# NCEES #107 vs #132



## Rei (Feb 22, 2010)

Both #107 and #132 deal with three-phase transmission line but one uses voltage/1.732 and one doesn't divided by 1.732; why is that? Thanks.


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## Flyer_PE (Feb 22, 2010)

#107: They're setting up a pu system with the base values being the line values. The base values in a per unit analysis are somewhat arbitrary. They happen to be chosing line values instead of phase values.

#132: The question is asking for short circuit current in amps. The calculation is done using phase values.


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## Rei (Feb 22, 2010)

Flyer_PE said:


> #107: They're setting up a pu system with the base values being the line values. The base values in a per unit analysis are somewhat arbitrary. They happen to be chosing line values instead of phase values.
> #132: The question is asking for short circuit current in amps. The calculation is done using phase values.


Yes, #107 ended up using line values to calculate the base, but then they find the pu value by dividing the phase resistance. I was thinking either find phase base value, or multiply 50 ohm/phase by 3. I totally disagree with the answer for this question.

For #132, they are asking for 3-phase fault, but the answer calculated using phase values. I'm going nut!


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## Flyer_PE (Feb 22, 2010)

^Welcome to the wonderful world of three-phase analysis. 

The thing to remember with calculating fault currents is that the first thing you want to do is convert the problem so that it can be analyzed as a single phase circuit. The voltages are usually given in Line values that will require conversion to phase values.

For problem 107, it's just the vagaries of pu analysis. Your base values for pu analysis are Voltage, Current, Power, and Impedance. Any two of these values can be chosen pretty much arbitrarily (It's best to pick values that make the math easier). The two values you choose will determine the values for the other two.


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## electric (Mar 23, 2010)

Rei said:


> Yes, #107 ended up using line values to calculate the base, but then they find the pu value by dividing the phase resistance. I was thinking either find phase base value, or multiply 50 ohm/phase by 3. I totally disagree with the answer for this question.


107:

Why Zpu=50/176.04 instead of 50*3/17.04, I still don't get it? All the calculations are done for line values and the value for Zactual used is 50/PHASE!!

I will never get the mystery of per unit calcs.


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## electric (Mar 23, 2010)

electric said:


> Rei said:
> 
> 
> > Yes, #107 ended up using line values to calculate the base, but then they find the pu value by dividing the phase resistance. I was thinking either find phase base value, or multiply 50 ohm/phase by 3. I totally disagree with the answer for this question.
> ...


^correction. I meant 50*3/176.04


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## Flyer_PE (Mar 24, 2010)

^Because the cables are not in series, the impedances don't add.


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## Insaf (Aug 2, 2011)

Instead of taking 65 KVA and 24 MVA base (even the problem suggested), take base KV 37.53 [(65/sqrt(3)] base MVA 8 MVA (24/3). Now you will get base impedance 176 (37.53^2/8) and PU impedance 0.284 (50/176).


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## Insaf (Aug 2, 2011)

Yes, #107 ended up using line values to calculate the base, but then they find the pu value by dividing the phase resistance. I was thinking either find phase base value, or multiply 50 ohm/phase by 3. I totally disagree with the answer for this question.

Explanation why you don't need to multiply by 3:

This problem is concerned with finding base value, so you need to follow the procedure finding base.

Suppose you don't like to choose the base KVA as 65 or base MVA 24, let choose Base KVA 37.53 (per phase voltage) and Base MVA 8 (per phase MVA), eventually you will get Base impedance 176 ohm ((37.53^2)/8). (Now you can think this 176 ohm is on per phase basis, even its not). So no matter what base value you choose. Now divide 50 by 176 and you will get the answer 0.284.

Please note that the question suggested to choose line rated values as base value. ( You need to find answer, NCEES will not see your work out)

Hope this will clarify little.

Thanks


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## lisfs (Apr 11, 2016)

Can this be solved using MVA Method?  It seems to work, but I'm not sure if it's right.  Please help check. 

System MVA = MVA1 = (60^2) / 13.25&lt;81 = 271.7 MVA

Z1 MVA = MVAZ1 = (60^2) / 16.75&lt;71 = 214.9 MVA

Z2 MVA = MVAZ2 = (60^2) / 13.4&lt;71 = 268.7 MVA

Total MVA = Add MVA1, MVAZ1, and MVAZ2 in series = 82.95 MVA

Isc = 82.95MVA / (sqrt3 * 60kV)  = 798Amp


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