sn+ gDrTroN
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* =.
J.
o UJ
F J
CS
o lrf F(
b 1f$ UJ
sJVON Z14L
ELECTRICAL TECHNOLOGY SJ VAN ZYL

5TH
EDITION
Published by:
lg rata P O Box 6201 Vandsrbijlpark
1900
Tel: 082 852 0340
@
LERATo 2011
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic,
mechanical, photocoqYing, recording and/or otherwise without the prior written permission of the publisher. This book may not be lent, resold, hireO out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of t[e publishers.
ISBN : 978098144833G
l*;
ELECTRICAL TECHNOLOGY SJ VAN ZYL

sTH EDITION
TABLE OF CONTENTS
CHAPTER ONE
1.1

ADVANCED ELECTRICAL MEASUREMENT
WATTMETER APPLICATIONS
1
1.1.1 ONEWATTMETER METHOD 2 1.1.2 BLONDELL'S THEOREM 4 1.1.3 TWOWATTMETER METHOD 6
1.1.3.1
1,1.4
TWOWATTMETER METHOD OF OBTAINING THE POWER FACTOR THREEWATTMETER METHOD 10
CHAPTER TWO
2.1 2.2 ??.t 2.2.2
 THREE.PHASE ELECTRICAL SYSTEMS
INTRODUCTION THREEPHASE
13
SYSTEMS
14 PHASE SEQUENCE OF THREE_PHASE
SYSTEMS
16
BALANCED STARCONNECTED, THREEPHASE SYSTEMS BALANCED, DELTA.CONNECTED, THREEPHASE SYSTEM UNBALANCED DELTACONNECTED, THREEPHASE LOAD 2.2.4.t STARDELTA (y_A) CONFTGURATTdN 26 2.2.4.2 DELTADELTA (A_A) CONFTGUMTTON 28 UNBALANCED STARCONNECTED, THREEPHASE LOAD 2.2.s.7 DELTASTAR (A_y) CONFTGURATiON 30 2.2.s.2 STARSTAR (y_y) CoNFTGURATTON 3s 2.2.5.3 MILLMAN'S THEOREM 35 2.2.5.4 DELTASTAR (A_y) CoNVERSTON 39 2.2.5.s STARDELTA (y_A) CONVERSTON 43 COMPLEX POWER 48 REACTIVE POWER 51
?.?.1 2.2.4
2.2.5
2.3 2,4 2.5
2.1 2.2
EXERCISE 54 POWER MEASUREMENT IN THREEPHASE EXERCISE 74
CHAPTER THREE
3.1

SYSTEMS
19
23 26
30
59
SYMMETRICAL COMPONENTS
INTRODUCTION
BO
3.1.1 POSTTTVE PHASE_SEQUENCE 3.1.2 NEGATTVE PHASE_SEQUENCE 8282 3.1.3 ZERO PHASESEQUENbE 82
3.2
RESOLUTION OF AN UNBALANCED, THREEPHASE SYMMETRICALCOMPONENTS 83
SYSTEM
OF
PHASORS
INTO
ITS
3.3 3.4 3.5
SiGNiFICANCE OF SYMMETRICAL COMPONENTS IN PROTECNVE SYSTEMS 85 DETECNON OF P.P.S AND I.I.P.S COMPONENTS OF CURRENT 86 DETECTION OF P.P.S AND N,P.S COMPONENTS OF VOLTAGE 88 E(ERCISE 101
3
CHAPTER FOUR
INTRODUCTION
4.2 4.3 4.4 4.5 4.6 4.7
REASONS FOR
THREEPHASEINDUCTION REGULATOR 108 SYNCHRONOUS PHASE MODIFIER 111 VALUE OFTHE SENDINGEND VOLTAGE 113 AUTOMATIC LOAD DISPATCHING TT4 POWER TMNSFER TL4
CHAPTER FIVE
5.1 5.2
5.2,1 5.2.2 5,2.3 5.2.4
5.3 5.4
5.4.1 5.4.2 5.4.3
5.5 5,6
6.6

T32
POWER ECONOMICS
MOST ECONOMICAL CROSSSECTIONAL AREA OF A CONDUCTOR OT TRANSMISSION KELVIN'S LAW T42 LIMITATIONS TO THE APPLICATION OF KELVIN'S LAW T43 TRANSMISSION VOLTAGE 144
GENEMTING COSTS L45 MOST ECONOMICAL POWER FACTOR 145
TARIFFS 146
CHAPTER SIX
6.5.1 6.5.2 6.5.3 6.5.4
4
INTRODUCTION 139 MAXIMUM DEMAND 139 ENCOUMGEMENT TO DIVERSIFY THE LOAD 139 ENCOUMGEMENT OF POWER FACTOR CORRECTION 140 LOAD AND FORM FACTORS I4O LOSS FACTOR 741
EXERCISE
6.1 6.2 6,3 6.4 6.5
TO4
INTERCONNECTIONS IO7
TAPCHANGINGTRANSFORMER 110
EXERCISE

5
160
PER UNIT SYSTEMS
INTRODUCTION 163 PERUNrT QUANTTTTES 163 ADVANTAGES OFTHE PERUNIT SYSTEM 165 DISADVANTAGES OF THE PERUNIT SYSTEM 165 THE PERUNIT SYSTEM 165 THREEPHASE EQUIPMENT 167 BASE SELECTTON FOR PER_UNrT QUANTTTTES 168 CHANGING BASE VALUES 168 PERUNIT IMPEDANCE OF A TRANSFORMER 169 APPLICATION IN NETWORK CALCULATIONS L7O EXERCISE

INTER.CONNECTED SYSTEMS
4.I
4.8
ii

6
IB7
LINE
141
CHAPTER SEVEN
7.I 7.2 7.3
7.3.T 7.3.2 7.3.3
7.4
7.4.L 7.4.2 7.4.3
7.5 7.5.1 7.6
 ALTERNATING
AND DIRECT CURRENT DISTRIBUTION
INTRODUCTION 191 REQUIREMENTS FOR GOOD DISTRIBUTION SYSTEMS !92 CLASSIFICATION OF DISTRIBUTION SYSTEMS T92 NATURE OF CURRENT T92 TYPE OF CONSTRUCTION 193 CONNECTION SCHEME 193 CONNECTION CIRCUITS OF DISTRIBUTION SYSTEMS 193 RADIAL DISTRIBUTION SYSTEMS 193 RING DISTRIBUTION SYSTEM 194 INTERCONNECTED SYSTEMS 195 DIRECT CURRENT DISTRIBUTION SYSTEMS 196 DIRECT CURRENT RING DISTRIBUTOR T97 ALTERNATING CURRENT DISTRIBUTION SYSTEMS 205 EXERCISE 2TB
7
CHAPTER EIGHT
_ ELECTRIC LIGHTING
8.1
INTRODUCTION 227
8.2
TROUBLE SHOOTING WITH
8.1.1 LAMPS 227 8.1.2 DICHROIC COLOUR CHANGE 222 8.1.3 FIXTURES AND PARTS 222 8.1.4 MECHANICAL LUMINAIRES 222 8.1.5 TECHNOLOGICAL DEVELOPMENTS 222 8.1.6 ELECTRONIC HIGHFREQUENCY BALLASTS 223 LUMINAIRES 224
8.2.1 FAULT CONNECTION OF LAMP WIRES 224 8.2.2 LAMP WIRES NOT CONNECTED 224 8.2.3 CATHODE BROKEN 225 8.2.4 COLD ENVIRONMENTS 225 8.2.5 POOR WIRE CONTACTS 225 8.2.6 TOO LONG LAMP WIRES 226 8.2.7 HUMIDITY IN THE AIR 226
8.3
8.3.1 8.3.2 8.3.3
8,4 8,5
8.5.1 8.5.2 8.5.3
8.6
8.6.1 8.6.2 8.6.3 8.6.4 8.6.5 8.6.6
LIGHTING DESIGN 226 VISUAL RESPONSES 227 VISUAL PERFORMANCE 227 CHARACTER OF LIGHT 228 PRODUCTION OF MDIATION 230 LAMP MATERIALS 23I
GLASSES 237
METALS 237 GASSES 232 DEFINITIONS 232 LIGHT
OUTPUT 232
AVEMGE
LIFE
232
EFFICACY 232 COLOURRENDING INDEX 232 LUMINOUS FLUX 233 LUMINOUS INTENSIW 233
8.6.7 8.6.8 8.6.9
8.6.10
8.7 8,7.1 8.7.2 8.7.3 8.7.4 8.7.5 8.7,6 8.7.7 8.8 8.8.1 8.8.2 8.8.3 8.8.4 8.8,5 8.9 8.9.1 8.9.2 8.9.3
LUMINOUS EFFICIENCY 233
ILLUMINANCE 233 SPECIFIC OUTPUT 233 SPECIFIC CONSUMPTION 234
LAMPS 234 INCANDESCENT LAMPS 234 COMPACT FLUORESCENT LAMPS 237 FLUORESCENT LAMPS 237 MERCURY VAPOUR LAMPS 243 METALHALIDE LAMPS 245 SODIUMXENON LAMPS 246 ELECTRODELESS INDUCTION LIGHTING CONTROLS 246
LOCALISED MANUAL SWITCHING 247 TIMEBASED SYSTEMS 247 DAYLIGHTLINKED SYSTEMS 247 OCCUPANCYLINKED SYSTEMS 247 LIGHTING MANAGEMENT SYSTEMS 247 EMERGENCY LIGHTING ESCAPE LIGHTING 248
248
SAFETY LIGHTING 248 STANDBY LIGHTING 248
CHAPTER NINE
9.1
LAMPS 246

