Camara Power Practice Problems 36-2 Prob #8

[Limamike]

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]

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]

 

[Limamike]

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

 

[rg1]

View attachment 9330

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

 

[mvsapre]

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]

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]

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]

@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]

@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]

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]

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.

Pardon my negligence Line Z is wrong.  

 

[mvsapre]

I got 892A after changing Z line to 0.43

 

[rg1]

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]

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]

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

 

[Limamike]

lol

 

[TNPE]

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]

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]

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