APPENDIX A

Standards for Drives

Miiiiiii!!ii!i
ii A1 IEC (INTERNATIONAL) STANDARDS 330

ii!iiiHi A2 CENELEC (EC) STANDARDS 332

ii~ A3 BRITISH STANDARDS 334

~+~;~~
' A4 IEEE (USA) STANDARDS 335

i~i~ii~ A5 UL (UNDERWRITERS' LABORATORIES, USA) STANDARDS 335

:--~.~:~-~

A6 OTHER STANDARDS 335

This list shows the main international, European and British Many IEC standards are identical in technical content with
standards which may be relevant to the design or application the CENELEC ( E N . . . ) standard of the same number; in this
of variable-speed drives. case, the CENELEC standard is listed here. There may be
discrepancies when revisions are ratified at different times,
International Electrotechnical Commission (IEC) standards
and in that case both are listed.
are generally preferred for the purposes of harmonisation to
facilitate world trade. They are not in themselves mandatory, IEC standards with four or less digits in the number have
whereas national standards and European harmonised stan- been renumbered by the addition of 6000. Where the stan-
dards may be adopted under legally binding regulations and dard has not been revised recently, the old number may still
become effectively or actually mandatory. appear on the document.
330 IEC (INTERNATIONAL)STANDARDS

A1 IEC (INTERNATIONAL) STANDARDS

CISPR 11 (1999-08) Ed. 3.1 Consolidated edition IEC 60146-2 (1999-11)

Industrial, scientific and medical (ISM) radiofrequency Semiconductor converters - Part 2: Self-commutated
e q u i p m e n t - electromagnetic disturbance characteristics - semiconductor converters including direct D.C. converters
limits and methods of measurement
IEC/TR2 60146-6 (1992-12)
CISPR 16-1 (1999-10) Semiconductor convertors - Part 6: Application guide for
Specification for radio disturbance and immunity measuring the protection of semiconductor convertors against over-
apparatus and methods - Part 1: Radio disturbance and current by fuses
immunity measuring apparatus
IEC 60204-1 (1997-10)
CISPR 16-2 (1999-08) Ed. 1.1 Consolidated edition Electrical equipment of industrial machines - Part 1: Gen-
Specification for radio disturbance and immunity measuring eral requirements
apparatus and methods - Part 2: Methods of measurement of
disturbances and immunity IEC 60249-1 (1982-01)
Base materials for printed circuits - Part 1: Test methods
IEC 50161
International electrotechnical vocabulary: electromagnetic IEC 60249-2-4 (1987-06)
compatibility Base materials for printed circuits - Part 2: Specifications.
Specification no. 4: epoxide woven glass fabric copper-clad
IEC 60034-5 (1991-02) laminated sheet, general-purpose grade
Rotating electrical machines - Part 5: Classification of
IEC 60249-2-5 (1987-06)
degrees of protection provided by enclosures of rotating
Base materials for printed circuits - Part 2: Specifications.
electrical machines (IP code)
Specification no. 5: epoxide woven glass fabric copper-clad
IEC 60034-6 (1991-11) laminated sheet of defined flammability (vertical burning
Rotating electrical machines - Part 6: Methods of cooling test)
(IC code) IEC 60249-2-12 (1987-04)
Base materials for printed circuits - Part 2: Specifications.
IEC 60034-8 (1972-01)
Specification no. 12: thin epoxide woven glass fabric copper-
Rotating electrical machines - Part 8: Terminal markings
clad laminated sheet of defined flammability, for use in the
and direction of rotation of rotating machines
fabrication of multilayer printed boards
IEC/TR2 60034-17 (1998-06)
IEC 60249-3-1 (1981-01)
Rotating electrical machines - Part 17: Cage induction
Base materials for printed circuits - Part 3: Special materials
motors when fed from converters - application guide
used in connection with printed circuits. Specification no. 1:
IEC 60034-19 (1995-07) prepreg for use as bonding sheet material in the fabrication
Rotating electrical machines - Part 19: Specific test methods of multilayer printed boards
for D.C. machines on conventional and rectifier-fed supplies IEC 60249-3-3 (1991-06)
Base materials for printed circuits - Part 3: Special materials
Future standard: IEC 60034-25 Ed. 1.0
used in connection with printed circuits. Specification no. 3:
Guide for the design and the performance of cage induction
permanent polymer coating materials (solder resist) for use
motors for converter supply
in the fabrication of printed boards
IEC 60068-2-64 (1993-05) IEC 60326-3 (1991-05)
Environmental testing - Part 2: Test methods - test Fh: Printed boards - Part 3: Design and use of printed boards
vibration, broadband random (digital control) and guidance
IEC 60326-4 (1980-01)
IEC 60146-1-1 (1991-04) Printed boards - Part 4: Specification for single and double-
Semiconductor convertors - general requirements and line sided printed boards with plain holes
commutated convertors - Part 1 - 1: Specifications of basic
IEC 60326-5 (1980-01)
requirements
Primed boards - Part 5: Specification for single and double-
IEC/TR 60146-1-2 (1991-04) sided printed boards with plated-through holes
Semiconductor convertors - general requirements and line IEC 60364-4-41 (1992-10)
commutated convertors - Part 1 - 2: Application guide Electrical installations of buildings - Part 4" Protection for
safety. Chapter 41" protection against electric shock
IEC 60146-1-3 (1991-04)
Semiconductor convertors - general requirements and IEC 60364-4-41 (1999-05) Ed. 3.2 Consolidated edition
line commutated convertors - Part 1 - 3: Transformers and Electrical installations of buildings - Part 4: Protection for
reactors safety. Chapter 41" protection against electric shock
 Appendix A 331

IEC 60364-4-444 (1996-04) IEC 60950 (1999-04)

Electrical installations of buildings - Part 4: Protection for Safety of information technology equipment
safety. Section 444: protection against electromagnetic
IEC 60990 (1999-08)
interference (EMI) in installations of buildings
Methods of measurement of touch current and protective
IEC 60364-5-54 (1980-01) conductor current
Electrical installations of buildings - Part 5: Selection and IEC 61000-2-4 (1994-02)
erection of electrical equipment. Chapter 54: earthing Electromagnetic compatibility (EMC) - Part 2" Environ-
arrangements and protective conductors ment. Section 4: compatibility levels in industrial plants for
low-frequency conducted disturbances
IEC 60364-5-523 (1999-02)
Electrical installations of buildings - Part 5: Selection and IEC/TR3 61000-2-6 (1995-09)
erection of electrical equipment. Section 523: current- Electromagnetic compatibility (EMC) - Part 2" Environ-
carrying capacities in wiring systems ment. Section 6" assessment of the emission levels in the
power supply of industrial plants as regards low-frequency
IEC 60364-5-548 (1999-04) Ed. 1.1 Consolidated edition conducted disturbances
Electrical installations of buildings - Part 5: Selection and
erection of electrical equipment. Section 548: earthing IEC/TR3 61000-2-7 (1998-01)
arrangements and equipotential bonding for information Electromagnetic compatibility (EMC)-Part 2: Environ-
technology installations ment. Section 7" low-frequency magnetic fields in various
environments
IEC 60384-14 (1993-06)
IEC 61000-3-2 (1995-03)
Fixed capacitors for use in electronic equipment-Part 14:
Electromagnetic compatibility (EMC) - Part 3: Limits.
Sectional specification. Fixed capacitors for electromagnetic
Section 2: limits for harmonic current emissions (equipment
interference suppression and connection to the supply mains
input current _< 16 per phase)
IEC 60417-1 (1998-08) IEC/TS 61000-3-4 (1998-10)
Graphical symbols for use on equipment- Part l" Overview Electromagnetic compatibility (EMC) - Part 3 - 4 " Limits.
and application Limitation of emission of harmonic currents in low-voltage
power supply systems for equipment with rated current
IEC 60417-2 (1998-08)
greater than 16 A
Graphical symbols for use on equipment - Part 2" Symbol
originals IEC/TR2 61000-3-5 (1994-12)
Electromagnetic compatibility (EMC) - Part 3: Limits.
IEC 61140 (1997-11) Section 5: limitation of voltage fluctuations and flicker in
Protection against electric s h o c k - common aspects for low-voltage power supply systems for equipment with rated
installation and equipment current greater than 16 A
IEC 60664-1 (2000-04) Ed. 1.1 Consolidated edition IEC/TR3 61000-3-6 (1996-10)
Insulation coordination for equipment within low-voltage Electromagnetic compatibility (EMC) - Part 3" Limits.
systems - Part 1: Principles, requirements and tests Section 6: assessment of emission limits for distort-
ing loads in MV and HV power systems. Basic EMC
IECITR3 60664-2-1 (1997-11)
publication
Insulation coordination for equipment within low-voltage
systems - Part 2 - l : Application guide. Dimensioning IECITR3 61000-3-7 (1996-11)
procedure worksheets and dimensioning examples Electromagnetic compatibility (EMC) - Part 3" Limits.
Section 7" assessment of emission limits for fluctuating
IEC 60664-3 (1992-10)
loads in MV and HV power systems. Basic EMC
Insulation coordination for equipment within low-voltage
publication
systems - Part 3: Use of coatings to achieve insulation
coordination of printed board assemblies IEC/TR3 61000-5-2 (1997-11)
Electromagnetic compatibility (EMC) - Part 5" Installation
IEC/TR3 60664-4 (1997-09)
and mitigation guidelines. Section 2" earthing and cabling
Insulation coordination for equipment within low-voltage
systems - Part 4: Consideration of high-frequency voltage IEC 61000-6-1 (1997-07)
stress Electromagnetic compatibility (EMC) - Part 6" Generic
standards. Section 1" immunity for residential, commercial
IEC 60695-1-1 (1999-11)
and light industrial environments
Fire hazard testing - Part 1-1: Guidance for assessing the
fire hazard of electrotechnical products. General guidelines IEC 61000-6-2 (1999-01)
IEC 60695-2-1/1 (1994-03) Electromagnetic compatibility (EMC) - Part 6 - 2 : Generic
Fire hazard testing - Part 2: Test methods. Section 1/sheet 1: standards. Immunity for industrial environments
glow-wire end-product test and guidance
IEC 61000-6-4 (1997-01)
IEC 60695-2-1/2 (1994-03) Electromagnetic compatibility (EMC) - Part 6" Generic
Fire hazard testing - Part 2: Test methods. Section 1/sheet 2: standards. Section 4" emission standard for industrial
glow-wire flammability test on materials environments
332 IEC (INTERNATIONAL)STANDARDS

IEC 61131-1 (1992-10) IEC 61508-7 (2000-03)

