V1-5-5264202

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[1]

CHAPTER V

FURTHER CONSIDERATION OF THE TWO-MACHINE SYSTEM

Pre-calculated swing curves. This chapter continues the consider

ation of the problem of the stability of two-machine systems. The

equal-area .criterion, which was discussed in Chapter IV, is a very

effectivemeans of determining whether a given two-machine system isstable when subjected to a given disturbance. The most important

type of disturbance is the occurrence of a fault and its subsequent

clearing by the opening of circuit breakers. If the clearing time is

known, as it usually is, the corresponding clearing angle must be found

before the equal-area criterion can be applied; or, conversely, if the

critical clearing angle for a given transmitted power is obtained from

the equal-area criterion, the corresponding critical clearing time must

be found in orderthat

the results of the stability study will be in themost useful form.Perhaps the simplest way to find the clearing angle corresponding to

a given clearing time, or the time corresponding to a given angle, is to

refer to the appropriate swing curve in a set of swing curves which have

already been calculated and plotted, and which may be appropriatelycalled pre-calculated swing curves. Such a set of curves, obtained

by solving the swing equation on the M.I.T. integraph, was published

by Summers and McClure2 and is reproduced here (Figs. 1 to 10) by

kind permission of Mr. McClure.The curves were derived for a sustained fault on a system consisting

of a synchronous machine of finite size connected through reactance

to an infinite bus. By the methods presented in Chapter IV, however,

the pre-calculated curves can be used with a system of two finite ma

chines connected through any linear network. The usual simplifying

assumptions are made, to wit: constant input, no damping, and con

stant voltage behind direct-axis transient reactance. To make the

curves generally applicable they are plotted in terms of a dimensionless

variable, the modified time T, defined by eq. 11 below.The swing equation of a two-machine system is

d2aM dt2 = Pa = Pi - Pu

149

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150 THE TWO-MACHINE SYSTEM

[3]

M1M2 equivalent inertia constant in megajoule-

where M = M1 + M2 = seconds per electrical degree. [2]

o= 01 - 02 = angular displacement in electrical degrees.

t = time in seconds.

P M2Pi l - M 1Pi2 • 1 · •i = M M = equrva ent Input In megawatts

1 + 2

which, either for a reactance network or for M 2 = 00 , reduces to

Pi = P i l = input of machine 1 (the generator) [4]

The equivalent power output, dependent on a, is given by the power

angle equation:

which, for a reactance network, reduces to

P« = PM sino

[5]

[6]

Expressions for Pc, Y, and PM are given by eqs. 28, 29, and 30 of

Chapter IV. For a reactance network the amplitude of the powerangle curve is

[7]

where X 12 is the reactance connecting machines 1 and 2, including the

direct-axis transient reactances of the machines themselves.

Substitution of eq. 5 into eq. 1 gives

d28

M dt2 = Pi - Po - PM sin (a.- 7)

or

where

p/ = Pi - Po

0' = 0 - 'Y

[8]

[9]

[10]

[11]

To put eq. 8 into dimensionless form, divide it by PM and then introduce a quantity r, defined by

r;;:P; /7tJPMT = t\)180 M = t\) GH

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PRE-CALCULATED SWING CURVES 151

[13]

[12]

where f = frequency in cycles per second.

GH = kinetic energy, in megajoules, of equivalent generator at

rated speed = 180fM.

The result is:

d P ,r u i .

180' dT2 = PM - sin 8 = P - sin 8

if a is in electrical degrees, or simply

u •p SIDu

d,,2

if a is in electrical radians. Here

p/ Pi - Pcp = PM = PM [14]

and 0' has been defined in eq. 10.

A differential equation has been obtained which is independent of the

inertia constants of the machines and of the constants of the network.

The solution of the equation depends on the ratio of the input to the

amplitude of the power-angle curve (both input and amplitude beingmeasured from the horizontal axis of symmetry of the sine curve if

the sine curve is displaced vertically) and on the initial angle 50' and

initial angular speed woo For the present purpose, swing curves for

a sustained fault are wanted; hence Pn, PM, and 'Y must be the con

stants of the power-angle equation for the faulted condition, and the

initial speed will always be zero. The solution then depends only upon

p (definedby eq. 14) and 00'.

Each family of curves in Figs. 1 to 10 is for a constant value of sin

00 the range covered being from 0 to 0.90 in steps of 0.10. Theindividual curves in each family are for constant values of p, ranging in

each family from a minimum value slightly larger than the value of

sin 00' for that family up to a maximum of 3.00.

The procedure for using the pre-calculated curves to determine

critical clearing time from a given critical clearing angle may be sum

marized as follows:

1. The power-angle curve (Pc, PM, and 1') for the faulted condi

tion, the power input Pi, and the initial angle 00 are assumed to-beknown, because they are needed for finding the critical clearing

angle Dc by the equal-area criterion.

2. Compute p = Pi - Pc, P = p PM , 00' = 00 - 1', sin 50

and oc = Dc - Y.

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160

10

60c

4 (

40

20

2 3 4 5 6Modified time T

FIG. 1. sin 00 = O.

10

10

160 t - - - - + - - __ t __

40

20

0

60c:c (

-120 t - - - + - - t - - f - 1 H 1 - - f - I - I + - i - 1 I ~ f - - - ; + - f - - - J

Q)

t:J

co 100 r---t---tt-.......... . . . , . - + + - - - H - ' - - W ' ~ ' - - -

E. . - - - - . - -1

U 80 t - - - + - - f t l - f 1 H + - 1 - ~ ~ ~ F - - : : : ' ' ' ' : = ; ' - 4 - ~ ~ - - + - ~ - - + - - - + _ - + - - - + _ - + - - - + ~ - t - - - i

4 5 6Modified time T

FIG. 2. sin 00 = 0.10.

FIGS. 1 and 2. Pre-calculated swing curves (copied from Ref. 2 by permission).

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105 6Modified timeT

FIG. 3. sin 80' = 0.20.

3

40

160

....ene120 J - - + - ~ - U - J . . I ~ - I - ~ ~ + ' ~ - b t c - t - - - t - - - i r - - t - - + - + - - ; - - t - - ; - - t - - t - - - 1l- 100 1 - - 4 - - V - I - I - I - : H - I - ~ ~ ~ ~ ~ - ~ ~ ~ - - - i - t - ; - - ; - - - - r - - - t - - t-a 80 t - - f - - I t - ~ -:0CD~ 60<

10

160

n~ 120 . . . . - - - + - - 4 - 1 - 1 - - I - I - + + + I - ~ ~ ~ ~ ~ ~ + - - - + - - t - - - + - - - + - - f - - + - + - - - + - - I - - - + - - - - 4bOQ)

'0

~ 100 t - - - + - - - I ' f - f - - N - - - I t - - - ' ] ~ ~ ~ = - - - - - - - t - - - - ~ ~ - - - - - - I - - - t - - - - - t:soQ)

a; 80 1 - - - + - - f . j f + B t I - I - . ~ ~ ~ ~__ - - - - - - - - - t ~ ~ - - - - - f - - t - - - P l i ~ - - - - - t~~ 60c::<

20

40


FIG. 4. sin 80' = 0.30.


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104 5 6Modified timeT

FIG. 5. sin 80' = 0.40.

2

160

40

20 t - - + - - + - . . . - - - + - - - + - t - - + - - - + - t - - + - - + - - + - - I - - - + - + - - + - - i - - + - - - + - - I

-)

~ 120 I - - . . . . . . . - - - + + + - + I + - I t - + - + - I - - i + - - t ~ ~ - t - - : - - ~ ~ - i - - t - - - t - + - - - + - t - - - - t - - i - - - ttoQ)

0

100 1---+--........# - I -......... - , j ~ ~ - f - - + : : . e - + - - - - + - - + - - - + - - + - - - + - - t - - - i - - + - - 1 f - - - + - - + - - - I.suQ) 80 1 - - ; - - t t - t H i H f i J ~ t - T . . . , . - t ~ ............. : : : - - t - - - t - - - - t ~ + - - t - - t - - - t - - t - - t - - t - - - - 1:0

~ 60c-c

10o,-O........-.._......-- _. . . . . . ._ •

.. ._oIoo---A.o_oI---I-.....&.-

--...._..&.--.._...........- . .o

160 t - - -+ - - -+ -

20 t--+--+-+---+---+--+---+-I---+---+--+---+--I---+--+-+---+---Io-o+----I

40

) 120 t---+--.......... 4 - - - ~ ~ ~ ~ - - - - - -bII l I ,--+-+--+--t---f---t---t--+---I '---+---f

~ 0

co 100 1 - - - + - - 1 l I - - - - j H f - , I - - I t - - - - . , I J - - , ~ L . - ~ ~U

:suQ)

:e 80 t - - - - ~ ~ ~~~ 60cc (


FIG. 6. sin 80' = 0.50.


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105 6Modified time T

FIG. 7.sin

00'

=0.60.

