sync generator
DESCRIPTION
Transient analysis of synchronous generatorTRANSCRIPT
EE xxxx - Laboratory Practice VI
SYNCHRONOUS GENERATOR TRANSIENT ANALYSIS
Semester 7
Instructed by: Mr. W.D. Prasad
Group Members:W.G.M. Amaradasa (Index)H.K.G. Amarasignhe (Index)(Name) (Index)(Name) (Index)
Name : W.G.M. AmaradasaIndex No. : 110027AGroup : G-08Field : Electrical EngineeringDate of Perform. : 11/08/2015Date of Submission : 25/08/2015
OBSERVATIONS
Experiment : Synchronous Generator Transient Analysis
Name : W.G.M. Amaradasa
Index No. : 110027A
Group : G-08
Date of Performance: 11/08/2015
Instructed By : Mr. W.D. Prasad
a) Obtaining of Short Circuit Armature Current oscillogram
Pre-short circuit line voltage : 77 V
Steady shot circuit current : 4.4 A
Generator Speed : 1497.8 rpm
No of Generator pole pairs : 2
b) Obtaining of field current oscillogram
Steady State Field Current : 0.20 A
c) Obtaining of Open Circuit Armature Voltage Waveform
d) Slip Test
Current Waveform
Voltage Waveform
Minimum Phase Current : 5 A
Maximum Phase Current : 5.1 A
Minimum Line Voltage : 46 V
Maximum Line Voltage : 43 V
Generator Speed : 1436 rpm
CALCULATIONS
i. Calculating Xd, Xd’, Xd’’, Td’, Td’’, Td0’, Td0’’ and Ta
Step 1:
Obtaining the half magnitude peak current value from short circuit current waveform and drawing the graph of that vs. time and calculate X d and X d
' '
Time (s) Ia(pk-pk) (A) Ia(pk) (A) Log(Ia(pk))0.01 44 22 3.09100.03 30 15 2.70810.05 24 12 2.48490.07 20.5 10.25 2.32730.09 19 9.5 2.25130.11 16.5 8.25 2.11020.13 16.5 8.25 2.11020.15 15 7.5 2.01490.17 15.5 7.75 2.04770.19 16 8 2.0794
From the Graph 1,
log ( A )=3.2526 A=antilog (3.2526 )=25.8575
log ( B )=2.1000B=antlilog (2.1000 )=8.1662
X d=√2V s
B=√2×77/√3
8.1662=7.6989 Ω
X d' '=
√2V s
A=√2× 77/√3
25.8575=2.4314 Ω
Step2:
Obtaining Δ x vs. time graph and calculate X d' and T d
'
Time (s)Log(A) A Δx (A-B) Log(Δx)
0.010 3.0634 21.3998 13.2336 2.5828
0.020 2.8951 18.0858 9.9196 2.2945
0.030 2.7466 15.5898 7.4237 2.0047
0.040 2.6166 13.6897 5.5235 1.7090
0.050 2.5040 12.2310 4.0648 1.4024
0.060 2.4074 11.1049 2.9387 1.0780
0.070 2.3257 10.2336 2.0674 0.7263
0.080 2.2576 9.5602 1.3941 0.3322
From the Graph 2,log (C )=2.3400C=antilog (2.3400 )=10.3812D=0.0436
X d' = 1
1Xd
+ C√2V s
= 11
13.3348+ 10.3812
√2× 77/√3
=4.1647 ΩT d
' =D=0.0436 s
Step3:
Obtaining Δ y vs. time graph and calculate T d' '
Time (s) Log(C) C Log(Δx) Δx Δy (Δx-C) Log(Δy)0.002 2.2044 9.0648 4.5847 97.9718 88.9070 4.48760.004 2.1588 8.6607 3.8596 47.4487 38.7879 3.65810.006 2.1132 8.2747 3.4355 31.0479 22.7732 3.12560.008 2.0676 7.9058 3.1346 22.9798 15.0740 2.71300.010 2.0220 7.5534 2.9012 18.1961 10.6427 2.36490.012 1.9764 7.2167 2.7105 15.0368 7.8201 2.05670.014 1.9308 6.8950 2.5493 12.7976 5.9026 1.7754