HARMONICS
IN POLYPHASE CIRCUITS
RELATIVE MAGNITUDES OF LINE AND PHASE CURRENTS AND OF LINE AND PHASE VOLTAGES POLYPHASE CIRCUITS WHEN THE CURRENTS AND VOLTAGES ARE NOT
OF BALANCED
SINUSOIDAL 249
9.1.1 STAR CONNECTION 25L 9.T.2 DELTA CONNECTION 254 9.1.3 EQUIVALENT STAR AND DELTA VOLTAGES OF BALANCED,
THREEPHASE SYSTEMS WHICH HAVE NONSINUSOIDAL WAVES AND THAT CONTAIN ONLY ODD HARMONICS 256 EXERCISE 262
9
ANSWERSTO EXERCISES 264
iv
CHAPTER ONE
THREE.PHASE ELECTRICAL MEASUREMENT
1.1
WATTMETERAPPLICATIONS
\o I(E
a
lo tl(1) l(^ lftt \_c
ao,
ldr
IE lc
lP /lio
Figure 1.1: Connection of a wattmeter between two lines of
a
threephase system
to measure ac power in threephase systems is the wattmeter, The wattmeter contains a lowimpedance current coil that is connected in series with the load, and which ideally has zero impedance. The wattmeter also has a highimpedance voltage coil that is connected across the load, and which ideally has infinite impedance. The connection of a wattmeter is shown in Figure 1,1, The current in the voltage coil and the resulting magnetic field in this coil are directly proportional to voltage applied to the circuit. The current in the current coil and the resulting magnetic field in this coil are propoftional to the current flowing into the circuit. Thus, the reading on the wattmeter is directly proportional to the active power: The basic instrument used
P Where:
=
V.Lcos 0
0 = the angle between the voltage and the current
= the load angle or the power factor angle = the angle of the voltage minus the angle of the current
The connections in Figure 1.1 will produce a reading of power delivered to the load. Since the two coils are completely isolated from one another, they could be connected anywhere in the circuit and the reading may or may not have any meaning. If one of the coils on the wattmeter is reversed, the equations for the power are the negative of what they were before the coil was reversed. This is due to the change in the variable reference as related to the + terminal.
'
Due to the physical construction of wattmeters, the + terminal of the voltage coil should always be connected to the same line as the current coil. Any one of the two coils can be reversed if becomes necessary to reverse a winding to produce an upscale
it
reading.
r0
li
P_
J
J
(J
(J
U
U
L
L
Figure 1.2: Wattmeter connections for the reversal of current
if the current coil is reversed, it
results in the network shown in Figure t.2.. tthe+terminal of thepotential coil isconnectedtothelinecontainingthe current coil and the meter is reading upscale, the power is flowing through the wattmeter from circuit A to circuit B. If a wattmeter indicates a reverse reading when it is normally phased, i.e. the + ends of the voltage and current coils are connected together, it means that the voltage and current are more than 90' out of phase' In sulh cases the terminals of either the current coil or the voltage coil can be reversed. for the reading to be fonruard and of the correct value'
For example,
1.1.1 ONE.WATTMETER
METHOD
A single wattmeter can be used to measure the power of any balanced threephase systeir without breaking the phases, The system can be starconnected or deltaconnected. Figure 1.3 shows tire connection of a single wattmeter in a starconnected, threephase lold when a neutral wire is available. The wattmeter measures the phase power that is then multiplied by three to find the total power in a balanced load' Figure 1.4 shows the connection of a single wattmeter in a deltaconnected, threephase load. The current coil of the wattmeter is connected in one line and the voltage coil is connected alternately between this line and the other two lines' The total power is then determined from the two readings on the wattmeter, This method is not of as much universal application as the twowattmeter method, because it is restricted to balanced loads onlY,
t
Figure 1.3: Single wattmeter connected in a balanced, threephase, starconnected system
Figure 1.4: Single wattmeter connected in a threephase, deltaconnected load to determine the total power The current Iu through the current coil of the wattmeter is the phasor difference of and I.u, i.e. the phasor sum of I35 ond I.u reversed.
Iu5
F
V.u
Figure 1.5: Phasor diagram for the voltages and currents of Figure 1.4 When the voltage coil of the wattmeter is connected to line b, the voltage will be V"u and the phase difference between this voltage and the current will be (30'  0). The reading on the wattmeter will be: W6
=
Vu6.Iu. cos lY"b
= Vao.Ia.cos
(30'
O)
= J3 .Von.Io6.cos (30'
Where:
= Iph =
Vor.,
S)
phase voltage in V Phase current in A
When the voltage coil of the wattmeter is connected to line c, the voltage will be V". and the phase difference between this voltage and the current will now be (30' + 6). The reading on the wattmeter will now be: Wu
=
Vu..Iu .cos
z!"'
= Van.Ia.cos (30' + 6)
= J3 .von.Ioh.cos (30' + O)
L.L.2
BLONDELL'S THEOREM In general it takes (n  1) wattmeters to measure power in a transmission system with n number of lines. This phenomenon was described by Blondell and is generally known as Blondell's theorem.
Lr
The voltage return for each wattmeter is connected to the line with no wattmeter in it. In singlephase systems one wattmeter is required to measure the power. It is clear that two wattmeters are needed to measure the total power in a threewire, threephase system. This method will work whether the system is balanced or not, The readings of the wattmeters are simply added together and the sum is the total power going down the threephase (or nphase) line, Using instantaneous values for an unbalanced, threewire, starconnected load can prove Blondell's theorem.
Figure 1.6: Two wattmeters connected in a starconnected system to prove Blondell's theorem
Wa+Wc = Vab,!a+vcb,ic But:
And:
V66
= V6nV56
V66 = V6nV6n wa * wc = (vun vbn)ia+ (v.n  v5n)i.
=
Yan,ia tbn,ia
=
Van.ra
*
In a starconnected system:
is*
i6
f i. =
Q
16 = 1616
* v.n.ir
v.n.i" + v6n(i,
v5n.i6
 i.)
And:
1.1.3 TWO.WATTMETER
METHOD
The twowattmeter method gives true power in the threephase circuit without regard to balance the waveform provided in the case of a starconnected load. The neutral of the load is isolated from the neutral of the power source. If there is a neutral connection, the neutral wire should not carry any current. This is possible only if the load is perfectly balanced and there are no harmonics present'
Figure 1.7: Twowattmeter method of measuring threephase power The twowattmeter method can also be used for a threephase, fourwire system in which the neutral wire carries the neutral current. In this method, the current coils of the wattmeters are supplied from current transformers insefted in the principal line wires in order to get the correct magnitude and phase differences of the currents in the current coils of the wattmeter, The reason being that in the threephase, fourwire system, the sum of the instantaneous currents in the principal line wires is not necessarily equal to zero/ as is the case in a threephase, threewire system.
0
The power delivered to a threephase, threewire, star or deltaconnected balanced or unbalanced load can be found by using only two wattmeters, The basic connections are shown in Figure 1.7. To show the application of the twowattmeter method to unbalanced loads, a starconnection is considered. Considering instantaneous values: Voltage across wattmeter
? = Van =
Voltage across wattmeter
Vun

Vbn

Vbn
c = Vcb
Total active power
=
V.n
=
Vun.Iu
+ V6n.I6 + V.n,I.
The current Iu does not pass through a wattmeter and can be eliminated. In any threephase system:
I.+16+I. .'' .'.
Ib Total active power =
W
Where:
Vun.Iu
0
= IuI. + Vnn( Iu * I.) +
= (Vun  Vnn)Iu + (V.n = Vab.Ia * Vs6.Is = Wu+W.
Vcn.Ic
Vun)I.
= total active power measured by two wattmeters = active power reading on wattmeter a w. = active power reading on wattmeter c W wa
in W
Thus, at any instant the total active power is given by the sum of the two wattmeter readings, This is true for balanced or unbalanced loads as well as star or delta
connected loads. To find the power factor from the twowattmeter readings in balanced loads, the starconnection of the three equal impedances shown in Figure 1.7 must be considered. The phasor diagram for the abc sequence is shown in Figure 1,8, A lagging current with phase angle 0 is assumed.
Vu.
V.u
V.n
V.u
Figure 1.8: Phasor diagram (abc phase sequence) for the voltages and currents of Figure 1.7
With the wattmeters in lines a and c (Figure 1.1), their readings are:
And:
W3
=
Vu5.I;.cos Zrvjb
w6
=
V.5.I..cos
ll'b
1.1.3.1TWOWATTMETER METHOD OF OBTAINING THE POWER FACTOR From the phasor diagram in Figure 1.8:
tY:'= And:
3oo+o
z{:o = 3ooo
When these equations are substituted into the previous equations:
=
V35.I6.cos (30o
+ 0)
W. =
V.6.I6.cos (30o
 $)
w6 And:
This will be the reading on the wattmeter if the twowattmeter method is used on balanced loads, Writing the expressions for W" and W" and using the cosine of the sum of the two angles: Wu
=
V1.I1(cos 30o.cos$

sin 30o.sin $)
And:
W. =
Then:
Wu
+ W.
And:
Wu
 W. =
=
V1.I1(cos 30o.cos$
J3
.VL.IL.cos
+ sin 30o.sin $)
O
JE .Vr.tr.sin O
Therefore:
tan
d
= Jtf*'*') (.W. + W.
J
Thus, the tangent of the impedance angle is J5 times the ratio of the difference between the readings on the two wattmeters and their sum. With no knowledge of the lines in which the wattmeters are connected nor of the phase sequence, it is not possible to distinguish between + $ and  $, However, when both the meter location and the phase sequence are known, the sign can be fixed by the following equations. For a positive phase sequence (abc): tan d
= nltfw' wt ] [W.+WoJ
or:
tan6=
or:
tan d
=
"[tr#) J3f
*'w')
[W. +W. J
For a negative phase sequence (cba):
tand
= J5f*t*.) (W. +Wb
J
*.
Or:
tan d
=
Jrf(.W5 *t ) +w, )
Or:
tan 6
=
Jrf
*. *.
l.W. + W.
) J
L.L,4 THREEWATTMETER METHOD The total power in a threephase system can also be found by connecting a wattmeter in each phase of the system to measure the power in each individual phase. These readings are then added together to find the total power in the system. When the power delivered to a threephase system is measured, each voltage coil may be connected across each load voltage. Similarly, each current coil may be connected in series with each load current. However, these connections are not always possible in practice, i.e,, a threewire, starconnected load would require access to the neutral point to make connections to the voltage coils. Similarly, in a threephase, deltaconnected motor winding, it would not normally be possible to open the windings internally, as would be necessary to inseft the current coils in series with each winding. Due to these factors, a threewattmeter method was introduced to measure the power in a threephase system using line voltages and currents instead of load voltages and currents.
!(o
a
!q.) P(J qJ
c o (J I
rO
E o !
L
o L o P (t I
Figure 1.9: Threewattmeter method of measuring power in a threephase system Figure 1,9 shows that the current coils are connected in series with the lines, which means that the currents in the wattmeters are \ne currents. One terminal of each voltage coil is connected to a line, and the othdr terminals of the voltage coils are connected together. The voltage coils are therefore connected in a star configuration. Thus, the voltage coils form a balanced, starconnected system. The voltage across voltage across each voltage coil lags the line voltage by each coil is therefore
5, *"
30" in a positive phase sequence system, and leads the line voltage by 30' in negative phase sequence system.
10
a
)
The power rneasured by each watbneter is then:
w +
f
30.:
_ 30"
=
# 1fi+l, Ji,''.".i;o #
Il'no'ii"';l?;"Jiil:'"xnf
'tr'.o,
1zl't
so"1
negative phase sequence positive phase sequence
g"? il:":" #".11x",iil!"k;!,1i.fl: se, ro u r w re, sta r. i
il:#H*:ln::u;"#i"T"#,'ffrr:,?,.1*:.;ml,jl
Figure 1'10; Three wattmeters
*nn:T;o
,o
u{ruohase,
fourwire, starconnected
From Figure 1.10: Wu
=
%n.Iu n.cos
Wo
=
V6n.I6n.Cos ZFn
z!"n Ian lbn
11
:
J
t__
I I
t t I
I
W. =
V.n.I.n,cos ,rY^ Im
Figure 1.11: Three wattmeters connected to a threephase, deltaconnected load Since the voltage coils form a balanced, starconnected system, the voltage across each voltage mil lags the line voltage by 30' for a positive phase sequence and leads the line
voltage
ry 30" for a negntive phase
sequence. Thus, the reading on the wattmeter in
each line:
wa
=
%= w. Where:
L2
=
f
.r".*.(zicbt3o.)
H:rm
(zft to")
f;+*(zi%i3o.)
+: rcgilhphffiqsEe  : FdiE Fl e serFlentB
CHAPTER TWO
THREE.PHASE ELECTRICAL SYSTEMS
2,L
INTRODUCTION The generation, transmission and distribution of electricity are accomplished by threephase alternating currents. An alternating current circuit having a single alternating current voltage source is called a singlephase circuit. Electrical power is delivered from a source/ such as an alternating current voltage generator, to a load by means of two wires, This arrangement is called a singlephase, twowire system.
Vun
.__r Threephase,
\/
LrrJ Singlephase
threewire supply
(V1'n")
,roor,
[I+.'l (J:;
Vo.
w Singlephase supply (V11n")
Figure 2.1: connection of different supplies to a threephase, fourwire system (equivalent circuit to Figure 2.4) Most consumers are fed from a singlephase alternating current supply. One wire is called the live conductor and the other wire is called the neutral conductor. The neutral conductor is usually connected to earth via protective gear. The standard voltage for a singlephase alternating current supply is 220 V. The majority of singlephase supplies
are obtained by connection to a threephase supply as shown in Figure 2.1. A polyphase circuit is a circuit containing more than one alternating current source and three or more wires, Upon these wires appear alternating current voltages having different phase angles. The most common polyphase circuits are those containing three alternating current sources and three or four wires.
13
These threephase circuits are widely used in the electrical power industry to transmit
power from generating stations to metropolitan areas and to distribute that power to individual consumers.
2.2
THREE.PHASESYSTEMS Threephase systems have some advantages over singlephase systems:
. o . .
More efficient use of copper wire for the distribution of power More constant power from generators and motors More constant torque on generators and motors Fewer ripples in the direct current output when alternating current is conveted to
direct current
A threephase supply is generated when three coils are placed 120' apaft and the whole rotated in a uniform magnetic field as shown in Figure 2.2. A threephase voltage is basically three singlephase voltages. Each voltage is separated from the next by a phase angle of 120'. The same basic structure found in the singleffise generator can therefore be used to generate the three voltages simply by equipping the rotor with three separate windings. If the windings are spaced 120" apaft, the voltages induced in these windings will then be shifted from each other by 120' of phase, as required. This concept is implemented in practical threephase generators, but the physical structure is somewhat different. Electromagnetic induction occurs when there is relative motion between a conductor and a magnetic field, In other words, either the conductor or the field may be moving while the other is stationary, In practical threephase generators however, the three windings (conductors) are stationary and the magnetic field is rotated, as shown in Figure 2.2.
.o
^t^.
a
Figure 2.2: Displacement of voltages in a threephase system
14
The windings are embedded in the stator and direct current (the excitation) is passed through brushes and slip rings to the field winding on the rotor, The field produced by the rotor as it turns, cuts the conductors of the three stator windings. Since the stator windings are 120" apart, the rotating magnetic field induces voltages that are separated in phase by 120'. A threephase generator commonly located in a power station produces threephase power. The rotor is driven by a prime mover, i.e. a turbine, and the rotor poles are excited by direct current. The stator has a threephase distributed winding. The axes of the phase windings are displaced from each other by 120 electrical degrees, as shown in Figure 2.2. Sinusoidal voltages are induced in the stator phases when the rotor is rotated, For a balanced system, the voltages have equal amplitudes and are 120" displaced in phase,
as shown in Figure 2.4. The equivalent circuit of the stator windings is shown in Figure 2.3. In this case the windings have a common connection labelled n, called the neutral, and the windings form a starconnected network. Since the neutral line n is an output, the output is said to be threephase, fourwire. The windings can also be deltaconnected.
Figure 2.3: Equivalent circuit of the stator windings of a threephase generator Figure 2.1 shows a circuit equivalent to Figure 2.3 and consisting of three alternating current generators. Shown in Figure 2.3 is the plot of the three voltages, v36, v6 dhd V"n. Each voltage is taken with respect to the neutral n. Three wires, called lines, therefore carry a threephase, alternating current supply. The currents in these wires are known as line currents and potential differences between the lines are known as line voltages. The fourth conductor, known as the neutral, is often used with a threephase supply,
15
V.n I
: 120.
i
"
t'' )(i "L20
Figure 2.4: Sinusoidal form of the phase voltages of a threephase generator, each with respect to the neutral
If the threephase windings shown in Figure 2.3 are kept independent, then six wires are needed to connect a supply source to a load. The three phases are usually interconnected to reduce the number of wires. This can be done in two ways, namely a starconnection and a deltaconnection. Sgglgg1_qilhlggpha5_9 Supdies, are usually c91n_eq!9^Q i0.St?Lwhereas threephase loads may be connected either in delta or star.
2.2.L
PHASE SEQUENCE OF THREEPHASE SYSTEMS The doublesubscript notation is used to avoid confusion in the direction of voltage and current. When the doublesubscript notation is applied to alternating current circuits, the sequence of the subscripts indicates the direction in which the current or voltage is assumed to be positive. Figure 2.5 represents an alternating current source connected in series with an impedance. The voltage across the impedance is designated V"o to Thus, if an arrow symbolise that the potential of a is positive with respect to representing the direction of this voltage is drawn alongside the impedance, the head of the arrow should point towards the end that is at higher potential, i.e. towards a in
b.
Figure 2.5,
Iao
H H
Vao

