force developed in the
device of Fig. 9.a is a force of interaction. The component of the torque given by (37) involving

74
mutual inductance is also due to interaction force. Forces of interaction between two current-
carrying coils are directed in a way that tends to increase the magnitude of the flux linkage of
each of the coil or in a way that tends to align the mmf axes of the coils.

The forces of interaction can be made proportional to electrical signals by supplying the
particular signals at one of the electrical inputs and their direction can be controlled.

11 Motional emf

The terms in the induced voltage equations (15), (33) and (34) for which inductances or mutual
inductances are constant with current i differentiated, are the transformer emfs. They give rise to
electrical/electrical power conversion. The other terms arise if the inductances or mutual
inductances vary as the movable members move. These voltages called motional or rotational
emfs give rise to mechanical/electrical power conversion. Expressions for torque or force in
electromagnetic devices can be derived by considering only these terms.

Let us consider the singly excited electromagnetic system. Electrical energy flow due to this emf
component caused by movement is
 dL 
d welec  i i dt  i 2 dL
 dt 

With current constant, the energy supplied to the field

1
d w fld  i 2 dL
2

The energy converted to mechanical energy is then

1 1
dwmech  d welec  d w fld  i 2 dL  i 2 dL  i 2 dL
2 2

Example 8
Considering only the motional emfs, derive an expression for torque developed in a doubly
excited electromagnetic device.

Solution
d w elec  e1i1dt  e 2 i2 dt  i1 i1dL1  i2 dM   i2 i2 dL2  i1dM   i12 dL1  2i1i2 dM  i22 dL2
1 1
d w fld  i12 dL1  i22 dL2  i1i 2 dM
2 2
1 1
dwmech  d w elec  d w fld  i12 dL1  i22 dL2  i1i2 dM
2 2
Hence
1 dL 1 dL dM
T fld  i12 1  i 22 2  i1i2
2 d 2 d d

75
12 Examples of Electromechanical Devices

12.1 Solenoid Relay (Fig. 10)

It is used to operate switches. Due to rotational movement about the pivot, the gap length is not
uniform over the cross-section of the gap but because the travel is short, this distortion is
sufficiently small to be negligible.

Fig. 10 Solenoid relay

A solenoid relay is operated from a 110-V, dc supply and the 5000-turn coil resistance is 5.5 kΩ.
The core diameter of the relay is 20 mm and the gap length is 1.5 mm, the armature being
stationary. The gap faces may be taken as parallel and the permeability of the ferromagnetic parts
as very high. Estimate (a) the gap flux density (b) the coil inductance (c) the pull on the armature

Solution
V 100
(a) I    20  10 3 A
R 5.5  10 3

F  NI  20  10 3  5000  100AT
F 100
H  3
 0.67  10 5 AT/m
l g 1.5  10
B   o H  4  10 7  0.67  10 5  84  10 3 Wb/m 2
(b)   BA  84  10 3  10 2    10 6  26.3  10 6 Wb

L
N

26.3  10 6  5000  6.56 H
I 20  10 3
1 dL
(c) F fld  i 2
2 dx
1.5  10 3 9.82  10 3
L( x)  6.56  
x x
where x is in meters

76
 dL 9.82  10 3

dx x2
1 9.82  10 3
Therefore F fld    20 2  10 6  2  0.88 N
2 1.5  10 6

12.2 Plunger relay (Fig. 11)

It is also used for relay purposes. It is used for tripping circuit breakers, for operating valves and
in other applications in which greater lengths of travel are required.

Fig. 11 Plunger relay

When the plunger relay is energized, the current remains constant at 2.0 A and the plunger
moves such that the air gap is reduced in length from 20 mm to 10 mm. If losses may be
neglected, calculate the average force experienced by the plunger and the energy taken from the
source. The diameter of the plunger is 50 mm. Neglect the air gap at the hole through which the
armature moves.