Programmable controllers - Part 1: General information Functional safety of electrical/electronic/programmable elec-
tronic safety-related systems - Part 7: Overview of techni-
IEC 61131-2 (1992-10)
ques and measures
Programmable controllers- Part 2: Equipment requirements
and tests IEC 61543 (1995-04)
Residual current-operated protective devices (RCDs) for
IEC 61131-3 (1993-03)
household and similar u s e - electromagnetic compatibility
Programmable controllers - Part 3: Programming languages
IEC 61800-3 (1996-06)
IEC/TR 61131-8 (2000-01)
Adjustable speed electrical power drive systems - Part 3:
Programmable controllers - Part 8: Guidelines for the
EMC product standard including specific test methods
application and implementation of programming languages
IEC 62326-4 (1996-12)
IEC 61508-1 (1998-12)
Printed boards-Part 4: Rigid multilayer printed boards with
Functional safety of electrical/electronic/programmable
interlayer connections. Sectional specification
electronic safety-related systems - Part 1: General
requirements IEC 62326-4-1 (1996-12)
Printed boards - Part 4: Rigid multilayer printed boards with
IEC 615O8-2 (2OOO-O5)
interlayer connections. Sectional specification. Section 1:
Functional safety of electrical/electronic/programmable
capability detail specification. Performance levels A, B
electronic safety-related systems - Part 2: Requirements for
and C
electrical/electronic/programmable electronic safety-related
systems
IEC 61508-3 (1998-12)
PLANNED FUTURE IEC 61800 STANDARDS
Functional safety of electrical/electronic/programmable
electronic safety-related systems - Part 3: Software
IEC 61800-4 Ed. 1.0
requirements
General requirements - rating specifications for A.C. power
IEC 61508-4 (1998-12) drive systems above 1000 V A.C. or 1500 V D.C. and not
Functional safety of electrical/electronic/programmable exceeding 38 kV
electronic safety-related systems - Part 4: Definitions and
IEC 61800-5 Ed.l.0
abbreviations
Adjustable speed electrical power drive systems - Part 5:
IEC 61508-5 (1998-12) Electrical, thermal and fuctional safety aspects
Functional safety of electrical/electronic/programmable elec-
IEC 61800-6 TR Ed.l.0
tronic safety related systems - Part 5: Examples of methods
Creation of technical report to replace IEC 61136-1 and
for the determination of safety integrity levels
withdrawal of IEC 61136-1 upon completion
l E e 61508-6 (2000-04)
Functional safety of electrical/electronic/programmable
electronic safety-related systems - Part 6: Guidelines on the
application of IEC 61508-2 and IEC 61508-3

A2 CENELEC (EC) STANDARDS

EN 81-1:1985 EN 954-1:1997
Lifts and service lifts. Safety rules for the construction and Safety of machinery. Safety-related parts of control systems.
installation of electric lifts General principles for design
EN 292-1:1991
Safety of machinery. Basic concepts, general principles for EN 1037:1996
design. Basic terminology, methodology Safety of machinery. Prevention of unexpected start-up
EN 292-2:1991
Safety of machinery. Basic concepts, general principles for EN 1050:1997
design. Technical principles and specifications Safety of machinery. Principles for risk assessment
 Appendix A 333

EN 12015:1998 EN 60146-1-3:1993 (IEC 60146-1-3:1991)

Electromagnetic compatibility. Product family standard for Semiconductor convertors. General requirements and line
lifts, escalators and passenger conveyors. Emission commutated convertors. Transformers and reactors
EN 12016:1998 EN 60204-1:1993
Electromagnetic compatibility. Product family standard for Safety of machinery. Electrical equipment of machines.
lifts, escalators and passenger conveyors. Immunity Specification for general requirements
EN 50081-1:1992 EN 60417-1:1999 ( IEC 60417-1:1998)
Electromagnetic compatibility. Generic emission standard. Graphical symbols for use on equipment. Overview and
Residential, commercial and light industry application
EN 50081-2:1994 EN 60417-2:1999 (IEC 60417-2:1998)
Electromagnetic compatibility. Genetic emission standard. Graphical symbols for use on equipment. Symbol originals
Industrial environment
EN 60439-1:1999 (IEC 60439-1:1999)
EN 50082-1:1998 Specification for low-voltage switchgear and controlgear
Electromagnetic compatibility. Generic immunity standard. assemblies. Type-tested and partially type-tested assemblies
Residential, commercial and light industry
EN 60445:1990
EN 50082-2:1995 Specification for identification of equipment terminals and
Electromagnetic compatibility. Generic immunity standard. of terminations of certain designated conductors, including
Industrial environment general rules for an alphanumeric system
EN 50160:2000 EN 60529:1992
Voltage characteristics of electricity supplied by public Specification for degrees of protection provided by enclo-
distribution systems sures (IP code)
EN 50178:1998 EN 60947-2:1996
Electronic equipment for use in power installations Specification for low-voltage switchgear and controlgear.
EN 55011:1998 (CISPR 11:1997) Circuit breakers
Limits and methods of measurement of radio disturbance EN 60947-3:1999
characteristics of industrial, scientific and medical (ISM) Specification for low-voltage switchgear and controlgear.
radiofrequency equipment Switches, disconnectors, switch disconnectors and fuse
EN 60034-1:1998 combination units
Rotating electrical machines. Rating and performance EN 60950:1992 (BS 7002:1992)
EN 60034-2:1999 Specification for safety of information technology equip-
ment, including electrical business equipment
Rotating electrical machines. Methods for determining
losses and efficiency of rotating electrical machinery from EN 61000-2-4:1995 (IEC 61000-2-4:1994)
tests (excluding machines for traction vehicles) Electromagnetic compatibility (EMC). Environment. Com-
EN 60034-7:1993 (IEC 60034-7:1992) patibility levels in industrial plants for low-frequency con-
ducted disturbances
Rotating electrical machines. Classification of types of
constructions and mounting arrangements (IM Code) EN 61000-3-2:1995 (IEC 61000-3-2:1995)
Electromagnetic compatibility (EMC). Limits. Limits for
EN 60068-2-6:1996 (IEC 60068-2-6:1995)
harmonic current emissions (equipment input current 16 A
Environmental testing. Test methods. Test Fc. Vibration
(sinusoidal) per phase)

EN 60068-2-21:1999 (IEC 60068-2-21:1999) EN 61000-3-3:1995 (IEC 61000-3-3:1994)

Environmental testing. Test methods. Test U. Robustness of Electromagnetic compatibility (EMC). Limits. Limitation of
terminations and integral mounting devices voltage fluctuations and flicker in low-voltage supply sys-
tems for equipment with rated current 16 A
EN 60068-2-27:1993 (IEC 60068-2-27:1987)
Environmental testing. Test methods. Environmental testing EN 61000-4-2:1995 (IEC 61000-4-2:1995)
procedures. Tests. Test Ea and guidance. Shock Electromagnetic compatibility (EMC). Testing and mea-
surement techniques. Electrostatic discharge immunity test.
EN 60068-2-29:1993 (IEC 60068-2-29:1987) Basic EMC publication
Environmental testing. Test methods. Environmental testing
procedures. Tests. Test Eb and guidance. Bump EN 61000-4-3:1997
Electromagnetic compatibility (EMC). Testing and mea-
EN 60068-2-32:1993 surement techniques. Radiated, radiofrequency, electro-
Environmental testing. Test methods. Test Ed. Free fall magnetic field immunity test
EN 60146-1-1:1993 (IEC 60146-1-1:1991) EN 61000-4-4:1995 (IEC 61000-4-4:1995)
Semiconductor convertors. General requirements and Electromagnetic compatibility (EMC). Testing and mea-
line commutated convertors. Specifications of basic surement techniques. Electrical fast transient/burst immu-
requirements nity test. Basic EMC publication
334 CENELEC
(EC) STANDARDS

EN 61000-4-5:1995 (IEC 61000-4-5:1995) EN 61000-4-28:2000 (IEC 61000-4-28:1999)

Electromagnetic compatibility (EMC). Testing and mea- Electromagnetic compatibility (EMC). Testing and mea-
surement techniques. Surge immunity test surement techniques. Variation of power frequency, immu-
nity test
EN 61000-4-6:1996 (IEC 61000-4-6:1996)
Electromagnetic compatibility (EMC). Testing and mea- EN 61000-6-2:1999 (IEC 61000-6-2:1999)
surement techniques. Immunity to conducted disturbances, Electromagnetic compatibility (EMC). Generic standards.
induced by radiofrequency fields Immunity for industrial environments
EN 61000-4-7:1993 (IEC 61000-4-7:1991) EN 61010-1:1993
Electromagnetic compatibility (EMC). General guide on Safety requirements for electrical equipment for measure-
harmonics and interharmonics measurements and instru- ment, control and laboratory use. General requirements
mentation, for power supply systems and equipment con-
EN 61136-1:1998
nected thereto
Semiconductor power convertors. Adjustable-speed electric
EN 61000-4-8:1994 (IEC 61000-4-8:1993) drive systems. General requirements. Rating specifications,
Electromagnetic compatibility (EMC). Testing and mea- particularly for D.C. motor drives
surement techniques. Power frequency magnetic field immu-
EN 61326:1998 (IEC 61326-1:1997)
nity test. Basic EMC publication
Electrical equipment for measurement, control and labora-
EN 61000-4-9:1994 (IEC 61000-4-9:1993) tory use. EMC requirements
Electromagnetic compatibility (EMC). Testing and mea-
EN 61496-1:1998 (IEC 61496-1:1997)
surement techniques. Pulse magnetic field immunity test.
Basic EMC publication Safety of machinery. Electrosensitive protective equipment.
General requirements and tests
EN 61000-4-10:1994 (IEC 61000-4-10:1993)
EN 61800-1:1998 (IEC 61800-1:1997)
Electromagnetic compatibility (EMC). Testing and mea-
Adjustable-speed electrical power drive systems. Rating
surement techniques. Damped oscillatory magnetic field
specifications for low-voltage adjustable-speed D.C. power
immunity test. Basic EMC publication
drive systems
EN 61000-4-11:1994 (IEC 61000-4-11:1994)
EN 61800-2:1998 (IEC 61800-2:1998)
Electromagnetic compatibility (EMC). Testing and mea-
Adjustable-speed electrical power drive systems. General
surement techniques. Voltage dips, short interruptions and
requirements. Rating specifications for low-voltage adjus-
voltage variations immunity tests
table frequency A.C. power drive systems
EN 61000-4-12:1996 (IEC 61000-4-12:1995)
EN 61800-3:1997 (IEC 61800-3:1996)
Electromagnetic compatibility (EMC). Testing and mea-
Adjustable-speed electrical power drive systems. EMC
surement techniques. Oscillatory waves immunity test.
Basic EMC publication product standard including specific test methods

EN 61000-4-14:1999 (IEC 61000-4-14:1999)

Electromagnetic compatibility (EMC). Testing and mea-
surement techniques. Voltage fluctuation immunity test

A3 BRITISH STANDARDS

BS 6651:1999 BS 7671:1992
Code of practice for protection of structures against Requirements for electrical installations, lEE Wiring
lightning Regulations. Sixteenth edition
 Appendix A 335

A4 IEEE (USA) STANDARDS

IEEE 519 (1992) IEEE C62 collection (1995)

Recommended practices and requirements for harmonic Surge protection
control in electrical power systems.

