Web for Power Systems II B.tech 6th Sem
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Transcript of Web for Power Systems II B.tech 6th Sem
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.
p (per unit)
Pmax = 2.4638
pm = 1.0
A1
pe = 2.4638 sin
(radians) π 3 0
Fig: 1 p– plot
A2
cr
Chapter 4: Load Flow Studies
Example 6
Solution:
30-bus IEEE sample system
G
G
C
C
C
C
Three Winding Transformer
Equivalents
13 9 10
4 12
4
1
2
4
3
13
14
12
15
16
18
23
10
20
21 19
17
6 5 7
9
11 22
24
30
25
27
29
28
26
8 G: Generators
G: Synchronous condensers
Bus data for IEEE-30 BUS
Clear % clears all variables from workspace.
Basemva = 100; accuracy = 0.001; accel = 1.8; maxiter = 100;
% IEEE 30-BUS TEST SYSTEM (American Electric Power)
% Bus Bus Voltage Angle --Load-- ---Generator---Injected
% N
o.
C
O
D E
Mag. Deg-
ree
MW Mvar MW Mvar Qmin Qmax M
v
a r
Bus-
data=
[1 1 1.06 0 0.0 0.0 0.0 0.0 0 0 0
2 2 1.043 0 21.70 12.7 40.0 0.0 -40 50 0
3 0 1.0 0 2.4 1.2 0.0 0.0 0 0 0
4 0 1.06 0 7.6 1.6 0.0 0.0 0 0 0
5 2 1.01 0 94.2 19.0 0.0 0.0 -40 40 0
5 0 1.0 0 0.0 0.0 0.0 0.0 0 0 0
7 0 1.0 0 22.8 10.9 0.0 0.0 0 0 0
8 2 1.01 0 30.0 30.0 0.0 0.0 -10 40 0
9 0 1.0 0 0.0 0.0 0.0 0.0 0 0 0
10 0 1.0 0 5.8 2.0 0.0 0.0 0 0 19
11 2 1.082 0 0.0 0.0 0.0 0.0 -6 24 0
12 0 1.0 0 11.2 7.5 0 0 0 0 0
13 2 1.071 0 0.0 0.0 0 0 -6 24 0
14 0 1.0 0 6.2 1.6 0 0 0 0 0
15 0 1.0 0 8.2 2.5 0 0 0 0 0
16 0 1.0 0 3.5 1.8 0 0 0 0 0
17 0 1.0 0 9.0 5.8 0 0 0 0 0
18 0 1.0 0 3.2 0.9 0 0 0 0 0
19 0 1.0 0 9.5 3.4 0 0 0 0 0
20 0 1.0 0 2.2 0.7 0 0 0 0 0
21 0 1.0 0 17.5 11.2 0 0 0 0 0
22 0 1.0 0 0..0 0.0 0 0 0 0 0
23 0 1.0 0 3.2 1.6 0 0 0 0 0
24 0 1.0 0 8.7 6.7 0 0 0 0 4.3
25 0 1.0 0 0.0 0.0 0 0 0 0 0
26 0 1.0 0 3.5 2.3 0 0 0 0 0
27 0 1.0 0 0.0 0.0 0 0 0 0 0
28 0 1.0 0 0.0 0.0 0 0 0 0 0
29 0 1.0 0 2.4 0.9 0 0 0 0 0
30 0 1.0 0 10.6 1.9 0 0 0 0 0];
% Line Data
%
% Bus bus R X ½ B 1 for Line code or
% nl nr pu pu pu tap setting value
Line data=
[1 2 0.0192 0.0192 0.02640 1
1 3 0.0452 0.1852 0.02040 1
2 4 0.4570 0.1737 0.01840 1
3 4 0.0132 0.0379 0.00420 1
2 5 0.0472 0.1983 0.02090 1
2 6 0.0581 0.1763 0.01870 1
4 6 0.0119 0.0414 0.00450 1
5 7 0.0460 0.1160 0.01020 1
6 7 0.0267 0.0820 0.00850 1
6 8 0.0120 0.0420 0.00450 1
6 9 0.0 0.2080 0.0 0.978
6 10 0.0 0.5560 0.0 0.969
9 11 0.0 0.2080 0.0 1
9 10 0.0 0.1100 0.0 1
4 12 0.0 0.2560 0.0 0.932
12 13 0.0 0.1400 0.0 1
12 14 0.1231 0.2559 0.0 1
12 15 0.0662 0.1304 0.0 1
12 16 0.945 0.1987 0.0 1
14 15 0.2210 0.1997 0.0 1
16 17 0.0824 0.1923 0.0 1
15 18 0.1073 0.2185 0.0 1
18 19 0.0639 0.1292 0.0 1
19 20 0.0340 0.0680 0.0 1
10 20 0.0936 0.2090 0.0 1
10 17 0.0324 0.0845 0.0 1
10 21 0.0348 0.0749 0.0 1
10 22 0.0727 0.1499 0.0 1
21 22 0.0116 0.0236 0.0 1
15 23 0.1000 0.2020 0.0 1
22 24 0.1150 0.1790 0.0 1
23 24 0.1320 0.2700 0.0 1
24 25 0.1885 0.3292 0.0 1
