SYNCHRONOUS GENERATOR TRANSIENT ANALYSIS

OBSERVATION SHEET

a) Obtaining of short circuit armature current oscillogram Pre –short circuit line voltage = 104.3 V Steady short circuit current = 5.77 AGenerator speed = 1500 rpmNumber of generator pole pairs = 2

Oscillogram 1b) Obtaining of field current oscillogram

Steady state field current = 0.3 A

Oscillogram 2

c) Obtaining of open circuit armature voltage waveform.

Minimum phase current = 4.8 A

Maximum phase current = 5.4 A

Minimum line voltage = 47.5 V

Maximum line voltage = 48.3 V

Generator speed = 1475 rpm

Oscillogram 3

d) Obtaining supply voltage waveform at slip test.

Oscillogram 3

Calculations

1.

STEP 1 From short circuit armature current wave form using oscillogram 1, computation for X d & X’’d

Time(ms) Half of peak to peak value. Ia (A) Log (Ia)0 27 1.43136376410 20.5 1.31175386120 18 1.25527250530 15 1.17609125940 14 1.14612803650 13.5 1.13033376860 12.5 1.09691001370 11.5 1.0606978480 11 1.04139268590 10 1100 9.5 0.977723605110 9.5 0.977723605120 9 0.954242509130 9 0.954242509140 9 0.954242509

Using the graph the calculated values of A & B in the ampere units

A = antilog (1.431363764) =27

B = antilog (0.954242509) = 9

According to the theory

X d=√2VsB

where Vs isthe rms pahse voltagebefor the S/C

= √2×104.39×√3

=9.462Ω

X ’’d=√2VsA

¿ √2×104.527×√3

=3.154Ω

Time(ms) vs Log (Ia)

STEP 2 From short circuit armature current wave form using oscillogram 1, computation for X’ d & T’d

Time(ms) Δx (A) log Δx0 17.838 1.2513461610 11.778 1.0710715520 8.318 0.92001891630 6.278 0.79782131140 5.078 0.70569269750 4.168 0.6199277160 3.348 0.52478544970 2.498 0.39759243480 1.728 0.23754373890 1.048 0.020361283100 0.539 -0.268411235110 0.172 -0.764471553120 0130 0140 0

Using the graph the calculated values of C & D in the ampere and ms units.

C = antilog (1.30044) =19.97

So C/e = 7.3469 so the log value of this is 0.866 hence correscopnding value obtained for D using the figure is 25 ms

According to the theory

X ’d=

11Xd

+C

√2V s

X ’d=

11

9.462+ 19.97×√3

√2×104.3

=2.93Ω

T ’d=D=25ms

Time(ms) vs log Δx

STEP 3 From short circuit armature current wave form using oscillogram 1, computation for T’’d & T’do

Time(ms) Δy (A) Log Δy0 14.24005858 1.15351177610 7.209749896 0.85792019920 3.704663695 0.56874878930 1.879807084 0.27411328240 0.910019998 -0.04094906450 0.398075891 -0.40003412460 0.140719611 -0.85164537470 0.028503233 -1.54510588280 0

Using the graph the calculated values of E & F in the ampere and ms units.

E = antilog (1.27955) =19.0348

So E/e = 7.0025 so the log value of this is 0.845 hence corresponding value obtained for F using the figure is 12 ms.

T ’ ’d=12ms

T ’do=T ’d×Xd

X 'd

T ’do=25×9.4622.93

=80.73ms

T ’ ’do=T ’’d×X ' dX ' ' d

T ’ ’do=12×2.933.154

=11.1477ms

Time(ms) vs log Δy

STEP 4 Graph between time vs Ia,(+ve peak + -ve peak)

Time(ms) Ia,(+ve peak + -ve peak)0 5410 4120 3630 3040 2850 2760 2570 2380 2290 20100 19110 19120 18130 18140 18

Using the graph the calculated values of G & H in the ampere and ms units.