A5 UL (UNDERWRITERS" LABORATORIES, USA) STANDARDS

UL 50 UL 508
Enclosures for electrical equipment Standard for industrial control equipment
UL 94 UL 508C
Test for flammability of plastic materials for parts in devices Standard for power conversion equipment
and appliances

A6 OTHER STANDARDS

ELECTRICITY ASSOCIATION, UK 485 (1998)

Standard for electrical characteristics of generators and
G.5/3 (1976) receivers for use in balanced digital multipoint systems
Limits for harmonics in the UK electricity supply system (RS-485 specification)
G.5/4 (2001 - proposed)
Planning levels for harmonic voltage distortion and the ANSI
connection of nonlinear loads to transmission systems and
public electricity supply systems in the United Kingdom X3.28 (1976)
Use of communication control characteristics of ASCII in
specified data communication links
EIA/TIA (PREVIOUSLY RS)
232 (1997)
Interface between data terminal equipment and data circuit -
terminating equipment employing serial binary data inter-
change (RS-232 specification)
 APPENDIX B

Symbols and Formulae

B1 SI UNITS AND SYMBOLS 336

B2 ELECTRICAL FORMULAE 338

B3 MECHANICAL FORMULAE 342

B4 WORKED EXAMPLES OF TYPICAL MECHANICAL LOADS 346

B1 SI UNITS A N D SYMBOLS

The following formulae are based on the International The system is independent of the effects of gravity, making a
System of Units, known as SI (Systeme Internationale clear distinction between the mass of a body (unit of
d'Unites) which is used throughout this book. SI was mass - kilogram) and its weight, i.e. the force due to gravity
adopted in February 1969 by a resolution of the CGPM (unit of force = Newton).
(Conference Generale de Poids et Mesures) as ISO
Recommendation R1000.
A base unit exists for each of the dimensionally independent Example:
physical quantities: length, mass, time, electric current,
A force of 1 N acting on a mass of 1 kg results in an accel-
thermodynamic temperature and luminous intensity. The
eration of 1 m s - 2.
SI unit of any other quantity may be derived by appropriate
simple multiplication or division of the base units without Conversion factors from non SI units to SI units are to be
the introduction of numerical factors. found in Appendix C.
 Appendix B 337

SI BASE UNITS Time-related Units

Symbol Quantity Unit Unit

Quantity Unit symbol Unit name
symbol name
Length m metre
t time S • second
Mass kg kilogram
r time constant S second
Time s second -1
u, v velocity ms metre per second
Electric current A Ampere -2
a acceleration ms metre per second
Temperature K Kelvin
per second
Luminous intensity cd candela
u; angular velocity rad S- 1 radian per second
a angular rad S- 2 radian per second
acceleration per second
DECIMAL MULTIPLES AND SUBMULTIPLES f frequency Hz Hertz
-1
n rotational frequency S (revolution)
per second
Factor Prefix Symbol

1012 tera T
10 9 giga G
Mechanical Units
10 6 mega M
10 3 kilo k
102 hecto h
Symbol Quantity Unit Unit
10 deca da
symbol name
10 -1 deci d
10 -2 centi c m mass kg kilogram
10 -3 milli m F force N Newton
10 - 6 micro la G (W) weight N Newton
10 -9 nano n J moment of inertia kg m 2 kilogram metre
10 -12 pico p squared
10 -15 femto f M (T) torque kgm kilogram metre
10 -18 atto a w (E) work (energy) J Joule
P power W Watt
P pressure Pa Pascal
E modulus of elasticity Pa Pascal
DERIVED UNITS Pascal
O" stress Pa
P density k g m -3 kilogram per
Geometrical Units cubic metre
6x rate of flow m 3 s -1 cubic metre
per second
Symbol Quantity Unit symbol Unit name k, k l, etc. any constant factor
1, s length, distance m metre
A area m2 square metre
V volume m3 cubic metre
a,/3, "7 etc. plane angle rad radian
o
degree
a,/3, "7 etc. solid angle steradian
338 ELECTRICAL
FORMULAE"Electrical Quantities

B2 ELECTRICAL FORMULAE

ELECTRICAL QUANTITIES
Total or apparent power in VA - VI - I2Z - - V2/Z
Active power in watts, W -- VI cos cp
Quantity Symbol Unit Unit
name symbol Reactive power in VAr - VI sin

Electromotive force E, e* Volt V

Potential difference V, v* Volt V THREE-PHASE INDUCTION MOTORS
Current I, i* Ampere A
Magnetic flux • Weber Weber All quantities r.m.s, values:
Frequency f Hertz Hz
Flux linkage A Weber-turns -- kWmech = horsepower × 0.746
Resistance R Ohm [2 kWetec = v/3 Vtll cos ~ at rated speed and load
Inductance L Henry H where Vt = supply voltage, Ii-- rated full load
Capacitance C Farad F current and c o s q ) - rated full-load power factor
Impedance Z Ohm [2
Reactance X Ohm f~ Efficiency, r / = (kWmech/kWelec) × 100 per cent
Power, D.C., or active P Watt W
Power, reactive Q Volt-Ampere VAr, var Phase current /p-// for star connection
reactive
Power, total or apparent S Volt-Ampere VA Ip = Il / v/3 for delta connection
Power factor angle cp __ o, deg.
Angular velocity aJ radians rad s- 1
per second
LOADS (PHASE VALUES)
Rotational velocity n revolutions s -1, rev s-1
per second Resistance R, measured in Ohms (no energy storage)
Revolutions •
mln --1
, r.p.m.
per minute Inductive reactance, Xt - coL - 27rfL Ohms
Efficiency r/ m (stores energy)
Number of pairs of poles p where f = frequency (Hz) and L - inductance (H)

*Capital and small letters designate r.m.s, and instanta- Capacitative reactance, X c - 1 / ( c o C ) - 1/(27rfC)
neous value, respectively
where f - f r e q u e n c y (Hz) and C - c a p a c i t a n c e (F)

A.C. THREE-PHASE (ASSUMING BALANCED

SYMMETRICAL WAVEFORM) IMPEDANCE

All quantities r.m.s, values: Impedance is the algebraic sum of the separate load
values thus:
VI = line-to-line voltage
Vp = Phase voltage (line-to-neutral) Z- v/(R 2 + X 2) or v/(R 2 + X~)
It - line current (star)
Ip = Phase current (delta) If R, XL and Xc are present in series in the same circuit then
XL and Xc may be summated, treating X c as negative, thus:
In a star connected circuit" Vp = lit~ v/3, Vl - ¢ 3 Vp, It - Ip
In a delta connected circuit" Ip = Il/v/3,It - v/3Ip, Vl - Vp z- v/(R + - Xc):)
Total of apparent power in V A - v/3 VlIt
Active power in watts, W = v/3 VtIl cos cp A.C. VECTOR AND IMPEDANCE DIAGRAMS
Reactive power in VAr - v/3 Vflt sin qo
Power factor (p.f.) - cos cp I f a voltage V is applied to an impedance Z, Figure B.la, the
current I will be phase displaced by an angle qS, Figure B. 1b.
= active power/apparent power
The current vector may be resolved into two component
= W/VAr
vectors at right angles, Figure B.2. The component in phase
with the voltage represents the value of the current due to
r e s i s t a n c e - - I cos 05, and the quadrature component repre-
A.C. SINGLE PHASE sents current due to reactance (which may be inductive or
capacitative) = I sin ~b.
All quantities r.m.s, values:
If each vector component is multiplied by the voltage V, the
V=/Z resulting triangle is similar, q5 is unchanged and the vector
 Appendix B 339

Zsin~b=X
V~ f

R
Z COS q~
=R

X
/ a

Figure B. 1 a impedance circuit Figure B.4 a impedance triangle

b voltage and current vectors b equivalent circuit

Isin~b

zT
I

Icos¢~ source I Z [-] load

Figure B.5 Elements of a power circuit

Figure B.2 Current vector diagram

Visin q~ and:
=12Zsin dp=12X
A cos ~ - R/z
sin ~b - X / Z
Vlcos$ P S tan ~b - X / R
=12ZCOSq~ Vl
=I2R Thus the impedance circuit Figure B.la, may be represented
by a circuit containing a resistance and a reactance in series,
Figure B.4b. All equations hold good whether X is inductive
or capacitive. If capacitive, the current, Figure B.lb, would
Figure B.3 Power vector diagram lead the voltage and the quadrature vectors of Figures B.2
and B.3 should be drawn in the opposite sense.
diagram of Figure B.3 shows the total power S, active power
P, and reactive power Q.
Since"
E.M.F., ENERGY TRANSFER

V-/Z E.m.f. is the internal driving force, symbol E, of an energy

source. It is equal to the sum of all the voltage drops in the
then: whole circuit, including any voltage drop attributable to the
energy source itself, Figure B.5.
S- VI - I 2 Z
E - I Z + Iz
similarly:
where E - s o u r c e e.m.f., I = current caused to flow bY E;
P = VI cos ~b -- I2Z cos ~b Z = s u m of all impedances external to the source and
z = impedance of the source itself.
and:
In the part of the circuit external to the source, the voltage
Q - VI sin ~b = I2Z sin ~b drop across the load is equal to the terminal voltage:

If these vectors are divided by/2, the resulting triangle is V-/Z

similar, x is unchanged, and the diagram of Figure B.4a
shows the impedance Z, and the resistance as the other two Substituting:
sides: E= V+Iz

Z cos ~b = R and:
Zsin$ = X
V = E-Iz
Further:
In A.C. circuits, quantities must be evaluated vectorially. Ifthe
z = x/(R: + x source is D.C., Z, z become R, r and evaluation is arithmetical.
340 ELECTRICAL FORMULAE: E.M.F., Energy Transfer

--1---I
i
i
i i
i i
i ~ i
i i
i i
i
I
i
i
i
i
'r
i
a
i
!

t equal areas
0 r R

Figure B.6 Maximum p o w e r transfer

n ,,,,,... ,,," 2 rc
Maximum power transfer from a supply source to a load
occurs when the resistance of the load is equal to the internal
resistance of the source. Varying values of load resistance
are plotted against power developed by the source (of fixed A~ equal areas
e.m.f, and internal resistance) in Figure B.6, showing that
...... 0.318 A
power dissipated reaches a peak value.
Note that practical power sources of low internal resistance
are usually unable to transfer maximum power due to current
overload.

