Kothari D.P., Nagrath I.J. Modern Power Systems Analysis

Transmission

line:

2o.5x3o

GT2

Transform

er

T,:

-l-!,?

I

l_0_

=

0.564

=

0.418

Flepresentation

of

power

System

eomponents

I

iliffij

I-

Example

4.I,

we now

calculate

the pu

values

of the

reactances

of

transfonners

and

generators

as

per

relation

(4.10):

Transformer

Z

:

0.209 x

Transformer

Tt:

Generator

1:

Generator

2:

Generator

3:

16x30

aiT

=

0.44

1.6

x

30

u

t7-

=

o'396

1.2

x

30

tazr

=

o'936

0.56

x

30

AZI-

=

0'437

Transformet

Tr:

Generator

1:

0.22 x

ff

=0.++

0.43s

x

(10'5i1

=0.3e6

(l

l)'

Generator2:

0.413 r*

*(6'612-

=0.936

I)

6.2)'

Generator

3:

0.3214 *

i9

x

(6'6)1

=

0.431

/.r

6.21'

obviously

these values

are

the

same

as

obtained

already

in

Exampre

4.r.

4.5

Complex

Power

Consider

a single-phase

load

fed from

a source

as

in

Fig.

4.9.

Let

v

-tvt

16

r

_tn

t

(6_

0)

The

reactance

diagram

of

the

system

is

shown

in

Fig.

4.g.

{-)frL_/-X-fX-)<1

JU

-_"

U000._--J--

64

0.44

I

i

Fig.

4.9

Reactance

criagram

of

the

system

of

Fig.

4.5 (roads

neglected)

Et'

Ez

and

E,

are

per

unit

values

of

voltages

to

which

the

generators

are

iJ"',ltl3;3llllrlt""

in

a

short

circuit

study,

these

wil

be

raken

ui

t

/.,"pu

(no

volta_ee

of

I

kV

in

the

circuit

of

generator

the

circuit

of generators

2

and

3

";

,r;

i;

Example

4.2

The

reactancc

data

of

gencrators

and

transtbrmers

are

usually

specified

in

pu

(or

per

cent)

values,

based

on

equiprnent

ratings

rather

than

in

actual

ohmic

values

as given

in

Exampl

e

4.7;

*iit"

rhe

transmirr;;;

hne

irnpedances

nray

be

;"";:,[J:Ti]#?l-et

us

rc-sotve

rJxarnpre

4.1

b;

assuming

rhe

rbuowing

Transformer

T,:

0.209

Transformer

T):

0.220

Generator

Gr:

0.435

Generator

Gr:

0.413

Generator

G3:

0.3214

With

a

base

MVA

of

30.

base

I

md

b:ise

voltage

of

6.1

k\;

in

Source

Fig. 4.9

Complex

power

flow

in

a single-phase

load

When

d

is

positive,

the

current

lags

behind voltage.

This

is

a

convenient

choice

of

sign

of

0 in power

systems

where

loads

have

mostly

lagging

power

factors.

Complex power

flow in

the

direction

of current

indicated

is

siven

bv

S=VI*

=lVllll

l0

=

lVl

l1l cos

d+

jlvl

l1l sin

0= P

+

ie

@.17)

or

(a)

tSl

=(p2+e\t,,

Modern

Power

system

Anatysis

I

Here

S

-

complex

power (VA,

kVA,

MVA)

lSl

=

apparent

power

(VA,

kVA,

MVA);

it

signifies

rating

of

I

P

=

lVl

l1l

cos

0

-

real

(active)

power

(watts,

kW,

MW)

Q

=

lVl

l1l

sin

0

=

reactive

power

-

voltamperes

reactive (VAR)

=

kilovoltamperes

reactive

(kVAR)

=

megavoltamperes

reactive

(MVAR)

It irnmediately

follows

from

Eq.

(4.17)

that

Q,

the

reactive

power,

is positive

for

lagging

current (lagging

power

factor

load)

and

negative

for

leading

cunent

(leading

power

factor

load).

With

the

direction

of current

indicated

in Fig.

4.9,

.9

=

P +

iQ

is

supplied

by

the

source

and

is absorbed

by

the

load.

Eqr-ration

(4.17)

can

be

represented

by

the

phasor

diagram

of

Fig.4.10

where

,n

0

=

Lan''

i1

=

positive

for

lagging

current

P

:gative

fbr

leading

current

nepresentation

ot

power

system

eompqlents

l.i; ffii

T-

As

per

Eq.(4.19),

Kirchhoff's

current

law

applies

to complex

power

(also

applies

separately

to real

and

reactive

powers).