3O ' - . . . - - - ' - - - - - - . . J . . . . . . . , j , - - - - - . J . - . . . . r . - . . . . l . - - - L - - - I - . . . . . . , j , _ ~ ' - - . . . - . - - L - - - I . . . - - ' - . . . . . . J _ _ _ 'o

160 I---+---t--

20 J - - + - - + - - + - - + - - + - - - lf - - f -- + - - + - - 1 - - - + - - + - I -- + - - + - - + - - + - - + - - - - -4 - - - 1

-120 1 - - - ~ - v - I - - W - - I - - I - - - - J ~ i L - - - - - - - J . ~ - - I : - - ~ 4 - - - - - - - 4 - ~

f-c(i j 100 1 - - - + - ~ ~ - I - + I - ~ ~ ~ + - - ~ ~ - + - - - + - - + - t - - - - + - - - + - - + - - - + - - - f - - - 4 - - - 4U

:a~ 80 ~ - - I - I - f I . I - I : - I - - I - I - ~ ~ ~ f = = - - - - - - 1 - - - l - ~ ~ I - - - f - - - - - - I - - - - - - - - I - - - - 4 ~~~ 60c<

n 120 t - - - - - ~ _ f _ l i - f _ _ t . i ~ t I - - - l _ _ - t - - - - f - - : : . ~ - - I - _ _ _ - f - - - I - - - - - - - ~ 1 - - - I~-c

B 100t - - t - - ~ ~ J t / _ - - I h ~ ~ ~ ~ + - - + - - + - - + ~ - + - + - - I - - I - - . J - - - - - I - ~ ~

:a1) 80 t - ~ f _ I r I _ I ~ ~ ~ ~ r - : : . - f _ + - - + - - - : - ~ - - I - f _ 1 - ~ ~ - - I - ~ - - l ~

° - - . .J .--- '_a.. .- ' - -- . . . .I .- . . . . . .L---J . , ;.--I . .--L_ ' --- -- ' ----&...-- ' ---L---I-J

o 2 3 4 5 6 1 8 9 10Modified time T

FIG. 8. sin 00' = 0.70.


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105 6Modified time T

3O - J . . . - . . L - - - . . . L - . . . L . . - - 1 . - . . . I . - - J . - - - L . - - . L - ~ - - - L - . . - . I - - . . . a . . - - - L - - - L . ~ - . l ~ . a . . . - - . . ao

FIG. 9. sin 00' = 0.80.

180

160

140

en~ 120 Q,)

'0

co 100-g'0Q,) 80

0

60c:

c: (

ltJt: ~ I ~ ~ 1/ j8°011) 00 II)

o 11).' ' .. ) /: ;;-;; ;; n: j- ~

J ; Q'11 7 Ji7 JI/

1111 III Ii / if 'I :Y r7I 1IJ 'I J '/JIf/ / V /11// I I I II r V . ~

/ IJvrI I if/7 V V

JJflli

rJjIfh '/ V V

I V/ V V

f-- ~ ~ /l ; »> 09 5 - _ -......-~ ,..... --- t--..r---. .......

40

20

oo 2 3 4 5 6 7 8 9 ro

Modified time T

FIG. 10. sin 00' = 0.90.


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3. Compute the equivalent inertia constant M by eq. 2.

4. Find the family of curves for the proper value of sin 00' and the

individual curve for the proper value of p. Enter this curve with theordinate 0' = Oe and read the corresponding abscissa T = Te•

Interpolation between curves or between families of curves may be

necessary.

5. By eq. 11 compute the critical clearing time te corresponding

to Te.

To determine the clearing angle corresponding to a given clearing

time, the order of the steps of procedure is altered in a way that

should be obvious.The procedure described above breaks down if the fault is of such

nature that there is no synchronizing power while the fault is on. Insuch a case PM = 0, from which it follows that p = 00 and T = 0 for

all values of t. The pre-calculated curve for this condition is the

vertical coordinate axis, and the relation between 0 and t cannot be

determined from it. However, this relation can be found by eq. 41

of Chapter IV, namely: ( )2M 0 - 00

t = [15]

Pa

Furthermore, the pre-calculated curves cannot be used to represent

conditions after clearing a fault because the curve for the proper value

of angle and speed (at the instant of clearing) does not have the proper

value of accelerating power or acceleration after clearing.

EXAMPLE 1

In Example 1 of Chapter IV, which deals with a machine connected

through reactance to an infinite bus, the critical clearing angle for the condi

tions of Example 4 of Chapter II was found by the equal-area criterion to be1380

• The corresponding critical clearing time, as determined from the

swing curve, is 0.61 sec. Check this value by use of the pre-calculated swing

curves.

Solution. From Example 4 of Chapter II, the power-angle equation

valid for the fault condition is

pu = 0.936sin 0 per unit,

whence

Po = 0, PM= 0.936, 'Y = 0;

the power input isPi = 0.80 per unit;

the inertia constant of the finite machine is

Ml = 2.56 X 10-4 per unit;

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and the initial angle is

whence sin = 0.310.

From the problem statement,

Oc = 1380

The following quantities are computed from the data:

Pi' = P, - Pc = Pi = 0.80

p = P/ = 0.80 = 0.854

PM 0.936

sin 00 = sin 00 = 0.310

= Dc = 1380

M = M1 = 2.56 X 10-4

~ = J1I PM = I 11 X 0.936 = 8 0t \ } 180M \} 180 X 2.56 X 10-4 ·

The most suitable pre-calculated curve is that for sin 00 = 0.30 andp = 0.85 in Fig. 4. The ordinate 0' = 1380 corresponds to the abscissa

If = 4.8. Hence T c = 4.8 and

T c 4.8tc = - = - = 0.60 sec.

8.0 8.0

This agrees reasonably well with the previously found value, 0.61 sec.

EXAMPLE 2

Find the clearing angle corresponding to a clearing time of 0.30 sec. on thesystem of Example 4, Chapter III, which consists of two finite machines

connected through an impedance network. (The equal-area criterion was

applied to this system in Example 2, Chapter IV.)

Solution. The following data are obtained from Example 2 of Chapter

IV:M = 2.64 X 10-4 per unit

Pi = 0.41 per unit

Pc

=-0.044 per unit

PM = 0.203 per unit

Y = -2.9°

80 = 12.7°

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1r X 0.203 = 3.66180 X 2.64 X 10-4

EFFECT OF FAULT-CLEARING TIME

and the following from the statement of this problem:

t = 0.30 sec.

The following quantities are computed from the data:

Pi = Pi - Pc = 0.41 - (-0.04) = 0.45

P = Pi = 0.45 = 2.2PM 0.203

00 = 00 - 'Y = 12.7 - (-2.9) = 15.6

sin 00 = 0.269

t= '1180 M=

T = 3.66t= 3.66 X 0.30 = 1.1

159

The most suitable pre-calculated curve is that for sin 00 = 0.30 and

p = 2.0 in Fig. 4. On this curve T = 1.1 corresponds to 0' = 76°.

o= 0' + 'Y = 76° - 3° = 73°

From Table 15, Example 4, Chapter III, at t = 0.3 sec. we find OAD

= 75.3°. This value was obtained by a point-by-point calculation. Theagreement is reasonably good.

Effect of fault-clearing time on transient stability limit. The

amount of power that can be transmitted from one' machine to the

other in a two-machine system without loss of synchronism when the

system is subjected to a fault depends on the duration of the fault.

The power limit can be determined as a function of clearing angle by

the equal-area criterion, and the relation between clearing angle and

clearing time can be found from the pre-calculated swing curves. It is

then possible to plot a curve of stability limit as a function of clearing

time. Such a curve is shown in Fig. 11. It shows that the transient

stability limit of the system can be greatly increased by decreasing the

time of fault clearing from 0.5 sec. or more to 0.2 sec. or less. The

time of fault clearing is the sum of the time that the protective relays

take to close the circuit-breaker trip circuit and the time required by

the circuit breaker to interrupt the fault current. Frequently a system

which is unstable for a particular type of fault and fault location can be

made stable by altering the existing relaying or by modernizing thecircuit breakers so as to decrease the clearing time. *

*Typical values of relay time and breaker time are given in Chapter VIII, Vol.

II . Modernization of breakers is discussed in the same chapter. Protective

relaying is discussed in Chapter IX, Vol. II .

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A curve like that of Fig. 11 can be obtained by the following proce

dure. First, the equal-area criterion isused to determine the stability

limit with instantaneous clearing. Truly instantaneous clearing is not

o

1..'2::s...(1)

c....

:cnJ

en h----......

---_....................

---_..............

--0.5 1.0 (X)

Fault- clearing time (seconds)

FIG. 11. Curve of stability limit as a function of fault duration.

obtainable in practice, but it may be regarded as the limit approached

as the clearing time is reduced. The stability limit for instantaneous

clearing is the same as that for disconnection of the faulted line when

there is no fault on it. This limit is determined as shown in Fig. 12.

o 00 01 r/2 Om

Angle 0

FIG. 12. Determination of stability limit for instantaneous fault clearing by useof the equal-area criterion.

After the power-anglecurves of pre-fault and post-fault (cleared) output have been plotted, the horizontal line representing the input,

which is equal to the initial output, is shifted up and down until areaA l equals area A2• It should be noted that moving this line changes

the initial angle 80 and thus moves the vertical line bounding area A l ·

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EFFECT OF FAULT-CLEARING TIME 161

The value of input determined by equality of the areas is plotted in

Fig. 11 as point 1 at zero clearing time.

Next the stability limit for a sustained fault is found as shown inFig. 13. The power-angle curve for the fault condition is used in place

of the post-fault curve; in other respects the determination of the

stability limit for a sustained fault is like that for instantaneous clear

ing. The value of stability limit so found is plotted as the asymptote

(point 2, Fig. 11) which the curve approaches at large clearing times.

