Short Circuit Calcs in new Text Book
Short Circuit Calcs in new Text Book
(OP)
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?
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?






RE: Short Circuit Calcs in new Text Book
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
RE: Short Circuit Calcs in new Text Book
RE: Short Circuit Calcs in new Text Book
DJR
RE: Short Circuit Calcs in new Text Book
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.
RE: Short Circuit Calcs in new Text Book
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