In

a series

RL

load

carrying

current

{

V=l(R+jxr)

P

=

I"R

=

active power

absorbed

by load

Q

-

IzXr

=

r-eactive

power

absorbed

by

load

In case

of

a series

RC

load

carrying

current

I

P

_I2R

O

-

IzX,

qreactive

power

absorbed

is

negative)

Consider

now

a balanced

three-phase

load

represented

in

the

form

of

an

equivalent

star as

shown

in

Fig.

4.L2.

The

three-phase

complex

power

fed into

load

is

given

by

S

=

3vpl-t

=

3

lvpt

l6pl;

:

JT

lvrl

zOrti

(4.20)

If

Ir

=llil

I

(6p-

A

Then

.S

=

,'5

lvLl

lILl

I

0

-

Ji

tvLt

vLt

cos

d +

iJT

t

0=P+iQ@.21)

Here

Fig.

4.12

Complex

power

fed

to

three-phase

load

tsl

=

Ji

tvrt

ttrl

P

-

Ji

tvLl

tILt

cos

d

e

=

Ji

lvLl

tILt

sin

d

-

power

factor

angle

lf

vL,

the

line

voltage,

is

expressed

in

kv;

and Iy,the

line

current

in

amperes,

s is in

kvA;

and if

the

line

current

is

in

kiloamperes,

s is

in

MvA.

(4.18)

(4.1e)

Fig.

4.10

Phasor

representation

of

complex

power

(lagging

pf

load)

If two

(or

more)

loads

are

in

parallel

as

in

Fig.

4.ll

s=vF_v(i+i)

f

Yri..r'r'!-

(pr+

pr)

+

j(et+

ez)

In terms

of load impedance

Z,

f

,

_

v,

_lvLll6P

rL-

z

Jiz

Substituting for I,

in Eq.

(4.20)

"

-

V'l'

r-

r

tr V,. is in kV, ,S is now

given

in MVA. Load

calculated

from

,_lvrl'_

lvl

"

-

Jt-

-T1O

4.6 SYNCHRONOUSMACHINE

The

synchronous machine is the

most important element

of a

power

system. It

converts

mechanical

power

into electrical form

and feeds it into the

power

network or, in the case of a motor,

it draws electrical

power

from

the network

and converts it into the mechanical

form. The machine excitation which is

controllable determines

the flow of VARs into

or out of the machine. Books on

electrical

machines

11-51

may be

consulted for a detailed account of the

synchronous machine.

We shall

present

here a simplified circuit

model of the

machine

which with

suitable modifications wherever

necessary

(under

transient

conditions) will be adopted throughout

this book.

Figure 4. 13

shows the schematic cross-sectional

diagram of a three-phase

synchronous

generator

(alternator)

having a two

pole

structure. The stator has

a balanced

three-phase

winding-aat,

bbt and

cct. The winding shown is a

concentrated one, while the

winding

in

an actual

machine is distributed across

the stator

periphery.

The rotor shown is a cylindrical"

one

(round

rotor or non-

salient

pole

rotor) with rotor winding

excited by the DC

source.

The rotor

winding is so arranged

on

rotor

periphery

that the field excitation

produces

nearly

sinusoidally

distributed flux/pole

(d)

in the air

gap.

As the rotor rotates,

three-phase emfs are

produced

in stator

winding.

Since the machine is a

balanced one and balanced loading will be considered,

it

can

be modelled on

per phase

basis for the reference

phase

a.

Tn o mqnhinc rrrifh mnra fhqn frrrn nnlec fhp

qlrnrre

ApfinpA cfnrnfrrro ronpafc

v

l/vrvo,

Lllv

quv

v v svrlllvu JLr uvLur v rvyvolo

electrically for every

pair

of

poles.

The frequency of induced emf is

given

by

f

=ffi

nz

where

.

High-speed

turbo-generators have cylindrical rotors and Iow spped hydro-generators

have salient

pole

rotors.

=

rotor speed

(synchronous

speed) in rpm

=

number of

poles

winding

Fig. 4.13 Schematic

diagram of a round rotor

synchronous

generator

On

no load the

voltage EJ induced in the reference

phase

a lags

90" behind

dywhich

produces

it and is

proportional

to

dyif

the magnetic

circuit is assumed

to be unsaturated.

This

phasor relationship is

indicated in Fig. 4.14.