From the Graph 3,log ( E )=4.6000E=antilog ( 4.6000 )=99.4843F=0.0047T d
' '=FT d' '=0.0047 s
T d 0 '=T d 'Xd
Xd '=0.0436 × 7.6989
4.1647=0.0806 sT d0 ' '=Td ' '
Xd 'Xd
' ' =0.0047 × 4.16472.4314
=0.0081 s
Step4:
Plot of envelop mean of current waveform with time and calculate armature time constant Ta
Time (s) Ia (+ve) peak -Ia (-ve) peak (Ia (+ve) peak) + (-Ia (-ve) peak)0.01 23 21 440.03 15 15 300.05 12 12.5 24.50.07 10 10.5 20.50.09 9 10 190.11 8.5 9 17.50.13 8 8 160.15 8 8 160.17 8 8 160.19 7.5 7.5 15
From the Graph 4,
G=34.4 Ge
=12.7
H=0.1728 sT a=H =0.1728 sii. Computing Short Circuit field current waveform
Field current variation following a sudden three phase short circuit at the armature given by,
I f =I f 0+ I f 0( Xd−Xd
' )X d ' [e−t
T d '−(1−T kd
T d' ' )e
−tT d ' '−
T kd
T d' ' e
−tT a cos (ωt )]Assuming no damper windings,
T kd=0Then,
I f =I f 0+ I f 0
( Xd−Xd' )
X d '[e−t
T d '−e−tTd
' ' ]I f 0=0.20 A ,I f =0.20+0.20 (7.6989−4.1647)4.1647
[e −t0.0436−e
−t0.0047 ]
I f =0.20+0.8486 [e −t0.0436−e
−t0.0047 ]
Field current waveform can be plotted by using Matlab. Following commands can be used to plot,>> x = 0:0.001:0.3;>> plot(0.20+0.8486*(exp(-x/0.0436)-exp(-x/0.0047)));
Computed filed current waveform is in the following figure,
Field Current waveform obtained by the oscilloscope is in the following figure,
iii. Computing and plotting the open circuit line voltage waveform
V a=√2V s cos (ωt+¿θ0)−√2V s(Xd−Xd '
Xd' ' )e
−tT d0 ' cos(ωt+¿θ0)−√2V s(
Xd '−Xd ' 'Xd
)e−t
Td 0 ' ' cos (ωt+¿θ0)¿¿¿
Assuming θ0=0 ,
V a=√2 V s cos (ωt ¿)−√2V s(Xd−Xd '
Xd' ' )e
−tT d0 ' cos(ωt¿)−√2V s(
X d' −Xd
' '
X d)e
−tTd 0
' '
cos (ωt ¿)¿¿¿
V a=√2×(77/√3)cos (2π × 50t ¿)−√2×(77 /√3)( 7.6989−4.16472.4314
)e−t
0.0806 cos (2 π ×50 t¿)−√2×(77/√3)( 4.1647−.43147.6989
)e−t
0.0081 cos(2 π ×50 t¿)¿¿¿
V a=62.87 cos(314.16 t ¿)−91.3857 e−t
0.0806 cos (314.16 t¿)−14.1543 e−t
0.0081 cos (314.16 t¿)¿¿¿
Armature voltage waveform can be plotted by using Matlab. Following commands can be used to plot,
>> y = 0:0.001:1;
>> plot(62.87*cos(314.16.*y)-91.3857*exp(-y/0.0806).*cos(314.16.*y)-14.1543*exp(-y/0.0081).*cos(314.16.*y))
Computed armature voltage waveform is in the following figure,
Armature Voltage waveform obtained by the oscilloscope is in the following figure,
iv. Compute Machine parameters from open circuit voltage waveform
Step 1:
Obtaining the half magnitude peak voltage value from open circuit current waveform and drawing the graph of that vs. time.
