# New NCEES #513 Question



## ndekens (Feb 27, 2009)

Ok, so I suck a little at this Per Unit analysis stuff....... so can someone explain this problem to me? Why does the solution use 100MVA as a base? Thanks!


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## Flyer_PE (Feb 27, 2009)

The system base for a PU analysis is arbitrary. You will get the same end answer regardless of what base you choose. Having said that, you need to put some thought into the base value you choose based on the system you are analyzing.

The reason they chose 100MVA is that when the 7.5% impedance of the 7.5 MVA transformer is converted to a 100MVA base, the pu impedance is 1. That leaves you with only having to convert the line impedance value to a pu value on the 100MVA base.


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## mudpuppy (Feb 27, 2009)

100 MVA is also the de facto standard power base in the transmission industry in the US.


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## ndekens (Mar 2, 2009)

Ok I got it on the base value of 100 MVA but how are the PU impedances found?


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## Flyer_PE (Mar 2, 2009)

The transformer impedance of 7.5% is on the transformer base of 7.5 MVA.

The conversion is punew=puold*(MVAnew/MVAold = .075 * 100/7.5 = 1

For the transmission line, the base impedance needs to be determined.

Zbase = (kVbase)2 / MVAbase

Where:

MVAbase = 100

kVbase = 12

Zbase = 122/100 = 1.44 ohms

Zactual = 20 Miles at j0.145 ohm/1000ft &lt;-- I'm too lazy to do the math. 

Zpu = Zactual / Zbase


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## indy-engineer (Mar 3, 2009)

Your explanation helps a lot. Thanks!


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## ndekens (Mar 4, 2009)

Ok so now my next issue with this problem is in the solution provided how did they determine that Ia(3phase)=Ibase(a)/(ZT1pu+ZLpu)? Is this a standard formula in the process of PU analysis that I(Actual) = Ibase/Zpu?


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## dzdave00 (Mar 4, 2009)

They skipped a couple steps and made it more confusing (as if they needed to do that!)

Start with I_actual = I_pu*I _base

for naming convention, say Z_pu = ZT1pu+ZLpu

I_pu = V_pu/Z_pu

V_pu = 1 (I assume this is the case, I do not have the question in front of me)

therefore, I_pu = 1/Z_pu

substitute into the first equation and you get:

I_actual = I_base/Z_pu


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## Flyer_PE (Mar 5, 2009)

^Nicely done, IMHO.


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## ndekens (Mar 10, 2009)

Ok, so I got this problem figured out but the only thing I dont understand as to why....and this may be a stupidly obvious answer......is why did they us the sqrt3 * 12KV in the solution when solving the problem. Isn't 12KV already a line to line voltage in this wye type problem?


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## Flyer_PE (Mar 10, 2009)

^It's a simple algebraic manipulation:

S = sqrt3*VLine*ILine

ILine = S / (sqrt3 * VLine)


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## phowardtx (Apr 14, 2010)

ndekens said:


> Ok, so I suck a little at this Per Unit analysis stuff....... so can someone explain this problem to me? Why does the solution use 100MVA as a base? Thanks!


NM


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## jbone123 (Apr 4, 2015)

Is the given 7.5% impedance of T1 for a 12KV/7.5MVA base? The problem solution seems to assume this, since when the 7.5% base is converted to a base of 100MVA, the voltage component of the conversion equation is left out.

Does this imply that the %Z of a transformer is always given on a base of the secondary voltage and the rated VA?

Thanks

-jbone


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## Oneeye0236 (Apr 6, 2015)

Did not get Per Unit at all for October exam. It started clicking this time around studying. Check out the Complex Imaginary Videos on Youtube. A couple times of watching these helped me out.

&gt;https://www.youtube.com/watch?v=dw30BAoHlYI&amp;index=3&amp;list=PL88ACDE903BDCC454


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