SYNCHRONOUS GENERATOR TRANSIENT ANALYSIS

Name:

Index No:

Group:

Date of Per: 2010 / 10 / 07

Date of Sub: 2010/ 12 / 10

Observation Sheet

Name:

Index No:

Group:

Date of Per: 2010 / 10 / 07

Instructed By:

a) Armature current when sudden short circuit the generator,

Pole pair number of the generator: 2Steady short circuit current: 2.25AGenerator Speed: 1500rpmPre Short circuit line voltage: 19V

Sudden short circuit armature current oscillogram.

b) Field current Ossillogram.

Steady State Field Current: 0.6A

c) Obtaining open circuit armature voltage waveform,

Calculations

Data taken from sudden short circuit armature current oscillogram,

T(ms) Armature current I( peak)(A)0 7.020 6.240 5.660 5.080 4.8100 4.6120 4.4140 4.2160 4.0180 3.8200 3.6220 3.5240 3.5260 3.4280 3.4300 3.4320 3.4

Armature Current Ipk (A) Vs T(ms)

From the graph,

The polynomial obtained from the graph is,

y = -1E-07x3 + 0.000x2 - 0.033x +6.843

Where y=Ia peak value & X=time in seconds.

Vs=19/√3 =10.97

From the graph,

A = 6.8A , B = 3.4A

By applying the theory,

X d=√2V s

B

X d=√2∗10.97

3.4

X d=4.563Ω

X d' '=

√2V s

A

X d' '=√2∗10.97

8.0

X d' '=2.281Ω

T(ms) Ia (FROM THE DRAWN GRAPH)

ΔΧ

0 6.657 3.25720 6.197 2.79740 5.777 2.37760 5.397 1.99780 5.057 1.657

100 4.757 1.357120 4.497 1.097140 4.277 0.877160 4.097 0.697180 3.957 0.557200 3.857 0.457220 3.797 0.397240 3.777 0.377

By from graph,

C=3.257

C/e =1.198

D=118ms

∴T 'd=D=0.118 s

X ' d=1

1Xd

+C

√2V s

=2.33Ω

T(ms) ∆Y Log(∆Y)

0 1.68 0.225

20 1.24 0.093

40 0.84 -0.076

60 0.67 -0.174

80 0.46 -0.337

100 0.29 -0.537

120 0.24 -0.619

By obtaining from the graph,

E = 1.680, E/e = 0.618, F = 61.5ms

T d' '=61.5ms

By the theory,

T do' =

Td' ∗X d

Xd' =110×4.563

2.33=215.42ms

T d0' ' =

T d' '∗Xd

'

Xd' ' =61.5×2.33

2.281=62.82ms

T(ms) Ia peak(Ia +ve peak)+(-Ia –ve peak)

A 0 7.0 14.0

20 6.2 12.440 5.6 11.260 5.0 10.080 4.8 9.60

100 4.6 9.20120 4.4 8.80140 4.2 8.40160 4.0 8.00180 3.8 7.60200 3.6 7.20220 3.5 7.00240 3.5 7.00

By obtaining from the graph,

G = 12.43, G/e = 4.57

H = T a=280ms

2) Arriving at short circuit field current wave form,

Short circuit field current expression,

I f=I f 0+ I f 0

(Xd−X d' )

Xd' [e−t /Td

'

−(1−T kd

T d' ' )e−t /T d

'

−T kd

T a

e−t /Ta cos (ωt )]

Assumed the damping is zero, Tkd

T a and

Tkd

T d' ' is zero.

Parameter value

X d 4.563Ω

X rsub d ¿ 2.281 Ω

X ' d 2.33 Ω

T ' d 118ms

T rsub d ¿ 61.5ms

T ' do 215.42ms

T rsub do ¿ 62.82ms

T a 280ms

¿0.6+0.6×4.563−2.33

2.33×(e

−t215.42−e

−t62.82 )

¿0.6+.58 (e−t /215.42−e−t /62.82)

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Time (ms)

Ia

The open circuit line voltage,

Va=√2V s cos (ωt+θ0 )−√2V s

[Xd−Xd' ]

Xd' '

e−tTd 0

'

cos (ωt+θ0 )−√2V s

[ Xd' −Xd

' ' ]Xd

e−tT d 0

'

cos (ωt+θ0 )

Assume that θ0 is zero,

V a=√2V scos (ωt )−√2V s

[X d−Xd' ]

Xd' '

e−t /T d 0'

cos (ωt )−√2V s

[X d' −Xd

' ' ]X d

e−t /T d0'

cos (ωt )

V a=(15.51−15.19 e−t

.21542−0.167e−t

.21542)cos (314.16 t)

Using the results of the slip test

Maximum line voltage 14V => maximum phase voltage=

14 /√3

Minimum line voltage 13V => minimum phase voltage =13/√3

Maximum phase current 1.3A

Minimum phase current 1.25A

Therefore according to the theory

X d=Maximum phase voltageMinimum phase current

XQ=Minimum phasevoltageMaximum phase current

X d=14 /√31.25

Ω=6.47Ω XQ=13/√3

1.3Ω=5.77Ω

Discussion

Q1) Compare the parameter values computed using the short circuit current oscillograme and the slip test.

The theoretical values are depend on the X and T values observed from the graphed based on the reading that have been taken during the practical. Calculated values are not accurate because of the errors occur during the practical. Therefore compare with the theoretical values there were differences in the readings. We can ultimately assume that the theoretical values are approximately 100% correct.

Q2) Compare the agreement of theoretical and observed oscillograme of short circuit field current and open circuit voltage.

The theoretically graph takes less time to come to steady but the practical graph isn’t it took much longer time and also the calculated field current there is no intermediate raise of the value. That because of the effect of damping which is assumed to be negligible as we arrive at the equation.

Q3) Features of short circuit oscillogram of phase and field currents.

There are 3 components take place from the moment of short circuit disturbance to steady state reach moment. They are as follows,

a. Sub Transient Component.b. Transient Component.c. DC offsetd. Steady state Component.

As the graph shows above transient and sub-transient states are decays with the time quickly but comparatively transient component has slower decay. The DC offset happens due to the armature reaction flux occurs at the moment of short circuit disturbance, when that happens short circuit field flux is act on the d-axis. Therefore there exist a sudden increase of the flux in the direction of the field.

ᵩf ᵩa (Armature reaction flux)

Q4) Importance of short circuit test.

A sudden short circuit at the terminals of a generator could cause damage to the generator and at times be sufficient to damage other equipment connected to it. To apply necessary protection schemes in synchronous generators, generator parameters such as synchronous reactance, transient reactance, sub transient reactance and etc. should be known. Only by performing sudden open short circuit test these parameters can be determined. Therefore it is important to do short circuit test.