Solution
For x 2  l g  10 mm, F  NI  1500  2  3000 AT
F 3000
H   300  10 3 AT/m
lg 
10  10 3

B   o H  4  10  300  10 3  0.377 T
7

  BA  0.377  25 2    10 6  0.74  10 3 Wb
N 1500  0.74  10 3
L2    0.56 H
I 2
N 2 N 2  o A 1500 2  4  10 7  25 2    10 6
Alternatively, L1     0.56 H
R1 x1 10  10 3
10 mm
For x1  l g  20 mm , L1  0.56   0.28 H
20 mm

77
 1 2 x2 dL 1 1 1
dx  I 2  dL  I 2 L2  L1    2 2  0.28  0.56 J
L2
Total work done  I 
2 x1 dx 2 L1 2 2

Total work done 0.56 0.56

Average force F fld ( ave )     56 N
Total displacement x 2  x1  0.01
1 1
Increase in stored energy  I 2 L2  L1    2 2  0.28  0.56 J
2 2
Energy supplied by the source = increase in stored energy + work done = 0.56 + 0.56 =1.12 J.

12.3 Horse shoe electromagnet

The electromagnet is made using a horse-shoe core as shown in Fig. 12. The core has an
effective length of 600 mm and a cross-sectional area of 500 mm2. A rectangular block of iron is
held by the electromagnet force of alignment and a force of 20 N is required to free it. The
magnetic circuit through the block is 200 mm long and the effective cross-sectional area is again
500 mm2. The relative permeability of both core and block is 700. If the magnet is energized by a
coil of 100 turns, estimate the coil current.

Fig.12 Horse shoe electromagnet

Solution
1 dR
f fld    2
2 dx

With the fixed and movable members separated by a distance x m, the total air gap would be 2 x .
Hence the total reluctance would be in the form
2x dR 2
R( x)  C  and 
o A dx o A

where A is the cross-sectional area of the air gap.

The force developed when x  0 is given by

1 2 B2 A
f fld    2 x  0   
2 o A o

78
where B is the flux density in the iron. We note that the cross-sectional area of the iron part of
the magnetic circuit is equal to the cross-sectional area of the air gaps.

Now
B2 A
 20 N .
o

Therefore
1
 20  4  10 7  2
B 6   0.222 T and
 500  10 

B 0.222
H   250 AT/m
μ o μ r 4  10 7  700

The total mmf required to create the above flux density in both the core and the block
F  Hl  250  600  200   10 3  200 AT

200
Hence I   2.0 A
100

12.4 Electrostatic voltmeter

The electrostatic voltmeter movement shown in Fig. 13 consists of three semi-circular metal
plates. The middle plate is attached to a taut suspension wire which acts as a torsional spring. (A
torsional spring will exert a restraining torque proportional to the angular displacement). The rest
position of the middle plate is at θ = 0, where it is just about to enter the space between the outer
plates. Suppose the plates have a radius of 4 cm and an air spacing of 1 mm between the centre
plate and the outer plates.
(a) Determine the capacitance of the system as a function of the angle θ
(b) Determine the torque on the centre plate as a function of the applied voltage e
(c) Suppose the voltmeter is to have a full-scale deflection of θ = 3 radians with an applied
voltage of 1000 V. What should be the spring constant of the suspension system?
(d) When used on alternating voltage, does this instrument measure average, rms or peak values?
(e) Suppose an alternating voltage of 600 V rmss at a frequency of 2000 Hz is applied to the
meter. Determine the deflection and also determine the input impedance of the instrument.

79
Solution
(a) The equivalent circuits can be indicated as shown below:

Fig. 13 Electrostatic voltmeter

This is a multi-plate capacitor system with n = 3 or two identical capacitances in parallel.

Hence
2 A
C    C1  C1  2C1  o
d

where d = the air spacing and A is the effective area of the plate
1
A( )  r 2
2

Hence the effective capacitance of the system

80
  o r 2
C ( )  where r = radius of the plates
d
1 dC  o r 2 e 2
(b) T fld  e 2   k e 2
2 d 2d
 r E 2 2
8.85  10 12  4 2  10 4  10 6
(c) T fld  o  3
 7.08  10 6 Nm
2d 2  10
Tmech  3 k  T fld .
70.8  10 7
Therefore k   2.36  10 6 Nm/rad
3
(d) T fld  k e . Therefore T fld (ave)  k   ave(e 2 )  k   E rms
2 2

Hence instrument measures rms values.

