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Coursework on Synchronous Generator Transient Analysis
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EE304
SYNCHRONOUS GENERATOR TRANSIENT
ANALYSIS
Instructed By: Mr. B.S. Madushanka Name : G.R. Raban
Index Number : 070384P
Field : EE
Group : 8
Date of Performance : 15/10/2009
Date of Submission : 13/11/2009
Observations
Variation of the current waveform
Variation of field current with time
Variation of phase voltage with time
The following readings were also obtained during the practical
Pre short circuit line voltage = 90 V
Steady short circuit current = 0.4 A
Generator speed = 1500 rpm
Number of generator pole pairs = 2
Steady state field current = 0.2 A
Calculations
(i) Step 1
Using the phase current oscillogram, the following values were obtained and the graph
of Ipk-pk versus time (graph-1) was drawn.
By the graph of Ipk-pk verses time, the following values were obtained;
(Since I have not used the peak-to-peak half value, I have divided the values by 2)
A = 79.2
2 A = 39.6 A
B = 17
2 A = 8.5 A
Pre S/C line to neutral voltage, Vs = 90
3 V
∴ Xd = 2Vs
B =
2 × 90
3
8.5 Ω = 8.64 Ω
X”d = 2Vs
A =
2 × 90
3
39.6 Ω = 1.86 Ω
T (ms) Ipk-pk (A)
5 68
15 56
25 41
35 36
45 32
55 29
65 26
75 24
85 23
95 22
105 21
115 20
125 19
135 19
145 18
155 17
165 17
175 17
185 17
Step 2
By using the ∆X components of graph-1, the following values were obtained and graph-2 was
drawn;
T (ms) ∆X (A)
0 62.2
8 45.0
16 33.0
24 25.0
32 20.2
40 16.6
48 13.8
56 11.4
64 9.4
72 7.4
80 6.6
88 5.8
96 4.6
104 4.2
112 3.4
120 2.6
128 1.8
136 1.4
144 1.0
152 0.6
160 0
By the values obtained by graph-2;
C = 62.2 A ⟹ C
e = 22.88 A
∴ D = 28 ms
Then, the following values were calculated;
X’d = 1
1
X d +
C
2V s
= 1
0.9622 = 1.04 Ω
T’d = 28 ms
Step 3
By using the ∆Y components of graph-2, the following values were obtained and graph-3 was
drawn;
By the values obtained by graph-3;
E = 101.68A ⟹ log 𝐸
𝑒 = 1.25
∴ F = 21.0 ms
Then, the following values were calculated;
T”d = 21.0 ms
T’do = T’d × Xd
Xd′ = 28 ×
8.64
1.04 ms = 232.62 ms
T”do = T”d × Xd′
Xd" = 21.0 ×
1.04
1.86 ms = 11.74 ms
T (ms) ∆Y (A) log(∆Y) (A)
0 49.8 1.697229
4 40.6 1.608526
8 33.0 1.518514
12 27.2 1.434569
16 21.8 1.338456
20 18.4 1.264818
24 14.6 1.164353
28 12.6 1.100371
32 10.2 1.0086
36 8.8 0.944483
40 7.4 0.869232
44 6.4 0.80618
48 5.4 0.732394
52 4.2 0.623249
56 3.4 0.531479
60 2.8 0.447158
64 2.2 0.342423
68 1.6 0.20412
72 1.0 0
Step 4
By using the values of graph-1, the following values were obtained and the envelope mean was
plotted against time to obtain the armature time constant;
T (s) Ipk-pk (A) log(Ipk-pk)
5 68 1.832509
15 56 1.748188
25 41 1.612784
35 36 1.556303
45 32 1.50515
55 29 1.462398
65 26 1.414973
75 24 1.380211
85 23 1.361728
95 22 1.342423
105 21 1.322219
115 20 1.30103
125 19 1.278754
135 19 1.278754
145 18 1.255273
155 17 1.230449
165 17 1.230449
175 17 1.230449
185 17 1.230449
By the graph;
G = 101.7 ⟹ log 𝐺
𝑒 = 1.27
∴ H = 140 ms
Ta = 140 ms
∴ The obtained X and T values are as follows;
Xd = 8.64 Ω
X’d = 1.04 Ω
X”d = 1.86 Ω
T’d = 28.0 ms
T”d = 21.0 ms
T’do = 232.62 ms
T”do = 11.74 ms
Ta = 140.0 ms
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 5 15 25 35 45 55 65 75 85 95 105 115 125 135 145 155 165 175 185
log
(Ip
k-p
k)
Time (ms)
Plot of envelope mean Vs time
(ii) Using short circuit field current expression, arriving at short circuit field current wave
form
If = If0+ If0
(Xd − Xd′ )
Xd′ e−t/Td
′− 1 −
Tkd
Td′′ e−t/Td
′′−
Tkd
Tae−t/Ta cos(ωt)
Since it is assumed zero damping, 𝑇𝑘𝑑
𝑇𝑎 and
𝑇𝑘𝑑
𝑇𝑑′′ is zero.
If = If0+ If0
(Xd − Xd′ )
Xd′ e−t/Td
′− e−t/Td
′′
If = 0.4 + 0.4 ×8.64 − 1.04
1.04× e−
t28 − e−
t21
If = 0.4 + 2.92 e−t/28 − e−t/21
The above function is plotted below;
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 50 100 150 200 250 300
I a (A
)
Time (ms)
(iii) Open circuit line voltage
Va = 2Vs cos ωt + θ0 − 2Vs
Xd − Xd′
Xd′′
e−t/Td 0′
cos ωt + θ0 − 2Vs
Xd′ − Xd
′′
Xde−t/Td 0
′
cos ωt + θ0
Assuming 𝜃0 is zero,
Va = 2Vs cos ωt − 2Vs Xd−Xd
′
Xd′′ e−t/Td 0
′
cos ωt − 2Vs Xd
′ −Xd′′
Xde−t/Td 0
′
cos ωt
Va = (73.48 − 293.29e−t/232.62)cos(314.16t)
The above function is plotted below;
-250
-200
-150
-100
-50
0
50
100
150
200
250
0 100 200 300 400 500 600 700 800 900
Va
(V)
Time (ms)
Discussion
Comparison of the theoretical and observed oscillogram of short circuit field current and
open circuit line voltage
The practically observed oscillogram and the theoretically calculated waveform are
relatively similar in shape. But the actual time they take to arrive at steady state are
different. This is because we have assumed that there is zero damping when calculating the
expression for the short circuit field current, whereas in the practical case, a small amount
of damping may exist.
We have used practically calculated X and T values to establish the theoretical expressions.
This may introduce some errors to the expression.
The theoretical graphs are smooth. But the practically obtained oscillogram may have been
affected by noise introduced by external devices. Therefore, we can observe a slight
difference in practical and theoretical observations.
Features of short circuit oscillogram of phase and field currents
In short circuit oscillogram we can identify four components in the field currents. They are,
a. Transient
b. Sub transient
c. DC offset
d. Steady state component
Generally sub transient fades away by completion of several cycles although transient is
quite unlikely to disappear rapidly and may last for quite a number of cycles.
The importance of short circuit study
Generator parameters such as synchronous reactance, transient reactance, sub transient
reactance and other relevant parameters can be determined through a short circuit study.
These parameters are necessary when designing protection schemes in synchronous
generators. Therefore, the short circuit study of a synchronous generator is essential.