0.955 A

MEAN AND R.M.S. VALUES, WAVEFORM

• • ~ sS

Principles
i "" "- .... , . , i .."
By definition, a symmetrical altemating quantity oscillates
about a zero axis and its mean value, therefore, is zero. Figure B.7 Sine wave
Each half cycle, however, has a definite mean value. For a instantaneous value
practical purposes, the mean of a half cycle represents the b rectified 1-phase full wave
D.C. value of a rectified alternating quantity. c rectified 1-phase half wave
The power delivered by a symmetrical alternating voltage or d rectified 3-phase full wave
current is proportional to the square of the quantity at any
instant. Negative values, when squared, are positive, so the
sum of the squares of the instantaneous values has a definite The mean height of one half cycle:
mean value. The square root of this mean represents that
value of a symmetrical quantity which produces the same
power or heating effect as if it were a direct quantity. It is
called the root mean square (r.m.s.) value, being the square
area/base -
(/o A sin a d a
)//
7r

root of the mean of the sum of the squares of the instanta- - (A/Tr) sin a d a
neous values of the voltage or current wave.
= 2A/Tr = 0.637A(3 sf)
The mean value of any variable quantity above (or below)
zero in a given period is the area enclosed by the variable,
If the half wave is not repeated in the next half cycle, from 7r
divided by the length of the base. The area enclosed is the
to 2 7r (i.e. half-wave rectification), the mean value is halved
sum of all instantaneous values in the period defined by the
since the area enclosed is the same but the period is double,
base length.
Figure B.7b:
half wave mean = A/Tr = 0.318A (3 sf)
Mean D.C. Value
If the supply is three phase, the effective value of the three
For a sine wave, Figure B.7a, the instantaneous value: phases added vectorially is 1.5 times the value of one phase,
Figure B.7c:
i = Asina
three-phase mean = 3 A / T r - 0.955A(3 sf)
The area is the sum of the instantaneous values in one half
cycle: Other waveforms (Figure B.8):

square wave mean = A B / B - A

~..,.i- A sinada
~--.,(
~0 ) f0 ~ semicircular wave mean = 7rA2/(2 x 2A) - 7rA/4

and period (base length) is 7r. triangular wave - A B / ( 2 x B) - A / 2

 Appendix B 341

mean value of
half cycle = A t
i 2~
t _J"
a
A2' \ aCr;a~mean square

f
0.785A ._.. /
1
1

Figure B.9 Sine wave

a instantaneous value
b squares of instantaneous values

qjlal areas
"''/ ....... 0.5A
The mean height of one half cycle:

area/base- (foTrA 2 sin2a da)/Tr

-- AZ/Tr

The representative value is the square root of this mean

Figure B.8 Other waveforms value:
a square wave
Ar.m.s. - v/(A2/2) = A/(v/2) - 0.707A(3 sf)
b semicircular wave
c triangular wave
Other Waveforms
Square wave Ar.m.s. -- A

Semicircular wave Ar.m.s.- ¢ 2 A / ¢ 3 - 0.816A(3 sf)

R.M.S. Value
Triangular wave Ar.m.s. -- A/v~3 - 0.577A(3 sf)
For a sine wave, Figure B.9a, where the instantaneous value:

i = Asina
Form Factor
The ratio of the r.m.s, to the mean value is called the form
the square of the instantaneous value, Figure B.9b is:
factor. A square wave has a form factor of unity. All others
i2 -- A2 sin2a are greater than unity and the sharper or more peaked the
wave shape, the higher the form factor:
The area is the sum of the squares of the instantaneous
values in one half cycle: sine wave form factor -- 7r/(2V/2) -- 1.11(3 sf)
square wave form f a c t o r - 1
Z(i) 2 - A 2 sin20l dc~ semicircular wave form f a c t o r - (4x/2)/(x/37r)
0 fo ~
- 1.039 (3 sf)
and the period (base length) is 7r. triangular wave form factor - 2/v/3 - 1.554 (3 sf)
342 MECHANICAL
FORMULAE:Laws of Motion

B3 MECHANICAL FORMULAE

Term Description Unit

d diameter m m
J
F force N
g acceleration due to gravity m s- 2
J total inertia kg m 2
JL load inertia kg m 2 Figure B. 10 The action o f a single force on a body
JM motor inertia kg m2
m mass kg The first law states that a force must be applied to a body in
M motor torque Nm
order to make it move from rest, or to change its speed or
Ma accelerating torque Nm
direction if it is moving.
Mt load torque Nm
n rotational frequency mln • --1"
The second law relates the applied force to the rate of change
nl -input mm • --1"
of momentum which it produces. Momentum is the pro-
n2 -output mln
• --1"

duct of mass and velocity, therefore for a constant mass,

An change of rotational frequency min- ~*
force is proportional to the rate of change of velocity, i.e. to
p pitch m
acceleration.
P motor power kW
Pa accelerating power kW The third law can be appreciated by considering, for
PL load power absorbed kW example, the way in which a rowing boat is propelled for-
r radius m ward by the rearward force exerted by the oars upon the
s distance m
water.
t acceleration time s
At acceleration period s
v linear velocity rn/min*
Av change of linear velocity m/min* Linear Motion
V traction capacity M 3 S--1
W energy J(Joule) Consider a body mass m acted upon by a single force F,
r/ efficiency Figure B.10. The body accelerates in the direction in which
# coefficient of friction - the force is acting, at a rate given by:

Note: for practical convenience, some o f the units in the A -F/m

formulae following are not SI units; for example, rota-
tional frequency is commonly measured in revolutions per After a time t has elapsed, the body has achieved a velocity
minute, although the SI unit is revolutions per second. In v, where:
these servo formulae, the terms used are as tabulated
above. Those which are in non-Sl units are marked*. v--u+at

(u is the initial velocity, before the force F was applied; if the

body was initially at rest, u is zero).
LAWS OF MOTION
The distance, s, travelled by the body during time t is:
Sir Isaac Newton (1642-1727) is well known for his work
s - ut + at 2/2
and discoveries in optics, gravitation and many other fields
of mathematics and the physical sciences. In 1687, Newton Distance and velocity are related by the following equation,
completed his monumental work on mechanics, in which the derived from the two previous ones:
concepts of force, velocity and acceleration were for the first
time accurately interrelated. 1,,2 - - U 2 -- 2as

The laws of motion, for which Newton is best known, are The work done by the force in accelerating the body is the
among the fundamental principles of mechanical engineer- product of force and distance:
ing, and are expressed as follows:
W-Fs

1 A body continues in its state of rest or uniform motion in The kinetic energy of the body, i.e. the energy which it
a straight line, unless impressed forces act upon it. possesses by virtue of its motion, is the product of its mass
2 The impressed force is proportional to the rate of change and the square of its velocity:
of momentum which it produces, and the change of
Ek -- mv2/2
momentum takes place in the direction of the straight
line along which the force acts. Furthermore, since energy is conserved, the work done by
3 Action and reaction are equal and opposite. the force is equal to the change in the body's kinetic energy
 Appendix B 343

(neglecting losses): After a time t has elapsed, the angular velocity ~ (rate of
change of angle) is given by:
W = m ( v 2 -- u2)/2
w=~vo +c~t
Power is the rate at which work is done, therefore it is the
product of force and velocity: (~o is the initial angular velocity, before the torque M was
applied; if the body was initially at rest, Wo is zero).
P= Fv
The angle, 7, through which the body rotates in time t is:

Rotational or Angular Motion 7 = coot + c~t2 / 2

A force acting perpendicular to a pivoted lever, Figure B. 11, Angle and angular velocity are related by the following
causes a turning effect or torque at the fulcrum. The torque is equation:
the product of the force and the radius at which it is applied:
03 2 -- CO2 -- 2c~7
M = Fr
The work done in accelerating the body is the product of
If a torque is applied to a body which is free to rotate, as in torque and angle of rotation:
Figure B.12, an acceleration results in a way which is ana-
logous to the example of linear motion above. Indeed, a W = M7
similarity will be noticed between the equations of motion.
The kinetic energy of the body is the product of its moment
Any body which is capable of rotating possesses a property of inertia and the square of its angular velocity:
known as moment of inertia which tends to resist accelera-
Ek -- J c o 2 / 2
tion in the same way as does the mass of a body in linear
motion. The moment of inertia is related not only to the mass
Since energy is conserved, the work done is equal to the
of the body, but also to the distribution of that mass with
change in kinetic energy (neglecting losses):
respect to radius.
The moment of inertia of a solid cylinder of radius r is W - J(w 2 - w2)/2
given by:
Power is the product of torque and angular velocity, i.e. the
J = mr2/2 rate at which work is being done:

By comparison, the moment of inertia of a hollow cylinder, P = Mw

of inner and outer radii respectively, is as follows:

J - m(r2oo - 4 ) / 2 Relationship Between Linear and

Angular Motion
It can be seen that, for a given outer radius, the moment of
inertia of a hollow cylinder is greater than that of a solid Consider a body of mass m moving in a circle of radius r
cylinder of the same mass. In Figure B. 12, a body having a with an angular velocity w, Figure B.13.
moment of inertia J is acted upon by a torque M. Its angular
When the body has rotated through an angle 7, it has
acceleration is:
covered a distance s along the circumference of the circle,
o~ = M / J where:

s=Tr
M= Fr

1 4
"1
,) m

F
Figure B. 11 The concept of torque

< M

Figure B.13 Relationship between linear and angular

Figure B. 12 The action of torque on a body motion
344 MECHANICAL
FORMULAE: Laws of Motion

Similarly, the tangential velocity or peripheral speed v, relative motion or tendency to relative motion. Consider a
being the quotient of distance and time, is given by: body of mass m at rest on a horizontal surface, Figure B. 15.
If a small force F is applied as shown, parallel to the surface
v - s / t - v /t on which the body is resting, an equal and opposite frictional
force FL is set up, preventing motion. If the applied force is
Angular velocity ~ is the quotient of angle and time:
gradually increased the opposing force increases with it, up
a; -- f / t to a point beyond which no further increase occurs, and the
body begins to move. The maximum value of the opposing
Therefore:
force is called the limiting frictional force. It can be shown to
V ~ OJr be independent of contact area, dependent on the nature of
the surfaces in contact, and proportional to the normal
Similarly, for acceleration: reaction (the force perpendicular to the surfaces in contact-
in this case the weight of the body, mg, where g is the
a - v/t- ~r/t
gravitational constant).
In general, where FL is the limiting frictional force and FN is
Therefore" the normal reaction, the ratio of the two forces is constant for
a - ar
a particular combination of surfaces in contact:

The moment of inertia is given by" F IFN - -

J- mr 2 The constant # is known as the coefficient of static friction

between the two surfaces. Friction is present whenever there
is relative motion between two surfaces in contact, although
The Effect of Gearing it may be considerably reduced by means of lubricants
which, by forming an interposing layer, keep the surfaces
When calculating the torque required to accelerate or
apart.
decelerate the moving parts of a machine, it is necessary to
take into account any gearing which introduces a ratio In all machines, part of the power supplied is used to over-
between the speeds of different parts. It is unusual to cal- come friction, thus the power available to do useful work is
culate the moment of inertia referred to the motor shaft, diminished. Other losses occur due to the viscous friction of
since this figure may be added arithmetically to the motor lubricants (e.g. the oil in a gearbox reduces friction between
inertia to arrive at a figure for the total inertia of the system. gears, but introduces other losses) air resistance etc.
Figure B.14 illustrates a motor, having a moment of inertia
The usable output power of a system is equal to the input
J1, driving a load with inertia J2, via a gearbox.
power minus the system losses. The ratio of output power to
If the gearbox has a ratio k, then the relationship between input power is called the system efficiency, 7/:
input and output angular velocities is as follows:
17 - - P O U T ~ P I N = ( P I N - - losses) ~ P I N
~01 ~--- k~2
The mechanical power lost in overcoming friction etc. is con-
Neglecting losses, the input and output torques are related verted into heat, and the disposal of this heat is an important
thus: consideration in large systems.
M1 = M 2 1 k Consider, for example, a machine driven by a 100 kW motor,
via a gearbox having an efficiency of 0.9 (usually expressed
The load inertia reflected back through the gearbox to the
motor shaft is reduced by a factor equal to the square of the as 90 per cent):
gear ratio. Therefore the total inertia which the motor has to l o s s e s - - P I N -- ~ P I N
overcome is given by:
- 1 0 0 - 90 - 10kW
J = J~ + J2/k 2
Therefore, 10 kW of heat is generated when the machine
works at full power. An oil cooler or heat exchanger may be
Friction and Losses
FN
Friction is the name given to the force acting tangentially to
the surfaces of two bodies in contact, which opposes their

motor gearbox load

M1 M2
/ ///// //////
k_, ,..,
o) 1 0) 2

mg
Figure B.14 The effect of gearing between motor and
load Figure B. 15 Friction between a body and a surface
 Appendix B 345

required to prevent overheating of the gearbox, and the actual pressure depends upon the density p of the liquid:
gearbox manufacturer will allow for this in the design.
p- pgh

