Electrical Technology

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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 : 978-0-9814483-3-G

l*;

ELECTRICAL TECHNOLOGY SJ VAN ZYL

-

sTH EDITION

TABLE OF CONTENTS

CHAPTER ONE

1.1

-

ADVANCED ELECTRICAL MEASUREMENT

WATTMETER APPLICATIONS

1

1.1.1 ONE-WATTMETER METHOD 2 1.1.2 BLONDELL'S THEOREM 4 1.1.3 TWO-WATTMETER METHOD 6

1.1.3.1

1,1.4

TWO-WATTMETER METHOD OF OBTAINING THE POWER FACTOR THREE-WATTMETER METHOD 10

CHAPTER TWO

2.1 2.2 ??.t 2.2.2

- THREE.PHASE ELECTRICAL SYSTEMS

INTRODUCTION THREE-PHASE

13

SYSTEMS

14 PHASE SEQUENCE OF THREE_PHASE

SYSTEMS

16

BALANCED STAR-CONNECTED, THREE-PHASE SYSTEMS BALANCED, DELTA.CONNECTED, THREE-PHASE SYSTEM UNBALANCED DELTA-CONNECTED, THREE-PHASE LOAD 2.2.4.t STAR-DELTA (y_A) CONFTGURATTdN 26 2.2.4.2 DELTA-DELTA (A_A) CONFTGUMTTON 28 UNBALANCED STAR-CONNECTED, THREE-PHASE LOAD 2.2.s.7 DELTA-STAR (A_y) CONFTGURATiON 30 2.2.s.2 STAR-STAR (y_y) CoNFTGURATTON 3s 2.2.5.3 MILLMAN'S THEOREM 35 2.2.5.4 DELTA-STAR (A_y) CoNVERSTON 39 2.2.5.s STAR-DELTA (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 THREE-PHASE 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 PHASE-SEQUENbE 82

3.2

RESOLUTION OF AN UNBALANCED, THREE-PHASE 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

THREE-PHASEINDUCTION REGULATOR 108 SYNCHRONOUS PHASE MODIFIER 111 VALUE OFTHE SENDING-END 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 CROSS-SECTIONAL 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

INTER-CONNECTIONS IO7

TAP-CHANGINGTRANSFORMER 110

EXERCISE

-

5

160

PER UNIT SYSTEMS

INTRODUCTION 163 PER-UNrT QUANTTTTES 163 ADVANTAGES OFTHE PER-UNIT SYSTEM 165 DISADVANTAGES OF THE PER-UNIT SYSTEM 165 THE PER-UNIT SYSTEM 165 THREE-PHASE EQUIPMENT 167 BASE SELECTTON FOR PER_UNrT QUANTTTTES 168 CHANGING BASE VALUES 168 PER-UNIT 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 INTER-CONNECTED 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 HIGH-FREQUENCY 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 COLOUR-RENDING 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 METAL-HALIDE LAMPS 245 SODIUM-XENON LAMPS 246 ELECTRODE-LESS INDUCTION LIGHTING CONTROLS 246

LOCALISED MANUAL SWITCHING 247 TIME-BASED SYSTEMS 247 DAYLIGHT-LINKED SYSTEMS 247 OCCUPANCY-LINKED 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 POLY-PHASE CIRCUITS

RELATIVE MAGNITUDES OF LINE AND PHASE CURRENTS AND OF LINE AND PHASE VOLTAGES POLY-PHASE 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,

THREE-PHASE SYSTEMS WHICH HAVE NON-SINUSOIDAL 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 l-c

lP /lio

Figure 1.1: Connection of a wattmeter between two lines of

a

three-phase system

to measure ac power in three-phase systems is the wattmeter, The wattmeter contains a low-impedance current coil that is connected in series with the load, and which ideally has zero impedance. The wattmeter also has a high-impedance 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 three-phase systeir without breaking the phases, The system can be star-connected or deltaconnected. Figure 1.3 shows tire connection of a single wattmeter in a star-connected, three-phase 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 delta-connected, 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 two-wattmeter method, because it is restricted to balanced loads onlY,

t

Figure 1.3: Single wattmeter connected in a balanced, three-phase, star-connected system

Figure 1.4: Single wattmeter connected in a three-phase, delta-connected 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 single-phase systems one wattmeter is required to measure the power. It is clear that two wattmeters are needed to measure the total power in a three-wire, 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 three-phase (or n-phase) line, Using instantaneous values for an unbalanced, three-wire, star-connected load can prove Blondell's theorem.

