# Camara Power Practice Problems 36-2 Prob #8



## Limamike (Apr 16, 2017)

Anyone has an idea why the generator and the Transformer are being eliminated from the short circuit method. I cant seem to come up with the 1030 via the MVA method, unless I just look at the line. Any help would be appreciated .


----------



## rg1 (Apr 16, 2017)

Limamike said:


> Anyone has an idea why the generator and the Transformer are being eliminated from the short circuit method. I cant seem to come up with the 1030 via the MVA method, unless I just look at the line. Any help would be appreciated .


Please provide the question


----------



## Limamike (Apr 17, 2017)




----------



## Limamike (Apr 17, 2017)

Why is the generator's contribution being ignored.  I've seen other problems exactly like this one and the generator is included


----------



## rg1 (Apr 17, 2017)

Limamike said:


> View attachment 9330


Generator contribution should be taken in this case. Is the answer not 45.76 MVA;or 800.6A?


----------



## mvsapre (Apr 17, 2017)

I agree. Generator contribution should be considered. This problem is similar to the last problem in NCEES practice test, example 540 except that problem does not consider line impedance.


----------



## rg1 (Apr 17, 2017)

rg1 said:


> Generator contribution should be taken in this case. Is the answer not 45.76 MVA;or 800.6A?


After rethinking- I guess the generator contribution is small and is neglected. An intelligent choice of options may help.


----------



## TNSparky (Apr 17, 2017)

It would seem the generator is the source but the maximum amount of energy available for the fault is dictated by the weakest link: in this case, the 25 MVA transformer. You can't push 50 MVA through a 25 MVA transformer.


----------



## Limamike (Apr 17, 2017)

@tnsparky.  Great explanation. The answers are a) 200 b) 820 c) 880 and d) 1030.  

The answer in the book is 1030.  I come up with 882.  with the @TNSparky approach.  So what am I still missing.  25/.06 = 417 and the line contribution is 33Kv^2/ (.12*5.28*30)=57.3

417 || 57.3  = 50.37 MVA.  So 50.37*1000/(sqrt 3* 33) = 882.    Book has 1030


----------



## rg1 (Apr 17, 2017)

Limamike said:


> @tnsparky.  Great explanation. The answers are a) 200 b) 820 c) 880 and d) 1030.
> 
> The answer in the book is 1030.  I come up with 882.  with the @TNSparky approach.  So what am I still missing.  25/.06 = 417 and the line contribution is 33Kv^2/ (.12*5.28*30)=57.3
> 
> 417 || 57.3  = 50.37 MVA.  So 50.37*1000/(sqrt 3* 33) = 882.    Book has 1030


I solved it by both methods pu as well as actual Z. The answer is same 800A.

1. pu method- Generator contribution=50/.1=500MVA; Xmer contribution=25/.06=417.7MVA and Line contribution=33**2/19.008=57.3 MVA

Net Fault MVA= 1/(1/500+1/417.7+1/57.3)=45.8; I=45.8/(33Xsqrt3)= 800.9A

2. Impedance method- Gen Z on 33kV side of the Transformer=(.1X(11**2)/50)X(33/11)**2= 2.18Ohms

                                    Xmer Z on 33 kV side= .06X(33**2)/25=2.61Ohms

                                    Line Z = 19.008 Ohms

Total Z= 2.18+2.61+19.008=23.8 Ohms

So the current is 33/sqrt3/23.8=800.6A.


----------



## mvsapre (Apr 17, 2017)

TNSparky said:


> It would seem the generator is the source but the maximum amount of energy available for the fault is dictated by the weakest link: in this case, the 25 MVA transformer. You can't push 50 MVA through a 25 MVA transformer.


I am totally confused now. 

This is what I got:

Z base= 33^2kv/25MVA= 43.56

30 miles=48km

48km=157480.31 feet

so Z for 30 miles long line is 157480.31*0.120= 18897.63 

Zpu line= 18897.63/43.56= 433.82PU

ZT=0.06

PU SC= 1/(0.06+433.82)=2.30

 Base Transformer fault current= 25*10^6/(sqrt 3 *33*10^3) = 437.38

So 437.38*2.30 = 1006

Nearest possible answer is 1030A.


----------



## mvsapre (Apr 17, 2017)

mvsapre said:


> I am totally confused now.
> 
> This is what I got:
> 
> ...


Pardon my negligence Line Z is wrong.


----------



## mvsapre (Apr 17, 2017)

I got 892A after changing Z line to 0.43


----------



## rg1 (Apr 17, 2017)

mvsapre said:


> I got 892A after changing Z line to 0.43


If you ignore Gen, yes the answer is correct 892 A. But on adding Genr contribution the current will reduce to 800, This is because the gen impedance is added to the circuit.


----------



## Limamike (Apr 17, 2017)

What spary said is correct, you have to ignore the Gen because you CANT get 50mva through a 25mva transformer.  BUT THAT said, it would still leave me with 882.  Which is no where near the 1030.  I think thats a mistake on PPI, they have a lot of mistakes in their books - NOT COOL!


----------



## trainrider (Apr 18, 2017)

I'm planning to :210:  the Camera book after this test. It's an oversize paper V: .


----------



## Limamike (Apr 18, 2017)

lol


----------



## TNPE (Apr 19, 2017)

No, what @TNSparky said is not exactly true, but I know what he means.  Without a source, the fault current would be goose eggs!  You have to get the generator and transformer on the same power base, then all other pu values will adjust accordingly. After obtaining all pu values, use the appropriate formula and multiply this value by the FLA of the transformer.


----------



## TNPE (Apr 19, 2017)

TNPE said:


> No, what @TNSparky said is not exactly true, but I know what he means.  Without a source, the fault current would be goose eggs!  You have to get the generator and transformer on the same power base, then all other pu values will adjust accordingly. After obtaining all pu values, use the appropriate formula and multiply this value by the FLA of the transformer.


I got 800.5 A using the pu method.  Keep in mind, the generator and XFMR are not on the same power base.  This has to be corrected, in turn, adjusting the impedance of the generator. 

It seems to be somewhat of a consensus here that the answer in the book is incorrect.  I'd agree.  If someone has anything to interject to validate the veracity of the answer, please do.


----------



## Limamike (Apr 19, 2017)

Hmmmmm....  So that was my original answer. 800.  Now I am confused!  So the gen does count?


----------



## Millerific (Apr 19, 2017)

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?


----------



## Millerific (Apr 19, 2017)

Millerific said:


> 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.


----------



## TNPE (Apr 19, 2017)

Limamike said:


> 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.


----------



## Millerific (Apr 19, 2017)

TNPE said:


> 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:
> 
> ...


Why is Vnew 11?  Shouldn't it be 33 to match the voltage base of the of the transformer and line?


----------



## Millerific (Apr 19, 2017)

Millerific said:


> Why is Vnew 11?  Shouldn't it be 33 to match the voltage base of the of the transformer and line?


nevermind...brain fart.  11 is right...


----------



## TNPE (Apr 19, 2017)

Millerific said:


> 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.


----------



## rg1 (Apr 19, 2017)

TNPE said:


> 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:
> 
> ...


Absolutely correct. These are the answers coming with actual Z values and MVA Method.


----------



## Millerific (Apr 19, 2017)

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


----------



## TNPE (Apr 19, 2017)

Millerific said:


> 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)


----------



## Limamike (Apr 19, 2017)

@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


----------



## Limamike (Apr 19, 2017)

Thanks


----------



## TNPE (Apr 20, 2017)

Limamike said:


> @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).


----------



## Limamike (Apr 20, 2017)

Yessir...got em.  Thanks @TNPE


----------

