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