Hi everyone,
I picked up a new text book the otehr day to brush up on industrial power. I came across soemthing i'm not sure is a mistake or not.

The example reads:

Utility: 120kV, 2000MVA, X/R=15
Base MVA = MVAb = 10
Base kV = 120, 13.8, 0.6

it then proceeds to work out base current for a voltage of 13.8kV and 0.6kV which is fine.

However it works out Per Unit Reactance (X) of the Utility by the following equation:

P.U.X = 10 / 2000 = 0.0050

I'm not sure where they are deriving this information? Has anyone seen something similar?

It would be very simply:
1) kV=sqrt(3)*Isc*Zsc
2) Ssc=sqrt(3)*Isc*kV
Then:
2a) Isc=Ssc/sqrt(3)/kV
Substitution in 1)
kV=sqrt(3)*Ssc/sqrt(3)/kV*Zsc
then:
Zsc=kV^2/Ssc
Zbase=kV^2/Sbase
Zscpu=Zsc/Zbase=Sbase/Ssc
If Sbase=10 MVA and Ssc=2000
Zscpu=10/2000=0.005

Ahh thank you for the help.

What is the name of the text book? Is it any good?

DJR

The name of the book is "Industrial Power Systems" by Shoaib Khan.

http://www.amazon.com/Industrial-Power-Systems-Sho...

I picked it up to brush up on some calcs I haven't done in what feels like 10 years!

It seems alright so far, a lot of references to the IEEE colour books and a couple of areas where they skip or dont' reference the formula (hence my confusion earlier). But not bad so far.

That´s the complete calculation:

Xold = 1; %pu at MVAold and kVold
kVold = 120; %kV
kVnew = kVold;
MVAold = 2000; % MVA
MVAnew = 10; %MVA

Xnew= Xold * (MVAnew/MVAold) * (kVold^2/kVnew^2)
Xnew=> 0.0050 pu at MVAnew and kVnew