The two extreme values of stability limit have now been found.

Any number of convenient values of initial power between these ex-

,.... C:J..D

Co.D~

2-r1P

m~ t - - ~ ~ ~ ~ ~ ~ ~ ~ ~ J - - - l -

Angle 0

FIG. 13. Determination of stability limit for sustained fault by use of the equalarea criterion.

tremes may now be,assumed, and the critical clearing angle for each is

found by the equal-area criterion as shown in Fig. 14. Power-angle

curves are drawn for the output before the fault, during the fault, and

after the fault, and the input line is drawn at one of the selected values

of initial power. The vertical line at the clearing angle a is shifted

from right to left until areas Al and A 2 are equal, thus fixing the

critical clearing angle. The clearing time corresponding to this clear

ing angle is determined from the appropriate pre-calculated swing

curve. Point 3, Fig. 11, is then plotted, its coordinates being theassumed power and the corresponding critical clearing time. Addi

tional points on the curve are determined in similar fashion.

The procedure described above, in which values of power are

assumed and the corresponding clearing times found, is simpler than

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the alternative procedure in which clearing times are assumed and the

corresponding power limits found. For in the latter procedure the

horizontal input line is shifted, resulting in a shift also of the verticalline at the initial angle 00, whereas in the former procedure only one

line, the vertical line at the clearing angle oc is shifted.The procedure which has been described for finding transient-stabil

ity power limit as a function of fault duration is applicable only to twomachine systems. Nevertheless the general conclusion, that decrease

of fault-clearing time improves stability and increases stability limits,

o s,Angle 0

FIG. 14. Determination of stability limit for fault cleared in finite time.

is just as valid for a multimachine system as for a two-machine sys

tem. The speeding up of relay and breaker operation is one of themost effective and important means of improving power-system

stability.

EXAMPLE 3

Plot stability limit in per unit as a function of the fault duration for a

three-phase short circuit (a) at the middle of one of the parallel transmission

lines of the power system shown in Fig. 15 and (b) at the sending end of one

of the lines. The fault is cleared in both cases by the simultaneous opening

of the circuit breakers at both ends of the line. The system consists of ahydroelectric station sending power over two parallel transmission lines at

generator voltage to a metropolitan system which may be considered an

infinite bus. The following data pertain to the system. The base power

is the aggregate rating of the hydroelectric generators.

Direct-axis transient reactance of hydroelectric generators: 0.35 per unit

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Stored energy of hydroelectric generators, H: Mj. per Mva. of rating

Frequency: 60 c.p.s,

Voltage behind transient reactance of hydroelectric generators: 1.00 perunit

Voltage of infinite bus: 1.00 per unit

Reactance of each transmission line: 0.40 per unit (neglect resistance)

Reactance of transformers at receiving end of lines: 0.10 per unit

Fault Fault

(6) 0.20 <e) 0.20 0.10 tJPGydro station -0 - . -40- - - -#A Me:OPolitan

H = 3.0 systemH=oo

FIG. 15. Two-machine power system. (Example 3.) Reactances are given in perunit on a common base.

~ 0.35. 0.20 • 0.10 ~ 0:5 ~ ------<@-.

(a) (b)

FIG. 16. Reduction of the network of Fig. 15, pre-fault condition. (Example 3.)

0.35. 0.40 • 0.10 ~ 0:5 ------ @-.

(a) (b)

FIG. 17. Reduction of the network of Fig. 15, post-fault condition. (Example 3.)

0.35 0.40 1 ~'V ~ 'V

(a) (c)

(d)~ • Q:. Q-rQ.:.lQ. • 1 ~'V I0.05 '\.i (b)

. @ ~ _ ...

FIG. 18. Reduction of the network of Fig. 15 with a three-phase short circuit at

the middle of one line. (Example 3.)

Solution. The network joining the two machines, which is considered to

consist of reactance only, is reduced as shown in Figs. 16, 17, and 18 for the

pre-fault and post-fault conditions and for a fault at the middle of the line,

respectively. For the first two conditions the reduction is accomplished by

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~/

' /Pre- fault......V

- ......) 4

J ~ , I ~ \ V Fault cieared

1/ J ~ 1\I ,

I ) ~/ V \

~ r\/ I \IJ

1\ ~ ~ ~ Fault at middle of l ine- r-, \II ~ < ,

r-, ~I f ; ~ ~ a u l t at sending end ~

1.08

1.00

--. 'c::s

8-

---

0.50

0.32

parallel and series combinations; for the third condition, by two yea con

versions and intervening series combinations. With a three-phase fault at

the sending end of one line it is obvious, without reducing the network, thatno power can be transmitted, The resulting reactances between machines,

1.50

oo 30 60 90 120 150 180

o(degrees)

FIG. 19. Determination of stability limit for instantaneous clearing of a fault at

either location and for a sustained fault at the middle of the line. (Example 3.)

and the amplitudes of the power-angle curves (PM = EAEB/XAB= l/XAB),are as follows: .

ConditionReactance PM(per unit) (per unit)

Pre-fault 0.65 1.54Post-fault 0.85 1.18Fault at middle 2.45 0.41Fault at end co 0

The power-angle curves are plotted in Fig. 19. The stability limit for

instantaneous clearing is found by the equal-area criterion in the upper part

of this figure. The value thus found, 1.08 per unit, is correct for both fault

locations. The stability limit for a sustained fault at the middle of the line

is found in the lower part of Fig. 19; it is 0.32 per unit. The stability limit

for a sustained three-phase fault at the end of the line is zero.

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TA

1

D

MIN

O

OC

A

N

TME

FR

M

C

A

N

A

EX

MP

3

sn

FuaMideoLn

FuaE

Ln

P

P

0

s

p.u.

d

T

tc=

T

50

-0

to

15

p

=Pi04

d

fompecc

se

deg.

d

s

fomFg2

cuv

fomFg2

10

06

4

25

5

04

00

46

6

00

09

05

3

22

60

07

01

5

1

01

08

05

3

20

7

10

02

5

2

01

07

04

2

17

8

13

02

.

.

·

06

03

2

15

9

16

03

7

5

02

05

03

1

12

1

22

04

.

.

·.

04

02

1

10

1

31

06

9

8

03

0302

1

08

1

60

11

.

.

·

02

01

8

·

.

·

·

1

1

05

01

00

4

·

.

·

·

1

1

08

0000

1

·

.

·

·

1

1

12

t a t3 d at

4t > Zo t

J t

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Pre-f au l t y

J 1 / I r-.......

IJ.

0.32 1-1..1-

Vi11111I 11j . ;

:--.: I - - Faultat middle of line I ............. \ I I I I I I I I I r-....1\

I , I I I I I I I 180120 : : 150

111 130 144i i 60

455130 : : 90:

70 82 956 (degrees)

FIG. 20. Determination of stability limit as a function of clearing angle by useof the equal-area. criterion, three-phase fault at the middle of the line. Areas for

P = 0.50 are shaded. (Example3.)

Stability limits between 0.32 and 1.08 per unit are assumed with the fault

at the middle of the line, and the corresponding critical clearing times are

found in Fig. 20. Stability limits between 0 and 1.08 per unit are assumed

with the fault at the sending end of the line, and the corresponding critical

clearing times are found in Fig. 21. In both figures curves of stability limit

as a function of clearing angle are plotted. The values of clearing angle are

entered in Table 1.The values of clearing time corresponding to these values of clearing angle

are found from the pre-calculated swing curves if the fault is at the middle

of the line and from eq. 15 if the fault is at the end of the line. In the first

case the numerical value of Tit is needed. Byeq. 11 it is

: = ~ 7 T P M = /7T' X 60 X 0,41 = 5.08t GH '\j 3.0 X 1

In the second case we have from eq. 15

2GH (<<5c - «50)

180 r,2 X 3.0 X 1 (<<5 c - «50) =

180 X 60 Pi - P..

Details of the determination of clearing time are given in Table 1.

Curves of stabil ity limit as a function of clearing time are plotted in Fig.

22

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Pre- fault-V-

J - 17\- ~

~ -- - - 0-\: 1: :___ ~ ~ u l t cleared-- -- , I , I ~-- I . . -- Stability limit _

, , I vs. clearing angle ~: II I I

{7 lY1I I -~ I ~ ~-- -- '- I - 11 - - -'- - - -

: I I I I , I I I I\.. I

if' , I I I

~ II I I~ I I , - - --7:; -t- - 1-,- - ' I -,- ~ ~: I ~I I I 1 0..I I~ I I . ' ..... ~ ~ ~ ~I Faut at end of line

, I ~ ~ I I ~/ r/ /i j j : . -

~ 1/ // 1 / / / /// / ~ v/ / ,

1.30

1.08

1.00

-=c:'~-e 0.50

oo 30 : : :60 90:

465258 75 95o(degrees)

120 150 180

FIG. 21. Determination of stability limit as a function of clearing angle by use ofthe equal-area criterion, three-phase fault at the sending end of the line. Areas

for P = 0.20 and P = 1.00 are shaded. (Example 3.)

