# Graffeo Example 31



## Dodgeviper1017 (Oct 1, 2016)

Find the Voltage regulation of a 25 kilovolt ampere, 480 V, 3 phase wye connected synchronous generator with the power factor of a) 0.81 lagging and b) 0.81 leading at rated voltage full load the winding resistance is 0.15 Ω per face, and the synchronous reactance is 6.1 Ω per face.

Why did they not use the 30 degree phase shift when converting from line to line to line to neutral? 480/1.733 = 277 angle -30 instead of 277 angle 0. Even assuming it was 0 why did they not keep the angles in VR formula. If they had I got 0.44 not 0.49. Using 30 degree phase shif I get 0.52. Why did they not keep angles in formula and why did they not use 30 degree phase shift? Ir appears they are working with phasors so I dont understand.


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## Dodgeviper1017 (Oct 1, 2016)

Another question what exactly is Ea and how did the know it added instead of subtracted. Where is this formula from?


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## Dodgeviper1017 (Oct 2, 2016)

I know this isnt a very scientific way to think about it but can I assume no 30 degree phase shift when given a power factor to be multiplied by a phasor? Or can I assume that if the answers are magnitudes that the 30 degree phase shift wont be used?


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## jmitch23 (Oct 7, 2016)

The phase shift only applies when converting between wye and delta. Also, the VR formula uses the voltage magnitudes.

Ea should be the generated voltage per phase.


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## AAGR2013 (Oct 12, 2016)

jmitch23 said:


> The phase shift only applies when converting between wye and delta. Also, the VR formula uses the voltage magnitudes.
> 
> Ea should be the generated voltage per phase.


not sure it is right. On Graffeo pg. 13, the 30 degree is in wye, Line-Line convert to Line-Neutral, not converting y to delta.


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## jmitch23 (Oct 12, 2016)

You're right. I don't know where my head was at the time.


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## Dodgeviper1017 (Oct 13, 2016)

Ok so what is the consensus on when to use it. I get that for Voltage Regulation only magnitudes are used but for voltage drop problems why isnt it used?


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## jmitch23 (Oct 19, 2016)

The phase angles are used for voltage drop equations, but are of particular importance if you have a complex impedance.


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## Dodgeviper1017 (Oct 20, 2016)

Do you have any examples so I can see what you are saying. I think you mean something like in this problem when you multiply a current by a impedance you need both angles but for 277 volts all by itself in the equation you dont use it.


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