How Does EasyPower Calculate Utility Impedance Values in Short Circuit Analysis?

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EasyPower considers the source impedance when performing a short circuit analysis. Below, we demonstrate two cases that generate the same short circuit currents.

Case 1: Utility  @ 115 kv
Utility data entry
Fault Current
Calculating the I_sc from the MVA values, we get:
3Ph = MVA / (SQRT(3)* I_sc_3ph*kV) = 12.551kA                               
1Ph = MVA / (SQRT(3)* I_sc_1Ph*kV) = 10.041kA

Assuming 2500MVA and 115kV as base values, we get Xp = 1pu. EasyPower assumes Xp=Xn 
  • p = Positive sequence    
  • n = Negative sequence    
  • z = Zero sequence 
Using the following sequence formula, we find that Xz=1.75pu.
I_f (SLG) = 3 *{V/(Xp+Xn+Xz)}

Case 2: Utility with Zero Impedance @ 115kV.

Here, we have increased the MVA to the maximum that EasyPower allows. However, it still has a small impedance value:

Utility Data Entry

Fault current displayed on one-line diagram

Using the same method, and assuming 2500MVA and 115kV as base values with the current utility value, we get:

Xp = Xn = .005pu
Xp + Xn + Xz = .015 => Xz = .005pu


To match the impedance values of the utility in the first case, we need to add an impedance, where the value plus the above values gives us an equivalent of the utility in the first case. Using a CL reactor, we can add the reactance (X/R in the utility is chosen to be very large, so the impedance is all inductive).

For 3 phase the Xp of the reactor should be:


1 - .005 = 0.995pu or 5.26355 Ohm. (Z_base = 115^2/2500 = 5.29Ohms)
Xz = 1.75 – 0.05 = 1.745 pu or 9.23105 Ohms, where Z_base = 5.29 Ohms:


Reactor impedances

Below, you can see the exact matches for the 3 phase and SLG faults:

Matching fault currents displayed on one-line diagram



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