25 26 0.2544 0.3800 0.0 1
25 27 0.1093 0.2087 0.0 1
28 27 0.0000 0.3960 0.0 0.968
27 29 0.2198 0.4153 0.0 1
27 30 0.3200 0.6027 0.0 1
29 30 0.2399 0.4533 0.0 1
8 28 0.0636 0.2000 0.0214 1
6 28 0.0160 0.0599 0.065 1]
Power Flow solution by Gauss-Seidel Method
Maximum power mismatch=0.000951884
Bus Voltage Angle ------Load-------- ---Generation----- Injected
No. Mag. Degree MW Mvar MW Mvar Mvar
1 1.060 0.000 0.000 0.000 -17.010 0.00
2 1.043 -5.496 21.700 12.700 40.000 48.826 0.00
3 1.022 -8.002 2.400 1.200 0.000 0.000 0.00
4 1.013 -9.659 7.600 1.600 0.000 0.000 0.00
5 1.010 -14.380 94.200 19.000 0.000 35.995 0.00
6 1.012 -11.396 0.000 0.000 0.000 0.000 0.00
7 1.003 -13.149 22.800 10.900 0.000 0.000 0.00
8 1.010 -12.114 30.000 30.000 0.000 30.759 0.00
9 1.151 -14.432 0.000 0.000 0.000 0.000 0.00
10 1.044 -16.024 5.800 2.000 0.000 0.000 19.00
11 1.082 -14.432 0.000 0.000 0.000 16.113 0.00
12 1.057 -15.301 11.200 7.500 0.000 0.000 0.00
13 1.071 -15.300 0.000 0.000 0.000 10.406 0.00
14 1.043 -16.190 6.200 1.600 0.000 0.000 0.00
15 1.038 -16.276 8.200 2.500 0.000 0.000 0.00
16 1.045 -15.879 3.500 1.800 0.000 0.000 0.00
17 1.039 -16.187 9.000 5.800 0.000 0.000 0.00
18 1.028 -16.881 3.200 0.900 0.000 0.000 0.00
19 1.025 -17.049 9.500 3.400 0.000 0.000 0.00
20 1.029 -16.851 2.200 0.700 0.000 0.000 0.00
21 1.032 -16.468 17.500 11.200 0.000 0.000 0.00
22 1.033 -16.455 0.000 0.000 0.000 0.000 0.00
23 1.027 -16.660 3.200 1.600 0.000 0.000 0.00
24 1.022 -16.829 8.700 6.700 0.000 0.000 4.30
25 1.019 -16.423 0.000 0.000 0.000 0.000 0.00
26 1.001 -16.835 3.500 2.300 0.000 0.000 0.00
27 1.026 -15.913 0.000 0.000 0.000 0.000 0.00
28 1.011 -12.056- 0.000 0.000 0.000 0.000 0.00
29 1.006 -17.133 2.400 0.900 0.000 0.000 0.00
30 0.994 -18.016 10.600 1.900 0.000 0.000 0.00
Total 283.400 126.200 300.950 125.089 23.30
Line Flow and Losses
--Line-- Power at bus & line flow --Line loss-- Transformer
From To MW Mvar MVA MW Mvar tap
1 260.950 -17.010 261.504
2 177.743 -22140 179.117 5.461 10.517
3 83.197 5.125 83.354 2.807 7.079
2 18.300 36.126 40.497
1 -172.282 32.657 175.350 5.461 10.517
4 45.702 2.720 45.783 1.106 -0.519
5 82.990 1.704 83.008 2.995 8.178
6 61.905 -0.966 61.913 2.047 2.263
3 -2.400 -1.200 2.683
1 -80.390 1.954 80.414 2.807 7.079
4 78.034 -3.087 78.095 0.771 1.345
4 -7.600 -1.600 7.767
2 -44.596 -3.239 44.713 1.106 -0.519
3 -77.263 4.432 77.390 0.771 1.345
6 70.132 -17.624 72.313 0.605 1.181
12 44.131 14.627 46.492 0.000 4.686 0.932
5 -94.200 16.995 95.721
2 -79.995 6.474 80.256 2.995 8.178
7 -14.210 10.467 17.649 0.151 -1.687
6 0.000 0.000 0.000
2 -59.585 3.229 59.945 2.047 2.263
4 -69.527 18.805 72.026 0.605 1.181
7 37.537 -1.915 37.586 0.368 -0.598
8 29.534 -3.712 29.766 0.103 -0558
9 27.687 -7.318 28.638 0.000 1.593 0.978
