# NCEES #539



## cruzy (Oct 23, 2010)

"A single phase transformer rated 11/7.2 kV has an impedance of 10.0%. During a factory short circuit test done at rated current, the voltage (kV) applied to the HV terminals will be most nearly:"

Answer: 10% of 100 kV = 10 kV.

Do you multiply by 10% because of the impedance or because 10% of transforer HV is the standard amount when doing a short circuit test on a transformer, and, if so, what standard is that? I don't see a specific value in any of my books, it just says in all of them that a low voltage is needed to be applied...


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## t5rrr (Oct 23, 2010)

cruzy said:


> "A single phase transformer rated 11/7.2 kV has an impedance of 10.0%. During a factory short circuit test done at rated current, the voltage (kV) applied to the HV terminals will be most nearly:"
> Answer: 10% of 100 kV = 10 kV.
> 
> Do you multiply by 10% because of the impedance or because 10% of transforer HV is the standard amount when doing a short circuit test on a transformer, and, if so, what standard is that? I don't see a specific value in any of my books, it just says in all of them that a low voltage is needed to be applied...


You only need the primary voltage to answer this question. Wildi's quoted less than 5% of rated primary voltage is typically applied during SC test, I have read in other books 5 or 10% is typically applied. Of the given answers 5% (5kV) is not aa choice so 10% (10kV) is the answer.

Not exactly a concise question but a clear example of how we have to assume through "experience" sometimes.


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## cableguy (Oct 23, 2010)

The way I worked this question is that we're doing rated current on the output with an impedance of .1 per unit.

1 per unit current times .1 per unit impedance = .1 per unit voltage.

On the high side, .1 per unit voltage = 10.


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## Flyer_PE (Oct 23, 2010)

If you do the math, you will find that, when you short one winding, the percentage of rated voltage applied to the non-shorted winding that results in rated current flowing in the shorted winding is pretty much the impedance of the transformer.

Example: 0.15 p.u. voltage applied to the primary generates 1.0 p.u. current in the shorted secondary: Your transformer impedance is 15%.


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## cruzy (Oct 28, 2010)

Wow so this is pretty much just a V = IR question (seems like a lot of these problems are when you break em down to p.u....just comes to knowing your source pu values, and that's where the tricky part for me is, to know which one is the 1 p.u. item...sometimes seems like it's the source, sometime seems like it's the load).


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## cableguy (Oct 28, 2010)

cruzy said:


> Wow so this is pretty much just a V = IR question (seems like a lot of these problems are when you break em down to p.u....just comes to knowing your source pu values, and that's where the tricky part for me is, to know which one is the 1 p.u. item...sometimes seems like it's the source, sometime seems like it's the load).


I've come to the conclusion that's exactly what the NCEES is trying to test. Do you understand the tricks and can you apply them?

A short circuit test is always about _RATED CURRENT_. "Rated" means 1.0 per unit current.

An open circuit test is about _RATED VOLTAGE_. 1.0 per unit voltage.


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## BamaBino (May 17, 2011)

t5rrr said:


> You only need the primary voltage to answer this question. Wildi's quoted less than 5% of rated primary voltage is typically applied during SC test, I have read in other books 5 or 10% is typically applied. Of the given answers 5% (5kV) is not a choice so 10% (10kV) is the answer.


That's not correct. If 5kV was a choice, it would not be the correct answer to this problem.


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## Insaf (Aug 3, 2011)

Question "A single phase transformer rated 11/7.2 kV has an impedance of 10.0%. During a factory short circuit test done at rated current, the voltage (kV) applied to the HV terminals will be most nearly:"

Answer: This problem is associated with % impedance of transformer

Formula: %Z= (Short Circuit Voltage, Vsc)/(Rated Primary Voltage, Vp)

Here given values, %Z=10%, Vp=100 kV, you need to find the Vsc (the voltage to be applied during SC test). So you will get Vsc= 100 kV *10%=10 kV.


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## BamaBino (Aug 11, 2011)

During this transformer short-circuit test,

Is = 13.9-kA and Ip=1-kA and the transformer is rated at 100-MVA correct?


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## BamaBino (Aug 11, 2011)

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## Jesse03 (Oct 22, 2011)

I got this one wrong because the Power Reference Manual from Camara states it's "typically" 5% or less of the rated voltage so I naturally chose 0.72kv. I think what Flyer_PE said is probably a better rule of thumb. That the impedance is pretty much equivalent to the voltage during a short circuit test. Thanks Flyer!


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## Insaf (Oct 23, 2011)

cruzy said:


> "A single phase transformer rated 11/7.2 kV has an impedance of 10.0%. During a factory short circuit test done at rated current, the voltage (kV) applied to the HV terminals will be most nearly:"
> 
> Answer: 10% of 100 kV = 10 kV.
> 
> Do you multiply by 10% because of the impedance or because 10% of transforer HV is the standard amount when doing a short circuit test on a transformer, and, if so, what standard is that? I don't see a specific value in any of my books, it just says in all of them that a low voltage is needed to be applied...


-----------------------------------------

In S.C. test secondary winding shorted and a voltage from variable source is applied and the voltage (Vsc) is increased until rated current flows in secondary (short circuited). Then Vsc is divided by primary rated voltage to get % impedance. (%Z=Vsc/Rated Primary Voltage)

In this problem, % impedance and rated primary voltage are given, so Vsc= %Z * Rated Primary Voltage= 0.10*100 KV=10 kV. There is no question of guessing or assuming something. To answer this question, no need to consider typical SC applied voltage (5%, 10% ...) used or not used in the industry. All information are provided, do the simple math.

Thanks,


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