Obviously

the terminal vclltage

V,

=

Er

I

l

-

Ef=Vt

Fig.4.14

Phasor relationship

between

fuand

E,

As

balanced steady

load is drawn from

the

three-phase stator winding,

the

stator currents

produce synchronously rotating flux

Q/poIe

(in

the direction

of

rotation

of the rotor).

This flux, called

armature

reaction

flux, is therefore

stationary

with respect

to field flux

Qy.It

intuitively fbllows

that

Qo

is in

phase

with

phase

c current

1o which causes it. Since the magnetic

circuit has

been

N

P

(4.22a)

impedance Z if required can be

(4.22b)

Qr

Field winding

-..

1)xr

\Xp>{

o

--F

vl

I

-T-

:t

-F

I

lot

-t-

I

\,

'.

\

'\r

\lQ

NN

Phasor

diagram

and

voltages

as

,liO*

|

Modern

powsr

Syglem_Anatygis

urrurnJd

to

be unsaturated,

the superposition

principle

is

applicable

so

that

the

resultant

air

gap

flux

is

given

by

the phasor

sum

d'

=

d1+

Q,,

@,3)

Further

assuming

that

the

armature

leakage

reactance

and resistance

are

re

eml

whtch

equals

the

termin

tage

V,.

Rapr"."n,",ion

of

po*"r'

Syrr"r

Cornp.on"n,,

NEffi

t-

The circuit

of Fig.

4.L6

can be

easily

modified

to include

the

effect

of

armature

leakage

reactance

and

resistance

(these

are

series

effects)

to

give

the

complete

circuit

model

of the

synchronous generator

as

in Fig.

4.I7.

The total

+xl=Xslsc

synchronous

reactance

of the

machine.

Equation

(4.24)

now

becomes

V,

=

Et

-

jlo

X,

-

IoRa

Fig.

4.16

This model

of

the synchronous

machine

can

be

further

modified

to

account

for the effect

of

magnetic

saturation

where

the

principle

of super-position

does

not

hold.

Fig- 4.17

circuit model

of round

rotor

synchronous generator

Armature resistance

Rn

is invariably

neglected

in

power

system

studies.

Therefore,

in

the

place

of the circuit

model

of Fig.

4.I7,

the

simplified

circuit

model of Fig. 4.18

will

be used

throughout

this

book.

The

corresponding

phasor

diagram is

given

in

Fig. 4.i9.

The

fieici

induceci

emi

Ey ieacis

the

terminal

voltage

by the

torque

(load)

angle

d

This,

in fact,

is

the

-condition

for acrive

power

to flow

out of the

generator.

The

magnitude

of

power

delivered

depends

upon sin

d

In the motoring

operation of a

synchronous

machine,

the

current

1,,

reverses

as shown in Fig. 4.20,

so that Eq.

@.25)

modifies

ro

Ef

=

V,

-

jIoX,

(4.26)

which is represented

by the

phasor

diagram

of Fig.

4.2I.It

may

be noted

that

V,

now leads

lby

d, This in

fact is the

condition

for

power

to flow

into motor

terminals.

under

loaded

(balanced)

conditions

showing

fluxes,

currents

phasors

is

drawn

in

Fig.

4.15.

(4.2s)

jlX"=

-

E"

Fig.

4.1S

phasor

diagram

of

synchronous generator

Here

d

-

power

factor

angle

6

-

angle

by which

Et leads

v,

called

load

angre

or torque

angle

We

shall

see in

Sec.5.10

that

dmainly

determines

the

power

delivered

by

the

generator

and

the

magnitude

of E,

(i.e.

excitation)

determines

the

VARs

delivered

by

it.

Because

of the

assumed

linearity

of

the

magnetic

circuit,

voltage

phasor

E,

Eo and

v, are

proportional

to

flux phasors

dr,

doand

d,

respectively;

furthei,

voltage

phasors

lag

90' behind

flux phasori.

It therefore

easily

follows

from

Fig.4.15

that phasor

AB

=-

Eois proportionalto

(o

(and

therefore

Io) and

is

90' leadin

g

d"

@r

1,).

With

the

direction

of

phasor

AB indicated

on the

diagram

AB

=

jlo

Xo

where

X"

ir the

constant

of

propotionality.

In terrnc nF thp

qlrnrro

rlofi-iri^- ^s v li-^^rr-- ---^rr- 4t-- r rt

quvvv

uvruulrurr

vL

,lra,

wE

ualt

urr€utly

wl-l[c)

ule l0ilowlng

expression

for voltages

without

the

need

of

invoking

flux

phasors.