f
\,_,/
_
E
Figure 2.5: Doublesubscript notation of voltage and current in alternating current circuits
IO
t
The current through the impedance flows from a to b and is therefore designated f"s. The phase sequence is the order in which the three phases attain theii maximum values. The phase sequence can be determined by the order in which the phasors representing the phase voltages pass through a fixed point on the phasor diagram if the phasors are rotated in an anticlockwise direction. The phase seque:nce in Figure 2.6 is positive or abc. The phase sequence is quite important in the thieephase distribution of power. In a threephase motor for example, if two phase voltages are interchanged, the sequence will change and the direction of rotation of the molor will be reversed.
V."
Figure 2.6: Phasor diagram for a positive phasesequence, showing phase and line voltages (Vu as reference)
The phase sequence can also be described in terms of the line voltages. Drawing the line voltages on a phasor diagram, the phase sequence can be determined by Jgain rotating the phasors in an anticlockwise direction. The sequence can be determined by noting the order of the passing first or second subscripts. In the system of Figure 2'6, the phase sequence of the first subscripts passing the fixed point is abc, itre phasor diagram is always started with the reference, from where the rest of the voltages are drawn according to the specified sequence. The voltages in Figure 2.6 will be:
%=
VpnZO'V
= Vp6Z120" V Vc = Vpnl120" V Van = VrZ3Oo V Vn. = VrZ9Oo V Vca = VrZ150" V V6
I
t7

Where:
= phrevolbge = line volbge
Vpr,
Vr
The phase sequence can also be negative or cba as shown in Figure 2.7. (D
V.u
Figure 2.7: Phasor diagram fgr a negative phasesequence, showing phase and line voltages (Vu as reference) The voltages in Figure 2.7 will be:
= VpnZ0o V Vo = Vpr.Z120'V Y, = Yp6l120" V Vau = VrZ30" V Yo, = VtZ150'V Vca = VlZ90" V Va
Remember: The voltage
giyen
The
voltage
The
referene
is always the
is always the is always
line voltage, unless othenvise stated.
reference, unless othenruise stated.
0", unless otherwise stated,
In a posiUve phase sequence system, the line voltage leads the corresponding phase voltage by 3O".
In a negaUve phase sequelrce system, the line voltage lags the corresponding phase voltage by 3O".
1B
L2.2 BALANCED STARCONNECTED,
THREE.PHASE SYSTEMS
Figure 2,8 shows^the windings of a three_phase generatol be three or rour output rines, whicrrL[::fl?;:rt":lTll?:.onnected in star. rhere can threephase, threewire .on"L.tui ou u
"'
ii''r""lprffi:?:ilT,.?:id*l
'nrllr'.un
Stator Rotor Field winding
Neutral
rnreePhase
output
[
1
I
n
Direct current excitation
c b a
Figure 2.g: Structure and wiring of a threephase, star_connected generator
The voltage induced
.
,?"lJ:f
,:,rl;: ;Iti:i"",""",:Eni?t:il#:'titi,,1:?itTffi :Ji;:",:ff fi rl";ffi :*iililI,ffii:lil"il:,..,,:""i';:tl'r'ttr#nril1+.r1hll$;tr
19
Eun
a(tr
za In
o c) V, (E
c oI
o OJ L
c I
o
F
/
Figure 2.9: Phase and line quantities in a symmetrical, threephase, fourwire, starconnected supply
Figure 2,4 shows the three phase voltages in sinusoidal form, The corresponding phasor diagram is shown in Figure 2.10. The maximum value of each phase voltage in Figure 2.4 is E^, so that: €an
enn
ecn
The
srn
dte
= =
Ean(m1
sin rrlt
E6n161Z0o
V
= =
Ebn('y sin (rot
= =
Ecn(m)

120")
[email protected])ZLZlo Y sin ( 7, = L6.92/6L7B Q
Za Zo
Conveft the load to an equivalent deltaconnected load to find the phase currents and then calculate the starpoint potential. Take V5 as reference.
Solution:
vulr2lo Eo,l30o
Ea6ZL50o
Y6200
vc/r20o Figure 2.33: Positive (abc) phase sequence with Vb as reference Deltacon nected im pedances:
7,x = 7.r
7x
*
Zu'70 zc
.35') (L9 .57 245, 16.92261.78"
(L2.01 zB7
= t2.01187.350 + 19.57145.540 + = 22,462287.90 O a a L^L.aLA 
r 1
r
")
77
C'd
zb
= 16.92'$1,78o + t2.0I287.350 = L9.42!l19.42o O
44
54
+
(16.92t
 6t.78.)(tz.0ttg7 .35") 19.57
245.54"
Current in each phase of the deltaconnected loadr
rVao rab_ d
=
3601150" 22.452287.9"
= L6.O27262.1o
A
rVca l^. 
/'
cA
360
t90"
19.42rt19.42
=
18.537270.58'A
Ia = IuoIca
Vu,
V'n
=
16.027162.10
=
31.675187.580 A

LB.537l70.58"
= Iu.Zu = (3I.675 zB7.5BoX12.0 = 380.4172174.93V =
VuVu'
= ll[:ooz]zo"l \J3) =
I tB7 .35o)
3*o.4t7tt74.s3"
311.5375228'030 V
Example 2.7 The load of a threephase, deltastar system consists of the following impedances:
= (L5.6 + j12.4) o = (I5.2  18.3) o Z" = (24.I + j16.7) o Zu 26
E6u
is 440 V and a negative phase sequence is used.
45
2.7.r 2.7.2
use stardelta conversion to calculate the currents drawn from the supply, Calculate the voltage across each impedance in the load.
2.7.t
E,"160o
Ev6lIB0o
vbz30
v^lL50"
o
E,orZ60
Figure 2.34: Negative (cba) phase sequence with Deltaconnected
im
7"n=
=
pedances
=
7^+2,6+'4" Lc 15.6 + j12.4
+ I5.2  j18.3
+
(15.5 +
jtz.4)(Ls.2 j18.3) 24.1+ j16.7
(1s,2
j18.3X2a.1+ j16.7)
Q
Zn+zc+t+ la 15.2
 j18.3 + 24.I + j16.7
56.919226.570 7_
as reference
:
= 45.2565222.270 7or=
Eba
+

15.6 + j12,4
C)
L+z*7''7u zb
z4.t + j16.7 +
15.6
56.457262.20 A
46
+ j12.4 +
(24.1+ j16.7x1s.6 + jt2.4)
ls.2  j18.3
l
r F
C.ru.r"ent
f
in each phase of the delta_connected load:
,i
=
Iao
=
%!Zao
4402780" as.ag22 _
22fi
= 9.6742_157.190 A T, 
Vu.
.DC 
zo,
4402_69.
66sl6z_16r;r
=
6.5752_33.430 A
TVca 
rca
z^_ k
440Zffi, id.os?
z6n"
= 7.8492_2.20 A Current in each line of the delta_connected load:
Iu = IuoI.u
= 9.6742_LS7.Ig" _ 7.B4gZ_2.2o = t7.tt2Z_169.37o A Io = Io.I"n
= 6.5752_33.430 _ 9.6742_757.tgo = I4.4OSZO.51o A
I. = I.uIb. = 7.84922.20  6.575Z_33.430 =
2'7.2
%s
4.O72ZS4.OSo A
= I6.Zu = (17.1122_168.37)(1s.6 + j12.4) = 341.0O64_t2g,ggo V
47
E
Vb, = Ia'Zo = (74.40s t0.s1ox ls.2