For a rate of flow ~, the power delivered by the pump is

Fluid F l o w
given by:
The term fluid is used to describe any medium capable
P - p~5 - pgh6
of flowing. Gases and liquids are typical fluids; others
include foams, slurries and some granular solids (e.g. bulk In practice, the pressure will be increased by viscous friction
grain). and turbulence within the pipe work, but under normal cir-
For flow to take place, a pressure gradient must exist, cumstances these effects are small in comparison with the
therefore fluid will flow through a pipe connecting two pressure due to the height (head) of the liquid column.
vessels only if a difference in pressure exists between them.
Flow takes place from the higher pressure vessel to the
lower. In the case of liquids, pressure is directly proportional f
> > > >
to depth, so in Figure B.16 liquid flows from vessel A to > > >
vessel B until the levels are equal, when the flow ceases. The > > >
rate at which the fluid flows through a pipe or duct depends > > >
on four factors: > > > > >

1 The pressure gradient.

2 The viscosity of the fluid.
3 The cross-sectional shape and area of the pipe or duct.
4 Whether the flow is laminar, Figure B.17a, or turbulent,
Figure B. 17b.

If turbulence occurs, a greater pressure is required to achieve

a given rate of flow than when the flow is laminar. Pipe work
and ductwork should therefore be designed to avoid abrupt
changes in direction or cross section, and without obstruc- Figure B. 17 Fluid flow
tions which might give rise to turbulence. Consider the
a laminar
arrangements shown in Figure B.18, where a fan forces air
b turbulent
through a heat-exchanger matrix. The fan inlet is open to the
atmosphere, as is the outlet from the matrix, therefore these
two points are at atmospheric pressure. The pressure
between the fan and the matrix exceeds atmospheric pres-
oO00
0
0 air flow
sure, therefore air flows through the matrix as shown. o 0 o0 >
air inlet air outlet
I f p is the pressure difference (due to the fan) which gives ooo
rise to a rate of flow ~5, the power by the fan is: 0 0
000
P - p~5

Similarly, Figure B.19 shows a pump being used to raise

t
fan
t
matrix
liquid from one tank to another at a higher level.
Figure B. 18 Example of air flow
The pressure difference between inlet and outlet of the pump
is proportional to the height h to which the liquid is being
pumped with respect to the level in the lower tank. The

final level

direction of
flow

>
pump
direction of flow
A B

Figure B. 16 Flow of liquid between two vessels Figure B. 19 Example of liquid flow
346 WORKED EXAMPLESOF TYPICAL MECHANICAL LOADS: C o n v e y o r

B4 WORKED EXAMPLES OF TYPICAL M E C H A N I C A L LOADS

Note: data is typical, not necessarily SI units. Calculations Calculate Speeds and Gearing Ratio
correct to 3 significant figures
Volume of the required delivery rate:

CONVEYOR required delivery rate per h o u r - volume × density

A horizontal conveyor carries a loose material (not sensitive From data:

to shock). The conveyor is required to start when fully
loaded. volume delivered - mass delivered/density (m 3s -1)

Find the rating of a standard three-phase 400V A.C. volume per second - 250/2000 - 0.125 (m 3s -1)
induction motor to drive the conveyor, assuming typically
four-pole, 50 Hz, 1475 m i n - 1 at full load. Linear velocity v of the conveyor belt:

From the same data, find the motor power if the conveyor v- volume per second/cross-sectional area
raises the load through a height of 20 m.
= 0.125 x 1/(0.8 × 0.1)

Data = 1.56ms -1

required delivery rate 900 tonnes (250kgs -1) Rotational speed co of the belt driving pulley:
per hour
density of load material 2 kg dm -3 (2000 kg m -3) co - linear speed/radius
width of conveyor belt 800 mm (0.8 m)
length of conveyor 150m (150m) = 1.56/0.2
maximum safe depth of load on 1O0mm (0.1 m)
belt to prevent spillage = 7.8 rad S- 1

weight of conveyor belt 15kgm -1

diameter of belt pulley 400 mm (0.4m) Converting this to revolutions per minute gives
coefficient of friction of 0.09
conveyor, static r.p.m. - aJ x 60/(27r) - 74.5 min -1
coefficient of friction of conveyor, 0.07
moving Therefore, gearbox ratio will be:
manufacturer's information:
1475" 74.5 ~ 20" 1
breakaway torque, fully loaded 6850 N m
operational requirement: desired
acceleration time 4s

Load, Force and Torque

Method
Notes:
1 Ensure all data is in SI units.
• Data typically states 'weight' of the conveyor belt, but
2 Determine the linear and rotational speeds and gearing
gives the value in kg m - l , which is mass per unit length
ratio.
and correct in the SI system.
3 Determine the loading and acceleration, and from them • A factor of two is applied to account for the retum run of
the forces and torques to accelerate and to run. the belt.
4 Calculate power ratings of motor and the drive.

l o a d - mass of charge plus mass of belt

= (volume of charge x density)

V + 2(conveyor length x mass of belt per metre)

... iiiiiiiiNiiliiiiHiiiiiiiiiiiiiiiiiiiiiiiiii = [(150 × 0.8 x 0.1)(2000)] + 2(150 x 15)

( =28.5x 103kg

Breakaway torque to start the conveyor from rest in the fully

i 0 0 0
I~ 150 rn __-.~ loaded state is given in the data as:

Figure B.20 Conveyor 6850 N m

 Appendix B 347

Horizontal force required to accelerate the loaded conveyor demand at the motor:
against rolling friction after breakaway:
P- Moo
Fa - (load mass x acceleration)
As power is proportional to speed, this seems to present a
+ (load mass x g x coefficient of rolling friction) problem because, at the instant of starting, speed is zero and
= ma 4- rag# so power is zero also.

= m ( a 4- g # ) Further, during acceleration the speed increases from zero to

the normal running speed, and therefore the power demand
Further, torque to accelerate the conveyor against rolling appears to have no particular value. But in fact it does,
friction is force x radius of the pulley: because the acceleration torque continues up to the point
where the motor reaches full speed, and so the accelerating
Ma - F a r torque at full speed can be used to calculate the accelerating
= [m(a 4 - g # ) ] r N m power demand.

According to the data given, the starting and acceleration

Now, determine the linear acceleration, from the calculated
period is to be 4 s. If a motor is selected which is typically
linear velocity (1.56 m s-l), and the required acceleration
rated for 150 per cent overload for 30 s, use can be made of
time (4 s):
this capability since, at 4 s, the demand period is comfortably
linear acceleration - 1.56/4 - 0.39 m s -2
within the overload period of 30 s.

Motor power to accelerate the conveyor is the accelerating

Substituting this into the torque equation:
torque at the conveyor pulley multiplied by the rotational
Ma = 28.5 x 10 3 x [0.39 + (9.81 x 0.07)1 x 0.2 speed of the pulley at full speed, taking the efficiency (r/) of
the gearing into account. For this example, r/is assumed to
= 6140Nm be 98 per cent, but in practice it must be verified for the
gearbox installed:
The horizontal force required to run the fully loaded con-
veyor at full speed against rolling friction (assuming wind Pa -- Maod/ ~7
resistance to be negligible) is = 6140 X 7.8/0.98

Fs - (load mass x g x coefficient of rolling friction) = 48.9 kW

= mg# This would be the rating of a motor able to deliver accel-

erating torque continuously. If full use is to be made of the
and, torque to run the conveyor against rolling friction is 150 per cent overload rating, the actual full load rating of the
force x radius of the pulley: motor is 48.9/1.5 - 32.6 kW"

Ms - F s r a s t a n d a r d m o t o r rated 37 k W will be satisfactory

= mg#r This rating must now be verified for the drive module
= 28.5 x 103 × 9.81 x 0.07 x 0.2 required to operate the motor. The reason is that the drive
must be able to deliver the current demanded by the break-
= 3.91 x 103Nm
away torque.
It is often helpful to plot the torque/speed demand as show in As current is approximately directly proportional to torque,
Figure B.21. the current demand for breakaway is in the ratio 6850/6140
of the current demand for acceleration.
Power Ratings for the Motor and Drive Note, however, that the current demand during acceleration
Since power is equal on both sides of a gearing mechanism, must be based on the 48.9 kW overload rating for accelera-
there is no necessity to refer torque and acceleration to the tion, not the selected motor rating (37 kW).
motor. The power demand at the belt pulley is the power Calculate the currents involved and, having found the
starting current demand, find rating of the drive. The starting
current is based on the motor rating of 48.9 kW and is cal-
culated from"
6 850 breakaway torque
6 140 I
accelerating torque P = v/3 x VL x IL x power factor kW
E
Z
(ff 3910 running torque .~1 where the working power factor is assumed to be the full-
ET
load, full-speed, full-voltage-rated power factor of the motor,
L
0 0.85 (from data):

total IL -- P / ( v / 3 x VL x power factor)

speed, n = 48.9 x 103/(v/3 x 400 x 0.85)
Figure B.21 Conveyor torque~speed demand =83A
348 WORKED EXAMPLESOF TYPICALMECHANICALLOADS"Conveyor

starting current = 83 × 6850/6140 Data

-- 92.6A diameter of winding drum 200mm (0.2 m)
The drive itself will have a short time over current rating in gearbox ratio, motor to drum 11 - 1
the same way as the motor has. For the Control Techniques coefficient of hoisting friction 0.095
Unidrive, this is 150 per cent for 60 s. The drive efficiency
must be taken into account (98 per cent). The required full-
load rated current of the drive is 92.6/(1.5 × 0 . 9 8 ) - 63 A. Velocity Ratio (VR)
In practice, the full-load current, the short-time rating and This is a simple numerical ratio determined by the total
the efficiency must be verified. It is inadvisable to attempt to number of falls (ropes passing between pulleys) or alter-
calculate, using the formula as for a motor, a theoretical natively by counting the total number of pulley wheels in
value for the rating of the drive module. Users of VSDs are both sheaves (Figure B.23).
recommended to consult the data supplied by the manu-
This example, with four falls, has a ratio of 4" 1.
facturers of drives.
An actual example (the Control Techniques Unidrive) shows Speed and Acceleration of the Hook
that the current demand will be within the capability of
the UNI 3405 (37 kW) module, which can deliver 70 amps The speed of the lifting hook is a function of the motor speed,
continuously. the gearbox ratio, the winding drum diameter and the VR.
drum speed -- motor speed/ (gear ratio × 60)min -1

