Complex Imaginary Variation 2 problem 68- transient current

[ElecPwrPEOct11]

This problem specifies a synchronous 3phase generator rated at 50kVA, 230V with per unit reactance of X'd= 0.4. With analysis performed at rated values, what is the magnitude of the transient current?

The solution uses:

Vbase = 230V

Sbase = 50kVA

I base = 50/ (sqrt (3) * V) = 125A

then

1) Vi=I' * X'd.

2) I (pu) = Vi / X'd so = 1 / 0.4 = 2.5 pu

3) On the 50kVA base I actual = Ipu * Ibase = 2.5 * 125 = 312.5A

For these types of problems I've always used the general generator equation Ea = Vt + j*I*X'd

At rated values Vt = 1pu = 230V. We know X'd, but don't know Ea or I. I'm not sure how to solve for I, but don't see how the equation in 1) above is valid. I'm looking to NCEES #135 for guidance but it simply assumes I = 1pu. 2 questions- is the CI answer correct, and if so where does the Vi= I' * X'd equation come from?

 

[ElecPwrPEOct11]

Bump. Any thoughts on CI's solution?

 

[Flyer_PE]

If you short the output terminals of the generator, Vt=0 so that term drops out.

Their solution is correct for determining the transient fault contribution for a generator.

 

[ElecPwrPEOct11]

^ O, that makes sense. Thanks Flyer for the consistently straightforward answers.

 

[Insaf]

If you short the output terminals of the generator, Vt=0 so that term drops out.

Their solution is correct for determining the transient fault contribution for a generator.

----------------------

Why output terminal need to be shorted? Any explanation will be appreciated.

Thanks,

 

[Flyer_PE]

The question is asking for the transient short circuit current for the generator. The output terminals are shorted by definition.

 

[Insaf]

We can solve another simple method:

KVA(sc)= 50/0.4=125

Then, I (trans)= (125*10^3)/(1.732*230)=313.77 A (very close to 312.5 A).

Thanks,