Graffeo Ex 51 - Using MVA Method

[TWJ PE]

Something seems to be tripping me up on this problem.

I've tried to solve using the MVA method but I'm missing something.

Ex. 51:

Two generators are connected through two transformers to a bus that is connected to a transmission line. The line is open at the far end, where there is a three phase fault. Prior to the fault, the voltage at the end of the transmission line is 230KV. The ratings are per unit reactances of the equipment based on the equipment ratings are shown below:

G1 = 700 MVA, 12KV, X''d=X2=.11, X0=.06

G2 = 600 MVA, 12 KV, X''d=X2=.14, X0=.08

T1 = 700MVA, 12/230KV, delta/wye, X=.14

T2 = 600MVA, 12/230KV, wye/delta, X=.12

Transmission line = base of 1000MVA, 230KV, X1=X2=.19, X0=.45

All wye winding neutrals are solidly grounded.

Find fault current.

 

[TWJ PE]

Thank you in advance!

 

[sebann]

I have an issue with that example as well. Everything is pretty straigh forward except I dont understand why the impedances for both generators are not recalculated based on new voltage base but just new 1000 MVA base.

Any advice?

Thank you

 

[ZcoreX29]

Using the MVA method, I came up with 6506A. Since you don't say how you are calculating the current using that method, I can only guess as to where you are going wrong. One possibility is that you are calculating the MVA for the transmission line using the per unit impedance instead of the actual ohmic impedance. What are the MVA values for each item in your circuit?

 

[TWJ PE]

I get Isc=(1000*1700))/(sqrt(3)*230)=4267

 

[ZcoreX29]

I have an issue with that example as well. Everything is pretty straigh forward except I dont understand why the impedances for both generators are not recalculated based on new voltage base but just new 1000 MVA base.

Any advice?

Thank you

For the per-unit method, you specify a MVA base for the entire system and a kV base for each section of the system where the transformer acts as the diving line between sections. The kV base for each section is typically the kV rating of the transformer (12 kV and 230 kV). Since the problem told you that the generator impedances were derived from the equipment ratings (12 kV), there is no need for a voltage conversion.

 

[ZcoreX29]

I get Isc=(1000*1700))/(sqrt(3)*230)=4267

Where do you get the 1000 and 1700?

 

[TWJ PE]

1700 from 7243 (from gen and xfmr) || 2222 (transmission)

 

[TWJ PE]

Originally thought it was the transmission line tripping me up but now I'm questioning everything.

 

[ZcoreX29]

I'm not sure how you obtained 7243 and 2222. What are your short circuit MVAs for each item in the circuit (G1, G2, T1, T2, Tline)?

You're using the zero sequence impedance for the t-line. For a symmetrical fault (3 phase fault) there is no zero sequence current therefore zero sequence impedance is not used. Try using the positive sequence.

 

[TWJ PE]

G1=6363

G2=4285

T1 & T2=5000

Tline=5263

 

[TWJ PE]

Nevermind... I was doing something totally off base.

G1||T1= 2799

G2||T2=2307

2307+2799=5106

TLine=1000/.19=5263

5106||5293=2584

Isc=(1000*2584)/(sqrt(3)*1000)=6505

Yeah... I'm not sure what the heck I was doing either. Thanks.

 

[sebann]

I have an issue with that example as well. Everything is pretty straigh forward except I dont understand why the impedances for both generators are not recalculated based on new voltage base but just new 1000 MVA base.

Any advice?

Thank you

For the per-unit method, you specify a MVA base for the entire system and a kV base for each section of the system where the transformer acts as the diving line between sections. The kV base for each section is typically the kV rating of the transformer (12 kV and 230 kV). Since the problem told you that the generator impedances were derived from the equipment ratings (12 kV), there is no need for a voltage conversion.

Thank you!!

 

[KatyLied P.E.]

For what it's worth Graffeo is very good in responding to questions. Nothing wrong with asking here either. Everyone benefits then.

 

[StinkyTofu]

Nevermind... I was doing something totally off base.

G1||T1= 2799

G2||T2=2307

2307+2799=5106

TLine=1000/.19=5263

5106||5293=2584

Isc=(1000*2584)/(sqrt(3)*1000)=6505

Yeah... I'm not sure what the heck I was doing either. Thanks.

W9TWJ,
Where did you get that 1000 from (in red above)? Shouldn't that be the line to line voltage (230kV)?

 

[TWJ PE]

Yes, you are correct - typo. Good catch.

 

[Limamike]

You know I have aske graffeo a few times on this one and 49 and no response.  Has anyone one been able to do 49 via MVA method?  Thanks 

 

[TWJ PE]

I don't have my book any more. Will you post the problem?

 

[bripgilb]

@doubleoinfo07 posted the question here 





I'm also interested in the MVA method for this question.

Thanks,

brip

 

[bripgilb]

Can anyone explain how to do Graffeo example 49 using MVA method.

The problem reads: The following system was operating at no load and rated voltage, when suddenly a tree fell causing a single line to ground fault at bus 2. Find the fault current. 

Parameters:

Generators: 200MVA, 13.8KV, X"= X2 = 25%, Xo = 4.2%

Transformers: 200MVA, 13.8/230KV, X = 12%

Transmission Line: X1 = X2 = 17%, Xo = 48%.

im having issues uploading the picture. Basically it's a generator, 2 transformers, and another generator, all in series. Buses 1 and 2 "sandwich" (so-to-speak) transformer one and buses 3 and 4 sandwich transformer 2. The fault occurs on bus 2. 

Thanks and good luck to everyone.