j 18.3)
342.685t49.780 V V., = Ir.7" = (4.0722s4.6sox24, 1
+ jL6.7)
LLg.394289.370 V
2.3
COMPLEX POWER Apparent power consists of real (active) power and reactive power, This
power ignores the phase relationship between the current and the voltage, When this phase angle is 0o, the apparent power equals the active power. When the phase angle is 90', the apparent power equals the reactive power, The apparent power is defined to be:
S=
E.I
Complex power also consists of real (active) power and reactive power, However, the phase relationship between the current and voltage is included in this power. Active power is unidirectional and reactive power reverses direction twice each field cycle and results in reactive energy near the oscillating source. This energy directly i
nfl uences, affects a nd im its operationa I characteristics. I
Ite
,tul
zz+
Figure 2.35: Circuit for the explanation of power relationships
To develop a relationship between complex power and other power
quantities, consider the circuit shown in Figure 2.35. The complex power is defined to be:
Sx
Where:
4B
= E.Ix
E = the rms value of the supply voltage Ix = the complex conjugate of the rms value
of the current
The conjugate of a quantity is the mirror image of that quantity. If the current is T. = l0 A, the conjugate of this current will then be I = l0 A, as shown in Figure 2.36.
I*:Ile Figure 2.36: Phasor diagram showing
I
and its conjugate
I*
The complex power is then:
S*
= EL\.IZj = E.Iz(9  0) = PljQ
P = E.I.cos$ = I2.R Q = E.I.sin$ = r2.x Where:
Where:
P
=
real or active power in W
Q
=
reactive power in VA'r
$
:
power factor angle
cos$:
Gos$
=
powerfactor
cos
Zl
The magnitude of the complex power is simply what was called apparent power, and the phase angle of the complex power is the power factor angle. The relationship between the complex power, active power and reactive power is shown in Figure 2.37 and in Figure 2.38. In Figure 2.37 it is shown that the phasor current in be split into two components. The active component that is in phase with the voltage and the reactive component that is 90' out of phase with the voltage:
= Ireactive = factive
I.cos $ Lsin d
49
figurd Z.Zl:
Representation of the active and reactive components of current
The inphase component produces the real (active) power and the quadrature component produces the reactive power, In Figure 2.38 it can be seen that, if the reactive power Q is positive, the load is inductive, the power factor is lagging and the complex power S lies on the positive side of the positive real xaxis.
Q (lagging power factor)
\J
Q (leading power factor)
Figure 2,38: Power triangles showing lagging and leading power factor
If the reactive power Q is negative, the load is capacitive, the power factor is leading and the complex power S lies on the negative side of the positive real xaxis. If the reactive power Q is zero, i.e. the load is resistive, the power factor is unity and the complex power S lies along the positive real xaxis. It is impoftant to know that complex power is conserved like energy. This means that the total complex power delivered to any number of individual loads is equal to the sum of the complex powers delivered to each individual load, regardless of how the loads are interconnected,
50
2,4
REACTIVE POWER
E
Figure 2.39: A pure resistive circuit
Figure 2.40: Waveforms generated when an alternating voltage is applied to the circuit in Figure 2,39
Figure 2,39 shows a pure resistive circuit with a switch. Figure 2.40 shows the resulting waveforms when an alternating voltage is applied to this circuit, The current waveform is exactly like the voltage waveform, which means that at each time instant Ohm's Law is obeyed. The power dissipated in the resistor is the instantaneous product of the current and voltage. The power is therefore a direct current component that alternates at double frequency. However, the power dissipation is always positive,
51
t Figure 2.4L= A simple alternating current transformer The waveforms of the voltage and current, shown in Figure 2.42, show that, due to the inductance of the coil, the current can't keep up with the voltage at the moment when the switch is closed. The current therefore lags the voltage for a few cycles and then
attains a steady alternating value, with its waveform lagging the voltage waveform by
90'.
Figure 242: Waveforms generated when an alternating voltage is applied to the transformer in Figure 2.41 The mean product of the voltage and current is zero, as the magnetising current in this case carries mo power. Power flow oscillates positively and negatively at twice system frequency amd hence the name reactive power. When an opencircuit transmission line is connerted tn a voftage source/ the line can absorb a large amount of electric charge dure b rapffiihnce. When the switch is closed, a surge of current into the line occurS.
52
This current reduces to a steady alternating value that is required to keep the line electrically charged atthe applied voltage. In ttris case, the current waveform leads the voltage waveform by 90'. There is theiefore again no transfer of real power. 1y5;. load is connected to a transformer or transmission line, the load current is added to the reactive current' This means that the total current iags or leads ilre vortaje tyan angle less than 90', depending on the ratio of the current corresponding to the active power loss in the load and the reactive power required to energise the transmission line or the transformer' The instantaneous power flow therefore cJnsists of a constant level plus an offset alternating reactive power. All practical electrical components have resistance' The current flowing in these components therefore causes resistive losses and heating. The heating can restrict the loadcarrying capability of a transmission line or transformer or the output capability of a generator. Reactive power involves current flow and therefore creates losses just like aitive power. However, the total current is the phasor sum of active and reactive components of the currents. Therefore, component ratings have to be carefully chosen to match the reactive requirements of a
circuit.
53
.d
EXERCISE 2.L 1.
An unbalanced, threephase, fourwire, starconnected load is connected to a 3BOV, symmetrical, threephase, starconnected supply. The value of the neutral current flowing away from the neutral point is 35.55176.40 A. The load consists of Za = 8.96133,20 f) and 26 = 10.12161.3" o. Z. is unknown. With a positive phase sequence, take$as the reference phasor.
1,1
Calculate tlre value of the impedance in phase c. Calculate the active reactive and complex power absorbed by this impedance.
t.2 2.
Refer
to
Figure 2.43 and calculate the line current
I..
The load is connected to
a
symmetrical, threephase, cbarotation supply with Eu = 24010o Y.
18256.3"O
7'O
Figure 2.43: Threephase starconnected load with line c shortcircuited 3.
An unbalanced starconnected load is connected to a symmetrical, threephase, threewire, abcrotation supply with Ea = 24BZ.IB0o V. The impedances of the load are the following: z" Z,o
z,
= (33,6  j42.8) fJ = (27.7 + j18.4) O = (9.7 + j14,5) o
If the
impedance of 4 sgddenly rises to infinity, calculate the voltage between the neutral point n of the load and phase c of the supply.
54
ll
A balanced, threephase, 360V, fourwire, negative phase sequence supply with E5 as reference, is connected to a starconnected load with the following load impedances:
= L2.Bz32.8o o = 2I'412I'4o e) 4 = 37.9287.90 f)
7u Zo
I
.r43
Calculate the line currents. Calculate the neutral current. Calculate the total complex, real and reactive power in the load.
=
The phases of an unbalanced, fourwire, starconnected load consists of the following
",u
4e
components: Phase Phase Phase
a: 28.47tF capacitor in parallel with a 45.2a resistor b: 39.6prF capacitor in parallel with a 22.4fl resistor c: 1B.BprF capacitor in parallel with a 31.6ft resistor
This load is connected to a thr,eephase, 400V, 50Hz supply, Take with a positive phasesequence.
i"2 53
Enu
as reference
Calculate the line currents. Calculate the complex, real and reactive power in each phase of the load. Calculate the total active, reactive and complex power in the load.
The phases of an unbalanced, fourwire, starconnected load consists of the following components: Phase Phase Phase
a: 28.4ttF capacitor in series with a 45.2A resistor b: 39.6pF capacitor in series with a 22.40 resistor c: 1B.BpF capacitor in series with a 31.6fi resistor
This load is connected to a threephase,3B0V,60Hz supply. Take with a negative phasesequence. 6"1 6"2 6"3
Eu5
as reference
Calculate the line currents.
Calculate the power absorbed per phase. Calculate the total active, reactive and complex power in the load.
55
7.
A
balanced, threephase, threewire, cbarotation supply
with V"r
connected to a starconnected load with the following load impedances: za 26
zc
=
360 V,
is
= 12.8132.80 Q = 2lAl2L4o e) = 37.9187.9" Q
7.t
Calculate the line currents.
7.2 7.3
Calculate the total power consumed. Draw a coniplete phasor diagram of the voltages and the currents.
B.
An unbalanced, threewire load is starconnected to a threephase, 440V symmetrical supply of which the neutral is eafthed. The phase sequence is positive and the reference phasor is E5u. The impedances of the load are: za 26
Z, 8.1 8.2
= (13.6  j12.8) o = (L7.7 + j19.4) O = (12.7 + j10.5) O
Use Millman's theorem and calculate the voltage between the neutral point of the load and eath. Calculate the line currents.
9.
The unbalanced, threephase, deltaconnected load in Figure 2.44 is connected to 360V ac supply.
9.1 9.2
Calculate the total input impedance of the circuit. Calculate the input power,
a
56
1il
(8.4 + j3.6)
(4.s + j9.6)
o
o
(8.4 + j3.6) o
(12.4
l
j7.s) o
(7.5 +j4.8) o
Figure 2.44t Circuit diagram for euestion 9
A balanced, threephase, con nected
starconnected with a negative phase sequence to a dertacon nected road with .source *re i"rr"*i,' g' r*o',:#["ou
n.u*
Zao
=
72.8232.80
o
Za, = 27.422L4o t) Zru = 37.92B7.go t)
The source emf is given by:
eb(t) = 339,411 sin (rot
_
#)V
Calculate the line currents. calculate the totalcomplex, rearand reactive power in the road.
57
IT
is
11.
The phases in the load of an unbalanced, stardelta system consists of the following components: Phase Phase Phase
ab: 28.4stF capacitor in series with a 45.2o resistor bc: 39.6pF capacitor in series with a 22.40 resistor ca: 1B.BpF capacitor in series with a 31.6f) resistor
This load is connected to a positive phase sequence, threephase, 60Hz supply with: e6(t) = 381.'b3B sin (rot
11.1
Lt.z
11.3
L2.
 #) v
Calculate the line currents. Calculate the power absorbed per phase. Calculate the total active, reactive and complex power in the load' A threephase, stardelta system with a negative phase sequence and V6. = 390l25o V at 50 Hz, has the following load impedances: Zac = 1B.B z75o Q' Zou = 15.62200 {' 7co = 2L'4250o Q
tz.L I2.2 t2.3 13.
currents. consumed,
Calculate the line Calculate the total power Draw a complete phasor diagram of the voltages and the currents
Consider Figure 2.45. Use deltastar conversion and calculate the current drawn from the supply.
73.25t42.25" O
B.tt57.L
14.5239.4" O
E
O
L0.22r0.2 0
16.55250.6'O
=
360V,50H2
Figure 2.45: Parallel network for Question 5B
t\ ti\
13
2"5
POWER MEASUREMENT IN THREE.PHASE SYSTEMS Example 2.8 A single wattmeter is used to determine the power in phase cof an unbalanced, threephase, deltaconnected load. The load is supplied from a balanced, threephase, 380:V, alternating current source. The voltage coil is connected to lines b and c. The load consists of the following impedances:
= (3.1+ j3.9) O = (6.2 + j9.6) CI Z,u = (4.8  j7.2) a 2"6 26.^
Take E.u as reference with
a
positive rotation and calculate the reading on the
wattmeter.
Figure 2.46(a): Circuit diagram for Example 2.8
59
Figure 2.45(b): Positive (abc) phase sequence with Consider the load:
rVn. Ihr 
Lb, 380
1L20"
6.2+ j9.6 = T_
rca
33.252262.86. A Vca
=
zca
38020
4.8W 43.914256.310 A
I.=
I.u

Iu.
= 43.914 t5631'
 33.252t62.86"
rL.52r237.O90 A
W.=
V.5.I..cos
zf
= (380X11.52l)cos( 60' =
50

540.367 W

37.09')
E.u as reference
Example 2.9 A balanced, threephase load is connected to a supply of 440 V with a positive phase sequence, The load has a power factor of 0.39 lagging, Two wattmeters are connected in lines a and c to measure the active power. The wattmeters show the input to be 64 kW. Determine the readings on the two wattmeters.
cos
0= 0=
0.39
57.050
= J5f*.*..] l.W. *W. l *') tan 67.050 = .;El*' , tand
lu)
W.Wu = 87.261*kW But:
W.+Wu
= 64
W.= 64Wa W.Wu
87.26r
64WaWu
87,26L
2Wu
W,
W.*Wu =
 23.261  11.6305 kW $4
= 64 Wc = 75.6305
W.11.6305
kW
Since the reading on Wu is negative, it means that the wattmeter in line a reads down scale. It is therefore obvious that, when the power factor is less than 0,5, the reading
on Wu must be negative.
61
Example 2.10
2.10.1 A 380V, threephase, deltaconnected induction motor has an output of 18.04 kW at a power factor of 0.82 lagging. The efficienry of the motor is B2o/o. Calculate the readings on each of the two wattmeters connected in lines a and c to measure the input. The motor is connected to a symmetrical, threephase supply with a positive phase sequence.
2.L0.2 Another starconnected load of 12.81 kW at a power factor of 0.86 lagging is then added in parallel to the induction motor. Determine the current drawn and the power taken from the line.
2.10.1
n
=
Pout D.
'ln 18.04
0.82
Pin
Pin
=
22 kW
W"+W. = 22kW cos0
= 0.82
0= tan0
=
tan34.e2o= W. But:

Wu
W.+Wu
=
34.920
n(w,  w" ) t'[w.
*.1
"[ry)
8.867 kW
= 22
W. = 22Wa And:
Or:
W.Wu =
8.867
DW=Wa =
8.867
2W_ And:
62
W=

13.133
=
5'5555 kW
Wc
=22 =22
Wc
=
Wa+Wc 6.5665 + And:
cJmotor
2.70.2
=
15.4335 kW D I motor
a"torotor 22
0s2
27.848234.920 kVA CoS
= 0.86 Otoao = 30'680 $636
loaa
Sload =
COS $1es6
72,8r 0,86
=
Stot =
L4.895230.680 kVA S6e1e1
*
S1s36
27 .B4B Z.34.92o
+ 14.895230.68'
= 42.7L6t33,440 kVA = D
'Stot
42.7L6
.'.
x
t03
Ir
(3s.644  j23.s41) kVA 35.644 kW
J3 ,v..I, (
Js Xseoxr,)
LL2.4LT A
Example 2.11 Refer to Figure 2.47 and prove Blondell's theorem. The load is supplied by a balanced, threephase, threewire supply system with a positive phase sequence at Eu. = 48010'V,
63
i
(8.2 + j17.4
a
(18,3 + j1s.1)
(24.2j16.4) f) c
b
Figure 2.47tTwo wattmeters connected to a deltaconnected load
E""
60'
Flgure 2.48: Positive (abc) sequence with O.rrert h €ad! ehffi of the load:
rVal r*=/aa
_ ffi1ffi" 18.3

115.1
: *?5l.LAA7o
64
A
Eu. as reference
o
T
rbc
Vn.

zn
4E0zffi" =
24.2 jr6.4 L6.4195225.880 A
T_
rca
Vca
=
za 4BOIIffi" 8.2 +
jIn
= 25.L9121L5.490 A Active power in each phase of the load: P65
=
V36.Iso.cos
Zls
= (480x20.231)cos = 7489.919 W Pb. = V6.,I5..cos
(60o

ZO.4to)
Z[h
(480X16.4195)cos
(
OOo
+
25.BBo)
= 6524.699 W D_ rca 
V.u.I.u.cos Zrv;
= (480X25.19l)cos (1800