V,

=

Ef

-

jloXo

(4.24)

where

Ef

=

uolrage

induced

by field

flux

Q,

alone

=

uo

load

emf

The

circuit

model

of

Eq.

(4.24)

is drawn

in

Fig.

4.16

wherein

X, is

interpreted

as

inductive

reactance

which

accounts

for

the effect

of armature

reaction

thereby avoiding

the

need

of

resorting

to

addition

of fluxes

l&q.@.23)1.

&

the'i

I

Modern

Power

System

Analysis

The

flow

of

reactive

power

ancl

terminal

voltage

of

a synchronous

machine

is

mainly

controllecl

by

means

of

its

excitation.

This

is

discussed

in

detail

in

Section

5.10.

Voltage

and

reactive

power

flow

are

often

automatically

regulated

uy

YUIL4S\/

IvSurqlvro

\uvv

vvve^v^r

vr

and

by

automatic

tap

changing

devices

on

transformers.

-

Representation

of

Power

System

Components

FIIS':

lvtl

ll,,l

cos

d

=

constant

=

active

power

output

Fig.4.22Synchronousmachineconnectedtoinfinitebus

It

rneans

that

since

lV,l

is

fixed,

the

projection

l/ol

cos

dof

the

phasor

Io

on

V'

remains

constant,

whiie

the

excitation

is

varied'

Phasor

diagrams

corresponding

to

high,

medium

and

low

excitations

are

presented

in

Fig'

4'23'

T\e

phasor

diagram

of

Fig.

4.23(b)colresponds

to

the

unity

power

factor

case'

It

is

obvious

frot

the

phasor

diagram

that

for

this

excitation

lEJl

cos

5=lV)

E1

-l

it"x'

---"'

/

0

(a)

Overexcited

c

I

|

--

-'-/

6666

lx"

I

(;

I

t

4.18

Simplified

round

roto

generator

E1

Fig.

ircuit

model

of

synchronous

Fig.

4.20

Motoring

oPeration

of

sYnchronous

machine

Ia

(b)Normal

excitation

(c)

Underexcited

Phasordiagramscfsynchronousgeneratorfeedingconstant

power

as

excitation

is

varied

jIJ'

Ef'

1

I

J

Vt',

Fig.

4.21

Phasor

diagram

of

motoring

oPeration

Normally,

a synchronous

generator

operates

in

parallel

with

other

generators

connected

to

the

power

system.

For

simplicity

of

operation

we shall

consider

a

generator

connected

to

an

inJinite

bus

as shown

in

Fig'

4'22'

As

infinite

bus

means

a large

system

whose

voltage

and

frequency

remain

constant

independent

of

the

power

exchange

between

the

synchronous

machine

and

the

bus'

and

independent

of

the

excitation

of

the

synchronous

machine.

consider

now

a synchronous

generator

feeding

constant

active

power

into

an

infinite

b's

bar.

As

the

machine

excitation

is

varied,

armature

current

In

and its

angle

g,

t.e.

power

factor,

change

in

such

a manner

as to

keep

Fi1.4.23

This

is

defined

as

normal

excittttittrt

Forthe

ovetexciterl

case

(Fig'

a'23a)'

i'e'

|,8,.1

cos

6>|v),1,

lags

behind

V,

so

thatthe

generator

feeds

positive

reactive

powerintothebus(ordrawsnegativereactivepowerfromthebus)'Forthe

Fig.

4.19

Phasor

diagram

of

synchro-

nous

generator

.,

I"

;iiii

I

Modern Power

system Analysis

t

underexcited case

(Fig.

4.23c),

i.e. lErl

cos 6 <

lV),

1o

leads V,

so that the

generator

feeds

negative reactive power

into the bus

(or

draws positive

reactive

Figure 4.24

shows

the overexcited

and underexcited

cases

of synchronous

motor

(connected

to infinite

bus) with constant power

drawn

from the

infinite

bus. In

the overexcited

case, Io leads

Vu i.e. the

motor draws

negative reactive

power

(or

supplies

positive

reactive

power);

while

in the underexcited

case

.Io

lags V, r.e. the motor

draws

positive

reactive power

(or

supplies negative

reactive

power).