INCLINED CONVEYOR hook s p e e d - drum speed

× drum circumference lifting ratio
An inclined plane adds a component of force to the forces
= [1475/(11 × 60)] × (0.27r/4)
required to start, accelerate and run a conveyor. This com-
ponent is the force required to raise the load (Figure B.22). = 0.351 m s -1
As the load to be raised is sensitive, acceleration from rest
The total force is the hypotenuse of the vector diagram
must be slow. An acceleration time of eight seconds should
(Figure B.22) of which the horizontal component is the force
be adequate. Derive the linear acceleration of the load from:
required either to start, accelerate or drive the load at con-
stant speed as already calculated, and the additional vertical
v -- u + at whereu-0
component is m g multiplied by the sine of the angle of the
a - v / t -- 0.351/8
slope, 7 (in this case, 20/150).
-- 4.39 × 10 -2 m s -2
The total force F is calculated from:

F - 4 [ [ ;2 + (mg. sin ,.)/)2] Lifting Force and Torque to Accelerate

from Rest to Full Speed
The linear force Fa required at the hook to raise the load
HOIST
from rest up to full lifting speed is the algebraic sum of the
forces required to:
Example: a hoist is to lift sensitive loads of 2 tonnes max-
imum. The block-and-tackle arrangement has two pulley • suspend the load at rest (mg)
sheaves at the top and two at the hook. The fixed end of the • accelerate the load from rest to lifting speed (ma)
rope is secured at the top sheave block. • and overcome rolling friction ( m g # )
The hoist is to be operated by a standard three-phase four-
Fa -- m g + m a + m g #
hole 400V 50Hz induction motor, full load speed
1475 m i n - 1 , controlled by a VSD. Select a suitable rating -- m ( g + a + glz)
for the motor. = m[a + g(1 + #)] N

F~ ~ F~ ~ F~

Fh ~ mg VR=2:1 =~~,/VR=3:1
mgsin?.LI _.~ F

Figure B.22 Forces on an inclined plane Figure B.23 Pulley ratios

 Appendix B 349

The equivalent tangential force Fda at the winding drum is where Mda is the accelerating torque at the drum (539 N ), COd
Fa multiplied by the inverse of the lifting ratio (the inverse, is the rotational speed of the drum in radians per second and
since the calculation is proceeding from the output to the r/is the efficiency of the gearbox.
input):
Note that the inverse of the efficiency is applied; greater
Fda -- Fa/4 -- m[a + g(1 + #)]/4 N torque and power are required from the motor than the cal-
culated torque at the winding drum.
The accelerating torque at the drum is:
drum speed - [motor speed/ (gear ratio x 60)] x 7r

M d a - Fda X d r u m radius = [1475/(11 x 60)] x 7r

= {M[a + g(1 + #)1/4} x d / 2 = 7.02 rad S - 1
= {2 × 103[(4.39 x 10 -2)
Accelerating power is:
+ 9.81(1 + 0.095)J/4} x 0.2/2
P a - 539 x 7.02/0.98
= 539Nm
= 3.86 kW

Lifting Force and Torque to Maintain This would be the rating of a motor able to deliver accel-
Full Speed erating lifting torque continuously. Standard motors are
usually designed with a short time overload rating of 150 per
The linear force Fs required at the hook to maintain the load cent of full load rating for 30 seconds. According to the data,
at full lifting speed is the algebraic sum of the forces accelerating time is eight seconds, comfortably within the
required to: maximum for which a standard motor is designed. If use is to
be made of the overload rating, the actual rating of the motor
• suspend the load at rest (mg)
is 3.86/1.5 = 2.58 kW.
• overcome rolling friction (mgl~)
From inspection of the figures calculated earlier, it is
obvious without calculation that the motor rating for accel-
Fs - mg + mg#
eration will be adequate for continuous lifting at full speed:
= mg(1 + t-z)
a standard motor rated 3 k W will be satisfactory
The equivalent tangential force Fds at the winding drum is Fs
multiplied by the inverse of the lifting ratio"
Drive Module Power Rating
Fds -- Fs /4
In hoisting applications, drive module ratings correspond
= mg(1 + # ) / 4 N typically to standard motor ratings, and the short time over-
load rating of a drive is typically 150 per cent for 60 s. A drive
The full speed torque at the drum is:
rating of 3 kW will therefore be appropriate. Unusual appli-
Mds -- Fds × drum radius cations sometimes require a drive with a higher rating than the
= [mg(1 + #)/4] x d / 2 rating of the motor, but such circumstances are not common.

= [2 × 103 x 9.81(1 + 0.095)/4] x 0.2/2 Notes:

= 537 N m 1 In this example, where the acceleration time is relatively
long, the motor rating has been calculated taking
advantage of the short time overload ratings. If the duty
Required Motor Power Rating
cycle is one that demands rapid repetition of acceleration
Power is the multiple of torque and speed. For the purpose of from rest, the overload factor is not an option for
finally calculating the required motor power it is immaterial either the motor or the drive due to the integration of 12
whether power is calculated at the winding drum or at the with time.
motor, provided that the efficiency, r/, of the gearbox is taken 2 A well designed VSD for A.C. motors will
into account. For this example ~7is assumed to be 98 per cent, possess a torque feedback control feature as well as a
but in practice it must be verified for the gearbox installed. programmable acceleration ramp. If the torque feedback
During acceleration the speed increases from zero to the is used to control the ramp, the motor can be made to run
normal running speed, and therefore the power demand at a very low speed as soon as the initial torque demand
appears to have no particular value. But in fact it does, appears. This enables the motor to collect the inevitable
because the accelerating torque continues up to the point slack and backlash in the winding system quickly and
where the motor reaches full speed, and so the accelerating then pick up the load slowly, before accelerating under
torque at full speed can be used to calculate the accelerating acceleration ramp control.
power demand:

P = M~/~ 7
SCREW-FEED LOADS
Motor power related to the accelerating torque:
Loads, such as that shown in Figure B.24 are, in principle,
Pa -- gda X COd/~] the same as the conveyor example: a mass is moved
350 WORKEDEXAMPLESOF TYPICAL MECHANICAL LOADS: Screw-Feed Loads

applications, acceleration times may be especially short, and

F [iiiiii•iiiiiiiiiiiiiiiiiii•iiiiiiiii•iiiii•iii•i!i•i!i!ii1 there may be repetitive duty cycles.
(M Heavy machine tools can exploit VSD control of convent-
ional A.C. squirrel-cage motors. Calculation of motor power
M1 M2 .......~[[~..... is essentially the same as for conveyors. Remember to:
L
Figure B.24 Screw feed 1 Ensure that all variables are converted to SI units.
2 Determine the rotational and linear speeds from the
motor full speed and the gearing ratio, taking note that
horizontally. The force F is related to acceleration and mass the displacement of the load is the lead L of the feed
as in a conveyor. In many instances the mass concerned may screw per single revolution of the feed screw.
be quite low, and the force required to overcome friction 3 Determine the loading and acceleration, and from them
may be an insignificant part of the total but, in machine tool the forces and torques to accelerate and to run.
 APPENDIX C

Tables

C1 MECHANICAL CONVERSION TABLES 351

C2 GENERAL CONVERSION TABLES 354

C3 POWER/TORQUE/SPEED NOMOGRAM 357

Cl MECHANICAL CONVERSION TABLES

LENGTH

mm cm m inch foot yard km mile

mm 1 10 -1 10 -3 3.937 x 10 -2 3.280 x 10 -3 1.093 x 10 -3 10 - 6 6.213 x 10 - 7

cm 10 1 10 -2 3.937 x 1O- 1 3.280 x 10 -2 1.093 x 10 - 2 10 - 5 6.213 x 10 - 6
m 1000 1 O0 1 39.3701 3.28084 1.09361 10 - 3 6.213 x 10 - 4
inch 25.4 2.54 2.54 × 10 - 2 1 8.333 x 10 - 2 2.777 x 10 -2 2.54 x 10 -5 1.578 x 10 -5
foot 304.8 30.48 3.048 x 10 -1 12 1 3.333 x 10 -1 3.048 × 10 - 4 1.893 x 10 - 4
yard 914.4 91.44 9.144 x 10 -1 36 3 1 9.144 × 10 - 4 5.681 x 10 - 4
km 10 6 105 1000 39370.1 3280.84 1093.61 1 6.213 x 10 -1
mile 1.609 × 10 6 160934 1609.34 63360 5280 1760 1.609 1
352 MECHANICAL CONVERSION TABLES: Area
? ? ?
x x x ~ x x x
14t3 ~ O0 O0 ~4~
~ ~ ~ ~ I ~ ~"
,.~ ,-., ,.--, .--~ ~ ~
X X X X X X X X
X
x x x ~ x - x -
,4 0,, 0", '~1'- o o ,--~ tr'~ X ~0 X
.,..~ ,--,~ ,..,,~ ~ 0 ~"4 ~r'b £ q e~'b £'q
o,-~ e, i e 4 m , , . 6 , - ~6~oo6,---
x~~xx_ ~ ~ X X
xx x'"~,,.,x,,., ~x
• ~ ~ ~ ~. C~ ,,.o ,,.q. ~ . . .,.~ ~,~
? T
xAA~xx x N
0 X " X '~" ~" '~" oo
tt'3 ,.-..~ OO OO ¢'4
O~ O~, t"~. ',,,O O O ¢'4
odo~o~ ~N
• o ~ . ~,,oo
.,-~ o,..~ O,..~ ~,.-~ O,~ ¢,q ,,.~ ,..~ ,,...~ ,,.-..~ ~"x,I
T l"~ O~ 0 0
0 0 0 ,--~ ,--~
x X
'7 T
X ~ X X
N X '~ X
. ~ ~r'~ ~'~ • tl'~ '~" £',,I O . ¢'4 .
•~I- ~1- X '° I'" o,I ¢'4
.,..~ 0
I I ~
~ X X
X X ~ ~ ~ ? ~ ???71
X X X [,..... X X X X X
% oo oo e*"~ o", oo ~O ,,~ ~ ~".,I
o'~ oq. ,.-, t'~ oo l"-.- O',, O', "~t"
X X __?
X X i.... 0,,, ,,,13 ??~ ??,, o
',,,0 e ~ £"4 0'~ oo
o X X X X X X X oo
0
~ ~ X
?
X X ~ ~
X ['-.-. O',, I~ O',, '~"
° ',,.O ~o o ,,~'- II'---
t.~ ~::) .,,~,. ,,~ '~.
~ ~ ~ ~ . . .
o
~ • t~ ~,~ . ° t'~ '~
? ?? T
.,..~
X X i,... X o~o
~
',,0 t'~'b ¢',,I x ~ x x ~ $ x
~ ~ ~ ~ 0 . .
o
t--- x x
• . . oo '~t" 0o
o ,.--~ , ~ ,..-~ ~",4 ["..- £-4 £"4 o'3 ",::t" tth
o o o~::, . ,,, ~ . , q . o .
o
N N
• __1 .&
t-.I ~,1 t-,.l
0
 Appendix C 353
I I
0 0 0 0 0 ~ 0 0
X X X ~ X X X X
,.D
0 0 0 0
X ~ ~ ~ X X X ~
~. . ~ ~ ~ ~ ~~ -
IT I ? T
0 0 ~ 0 0
X X ~ ~ X ~ X X
~ ~ ~ X
I ?
o o
x~x~
o~ o,,i i.rb ~
X X X I....,
¢'N o,,I ~r'~ ,,_~
0 0 c",l
,,,--i ,,--i ~ X
,,,-,,,-i " ~ ~ o
,- ,~ .,--~ X 0 oo 0
IT? T ? ? ?
? ? 0 0 0 0
o o o
o
o X X X X
oo oo £-4 x ~ ~ ~ x x
. ~ x oo (:~ ,~I" ,-- • ~ ~ ~
~1 ~1 ',d ,,.-, ¢N
IN
Q
?
0
o
x
X ~ ,~r'b ~,~ T i ~ ~I ?I - I O I
t~ ~
~ t'-,l • 1",1 0 X X X X X X X
~ ,--~ ,---~ X ~ 7
?, IN
x ~) x x ,~i. X
o o ~ _.9. E o£. oo oo o o~ t,4 e4 m
x x~ X X X
? I?1 T
x ~ xxx~xN
• o ~ ~ ~
,.F'I
~) 00 o
~,~ . ,....~ 0 o
,~q it3 X tth ~t'~ ,-.~ £'4
oo O oo . ~:::~ . ,-- ~'N m
>,,,
m
I-- IN
IN
~ IN
0 i_u I, LI
Z Z _~_~.~ _~ ~ ~_~_~_~_~
ILl m
354 CONVERSIONTABLES:Torque
MECHANICAL