1.4 ex>.5 1.0Fault- clearing time ( seconds)

'\1\

1\\\ f\

' \1\

\ .< ,

\ r - Fault at middle of line11 --

t\. t'--.... f l l - --, II ==

'L- Fault at sending end of line

- -

-: . . . -o

o

~ 0.5

1.0

FIG. 22. Curve ofstability limit as a function of fault-clearing time for the systemof Fig. 15. (Example 3.)

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Curves for determining critical clearing time. A more direct way of

determining the critical clearing time of a fault on a two-machine

system than that described in the first part of this chapter has beendeveloped and described by Byrd and Pritchard.f Their method

facilitates the plotting of curves of stability limit as a function of fault

duration. It takes cognizance of the fact that the voltages behind

transient reactances, and hence the amplitudes of the power-angle

curves, usually vary with the power transmitted in a way determined

by operating practice. This fact was disregarded in our previous

discussion, although allowance could have been made for it.In Byrd and Pritchard's method two assumptions are utilized in

addition to those which were made in developing the equal-area criterion. They are:

1. That the network is purely reactive.

2. That all circuit breakers which open to clear the fault do '80

simultaneously.

Both these assumptions are commonly, although not necessarily, made

in using the equal-area criterion in order to simplify the calculations.

The derivation of Byrd and Pritchard's method follows: Let thepower-angle curves of the two-machine reactance system be:

Before the fault: Pu, = Pm sin 0

During the fault: Pu, = T1Pm sin a

After the fault: Pu, = T2Pm sin 0

[16]

[17]

[18]

In other words, Tl and T2 are the ratios of the amplitudes of the power

angle curves during and after the fault, respectively, to that before the

fault. Inasmuch as the amplitude of the power-angle curve of areactance network is given by eq. 7,

p _ E1E2

m - X12

[19]

[20]

it isapparent that f l and T2 may be expressed in terms of the reactances

between machines before, during, and after the fault, thus:

X 12 before faultT1 ==X 12 during fault

X12 before faultT2 ==

X 12 after fault[21]

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CUltVES DETEltMINING CRITICAL CLEAItING TIME 169

It is also apparent that, if the internal voltages E1 and E2 should vary,

the amplitudes of all three power-angle curves would be changed

proportionately, and the ratios rl and r2 would not be affected.The three power-angle curves and the input line Pi are drawn in

Fig. 14 so that the equal-area criterion may be used to find the critical

clearing angle oc. Ordinarily the two shaded areas are equated. It is

just as correct, however, to equate the irregular area under the heavy

line to the area of the rectangle below the input line. If this is done,

the following equation results, from which Oc can be found:

[22]

Upon setting Pi = Pm sin 00 and evaluating the integrals, eq. 22 becomes:

(Om-OO)Pm sin 80= -rIPm(COSOc-coSOO)-r2Pm(COSom-cosoc) [23]

(om - 00) sin 00 = (r2 - rl) cos Oc + rl cos 00 - T2 cos Om [24]

(om - 00) sin 00 - Tl cos 00 + r2 cos Omcos Oc = [25]

r2 - rlwhere

[26]

Thus, if TI, r2, and 00 are known, the critical clearing angle Oc can be

found from eqs, 25 and 26. The corresponding critical modified time

Tc can then be found from pre-calculated swing curves (Figs. 1 to 10)

which are solutions of the swing equation (13) in which 0' now is

simply 0 and p = (sin oO)/rl. The actual clearing time to in seconds

can be found from the equation

~ 8 M I GHtc = T

c 1rrlPm = Tc \ j ~ f r l m [27]

which differs from eq. 11 in that PM has been replaced by TIPm, the new

symbol for the amplitude of the power-angle curve during the fault.

The steps which serve to determine T c as a function of Tl, T2, and sin

00 were carried out for many values of the independent variables, and

the results were plotted in the form of curves of Tc versus r2 for constantTI and sin 00. See Figs. 23 to 39. Each family of curves is for a con

stant value of sin 00, the range covered being from 0.10 to 0.90 in steps

of 0.05. The individual curves in each family are for constant values

of Tl.

8/13/2019 V1-5-5264202


170

1.0

0.8

0.6

0.4

0.2

THE TWO-MACHINE SYSTEM

J I

~ ~I I Tl I J rr

0 , l- I 8 &t)1

8 .....1-0 ..... .....

0 1-0 0 0

1 -1 - II

II j.......

c: ,, 'rJ ..,.... II I

I

I' J J J JII III I I r,

--If..r . ~ ~I I I r1 I

( ) .'I I.J //

I < [ ; ; j j l ; t ~I Il...l

1...oIl L..I .... L. .1iiJ I.....l ..... looII .... L I . I I I ~ L-L- -. ...

oo 1.0 2.0 3.0 4.0 5.0

1.0

0.8

0.6

0.4

0.2

FIG. 23. sin 80 = 0.10.

I I I I I I J I , l I I I I I

i-H &t) - - ~ o l I I I0 i- t)

8lI t ' ) ..... , f-1- C\I f- 1 ( ' ) , ..... 0 .......... i - f - ..... - : IJd , 0 _0 _ 1 - 0 1- ' -

_ _ 00

f- f- II ~ 1 1 - .....

I- :: ....... f- ........ J - :: c: f - I - ,-- ........, ........JI'-I- I- I - i -- - ,

J I r 1j I' II, , I -,

I' j If 'I.I I I IJ ]

J ,II IJ ..oil

v II' 1...oIl

v l....l

II' L...II I....ll ~ L -

L...o ..... - 1 -1 - ....

-oo 1.0 2.0 3.0 4.0 5.0

FIG. 24. sin 80 = 0.15.

FIGS. 23and 24. Curves for determination of clearing time (copied from Ref. 3by permission).

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CURVES DETERMINING CRITICAL CLEARING TIME 171

I I I I I I ,• •

I • I ,

1-+-1 - - 0 i - I - - --fir) : I ,I t)

If? 'raj

0 ..... --- - ....0 1 - 1 - ' - ~ J i - ' - - ---0 0 '0 '

11

~ ~ 1 -1 - - II J IIi-i-i-

.. .0-#i - I - -

= -' 4,. I 1...i-i-i- i-i-

i-i-r- i - I - --- : II I

, J J J, I I I-I-, J /2

I , o ~ v v f ~j ,,- l. ,lI ilI

~ ..... '. ,.- i.oooioooo .... ~ ., - , 10.-.... .- -- ...- ..-- ...... ,..-

t - 0.25_

Tc

FIG. 25. sin 80 = 0.20.

1.0

0.8

.. .N 0.6

0.4

0.2o 1.0 2.0 3.0 4.0 5.0

1.0

0.8

a ' 0.6

0.4

I, ~ i - ~ ~ ~ ~ i - i-

I II I I I ~ i=--l l )

1:fJt ~ K > ~ ~ ~C? - f- ..... t- - i-

I-d I - - - -1 ° o l- e - l- e:) -....--I- CV i- C\J

~ /IT II II ~ - J ~ I- C:i J

.. .M~ 0-# ~

a..' J - a..... 1 a.... I-:-- - II - II II

i- i- I -- -a.. - i- J...

I I t-- i-

I/ / 'I I I t ) ; ~ ~ lJ III J J ,,-

r-J'I.IJ

(; i ~ 1 -

J II ~ ~I J , , ~ ~, ~ ~ ~ ~ [. II i.lIII .....

~ i l I ~ i l I -... ..... ........ 1.1--...

.- 1.1...._.. -... -- --- t

Tc

FIG. 26. sin 80 = 0.25.

0.2o 1.0 2.0 3.0 4.0 5.0

1.0

0.8

... 0.6

0.4

1--- I - - ~ a . n tW II I I r I I , t I I I , I I , I

ttC:fJI ~ I

::r'/fI-- I-

0 I-S lili-r - - i - 0 OJ 10·

>- °ti- .... f - ~ c : : ) l 10. 'O·t-_

- f- II I

II :1-11 ~ i - -i- II I II 1 II- ...

- a.....~ ~ . . . , , , c: j

I-i- - i.: 1 ... t : i- Ji- 1-- ,1 I I ,II I I IJr II v

I I I v ~

', ~ .,, ~ ~ I - ~~ ~ - ---,-

--, - -

p

0.2

o 1.0 2.0 3.0 4.0 5.0

FIG. 27. sin 80 = 0.30.

FIGS. 25, 26, 27. Curves for determination of clearing time (copied from Ref. 3by permission).

8/13/2019 V1-5-5264202



-- - ~ ; U ~ ; ~ t ~ I t ~ t t ~ i ~ I I I I .I- t - ~-- - 0 2 2 9i>-fil 1-- ~ :1

I r l ~ ........-- - 0 1-- ~ i ~......- - .... 11

/I II II II

t-II ~

1-- O' J 10

,

--....

---- .... - ---- II ~ : _ If

rj'------__ a..;

:-c: 4.' 'f I-4.' 'f

j'- _

i--- 4.' 'f.,-._ .....,/J .....,I II I 1/ J vII I J

If' v

v I;' I L. l

J v ~ I...... I.....

I I J v l,;Ill IlllI' L... 1...- ......

J L. l ~ ~ . . - ~ i i I ~ .... ~- - - ~ 1 1 1 ~ ...... _I.- ...

I I

I III I

1.0

0.8

...... 0.6

0.4

0.2o 1.0 2.0 3.0 4.0 5.0

1.0

0.8

....0.6

0.4

FIG. 28. sin 80 = 0.35.