G = 40.73 A from the graph

H = 130.1 ms from the graph

So T a=H=130.1ms

Time(ms) vs Ia,(+ve peak + -ve peak)

2. Field current variation fallowing a sudden three phase short circuit

I f=I f 0+ I f 0Xd−X 'd

Xd

¿

Ifo is the field current before the short circuit which observed as 0.3 ATkd is the direct axis damper time constant and assuming no damper sField current variation

I f=I f 0+ I f 0Xd−X 'd

Xd

¿

I f=0.3+0.3(9.462−2.93)

9.462[e−t25−e

−t11.1477 ]

I f=0.3+0.2071 [e−t25−e

−t11.1477 ]

Field current variation

Field current variation

3) Armature line voltage on sudden open circuit

V a=√2V s cos (ωt+θ0 )−√2V s

[Xd−X 'd ]X ' ' d

e−tT do cos (ωt+θ0 )−¿√2V s

[Xd−X ' ' d ]X ' ' d

e−tT ' do cos (ωt+θ0 )¿

Assuming θ0=0 ;

V a=√2V s cos (ωt )−√2V s

[Xd−X ' d ]X ' ' d

e−tT ' do cos (ωt )−¿√2V s

[ X ' d−X ' 'd ]Xd

e−tT ' ' do cos (ωt ) ¿

V a=√2104.3√3

cos (2 πft )−√2 104.3√3[9.462−2.93 ]3.154

e−t80.73 cos (ωt )−¿√2 104.3√3

[2.93−3.154 ]9.462

e−t

11.1477 cos (2 πft )¿

V a=85.16cos (314.16 t )−176.37 e−t80.73 cos (314.16 t )+2.016 e

−t30.172 cos (314.16 t )

Armature line voltag

4) Value of Xd and XQ from the slip test,

From the graphs of armature current and armature voltage of the slip test,

X d=(Va (pk−pk )max)(Ia( pk−pk )min)

=18V9 A

=2Ω

XQ=(Va( pk−pk )min)(Ia (pk−pk )max)

= 12V12.8 A

=0.9378Ω

Discussion

Compare the parameter values computed using Short circuit current oscillogram, open circuit voltage

oscillogram and slip test

Since the two methods to find the parameter values of the generator use graphical decomposition methods,

and each method having certain assumptions, which introduces to neglect some transient components like

double frequency component and displaced angle, there should be some deviation between the parameter

values obtained by the two methods.

And the graphical approach having more than one step leads to some inherent error in interpolating and curve

fitting.

Compare the agreement of theoretical and observed oscillogram, of short circuit field current and open

circuit line voltage

When we compare the theoretical and practical graph of the variation of field current at short circuit test, it

seems both are having same shape. But the practical curve takes more time to acquire the steady state,

compare with theoretical curve. This may occur due to the effect of damping effect in real situation.

Features of Short Circuit Oscillogram

In short circuit oscillogram we can identify 4 components in the transient behaviour.

(1) Transient component

(2) Sub transient component

(3) DC offset component

(4) Steady state component

Transient and sub transient components occure for very short period of time and transient

component takes much time to decay than sub transient component

When we observe the short circuit oscillogram , four components in field current can be

noticed. They are sub transient , transient , DC offset and steady state componenets. Generally sub

transient period lasts for 3-4 cycles of current. However transient period is long and lasts for some

hundreds of cycles.

DC offset current is caused due to the armature reaction. Before the short circuit, armature

flux is zero. With the short circuit, armature reaction flux ?a is forced to be established. i.e. sudden

armature current tends to be established. Depending on the field flux, the phase 'a' winding is

receiving at the instant of the short circuit, phase 'a' coil establishes a DC current to maintain the

same flux following the short circuit causing the DC offset term.

Importance of short circuit study on synchronous generators.

To apply necessary protection schemes in synchronous generators, generator parameters such as

synchronous reactance, transient reactance , sub transient reactance and etc. should be known. By

performing sudden open short circuit test only, these parameters can be determine. Therefore it is

important to do short circuit test.