E1

(a)

Overexcited

V1

--

(b)

Underexcited

Fig.4.24

Phasor

diagrams of synchronous

motor

drawing constant

power

as

excitation is varied

From the above

discussion we can

draw the

general

conclusion that

a

synchronous

machine

(generating

or motoring) while

operating at

constant

power

supplies

positive

reactive

power

into

the bus bar

(or

draws

negative

rcqettvc nrtrx/cr frnrn fhc hrrc hqr\ rrrhan nrrcrcvnifarl An rrnrlarownifprl mqnl"i-o

I

v

vv

vsU Yv rrvrr

v

Y

vlvl\vlLvv. / lll ulluvtvnvltvu

lll4vllltlv

on

the

oih.. hand, feeds negative

reactive

power

into the bus bar

(or

draws

positive

reactive

power

from the bus

bar).

Consider now the

power

delivered by a

synchronous

generator

to an infinite

bus.

From Fig. 4.19

this

power

is

P

=

lVtl llol cos 0

The above

expression.can be written in

a more useful form

from the

phasor

geometry.

From Fig. 4.19

lnA

_

rl,lx,

sin

(90"

+

0)

-

sin 6

^ #.......'@,^

^

lL^

"

j

a

^

{ -

A l4

^

li^/\^^

AaAu

t

_

^

-

Representation

of

Power System

Components

I

iilis'.$

T.._

or

ln,l

l1,l

cos

e:

E:

sin

6

(4.27)

(4.2e)

(4.28)

The

plot

of

P versus

{

shown

in Fig. 4.25,

is called

the

power

angle curve.

The maximum

power that

can be

delivered

occurs at 6

=

90"

and is

given

by

For

P

) P** or for

6> 90'

the

generator falls out of

step. This

problem

(the

stability)

will be

discussed

at length

in Chapter

12.

Fig.

4.25

Power

angle curve

of a synchrcnous

generator

Power

Factor

and

Power Control

While

Figs

4.23 and

4.24

illustrate how

a synchronous machine

power

factor

changes

with

excitation

for fixed

power

exchange,

these

do not

give

us a clue

regarding

the

quantitative

values of llnl

and d This

can easily

be

accomplished

by recognizing

from Eq.

(4.27)

that

lEll

sin

6

-llolX,

cos d

PX"

=

#

=

constant

(for

constant

exchange of

power

to

lyrl

,

infinite bus

bar)

(4.30)

Figure

4.26

shows

the

phasor diagram

for a

generator

delivering

constant

power to infinite

bus

but

with varying

excitation.

As lEtl sin dremains constant,

the

tip of

phasor

Ermoves

along

a line

parallel

to

y,

as excitation is varied.

The

direction

of

phasor 1o is

always 90o

lagging

jI"X,

and

its magnitude is

obtained

from

(l1olX5)/X5.

Figurc

4.27 shows

the

case of limiting excitation

with

d= 90".

For

excitation

lower

than

this

value the

generatff

becomes unstable.

Modern

Power System

Analysis

Ftepreseniation

of

Fqwer System

Components

|

1!1

Salient

Pole Synctrronous

Generator

A salient

pole

from

a round

synchronous

machine,

as shown

in Fig.

rotor

machine

by

constructional

features

4.29,

is distinguished

of

field

poles

which

/.{1

\3

Iaz

Effect

of

varying

excitation

of

generator

delivering

constant

power

to infinite

bus

bar

employed

in machines

coupled

to hydroelectric

turbines

which are inherently

slow-speed

ones

so

that

ttre synchronous

machine

has

rnultiple

pole

pairs

as

different

from

machines

coupled

to

high-speed

steam

turbines

(3,000/1,500

rpm)

which

have

a

two-

or four-pole

structure.

Salient

pole

machine

analysis

is

made

through

the

two-reaction

theory

outlined

below.

Direct axis

I

'l

Fig.

4.29 Sallent

pole

synchronous

machine

(4-pole

structure)

In

a

round

rotor

machine,

armatLlre

current in

phase with field

induced

emf

Ey

or

in

quadrature

(at

90")

to

S,

produces

the

same

flux

linkages

per

arnpere

ai

the air

gap

is

uniform

so that

the

armature

reaction

reactance

offered to

in-

phase or

quadrature

current

is the

same

(X,,

+ X1

=

Xr), In

a salient

pole

l-,- --1-- -^^-:-L^--. ft i^ +L^ l^^^+ ^l^-^ +L^

machrne

at

gap

ls

non-unllorTn

arong

IULOI'ljcrlPilury.

rL ls Lllc rtrilsL .lruug trrtr

axis

of

main

poles

(called

direct

axis)

and

is the

largest

along

the

axis

of the

interpolar

region

(called quadrature

oxis). Armature

current

in

quadrature with

El

produces

flux

along

the direct

axis

and

the reluctance

of

flux

path

being low

(because

of

small

air

gap), it

produces

larger

flux

linkages

per

ampere and

hence

the

machine

presents

larger

armature

reaction

reactance

X,

(called

direct

axis

reactance)

to the

flow

of

quadrature

component

Il of

armature current

1o.