I I I I I II I I I I 1- 1/ I I I 1/ I II IIT:J:,.

1-1- ~ 8 r ~ [ .... W ~ f 1- 0 ~ ~ ~ r i l ~ t ~ ~ ~ ~ I I ~ ~ ~ ~ - t:i tt; ~ ~ O' ...- O' 1-..... O' II O' - 1 -1 -

1-1- ~ ~ ~ j ~ II t . . ~ II.... _ ~ --1- 1/ ~ .... P 1/

~ .... .... ~ II J tt I j ~ ~ ~ '''''1} ... -\ if

... c:1- - 4. f ~ f-- ~ 1 ~ 1 -

I-~ ,

I , ~ ,I I ~ ,. ~~ ~ ~ ~ J

J

,if ,

J ~. .l....IIlI l . . . o I ~ ,- ~ .... _..... - ...

-- - '---

Tc

FIG. 29. sin 80 = 0.40.

0.2o 1.0 2.0 3.0 4.0 5.0

T

cFIG. 30. sin 80 = 0.45.


8/13/2019 V1-5-5264202


CURVES DETERMINING CLEARING TIME 173

J II II I 1 II J IJI J I '(

II J J ] v J11

I II I 1/ II II 1 ~

5.0.0.0.0.00.5 '-'-__ ---..........,,- - . . I o . . . I ~ - - I - I o . . . . - . . ...r....I

o

FIG. 31.. sin ao = 0.50.

...

1.0

0.9

0.8

0.7

I- I I

fS; f1- ~ g l ~ , I I II L ~ ~ j I I II t )

~ ~ 'O J L -L - t ~ 1-1- ~ 'I I f II- 0

I I ~- 0 0 o O i 'I- t;;:)' tr ~ I I J ~ ,...

I- l- II : II II ~ I- III-

.. ...1 - 1 - ~ ti- II I I c::r

I- .. ..............., ......., I-1 - 1 -

(,.....,J l- I- .... fo

'I 'I I I If 'f( . . . , l - I-

J J IJ IJ rt-fo- ....I I I I T 1

I

, i J vj , , vI I II , ~

'I J J, ~

1.11 ~ ..,l...oillill,

J ~ I..- 'I IlIlII' ..,--

1...- - _.......I, ~ ~ ,,- L-'-'

_..._L- ... '

0.6

0.5o 1.0 2.0 3.0 4.0 5.0

Tc

FIG. 32. sin ao = 0.55.

FIGB. 31 and 32. Curves for dete rmination of clearing time (copied from Ref. 3

by permission).

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174

...e-t 0.8

THE TWO-MACHINE SYSTEM

I I , ~ ~ ~ ~ ~ ~ __ ~

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ -

FIG. 33. sin 80 = 0.60.

I , O ~ I ~ I ~ 1 4 - ~ L ~ _ ~ I ~ I I I I I I

s ~ ~ ~ ~ - t8 /f{ ~ ~ ~ ~ ~ ~ ~ > o - F f 8 l-1 0 0 c ; - c ; ~ ~ . - () ~ <:>. f- ~ t : : ~ t::)

~ /I I /I ~ zr q I ... II /t t.- ~ _ /t ~ I I . : ->0-- 4 ~ -

c:I.;.J , .... 1... J 1... - ~ -1 ~ - ~ I I - ...,-:: ' \ ~ 2 ~ : ~ ,\'\0

- l- '('\ ~~ ~ I, I II I ~ lA I 1:1 0 _

J J I ) ~ ~ ~ ~ ~I I ~ ~

1\L..;;;..... 1 I I I ~I , ~ ,,- L . . o o ~

I f I I v~ ~ ~ ...... ~ - - - - ~~ J If' v: --' ..... . - .... ....- ...... 10-- ....

i I o o o l I ~ .... ..

. ,

1.0

0.9

... 0.8

0.7

0.6o 1.0 2.0 3.0 4.0 5.0

1.Q

0.9

0.8

FIG. 34. sin 80 = 0.65.

I

~ ~ & ~ l qr J > - ~ ~ W IJ

I , I I I J , I I , , ,- 0

Ort' I I I 1 1 ~ < i ) I- -- _0 o c::i Ci O' I I ()-:

-- I /Ht Ht : :r-I/O li- J

I If' I ~ I I

- ...... 'J 14,. J 4,...tr-- - v- v- 0( , ~ 'D,I-~ ( ) ~ ~ ~t I , ~ .C \ ___ .... -I I I II 1...- ~I I ~ ~ ~ , ~ ~ ~ _... -, ~ 1 . . . o I l I ~ ....- - ~ ... ~

_...11 -

0.7o 1.0 2.0 3.0 4.0 5.0

FIG. 35. sin 80 = 0.70.


8/13/2019 V1-5-5264202



1.0

0.9

0.8

~ ~ I ~ t ~ ~ t f l l ~ ~ X i ~ ~ W d ' l IIi} I I ~ ~~ 0'Yo 0:' If' T T ~ ~ 1 - 1 -~ O 0 t::i O' O' o 0'// 1 ~ : t 1 -1 - ~ T o ~

u: : J ~ ~ I - ~ ,<, ~ ~'h...-- I - ~ .,

11 } I I v lIi ..... I.....

I . ~ 1..-.....

J J I/' '

l . . . . o l I ~l..o

~ ~ ...'I ~ .,. ~ jllIII L . o o ~ '-

io _I-

0.7o 1.0 2.0 3.0 4.0 5.0

FIG. 36. sin 00 = 0.75.

030

r'= .'1 = O. 0 rrl =0.40

o J1/ ~ r ~ J ~ ~ a ~ \ f f i 1 < j - ~ B : - t \ i P 17;L ~ ~ . ~ ~ ~ ~ o ' f . ~ a:t a : l , ; ~ ~ o . ~ ~ Iio \ ) ' b \ < ) ~ , 'r .1:- ; I i • Ci

J -/t ' - ' h ~ ~ - t \ ~ , , ~ 1 . t ~ . , , ~ - 1 ' \ I ' ~ ~ ~ - ~ ~ . . - ~ ~ o ~ z ; ; . . - e

t-, 11 ~ - I ~ I - f - < ~ tt . ~ --- ~ \ - , , ' - ~ - ~

11'1 I.' 1....1 I..- 1.1I ...... ... 1 - ....... L. - .... 11I

r:r/: ~ ~ ~ l..o.IlIl-- ~ ~ ...-...... '-- ... 1-- ..... 1.-:- ~ i I - '

1.0

~ 0.9

Tc

FIG. 37. sin 00 = 0.80.

0.8o 1.0 2.0 3.0 4.0 5.0

1.00

0.95

0.90

o 0 ~ ~ ~ ~ ~ O ~ I - ~ ~ ~ ~ I Wo'>v

- n,F-f- rt> - ~ v~ I - ~ o· f-f- oep co' Alo,,,

- I I- 1/ 1/ rlf- f-I- // < ~ ~ < . t ; ~ ~ ~ - ?I- 4.. .... - r- ....jf - ..... < ) I . ~ (,

............ ~ . . . ~ ' J ~ l-I..;

r J I;' I;l' I.l'

II IJ l' I.... ~ I...- ~ ' - 'j J I.l' i...ll ijll'

I v ~ ~--

..... jllIII t -

IJ ~ 1....1 L....o ..... ~ , . . . . II ....... JI 1....1 l.olII

Tc

FIG. 38. sin 00 = 0.85.

0.85

o 1.0 2.0 3.0 4.0 5.0

~ .... ~ ~ * ~ Y / ) 4 : < § l - l l i ~ ' j - t 1 ~ ~ : 1 C i L... II 0 .....

~ ~ D ~ 0 9 ..... ~o l- e::; O· O· + 0 ' ~ ~ ( ) . t . ~ ~ ~ ( ) . O ~ ~ ~ . , . - II 1/ ' ( ~ ~ L . , - -f- ~ \ . . . . - I ~ - ' - t ~ ~ ~ h ~ -'

1 \ ~.....

... -) ...; 4..' ..... ~ . . . . . _ l0- ll

, I v: - ' ',- ... , Iiiiiii

I I.. - - ----

- ~ ......

IT

1.00

... 0.95

0.90

o 1.0 2.0 3.0 4.0 5.0

FIG. 39. sin 80 = 0.90.

FIGS. 36, 37, 38, 39. Curves for determination of clearing time (copied from Ref. 3by permission).

8/13/2019 V1-5-5264202



1.0.9.8.7.4 0.5 0.6rJ or r2

0.3.2.1

I iJ

v1/

iJ

II'1 /

J

VII

/

vV

WI

I;

v:,.,

l.',.,,.,

I'

'-'l;l

I#

oo

0.1

0.3

0.4

0.7

0.9

0.8

0.6

0.2

It is convenient to have an additional curve for determining the

stability limits for instantaneous clearing and for sustained faults. To

obtain the equation of such a curve refer to Fig. 12, which shows theapplication of the equal-area criterion to the case of instantaneous

1.0

c 0.5

FIG. 40. Curve for determining the stability limit for a sustained fault or for afault cleared instantaneously (copied from Ref. 3 by permission).

clearing, and equate the area of the rectangle under the Pi line between

ao and 6m to the area under the post-fault power-angle curve between

the same limits.

