(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