Complex Imaginary Volume 2 #36

[jcbabb]

This problem involves computing the power output of a 3-phase induction generator which is connected to a wye system. You are given the full load output of 325A and the generator nameplate rating of 2.4 kV.

My thought was that the apparent power should be calculated as such:

(1) S=Vp*I*sqrt(3)

But the solution uses:

(2) S=3*Vp*I

In both equations Vp= phase voltage and I= line current = phase current.

I based my formula on the fact that we calculate the size of a transformer similarly. If I need a 480VD to 208Y/120V transformer size, I will take the amps I have on the 208Y/120V panel and multiply that by 208*sqrt(3) or:

(3) KVA=Vp*I*sqrt(3) which is the same as (1)

Can anyone explain why I would use (2) instead of (1)???

 

[DK PE]

The equation to commit to memory in my opinion is S = √3 x Vline x I line which is alway true wye or delta. In this case of wye system V phase = V line / √3 and if you substitute that into equation 2 you should get the equation I have above. Your equation 1 is incorrect in general.

 

[DK PE]

I guess to further illustrate my point with an example: Assume we have a 208Y/120 V system and a 3 phase motor with a line current of 10 A. What is the VA drawn by the motor?

S = √3 x Vline x I line = √3 x 208 x 10 ~ 3600 VA is correct

equation (2) S=3*Vp*I = 3 x 120 x10 is correct as well

your equation 1 gives incorrect solution.

 

[jcbabb]

I guess to further illustrate my point with an example: Assume we have a 208Y/120 V system and a 3 phase motor with a line current of 10 A. What is the VA drawn by the motor?

S = √3 x Vline x I line = √3 x 208 x 10 ~ 3600 VA is correct

equation (2) S=3*Vp*I = 3 x 120 x10 is correct as well

your equation 1 gives incorrect solution.

Hey, thanks for taking the time to reply!

I guess my confusion is above where you state that eq (2) would get the same result. Actually eq (2) would use the phase voltage, which is 208V or:

S=3*VP*I = 3*208*10 = 6240 VA.

 

[jcbabb]

Also, I reworked the problem and found where eq (2) comes from. The proper formula is as DK PE stated:

S = VL * IL * sqrt(3)

Where VL is line Voltage and IL is line current.

Because we are given the phase voltage (Vp) in the problem,

VL = Vp * sqrt (3)

Then plug that into the proper apparent power formula:

S = [Vp * sqrt (3)] * IL * sqrt(3)

Which simplifies to:

S = 3 * Vp * IL

Which is eq (2) from my first post.

 

[DK PE]

I guess my confusion is above where you state that eq (2) would get the same result. Actually eq (2) would use the phase voltage, which is 208V or:

S=3*VP*I = 3*208*10 = 6240 VA.

I assume you are no longer confused given your last post but the phase voltage is not 208 that is the line to line or normally termed the line voltage. The line to neutral voltage is sometime referred as the phase voltage and that is 120V

 

[jcbabb]

You are correct, which is what mixed me up for eq (3) on my first post as well. It really helps to hash these terms out. I use them every day and still get confused with the wording of the problems.

Thanks!