(e) Deflection  Tfld  V2 being dc or rms value. Therefore for V = 600 rms.
2
 600 
    3  1.08 rad
 1000 

The effective capacitance when   1.08 rad is

8.85  10 12  4 2  10 4  1.08
C 3
 1.529  10 11 F
10
1 1
Input impedance    5.2 MΩ
C 2  2000  1.529  10 1

12.5 Electrostatic loudspeaker

The force between charged plates is exploited in the electrostatic loudspeaker shown in Fig. 14.
Two circular metallic plates are separated by a compressible ring of insulating material. When a
voltage et is applied between the plates, the resulting force causes a change in the plate spacing x,
which in turn results in an acoustic wave from the plate surfaces. For accurate sound
reproduction, the variation in spacing x should be proportional to the output voltage e of the
audio amplifier. Unfortunately, the electrostatic force is proportional to the square of the terminal
voltage et. To overcome this difficulty, a terminal voltage et = E + e is applied, where E is a
constant voltage much larger than the audio frequency signal voltage e.
(a) Suppose the voltage E is 1000 V. Assuming the spacing x between the plates does not vary
appreciably from a value of 0.5 mm. If the relative permittivity of the insulating ring is assumed
to be 1.0, show that the force between the plates is given approximately by
f = 0.139 + 2.78 x 10-4e newtons
(b) Suppose the ring provides a spring constant of 300 N/m. If e = 100sinωt, determine the peak-
to-peak oscillation in the plate spacing.

81
 Fig. 14 Electrostatic loudspeaker

Solution
1 2 dC
(a) f fld 
et
2 dx
  A
C o r
x

Hence
1  o  r Aet2 1  o r A
f fld   2
 2
E  e2   1  o 2r A E 2  2 Ee  e 2
 
2 x 2 x 2 x

1  o r A 2
f fld  
2 x2
E  2 Ee  if e 2  0

Now
1  o r A 2 1 8.85  10 12  1   0.12  106
E    0.139 and
2 x2 2 0.52  4

1  o r A
2
2 E   2  0.139  2.78 10 4
2 x E


Hence f fld   0.139  2.78  10 4 e Nm 
(b) 300 x pp  2  2.78  10 4  100  x pp  0.185 mm

82
13 Further exercises

1. A machine with two coils has inductances as follows: (on rotor) L1 = 0.1 H, (on stator) L2 =
0.5 H, M = 0.2cosθ H, where θ is the angle of the rotor coil axis displaced anti-clockwise
with respect to the stator coil axis. Coil 1 (on rotor) is short circuited. Coil 2 (on stator) is
energized from a 60-Hz sinusoidal voltage source of 110 V. Resistances of the coils may be
neglected. Assume the circuit operates in sinusoidal steady state. θ is set to 30°.
(a) Find an expression for the instantaneous torque on the rotor.
(b) Find the value of the average torque on the rotor
(c) Determine the direction of this torque.

2. Fig. E.1 shows the essential components of a loudspeaker. The permanent magnet produces a
uniform radial magnet flux density of 0.8 Wb/m2 across a cylindrical air gap. The coil of 30
turns is wound on a fiber cylinder of diameter d = 2 cm. When assembled, the coil is inserted
into the air gap of the magnet.
(a) Determine the force of the cone as a function of the current i.
(b) Determine the induced voltage in the coil per unit of coil velocity
(c) Neglecting coil resistance, show that the electrical power input to the coil is equal to the
mechanical power delivered to the cone.
(d) Over most of the audio frequency range, the force is absorbed by the damping action of
the air being driven by motion of cone. Suppose the damping coefficient of the cone is
0.3 N per m/s of velocity. If the current in the coil is imsinωt, determine the velocity of
the coil and the voltage induced in the coil. Show that the impedance looking into the coil
terminals is resistive and has a value of 7.6 Ω

Fig.E.1 Magnetic loudspeaker

3. A rotating machine of the form shown in Fig. E.2 has a reluctance which can be expressed
approximately by
R = 5.06 x 104(2.5 + 1.5 cos2θ) AT/Wb
where θ is the angle between the axis of the rotor and of the stator.
(a) Suppose a voltage of 110 V rms at 60 Hz is applied to the 15-turn winding. If the winding
resistance is negligible, determine the magnetic flux in the machine.
(b) At what angular velocity must the rotor rotate to produce an average torque?

83
(c) What is the maximum value of the average torque this machine can produce when
connected to a 110-V, 60-Hz supply? What is the maximum mechanical power?

Fig.E.2 Reluctance motor

84