115.49o)
= 5203.698 W D
Ps6*P6s*P63 = 7489.9L9
+ 6524.699 + 5203.698
19218.316 W Current in lines a and c:
Iu=
Ian

Ica
= 20.23I t20.47 0
 25.I9LzLL5.4go
33.66t227.730 A
65
I.= =
I.u

Io.
25.t9t tLL5.49"

16.4195 tZ\.BB"
39.3752L30.580 A Readings on wattmeters in lines a and c:
W3
= V65.I3.cos zi%b = (480X33.661)cos (600 + 27.73o) = 639.967 W
W.=
V.5.I.,cos z['o
(480X39,375)cos (120o

130.58o)
18578.591W
W=
Wa+Wc
= 639.967
+ 18578.691
19218.658 W
P = W, which
proves Blondell's theorem.
Example 2.12 unbalanced, threephase, threewire, starconnected load is supplied from a balanced, threephase source with a negative phase sequence at 380 V, 50 Hz. The phase impedances of the load are the following:
An
za 26
zc
= 24.8/36" Q = 72175" Q = 18.61840 Q
Use Millman's theorem and determine the power absorbed by the load using the twowattmeter method if the wattmeters are connected in lines a and c respectively. Take E" as reference.
66
vbclgjo
vb1l20o
valjo vubl30"
vcazrs}" vclIzgo
Figure 2.49: Negative (cba) system with
Eu
as reference
Millman's theorem to find neutral displacement voltage: Vrn =
%.%+V6.Y6+V..1 Yu +Y5 +Y.
38020",3802120",3901720
(43)(24.8236") (J3)(7227s) (J:xra.6ze+.) =1rl_!1 24.8236"' I2l75"' 18.6zg4" = 356.5112169.59"V Voltage in each phase of the load:
Vu,
=
=
VuV.n 3Bo
lo" 
J3
3s6.6rrt169.59.
= 573.76426.45"V %, = V.Vrn 380 z I20" _ 356.6Lr2_169.s9" = __E
= 27L,788227.52V Current in each phase of the load: Vu,
=
Iu.Zu
= (Iu)(24.8t36.) Iu = 23.136229.55
573.764t6.45.
A
67
V., = Ir.Z" 27I.7BBZ2752 = (IcX1B.6ZB4")
I.=
[4.6L2ZL11.52' A
Readings on the wattmeters in lines a and c:
Wu=
Vu6.Iu.cos z1%o
(380X23.136)cos( 30' + 29.55')
8791.409 W
w. =
vgs.Is.cos zYd
= (380x14.612)cos( 90" + 111.52') = 5165.489 W !{= Wa+W. 879L.409 + 5165.489 =
13955.898 W
Example 2.13 deltaconnected load is connected to a 380V, phase positive sequence. The load impedances are the 50Hz supply with a following:
2.L3.t An unbalanced, threephase,
Zao
7n, 7ru
= (11.45 + j 14.44) a = (17.53 + j24.22)o = (20.08 + j L6.77) a
Take
Eu5
as reference and calculate the total power delivered to the load by
using the load voltages and current.
2.13.2
6B
If a wattrneter
is connected in each line of the load in Question 2.I3.L, calculate the readings on the wattmeters and show that the sum of these readings is equal to the total power calculated in Question 2.13.1.
Eo"
Figure 2.50: positive (abc) phase sequence with
2.13.1 Current in each phase of the load:
rVao 
raD
=laa

36010" 11.45 + j14.44
=
= 19.535251.59. I5.
= =
A
Ss
lb,
[email protected]"
17.53 + j24.22
= L2.O4lZl74.l
A
TVca rca 
=lca
= =
3ffi2120" 20.08 + jI6.n 13.761280.13. A
Active power in each phase of the load: P35
=
= =
V66.I6 b.COS
/.y"b rab

(360X19,535)cos(0. + 51,59.)
4369.246W
Eab
as reference
Pb. = V5..16.,cos Z[tr
(360X12.04l)cos( 120" + 174.t')
254t,783 W P.u
=
V.u.I.u .cos
zl'
= (360x13.761)cos(120"  80.13") = 3802.159 W P =Pan*P66*P66 = 4369.246 + 254I.783 + 3802.169
= 10713.198 W 2.13.2 Current in each line: 16=I36Is6
= 19.535251.59'  l3.76tl90.l3 = 3O.47627t.29" A = Iu. 
Iub
= I2.04ILI74.L  19.535151.59' = 27.9182149.74'A = I.u  Ib.
= 13.761 t80, 13'  I2.04LIL7 4.I' = 2O.5OtZ45.9o A Since the voltage coils form a balanced, starconnected system, the voltage across each voltage coil lags the line voltage by 30" (positive phase sequence). Thus, reading on 6:;." d!*e++'r""**'..w wattmeter in each line:
w.
=
u+ .Iu,.o, J3
=
fry.l(30.476)cos (0, + 7r.zs
=
70
'. r,3
1
zf,o ra
./
4759.475W
t :0") 
30.)
wn
= $.t6..o, J3
( zr;%.r 3o') rr
= (#),r
r18)cos( 120"
=
w.=
=
2878.49 W
f.t.,.or1zl"tso";
[#),m
= 3O74.9 W
 14s.74. 30")
ro1)cos(120.

4s,e.
 30.)
W
=
Wa
= =
4759.475 + 2878.49
*W5 +W. + 3074.9
LO7L2.865 W
Example 2.14
36.5t36.6 A
25.5t25.5"a
Figure 2.51: Three wattmeters connected to an unbalanced starconnected load
77
Refer to Figure 2.51. The unbalanced, threephase, fourwire load is connected to a symmetrical, threephase supply of 440 V. Calculate the readings on the wattmeters to
find the total power drawn by the load, Use E5. as reference with a negative phase sequence.
Figure 2.52: Negative (cba) phase sequence with Current in each phase of the load: Vun
=
440190'
(
J5 X1.,Xs6.6236.6')
I"n
6.94'Z126.60 A
Vun
Iun'Zon
(.''6 Xro.X+s.sz4s.5")
440/,34'
5.583275.5'A
Iun
T2
Iun,Zun
V.n
=
Irn'Zrn
4402150'
=
dt)e,)(25.522s.5.)
I"n
=
9.9622124,5" A
Ebc
as reference
Wu=
Vun,Iun.cos
z{an
(#)(6.e41)cos( eo' + 126.6") 1415.s69 W
Wu=
=
W.
V6n.I6n,cos
ZIM
t#)(5.583)cos(30'
 7s.s')
=
994.08 W
=
V.n.I.n.cos ZtV;
=
(#)(e.e62)cos(lso' 
124.s")
= zzellszw W
= Wu *Ws +W. = 1415.569 + 994.08 + 2284.L62 = 4593.811W
73
EXERCISE 2.2
1.
Calculate the readings on two wattmeters connected in lines a and c of a threephase system. The load is starconnected to a symmetrical, threephase, deltaconnected source with eu5(t) = 622.254 cos (ot + 90o) V with a positive rotation. The impedances of the load are: za 26
zc
2.
= L2.4139" A = 18.6l56'A = 2L.726I" {l
Two wattmeters are connected to an unbalanced, threephase, stardelta system with a positive phase sequence. The following readings were taken: T
10,55231.50 A at a power factor of 0.85264 lagging
aLbc 
20.23t25.20 A
rab 
Zr" = 30.34t90.90 0
The load is connected to a 360V, 50Hz supply. Prove Blondell's Theorem if the current coils of the wattmeters are connected to lines b and c.
3,
The following impedances are connected in delta in the load of a threephase sytem: Zou Zco
zu,
= (I2.4 + j15.5) O = (15.6 + j9,3) O = (9.5  j18.2) o
The coils are connected across a symmetrical, 480V, threephase supply with phase sequence and Eu. =  480 V,
a
negative
3,1 Calculate the line currents. 3.2 Calculate the readings on each of the two wattmeters connected in lines a and b to measure the total power. 3.3 Convert the delta to an equivalent starconnected load, 3,4 Calculate the readings on each of the two wattmeters connected in lines a and b of the starf the powe, system is the one line diagram. The ca rcu ate I
i
A 66kv, transmission rine transmi"..iif,.,IfTtrJff*1"$"j1fi1"":r#t:.ff,i'J:il the year as perunit system simplifies numerical analysis and representations are very 4s00 kvA for 75 ffi5f:Jl:fi'ol.lx,j["r]" 3200 kVA for 90 days 900 kVA for the remainder of the year
follows:
days
The total interest and depreciation R0.71lkwh and, The copper conductor Percentage system except that all quantities R52,40/kg. rne coppli r.,uru a"niitv  ' o1S orsinsre lonductor i, ize ;f,/k'.
are
ffiT88:ij"ii Jn"":ff.1,,?JIffi$"JrL:H
6.1 6.2 6.3
facile in the use of the system because of its Calculate the most economical conducto/d They also take advantage of its analytical Calculatethediameterofthisconductor.' 'xpressed on a per unit base by the equation: Calculate the load anO form ia;;.
le 6
163
5,
A ceftain load varies as follows for 248 days per annum: 270 kVA at a power factor of 0.809 lagging for eight hours per day 96 kVA at a power factor of 0.707L lagging for eight hours per day 60 kVA at a power factor of 0.9455 lagging for eight hours per day For the remaining time of the year the load varies as follows:
72kVA at a power factor of 0.809 lagging for eight hours per day 20 kVA at a power factor of 0.707I lagging for eight hours per day No load for eight hours per day
5.1 5.2
Calculate the annual load and form factors, Calculate the efficiency at full load and a power factor of 0.809 lagging, the two allday efficiencies and the allyear efficiency for each of the following 240kVA transformers:
Initial cost
Transformer
Iron loss
A
1.2 kW
4.8
KW
R21 000
B
2.5 kW
3.6 kW
R1B 000
Fullload copper loss
5.3 Calculate the total annual running cost of each transformer. The cost of electrical energy is R0.72/kWh and the annual interest and depreciation charges are 10olo. 5.4 Calculate the annual lost factor for each transformer. 6. A 66kV, transmission line transmits a threephase, balanced load that varies throughout the year as follows: 4500 kVA for 75 days 3200 kVA for 90 days 900 kVA for the remainder of the year
The total interest and depreciation charges is l0o/o and electrical energy costs R0.7llkwh and. The copper conductor has a crosssectional area of 10 mm2 and costs R52.40/kg. The copper has a density of 8.9 Mg/mt and the resistance of one kilometre of single conductor is 178 ma/km,
6.1 6.2 6.3
Calculate the most economical conductor c.s.a. Calculate the diameter of this conductor. Calculate the load and form fadors.
161
7.
A substation transformer supplies 1.2 MW at a power factor of 0.707I lagging.
7.L
Calculate the kVA'r rating of lossfree static capacitors required for constant kW correction to 0.9563 lagging , Calculate the kVA'r rating of a synchronous motor required for constant kVA correction to 0.9563 lagging.
7.2
B.
An industrial consumer has a constant load of 1800 kW at a power factor of 0,8 lagging for eight hours per day for 25 days in a month of 30 days. For the remaining time there is a constantload of 180 kW at a power factor 0.9 lagging.
8,1
Use the following tariff and calculate the monthly cost:
Unit (kWh) charge per month: The first 200 kwh per kVA of maximum demand in the month The next 200 kwh per kVA of maximum demand in the month Additional kWh supplied
69c/kWh
6lc/kWh 52clkWh
Maximum demand charge per month:
The first 240 kVA of maximum demand in the month The next 480 kVA of maximum demand in the month The next 720 kVA of maximum demand in the month Additional kVA of maximum demand
R13.20lkVA R12.10/kVA R10.80/kvA R9.00/kvA
8.2
The consumer now improves the power factor of the main load to 0.96 lagging using lossfree static capacitors with a cost of R1310.00 per month. Calculate the monthly saving due to this action.
9,
A load with a maximum demand of 600 kVA at a power factor of 0.8 lagging is to be improved to the most economical power factor. The annual tariff is R40,00 per kVA and the annual interest and depreciation charges is 10olo total. The initial cost of the lossfree static capacitors is R110,00 per kVA'r.
9.1 Calculate the most economical power factor. 9.2 Calculate the kVA'r rating of the capacitors required. 9.3 Calculate the annual net saving. 9.4 Calculate the time taken to save the initial cost of the capacitors,
152
CHAPTER SIX
PER.UNIT SYSTEMS
6.1
INTRODUCTION Answers to problems pertaining to electrical power systems are almost always required in terms of volts, amperes, ohms and kVA. In the process of computation, it is more convenient to express voltage, current, impedance and power in terms of percent or per unit, of a selected base or reference value of each of these quantities. The perunit value of any quantity is defined as the ratio of the quantity to its base value expressed as a decimal. The ratio in percent is 100 times the value in per unit. The
electrical characteristics of machines are usually specified by the designers and manufacturers in terms of percent or per unit.
Any attempt to mathematically model the power system is heavily dependent on circuit concepts. The fact that power systems are threephase, is a major complication. Another important complicating factor is the large number of components. Typical
systems can consist of tens of generators and hundreds of transmission lines and transformers. Another factor to consider is that transformers distribute the system into many different voltage sections. These methods of representation must therefore particularly deal with these factors. Their complicating effects have to be minimised as much as possible. The basic picture of the power system is the one line diagram, The' diagram communicates the essential interconnection information with maximum simplicity, The perphase equivalent circuit takes advantage of the symmetry inherent in balanced threephase circuits, The perunit system simplifies numerical analysis and eliminates the paftitioning effect of transformers. All these representations are very useful in displaying and formulating power system problems.
6.2
PERUNTTQUANTTTTES The perunit system is similar to the percentage system except that all quantities are expressed as decimal fractions instead of percentages. The base quantities then have the value of unity (one per unit) instead of 100o/o. It is necessary for power system engineers to become familiar with and facile in the use of the system because of its wide industrial acceptance and use. They also take advantage of its analytical simplifications. Any quantity can be expressed on a per unit base by the equation: Per unit
value =
actualvalue base value
163
The actual value is the actual value of the voltage, current, power or impedance as it appears in the power system. The new base value is determined and is usually the value that leads to confusion in the early stages of applying the per unit system, To help prevent this confusion, it will help to remember the following rules; The value of
Sno
is the same for the entire system concerned once it's been chosen.
The ratio of the voltage bases on either side of a transformer selected, to the same as the rbtio of the transformer voltage ratings. The value of Vn5 is a chosen value, but will vary from one zone to another zone, Once these rules are obeyed, all other base values are related to the power quantities chosen as base values. This means that the usual electrical laws, as they are known, still applies. Voltage, current, impedance and power are so related that the selection of base values for any two of them determines the base values of the remaining two. The base impedance is that impedance which will have a voltage drop across it equal to the base voltage when the current flowing in the impedance is equal to the base value of the current. The base apparent power in singlephase systems is the product of the base voltage and the base current. Base voltage and base apparent powers are the quantities usually selected to specify the base.
The actual value is also a value in volts, amperes, ohms, etc. In a power system, a base power and voltage are selected at a specific point in the system, A transformer has no effect on the base apparent power of the system, The reason for this being that the apparent power into the transformer equals the apparent power out of the transformer. On the other hand, voltage changes when it goes through a transformer, so the value of V5ur. changes at every transformer in the system according to it turns ratio. Because the base values change in passing through a transformer, the process of referring quantities to a common voltage level is automatically taken care of during perunit conversion. The perunit system has the distinct advantage that, with it, all basic circuit relations apply. Suauut
=
Spu.Snu
The perunit system simplifies many of the problems of circuit analyses. In the conventional form of calculation using volt and ampere, the solution of a system involving power lines of several different voltage levels, requires that all impedances that are to be added, to be transferred to a single voltage level. In the per unit system, the different voltage levels entirely disappear and a power network involving generators, transformers and lines (of different voltage levels) reduces to a system of simple impedances. Further more, machines such as generators and transformers, when described in the per unit system, have their characteristics specified by almost the same number, regardless of the rating of the machines.
164
ADVANTAGES OF THE PER.UNIT SYSTEM In many engineering situations it is useful to normalise dimensioned values. As said, it is commonly done in power system analysis and the standard method used is the perunit system. Advantages include the following:
. .
Device parameters tend to fall in a relative narrow range, making inaccurate values prominently,
The method is defined so as to eliminate ideal transforme?s as circuit components. Since the typical power system contains hundreds of transformers, this is an inconsiderable saving.
. .
to this advantage, the voltage throughout the power system is normally to unity.
Related close
Both the percent and perunit methods of calculation are simpler than the use of actual volts, amperes and ohms.
6.4
DISADVANTAGES OF THE PER.UNIT SYSTEM The perunit system also has some disadvantages. Disadvantages may include the following:
. .
The system modifies component equivalent circuits, making them somewhat more abstract. Sometimes phase shifts that are clearly present in the unscaled circuit, vanish in the perunit circuit. Some equations that hold in the unscaled case are modified when scaled into per
unit. Factors such as
6.5
J:
and 3 are removed or added by the method.
THE PER.UNIT SYSTEM As discussed, the perunit system is very handy to use in the analysing of large power systems with different voltage levels. In the per unit system, the voltages, currents, powers and impedances are not measured in there actual SI units as we know them i,e.
volts, amperes, VA or ohms. Instead, each electrical quantity is measured and expressed as a decimal fraction of some base level. A given perunit value for an impedance is the ratio of the voltage drop across the impedance when it is carrying the rated current of the section of the circuit in which it is connected, and the rated voltage of that section of the circuit.
165
a Lpu 
Zactua
l'Irated
,
,.
,
..,.., (i)
Vrated
As shown, a perunit quantity is the ratio between the actual quantity and the chosen base quantity. It therefore follows that:
uou=
n
#
(
ii)
Alsol
Ipu
= grnb
.,........ (iii)
And:
Zpu
= '+ Lnb
..........(iv)
It
is usual to take the rated values, i,e, the nameplate values, as the base values. Ohm's law:
7 apu(nb) 
Vno
..,..,....(v)
L"
Substitute Equation (v) into Equation (iv):
a Pu It
Zacual'Ibase(rated;
(vi)
vbase(rateO
is known that:
rrbase(rated)  Sbaselrated; vr._a.*
(vii)
Substitute Equation (vii) Equation (iv):
"' Lbase = l!t* 5brr"
(viii)
Substitute Equation (viii) into Equation (iv):
7
'
l)U
166

