Web Answers for Power Systems - II Chapter 2: Nature of Faults in Electrical System Question 5 Answer: By percentage reactance we mean the percentage of the total phase voltage dropped in the circuit when full load current is flowing i.e. % 100 IX X V ….(1) where I = full load current, V = phase voltage and X = reactance in ohms per phase. It can also be expressed in terms of KVA or KV as under: From equation (1) ……….(2) where X is the reactance in ohms. If X is the only reactance element in the circuit then short circuit current is given by SC V I X . By putting the value of X from equation (1), 100 % SC I I X i.e. short circuit current is obtained by multiplying the full load current by 100/% X. In view of what have been stated above it is worthwhile to mention here the advantage of using percentage reactance instead of ohmic reactance in short circuit calculations. Percentage reactance values remain unchanged as they are referred through transformers, unlike ohmic reactances which become multiplied or divided by the square of transformation ratio. Further, as the various equipments used in the power system have different KVA ratings, it is necessary to find the percentage reactance of all the elements on a common KVA rating which is known as base KVA and the conversion is effected by using the relation as %age reactance of base KVA = Base KVA Rated KVA %age reactance at rated KVA. These all led to conclude that it is preferable to express the reactances of various elements in percentage values for short circuit calculations.
Transcript
Web Answers for Power Systems - II
Chapter 2: Nature of Faults in Electrical System
Question 5
Answer:
By percentage reactance we mean the percentage of the total phase voltage dropped in the circuit
when full load current is flowing i.e.
% 100IX
XV
….(1)
where I = full load current, V = phase voltage and X = reactance in ohms per phase.
It can also be expressed in terms of KVA or KV as under:
From equation (1)
……….(2) where X is the reactance in ohms.
If X is the only reactance element in the circuit then short circuit current is given by SC
VI
X.
By putting the value of X from equation (1), 100
%SCI I
X
i.e. short circuit current is obtained by multiplying the full load current by 100/% X.
In view of what have been stated above it is worthwhile to mention here the advantage of using
percentage reactance instead of ohmic reactance in short circuit calculations. Percentage reactance
values remain unchanged as they are referred through transformers, unlike ohmic reactances which
become multiplied or divided by the square of transformation ratio.
Further, as the various equipments used in the power system have different KVA ratings, it is
necessary to find the percentage reactance of all the elements on a common KVA rating which is
known as base KVA and the conversion is effected by using the relation as %age reactance of base
KVA = Base KVA
Rated KVA%age reactance at rated KVA.
These all led to conclude that it is preferable to express the reactances of various elements in
percentage values for short circuit calculations.
Question 6
Answer:
Whatever may be the value of base KVA, short circuit current is the same i.e., the value of short
circuit current does not change if different base KVAs are taken.
In order to explain the answer, a single line diagram of the system is shown below in which a 3 phase
transmission line operating at 66 KV and connected through a 1000 KVA transformer with 5%
reactance to a generating station bus bar is considered.
The generator is of 2500 KVA with 10% reactance.
i) If 2500 KVA is chosen as the common base KVA, the reactance of the various elements in the
system on this base value will be:
Reactance of transformer at 2500 KVA base
2500
5 12.5%1000
Reactance of generator at 2500 KVA base =2500
10 10%2500
Total percentage reactance on the common base KVA
% 12.5 10 22.5%X
The full load current corresponding to 2500 KVA base at 66 KV is given by:
2500 1000
21.873 66 1000
I A
Short circuit current 100 100
21.87 97.2% 22.5
SCI I AX
ii) Now if 5000 KVA is chosen as the common base KVA, then:
Reactance of transformer at 5000 KVA base
5000
10 25%1000
Reactance of generator at 5000KVA base
5000
10 20%1000
Total percentage reactance on the common base KVA
%X = 25 + 20 = 45%
~ 2500 KVA 10%
1000 KVA 5%
11/66 KV
66 KV Line
Single line diagram of the system
Full load current corresponding to 5000 KVA base at 66 KV is
5000 1000
43.743 66 1000
I A
Short circuit current, 100 100
43.74 97.2% 45
SCI I AX
If the two cases as shown above are examined, it may be observed that for two different base KVA,
short circuit current is the same.
Example 8
Solution:
Let base KVA be 35,000 KVA
% Reactance of alternator A at base KVA
35,000% 30 70%
15,000AX
% of reactance of alternator B at base KVA
35,000% 50 87.5%
20,000BX
Line current corresponding to 35,000 KVA at 12 KV
A,
I 168410123
10000353
3
Fig. (ii) shows the reactance diagram of the network at the selected base KVA.
Total % reactance from the generator neutral upto fault point,
% A BX X X70 87.5
38.89%70 87.5
A B
A B
X X
X X
Short circuit current 100 100
1684 4330% 38.89
SCI I AX
.