On

the

other

hand,

armature

current

in

phase with

{

produces

flux along the

quadrature

axis

and

the

reluctance

of the

flux

path

being

high

(because

of large

62

Fig.

4.26

/,"

V1

Fig.

4.27 Case

of limiting

excitation

of

generator

delivering

constant

power

to infinite

bus bar

Similar

phasor diagrams

can

be

drawn

for synchronous

motor

as

well

for

constant

input

power

(or

constant

load

if copper

and

iron

losses

are neglected

and

mechanical

ioss is

combined

with load).

Another

important

operating

condition

is variable

power and

fixed

excita-

tion.

In this

case lV,l

and lE1tr

are

fixed,

while d

and

active

power vary in

accordance

with Eq.

(a.28).

The

corresponding

phasor diagram

for

two values

of

d is

shown

in

Fig.

4.28. It is

seen from

this

diagranr

that

as d

increases,

current

magnitude

increases

and

power

t'actor

improves.

lt

will be shtlwn

in

Section

5.10

that as dchanges,

there

is no

significant

change

in the

flow of

reactive

Power'

Locus

of Er

-

Er',

--4,

,--' l\---jluzX"

,

n'

l/-

Operation

of

synchronous

generator

with variable

power

and

fixed

excitation

Fig.

4.28

4

r

i5

I

I rl,l-r-

ffil

Mod"rn

Po*r.

syrt"t

Rn"tyri,

interpolar air

gap),

it

produces

smaller flux linkages

per

ampere

and hence

the

machine

presents smaller armature reaction reactance

Xu

(guadrature

axis

reactance a X) to the flow of inphase component Io

of armature current

/o.

Since a salient

pole

machine offers different reactances to the flow of Il and

1o

cornponents of armature

current Io, a circuit model cannot be drawn.

The

phasor

diagram of a

salient

pole generator

is shown in Fig. 4.30.It can be easily

drawn by following the steps

given

below:

Fig. 4.30 Phasor diagram of salient

pole

synchronous

generator

1. Draw

%

*d Io at

angle 0

2. Draw

IoRo. Draw CQ

=

.il,X,t(L

to 1,,)

3.

Make lCPl

-

llol

Xq and draw

the line OP which

gives

the direction of

Ey

phasor

4. Draw

a I from

Q

to the extended

line OP such that OA

=

Ef

It can be shown

by the above theory that the

power

output

of a salient

pole

generator

is

given

by

lv,l'

(xo

-

xn)

sin 26

(4.31)

2XdXq

The first term

is the same as for a round

rotor machine

with X,

=

Xa and

constitutes

the major

part

in

power

transfer. The second term

is

quite

small

(about

I0-20Vo)

compared

to the first term and is known

as reluctance

power.

P

versus

d

is

plotted

in Fig. 4.31. It

is noticed that

the maximum

power

output occurs

at 6

<

90'

(about

70').

Furt1t"r

34

(change

in

power per

unit

'

d5'

change

in

power

angle for small

changes in

power

angle),

called the

synchronizing

power.cofficient,in the operating

region

(r<

70')

is larger in

a salient

pole

niachin.:

than in

a round rotor machine.

Representation

of

power

Resultant

Fig.

4.31

power

angre

curve

for

sarient

pore

generator

In this

book

we

shall

neglect

the

effect

of

sariency

and

take

X'=

X't

in

all

types

of

power

system

studies

considered.

During

a

machine

transient,

the

direct

axis

reactance

changes

with

time

acquiring

the

following

distinct

values

during

the

complete

transieht.

X/

=

subtransient

direct

axis

reactance

Xh

=

transient

direct

axis

reactance

X,r

=

steady

state

direct

axis

reactance

The

significance

and

use

of

these

three

values

of

direct

axis

reactance

will

be

elaborated

in

Chapter

9.

Operating

Chart

of

a

Synchronous

Generator

while

selecting

a

large

generator,

besides

rated

MVA

and power

factor,

the

greatest

allowable

stator

and

rotor

currents

must

also

be considered

as they

influence

mechanical

stresses

and

temperature

rise.

