Utility Zpu

[bjenks]

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.

 

[dpc]

The utility MVA fault duty has to be for some specific point in their system. This takes the voltage into account. But if all you need is Zpu, you don't need to involve the voltage in your calculation. You'd need the voltage if you want to determine the actual fault current in amperes.

 

[odlanor]

4.18-short-circuit MVA and impedance equivalent
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

 

[bjenks]

I believe I am able to answer my own question for how Z_pu= kVA_base/kVA:
1) Z_pu= Z/Z_base
2) Z= V²/VA
3) Therefore Z_pu= [V²/VA]/[V_base²/V_baseA_base]
4) Since we are using the utility fault data, we are setting V_base=V=V_LL
5) Substituting into formula 3 we get Z_pu= [V²/VA]/[V²/V_baseA_base]
6) The V²/V² will equal 1
7) Thus Z_pu= VA_base/VA or kVA_base/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.