Zactual'Sbase
V#'"
(ix)
6.5.1
THREEPHASE EQUTPMENT Since threephase systems are solved as a single line with a neutral return, the bases for quantities in the impedance diagram are kVA per phase and volts from line to line. Although a line voltage may be specified as a base, the voltage in the singlephase circuit is still the voltage to neutral. The base voltage to neutral is the base voltage
from line to line divided bV JS . This is also the ratio between line to line and line to neutral voltages of a balanced, threephase system, As a result of this, if the system is balanced, the perunit value of a line to neutral voltage on the line to neutral voltage base is equal to the perunit value of the line to line voltage at the same point on the line to line voltage base. Similarly, the threephase kVA is three times the kVA per phase and the base value of the threephase kVA is three times the perphase value of the base kVA. The perunit value of the threephase kVA on the threephase kVA base is therefore identical to the perunit value of the kVA per phase, on the kVA per phase base. The impedance of threephase equipment is always given as perphase quantities. From Equation (i):
a pu But:
Zactual/ph'Irated/ph Vated/
,.........(x)
ph
Ipr' = IL S
b.r"
.,........(xi)
JE.v' And:
Vph =
VL
..,....... (xii)
J3
Substitute Equations (xi) and (xii) into Equation (i): Zactual'Sbase 7  Vfu**,
4PU
(xiii)
Substitute Equation (xiii) into Equation (v): V'irtin"t Lnb = =

5nn
t67
6.5.2
BASE SELECTTON FOR PERUNrT QUANTTTTES The selection of base values is made to reduce the work required by calculations as much as possible, A base is first selected for some part of the circuit. The base selected should be one that yields perunit values of rated voltage and current approximately equal to unity to simplify calculations. When the manufacturer gives in percent or per unit the reactance and resistance of a component, the base is understood to be the rated voltage and kVA of the component, A great advantage in making perunit calculations is realised by the proper selection of different bases for circuits conhected to each other through transformers. To achieve the advantage in a singlephase system, the voltage bases for the circuits connected through transformers must have the same ratio as the turnsratio of the transformer, With such a selection of voltage bases and the same kVA base, the perunit value of an impedance will be the same when it is expressed on the base selected for its own side of the transformer, as when it is referred to the other side of the transformer and expressed on the base of that side of the transformer.
6.5.3 CHANGING BASE
VALUES
if
network calculations need to be done using perunit values, all the perunit values must be caleulated using the same base values. The base units for any electrical equipment are Sn5, Vn5 ond Zn5. Let:
gb = given base (base at which Zo, is given) nb = new base (base at which the new Zp, has to calculated) From Equation (xiii):
7
Aactual
And:

7 Aactual 
Zpr1q51'Vfi
=
5sn
Zpuln6y'V'fu
=5nn
Therefore: Z prtgul
'V;b
Ssu =
168
Z pu(nul 'Vnzu
Snn
be
If Vn6 and
Snu
change, then:
Zpulnb)
If
Sn6
changes and
Vnu
apu(nb)
Vn6
changes and
a
Sn6
r,,.r[*)[#),
stays the same (Vn5 = Vsu):
a_
If
=

',*'[+)
stays the same (Sn5 = Ssn):
epu(nb)

'rr,rr[*)
The equation for the new base impedance shows that the same equation is valid for either singlephase or threephase circuits. In the case of threephase, linetoline voltage must be used with kVA per phase.
6.5.4 PER.UNIT
IMPEDANCE OF A TRANSFORMER
All impedances in any part of a system must be expressed on the same impedance. base when calculations are done. Sometimes the perunit impedance of a system component is expressed on a base other than the one selected as base for the pad of the system in which the component is located, It is therefore necessary to have some means of convefting perunit impedances from one base to another. Reference will always be made to the high voltage (HV) and low voltage (LV) side of the transformer.
Example 5.1 Consider a 24kVA, 4801220 volt transformer with a leakage impedance of (0 + j0.055) ohm referred to the low voltage winding of the transformer. Nameplate or rated values are used as base values,
169
High voltage (HV) side
Vso
= 24 kVA =480V
7sa
=
Ssn
Low voltage (LV) side Sbur. Vs35s
v&("nt)
= 24 kVA = 220 V
a abase 
Ssu
,'2
vgb(line) 
^
5sn
.t
=
(4Bo)2
= =
24 xL03
= 9.6 C) Znv
= (0.055)fgl'
0.055
f)
\220 )
= 0.262 O 7ru
Zlr, =
Q2O)2
24 xL03 2.O17 Q
7pu =
0.0s5 2.017
O.O273Z9O'pu
0.262 9.6
= O.O273190"
pu
This shows that the perunit impedance of a transformer is the same when referred from one winding to the other. If the perunit values are used, the equivalent circuit of the transformer can now be drawn as shown in Figure 6.23. Zpu
Vou=1pu
= 0.0273190. pu
Vor=1Pu
Figure 5.1: Equivalent transformer circuit of Example 6.1
6.6
APPLICATION IN NETWORK CALCULATIONS The selection of base values in a power system is used in network calculations.
t70
Example 6.2 Figure 6.2 shows a schematic diagram of a radial transmission system. Use the nameplate values of the generator as base values and draw the equivalent circuit for the system.
Generator Transformer
11 kV
r32lrl kv
60 MVA
40 MVA
X=20o/o
Transformer 2
Line
1
(0 + j2.9)
o
732166 kV
30 MVA X = 7o/o
X=8Vo
Figure 6.2: Singleline diagram for the radial transmission system of Example 6.2
Transformer 1:
Generator: Sbur"
= 60 MVA
Sbur.
V565g
=
Vbur.
11 kV
= 60 MVA = 132 kV
Transformer 2: Sbu.. V62ss
= 60 MVA = 66 kV
Refer all impedances to the common base values: Generator:
Sn6
=
Sso
And:
Vn6
=
Vsn
Zpu(nb)
Transformer
And:
1:
= O'2ZSO Pu
Sn6
r
Ssn
V65
=
Vs5
Zpulnb)
=
,r".r[Fl [5su
/
= (ooB)(#) = O.L2Z9O
pu
L7T
Line:
Z tine,S nb
Zpu =
v& (2.9X60 x 106 )
(t32 x t03)2 = O.OlZ9Oo pu Transformer 2:
Snb
#
Sso
Andr
Vnb
=
Vsn
apu(nb)