Area of rectangle = Pi(8m - 60) = Pm(6m - 60) sin 60

Area under power-angle curve=T2Pm16m

sin as=T2Pm (cos - cos 8m)

Equating the two areas, we get

am - ao) sin = T2(COS ao - cos ~ m [28]

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where 8m is as given by eq. 26. Equation 28 expresses implicitly a

relation between T2 and sin 80 which is plotted in Fig. 40. To find the

stability limit for instantaneous clearing, enter this curve with T2 andread sin 80• Then the stability limit is Pm sin 80• The same curve can

be used to find the stability limit for

a sustained fault by entering it with P

the value of TI instead of T2; this be

comes evident when Fig. 13 is com

pared with Fig. 12.

It has already been mentioned

that, as the transmitted power (Pi ==

Pm sin 80 ) varies, the internal voltagesE1 and E2 usually vary, and hence

Pm = E1E2 X 12 also varies. For the

present purpose the most useful way a0------------ ' ·to show these relations is to plot Pi sin 00

and Pm against sin 80, as shown in F 41 T · 1 f P d0. . ypica curves 0 , an

Fig. 41. r; versus sin

The procedure for obtaining a curve

of power limit as a function of fault duration by Byrd and Pritchard's

method will now be summarized.

1. Reduce the reactance network to an equivalent between the

two machines and neutral for each of the three conditions:

a. Before the fault.

b. During the fault.

c. After clearing the fault.

Only the reactances X 12 between the two machines are used in what

follows.

2. Compute the equivalent inertia constant M by eq. 2.

3. Calculate and plot curves (like those of Fig. 41) of

a. Pm = EIE2 X I 2 versus sin 80

b. Initial power Pi versus sin 80

maintaining the bus voltages at the values which would be held in

actual operation.

4. Compute TI and T2 by eqs. 20 and 21.

5. a. Enter the curve of Fig. 40 with Tl and read sin 80• From

Fig. 41 read Pi corresponding to this value of sin 80• This is the

stability limit for a sustained fault.

b. Repeat, using T2 instead of TI. The value of Pi thus found isthe stability limit for instantaneous clearing.

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6.t Select values of sin 00 which are multiples of 0.05 and which

are between the values found in steps 5a and 5b. For each such

value of sin 00 find the proper family of curves from Figs. 23 to 39;find the curve for the value of Tl; enter this curve with T2 and read

Te•

7. For each value of sin 00 used in step 6 read the corresponding

values of Pi and Pm from the curves of step 3 (like Fig. 41).

8.t For each value of Tc found in step 6 compute the clearing time

teby eq. 27, using the proper value of Pm determined in step 7.

9. Plot stability limit Pi as a function of clearing time te• This

curve will look like the one in Fig. 11.

The method which has been described fails if Tl = 0, since for this

caser = 0 and t is indeterminate from eq. 27. To avoid this difficulty

a new modified time p is now introduced, related to the actual time t by

t = p J180M = p IGH . [29]7rPm ~ r f m

and differing from Tin eq. 27 in that Pm, the amplitude of the pre-fault

power-angle curve, is used instead of T1P

m, the amplitude of the fault

power-angle curve. Hence p = T/Vi). The swing equation now

becomes

p =

with 8 in electrical radians, and it has the solution

2(0 - 00)

.sin00

[30]

[31]

which will be recognized as eq. 15, transformed by the substitution of

eq. 29 for t, and with 0 - 00 expressed in electrical radians.

Given Tl = 0 and the values of T2 and sin 00, Pc may be found by

using eqs. 25 and 26 to get oc, and then eq. 31 to get Pc. From the

results of such calculations Pc has been plotted in Fig. 42 against sin 00

for various values of T2.

If Tl = 0, steps 6 and 8 of the procedure are replaced by steps 6A

and 8A, respectively, which are as follows:

6A. Select values of sin 00 which are between zero and the value

found in step 5b. Find the curve in Fig. 42 corresponding to the

[If r l = 0, substitute steps 6A and 8A, described on this page and page 180,

for steps 6 and 8, respectively.

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a a t td a z zo § : a ; Zo sl j

70

65

60

55

50

45

40

30

25

20

1.5

35

P

Cvfodemnncentmf=

0 coedfomR3bpmso

FG4

05

10

o o

10 08 060 c

04 2

c

8/13/2019 V1-5-5264202



value of r2. Enter it with the selected values of sin ao and read the

corresponding values of Pc.

8A. For each value of Pc found in step 6A compute the clearingtime tcby eq. 29, using the proper value of Pm determined in step 7.

EXAMPLE 4

Plot a curve of stability limit as a function of fault duration .for a three

phase fault at the middle ofone of the 132-kv.transmission linesof the power

system of Fig. 43. The two hydroelectric generators i and k are to be ra.

0.175

3.0/13.2O l 7 ~33.0/13.2

0.30

0-111:0.35

0.35 )(

6.6/132Fault

.6. A-

0.140 33.0/13.2

0.140 f----®33.0/13.2

FIG. 43. One-line diagram of power system, with reactances of lines and trans

formers in per unit on a lOO-Mva. base (Example 4). Data on generators and loadsare given in Table 2. (From Ref. 3 by permisslon.)

garded as one machine of a two-machine system, and the steam turbo

generators, a, b, g, and h, together with loads c, d, e, and J as the other

machine. Each load is assumed to consist of three equal parts, one of

resistance, one of synchronous motors operating at half load, and one of

induction motors at 1/2.75 load. The reactances and inertia constants ofthe generators and of the composite loads are given in Table 2, expressed onabase power equal to the rating of the individual generator or load. In Fig.

43 the reactances of the lines and transformers are given on a base of 100Mva. and nominal voltage. The system frequency is 60 c.p.s,

Solution. The procedure described on pp. 177-8 will be followed.

1. Network reduction. All reactances will be expressed on a system base

of 100 Mva, The reactances in Fig. 43 are already on this base. The

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machine reactances in Table 2, given on the machine base, are converted to

the system base by multiplying them by (100 Mva.)/ (machine rating in

megavolt-amperes).TABLE 2

MACHINE REACTANCES AND INERTIA CONSTANTS (EXAMPLE 4)

Rating,

Hd

Machine Kind Mva. P.F.

Machine System Machine SystemBase Base Base Base

a Turbogenerator 40 0.8 0.18 0.45 6.0 2.40b

c Composite load 55 0.9 0.56 1.01 1.95 1.07d 30 u 1.86 0.58e

u

f u 55 1.01 1.07g Turbogenerator 50 0.8 0.18 0.36 6.0 3.00h i Hydro generator 29. 4 0.85 0.30 1.02 3.0 0.88k u u

The network is so nearly symmetrical about a horizontal axis that the

reduction may be facilitated with very little sacrifice in accuracy by assuming that the two sections of the long receiving bus are tied together and also

that the load busses d and e are tied together. In addition, machines i and

k are assumed to be tied together behind their transient reactances; ma

chines a, b, c, d, e, j, g, and h are assumed to be tied together similarly.

Then, after simplifying the hydroelectric-station circuits and the receiving

system by series and parallel combinations, the circuit of Fig. 44a results.

The reactances of the three parts which are separated from one another by

transformers are given on a common power base of 100 Mva, and on voltage

bases equal to the nominal voltages of 6.6, 132.0, and 33.0 kv, It will benoticed that the ratio of the receiving-end transformers, 125.4 kv./33.0 kv.,is not equal to the ratio of base voltages, 132.0 kv./33.0 kv, Therefore the

base voltage for the receiving system should be changed from 33.0 kv. to

33.0 X 132.0/125.4 = 34.7 kv.; the reactance of that system on the new

base is

0.13 X (33'0)2 = 0.12 per unit34.7

The new value is shown in Fig. 44b. Parallel and series combinations give

the result shown in Fig. 44c for the pre-fault network. Xu = 1.10 per unit.The faulted network is obtained by grounding the midpoint X of one

line in Fig. 44b. Then a ~ Y conversion, series combinations, and a ~conversion lead to the ~ network of Fig. 44d. X12 = 7.20 per unit.

The post-fault network is obtained by omitting one of the parallel lines of

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Fig. 44b. Then a series combination leads to Fig. 44e. X 12 = 1.28 per

unit.

2. Inertia constant. The inertia constants in Table 2, given on the machine base, are converted to the system base by multiplying them by (rna-~ ~ ~ a

6.6/132 kv. 0.35 125.4/33.0 kv.

051 O . ~ 1 5 2~ ~ 3 ~ b0.35

/>. ~vvv (c)

(d )

~ ~ (e)

FIG. 44. Reduction of the network of Fig. 43 (Example 4). The reduced networksare given by c for the pre-fault condition, by d for the fault condition, and by e for

the post-fault condition.

chine rating)/(100 Mva.). Then

HI = Hi+ H k = 1.76

H2 = Ha+ Hb+ H c+ Hd+ He+H/+ H, + Hh = 14.1

H = H1H 2 = 1.76 X 14.1 = 1.56

HI + H2 15.9

3. Curves ofP« andPi versus sin 00. Assume that 132kv. (1.00 per unit)

is maintained on the high-voltage side of the step-up transformers, and 33.0kv. (0.95 per unit) is maintained on the low-voltage side of the step-down

transformers. The voltages behind transient reactance, E1 and E21 as

functions of the initial angle 00 between them, can be found from the vectordiagram of Fig. 45 by assigningarbitrary values to the current I and solving,

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either graphically or by trigonometric computation, the resulting triangles

for E ll E2, and 00. Then P« and Pi are calculated from the relations

P _ E1E2 _ E1E2

m - Xl2

- 1.10

and

Pi = Pm sin 00

0.661 0.121

FIG. 45. Vector diagram for determining E1 and E2 as functions of (Example 4).

