# NCEES PE electrical and Computer practice exam 528



## Rajan (Oct 15, 2020)

A test of a 10 KVA, 2300/230V, single phase transformer gives an open-circuit test power consumption of 69W, and a short circuit test reveals a resistance referred to the high voltage side of 10.9 ohm. The efficiency of this transformer, operated at full load and unity power factor is most nearly:

I am confused with the steps in the solution.  Do we have any other methods?


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## LyceeFruit PE (Oct 15, 2020)

hey @leggo PE can you move this into the Power PE forum?


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## leggo PE (Oct 15, 2020)

LyceeFruit PE said:


> hey @leggo PE can you move this into the Power PE forum?


Done!


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## BebeshKing PE (Oct 16, 2020)

Rajan said:


> A test of a 10 KVA, 2300/230V, single phase transformer gives an open-circuit test power consumption of 69W, and a short circuit test reveals a resistance referred to the high voltage side of 10.9 ohm. The efficiency of this transformer, operated at full load and unity power factor is most nearly:
> 
> I am confused with the steps in the solution.  Do we have any other methods?


Which part of the solution you don't understand, though? The formula for efficiency is efficiency=Pout/(Pout+Core loss+Copper loss).

Pout=KVA output(pf)=10kVA(1.0)=10kW

Core loss=No load loss=Open ckt power consumption=69W

Copper loss=(I^2)R  where I is the current at the high-voltage side. I=Spri/Vpri=10,000kVA/2,300V =4.35A

Copper Loss=(4.35)^2(10.9)=206.05W

So,

efficiency=10,000/(10,000+96+206.05) x100 =97.3% (answer)


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## yaoyaodes (Dec 8, 2021)

Why the current is not 2300/10.9?


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## xnuke (Dec 19, 2021)

That is the primary current with rated primary voltage applied and the secondary shorted. The current you need to use for I^2R loss is the primary current when the secondary is at full load, not shorted, i.e., primary full load current.


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