TORQUE

N cm Nm kp cm kp m p cm oz in inlb ftlb

Ncm 1 10 . 2 1.019 x 10 -1 1.019 × 10 -3 101.972 1.41612 8.850 x 10 -2 7.375 x 10 - 3

Nm 100 1 10.1972 1.019 × 10 -1 10197.2 141.612 8.85075 7.375 x 10 -1
kpcm 9.80665 9.806 x 10 . 2 1 10 -2 1000 13.8874 8.679 x 10 -1 7.233 x 10 - 2
kpm 980.665 9.80665 100 1 105 1388.74 86.7962 7.23301
pcm 9.806 × 10 -3 9.806 × 10 -5 10 -3 10 -5 1 1.388 x 10 - 2 8.679 x 10 - 4 7.233 x 10 . 5
ozin 7.061 × 10 -1 7.061 × 10 -3 7.200 x 10 . 2 7.200 x 10 -4 72.0078 1 6.25 x 10 . 2 5.208 x 10 .3
inlb 11.2985 1.129 × 10 -1 1.15212 1.152 x 10 . 2 1152.12 16 1 8.333 x 10 . 2
~lb 135.582 1.35582 13.8225 1.382 x 10 -1 13825.5 192 12 1

FORCE

N kp p OZ lbf

N 1 1.019 X 10 -1 101.972 3.59694 2.248 x 10 - 1

kp 9.80665 1 1000 35.274 2.20462
p 9.806 x 10 -3 10 -3 1 3.5274 x 10 - 2 2.204 x 10 - 3
oz 2.780 x 10 -1 2.835 x 10 -2 28.3495 1 6.25 x 10 -2
lbf 4.44822 4.536 x 10 -1 453.592 16 1

POWER

k p m s -1 -1
kW PS hp kcal s

kW 1 1.35962 1.34102 101972 2.388 × 10 -1

PS 7.355 × 10 -1 1 9.8632 x 10 -1 75 1.756 x 10 -1
hp 7.457 x 10 -1 1.01387 1 76.0402 1.781 × 10 -1
k p m s -1 9.806 x 10 -3 1.333 x 10 -2 1.3515 x 10 -2 1 2.342 × 10 - 3
kcal s - 1 4.1868 5.69246 5.61459 426.935 1

C2 GENERAL CONVERSION TABLES

LENGTH AREA

SI unit - metre (m) SI u n i t - square metre (m 2)

to convert from: to: multiply by: to convert from: to" multiply by:

mile m 1609.344 square miles m 2 2.59 x 10 6

nautical mile m 1853 acre m 2 4047
km m 103 hectare (ha) m 2 104
cm m 10 - 2 k m 2 (sq. k m ) m 2 106
mm m 10 - 3 cm 2 m 2 10 - 4
yd m 0.9144 mm 2 m 2 10 - 6
ft m 0.3048 yd 2 m 2 0.8361
in m 2.54 x 10 - 2 ft 2 m 2 9.29 x 10 - 2
mil m 2.54 x 10 - 5 in 2 m 2 6.45 x 10 - 4
mil 2 m 2 6.45 x 1 0 - l O
 Appendix C 355

VOLUME VELOCITY (LINEAR)

SI unit - cubic metre (m 3) SI u n i t - metre per second ( m s - l )

to convert from: to: multiply by: to convert from: to: multiply by:

yd 3 m3 -1
0.765 m.p.h. (mile per hour) ms 0.4470
ft 3 m 3 2.83 x 10 - 2 flmin- 1 ms- 1 5.08 x 10 - 3
in 3 m3 1.64 x 10 . 4 fts -1 ms -1 0.3048
dm 3 m3 10 . 3 kmh -1 ms -1 0.2778
• -1 -1
litre m3 10 . 3 m mln ms 1.67 x 10 - 2
-1
gallon (imperial) m3 4.55 x 10- 3 knot ms 0.5145
gallon (US) m3 3.79 x 10 - 3
pint (imperial) m3 5.68 x 10 - 4
pint (US) m3 4.73 x 10 - 4

VELOCITY (ANGULAR)
MASS
SI unit - radians per second ( r a d s - 1 )

to convert from: to" multiply by:

SI u n i t - kilogram (kg)
r.p.m. (revolutions per min) rad s - 1 0.1037 (27r/60)
to convert from: to: multiply by: --1
r s - 1 (revolutions per sec) rad s 6.283 (270
ton (imperial) kg 1016 o s - 1 (degrees per sec) rad s - 1 1.75 x 1 0 - 2 (27r/360)
ton (US) kg 907.2
tonne (metric) kg 103
slug kg 14.59
lb kg 0.4536 TORQUE
oz kg 2.84 x 10 - 2
g kg 10 - 3
SI unit - Newton meter (N m)

to convert from: to: multiply by:

FORCE AND WEIGHT
lbf ft Nm 1.356
lbf in Nm 0.1129
SI unit - Newton (N) ozf in Nm 7.062 x 10 - 3
kgf m Nm 9.8067
to convert from: to: multiply by:
kp m Nm 9.8067
tonf (ton wt) N 9964
lbf (lb wt) N 4.448
poundal N 0.1383
ozf (oz wt) N 0.2780 ENERGY
kp y 9.807
p N 9.81 x 10 . 2
kgf (kg wt) N 9.807 SI unit - Joule (J)
gf (g wt) N 9.81X 10 - 2
10 - 5 to convert from: to: multiply by:
dyn N
btu 1.055 × 103
therm (105 btu) 1.055 × 108
PRESSURE AND STRESS cal 4.187
ft lbf (ft lb wt) 1.356
ft poundal 0.0421
SI u n i t - Pascal (Pa)

to convert from: to: multiply by:

at (technical atmosphere) Pa 9.81 x 103 POWER

in WG Pa 248.9
mm WG Pa 10.34
in HG Pa 3385 SI u n i t - kilowatt (kW)
mm HG (torr) Pa 131.0 to convert from: to: multiply by:
-2
kpcm Pa 9.81 x 103
-2
Nm Pa 1.0 HP kW 0.7457
bar Pa 105 ps kW 0.7355
lb fft - 2 Pa 47.88 ch, CV kW 0.7355
-1
lb fin - 2 Pa 6895 Btus kW 1.055
-1
kg f m - 2 Pa 9.807 kcal s kW 4.1868
-2 -1 1.36 x 10 - 3
kg f c m Pa 9.81 x 104 ft l b f s kW
356 GENERAL
CONVERSIONTABLES: Moment of Inertia

MOMENT OF INERTIA FORCE

SI unit - kilogram metre 2 (kgm 2) SI unit - N e w t o n (N)

to convert from: to: multiply by: to convert from: to" multiply by:

k g f m 2 (GD 2) kg m 2 0.25 lb(f) N 4.4482

lbfft 2 ( W K 2) kgm 2 4.21 × 10 - 2 N lb(f) 0.22481
kp m s 2 kg m 2 9.807
ft l b f s 2 kgm 2 1.356
lbf in 2 kg m 2 2.926 x 10 - 4
o z f in 2 kg m 2 1.829 x 10 - 5
MOMENT OF INERTIA

SI unit - kilogram metre 2 ( k g m 2)

TEMPERATURE to convert from: to: multiply by:

lb in s 2 kg m 2 0.113
SI u n i t - Kelvin (K) ozins 2 kgm 2 7.06155 x 10 - 2
to convert from: to: factor: kg m 2 lb in s 2 8.85075
kgm 2 ozins 2 141.612
°C K ×1 kg cm 2 kgm 2 10 - 4
t°C K t+273.15
°F K × 0.5555
t°F K ( t - 32) x 0.5555
LINEAR ACCELERATION

FLOW SI unit - metre per second 2 (m s-2)

to convert from: to" multiply by:
SI unit - cubic metre per second ( m 3 s - 1) -2 -2
in s ms 2.54 x 10 - 2
--2 --2
to convert from: to: multiply by: fts ms 0.3048
--2 -2
ms in s 39.37
m 3 S-1 -2 -2
gallon per hour (imp) 1.26 × 10 - 6 ms fts 3.2808
gallon per hour (US) m 3 S-1 1.05 x 10 - 6
litre per hour m3s - 1 1.67 x 10 - 5
litre per second m3s - 1 10 - 3
cfm m 3 S-1 4.72 × 10 - 4
m3h -1 m 3 S-1 2.78 × 10 - 4
m 3min- 1 m 3 S-1 1.67 × 10 - 2

TORQUE

SI unit - N e w t o n metre (N m)

to convert from: to: multiply by:

lb ft Nm 1.356
lb in Nm 0.1129
oz in Nm 7.062 x 10 - 3
Nm lbft 0.7375
Nm lbft 8.857
Nm oz in 141.6
 Appendix C 357

