Utility Zpu
Utility Zpu
(OP)
In the IEEE Red book, when you are calculating the utility Zpu they just use the simple formula: (KVAbase)/(KVAsource) which is a unit less number.
Thus for a base of 100MVA and a fault MVA of 400 you would get Zpu=100/400=.25pu.
I don't understand the math behind that simple formula as I would think the formula would be based on kV*kV/MVA. So with the utility at 69kV the Zbase would be69*69/100= 47.61 which would then be converted to Zpu somehow or other. I understand all the other pu calculation, just not the utility.
Thus for a base of 100MVA and a fault MVA of 400 you would get Zpu=100/400=.25pu.
I don't understand the math behind that simple formula as I would think the formula would be based on kV*kV/MVA. So with the utility at 69kV the Zbase would be69*69/100= 47.61 which would then be converted to Zpu somehow or other. I understand all the other pu calculation, just not the utility.






RE: Utility Zpu
RE: Utility Zpu
ref: Symmetrical components ... - Blackburn
4.18.1-MVAsc = three-phase short-circuit MVA
IkA3ph = total three-phase fault current in kA
kV = phase-to-phase voltage
IkA3ph = MVAsc / (sqrt(3) * kV)
Zohm= kV/ (sqrt(3) * IkA3ph) = kV^2/MVAsc
Zpu= Zohm / (kV^2/MVAbase)
Zpu= (kV^2 /MVAsc) * (MVAbase/kV^2)
Zpu= MVAbase/MVAsc
4.18.3- a short-cicuit study indicates at bus X in the 69kV system:
MVAsc=594 MVA
MVAgsc= 631 MVA
on a 100 MVA base. Thus the total reactance to the fault is:
X1=X2=100/594 = 0.1684 pu
Xg= (3*100)/631=0.4754 pu
X0= Xg-X1-X2
X0=0.4754-0.1684-0.1684 = 0.1386 pu
RE: Utility Zpu
1) Zpu= Z/Zbase
2) Z= V2/VA
3) Therefore Zpu= [V2/VA]/[Vbase2/VbaseAbase]
4) Since we are using the utility fault data, we are setting Vbase=V=VLL
5) Substituting into formula 3 we get Zpu= [V2/VA]/[V2/VbaseAbase]
6) The V2/V2 will equal 1
7) Thus Zpu= VAbase/VA or kVAbase/kVA
This only works if the actual voltage and the voltage base are the same. I am not sure if there would ever be a cause for me that it isn't true.