Camara Power Practice Problems 36-2 Prob #8

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I'm getting 882 with the PU method and 800.6 with the MVA method.  I don't know what I'm doing wrong to get different answers. 

Can someone post their work for the PU method?

 
I'm getting 882 with the PU method and 800.6 with the MVA method.  I don't know what I'm doing wrong to get different answers. 

Can someone post their work for the PU method?
I just noticed if I don't use the generator in the MVA method, I get the same answer as my PU method answer, which is an answer of 882.  I'm totally confused now. 

I'm still not sure whether the answer is 882 or 800.6 though.  The answer can't be 1030 though.

 
Hmmmmm....  So that was my original answer. 800.  Now I am confused!  So the gen does count? 
Yes, the generator counts.  You have to sum all impedances up to the point of the fault.  However, as previously mentioned, the generator and XFMR are on different power bases.  This has to be resolved, whereby the Znew for G = 0.05, not 0.1.  This comes from the change of base formula, Znew=Zold(Vold/Vnew)^2(Snew/Sold)=0.1(11/11)^2(25/50).  Get everything in terms of the XFMR base cause it has a "let through" of 25 MVA, and the line is connected to the XFMR of the same base (i.e. Zbase=(33kV)^2/(25 MVA)).

Therefore, you have:

1pu/(0.05+0.06+0.4364) = 1.83 pu

Ifault=(1.83)(437.3866A)=800.4879A

Per the question and what's asked for, this is how you solve it (but can be solved multiple ways, I just prefer the pu method due to it being more uniform and easier to spot errors).

If I'm wrong, someone correct me; but I am confident in what's presented above.  I hammered hard on fault current problems before I took the exam and feel it is one of my strongest areas.

 
Yes, the generator counts.  You have to sum all impedances up to the point of the fault.  However, as previously mentioned, the generator and XFMR are on different power bases.  This has to be resolved, whereby the Znew for G = 0.05, not 0.1.  This comes from the change of base formula, Znew=Zold(Vold/Vnew)^2(Snew/Sold)=0.1(11/11)^2(25/50).  Get everything in terms of the XFMR base cause it has a "let through" of 25 MVA, and the line is connected to the XFMR of the same base (i.e. Zbase=(33kV)^2/(25 MVA)).

Therefore, you have:

1pu/(0.05+0.06+0.4364) = 1.83 pu

Ifault=(1.83)(437.3866A)=800.4879A

Per the question and what's asked for, this is how you solve it (but can be solved multiple ways, I just prefer the pu method due to it being more uniform and easier to spot errors).

If I'm wrong, someone correct me; but I am confident in what's presented above.  I hammered hard on fault current problems before I took the exam and feel it is one of my strongest areas.
Why is Vnew 11?  Shouldn't it be 33 to match the voltage base of the of the transformer and line?

 
Why is Vnew 11?  Shouldn't it be 33 to match the voltage base of the of the transformer and line?
The Vnew/old would come into affect if you had a generator producing at, say, 13.2 kV, but the XFMR was rated at 12.47 kV on that side (i.e. the primary side connected to the generator).  In this case, we aren't given any info about the voltage of the generator, therefore, it's safe to assume it is producing voltage at the rated input of the XFMR.

 
Yes, the generator counts.  You have to sum all impedances up to the point of the fault.  However, as previously mentioned, the generator and XFMR are on different power bases.  This has to be resolved, whereby the Znew for G = 0.05, not 0.1.  This comes from the change of base formula, Znew=Zold(Vold/Vnew)^2(Snew/Sold)=0.1(11/11)^2(25/50).  Get everything in terms of the XFMR base cause it has a "let through" of 25 MVA, and the line is connected to the XFMR of the same base (i.e. Zbase=(33kV)^2/(25 MVA)).

Therefore, you have:

1pu/(0.05+0.06+0.4364) = 1.83 pu

Ifault=(1.83)(437.3866A)=800.4879A

Per the question and what's asked for, this is how you solve it (but can be solved multiple ways, I just prefer the pu method due to it being more uniform and easier to spot errors).

If I'm wrong, someone correct me; but I am confident in what's presented above.  I hammered hard on fault current problems before I took the exam and feel it is one of my strongest areas.
Absolutely correct. These are the answers coming with actual Z values and MVA Method.

 
So basically the Camara book not only gave the wrong answer, but provided four incorrect choices as well... No wonder we were all confused

 
So basically the Camara book not only gave the wrong answer, but provided four incorrect choices as well... No wonder we were all confused
820 is the closest and it wouldn't be unheard of to see something similar on the test.  Remember, this exam will test your gut (and comprehension) as much or more than your subject knowledge and engineering competency.  That said, I didn't have/use the Camara text when I took it, and it seems the rest of you should be cautious about some topics it presents.  Ideally, this very problem, for one, and I've seen other erroneous representations on this very forum (power factor correction comes to mind... how the hell do you mess that up? - these are the softballs)

 
@tnpe OK man. I'm rolling with that. That was my original answer and that's what make sense. So rolling in with that process 

 
@tnpe OK man. I'm rolling with that. That was my original answer and that's what make sense. So rolling in with that process 
Good luck!  Just use what works for you, and as I stated previously, I prefer the pu method cause errors will jump off the page at you.  If you have fractional impedances but end up with an odd ball 10 pu, you know you've made a mistake somewhere.  With that said, it wouldn't be unheard of to have a fault magnitude less than the full load current of the XFMR, depending on impedances, fault location and fault type.  However, you most likely won't see it, but it is possible and fair game.

The biggest hurdle with this exam is the concepts, ideally with regards to motor applications.  If you have a good motor book, be sure to take it with you!  I would recommend Wilde or Chapman (if you have each, take both of them).

 

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