Such

timiting

parameters

in

the

operation

are

brought

out

by

means

of an

operating

chart

or-

performance

chart.

For

simplicity

of analysis,

the

saturation

effects,

saliency,

and

resistance

iue

ignored

and

an

unsaturated

value

of

synchronous

reactance

is

considered.

Consider

Fig.

4.32,

the

phasor

diagram

of

a cylindrical

rotor

machine.

The

locus

of constantllolx,V)

and

hence

MVA

is

a circle

centered

at

M.

The

locus

of

constant

lEtl

(excitation)

is

also

a

circle

centered

at O.

As

Mp

is proportional

to

MVA,QP

is

proportional

to

MVAR

and

Me

to

MW,

all

to

the

same

scale

which

is

obtained

as follows.

lv,l lE,l

,=-1;-sind+

I\,/lnrlorn Dnrrrar Qrrciam Analrrcic

rYrvu!rrr I

vYYvr v)'srvrr|,rrrqrturu

Fig.4.32

Phasor

diagram

of synchronous

generator

For

zero excitation.

i.e. lE.l

=

0

-

iIoXr'=

Y,

or

Io

=

jV,lX,

i.e.

llol

=lV)lXr

leading

at

90"

to OM

which corresponds to

VARs/phase.

Consider

now the chart shown in Fig.

4.33 which is drawn for

a synchronous

machine having Xt

=

1.43

pu.

For zero excitation, the

current is 1 .01I.43

-

0.J

pu,

so that the length

MO conesponds to

reactive

power

of 0.7

pu,

fixing both

active and reactive

power

scales.

With

centre

at

0

a number of semicircles

are drawn

with

radii

equal to

different

pu

MVA

loadings. Circles of

per

unit excitation are drawn

from centre

M with 1.0

pu

excitation

corresponding to the

fixed terminal voltage OM

.

Lines

may also

be

drawn from 0

conesponding to

various

power

factors

but for

clarity only 0.85

pf

lagging line is shown. The operational

limits

are fixed

as

fbllows.

Taking 1.0

per

unit

active

power

as the rnaximum

allowable

powel', a

horizontal limirline

ubc is drawn through

b at 1.0

pu.

It is assumed

that the

machine is rated

to

gire

1.0

per

unit active

power

at

power

factor 0.85

lagging

and this tixes

point

c.

Limitation of the stator current

to the corresponding

value

requires

the limit-line

to become a circular

arc cd about centre 0.

At

point

d the

rotor

heating

becomes more important

and the arc de

is fixed by the

maximum

excitation current

allowable, in this case

assumed to be lEtl

=

2.40

pu

(i.e.2.4

times

ly,l).

The remaining

limit is decided by

loss of synchronism

at

leading

power

factors. The

theoretical lirnit is

the line

perpendicular to MO

at M

(i.e.

d=

90o),

but in

practice

a safety margin

is brought in to

permit

a

further

small

increase

in load

belore instability.

ln Fig.

4.33, a

0.1

pu margin is

employed

and is

shown

by the curve

afg which is drawn

in the following

way.

Fig.

4.33

operating

chart

for

rarge

synchronous

generator

,

Consider

a

point

h

onthe

theoretical

limit

on the

lETl

=

1.0 pu

excitations

arc,

the pcrwer

Mh

is reduced

by

0.1

pu

to

Mk;

the

op"ruiing

point

must,

however,

still

be

on

rhe

on

the

desired

limiting

curve.

This

is

repeated

for

other

excitations

giving

the

curve

afg.

The

complete

working

area.

shown

shaded.

is

gfabcde.

,{

working

point

placed

within

this

area

at

once

defines

the

MVA,

Mw,

MVAR,

current,

power

factor

and

excitation.

The

load

angle

6 can

be

measured

as

shown

in

the

figure.

4.7

REPRESENTATION

OF

LOADS

Load

drawn

by

consutners

is

the

toughest

parameter

to

assess

scientifically.

The

magniiude

of

the

ioad,

in

iact,

changes

continuously

so

that

the

load

forecasting

problern

is

truly

a

statistical

one.

A

typical

daily

load

curve

is

shown

in

Fig'

1.1.

The

loads

are generally

composed

of

industrial

and

domestic

components.

An industrial

load

consists

mainly

of large

three-phase

induction

nlotors

with

sulficient

load

constancy

and predictable

duty

.y.lr,

whereas

the

domestic

lclad

mainly

consists

of

lighting,

heating

and

many

single_phase

devices

used

in

a

random

way

by

houscholders.