Example 9
Solution:
Assuming base MVA as 1200, the percentage reactance of one generating station is 100% and that of
the other is
%150100
800
1200
The % reactance of the cable is 0.5 1200
100 496%11 11
When a 3- fault takes place at 1200 MVA capacity plant the equivalent circuit will be as follows.
When the fault is F, fault impedance between F and the neutral bus will be 86.59%
The short circuit MVA of this bus will be as follows 1200
100 138686.59
MVA
For fault at the other station, the equivalent circuit will be as follows:
The equivalent fault impedance between F and neutral bus will be 119.84%
The short circuit MVA will be 1200
100 1001119.84
MVA .
Example 10
Solution:
Let 10,000 KVA be the base KVA
% Reactance of alternator A at base KVA
3
10,000% 10 10%
10 10AX
% Reactance of transformer or base KVA
3
10,000% 5 10%
5 10TX
Since the line impedance is given in ohms, conversion into % impedance is made by using the
following relation.
% impedance =
~ ~
10MVA
10%
Load
5MVA
5% F1 1Ω 4Ω F2
Fig: 1
ZPB
150% 100%
496%
F
150%
496%
100%
F
% reactance of transmission line using the above relation comes to
% resistance of transmission line,
i) The reactance diagram of the network on the selected base KVA is shown in
fig.2
Total % reactance
2% % %
10 10 40
60%
A TX X X
% Resistance = 10%
% impedance from generator neutral upto fault point 2F
short-circuit KVA=100
10,000 16,44060.83
KVA .
ii) For a fault at the high voltage terminals of the transformer (point 1F ), total % reactance from
generation neutral upto fault point
1 % % 10 10 20%A TF X X
Short – circuit KVA =100
10,000 50,00020
KVA .
Neutral
XA = 10%
XT = 10%
F1
RL = 10%
XL = 40%
F2
Fig: 2
Chapter 3: Power System Dynamics
Example 2
Solution:
As the system is operating initially under steady state condition, a small perturbation in power will
make the rotor to oscillate. The natural frequency of Oscillation is given by:-
1 2
0 /Cn
Pf M
For 60% loading:
cos21
X
VVPe
0
1.1 1 1.1cos .8 1.76
.5 .5 [Since sin 0
0.60.6
1]
50
31
f
GHM
1.76
50 9.6 rad / sec. 1.533
nf Hz.
Example 8
Solution:
a) The equivalent circuit is shown in figure (b), from which the equivalent reactance between the
machine internal voltage and infinite bus is
per unit
~
~
G
B11
B12
1 2
1 2
3
3
+
–
B21
B22 F
B13
+
–
0.30dX
0.10TRX 12 0.20X
1.0busV
13 0.10X 23 0.20X
E
djX
0.30j
ep TRjX
0.10j
12jX
0.20j 13jX
0.10j
23jX
0.20j 1.0 0
Fig: (a) Single-line diagram
for example 8
Fig: (b) Equivalent circuit
The current into the infinite bus is
1.05263 18.195 per unit
and the machine internal voltage is
bus eqE E V jX I
1.0 0 0.54737 71.805
1.1709 0.5200 1.2812 23.946j per unit
b) From per unit.
Example 9
Solution:
Plots of ep and mp versus are shown in figure (i). From example 8 the initial operating point is
per unit and radian. At 0t , when
the short circuit occurs, ep instantaneously drops to zero and remains at zero during the fault since
power cannot be transferred past faulted bus 1.
with ,
Integrating twice with initial conditions and 0
0d
dt,
At 3t cycles = 0.05 second,
The accelerating area A1 shaded in figure (i), is
At 0.05t s the fault extinguishes and ep instantaneously increases from zero to the sinusoidal
curve in Fig. (i). continuous to increase until the decelerating area A2 equals A1. That is,
Integrating,
The above nonlinear algebraic equation can be solved iteratively to obtain
2 0.7003 radian 40.12
Since the maximum angle 2 does not exceed , stability is maintained. In
steady-state, the generator returns to its initial operating point 1.0ess mp p per unit and
0 23.95ss .
Note that as the fault duration increases, the risk of instability also increases, The critical clearing
time, denoted crt , is the longest fault duration allowable for stability.
Example 10
Solution:
The p plot is shown in figure 1. At the critical clearing angle, denoted cr , the fault is
extinguished. The power angle then increases to a maximum value
3 0180 156.05 2.7236 radians, which gives the maximum decelerating area. Equating
the accelerating and decelerating
p (per unit)
Pmax = 2.4638
pm = 1.0
A2
A1
pe = 2.4638 sin
(radians) π 3 2 1 0
2
Fig: (i) p– plot
areas,
Solving for cr ,
2.4638cos 0.05402cr
1.5489 radians 88.74cr
From the solution to the swing equation given in Example 9,
Solving
Using radian,
cycles.
If the fault is cleared before 11.38crt t cycles, stability is maintained. Otherwise, the generator
goes out of synchronism with the infinite bus; that is, stability is lost.
