I need some help with NCEES #513.

[danderson]

It's a 12KV fault 20 miles down line from the substation transformer. Distribution line has a Z of j.145 ohms per 1000'. Sub transformer is 7.5MVA with 7.5% impedance. Wants to know the fault current.

After looking in the back of the book, I'm confused. Did they arbitrarily pick 100 MVA as their S base, or was it because 7.5M /.075 is 100M?

When they calculated the transformer impedance they said...

.075 pu x (100MVA/7.5MVA) = 1.0 pu

I know where .075 pu came from, but what is 100MVA/7.5MVA? I should know this but I haven't done pu in a long time. TIA

 

[danderson]

I backed up and punted. I used the MVA method. Much easier for me. Pretty good explanation of MVA method in PRM chpt 36.

 

[Lielec11]

I just did the same question...MVA method is so much easier than how they do it

 

[katag]

To add to this, why is the 60kv Utility not included in the calcuation? I realize that there is not enough information given about the utility but I guess I am kind of comparing it to Problem 132 where the system impedance must be included in the calculation. In theory, doesn't the utility/system contribute to the short circuit/fault or are these different things?

 

[Flyer_PE]

The 100MVA base was chosen based on convenience. For any PU analysis, you get to pick two base values (usually MVA and Voltage) with the rest of the base values being determined from those two. The same calculation can be performed on any MVA base and the answer will come out the same.

Since no impedance information is given for the 60 kV utility, it is treated as an infinite bus (impedance=0) in the problem.

 

[Insaf]

MVA base can be choosen based on convenience; suppose we choose Base MVA =7.5 MVA and base KV =12 kV (trans sec)

Base Imp =12^2/7.5 = 19.2 ohm

Base Current =7.5x1000/(1.732x12) = 361 A

Line imp=20x5280x0.145/1000 = 15.3 ohm

= 15.3/19.2

= 0.798 pu

So, I(pu)=1/(0.075+0.798)

= 1.145

Finally,Isc = 1.145x361

= 415 A

 

[katag]

When using the MVA method for this problem and other problems we always do the calculations in 3 phase, right? I was reworking this problem and hesitated for a second because when trying to calculate the LINE MVA I wasn't sure whether to use V-LL or V-PH voltage because it deals with line impedance. KV^2/%Z for the line is always done using V-LL, correct? Been studying all day and I'm starting to over analyze everything. Thanks for the help.

 

[Lielec11]

Yes you are correct, V-LL is always used when calculating the line MVA.

 

[patkt]

MVA base can be choosen based on convenience; suppose we choose Base MVA =7.5 MVA and base KV =12 kV (trans sec)

Base Imp =12^2/7.5 = 19.2 ohm

Base Current =7.5x1000/(1.732x12) = 361 A

Line imp=20x5280x0.145/1000 = 15.3 ohm

= 15.3/19.2

= 0.798 pu

So, I(pu)=1/(0.075+0.798)

= 1.145

Finally,Isc = 1.145x361

= 415 A

Excellent Explanation for the Per Unit calculation.

 

[patkt]

Using MVA Method:

SfT1: 7.5MVA / 7.5% = 100 MVA

We have to convert Z per given length: length is given in miles and Z is given per feet. So we have to convert unit.
Z = j0.145*20*5280 = j15.312

Now Find Sfz = 12KV^2 / j15.312 = 9.404

Here, we are using MVA method, so consider Series as parallel to find equivalent Sfeq = (1/100MVA)+(1/9.404) = 8.59 MVA

Now, we have total Sfeq; lets find Isc = Sfeq/(sqrt3*12kv) = 8.59mva / (1.732*12kv) = 413.5 A

Nearest Answer is 415A.