pu and X/R Ratio

Is there any relation between pu and X/R ratio. From the info given in problem, can someone find pu Z of Gen. Any material on this part of the subject? Thanks

Impedance or pu impedance is a complex number having both terms X and R.

X/R = tan(theta) .. the impedance angle.

if for two equipment, say generator and transformer, having same X/R ratio, then both have the same impedance angle and you can simply add the pu impedance.

if X/R ratio is different, then you calculate impedance angle for both and add the pu impedances as vectors.

For the above question, I couldn't see how he get the pu of the generator by tis simple division <_< . How was the original question?

Omer, Thanks for response and you are right. I am sorry, I wanted to post the question, but posted the answer. Now I post the Question. 

 I couldn't see from where he got this relationship.

 Impedance of the generators could be any value. 

If I would solve this question is and the impedance of the generator is not given,  I would consider it as in finite bus.

Omer said:

 I couldn't see from where he got this relationship.

 Impedance of the generators could be any value. 

If I would solve this question is and the impedance of the generator is not given,  I would consider it as in finite bus. I think you meant Infinite bus.

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I however, though I do not have a reasoning for it, or say another way to look at it is take same pu of 5% for Generator too. The aim to throw this question to forum was to have second opinion, least I miss something. Anyways I thank you, Omer for the engagement. You feel good when you come to know that many are on same page.

After reading the question we can tell: Vbase=460V; MVA base= 2MVA which means Ib=2MVA/(SQRT(3)x460)=2510,218562Amps; Zbase= 0.1058

Let's do it using P.U. first then we will use Ohms:

1) P.U

Xgenerator is given = 0.1p.u.

Xtransformer is given = 0.05p.u.

Then Ifault=1/(0.1+0.05)= 6.666... times Ibase

Ifault=6.6Ib=16734,79041Amps

2) Ohms

Xgen=460**2/20MVA = 0.01058 Ohms

Xtransformer=0.05xZbase= 0.00529 Ohms

Then Ifault = (460/SQRT(3))/(0.01058+0.00529)=16734.79041

P.S. I put as much decimals as I could to get the exact same answer

Can we revisit this question?

@FPar I don't see how you went from Zbase_gen = 0.1058 ohms to a p.u. value of 0.1.

rg1 said:

Omer, Thanks for response and you are right. I am sorry, I wanted to post the question, but posted the answer. Now I post the Question. View attachment 10173

Click to expand...

Isc  502.04 KA