L/L fault question

[Omer]

 

[rg1]

Is it 1 pu? Base 34.5kV.

 

[FPar]

The answer I'm getting is Ia=0; Ib=-2,17J; Ic=2,17J. Is this the same answer you getting?

 

[rg1]

Is it 1 pu? Base 34.5kV.

 I was wrong. It is 0.5pu. @FParit does not ask about currents, It asks about +sequence Voltage. IMHO your currents are also wrong. The Positive and negative sequence currents are 1.6pu and -1.6pu respectively. The 1 pu I was mentioning earlier, was Vf ie. Fault Voltage - that is also positive sequence but that was Voltage at source. This is quite tricky here.

One will have draw the sequence diagram for LL fault to get the answer.

 

[Omer]

Yes, it is .5 pu

and you will have draw the sequence diagram for LL fault to get the answer.

 

[FPar]

My mistake was that I used the wrong value (0.2p.u.) for the generator's Z+. I wonder if you really need to make any calculation at all?  

 

[Omar Hadid]

Would you please provide the steps of solutions for this question?

 

[Bonsai]

The sequence network for a line-to-line fault is the positive sequence in parallel with the negative sequence. The positive sequence contains the voltage source, assumed to be 1 p.u. if not otherwise given, and Z1 in series. The negative sequence contains Z2. Using the voltage divider, we can see that the voltage across Z1 is (1pu voltage) * .2/(.2+.2) = 0.5 pu. Subtracting that, we have the voltage across the positive sequence, which has the 1 pu source minus the 0.5 pu drop across Z1.

 

[Bonsai]

Edit: I neglected to include the sequence impedances for the generator. Z1 and Z2 should both be 0.2 + 0.1 = 0.3 pu, I believe. The outcome for the voltage across the positive sequence is the same.

 

[rg1]

Edit: I neglected to include the sequence impedances for the generator. Z1 and Z2 should both be 0.2 + 0.1 = 0.3 pu, I believe. The outcome for the voltage across the positive sequence is the same.

If this makes sense?

 

[rg1]

The sequence network for a line-to-line fault is the positive sequence in parallel with the negative sequence. The positive sequence contains the voltage source, assumed to be 1 p.u. if not otherwise given, and Z1 in series. The negative sequence contains Z2. Using the voltage divider, we can see that the voltage across Z1 is (1pu voltage) * .2/(.2+.2) = 0.5 pu. Subtracting that, we have the voltage across the positive sequence, which has the 1 pu source minus the 0.5 pu drop across Z1.

Please correct. Positive sequence voltage is not Voltage across Z1. In fact V1= V1-I1Z1. In this particular case you are getting the answer right but this may not happen always. See my diagram below. That is the right one.

 

[rg1]

A small correction V1=Vf-I1Z1

 

[Bonsai]

I think I was maybe a little unclear, and probably should have included a drawing, but you are correct, and that same solution was my intention.

 

[Omar Hadid]

Gentlemen, Is much appreciated, is make more sense now.

 

[Stephen2awesome]

If this makes sense?

View attachment 10235

Would not include the Z1 and Z2 from the generator?

 

[rg1]

Would not include the Z1 and Z2 from the generator?

Included.

0.2 plus 0.1 make 0.3.

 

[Stephen2awesome]

Included.

0.2 plus 0.1 make 0.3.

So, Z1 and Z2 from Xfmr + Z1 and Z2 from the Generator? 

 

[Stephen2awesome]

I thought L-L faults were Ea/ Zeq?  Zeq = Z1+Z2 for both generator and xfmr. Maybe I am missing something? So, .2+.2 from Xfmr, and .1+.1 from the Gen.

 

[rg1]

I thought L-L faults were Ea/ Zeq?  Zeq = Z1+Z2 for both generator and xfmr. Maybe I am missing something? So, .2+.2 from Xfmr, and .1+.1 from the Gen.

IMHO you are mixing two issues. Symmetrical and Non symmetrical faults . This question is about non symmetrical faults and asks about Positive sequence Voltage .

 

[Stephen2awesome]

IMHO you are mixing two issues. Symmetrical and Non symmetrical faults . This question is about non symmetrical faults and asks about Positive sequence Voltage .

For line to line faults you use positive and negative sequence impedances right?My question remains, why wasn't both positive and negative sequence impedances used for both Xfmr and generator? The bus shown is on the high side of the transformer closest to the generator.