v. Compute Xd and Xq by Slip Test
According to the observations,
VLine ( pk−pk ) max=2× 46√2 VVLine ( pk−pk ) min=2× 43√2V
Ia( pk−pk)max ¿=2× 5.1√2 A Ia( pk−pk)min ¿=2×5 √2 A
Xd=¿
(Va (pk−pk)max )(Ia ( pk− pk)min)
=2× 46√2/√32× 5√2
=5.3116 Ω¿X
d=¿(Va( pk−pk)min)(Ia (pk−pk)max )
=2× 43 √2/√32 ×5.1√2
=4.8679 Ω¿
GRAPHS
Table 1
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.21.0000
1.1000
1.2000
1.3000
1.4000
1.5000
1.6000
1.7000
1.8000
1.9000
2.0000
2.1000
2.2000
2.3000
2.4000
2.5000
2.6000
2.7000
2.8000
2.9000
3.0000
3.1000
3.2000
3.3000
3.4000
3.5000
3.6000
3.7000
3.8000
3.9000
4.0000
f(x) = − 203.238282271908 x³ + 110.934436151537 x² − 20.0106412066632 x + 3.25258356304397
Graph1 : Ia(pk)(log scale) vs. Time
Time (s)
Ia(p
k)(lo
g sc
ale)
A
B
Time (s) Log(Ia(pk))
0.01 3.0910
0.03 2.7081
0.05 2.4849
0.07 2.3273
0.09 2.2513
0.11 2.1102
0.13 2.1102
0.15 2.0149
0.17 2.0477
0.19 2.0794
Table 2
0.000 0.010 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.0900.0000
0.1500
0.3000
0.4500
0.6000
0.7500
0.9000
1.0500
1.2000
1.3500
1.5000
1.6500
1.8000
1.9500
2.1000
2.2500
2.4000
2.5500
2.7000
2.8500f(x) = − 1.04648678276478 ln(x) − 1.9158235816487
Graph2 : Δx vs Time
Time (s)
Δx (l
og sc
ale)
Time (s) Log(Δx)
0.010 2.58280.020
2.2945
0.0302.0047
0.0401.7090
0.0501.4024
0.0601.0780
0.0700.7263
0.0800.3322
Table 3
0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.0160.00000.10000.20000.30000.40000.50000.60000.70000.80000.90001.00001.10001.20001.30001.40001.50001.60001.70001.80001.90002.00002.10002.20002.30002.40002.50002.60002.70002.80002.90003.00003.10003.20003.30003.40003.50003.60003.70003.80003.90004.00004.10004.20004.30004.40004.50004.6000
Graph3 : Δy vs. Time
Time (s)
Δy (l
og sc
ale)
Time (s)Log(Δy)
0.002 4.4876
0.0043.6581
0.0063.1256
0.0082.7130
0.0102.3649
0.0122.0567
0.0141.7754
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
0.81.62.43.2
44.85.66.47.2
88.89.6
10.411.2
1212.813.614.415.2
1616.817.618.419.2
2020.821.622.423.2
2424.825.626.427.2
2828.829.630.431.2
3232.833.634.435.2
3636.837.638.439.2
4040.841.642.443.2
4444.8
Graph4 : (Ia (+ve) peak) + (-Ia (-ve) peak) vs. Time
Time (s)
Ia (+
ve) p
eak)
+ (-
Ia (-
ve) p
eak)
Table 4
Time ms) (Ia (+ve) peak) + (-Ia (-ve) peak)
0.0144
0.03 30
0.05 24.5
0.07 20.5
0.09 19
0.11 17.5
0.13 16
0.15 16
0.17 16
0.19 15
DISCUSSION
1. Comparison of the parameter values computed using the short circuit current
Oscillogram and the slip test.
2. Comparison of the agreement of theoretical and observed Oscillogram of short circuit
field current and open circuit line voltage.
3. Features of Short Circuit Oscillogram of phase and field current
4. Importance of short circuit study on synchronous generators