# NCEES Problem #118



## bfingland (Oct 3, 2010)

This problem describes a secondary transmission line with a current of 75.93A and voltage of 132kV fed from a generator on the primary (at 13.2 kV) thru delta-wye transformer. The question is: "The load current in the generator is most nearly..."

When I solved the problem, I assumed it was asking for the line current at the generator and used the following, Ip = Is*sqrt(3)/a, since it was a delta-wye configuration. This gave me approximately 1320A

However, the solution gives Ip = (Vs/Vp)Is = (132kV/13.2kV)(75.93A) = 760A, which appears to solve for the phase current on the delta primary.

Both the 1320A and the 760A are listed as possibilities. Next time, how can I determine which current I should be calculating when the question doesn't specifically call for the "phase" or "line" current? Thanks.


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## Flyer_PE (Oct 3, 2010)

The only place I can think of that you will ever be asked for a phase current is if you are solving for an individual transformer in a delta configuration.

For this problem, the fact that the transformers are wye-delta doesn't contribute to the solution at all since you are given the turns ratio for the entire bank.


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## LMAO (Oct 3, 2010)

bfingland said:


> This problem describes a secondary transmission line with a current of 75.93A and voltage of 132kV fed from a generator on the primary (at 13.2 kV) thru delta-wye transformer. The question is: "The load current in the generator is most nearly..."
> When I solved the problem, I assumed it was asking for the line current at the generator and used the following, Ip = Is*sqrt(3)/a, since it was a delta-wye configuration. This gave me approximately 1320A
> 
> However, the solution gives Ip = (Vs/Vp)Is = (132kV/13.2kV)(75.93A) = 760A, which appears to solve for the phase current on the delta primary.
> ...


You should *always* assume they are asking for line values (voltage/current) unless told otherwise.


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## GabeM (Oct 3, 2010)

You aren't given the turns ratio, it just says "13.2 - 132 kV" transformer. However, if you were given a turns ratio of 1:10, then the line current would be 1320 A right?


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## bfingland (Oct 3, 2010)

GabeM said:


> You aren't given the turns ratio, it just says "13.2 - 132 kV" transformer. However, if you were given a turns ratio of 1:10, then the line current would be 1320 A right?



GabeM, this is what I thought, however the correct answer was 760A.

Flyer, I wasn't given the turns ratio, but calculated it from the voltages easily enough (a=Vp/Vs). I'm still not clear though on why the delta-wye configuration doesn't come into play. Can you elaborate on this a bit more? How would a question be asked where the winding configuration would need to be considered?


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## GabeM (Oct 3, 2010)

bfingland said:


> GabeM said:
> 
> 
> > You aren't given the turns ratio, it just says "13.2 - 132 kV" transformer. However, if you were given a turns ratio of 1:10, then the line current would be 1320 A right?
> ...


I believe the correct equation is a=Vp*1.73/Vs for a delta/wye transformer.


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## agabee (Oct 3, 2010)

bfingland said:


> GabeM said:
> 
> 
> > You aren't given the turns ratio, it just says "13.2 - 132 kV" transformer. However, if you were given a turns ratio of 1:10, then the line current would be 1320 A right?
> ...


look at it this way. The ratio of line-to-line voltages would give you line to line currents. So forget about the sqrt 3, if you are dealing with line currents on both sides which is the case here. The transformer turns ratio is by definition phase quantity for prim and sec and this creates confusion if the connection is provided as typically given in transformer bank problems. The bottomline is to use line-to-line ratio to calculate line to line currents and the apply the necessary sqrt3 to determine phase current of the delta (if required)


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## bfingland (Oct 3, 2010)

GabeM said:


> bfingland said:
> 
> 
> > GabeM said:
> ...



Good call Gabe. That is the mistake I made; I was factoring in the delta-wye configuration when I was calculating the line current, but not when calculating the turns ratio first. Without skipping that step I would factor in the sqrt(3) in to the turns ratio calculation, and then it ends up canceling when you calc the line current on the primary. Thus you get the simplified equation listed in the solution: Ip = (Vs/Vp)Is

Thanks for the help!


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## electric (Oct 6, 2010)

Why can't I solve this with 50,000/[sqrt(3)*13.2]=2186A

Is it because, the MVA and KV are just rated values of the Generator or is it because of the impedance of T.L. between the Gen &amp; Transformer?


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## bfingland (Oct 7, 2010)

electric said:


> Why can't I solve this with 50,000/[sqrt(3)*13.2]=2186A
> Is it because, the MVA and KV are just rated values of the Generator or is it because of the impedance of T.L. between the Gen &amp; Transformer?


Correct, your calculation based on the rated values of the generator gives the rated current. The 'load current in the generator' that the question is looking for is the 75.93A reflected across the transformer. The impedance of the transmission line does not come in to play on this problem, and you assume the impedance between the generator and the first transformer is negligible as nothing was given.


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