TABLE 3

CALCULATION OF Pm AND Pi VERSUS sin ao (EXAMPLE 4)

P, EI E2 sin Pm

0.125 1.106 0.933 7.7 0.134 0.9350.250 1.121 0.934 15.3 0.264 0.9480.375 1.145 0.936 22.7 0.386 0.971

0.500 1.177 0.939 30.0 0.499 1.0010.750 1.267 0.950 43.6 0.689 1.0901.000 1.373 0.972 55.9 0.828 1.221

The results of the calculations are given in Table 3. Curves of Pm and Pi as

functions of sin 00 are plotted in Fig. 46.

4. Computation of f l and f2 . By eqs. 20 and 21,

1.10 3f l= 0.15

7.20

1.10r2= -= 0.86

1.28

5. Stability limits for sustained fault and [or instantaneous clearing.

a. Sustainedfault. Entering the curve of Fig. 40 with f l = 0.153, we read

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sin 80 = 0.117. Entering the Pi curve of Fig. 46 with this value of sin 00, we

read Pi = 0.110.

b. Instantaneous clearing. Entering the curve of Fig. 40 with '2 = 0.86,

I

/Pm /

v1----- J

/

J V

//1>;

/1/

V

0.5

1.0

oo 05 IDsin 00

FIG. 46. Curves of Pm and Pi as functions of sin 80 (Example 4).

TABLE 4

CALCULATION OF POWER LIMIT Pi AS A FUNCTION OF CLEARING TIME te(EXAMPLE 4)

1 2 3 4 5 6

sin 80 Pi Pm Tc/tcteTe (sec.)

0.80 0.933 1.18 4.68 0.13 0.030.75 0.843 1.14 4.57 0.24 0.050.70 0.763 1.10 4.50 0.35 0.080.65 0.690 1.07 4.44 0.45 0.100.60 0.623 1.04 4.37 0.55 0.130.55 0.560 1.02 4.33 0.67 0.150.50 0.500 1.00 4.29 0.78 0.180.45 0.443 0.986 4.26 0.90 0.22

0.40 0.390 0.973 4.23 1.07 0.250.35 0.336 0.963 4.21 1.26 0.300.30 0.285 0.954 4.19 1.50 0.360.25 0.236 0.946 4.17 1.79 0.430.20 0.188 0.940 4.16 2.29 0.550.15 0.141 0.936 4.15 3.31 0.80

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SUMMARY OF CALCULATING TRANSIENT STABILITY 185

1\\,

\ 'i\

,' r-,

............ ............ r - - . . ::::-

we read sin 00 = 0.825. Entering the Pi curve of Fig. 46 with this value of

sin 00, we read Pi = 0.995.

6, 7, and 8. Stability limii« [or finite clearing times. These three steps arecarried out in Table 4. In col. 1 are entered values of sin 00 from 0.15 to 0.80

at intervals of 0:05. In cola. 2 and 3 are the corresponding values of Pi and

1.0

1.0

1r X 60 X 0.153 v p = 4.3 v p1 X 1.56 m m

oo 0.5

Clearing timet, (seconds)

FIG. 47. Curve of stability limit as a function of clearing time, three-phase shortcircuit at the middle of one of the 132-kv. lines of the system of Fig. 43 (Example 4).

PM read from the curves of Fig. 46. In col. 4 is the value of rcltc,which by

eq, 27 is

In col. 5 is the value of Te read from the curves. In col. 6 is fe, obtained by

dividing Te from col. 5 by Tc/ tcfrom col. 4.

9. Curve oj stability limit as a Junction oj clearing time is plotted in Fig. 47.

Summary of methods of calculating transient stability. Byrd and

Pritchard's method, which has just been illustrated, is probably the

best way of finding the critical switching time corresponding to a given

transmitted power, or of finding the transient stability limit for a given

switching time, for a two-machine pure-reactance system with only one

instant of switching after inception of the fault.

If the network is not purely reactive (for example, if line resistance

or shunt loads are to be taken into account), this method is not applicable. However, the equal-area method may be used in conjunction

with pre-calculated swing curves as described on pp. 149-57 of this

chapter.

If the system has three or more machines, their swing curves must be

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TABLE 5

RECOMMENDED METHODS OF CALCULATING TRANSIENT STABILITY OF A TwoMACHINE SYSTEM

Number of Recommended Method of Calculating

Instants of A BSwitching Examples Critical Switching Time Stability Limit Corre-or Circuit Corresponding to a sponding to a GivenChange Given Power Switching Time

Sustained fault Use equal-area crite-

1Fault cleared instantly rion or Byrd and

Switching out a sound Pritchard's methodline (steps 1 to 5).

(Pure-reactance network)

Plot part of curve by Byrd and Pritchard'smethod.

Fault cleared in finite (Linear impedance network)time by simultane-

Use equal-area method Cut and try by assum-2 ous opening of all

breakersfor finding critical ing power and pro-

switching angle, then ceeding as in A.

pre-calculated swingcurve for findingswitching time.

Fault cleared by se.,quential opening of2 breakers

3 Fault cleared by si- Dse a combination of swing curves, equal-areamultaneous opening criterion, and successive trials.of all breakers, fol-lowed by simultane-ous reclosing

calculated to a value of time when one can determine whether the sys

tem is stable for the assumed power and switching time. The pro

cedure is then repeated for different values of power or of switching

time.

If there is more than one instant of switching after inception of the

fault on a two-machine system (for instance, if a fault is cleared by

successive opening of two or more circuit breakers), it is still possibleto use the equal-area criterion as an adjunct to swing curves, making

continuation of the swing curves beyond the instant of the last switch

ing operation unnecessary. Suppose, for example, that a fault is

cleared by successive opening of two breakers at known times and that

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CERTAIN FACTORS AFFECTING STABILITY 187

the power limit is sought. A value of power is assumed, and the swing

curve is calculated till the time of opening of the second breaker.

The equal-area criterion for stability is then used. The procedure isrepeated for other values of power until one is found which makes the

positive and negative areas equal. This value is the stability limit.

A plot of net area against power enables one to find the stability limit

by interpolation after two or three trials.

Table 5 lists the foregoing methods of calculation for two-machine

systems with recommendations concerning which to use in each of

various situations.

For estimating purposes two sets of curves which are not reproduced

here will be found useful. One of them gives, for a two-machinereactance system, curves of critical clearing time tc against sin 00 for

constant rl and r2. Four families of such curves are given, each for a

value of r l corresponding to a different type of short circuit at the

sending end of a transmission line and for several values .of r2. The

equivalent inertia constant, H = H1H2 H 1 + H2 ) , is assumed to be

1.5. If it has a different .value, the switching time read from the

curves must be multiplied by a correction factor VH/1.5.

The other set of curves gives the critical clearing time of a fault on ametropolitan power system having generators of H = 8 operating at

full load and 0.85 power factor, based on the assumption that the

generator or group of generators nearest the fault swings with respect

to the remaining generators.

Certain factors affecting stability. With a knowledge of the meth

ods ofanalyzing the stability of the two-machine system that have been

described in the preceding pages, we can proceed to draw a number of

general conclusions regarding the effect on stability of certain features

of apparatus design, system layout, and operation. The effect of eachfeature must be considered under all three conditions-before the

fault, during the fault, and after clearing the fault. Some features of

layout or design promote stability during all three conditions, whereas

others are beneficial during one condition but detrimental during

another.

In the equal-area criterion for stability, the power-angle curves for

each of the three conditions are used. (Refer to Fig. 14.) The factors

determining the relative sizes of the two compared areas Al and A 2 are

(1) the clearing angle Oc and (2) the amplitudes of the three power-angle curves, Pm, rIPm, r2Pm» relative to the height of the input

(initial power) linePi. Stability isaided by decreasing area A 1 and by

increasing area A 2 • For a given Pi this may be accomplished, from

the geometrical viewpoint, chiefly by decreasing the clearing angle Oc

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and by increasing the amplitudes of the fault and post-fault curves,

T1Pm and T2Pm. It is also helpful to decrease the amplitude of the

pre-fault power-angle curve if such reduction is possible without at thesame time decreasing the amplitudes of the other two power-angle

curves, because so doing increases the initial angle 00- If any feature

changes the amplitudes of all three curves proportionally, however, it is

beneficial to increase rather than to decrease the amplitudes.

The clearing angle s, depends on the clearing time and on the equiv

alent inertia constant (eq. 2). The importance of rapid fault clearing,

obtainable by the use of high-speed circuit breakers and fast relaying,

has already been stressed. For a given clearing time the clearing

angle is decreased by increasing the inertia constant. If the inertiaconstants of the two machines are far from equal, the equivalent

inertia constant of the system is very nearly that of the lighter ma

chine; hence it is more effective to increase the inertia of the light

machine than of the heavy one. Seldom, however, has it proved

economical to increase the inertia of a generator beyond its normal

value. t But a consideration of the role of inertia sheds some light on

the relative critical clearing times for hydroelectric generators (average

H = 3) and steam turbogenerators (average H = 6) if the circuit

reactances are about the same. Since the time varies as the square root

of the inertia constant Ceq. 27), a fault on a hydroelectric system must

be cleared in about 70% of the critical clearing time of the correspond

ing steam system.

