# NCEES #513



## cruzy (Oct 26, 2010)

Since when is dividing amps by impedance give you amps? The way I did it was to take 1pu (volts) / total Z to get the Ipu. Then multiply that by the Ibase. I just assumed a 1pu volts, I'm not sure that was even correct. But nonetheless, I still don't understand how they got real amps by doing Ibase/Zeq. The only way I see that they would get Ireal = Ibase/Zeq is if they actually did Ibase * (1pu volts/ Zeq) = Ibase/ Zeq and they didn't show all the work.


----------



## vinnius (Oct 27, 2010)

cruzy said:


> Since when is dividing amps by impedance give you amps? The way I did it was to take 1pu (volts) / total Z to get the Ipu. Then multiply that by the Ibase. I just assumed a 1pu volts, I'm not sure that was even correct. But nonetheless, I still don't understand how they got real amps by doing Ibase/Zeq. The only way I see that they would get Ireal = Ibase/Zeq is if they actually did Ibase * (1pu volts/ Zeq) = Ibase/ Zeq and they didn't show all the work.


it might help to check out the errata sheet for those extra 2 lines that might explain why you get amps


----------



## cruzy (Oct 27, 2010)

vinnius said:


> cruzy said:
> 
> 
> > Since when is dividing amps by impedance give you amps? The way I did it was to take 1pu (volts) / total Z to get the Ipu. Then multiply that by the Ibase. I just assumed a 1pu volts, I'm not sure that was even correct. But nonetheless, I still don't understand how they got real amps by doing Ibase/Zeq. The only way I see that they would get Ireal = Ibase/Zeq is if they actually did Ibase * (1pu volts/ Zeq) = Ibase/ Zeq and they didn't show all the work.
> ...


All those extra lines show is how they got the impedances on their selected bases, but they go right into I = I/Z after that, which doesn't answer my question yet


----------



## GabeM (Oct 28, 2010)

This might sound crazy this close to the exam, but I would forget about per unit and learn how to use the MVA method linked in the following thread:

MVA method

I can't answer your question because I don't even know how to use per unit. I don't know if this is going to bite me in the exam, but I was able to do all the short circuit problems in the NCEES sample exam using the MVA method.

See the paper on how to use the MVA method, but in summary, convert transformers to MVA by taking their MVA rating and dividing by its per unit impedance, and convert transmission lines to MVA by V^2/impedance.

Then, combine the MVA's like you would capacitors: in parallel you just add them, in series they are the inverse of the sum of the inverses.

Finally, divide by the voltage of the line where the fault occurs and you get the short circuit current.

I guess it's hard to explain HOW it works, but I found it is easy to do after I took about a half hour to learn it. I can't figure out the per unit method at all.


----------



## cableguy (Oct 28, 2010)

I agree with GabeM, once you go MVA method, you won't want to go back.

It made it to my formula workbook as a full page.

NCEES sample exam problems 513, 530, and 540 can easily be solved with this method.

I'll still do ground faults via the classical method, but for 3 phase / short circuits, MVA is the way to go. Take the MVA method paper, make a few notes (write down the ABCs), then work those 3 NCEES problems. Note that for 513 you'll need to compute MVA of the transmission line. This is done with line-line kV. It's so straightforward, it's worth 30-45 minutes today if you don't know it already.


----------



## GabeM (Oct 28, 2010)

I believe the solution to 513 is dividing volts per unit by ohms per unit to get amps per unit. It then multiplies by 4811 amps to convert per unit amps to actual amps. The solution is confusing because it omits volts per unit. It can do this because volts per unit is 1. However, this means that volts per unit base is 12 kV/1.73, not 12 kV. This trips me up because I thought you typically set the line voltage as the per unit base. I think where the square root of 3 gets absorbed is my main problem with the per unit method.

EDIT: I meant 12 kV/1.73, not 12kV*1.73.


----------



## cableguy (Oct 28, 2010)

For the per unit version of the solution to 513, I didn't do it "their way", but I came up with the same answer.

I didn't convert to 100MVA base; I left the transformer at 7.5MVA base and converted the line. Saved a step. Zbase for the line was 12kv^2/7.5M = 19.2 ohms. This led to j.7975 per unit for the line. Then current = (1 pu volt)/(j.7975+j.075) = 1.146 per unit amps. Base amps = 7.5M / (sqrt(3) * 12kV) = 360.84 amps.

Fault amps = 1.146 * 360.84 = 413.52


----------



## Flyer_PE (Oct 28, 2010)

For problems like this, the best method will be the one that you're most comfortable with. My advice is to stick with the methodology or methodologies that you are used to and will get you to the correct answer on a consistent basis. This is especially true when the clock is ticking during the exam.


----------



## GabeM (Oct 28, 2010)

cableguy said:


> For the per unit version of the solution to 513, I didn't do it "their way", but I came up with the same answer.
> I didn't convert to 100MVA base; I left the transformer at 7.5MVA base and converted the line. Saved a step. Zbase for the line was 12kv^2/7.5M = 19.2 ohms. This led to j.7975 per unit for the line. Then current = (1 pu volt)/(j.7975+j.075) = 1.146 per unit amps. Base amps = 7.5M / (sqrt(3) * 12kV) = 360.84 amps.
> 
> Fault amps = 1.146 * 360.84 = 413.52


Is the per unit voltage base always phase voltage? That's the only way your equation (1 pu volt)/(j.7975+j.075) = 1.146 per unit amps could work.


----------