C3 POWER/TORQUE/SPEED NOMOGRAM

Nm 2000
Ibft
- -

kW - - 10 hp - - 30 min -1
7 - - -
_ - - 9
2000 - -
m

6------8
B
-- 1000 --- 40
5---
---7
1000 --
--6 -- 50
4 - -
- . 500
----5 - - 60
- - 400 =
500 - -
-- 70
3 4 400 -- - - 300
-- 80
300 --
-- 200 90
----3 200 - - 100
-
2.,-..-

--. 100

1.5'-- - - 2
100 -- --- 150
m

-- 50
50 - - - - 40 - - 200
1--
4O - - - 30
0.9---
30--
0.8-- - - 20
--1 - - 300
0.7-- 20.---
-- 0.9
0.6--
-- 0.8
--10 - - 400
- - - 0.7
0.5'--
10--_
- - - 0.6 - - 500
0.4-- " ~ ' 5

- - - 0.5 - - 600
.-4 5252 x hp
5"- M (Ib ft) =
min -1 . i 700
0 . 3 . - - - - - 0.4 4 3
_=
800
3--
"--2 - - 900
-- 0.3 1000
2 --- . =

0 . 2 " "
9550 x kW
M(Nm) =
--1 min -1

-- 0.2 1 -_ - - - 1500
.,=

" - - 0.5
.= - - 2000
0.5 - - ' - - 0.4
0.1-.- 7124 x hp
i 0.4 0.3 M(Nm) =
min -1
0.3--
'-" 0.2
- - 3000
--0.1

power torque speed

 APPENDIX D

World Industrial Electricity Supplies (<1 kV)

Country Industrial three-phase Supply frequency (Hz) Comments

supply voltages below
1000 V (V)
Afghanistan 380/220 5O
280/220
Algeria 380/220 50
Angola 380/220 50
Anguilla 400/230 50
Antigua and Barbuda 400/230 60
208/120
Argentina 380/220 50
Armenia 380/220 50
Australia 415/230 50
Austria 380/220 50
Azerbaijan 380/220 50
Bahamas 240/120 60
Bahrain 400/230 50,60
Bangladesh 400/230 50
Barbados 400/230 50 three-phase delta - earthed mid point of
230/115 phase; one-phase, three-wire- earthed
230/115 mid point; three-phase star - earthed star point
Belarus 380/220 50
Belgium 380/220 50
Belize 480/277 60
Benin 380/220 50
Bermuda 208/120 60 three-phase star - earthed star point
240/120 one-phase, three-wire - earthed mid point
Bolivia 400/230 50
Bosnia and Herzegovinia 380/220 50
Botswana 380/220 50
Brazil 380/220 60
Brunei 415/240 50
Bulgaria 380/220 50
Burkina Faso 380/220 50
Burundi 380/220 50
Cambodia 400/230 50
Cameroon 380/220 50
Canada 600/347 60
416/240
Canary Islands 380/220 50
Cape Verde Islands 380/220 50
Cayman Islands 480/240 60 three-phase delta - earthed mid point of phase
 Appendix D 359

Table (continued)
Country Industrial three-phase Supply frequency (Hz) Comments
supply voltages below
1000 V (V)

Central African Republic 380/220 50

Chad 380/220 50
Chile 380/220 50
China 380/220 50
Colombia 240/120 60 three-phase delta - earthed mid point of phase
Comoros 380/220 50
Congo 380/220 50
Costa Rica 240/120 60 three-phase delta - earthed mid point of phase
Croatia 400/230 50
Cuba 380/220 50
Cyprus 415/240 50
Czech Republic 380/220 50

Denmark 400/230 50
Djibouti 400/230 50
Dominica 400/230 50
Dominican Republic 230/110 60 three-phase delta - earthed mid point of phase

Ecuador 440/220 60 one-phase, three-wire - earthed mid point

Egypt 380/220 50
E1 Salvador 440/220 60 three-phase d e l t a - earthed mid point of phase
240/120 one-phase, three-wire - earthed mid point
Estonia 380/220 50
Ethiopia 380/231 50

Falkland Islands 415/230 50

Faroe Islands 415/240 50
Fiji 415/240 50
Finland 690/400 50
France 400/230 50
690/400
400/230
French Guiana 380/220 50
French Polynesia 220/127 60

Gabon 380/220 50
Gambia 380 50
Georgia 380/220 50
Germany 400/230 50
690/400
Ghana 440/250 50
Gibraltar 415/240 50
Greece 380/220 50
Greenland 380/220 50
Grenada 400/230 50
Guadeloupe 380/220 50,60
Guam 480/227 60
480 three-phase - nonearthed neutral
240/120 three-phase open delta - earthed mid point of phase
208/120
Guatemala 220/110 60
Guinea-Bissau 380/220 50
Guyana 380/220 50,60

Haiti 380/220 60
230/115 one-phase, three-wire - earthed mid point
Honduras 480/277 60
240/120
Hong Kong 380/220 5O
Hungary 400/230 5O

Iceland 400/230 50
India 440/250 50
360 WORLDINDUSTRIALELECTRICITYSUPPLIES

Table (continued)
Country Industrial three-phase Supply frequency (Hz) Comments
supply voltages below
1000 V (V)

India 400/230
Indonesia 380/220 50
Iran 380/220 50
400/231
Iraq 380/220 50
Ireland, Northern 400/230 50
380/220
Ireland, Republic of 400/230 50
Israel 400/230 50
Italy 400/230 50
Ivory Coast 380/220 50

Jamaica 220/110 50 three-phase delta - earthed mid point of phase

Japan 380/220 50,60 one-phase, three-wire - earthed mid point
200V+20%
200/100
Jordan 400/230 50

Kazakhstan 380/220 50
Kenya 415/240 50
Korea, North 380/220 60
Korea, South 380/220 60
Kuwait 415/240 50
Kyrgyzstan 380/220 50

Laos 380/220 50
Latvia 380/220 50
Lebanon 380/220 50
Lesotho 380/220 50
Liberia 416/240 60 three-phase, three-wire
240/120 one-phase, three-wire - earthed mid point
208/120 three-phase - nonearthed neutral
Libya 400/230 50
220/127
Liechtenstein 380/220 50
Luxembourg 380/220 50
220/127
208/120

Macedonia 380/220 50
Madagascar 380/220 50
Malawi 400/230 50
Malaysia 415/240 50
Mali 380/220 50
220/127
Malta 415/240 50
Martinique 220/127 50
Mauritania 380/220 50
Mauritius 430/230 50
Mexico 480/227 60
220/127 three-phase three-wire
Moldova 380/220 50
Monaco 400/230 50
Mongolia 380/220 50
Montserrat 400/230 60
Morocco 380/220 50
Mozambique 380/220 50
Myanmar 400/230 50

Namibia 390/220 50
Nauru 415/240 50
Nepal 440/230 50
Netherlands 400/230 50
 Appendix D 361

Table (continued)
Country Industrial three-phase Supply frequency (Hz) Comments
supply voltages below
1000 V (V)

Netherlands Antilles 380/220 50, 60

220/127 three-phase delta - earthed mid point of phase
208/120
New Caledonia 380/220 50
New Guinea 415/240 5O
New Zealand 400/230 5O
Nicaragua 480/240 60
240/120 three-phase delta - earthed mid point of phase
Niger 380/220 50
Nigeria 400/230 50
380/220
Norway 400/230 50
690
Oman 415/240 50
Pakistan 400/230 50
Panama 480/277 60
208/120
Papua New Guinea 415/240 50
Paraguay 380/220 50
Peru 380/220 60
Philippines 440/240 60 three-phase, three-wire
Poland 400/230 50
690/400
Portugal 380/220 50
Puerto Rico 415/240 60
Qatar 415/240 50
Reunion 415/240 50
Romania 400/230 50
690/400
440/220
Russian Federation 380/220 50
Rwanda 380/220 50
St Helena 415/240 50
St Kitts & Nevis 400/230 60
St Lucia 415/240 50
St Vincent and the Grenedines 400/230 50
San Marino 380/220 50
Saudi Arabia 380/220 50,60
Senegal 380/220 50
220/127 one-phase - earthed neutral
Seychelles 400/230 50
Sierra Leone 400/230 50
Singapore 400/230 50
Slovak Republic 400/230 50
Slovenia, Republic of 380/220 50
Solomon Islands 415/240 50
Somalia 440/220 50 three-phase delta - earthed mid point of phase
220/110 three-phase delta - earthed mid point of phase
South Africa 525 50
400/230
Spain 4O0/230 50
Sri Lanka 400/230 50
Sudan 415/240 50
Suriname 220/127 50,60
220/110 three-phase delta - earthed mid point of phase
Swaziland 400/230 50
Sweden 400/230 50
Switzerland 400/230 50
Syrian Arab Republic 380/220 50
Taiwan 400/300 60
362 WORLD INDUSTRIALELECTRICITYSUPPLIES

Table (continued)
Country Industrial three-phase Supply frequency (Hz) Comments
supply voltages below
1000 V (V)

Tajakistan 380/220 50
Tanzania 400/230 50
Thailand 380/220 50
Togo 380/220 50
Tonga 415/240 50
Trinidad and Tobago 400/230 60
Tunisia 380/220 50
Turkey 380/220 50
Turkmenistan 380/220 50
Uganda 415/240 50
Ukraine 380/220 50
United Arab Emirates 415/240 50
380/220
United Kingdom 400/230 50
United States of America 480/277 60
575 three-phase delta
575 three-phase delta- one phase earthed
460/265 three-phase delta
240/120
208/120
Uruguay 380/220 50
Uzbekistan 380/220 50
Vatuatu 500 50
380
Venezuela 208/120 60
240/120 three-phase delta - one phase earthed
Vietnam 380/220 50
Virgin Isles 208/120 60
Yemen, Republic of 440/250 50
Yugoslavia, Federal Republic of 380/220 50
Zaire 380/220 50
Zambia 400/230 50
Zimbabwe 390/225 50
 APPENDIX E

Bibliography and Further Reading

The documents listed here have been selected to provide the Improvement of induction machine stability by modulation
reader with useful sources of information and further read- techniques by M.Cade. lEE Proceedings Part B, November
ing relating to electrical variable speed drives and their 1994, pp. 347-352. Detailed description of space vector
application. modulation.
Converter engineering by G. Moltgen. John Wiley, 1984,
P WM rectifier using indirect voltage sensing by P. Barrass
ISBN 0 471 90561 5. A reference for fundamental power
and M. Cade. lEE Proceedings Part B, September 1999, pp.
converter operations and relationships.
539-544. Detailed description of PWM rectifier control.
Electric motor handbook by B.J. Chalmers. Butterworths,
1988, ISBN 0-408-00707-9. A practical reference book Electric fuses by A. Wright and P.G. Newbury. 2nd Edition.
covering many aspects of characteristics, specification, IEE Power Series 20, ISBN 0 85296 825 6. A clear guide to
design, selection, commissioning and maintenance. fuse design, performance, application and good practice.

Sensorless vector and direct torque control by P. Vas. Electrical engineer's reference book by G.R. Jones,
Oxford University Press, 1998, ISBN 0198564651. General M.A. Laughton and M.G. Say. Butterworth-Heinemann,
background to the theory of vector control of motors. 1993, ISBN 7506 1202 9. Everything in one book.