Figure 1.6: Two wattmeters connected in a star-connected system to prove Blondell's theorem

Wa+Wc = Vab,!a+vcb,ic But:

And:

V66

= V6n-V56

V66 = V6n-V6n wa * wc = (vun -vbn)ia+ (v.n - v5n)i.

=

Yan,ia- tbn,ia

=

Van.ra

*

In a star-connected system:

is*

i6

-f i. =

Q

16 = -16-16

* v.n.ir-

v.n.i" + v6n(-i,

v5n.i6

- i.)

And:

1.1.3 TWO.WATTMETER

METHOD

The two-wattmeter method gives true power in the three-phase circuit without regard to balance the waveform provided in the case of a star-connected 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: Two-wattmeter method of measuring three-phase power The two-wattmeter method can also be used for a three-phase, four-wire 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 three-phase, four-wire system, the sum of the instantaneous currents in the principal line wires is not necessarily equal to zero/ as is the case in a three-phase, three-wire system.

0

The power delivered to a three-phase, three-wire, star- or delta-connected 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 two-wattmeter method to unbalanced loads, a star-connection 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

= -Iu-I. + 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 two-wattmeter readings in balanced loads, the star-connection 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

1.1.3.1TWO-WATTMETER METHOD OF OBTAINING THE POWER FACTOR From the phasor diagram in Figure 1.8:

tY:'= And:

3oo+o

z{:o = 3oo-o

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 two-wattmeter 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 THREE-WATTMETER METHOD The total power in a three-phase 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 three-phase 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 three-wire, star-connected load would require access to the neutral point to make connections to the voltage coils. Similarly, in a three-phase, delta-connected 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 three-wattmeter 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: Three-wattmeter method of measuring power in a three-phase 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, star-connected 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{ru-ohase,

four-wire, star-connected

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 three-phase, delta-connected load Since the voltage coils form a balanced, star-connected 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:r-m

(zf-t 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 single-phase 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 single-phase, two-wire system.

Vun

-.__r Three-phase,

\/

L--rrJ Single-phase

three-wire supply

(V1'n")

,roor,

[I+.'l (J:;

Vo.

w Single-phase supply (V11n")

Figure 2.1: connection of different supplies to a three-phase, four-wire system (equivalent circuit to Figure 2.4) Most consumers are fed from a single-phase 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 single-phase alternating current supply is 220 V. The majority of single-phase supplies

are obtained by connection to a three-phase 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 poly-phase circuits are those containing three alternating current sources and three or four wires.

13

These three-phase 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 Three-phase systems have some advantages over single-phase 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 three-phase 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 three-phase voltage is basically three single-phase voltages. Each voltage is separated from the next by a phase angle of 120'. The same basic structure found in the single-ffise 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 three-phase 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 three-phase 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 three-phase generator commonly located in a power station produces three-phase 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 three-phase 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 star-connected network. Since the neutral line n is an output, the output is said to be three-phase, four-wire. The windings can also be deltaconnected.

Figure 2.3: Equivalent circuit of the stator windings of a three-phase 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 three-phase, 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 three-phase generator, each with respect to the neutral

If the three-phase 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 star-connection and a delta-connection. Sgglgg1_qilhlggpha5_9- Supdies, are usually c91n_eq!9^Q i0-.St?Lwhereas three-phase loads may be connected either in delta or star.

