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# Short Circuit Calcs in new Text Book

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 Jereb (Electrical) 16 Jul 12 1:59
 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?
 7anoter4 (Electrical) 17 Jul 12 0:26
 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
 Jereb (Electrical) 17 Jul 12 0:51
 Ahh thank you for the help.
 djr3203 (Electrical) 18 Jul 12 15:43
 What is the name of the text book? Is it any good? DJR
 Jereb (Electrical) 18 Jul 12 17:36
 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.
 odlanor (Electrical) 25 Jul 12 11:52
 ThatÂ´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

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