rg1
Well-known member
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
pu and X/R Ratio
- Thread starter rg1
- Start date
Help Support Professional Engineer & PE Exam Forum:
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?
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?
rg1
Well-known member
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. 
rg1
Well-known member
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.
FPar
Member
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
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