2.2.L

PHASE SEQUENCE OF THREE-PHASE SYSTEMS The double-subscript notation is used to avoid confusion in the direction of voltage and current. When the double-subscript 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: Double-subscript 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 anti-clockwise direction. The phase seque:nce in Figure 2.6 is positive or abc. The phase sequence is quite important in the thiee-phase distribution of power. In a three-phase 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 phase-sequence, 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 anti-clockwise 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

= Vp6Z-120" V Vc = Vpnl120" V Van = VrZ3Oo V Vn. = Vr-Z-9Oo 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 phase-sequence, showing phase and line voltages (Vu as reference) The voltages in Figure 2.7 will be:

= VpnZ0o V Vo = Vpr.Z120'V Y, = Yp6l-120" V Vau = VrZ-30" V Yo, = VtZ-150'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 STAR-CONNECTED,

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 three-phase, three-wire .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 three-phase, 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';:t-l'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, three-phase, four-wire, 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")

Em@)Z-LZlo Y sin ( 7, = L6.92/-6L7B Q

Za Zo

Conveft the load to an equivalent delta-connected load to find the phase currents and then calculate the star-point potential. Take V5 as reference.

Solution:

vulr2lo Eo,l30o

Ea6ZL50o

Y6200

vc/-r20o Figure 2.33: Positive (abc) phase sequence with Vb as reference Delta-con nected im pedances:

7,x = 7.-r

7x

*

Zu'70 zc

.35') (L9 .57 245, 16.922-61.78"

(L2.01 zB7

= t2.01187.350 + 19.57145.540 + = 22,462287.90 O a a L^--L.aL-A -

r 1

r

")

77

-C'-d

zb

= 16.92'-$1,78o + t2.0I287.350 = L9.42!l-19.42o O

44

54

+

(16.92t

- 6t.78.)(tz.0ttg7 .35") 19.57

245.54"

Current in each phase of the delta-connected loadr

r-Vao rab_ d

=

3601150" 22.452287.9"

= L6.O27262.1o

A

r-Vca l^. -

/'

cA

360

t-90"

19.42rt-19.42

=

18.5372-70.58'A

Ia = Iuo-Ica

Vu,

V'n

=

16.027162.10

=

31.675187.580 A

-

LB.537l-70.58"

= Iu.Zu = (3I.675 zB7.5BoX12.0 = 380.4172174.93V =

Vu-Vu'

= l-l[:ooz]zo"l \J3) =

I tB7 .35o)

3*o.4t7tt74.s3"

311.5375228'030 V

Example 2.7 The load of a three-phase, delta-star 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 star-delta conversion to calculate the currents drawn from the supply, Calculate the voltage across each impedance in the load.

2.7.t

E,"160o

Ev6lIB0o

vbz-30

v^l-L50"

o

E,orZ-60

Figure 2.34: Negative (cba) phase sequence with Delta-connected

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

56.9192-26.570 7_

as reference

:

= 45.25652-22.270 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

T-Vca -

rca

z^_ k

440Zffi, id.os?

z6n"

= 7.8492_2.20 A Current in each line of the delta_connected load:

Iu = Iuo-I.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.u-Ib. = 7.8492-2.20 - 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 = l-0 A, as shown in Figure 2.36.

I*:Il-e Figure 2.36: Phasor diagram showing

I

and its conjugate

I*

The complex power is then:

S*

= EL\.IZ-j = 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 in-phase 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 x-axis.

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 x-axis. 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 x-axis. 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 open-circuit 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 ty-an angle less than 90', depending on the ratio of the current corresponding to t-he 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 load-carrying 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, three-phase, four-wire, star-connected load is connected to a 3BO-V, symmetrical, three-phase, star-connected supply. The value of the neutral current flowing away from the neutral point- is 35.55176.40 A. The load consists of Za = 8.961-33,20 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, three-phase, cba-rotation supply with Eu = 24010o Y.

18256.3"O

7'O

Figure 2.43: Three-phase star-connected load with line c short-circuited 3.

An unbalanced star-connected load is connected to a symmetrical, three-phase, threewire, abc-rotation 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, three-phase, 360-V, four-wire, negative phase sequence supply with E5 as reference, is connected to a star-connected load with the following load impedances:

= L2.Bz-32.8o o = 2I'412I'4o e) 4 = 37.92-87.90 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, four-wire, star-connected load consists of the following

",u

4e

components: Phase Phase Phase

a: 28.4-7tF capacitor in parallel with a 45.2-a resistor b: 39.6-prF capacitor in parallel with a 22.4-fl- resistor c: 1B.B-prF capacitor in parallel with a 31.6-ft resistor

This load is connected to a thr,ee-phase, 400-V, 50-Hz supply, Take with a positive phase-sequence.

