# NCEES Power Afternoon Problems 525 & 531, Need some clarification



## chicago (Sep 14, 2007)

Problem # 525) In the solution, the first phrase states "The transformers will load in inverse proportion to their impedances."

Problem # 531) In the solution, the first phrase states "The vars vary as the square of the applied voltage."

For both problems, I am not looking for help in deriving the solution they have posted, but rather an understanding of what they mean by the respective statements.

Appreciate your efforts in advance.


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## Flyer_PE (Sep 14, 2007)

Problem 525: The parameter that determines how much load is carried by each transformer is the ratio of transformer impedances rather than the power rating of the transormers. It becomes essentially a current division problem. The lower impedance transformer will carry a greater portion of the total load.

Problem 531: The capacitor has a kvar rating at 240 volts. The application is on a 208 volt system. Power = V2/Z. Since Z in this case is fixed, the power will vary by the square of the voltage.

Let me know if this helps.

Jim


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## chicago (Sep 24, 2007)

Jim, thanks for the feedback. I understand the problems better now.

For Problem 525, if I understand correctly, I could setup the equation as follows:

I2/I1 = Z1/Z2

correct?


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## Flyer_PE (Sep 24, 2007)

The ratios are correct. The only thing to remember is that, since the impedances are given in percentages, they have to be converted to a common base. It doesn't matter if you convert Transformer 2 impedance to the Transformer 1 base or the other way around.

You can definitely expect to find a few problems that involve converting pu values from one base to another on the test.

Jim


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## shellbell500 (Oct 11, 2007)

i had the exact same question. thanks for clearing it up. i guess a lot of problems go back to that relationship, z=v^2/s!


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## Art (Oct 11, 2007)

525)

let's assume the PU V is 1

it will be the same for any xfmr in parallel

let's assume the following PU Z for each xfmr

T1 1

T2 1/2

T3 1/3

Tx 1/x

P=V^2/Z and V^2 = 1

P1 = 1/1 = 1

P2 = 1/(1/2) = 2 or 2 times (inverse of Z) P1

P3 = 1/(1/3) = 3....

Px = 1/(1/x) = x or x times (inverse of Z) P1

in other words they load inverse to their Z

531)

for any power (real P watts, complex S va, reactive Q var) the magnitude is proportional to V^2/Z

power is ALWAYS proportional to the square of the voltage (and the current I^2 Z)


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## jdd18vm (Oct 12, 2007)

Art said:


> 525)let's assume the PU V is 1
> 
> it will be the same for any xfmr in parallel
> 
> ...


Very good example representation Art. Put that in your notebooks guys.

John


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## Art (Oct 13, 2007)

jdd18vm said:


> Very good example representation Art. Put that in your notebooks guys.
> John


thank you sir

I was taught well by good teachers...

do NOT memorize anything...

learn to derive the basics from the physics

and convince yourself of the rest mathematically...

know the basics and build on them...it's not a coincidence these tests are open book...

the FE a bit less though, because it tests the basics...


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