The

design

un6

upJrotion

of

power

systems

both

economically

and

electrically

are

greatly

influenced

by

ttrp

nature

and

magnitude

of

loads.

-

2.O

pu

excitation

.

0.85

pf

lagging

i\

E

a

-o

(E

C]'

(U

o

c)

F

0.7

\

I

,

tl

,

-,---

M

9o.s

Leading

0.5

1.0

>-

Reactive power

(pu)

lagging

l

N

I

Locus

11, I

X"

(circle

centre

M)

.sa I

._IaZ

I

Modern

Power

Svstem Anatr-rsic

In representation

of

loads

for

various

system

studies

such

as

load

flow

and

stability

studies,

it is

essential

to

know

the variation

of real

and

reactive

power

with

variation

of

voltage.

Normatly

in

such

studies

the load

is

of composite

nature

with

both

industrial

and

domestic

components.

A

typical

composition

of

Representation

of

p

JlfZJ.t

Sotution

Base

MV

A

=

645,

3_phase

Base

kV

=

24,

line-to_line

Load

volt

)L

ASa=

i

=

1

pu

Synchronous

reactance

X,

=

+#

=

1.344

pu

"

(24)z

-'r-

Full

load

(MVA) =

I pu,

0.9

pf

lagging

Load

current

=

generator

current

Io=

7

pu,

0.9 pf

lagging

=

0.9

-

7

0.436

pu

(a)

Excitation

emf

(see

Fig.

4.Ig)

Ef

=

V,+

j

XJ"

=110"+

j1.344

(0.9-

j0.436)

=

1.586

-

j

l.2l

=

199

137.1"

E,

(actual) =

1

.99 x

24

=

47.76

kV

(line)

6=

3j.1"

(leading)

(b)

Reactive

power

drawn

by

load

Q

=

VJ,,

sin

r/

=

1 x

I x

0.436

=

0.436

pu

or

0.436

x

M5

=

281

MVAR

The generator

of

Example

4.3

is

carrying

full

load

at

rated

voltage

but

its

excitation

emf

is

(i)

increased

by

20vo

and

(ii)

reducedby

20vo.

Calculate

in

each

case

(a)

load

pf

(b)

reactive

power

drawn

by

load

(c)

load

angle

6

Solution

Full

load,

1x0.9=0.9pu

r.99

Induction

motors

Synchronous

motors

Lighting

and

heating

55-757o

5:75Vo

20-30Vo

Though

it

is

always

better

to consider

the

P-V

and

Q-V

characteristics

of

each

of

these

loads

for

simulation,

the

analytic

treatment

would

be

very

cumbersome

and

complicated.

In

most

of the

analytical

work

one

of

the

following

three

ways

of load

representation

is used.

(i)

Constant

Power

Representation

This

is used

in

load

flow

studies.

Both

the

specified

MW

and

MVAR

are

taken

to

be

constant.

(ii)

Constant

Current

Representation

Here

the

load

current

is

given

by

Eq.

(4.17),

i.e.

I=P:iQ-tn

'

V{<

-"'l(6-0)

where

V

=

lVl

16and

0=

tan-l

QlP

is the power

factor

angle.

It is

known

as

constant

current

representation

because

the

magnitude

of current

is regarded

as

constant

in

the

.study.

(iii)

Constant

Impedance

Representation

This

is

quite

olten

used

in

stability

studies.

The

load

specified

in

MW

and

MVAR

at nominal

voltage

is

used

ro

compure

the

load

impedance

(Eq.

(4.?2b)).

Thus

I

"

z=!:-w*

-

lvl2

-t

I

-P=JO-

P-JQ:T

which

then

is

regarded

as

constant

throughout

the

study.

f l-

---

.---l

l Fvattrt li a e I

I

F^sr..lsrv

zrv

I

t--

T

A

synchronous

generator

is rated

645

MVA

,

24

kv,0.9

pf

lagging.

It has

a

syrrchronous

reactance

l.z

o.

The

generator

is

feeding

full

load-at

0.9 pf

lagging

at

rated

voltage.

Calculate:

(a)

Excitation

emf

(E1)

and power

angle

6

(b)

Reactive

power

drawn

by

the

load

Carry

out

calculations

in pu

form

and

convert

the

result

to

actual

values.

P_

Ef=

Kothari D.P., Nagrath I.J. Modern Power Systems Analysis

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