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, four-wire, star-connected load consists of the following components: Phase Phase Phase

a: 28.4-ttF capacitor in series with a 45.2-A resistor b: 39.6-pF capacitor in series with a 22.4-0 resistor c: 1B.B-pF capacitor in series with a 31.6-fi resistor

This load is connected to a three-phase,3B0-V,60-Hz supply. Take with a negative phase-sequence. 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, three-phase, three-wire, cba-rotation supply

with V"r

connected to a star-connected load with the following load impedances: za 26

zc

=-

360 V,

is

= 12.81-32.80 Q = 2lAl2L4o e) = 37.91-87.9" 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, three-wire load is star-connected to a three-phase, 440-V 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, three-phase, delta-connected load in Figure 2.44 is connected to 360-V 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, three-phase, con nected

star-connected with a negative phase sequence to a derta-con nected road with .source *re- i"rr"*i,' g' r*o',:#["ou

n.u*

Zao

=

72.82-32.80

o

Za, = 27.422L4o t) Zru = 37.92-B7.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, star-delta system consists of the following components: Phase Phase Phase

ab: 28.4-stF capacitor in series with a 45.2-o resistor bc: 39.6-pF capacitor in series with a 22.4-0 resistor ca: 1B.B-pF capacitor in series with a 31.6-f) resistor

This load is connected to a positive phase sequence, three-phase, 60-Hz 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 three-phase, star-delta 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 delta-star 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, delta-connected load. The load is supplied from a balanced, three-phase, 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:

r-Vn. Ihr -

Lb, 380

1L20"

6.2+ j9.6 = T_

rca

33.252262.86. A Vca

=

zca

38020

4.8-W 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, three-phase 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.= 64-Wa W.-Wu

87.26r

64-Wa-Wu

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 380-V, three-phase, delta-connected 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, three-phase supply with a positive phase sequence.

2.L0.2 Another star-connected 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. = 22-Wa And:

Or:

W.-Wu =

8.867

D-W=-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

27.8482-34.920 kVA CoS

= 0.86 Otoao = 30'680 $636

loaa

Sload =

COS $1es6

72,8r 0,86

=

Stot =

L4.8952-30.680 kVA S6e1e1

*

S1s36

27 .B4B Z.-34.92o

+ 14.8952-30.68'

= 42.7L6t-33,440 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, three-phase, three-wire supply system with a positive phase sequence at Eu. = 48010'V,

63

i

(8.2 + j17.4

a

(18,3 + j1s.1)

(24.2-j16.4) f) c

b

Figure 2.47tTwo wattmeters connected to a delta-connected load

E""

60'

Flgure 2.48: Positive (abc) sequence with O.rrert h €ad! ehffi of the load:

r-Val r*=/aa

_ ffi1ffi" 18.3

-

115.1

: *?5l.LAA7o

64

A

Eu. as reference

o

T-

rbc

Vn.

-

zn

4E0z-ffi" =

24.2- jr6.4 L6.41952-25.880 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

33.66t2-27.730 A

65

I.= =

I.u

-

Io.

25.t9t tLL5.49"

-

16.4195 t-Z\.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, three-phase, three-wire, star-connected load is supplied from a balanced, three-phase 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 = 721-75" 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 vubl-30"

vcaz-rs}" vcl-Izgo

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",3901-720

(43)(24.8236") (J3)(722-7s) (J:xra.6ze+.) =1rl_!1 24.8236"' I2l-75"' 18.6zg4" = 356.5112-169.59"V Voltage in each phase of the load:

Vu,

=

=

Vu-V.n 3Bo

lo" -

J3

3s6.6rrt-169.59.

= 573.76426.45"V %, = V.-Vrn 380 z -I20" _ 356.6Lr2_169.s9" = _-_E-

= 27L,7882-27.52V Current in each phase of the load: Vu,

=

Iu.Zu

= (Iu)(24.8t36.) Iu = 23.1362-29.55

573.764t6.45.

