# NCEES #132



## cruzy (Oct 23, 2010)

Why do they divide the transmission system voltage of 60 by square root of 3 when dividing by the impedances to find the fault current? How do they know if that transmission system is not a delta and then the line-line voltage just equals the phase voltage (so that it would be just 60/Z)??


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## nmh0408 (Oct 23, 2010)

cruzy said:


> Why do they divide the transmission system voltage of 60 by square root of 3 when dividing by the impedances to find the fault current? How do they know if that transmission system is not a delta and then the line-line voltage just equals the phase voltage (so that it would be just 60/Z)??



It is the Line to Neutral fault current.


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## cableguy (Oct 23, 2010)

Yup, currents are based on the line to neutral voltage. Take V line-neutral and divide by impedance... there's your current. Are you taking the test on Friday?


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## cruzy (Oct 23, 2010)

cableguy said:


> Yup, currents are based on the line to neutral voltage. Take V line-neutral and divide by impedance... there's your current. Are you taking the test on Friday?


Yep. I think I've studied so much that I'm at the point where I'm making things more confusing to me than they used to be.


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## Benee (Oct 23, 2010)

cruzy said:


> cableguy said:
> 
> 
> > Yup, currents are based on the line to neutral voltage. Take V line-neutral and divide by impedance... there's your current. Are you taking the test on Friday?
> ...



Well, You're not the only one that have that thought. So don't feel bad.


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## Benee (Oct 23, 2010)

cableguy said:


> Yup, currents are based on the line to neutral voltage. Take V line-neutral and divide by impedance... there's your current. Are you taking the test on Friday?



Since you're very knowledgeable, Let me ask this question, IF the problem state that the Transmission lines were connected from the Delta source, then what Voltage do you use and why ? Can you explain, Master ?


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## cruzy (Oct 23, 2010)

Let me take a crack at and see if I'm finally right. Anyone, please correct me if I'm wrong. So since the formula calls for L-N voltage, if you had a delta source, then your L-N voltage is just your L-L voltage, but then your short circuit current is the one that would need to be divided by square root of 3 since in delta I(l-n) = I(l-l)/ square root of 3. So you would get the same answer.


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## Flyer_PE (Oct 23, 2010)

First: It's not a line to neutral fault. It's a three-phase balanced fault (All three lines are shorted together). If you start shorting just one or two conductors to each other, neutral, or ground, you're going to have unbalanced currents and will be using symmetrical components to determine the individual line currents.

Second: It doesn't matter whether the source is delta or wye. Regardless of which configuration you are analyzing, the equations to reduce the circuit to a single loop are the same. In order to determine the fault current, you're going to start with a phase voltage that is VLine/sqrt3. Whether you actually have a neutral or not, the math works the same.


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## Flyer_PE (Oct 23, 2010)

cruzy said:


> Let me take a crack at and see if I'm finally right. Anyone, please correct me if I'm wrong. So since the formula calls for L-N voltage, if you had a delta source, then your L-N voltage is just your L-L voltage, but then your short circuit current is the one that would need to be divided by square root of 3 since in delta I(l-n) = I(l-l)/ square root of 3. So you would get the same answer.


That's pretty much it. Whether it is the current or the voltage that is being reduced by the square root of 3 doesn't matter. The equation you wind up with is the same either way.


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## Benee (Oct 24, 2010)

Flyer_PE said:


> cruzy said:
> 
> 
> > Let me take a crack at and see if I'm finally right. Anyone, please correct me if I'm wrong. So since the formula calls for L-N voltage, if you had a delta source, then your L-N voltage is just your L-L voltage, but then your short circuit current is the one that would need to be divided by square root of 3 since in delta I(l-n) = I(l-l)/ square root of 3. So you would get the same answer.
> ...


Thanks Jim, your statement is true, Final equation is the same but I think in order to derive the final equations, the method is diffrence. I try to figure it out how to come up with the final equation but don’t know how to do it yet. Of course for the PE exam, I just accept the fact and use it


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## cruzy (Oct 24, 2010)

Benee said:


> Flyer_PE said:
> 
> 
> > cruzy said:
> ...


I think that the equation is V(L-N)/Z(total,equivalent) = I(fault).

Jim,

I have in my notes that for the system source Zth(pu) = V/Isc = 1/MVA. I don't get the 1/MVA part (i think by 1 they mean 1 volt pu), but it seemed to work for NCEES # 530, where I did 1/40MVA = 0.025 which lead to the right answer. Maybe it was just a lucky mistake? Or are my equations right?


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## cableguy (Oct 24, 2010)

cruzy said:


> I have in my notes that for the system source Zth(pu) = V/Isc = 1/MVA. I don't get the 1/MVA part (i think by 1 they mean 1 volt pu), but it seemed to work for NCEES # 530, where I did 1/40MVA = 0.025 which lead to the right answer. Maybe it was just a lucky mistake? Or are my equations right?


The system impedance is MVAbase / Fault Duty MVA. So if the MVAbase were 10MVA, with 40 MVA fault duty, we'd have .25 pu impedance. In this case, we had 1 MVA for the base.


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## cruzy (Oct 24, 2010)

cableguy said:


> cruzy said:
> 
> 
> > I have in my notes that for the system source Zth(pu) = V/Isc = 1/MVA. I don't get the 1/MVA part (i think by 1 they mean 1 volt pu), but it seemed to work for NCEES # 530, where I did 1/40MVA = 0.025 which lead to the right answer. Maybe it was just a lucky mistake? Or are my equations right?
> ...


So if they had asked for the phase angle as well in the answer and not just the magnitude, would we have had to include the 30 degree phase shift when dividing 60 by sqrt 3 to go from line-line to line-neutral?


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## cableguy (Oct 24, 2010)

If they had given us the phase rotation (abc or cba), and if they had given us a fixed reference (Vab=12.47 angle 0 kV, for example), then we could have accounted for the phase shift, yes. But without those 2 pieces of information, you can't solve the problem with a phase shift unless you make some serious, problem-altering assumptions.

Unless they assign a reference, I can put *my* reference anywhere I want. I can say that Vbn = 7.2 angle 0 kV and solve the problem from there. So they have to give us a reference angle, along with phase rotation, in order for us to arrive at their answers.

Something that helps me with phase shift is I took one page of my formula workbook and on the left half of the page, I drew the voltage and current phasors for ABC rotation, and on the right half of the page, I drew out CBA rotation. Now, whenever I hit a problem that needs to account for the angle, I can quickly look to my diagrams and see when to apply the 30 degree phase shift.


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