CI V1-30

[EEpowerOK]

I could not find previous discussions on this problem.

I get a max power (Watts) of 4.1 MW

I get the answer in the book of 17 MVAR or apparent power of 17 MVA but not watts.

3 * 8660 * 9140 @ 14 degrees

--------------------------------------- = 4.1 MW

14 j

 

[jcbabb]

Can you post the question?

 

[EEpowerOK]

A short circuit study is being performed on part of a system that includes a Y connected 3 phase synchronous generator rated for 15 KV, 10 MVA, .78 lagging power factor and a synchronous reactance of 14 ohm per phase. If the internal generated voltage at rated conditions is 9.14 KV at 14 degrees, what is the maximum power output of this generator?

 

[iahim]

The formula for power transmitted is:

P=(E generated LL) x (E terminal LL) x sin(delta) / X

This is maximum when sin(delta)=1, or delta=90 (angle between the two voltages).

So Pmax = 9.14k x sqrt(3) x 15k / 14 = 19.96M

The generated voltage in the synchronous generator model is LN, so you need to add the sqrt(3). Or you could convert the terminal voltage to LN, but then you need to multiply the power by 3.

I don't have this exact problem in my version of the CI. Does this match their solution?

 

[sghataurah]

you cannot generate more real power than apparent; using 10MVA at 0.78pf, I would guess a maximum power of 7.8MW.

what is the answer in the book?

 

[EEpowerOK]

17 MW is their answer.

 

[iahim]

The formula for power transmitted is:

P=(E generated LL) x (E terminal LL) x sin(delta) / X

This is maximum when sin(delta)=1, or delta=90 (angle between the two voltages).

So Pmax = 9.14k x sqrt(3) x 15k / 14 = 19.96M

The generated voltage in the synchronous generator model is LN, so you need to add the sqrt(3). Or you could convert the terminal voltage to LN, but then you need to multiply the power by 3.

I don't have this exact problem in my version of the CI. Does this match their solution?

Sorry, I typed the result wrong. Should be 16.96MW (not 19.96).

 

[iahim]

you cannot generate more real power than apparent; using 10MVA at 0.78pf, I would guess a maximum power of 7.8MW.

what is the answer in the book?

The 10MVA is the rating, not the maximum apparent power.

For delta = 90, Q = (E terminal)^2) / X = 16.07MVAR, so with P from above, S = Sqrt (P^2 + Q^2) = 23.37MVA. So you still generate more MVA's than MW's.

 

[JB66money]

The answer of 17MVAR is correct, because the question asks for the maximum power that the generator can deliver. The maximum power can only occur when the phase angle (delta) is 90 degrees and not 14 degrees, so =3x(E ratedLN) x (E internal LN) x sin(90) / X = P=3x(8660x9140)/14=16.95MVAR.

 

[JB66money]

The answer of Pmax ~= 17 MVAR is correct because the question asks for the maximum power that the generator can deliver. The generator can only deliver maximum power when the phase angle (delta) is 90 degrees and not 14 degrees. So Pmax = 3x(Einduced)x(Einternal)xsin(90)/Xs = 3x(8660)x(9140)/14 = 16.95 MVAR.