A

67

V., = Ir.Z" 27I.7BBZ-2752 = (IcX1B.6ZB4")

I.=

[4.6L2Z-L11.52' 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 delta-connected load is connected to a 380-V, phase positive sequence. The load impedances are the 50-Hz supply with a following:

2.L3.t An unbalanced, three-phase,

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:

r-Vao -

raD

=laa

-

36010" 11.45 + j14.44

=

= 19.5352-51.59. I5.

= =

A

-Ss-

lb, 3@t-L20"

17.53 + j24.22

= L2.O4lZ-l74.l

A

T-Vca 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=I36-Is6

= 19.5352-51.59' - l3.76tl90.l3 = 3O.4762-7t.29" A = Iu. -

Iub

= I2.04IL-I74.L - 19.5351-51.59' = 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, star-connected 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 star-connected load

77

Refer to Figure 2.51. The unbalanced, three-phase, four-wire load is connected to a symmetrical, three-phase 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

=

4401-90'

(

J5 X1.,Xs6.6236.6')

I"n

6.94'-Z-126.60 A

Vun

Iun'Zon

(.''6 Xro.X+s.sz-4s.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 three-phase system. The load is star-connected to a symmetrical, three-phase, delta-connected 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 = 18.6l-56'A = 2L.726I" {l

Two wattmeters are connected to an unbalanced, three-phase, star-delta 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" = 30.34t-90.90 0

The load is connected to a 360-V, 50-Hz 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 three-phase 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, 480-V, three-phase 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 star-connected 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 66-kv, transmission rine transmi"..iif,.,IfTtrJff*1"$"j1fi1"":r#t:.ff,i'J:il the year as per-unit 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 R0.71lkw-h and, The copper conductor Percentage system except that all quantities R52,40/kg. rne coppli r.,ur-u a"niitv - -' o-1S 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 all-day efficiencies and the all-year efficiency for each of the following 240-kVA transformers:

Initial cost

Transformer

Iron loss

A

1.2 kW

4.8

KW

R21 000

B

2.5 kW

3.6 kW

R1B 000

Full-load copper loss

5.3 Calculate the total annual running cost of each transformer. The cost of electrical energy is R0.72/kW-h and the annual interest and depreciation charges are 10olo. 5.4 Calculate the annual lost factor for each transformer. 6. A 66-kV, transmission line transmits a three-phase, 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 R0.7llkw-h and. The copper conductor has a cross-sectional 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 sub-station transformer supplies 1.2 MW at a power factor of 0.707I lagging.

7.L

Calculate the kVA'r rating of loss-free 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 (kW-h) charge per month: The first 200 kw-h per kVA of maximum demand in the month The next 200 kw-h per kVA of maximum demand in the month Additional kW-h supplied

69c/kW-h

6lc/kW-h 52clkW-h

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 loss-free 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 loss-free 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 three-phase, 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 per-phase equivalent circuit takes advantage of the symmetry inherent in balanced three-phase circuits, The per-unit 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

PER-UNTTQUANTTTTES The per-unit 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 single-phase 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 per-unit conversion. The per-unit system has the distinct advantage that, with it, all basic circuit relations apply. Suauut

=

Spu.Snu

The per-unit 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 per-unit methods of calculation are simpler than the use of actual volts, amperes and ohms.

6.4

DISADVANTAGES OF THE PER.UNIT SYSTEM The per-unit 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 un-scaled circuit, vanish in the per-unit circuit. Some equations that hold in the un-scaled 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 per-unit 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 per-unit 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 per-unit 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

