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2. Calculation Short Circuit Current (Base KVA Method)

SHORT CIRCUIT CURRENT CALCULATION (BASE KVA METHOD)

EXAMPLE:

Calculate Fault current at each stage of following Electrical System SLD having details of.

  • Main Incoming HT Supply Voltage is 6.6 KV. * Fault Level at HT Incoming Power Supply is 360 MVA. * Transformer Rating is 2.5 MVA. * Transformer Impedance is 6%.

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  • Let’s first consider Base KVA and KV for HT and LT Side. * Base KVA for HT side (H.T. Breaker and Transformer Primary) is 6 MVA * Base KV for HT side (H.T. Breaker and Transformer Primary) is 6.6 KV * Base KVA for LT side (Transformer Secondary and down Stream) is 2.5 MVA * Base KV for LT side (Transformer Secondary and down Stream) is 415V

FAULT LEVEL AT HT SIDE (UP TO SUB-STATION):

(1) FAULT LEVEL FROM HT INCOMING LINE TO HT CIRCUIT BREAKER

  • HT Cable used from HT incoming to HT Circuit Breaker is 5 Runs , 50 Meter ,6.6KV 3 Core 400 sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0990 Ω/Km. * Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable. * Total Cable Resistance=(0.05X0.1230) / 5 * Total Cable Resistance=0.001023 Ω * Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable. * Total Cable Reactance=(0.05X0.0990) / 5 * Total Cable Reactance =0.00099 Ω * Total Cable Impedance (Zc1)=√(RXR)+(XxX) * Total Cable Impedance (Zc1)=0.0014235 Ω——–(1) * U Reactance at H.T. Breaker Incoming Terminals (X Pu)= Fault Level / Base KVA * U Reactance at H.T. Breaker Incoming Terminals (X Pu)= 360 / 6 * U. Reactance at H.T. Breaker Incoming Terminals(X Pu)= 0.01666 PU——(2) * Total Impedance up to HT Circuit Breaker (Z Pu-a)= (Zc1)+ (X Pu) =(1)+(2) * Total Impedance up to HT Circuit Breaker(Z Pu-a)=0.001435+0.01666 * Total Impedance up to HT Circuit Breaker (Z Pu-a)=0.0181 Ω.——(3) * Fault MVA at HT Circuit Breaker= Base MVA / Z Pu-a. * Fault MVA at HT Circuit Breaker= 6 / 0.0181 * Fault MVA at HT Circuit Breaker= 332 MVA * Fault Current = Fault MVA / Base KV * Fault Current = 332 / 6.6 * Fault Current at HT Circuit Breaker = 50 KA

(2) FAULT LEVEL FROM HT CIRCUIT BREAKER TO PRIMARY SIDE OF TRANSFORMER

  • HT Cable used from HT Circuit Breaker to Transformer is 3 Runs , 400 Meter ,6.6KV 3 Core 400 sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0990 Ω/Km. * Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable. * Total Cable Resistance=(0.4X0.1230) / 3 * Total Cable Resistance=0.01364 Ω * Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable. * Total Cable Reactance=(0.4X0.0990) / 5 * Total Cable Reactance =0.01320 Ω * Total Cable Impedance (Zc2)=√(RXR)+(XxX) * Total Cable Impedance (Zc2)=0.01898 Ω——–(4) * U Impedance at Primary side of Transformer (Z Pu)= (Zc2 X Base KVA) / (Base KV x Base KVx1000) * U Impedance at Primary side of Transformer (Z Pu)= (0.01898X6) /(6.6×6.6×1000) * U Impedance at Primary side of Transformer (Z Pu)= 0.0026145 PU——(5) * Total Impedance(Z Pu)=(4) + (5) * Total Impedance(Z Pu)=0.01898+0.0026145 * Total Impedance(Z Pu)=0.00261——(6) * Total Impedance up to Primary side of Transformer (Z Pu-b)= (Z Pu)+(Z Pu-a) =(6)+(3) * Total Impedance up to Primary side of Transformer (Z Pu-b)= 0.00261+0.0181 * Total Impedance up to Primary side of Transformer (Z Pu-b)=0.02070 Ω.—–(7) * Fault MVA at Primary side of Transformer = Base MVA / Z Pu-b. * Fault MVA at Primary side of Transformer = 6 / 0.02070 * Fault MVA at Primary side of Transformer = 290 MVA * Fault Current = Fault MVA / Base KV * Fault Current = 290 / 6.6 * Fault Current at Primary side of Transformer = 44 KA

(3) FAULT LEVEL FROM PRIMARY SIDE OF TRANSFORMER TO SECONDARY SIDE OF TRANSFORMER:

  • Transformer Rating is 2.5 MVA and Transformer Impedance is 6%. * % Reactance at Base KVA = (Base KVA x % impedance at Rated KVA) / Rated KVA * % Reactance at Base KVA = (2.5X6)/2.5 * % Reactance at Base KVA =6% * U. Reactance of the Transformer(Z Pu) =% Reactance /100 * U. Reactance of the Transformer(Z Pu)= 6/100=0.06 Ω—–(8) * Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=(Z Pu)+(Z Pu-b)=(7)+(8) * Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=0.06+0.02070 * Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=0.0807 Ω—–(9) * Fault MVA at Transformer Secondary Winding = Base MVA / Z Pu-c * Fault MVA at Transformer Secondary Winding = 2.5/0.0807 * Fault MVA at Transformer Secondary Winding =31 MVA * Fault Current = Fault MVA / Base KV * Fault Current = 31 / (1.732×0.415) * Fault Current at Transformer Secondary Winding = 43 KA

FAULT LEVEL AT LT SIDE (SUB-STATION TO DOWN STREAM):

(4) FAULT LEVEL FROM TRANSFORMER SECONDARY TO MAIN LT PANEL

  • LT Cable used from Transformer Secondary to Main LT Panel is 13 Runs , 12 M…