# NCEES Power Practice Prob # 511????



## ndekens (Feb 19, 2008)

Is it me or am I just doing the math wrong? I keep getting an answer of 13.84KV. Can someone double check? The book say's 12.95KV.


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## Dark Knight (Feb 19, 2008)

ndekens said:


> Is it me or am I just doing the math wrong? I keep getting an answer of 13.84KV. Can someone double check? The book say's 12.95KV.


I will check when back at home. Don't have the EERM or the Sample test with me. I would not be surprised if it is a typo error.

I am almost sure you will have your answer before I go home tonight but just in case I hope to come back with the answer at 8:30PM.

G'luck mate


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## Flyer_PE (Feb 19, 2008)

The answer to your question can probably be found in this thread. Let me know if it isn't in there.

Jim


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## ndekens (Feb 19, 2008)

IFR_Pilot said:


> The answer to your question can probably be found in this thread. Let me know if it isn't in there.
> Jim


In that thread you state the following:

****************************************

This is pretty much just an expansion on how the NCEES solution is put together:

Known Values:

VLL=VAB=12.5kV at 0 deg.

IL=IaA=70A at -20 deg.

ZLine=5+j10 ohms

Step one, convert line voltage into phase voltage: VAN=VAB/sqrt 3 at -30 deg. = 7217 V at -30 deg

IL will remain at 70A at -20 deg. regardless of if you are using line or phase voltage at the load.

The next thing you need to know is VaA. Since we are now working on a phase voltage basis we don't need to concern ourselves with VbB.

VaA=IL*ZLine=783 V at 43.4 deg.

Now we can determine VaN.

VaN = VaA + VAN = (7217 ang. -30) + (783 ang. 43.4 deg.) = 7.48kv ang. -24.3 deg.

Converting back to line voltage values:

Vab = sqrt 3 * VaN = 12.95 kV

*******************************************************

I get all that except the "Vab = sqrt 3 * VaN = 12.95 kV"

My value for VaN is 7.992. Im thinking my problem lies in the adding in phasor for of 7.217KV ang -30 and 783V ang 43.3. DO I have to convert back to rectangular form to accomplish this? Or is there another way?


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## Flyer_PE (Feb 19, 2008)

ndekens said:


> I get all that except the "Vab = sqrt 3 * VaN = 12.95 kV"
> My value for VaN is 7.992. Im thinking my problem lies in the adding in phasor for of 7.217KV ang -30 and 783V ang 43.3. DO I have to convert back to rectangular form to accomplish this? Or is there another way?


I can expand it like this:

VaN = VaA + VAN = (7217 ang. -30) + (783 ang. 43.4 deg.)

VaA = (7217 ang. -30) = 6250 - j3608

VAN = (783 ang. 43.4) = 569 + j539

VaN= 6250 - j3608 + 569 + j539 = 6819 - j3069 = (7478 ang.-24.2 deg.)

The phase voltage value is 7478. Therefore the line voltage is: sqt3*7478=12.95kV

Let me know if this helps.

Jim


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## ndekens (Feb 19, 2008)

IFR_Pilot said:


> I can expand it like this:
> VaN = VaA + VAN = (7217 ang. -30) + (783 ang. 43.4 deg.)
> 
> VaA = (7217 ang. -30) = 6250 - j3608
> ...


I got it. Thanks......I was thinking I could just add in phasor form straight across.


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## yaoyaodes (Nov 29, 2021)

ndekens said:


> In that thread you state the following:
> 
> ****************************************
> 
> ...


it's a Delta connection. The phase voltage should equal the line voltage, why do we need to divide an sqrt of 3?


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