# NCEES 514



## GTOShoota (Oct 20, 2013)

I havent' seen any other threads on this problem, but I'm really dumbfounded on their solution. The diagram in the back hasn't helped me much either. Maybe I'm not as strong on the PU as I thought. Any suggestions?


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## iahim (Oct 20, 2013)

You need to understand symmetrical components to solve this problem. The trick is constructing the correct sequence network for the single line to ground fault. If you have Grainger/Stevenson, look on page 524. Once you built the network, you just use KVL to solve for positive seq. voltage.

Another option, is to simply use the equations on page 526 for single line to ground fault. With those you solve this problem in two lines, without drawing any circuit.


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## GTOShoota (Oct 20, 2013)

Thanks! I have that book and will be reading through that section tonight. Thanks again for the quick reply!


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## kris7o2 (May 24, 2021)

Hi, I also can't understand this problem.

I don't understand why Ia1 = 1.0/(2*Z1+Z0). Can anyone explain why?


A phase-to-ground fault occurs on the 33-kV transmission system shown in the figure...


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## akyip (May 24, 2021)

kris7o2 said:


> Hi, I also can't understand this problem.
> 
> I don't understand why Ia1 = 1.0/(2*Z1+Z0). Can anyone explain why?
> 
> ...


For a single-line-to-ground fault (also called a single-phase-to-ground fault), the overall sequence network diagram has the positive (Z1), negative (Z2), and zero (Z0) sequence impedances all in series. So here:

Z = Z1 + Z2 + Z0

Z = Z1 + Z1 + 10Z1 = 12 * Z1

Since the sequence networks are in series, that means that the positive, negative, and zero-sequence currents are all equal for a single-line-to-ground fault:

I1 = I2 = I0 = V / Z = V / (12 * Z1)

I attached an illustration to explain this better. Note that ZF in the circuit diagram is the fault impedance, which is typically zero (usually a bolted short-circuit fault is assumed if the type of fault is not stated as bolted or arcing, and a bolted fault means that ZF = 0).


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## djl PE (May 24, 2021)

Thank God I'm not EE.

Good luck!


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## Zach Stone P.E. (May 25, 2021)

kris7o2 said:


> Hi, I also can't understand this problem.
> 
> I don't understand why Ia1 = 1.0/(2*Z1+Z0). Can anyone explain why?
> 
> ...


Nonsymmetrical faults such as unbalanced three-phase faults, single line to ground faults, double line to ground faults, and line to line faults all have different fault circuit compared to each other that are a different combination of the positive, negative, and zero sequence network. These fault circuits can be found in the CBT Reference Handbook.

Since each fault circuit is different, the resulting circuit analysis such as KVL and KCL formulas come out different.

This problem is for a phase-to-ground (single line to ground) fault. Single line to ground faults includes the positive, negative, and zero sequence network all in series.

Looking at the single line to ground fault circuit in the handbook, you can solving for Ia1 using Ohm's law Ia1 = Vp/Zeq where Vp is the pre-fault voltage (1 pu in this case) and Zeq is the equivalent impedance. Since the positive, negative, and zero sequence network are in series for this fault type, the equivalent impedance (Zeq) is the series sum of the positive sequence impedance (Z1), the negative sequence impedance (Z2), and the zero sequence impedance (Z0).


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## Zach Stone P.E. (May 25, 2021)

djl said:


> Thank God I'm not EE.
> 
> Good luck!


It's not that bad! 

It's actually pretty neat the way the math works out in electrical power engineering. Many neat coincidences.


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