The amplitude of the power-angle curves is E1E2 X 12 , where E1 and

E2 are the internal voltages of the two synchronous machines and X 12

is the reactance between these voltages, which is, in general, different

for each of the three circuit conditions. Increasing the internal volt

ages increases the amplitude of all three power-angle curves and aidsstability. An increase in internal voltages usually accompanies an

increase of load on the machines, but this does not necessarily mean an

increase of the equivalent input Pi. If both generators have local

loads which are increased in the proper ratio, Pi is not affected. (See

eq.3.)

The amplitude of all the power-angle curves is increased by decreas

ing the reactance X 12 between the machines. This reactance consists

principally of the reactances of the two synchronous machines, trans

former reactance, and line reactance. A large part of it is in the machines. The transient reactance of each class of large synchronous

machines (steam turbogenerator, water-wheel generator, condenser,

etc.) has a characteristic value and does not vary much in normal

tA notable exception is the Boulder Dam station.

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CERTAIN FACTORS AFFECTING STABILITY 189

designs. A lower value of reactance (in per unit) is obtained essentially by building a larger machine and under-rating it. This has

seldom proved an economicalway to aid stability.§ The reactances oftransformers also have, for a given size and voltage, normal values

belowwhich it is difficult to go. The reactance of an overhead trans

mission line is only slightly affected by any change of spacing or con

ductor size which is practical from other standpoints than stability.

A decrease in system frequency, of course, decreases the reactances of

all parts of the system and thereby aids stability. Therefore a fre

quency of 25 or 50 c.p.s. is preferable to 60 c.p.s. from the standpoint of

stability. Nevertheless, because of other advantages, 60 c.p.s. has

become the standard power-systemfrequency in the United States, anda transmission project operating at a lower frequency would necessarily

include frequency-changing equipment at one or both ends. This ad

ditional equipment might offset any advantage of the lower frequency

in regard to stability. Up to the present time the stability limitations

of 60-c.p.s. systems have not been serious enough to warrant lower

frequencies. Series capacitors have been used on some transmission

lines to partially compensate for the inductive reactance of the lines.

However, they have not been used on any major project where stability

was an important factor, although they have been seriously considered.6,7,g

The most important means of reducing the reactance are (1) to

connect more lines in parallel and (2) to raise the transmission volt

ages.The ratio r l of the amplitude of the fault power-angle curve to the

amplitude of the pre-fault power-angle curve depends upon the type

and location of the fault. The effect of the type of fault will be dis

cussed in Chapter VI. The effect of location will be considered

briefly now. A fault on a bus or on a line close to a bus is more severe

than a fault of the same type near the middle of a line. Most severe

of all is a three-phase short circuit at some point (for example, fault b

in Fig. 15) where it entirely blocks the transfer of power from one

generator to the other; then r l = o. Although a fault near the end

of a line and one near the middle of the line are equally probable, the

most severe location is usually assumed in a stability study so that the

results will be conservative.

The ratio T2 of the amplitude of the post-fault curve to the amplitudeof the pre-fault curve depends on the number and location of lines

which are opened to clear the fault; therefore it depends upon the

fault location and upon the relaying scheme. The most favorable

§Again the Boulder Dam station is a notable exception.

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value of T2 encountered in practice is 1. This value applies to a fault

on an unloaded radial feeder cleared by disconnection of the feeder.

It is also attained with quick-reclosing schemes which restore thefaulted line to service after a brief time to allow for extinction of the

arc. II A fault on a bus, although electrically equivalent before clearing

to a fault on a line adjacent to a bus, is more severe in its effects after

clearing than a line fault is because several lines must be opened to

clear it. Sometimes a comparative study is made between a small

number of high-voltage lines and a large number of low-voltage lines

having the same parallel reactance in per unit. The large number of

lines is preferable to the small number from the standpoint of stability,

because to clear a fault requires the opening of one line in either case,

24

24

16

J

16 16

FIG. 48.System forwhichthe effectof high-voltagebussingis considered in Prob, 4.

but one line is a small fraction of the entire number of lines when thisnumber is great. For other reasons, however, a smaller number oflines may be preferable.

When two or more parallel lines are used, the number of intermediatebusses is a factor affecting stability. For example, in the system ofFig. 48 it is worth inquiring whether the addition of the two high-voltage busses shown in broken lines is beneficial or detrimental. If the

high-voltage busses and circuit breakers are provided, a line fault can

be cleared by switching out one line while leaving all the transformersconnected, whereas, without the high-voltage busses, the transformers

would be switched out with the line. Thus the busses increase the

value of T2 and are beneficial after clearing of the fault. On the other

hand, during the fault the shock to the system is increased by the

busses; that is, Tt is lessened. Which effect predominates depends

upon the speed of clearing. With fast clearing the busses are benefi

cial; with slow clearing they are detrimental.

The same principle may determine, although less obviously, the

effectupon stability of changes of layout inmore complicated networks.

The effect of a given change may be either beneficial or detrimental,

depending upon the speed of fault clearing. Needless to say,_ many

USee Chapter XI, Vol. II.

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PROBLEMS 191

contemplated changes of connections are conceived from other motivesthan the improvement of stability. They may be intended, for

example, to improve voltage regulation, to increase the reliability ofsupply to certain loads, or to relieve overloading of certain lines.

Nevertheless the effect of such changes on stability conditions should

be considered.Most of the conclusionswhich have been drawn from a study of the

two-machine system apply equally well to a multimachine system.

When a muItimachine system is unstable, as a rule, it is split into two

groups of machines whichgoout of synchronism with each other, while

the machines within each group stay in synchronism. The grouping

may differ with the fault location. Still, for a given fault location,the general behavior of the multimachine system is similar to that of a

two-machine system. It could be analyzed as a two-machine system

except that the machines within a group may swing so far with respect

to each other (yet without goingout of step) as to have a marked effect

on the relations between the two groups. In addition, the grouping

for a given fault location is not always apparent in advance.

REFERENCES

1. R. H. PARK and E. H. BANCKER, System Stability as a Design Problem,

A.l.E.E. Trans., vol. 48, pp. 170-93, January, 1929.

2. I. H. SUMMERS and J. B. MCCLURE, Progress in the Study of System Stabil

ity, A.I.E.E. Trans., vol. 49, pp. 132-58, January, 1930.

3. H. L. BYRD and S. R. PRITCHARD, JR., Solution of the Two-Machine

Stability Problem, Gen. Elec. Rev.,vol. 36, pp. 81-93, February, 1933.

4. R. D. EVANS and W. A. LEWIS, Selecting Breaker Speeds for Stable Opera

tion, Elec. Wid., vol. 95, pp. 336-40, Feb. 15, 1930.

5. S. B. GRISCOM, W. A. LEWIS, and W. R. ELLIS, Generalized Stability Solu

tion for Metropolitan Type Systems, A.I.E.E. Trans., vol. 51, pp, 363-72, June,

1932; disc., pp. 373-4.6. E. C. STARR and R. D. EVANS, Series Capacitors for Transmission Circuits,

A. I. E. E. Trans., vol. 61, pp. 963-73, 1942; disc., pp. 1044-6.

7. R. B. BODINE, C. CONCORDIA, and GABRIEL KRON, Self-Excited Oscillations

of Capacitor-Compensated Long-Distance Transmission Lines, A.I.E.E. Trans.,

vol. 62, pp, 41-4, January, 1943; disc., pp. 371-2.

8. J. W. BUTLER, J. E. PAUL, and T. ·W. SCHROEDER, Steady-State and Tran

sient-Stability Analysis of Series Capacitors in Long Transmission Lines, A.I.E.E.

Trans., vol. 62, pp. 58-65, February, 1943; disc., pp. 377-80.

PROBLEMS ON CHAPTER V

1. Check the results of Example 3, part a, by Byrd and Pritchard's

method.

2. Work Example 4 for a three-phase fault on a 132-kv. line near thehydroelectric-station high-voltage bus.

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3. WorkExample 4, assuming that the voltages behind transient reactanceare independent of the initial power transmitted.

4. Find the effect of high-voltage bussing on the stability of the 6o-cyclesystem shown in Fig, 48 (consisting of a generator G1 feeding over a double

circuit high-voltage line to the infinite bus G2) by plotting stability limit in

per unit against fault duration in seconds for a three-phase fault at the sending end of one circuit (a) with no high-voltage bussing and (b) with high

voltage bussing at both ends, as indicated by the broken lines. The re

actances of the circuit elements are given in per cent based on the rating of

G1. Assume that the voltage of the infinite bus and the sending-end

voltage of the high-voltage lines are 1.00 per unit for all values of initial

power. Explain why the curves cross.

5. Find the effect of fault location on the system of Fig. 48 with high

voltage bussing, by plotting stability limit in per unit as a function of fault

location as the three-phase fault moves from the sending end to the receiving

end of the line. The clearing time is constant at 0.15 sec.

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