# Grainger Unsolved Problem 2.11



## electric (Oct 18, 2010)

Can someone solve the following problem?

'A balanced delta connected resistive load of 8000KW is connected to the low-voltage, delta connected side of a Y-Delta transformer rated 10,000KVA, 138/13.8 KV. Find the load resistance in ohms in each phase as measured from line to neutral on the high voltage side of the transformer. Neglect transformer impedance and assume rated voltage is applied to the transformer primary.'

I am coming up with 7144 ohms.

P per phase = 2666.67, Vphase=13.8KV (as it is delta) hence Iphase= 193.2A

Rphase = Vphase/Iphase= 71.44 ohms.

Transferring to H.V. side = a square*71.44 = 7144 ohms.


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## Flyer_PE (Oct 19, 2010)

Here's what I get:

ILine-138kV = S/(sqrt3*VLine) = 8MVA/(sqrt3*138kV) = 33.47 Amps

ZPhase = VPhase/IPhase = (138kV/sqrt3)/33.47 = 2.38k Ohms

OR

RPhase = VPhase2/PPhase

Where:

PPhase = 8MW/3 = 2.67 MW

VPhase = 138kV/sqrt3 = 79.7 kV

RPhase = (79.7kV)2/2.67MW = 2.38k Ohms


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## cableguy (Oct 19, 2010)

I got what Flyer got (2.38k) - didn't look at his solution first either. I did the easy way out, calculate high side current based on 8MW and then do R=V/I where V=138/sqrt(3)


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## mull982 (Oct 19, 2010)

electric said:


> Can someone solve the following problem?'A balanced delta connected resistive load of 8000KW is connected to the low-voltage, delta connected side of a Y-Delta transformer rated 10,000KVA, 138/13.8 KV. Find the load resistance in ohms in each phase as measured from line to neutral on the high voltage side of the transformer. Neglect transformer impedance and assume rated voltage is applied to the transformer primary.'
> 
> I am coming up with 7144 ohms.
> 
> ...


This is the same method I used to calculate the impedance and came up with the same answer.

I understand Flyer PE's method below as well and follow how he came up with the answer. I'm not sure now which methond is correct since they both provide different answers.

I just looked at the solutions that I had for this problem and the solution gives an answer of 2380ohms solved the same way Flyer did in his first method.

Why does this method come up with a different answer then reflecting the calculated secondary impedances?


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## cableguy (Oct 19, 2010)

Because it's a Y-Delta transformer, and you're not doing the final delta to wye conversion step. It's asking for the line to neutral reflected resistance.

Rwye=(7144^2)/(3*7144)=2380


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## DK PE (Oct 19, 2010)

mull982 said:


> I understand Flyer PE's method below as well and follow how he came up with the answer. I'm not sure now which methond is correct since they both provide different answers.
> I just looked at the solutions that I had for this problem and the solution gives an answer of 2380ohms solved the same way Flyer did in his first method.
> 
> Why does this method come up with a different answer then reflecting the calculated secondary impedances?


Another possible solution method is to convert the delta load of 71.4 ohms to a wye load of 71.4/3 = 23.8 ohms from line to neutral on that side and then reflect that to the primary side = 2380 ohms.


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