# L/L fault question



## Omer (Oct 16, 2017)




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## rg1 (Oct 16, 2017)

Is it 1 pu? Base 34.5kV.


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## FPar (Oct 17, 2017)

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


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## rg1 (Oct 17, 2017)

rg1 said:


> 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.


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## Omer (Oct 18, 2017)

Yes, it is .5 pu

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


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## FPar (Oct 18, 2017)

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?


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## Omar Hadid (Oct 25, 2017)

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


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## Bonsai (Oct 25, 2017)

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.


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## Bonsai (Oct 25, 2017)

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.


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## rg1 (Oct 25, 2017)

Bonsai said:


> 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?


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## rg1 (Oct 25, 2017)

Bonsai said:


> 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.


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## rg1 (Oct 25, 2017)

A small correction V1=Vf-I1Z1


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## Bonsai (Oct 25, 2017)

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.


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## Omar Hadid (Oct 25, 2017)

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


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## Stephen2awesome (Oct 26, 2017)

rg1 said:


> If this makes sense?
> 
> View attachment 10235


Would not include the Z1 and Z2 from the generator?


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## rg1 (Oct 26, 2017)

Stephen2awesome said:


> Would not include the Z1 and Z2 from the generator?


Included.

0.2 plus 0.1 make 0.3.


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## Stephen2awesome (Oct 26, 2017)

rg1 said:


> Included.
> 
> 0.2 plus 0.1 make 0.3.


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


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## Stephen2awesome (Oct 26, 2017)

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.


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## rg1 (Oct 26, 2017)

Stephen2awesome said:


> 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 .


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## Stephen2awesome (Oct 26, 2017)

rg1 said:


> 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.


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## rg1 (Oct 26, 2017)

It is used. Total Positive sequence z = Positive of Gen+Positive of Xmer= .2+.1=.3

similarly Total negative sequence z= Neg of Gen+Neg of Xmer=.2+.1=.3

What is left now?


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## Stephen2awesome (Oct 26, 2017)

rg1 said:


> It is used. Total Positive sequence z = Positive of Gen+Positive of Xmer= .2+.1=.3
> 
> similarly Total negative sequence z= Neg of Gen+Neg of Xmer=.2+.1=.3
> 
> What is left now?


To find Ia1, 1pu/(.3+.3)  then use 1pu - Ia1*(Z1 equivalent for both).....ah okay. Not sure where I got confused.


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## rg1 (Oct 26, 2017)

No worries . Enjoy the Exam.


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## navixv (Oct 26, 2017)

Stop studying and relax guys.


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## Stephen2awesome (Oct 26, 2017)

navixv said:


> Stop studying and relax guys.


lol, I just wanted to make sure I was understanding it correctly. I'm just getting my things together right now.


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## knight1fox3 (Oct 26, 2017)

Best of luck to you all tomorrow! :thumbs:


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## navixv (Oct 26, 2017)

Yup. Good luck guys. 

Just be confident and don't make the exam bigger than it is. It's not THAT difficult. I failed the first time because i was just unprepared in a lot of subjects, and I still got within 5 or so questions of passing. Turns out when they give you a list of topics for a test, you should pay attention to it. And more than likely you will leave the exam feeling down, don't fret, that's normal.


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## Stephen2awesome (Oct 26, 2017)

navixv said:


> Yup. Good luck guys.
> 
> Just be confident and don't make the exam bigger than it is. It's not THAT difficult. I failed the first time because i was just unprepared in a lot of subjects, and I still got within 5 or so questions of passing. Turns out when they give you a list of topics for a test, you should pay attention to it. And more than likely you will leave the exam feeling down, don't fret, that's normal.


I failed the first time as well. I took a review course this time (actually two) so I believe I'm more prepared now than I was back in April. Good luck.


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## rg1 (Oct 26, 2017)

Best of luck everyone. We discussed a lot of questions here in forum, this time, and I am sure we are going to do well. Mine is first time and it is my mid life rehabilitation. If I do not pass, my kids and wife will pull my leg and so it is more important for me to prove myself. My wife is all set to give me a lot of household work from tomorrow. Studying was a great excuse, whether studied or not is a different issue. So in all ways, the cover, the  exam was providing is going away. Very sad. However I enjoyed the preparation.


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