CI 3-34

[iahim]

What is the internal generated phase voltage for a 3-phase synchronous generator with per unit reactance of 1.1 and steady-state current of 0.97 pu. Base values: 250kV, 100MVA.

The CI solution is V=iX=1.07pu, or 268V. Is that correct?

I think the internal generated phase voltage should be the terminal voltage (L-N) plus iX = 1+1.07 = 2.07 pu. Vbase LN = 144V, so the solution should be 298V.

Thanks!

 

[EEpowerOK]

Vout + I * X = 1pu + .97 * 1.1j = 1.462 pu

Vout would have to be the base voltage to call it 1 pu.

 

[iahim]

Vout + I * X = 1pu + .97 * 1.1j = 1.462 pu

Vout would have to be the base voltage to call it 1 pu.

You are right. I missed the j in the reactance, so the solution would be 210.58V. But I still don't agree with CI's answer.

 

[saberger_vt]

What is the internal generated phase voltage for a 3-phase synchronous generator with per unit reactance of 1.1 and steady-state current of 0.97 pu. Base values: 250kV, 100MVA.

The CI solution is V=iX=1.07pu, or 268V. Is that correct?

I think the internal generated phase voltage should be the terminal voltage (L-N) plus iX = 1+1.07 = 2.07 pu. Vbase LN = 144V, so the solution should be 298V.

Thanks!

If I am reading this correctly, 1pu for voltage is 250,000V, since the base voltage is 250KV. The keyword I see here is "Phase", there is no indication they are going from L-N, it's specifically based on phase. With that in mind, the way I read this is the impedance for the phase is 1.1pu and current through that phase is 0.97pu. So the phase voltage is the phase impedance multiplied by the phase current (1.1pu)(0.97pu) = 1.07pu. So I would come up with a phase voltage of (1.07pu)(250KV) = 267,500V, or rounding to 268KV.

Remember that there is no "j" in per unit, that's already accounted for when you do the impedance per unit base calculation.

 

[iahim]

saberger_vt,

I agree with you, if the problem asks for the voltage drop over the synchronous reactance. But they ask for "internal generated phase voltage". I think the word phase is there because the sync gen model is per phase, so the the internal generated voltage and terminal voltage are L-N quantities.

 

[saberger_vt]

saberger_vt,

I agree with you, if the problem asks for the voltage drop over the synchronous reactance. But they ask for "internal generated phase voltage". I think the word phase is there because the sync gen model is per phase, so the the internal generated voltage and terminal voltage are L-N quantities.

Iahim,

Maybe I do not have all the information for the problem: Is this a delta or wye generator? Where is the neutral on the generator connected? Is the base voltage line to neutral or is it line to line? If their is no answer to these questions, then you need to use the information that is given in the problem, i.e. "phase" voltage. Remember, the voltage they are looking for is the "induced" voltage on the phase, which of course will be greater than the base voltage since the you have loss in the phase coil impedance. In this case, I believe there is a 17,500V drop over the coil, to get the base voltage value. Since you know the phase current and phase impedance (given in per unit), that's all you need to find the induced voltage on the phase coil.

If there is missing information in the above problem, please let me know.

 

[iahim]

saberger_vt,

I don't have the problem with me, but when I get home tonight I will scan it and post it here. Thanks for the help!