trrar[Fl
1_
/ (o oD(#) l)su
0.056290" pu Zpulsen)
= 0'2190"
Zpufi1) Vpu(sen)
Zpu(tine)
Pu
= 0.72190
= 0'01290"
PU
Zp,r(rz)
Pu
= 0'L4290'
= 1 Pu
Figure 5.3: Equivalent circuit for the network of Example 6.2 Example 6.3 Figure 6.4 shows a schematic diagram of a radial transmission system. Use the nameplate values of Transformer 1 as base values and draw the equivalent circuit for the system. Generator
13.8 kV
60 MVA X = 20c/c
Transformer
132/11 kV 40 MVA X=
8olo
Transformer 2
1
(0 + j2.9)
o
t32/66 kV 30 MVA
X=
7o/o
Figure 5.4: SinEleline diagram for the radial transmission system of Example 6.3 772
Transformer 1:
Generator: Sbuse
= 40 MVA
Sbu..
V535s
=
Vbr."
11 kV
= 40 MVA = 132 kV
Transformer 2: Sbase
Vh5s
= 40 MVA = 66 kV
Refer all impedances to the common base values: Sn6
+
Ssu
Vn5
+
Vsn
apu(nb)
_
Generator:
And:
d
a_
=
''*'[+J[*)'
,",(#l#)' O.2t,/9Oo pu
Transformer
1:
Snb = Ssn
And: .a
Line:
Vnb = Vso apu(nb) _ O,O8Z9O. a_
apu(nb)
=
pu
Z line,Snu
v& (2.9X40 x106)
(t32 x t03)2 O.OO67290'pu
Transformer 2:
Sn5
+
Seu
And:
Vng
=
Vsn
Zpulnb)
=
t."[+J
=
(o
=
0.0933290. pu
oa[#)
173
Zpulsen)
= 0.21190"
Zpu(rine)
Pu
Zpu[t) = 0.08290" Pu Vpulsen)
=
1
= 0'0067290'
Pu
Zpug2)
= 0'0933290"
Pu
PU
Figure 6.5: Equivalent circuit for the network of Example 6.3 Example 5.4 Determine the Thevenin equivalent circuit for the network shown in Figure Sbur" = 150 MVA ahd Vsssg = I32 kV in the transmission lines.
11 kV lOO MVA
X
=
1B%o
13.8 kV 150 MVA X = 22o/o
t32l7t kv 1OO MVA
X=
9o/o
t3zltr
kv
150 MVA X = 9o/o
22 kV
r32l22kV
250 MVA X = 25o/o
250 MVA X = 10%o
Figure 6.6: Line diagram of the network for Example 6.4
174
6.6.
Take
Refer all impedances to the common base values:
Generator
1:
Zpulnb) =
"*"[+) (o
1s)[i#)
= O.27Z9O'pu 6
Transformer
1:
Zpulnb) =
'rrar[+) (oor)(i#) 0.135290" pu
Line
1;
Zpu =
Z ttn"'S nu
Vrt (2.9X1s0 x
106
)
(132 x 103)2
0.02529O'pu
'
Generator
2:
Zpulnb)

(vno)'
= znurool[*
J
rc.22\fE!)' ' '[
11 ,l = O.24629O pu Transformer Line
2:
2:
Zpu(nb)
= 0.0829O" pu
7p, =
Ztin"'Sno
vto (4X1s0 x to6)
Trl, " lo1f
O.O34419O" pu
t75
Generator
3:
Zpulnb) =
"".,'[+) =
(o2s)t#) 0.1529Oo pu
Transformer
3:
Zpplnb) =
t,,r[+) (0
1)[#)
0.O629O'pu 7_
Line 3:
Z tine.S nb
v& (sx1s0 x t06)
lrgtlo'tr O.O43I9O" pu
0.27190 pu 0.135t90" pu 0.025290" pu
0.346190" pu 0.08290' pu 0.0344/90' pu
t'ry
0,15290.
pu 0.06Z90. pu 0.043290. pu
Figure 6,7: Equivalent circuit for the network of Example 6.4 7_ zpg(TH)

(0.
27
+ 0.
1
35 + 0.
02s)//(0. 346+0.
0B +
0.0344)I I Q.1 s +0.
06
+0.043)
= 0.11829O'pu 1pu
0.118290" pu
Figure 6.8: Thevenin equivalent circuit for the network of Example 6,4 176
I
Example 6.5 Figure 6,9 shows
a schematic
diagram of
a
radial transmission system. Use the
nameplate values of the transformer as base values and calculate the terminal voltage on the generator if the load takes full load current at 0.9 pu volts.
Generator Transformer
kV
(o'48 + j2'B)
CI
600 kw
11 kV
11/3.3
2
1 MVA
cos 0
X=7o/o
lagging
MVA X=18o/o
= 0.8
Figure 6.9: Line diagram of the network for Example 6.5
Pload =
J5.V1ou6.I6u6.cos $
600 x 103 = ( Jl t
L3L.2L6Z36.87' A
rload
T
rbase
Xt,s x lo3xlbidxo,B)
S

b.r.
J3,Vour.
1x106 (J3X3,3 x 103)
L74.955 A T_
rpu
=
Itoad I
b.."
131.216
L74.955
O.75236.87'pu Transformer: Zpulnb)
= O'O7Z9O
Pu
177
Line:
=
Zpu
e (0.48 + i12.8)(1 x (3.3 x to3)2
=
106
)
= O.25Ll8O.27o pu Vpulsen)
=
Vpu(toad)
*
Vpufi)
0.9 a0. + (0.7
5
*
Vpu(tine)
236.87')(0.07 t90.)+(0.7 5 136.87 ")(0.26t tBO.27 ")
1.08829.335'pu Vterminal
=
Vpu1ggny,V535g
=
(L088t9.335"X11)
= tL.95819.335'kV Example 6.6 Figure 6.10 shows a schematic diagram of a radial transmission system. Use a base of 120 MVA and calculate the terminal voltage of the generator in per unit and in kV if the voltage on the load is to be maintained at 33 kV.
Transformer
Generator
11/66 kV 60 MVA X = 9o/o
Transformer
Line
(1.86 + j12.8)
o
66/33 kV 75 MVA X = I2o/o
60 MW cos $ = Q.31 lagging
Figure 5.1O: Singleline diagram for the radial transmission system of Example 6.6
Ptoad
60
x
106 Itoao
178
J3
.V6u6.I6u6,cos $
"6 X:: t295.96Z35.9" A
(
x 1o3XIbadXo.B1)
rrbase  Sbu..
J3.Vbase
120 x 106

(J3X33 x 10')
= r tse
2099.455 A Itoad
lb.r"
L295.962
 35.9'
2099.456
O.6L7l35.9o pu Transformer
2:

(,s^o
Zpulnb) = apu(sb)
^
l)so
) I
J
(0.12)f!q) ' '17s,/ O.L92Z9O'pu Vpulno) = Zpulno)'Ipu
(o.r92 t9o)(0.6r7 /35.9") O.LL85Z54.1o pu Line:
a
Lpu 
=
Ztin"'Snb
v& (1,86 + j12,8X120 x 106 )
(66x103)2
O356ZaL73'pu Vpu(tine)
=
Zpultine).Ipu
(0.3s6 tBL.7 3')(0.677
t3
5,
9")
O.2L97Z45.83o pu
179
Transformer
1:
Zp.,1nu1
=
trr,rr[+j
= (o.oe)fEq.) ' '[ 60
,
,
Vpu(nb)
=
0.1829O. pu
=
ZpulnU;'Ipu
= (0.18290')(0.617 t35.9') = O.tLtLZS4.l'pu
Since
the load is used as reference, the perunit voltage across the load will
Lzj"
pu. Vpulgen)
+
Vpt,(rz)
*
*
Vpufil)
=
Vpu(load)
=
LZj" + 0.LL85t54.1 + 0.2L97t45.83' + 0.1tIIz54.I L.3332L4.94'pu
Vpu(line)
be
Vlterminat) = Vpgqsgn;'V535s
= (r.3332t4.94"X11) = 14.653214.940 kV
Example 6.7 Consider the radial transmission system in Figure 6.11. Use the nameplate values of Transformer 2 as base values and calculate the terminal voltage on the generator if the load takes full load current at0.942 pu volts.
Generator
Transformer
1
kV (1.69 + j11.7) o MVA X=9o/o
66/33 24.5
kV MVA X=I2o/o
33/11 27.5
36 MW cos g = Q,796 lagging
Figure 6.11: Singleline diagram for the radial transmission system of Example 6.7
180
Ptoad
t
=
J3
,V1ou6'I6u6.cos $
= ( J3 )(11 x 103XIbadX0,7B6) Iroad = 2403'957138'19' A
J6 x 106
+!ttt
rbase=
J3.Vbase
=
= rrpu 
27.5 xL06
(J:Xtt
103 ) " 1443.376 A Iload r base
Z38.L9" = 2403.957 rqa376
= Transformer
2:
Line:
Zpulnb)
7ou
= O,!2l9O
= = =
Transformer
1:
Zpulnb)
I.666238.19o pu Pu
Ztint:snu
v6 (1'69 + j1l,7X?.5 x 106)
(33x103)2 O.2985281.78o pu
= ,rr,rr,[+)
=
(o.oe)
'
eE\
'\24.s )
= jo.lol
pu
The perunit voltage across the load is given as 0.94210 pu.
181
Vpu(sen)
Vlgen)
=
Vpu(load)
=
0.94220" + (0.L2t90,)(t.666t38. 19") + (0.2985 tB 1, 7B.X 1. 66 6 Z.38. L9.) + (0. L}r t90")(L666 t38. 19")
=
I.655222.455. pu
=
Vpu1gg6y.Vg65s
+Vpu6z) *Vpu(tine) *VpuG1)
(L.655t22,45S.X66) o = = 109.23222.455. kV Example 6.8 Figure 6.12 shows a schematic diagram of a radial transmission system working at 50 Hz. Use a base of 120 MVA and calculate the resistance and inductance of the line when the load takes fullload current. Generator
157.537
Transformer
z\0.97. kV
1
Transformer
Line
132/BB kV
BB/33 kV
144 MVA
X=
96 MVA X=
10.8olo
B.4o/o
2
48 MW cos $ = Q.991 lagging
Figure 6.12: Singleline diagram for the radial transmission system of Example 6.8
Ptoad
48 x 106 Iroad
= J3 ,V1ou6.I1ou6.cos
= (J5)(33 x 103XIbadX0.B91)
= 942.5L7127"
rud\e 
Sb.r.
J3.vor.. 120 x 106
=
TB2
$
2099.455 A
A
t
Ipu =
I toad I
b.r" 942.517
Z27"
2099.456 =
Transformer
1:
O.4489127" pu

(s.o )  / l)su
Zpulnb) = zpu(qb) I
I
o.LoBzsoofgq) \r44 )
jo.o9 pu Transformer
2:

Zpulnb) = 1e(sb)
(s''o ) [sqb
J
o.os4tsoofEql
\.e6i
j0.105 pu 757.537 2L0,97"
Vpu(sen)
r32
I.L9351LO.97" pu \,vpg(gen) _ LL9351L0.97" = \,v pu(line) Vpu(tine)
Vpu(load) 7
+ Vpugr; + Vpu6zl *
Vpu(line)
Z0o + (0.09290"X0.4489 t27 o) + (0. 105290)(0.4489 l27o)
+
Vpu(rine)
 O.lg9LZ48.5o pu =
Ipg.Zpultine)
= (0.4489 Z27.)( Zpulrin.)) Zpu(rine) = O.4436275.5o Pu
0.I99I 248.5.
7
apu(rine)
0.4436275.5. Ztin"
Ztin.'Snb _  _ril_
I lo6) = (2,,,.'"X120 (BB x 103)2
=
28.625275.5" O
183
R
=
7.169
C)
L = 88.213
And:
mH
Example 6.9 Refer to the single line diagram of a radial transmission system in Figure 6.13. Use a voltagebase of 273 kV and a kVAbase of 210 MVA and calculate the actual voltage on the termindls of the generator. The impedance of the transmission line conductors is
(0.05 + j0,1s) o/km.
Generator Transformer 1 Transformer Line
2
Transformer 3
1
Line 2
48 km
24slr32kv
132/BB kV
125 MVA
96 MVA X = 9o/o
X = I2o/o
12 km BB/11
kV
72MVA X = 10o/o
36 MW cos $ = Q.7gg lagging
Figure 6.12: Singleline diagram for the radial transmission system of Example 6.9
Pload =
36x106= T
J3 (
Vrouo Ilou6 CoS $1s66
J: Xrr x lo3XIbadXo.TBB)
2397.855238'A
rload
Tlbase 
Snu
J5.vno 210 x 106
(J3X11"103)
tto22,t4L5
A
Itoaa
T
Ibur"
= =
LM
?397.B5st38" 17022.1415
O.2L75Z38. p.u.
=7go*t*l _
(o',)[#l#l
=
jO.1624 P.u.
[#),"',
Vgb =
147.086 kv
z
(0.0s +j0.1sx48)
7.589527'.s'ss6'o ZIine'Sno Lnb
v;b (7.s8ss z71.s6q')919
=
7no
"10
)
1r+Z.OaOxfO'1'
=
Transformer 2:
u
O.O7g7l7t'565" P.u'
= zgo'+t*l
(ooe)[#l#)' = j0.1586 vsu
=
[FJ*r,
= 98.057
z
=
P.u.
kV
(0,0s +j0.1sX12)
= 1,897271.565"
O
185
\_( Ztine,Snu
a_ Lnb
va =
= a_ Lnb
Transformet 3:
(1.897 z7
.565 ") (2IO xLO 6 )
(98.057xtO3)2
0.04t427L555" p.u. z^".
""
=
L
snb
ryq)' j
Ssb [%o
(o1)(#)[*b)' jo.2349 p.u.
Load:
vgb 
[tr),",
= t2.257 kV
Vo,
( tt
= t_t
)
Itz.zst )
o.8972o" p.u. Vp.u.(rt)
*
Vp.u.(tine1)
*
Vp.u.gz)
*
Vp.u.(tine2)
*
Vp.u.63)
(0.217 5 z38" )(0.1624 t90 " + 0.07 37 27 1.565' + + 0.0414t71.565" + 0.2349t90") + 0.89720"
0.998326.28' p.u. Vgen = (0.9983/:6.28"X24s)
244.583525.28" kV
.i
.
186
*
0.
1
Vp.u.ltoad)
586290'
'.(
EXERCISE 6
a
Figure 6.13 shows a schematic diagram of a radial transmission system. Use a base of 96 MVA and calculate the terminal voltage of the generator if the terminal voltage on the load is to be maintained at 32 kV.
Generator Transformerl
66/88
Transformer 2
kv
(o'92 + i9'7s)
o
54 MVA X = ILo/o
Load
BB/32 kv
48 MW
69 MVA
cos $
X
lagging
= L2o/o
=
Q'91
Figure 6.13: Singleline diagram for the radial transmission system of Question 2.
Refer to Figure generator.
6.14,
Use
a base of 55 MVA and calculate the terminal voltage of the
Generator Transformerl
Transformer 2
('2 + j9'6) o
kV
33/66 75 MVA
X=0,1
pu
Load
66/11 kV 90 MVA
24 MW cos 0 = 0'707L
X=
lagging
BVo
Figure 6.14: Singleline diagram for the radial transmission system of Question
{3
1
2
Consider the network in Figure 6.15. The load takes full load current at 0.936 pu volts, Use a base of 700 kVA and calculate the terminal voltage on the generator.
Generator Transformer
6.6/11
kv
1.2 MVA
X=
9o/o
1
Transformer
Line
(2.24 + i0.36)
o
2
Load
11/3.3 kv
7BO KW
1.8 MVA
cos $
X=
lagging
7.5o/o
= 0'7193
Figure 6.15: Singleline diagram for the radial transmission network of Question 3
187
4"
Consider the radial transmission system in Figure 6.16. Use the nameplate values of Transformer 2 as base values and calculate the terminal voltage on the generator if the load takes full load current at 0.92 pu volts.
Generator Transformerl
kV
33/48 28 MVA X = l2o/o
Transformer 2
Line
(1.05 + j1o.s)
o
Load
32.4 MW cos $ = Q.399 lagging
4Bl11 kV 42 MVA X
=
10olo
Figure 6,16: Singleline diagram for the radial transmission system of Question 4 Figure 6.17 shows a schematic diagram of a radial transmission system. Use a base of {r5. '\/1 210 MVA and calculate the resistance and inductance of the line.
Line
Generator Transformerl
Vsen
= 15619.7"
kV
Transformer 2
132lBB kV 120 MVA
BB/33 kV 150 MVA
X=
X
B.Bolo
=
Load
51 MW cos $ = Q.916 lagging
10.4o/o
Figure 6.17: Singleline diagram for the radial transmission system of Question 6.
5
Consider the radial transmission system in Figure 6.18. Use the nameplate values of Transformer 1 as base values and calculate the resistance and reactance of the line if the load takes full load current at 0.945 pu volts. Generator Transformer
Vsen
1
Line
Transformer
2
Load
kV 60 MVA
33/5,6 kV 45 MVA
18 MW cos 0 = 0.8
X=
X=
lagging
= 21,34623.7'kV 11/33
9o/o
8olo
Figure 5.18: Singleline diagram for the radial transmission system of Question 6
1BB
l
7.
Refer
to the threephase network shown in Figure 6.19.
Use
a
base of 48 MVA and
determine the value of the generator terminal voltage.
Generator Transformer
BBi66
,
kV
Transformer 2
Line
1
(1'15 + j11.s)
f,
78 MVA
54 MW cos $ = Q.7gg
X=
lagging
66/11 kV
64 MVA X = LZo/o
Load
9o/o
Figure 6.19: Singleline diagram for the threephase network of Question 7 B.
Figure 6,20 shows
a schematic
diagram
of a radial
transmission
system. Use the
nameplate values of the transformer as base values and calculate the terminal voltage on the generator if the load takes full load current at 0.915 pu volts.
Generator
Transformer
kV Bolo
33/11 75 MVA X=
Line
@.22
+ j2'2) o
6MW cos $ = Q.399 lagging
Figure 5.20: Line diagram of the network for Question 9,
B
Figure 6,21 shows a schematic diagram of a radial transmission system, Use the nameplate values of the generator as base values and draw the equivalent circuit for the system.
Generator Transformerl
66 kV 48 MVA
11/132 kV 32 MVA
X
X
=
18o/o
=
100/o
Transformer 2
Line
(0.24 + j2.4)
o
132/33 kV 60 MVA X = 9o/o
Figure 5.21: Singleline diagram for the radial transmission system of Question 9
189
a
10. Figure 6.22 shows a schematic diagram of a radial transmission
system. Use the
nameplate values of Transformer 1 as base values and draw the equivalent circuit for the system.
Generator Transformerl
13.8 kV 7s MyA
11/BB
X=15o/o
kV
Transformer 2
Line
(0.18 + j1'B)
o
50 MVA
88/6,6 kV 90 MVA
X=11olo
X=
Bolo
Figure 6.22: Singleline diagram for the radial transmission system of Question 11.
Determine the Thevenin equivalent circuit for the network shown in Figure = 90 MVA dfld V6u." = BB kV in the transmission lines.
10
6.23.
Take
Sbur"
11 kV
BB/11 kV
60 MVA
60 MVA X = 8o/o
X
=
l5o/o
13.8 kV
BB/11 kV
90 MVA
90 MVA X = 100/o
X
=
18o/o
24 kV 150 MVA X = 27o/o
BBl24 kV 150 MVA X = L2o/o
(0,36 + j2.4)
j4.B
a
o
0.54 + j5.9
o
Figure 5.23: Line diagram of the network for Question
11
i
I
190
EXERCISE
1.1 1.2
z.
2.t
EXERCTSE
32
9.643r{54.8" f,) 4991.6321s4.8" VA
^#Stazvs.cz
4.2 4.3 5.1
n
9.789Z4L.340 3.4
1000.106 w 22.37 kw'...,,
9.712t21.4" A 5..3ea;rsz.tp n
4.1
9.2s9176.73 4
4758.114 W
12.5551169.42 A
12.22t33.28 4.2 5.1
"
5.5091158.04" A 10.703275.57, A
7.434t4933
A
1179.941 W
88r.047W
*571.105 W
4067.849/.42.69"
5.3
VA
2989.883 \jv *2758,26 VA?
3.106r*138.49" A 1.5172t2.62' A 202.00s
6.2
w
216.11 W 72.716 W
7472.73/r7A.53'
6.3
VA
490.83 W
1388.532 VA?
76.2382tt7.7 9.7r2t51.4. A
7,1
5.484t177.9" 7.2 8.1 8.2
A
796.262trr2.55
29.5522t46.27" 8.05 {)
V
21.9125 mH 19.468 kW 2156.844 W 21.4805 W 6.1195 W
922W
18.2982103.060 A
15. 16.
A
17.
6.3535283.62" A
0
18.
9376.598 W
29.543t722,55a
A
19.
27.246/L20.97" A 21.A2rz25.r20 AlA
kVA?
30.341242.74"
:
70. 21.
16945.922W 849.3115 W 1323.077 W 1186.937 W 6008.s31 W 2907.627 W 10,679 kW 10.679 kW
:
A
EXERCISE 2.2
L
948.1155 W 9209.2155 W
2. 3.1
8967.Jg4W 7.814tL37.76" A 47.646213;L98" A
264
A
A
6670,881 W 2919.3795 W 6860.620s W 16945.575 W
EXERCISE
47.7L6t44.64.
h
13.
78.2352158.41" A
19.033 kW * 8.923 11.1
17V5.246W, 3250.254 W 5025.5 W 8.021 kW 3S06,933 W
17.775276.38 4
3Q.24621.940 A 10.2
A
,
0,9326 lagging 1.297 A 480 W 0.1985 lagglng 3.173 A
14.
72.266t27.44
W
o.55229s.29
4308,114 W
, .
4955.35
11.1 11.2 11.3 12.1 12.2 12.3
A
16.2552t.39 9.1 9.2 10.1
5.2 6. 7. 8.1 4.2 8.3 9. 10.
2.114t94.L9'A
6.1
o
0
228.338193.63" V 15.238287.2: A
22.879177.405. A
s.2, ,,:,
999.986 W 22.372]y.i$
10.699220.8050 9.4435t72.190 Q
2877.338 W 4078.8865 VA'r
3. 4,t
2.2 {CONTTNUED}
3
1.
635,473 VA'r
2.
9.672t0.480 E 50.17351*88.03" A 11,5.416.tL5.78 A
,
EXERCISE 3 (CONTINUED) 75133" V 10.09.
EXERCISE 5.1 5.2
v
201.259t109.91'V 201.259z130.09' V
6.1
97,781tr34.39"
6.2 6.3 7.1 7.2
V 97.7811=105.61" V
s7]81tL439' 4.1
V
'
98,45282.69'n
,
60.631t36.42'A 4.2
29.74222.89 0 r353.472tt2.36"
5,1
37.375231.55' A
29.392t147.51 4
7.3 8.1 8.2 8.3 8.4
585.2822123.3'V s06.083143.3'v
9.1
VA
19,801143.39'A 5.7.
7.7
7.2
29.513128.86'A 31.501r103.06' A r4.L66t123.925' A 33.598r*155.33'A 43.3865136.04'A 26.40L229.98'A
3.490
182.9805223.43' MVA 105.632263i34r MVA 2.55" 19.347 MW
0.162
lagging
:
'
76.74
441,288258.76' MVA 623.242230.95" MVA 8.t)71 MW 11.32"
5.723 MW
1208.1294' 925.7t6t34.95 MV4 s12.136260.85'MVA
840.789293.05" V o.
4 (CONTINUED)
120.343 MW LLV726 MW 63.011 KVA'r
24.959
MVA',?
1243.8031150.18'V
10.1
2.617 MW 2s.022 MVA"r 930.18243'MVA 381.704250.6" MVA 6.748177" A
856.644216.A4'V
t0.2
7.275
LL87.164271. V
10.3 11.1
353.817278.23 ,v1V4 40.875t4.73 MVA
9.2 9.3
416.8935138.64' MVA TL,2
EXERCISE 4 1.1
12.1
171.3145 MW 164.999 MW
52.499 MVA?
1.2
109.7482103.97' MVA
3s8.75t26.97" MVA
2s.145 MVAI 1.2
175.005 MW
A:
receives 608.184 MW and receives 292.259 MVA'r B: sends 237.627 YIW and sends 313.926 MVA? 0.9013 lagging 0.6035 lagging
2.2
5.935. 201.606274.89' MVA 552.7005111.45" MVA
2.3
5.1535 MW
3.1
89.90s MW 10.88 MVA?
L2.2
3,6915 MW
13.1
88t29.21
L3,2 13.3
24.315 MW
t4.
6670.881 W
15.1 15.2 15.3 15.4 16.1
253.722255.47" MVA
kV
600.2252t02.r45 MV{
14.760/o
253.725t55.465" MVA 74.75t/o 37.563 I\4W
177.92AMVA'I 12.355 MW L6.2
242.152MVA"r 25.208 MW 64.224 MVA"
t6.4
667.25260.65' MVA 654.371225.805' MVA 22.92.
17.3
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560.6s245.01' MVA
16.3
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MVA
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kVA'r
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EXERCISE 5
pt
228.27 V. 231.873 V:
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EXERCISET
kv
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