10 15.828 0.656 15.842 0.000 1.279 0.969
28 18.840 -9.575 21.134 0.060 -13.085
7 -22.800 -10.900 25.272
5 14.361 -12.154 18.814 0.151 -1.687
6 -37.170 1.317 37.193 0.368 -0.598
8 -30.000 0.759 30.010
6 -29.431 3.154 29.599 0.103 -0.558
28 -0.570 -2.366 2.433 0.000 -4.368
9 0.000 0.000 0.000
6 -27.687 8.911 29.086 0.00 1.593
11 0.003 -15.653 15.653 -0.000 0.461
10 27.731 6.747 28.540 0.000 0.811
10 -5.800 17.00 17.962
6 -15.828 0.623 15.640 0.000 1.279
9 -27.731 -5.936 28.359 0.000 0.811
20 9.018 3.569 9.698 0.081 0.180
17 5.347 4.393 6.920 0.014 0.037
21 15.723 9.846 18.551 0.110 0.236
22 7.582 4.487 8.811 0.052 0.107
11 0.000 16.113 16.113
9 -0.003 16.114 16.114 -0.000 0.461
12 -11.200 -7500 13.479
4 -44.131 -9.941 45.237 0.000 4.686
13 -0.021 -10.274 10.274 0.000 0.132
14 7.852 2.428 8.219 0.074 0.155
15 17.852 6.968 19.164 0.217 0.428
16 7.206 3.370 7.955 0.053 0.112
13 0.000 10.406 10.406
12 0.021 10.406 10.406 0.000 0.132
14
-6.200 -1.600 6.403
12 -7.778 -2.273 8.103 0.074 0.155
15 1.592 0.708 1.742 0.006 0.006
15 -8.200 -2.500 8.573
12 -17.634 -6.540 18.808 0.217 0.428
14 -1.586 -0.702 1.734 0.006 0.006
18 6.009 1.741 6.256 0.039 0.079
23 5.004 2.963 5.815 0.031 0.063
16 -3.500 -1.800 3.936
12 -7.152 -3.257 7.859 0.053 0.112
17 3.658 1.440 3.931 0.012 0.027
17 -9.000 -5.800 10.707
16 -3.646 -1.413 3.910 0.012 0.027
10 -5.332 -4.355 6.885 0.014 0.037
18 -3.200 -0.900 3.324
15 -5.774 -1.661 6.197 0.039 0.079
19 -6.703 0.787 2.888 0.005 0.010
19 -9.500 -3.400 10.090
18 -2.774 -0.777 2.881 0.005 0.010
20 -6.703 -2.675 7.217 0.017 0.034
20 -2.200 -0.700 2.309
19 6.720 2.709 7.245 0.017 0.034
10 -8.937 -2.389 9.558 0.081 0.180
21 -17.500 -11.200 20.777
10 -15.613 -9.609 18.333 0.110 0.034
22 -1.849 -1.627 2.463 0.001 0.180
22 0.000 0.00 0.000
10 -7.531 -4.380 8.712 0.052 0.107
21 1.850 1.628 2.464 0.001 0.001
24 5.643 2.795 6.297 0.043 0.067
23 -3.200 -1.600 3.578
15 -4.972 -2.728 5/756 0.031 0.067
24 1,771 -1.270 2.186 0.006 0.012
24 -8.700 -2.400 9.025
22 -5.601 -2.728 6.230 0.043 0.067
23 -1.765 -1.270 2.174 0.006 0.012
25 -1.322 1.604 2.079 0.026 0.014
25 0.000 0.000 0.000
24 1.330 -1.590 2.073 0.008 0.014
26 3.520 2.372 4.244 0.444 0.066
27 -4.866 -0.786 4.929 0.026 0.049
26 -3.500 -2.300 4.188
25 -3.476 -2.306 4.171 0.044 0.066
27 0.000 0.000 0.000
25 4.892 0.835 4.963 0.026 0.049
28 -18.192 -4.152 18.660 -0.000 1.310
29 6.178 1.675 5.401 0.086 0.162
30 7.093 1.663 7.286 0.162 0.304
28 0.000 0.000 0.000
27 18.192 5.463 18.994 -0.000 1.310 0.968
8 0.570 -2.003 2.082 0.000 -4.368
6 -18.780 -3.510 19.106 0.060 -13.085
29 -2.400 -9.00 2.563
27 -6.093 -1.513 6.278 0.086 0.162
30 3.716 0.601 3.764 0.034 0.063
30 -10.600 -1.900 10.769
27 -6.932 -1.359 7.064 0.162 0.304
29 -3.683 -0.537 3.722 0.034 0.063
Total loss 17.594 22.233