Calculating Current Limiting Reactor Size in EasyPower

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EasyPower reports the sequence impedance values in PU:

Base value is 10MVA, and in this case the base voltage is 115kV. This source has 2500MVA SC MVA. That is equal to 250 PU, which means Z_pu will be 0.004pu (1/MVA_pu). See below for the EZP report:

User-added image

The short circuit contribution comes from the 2500MVA fault MVA.


2500/ (sqrt (3) * 115kV) = 12.551 kA                                                                                                  

Fault current on the oneline 

To limit the current, let’s say to 8kA, we need to see how much additional impedance is needed:

8kA will be ~ 159.34 pu (See the following calculations)  


I_base = 10MVA/ (sqrt (3))*115 = 50.20437A



Approximate I pu = 8 kA/I_base = 8,000A/50.20437A = 159.34 pu



Source voltage is 1pu, therefore, the new impedance will be: 1 / 159.34 pu ~ 0.00627554 pu



Z_new - Z_old = Z_CL => Z_CL = 0.00627554 pu – 0.004 pu = 0.00227554 pu (CL value)



Z_base = V2/S = (115)2/10 MVA = 1322.5 Ohms


Converting to Ohms:


ZCL = Z_base x pu = 1322.5 x 0.00227554 = 3.0094 Ohms


Putting this value in EZP:

Insert Current Limiting Reactor

Enter calculated impedance

Completed Current-Limiting Reactor



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