THREE-PHASE EQUTPMENT Since three-phase 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 single-phase 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, three-phase system, As a result of this, if the system is balanced, the per-unit value of a line to neutral voltage on the line to neutral voltage base is equal to the per-unit value of the line to line voltage at the same point on the line to line voltage base. Similarly, the three-phase kVA is three times the kVA per phase and the base value of the three-phase kVA is three times the per-phase value of the base kVA. The per-unit value of the three-phase kVA on the three-phase kVA base is therefore identical to the per-unit value of the kVA per phase, on the kVA per phase base. The impedance of three-phase equipment is always given as per-phase 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 PER-UNrT 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 per-unit 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 per-unit calculations is realised by the proper selection of different bases for circuits conhected to each other through transformers. To achieve the advantage in a single-phase system, the voltage bases for the circuits connected through transformers must have the same ratio as the turns-ratio of the transformer, With such a selection of voltage bases and the same kVA base, the per-unit 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 per-unit values, all the per-unit 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 single-phase or three-phase circuits. In the case of three-phase, line-to-line 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 per-unit 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 per-unit 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 24-kVA, 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 per-unit impedance of a transformer is the same when referred from one winding to the other. If the per-unit 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: Single-line 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: SinEle-line 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_ z-pg(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.2L6Z-36.87' 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.752-36.87'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)

2-36.87')(0.07 t90.)+(0.7 5 1-36.87 ")(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: Single-line diagram for the radial transmission system of Example 6.6

Ptoad

60

x

106 Itoao

178

J3

.V6u6.I6u6,cos $

"6 X:: t295.96Z-35.9" 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.6L7l-35.9o 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

t-3

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 t-35.9') = O.tLtLZS4.l'pu

Since

the load is used as reference, the per-unit 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: Single-line diagram for the radial transmission system of Example 6.7

180

Ptoad

t

=

J3

,V1ou6'I6u6.cos $

= ( J3 )(11 x 103XIbadX0,7B6) Iroad = 2403'9571-38'19' A

J6 x 106

+!ttt-

rbase=

J3.Vbase

=-

= rrpu -

27.5 xL06

(J:Xtt

103 ) " 1443.376 A Iload r base

Z-38.L9" = 2403.957 rqa376

= Transformer

2:

Line:

Zpulnb)

7-ou

= O,!2l9O

= = =

Transformer

1:

Zpulnb)

I.6662-38.19o 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 per-unit voltage across the load is given as 0.94210 pu.

181

Vpu(sen)

Vlgen)

=

Vpu(load)

=

0.94220" + (0.L2t90,)(t.666t-38. 19") + (0.2985 tB 1, 7B.X 1. 66 6 Z.-38. L9.) + (0. L}r t90")(L666 t-38. 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 full-load 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: Single-line diagram for the radial transmission system of Example 6.8

Ptoad

48 x 106 Iroad

= J3 ,V1ou6.I1ou6.cos

= (J5)(33 x 103XIbadX0.B91)

= 942.5L71-27"

r-ud\e -

Sb.r.

J3.vor.. 120 x 106

=

TB2

$

2099.455 A

A

-t

Ipu =

I toad I

b.r" 942.517

Z-27"

2099.456 =

Transformer

1:

O.44891-27" 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 t-27 o) + (0. 105290)(0.4489 l-27o)

+

Vpu(rine)

- O.lg9LZ48.5o pu =

Ipg.Zpultine)

= (0.4489 Z-27.)( 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 voltage-base of 273 kV and a kVA-base 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: Single-line diagram for the radial transmission system of Example 6.9

Pload =

36x106= T-

J3 (

Vrouo Ilou6 CoS $1s66

J: Xrr x lo3XIbadXo.TBB)

2397.8552-38'A

rload

Tlbase -

Snu

J5.vno 210 x 106

(J3X11"103)

tto22,t4L5

A

Itoaa

T-

Ibur"

= =

LM

?397.B5st-38" 17022.1415

O.2L75Z-38. 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

=

[F-J*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.(t-ine1)

*

Vp.u.gz)

*

Vp.u.(t-ine2)

*

Vp.u.63)

(0.217 5 z-38" )(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: Single-line 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: Single-line 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: Single-line 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: Single-line 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: Single-line 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: Single-line diagram for the radial transmission system of Question 6

1BB

l

7.

Refer

to the three-phase 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: Single-line diagram for the three-phase 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: Single-line 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: Single-line 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'...,,

9.712t-21.4" A 5..3ea;-rsz.tp n

4.1

9.2s91-76.73 4

4758.114 W

12.5551169.42 A

12.22t33.28 4.2 5.1

"

5.5091-158.04" 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 1.5172-t2.62' 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?

76.2382-tt7.7 9.7r2t-51.4. A

7,1

5.484t177.9" 7.2 8.1 8.2

A

796.262t-rr2.55

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.

6.35352-83.62" A

0

18.

9376.598 W

29.543t-722,55a

A

19.

27.246/L20.97" A 21.A2rz-25.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 47.6462-13;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

17.7752-76.38 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.5522-9s.29

4308,114 W

, .

4955.35

11.1 11.2 11.3 12.1 12.2 12.3

A

16.2552-t.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

228.3381-93.63" V 15.2382-87.2: A

22.8791-77.405. A

s.2, ,,:,

999.986 W 22.372]y.i$

10.6992-20.8050 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.t-L5.78 A

,

EXERCISE 3 (CONTINUED) 75133" V 10.09.

EXERCISE 5.1 5.2

v

201.259t-109.91'V 201.259-z130.09' 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

29.392t-147.51 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 31.501r-103.06' A r4.L66t-123.925' 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

840.7892-93.05" V o.

4 (CONTINUED)

120.343 MW LLV-726 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

3.208%

r7.4

560.6s245.01' MVA

16.3

86.831 MW

-25.721 lt'lvA? 3.2

v9.801229.56"

J.J

3.4

363.6445t48.41" MVA 0.2028 lagging 3.074 MW

4,L 4.2

14.4725 MVt

MVA

'

:

393.872t42.99 MV{

7j3

265

EXER.CTSE4 (CONTTNUED) LV.7

0.6945 lagging

L'1:,:::=,

,

''.;;:;;1;;., .,

0.7315 lagging 77:2

6,2340/0

352.217240:52? MYE:

19,1 19,2

..:-:.-::.:.:,.=..,,"

:.::,=.'':'

382.3Kw,''',

::

2"r ,:,.:.:-

IrIUA

R243140.54

4.

0;,7823 crn?

qt

5.2

:.

R248327.44

2.1 7.2

.

3.1 :, ,,,,, :::::: :

5;r'.. ::::=:::: ,,.._:

7J E_2

J.L =t

1,331 0.9733

0.97 0.948 o.9167 0.3599 cm2 0.6769 cm

4.2

::.,

,

.1j5.364 i;_, 182.729 V :

170.802V 3116.11321.385" V

0.11

5.414s 833.123 821.12052*590 kVA

kVA'r

R341345.00

Rl1473,00:' 0.9614 lagging 227.708 kVA?

an579.33:.'

795 31_380 kv 5-tri3j_120kY

* -*l f{-E4{! LJITS

sf; qf7

s-g?3c tr13-4kl:J1e

41 .is-E-F 1Y

2842.65824.26V 3173,4M2L.O8V 9042.838 W

5.

a.&97_4,4r kY

]1E6

210,437 V a::,204,947, V

2948.781/?.LLV

€-#{17.560kv

*_

:,:a,a..:

221,.63v

:

4.L

ErmExsE6

7".

2L4.96f.V,,. 92.6345 A 30,6345 A 77;3655 A 209,3655 A

.' :

:

4.7185 years

7-

,:

3.3, ,:::,::::,:=184;d98 V .::

0.9736

*1 6:

:::.:) ) 2L6,362:ll:-

I --.:::219.801 V

0.394

0;9746

o,t

v

213;061 V l:213;154 V

2.:2=.....:: zfiAA3V :t,

559.7832-14.79e kVA R5738.28 -t=,.;:,,.-

2r6.ore

, ::,:,:='::: t,, ',':::::, -

EXERCISE 5

pt

228.27 V. 231.873 V:

:'::

542:006239.205' MVA

1g

236.672V

59.615 MVA1

.:*

551571 A :,, 80,429 A

:,',lQfi,

=':,:=:

L.7

S:;W7-i

v29.429/4L.3'

1.1 11

'

,

18.1 18.2 18.3

2,978o/o.

EXERCISET

kv

1U76.561:{-0,01" V 1064.710:25e

,

.

6=

V

:::

1071,8410;18'V ,, 8.588/-179.39: A,,, 59:3181*40.84o A,l

Y

::,

7.1

47:;$$$2:9.4o

7.2

28i.3792-r7;29e Y'