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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%.

UntitledCalculation:

  • 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.mmAluminium 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 Meter , 1KV, 3.5C x 400 Sq.mmAluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0618 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.012X0.1230) / 13
  • Total Cable Resistance=0.00009 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.012X0.0618) / 13
  • Total Cable Reactance =0.00006 Ω
  • Total Cable Impedance (Zc3)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc3)=0.00011 Ω——–(10)
  • U Impedance at Main LT Panel (Z Pu)= (Zc3 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Main LT Panel (Z Pu)= (0.00011X2.5×1000)/(0.415×0.415X1000)
  • P P.U Impedance at Main LT Panel (Z Pu)= 001601 Ω ——(11)
  • Total Impedance up to Main LT Panel (Z Pu-d)= (Zc3)+ (Z Pu-c) =(11)+(9)
  • Total Impedance up to Main LT Panel (Z Pu-d)= 0.001601 +0.0807
  • Total Impedance up to Main LT Panel (Z Pu-d)=0.082306 Ω.——(12)
  • Fault MVA at Main LT Panel = Base MVA / Z Pu-a.
  • Fault MVA at Main LT Panel = 2.5 / 0.082306
  • Fault MVA at Main LT Panel = 30 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 30 / (1.732X0.415)
  • Fault Current at Main Lt Panel = 42 KA

(5) Fault Level from Main LT Panel to Sub Panel:

  • LT Cable used from Main LT Panel to Sub Panel is 2 Runs , 160 Meter , 1KV, 3.5C x 400 Sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0618 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.160X0.1230) / 2
  • Total Cable Resistance=0.008184 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.160X0.0618) / 2
  • Total Cable Reactance =0.004944 Ω
  • Total Cable Impedance (Zc4)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc4)=0.0095614 Ω——–(13)
  • U Impedance at Sub Panel (Z Pu)= (Zc4 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Sub Panel (Z Pu)= (0.0095614 X2.5×1000)/(0.415×0.415X1000)
  • P P.U Impedance at Sub Panel (Z Pu)= 13879 Ω ——(14)
  • Total Impedance up to Sub Panel (Z Pu-e)= (Zc4)+ (Z Pu-d) =(14)+(12)
  • Total Impedance up to Sub Panel (Z Pu-e)= 0.13879 +0.082306
  • Total Impedance up to Sub Panel (Z Pu-e)=0.2211 Ω.——(15)
  • Fault MVA at Sub Panel = Base MVA / Z Pu-a.
  • Fault MVA at Sub Panel = 2.5 / 0.2211
  • Fault MVA at Sub Panel = 11 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 11 / (1.732X0.415)
  • Fault Current at Sub Panel = 16 KA

(6) Fault Level from Sub Panel to Motor Control Panel:

  • LT Cable used from Sub Panel to Motor Control Panel is 6 Runs , 150 Meter , 1KV, 3.5C x 400 Sq.mm Aluminium Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0739 Ω/Km.
  • Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
  • Total Cable Resistance=(0.150X0.1230) / 6
  • Total Cable Resistance=0.003075 Ω
  • Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
  • Total Cable Reactance=(0.150X0.0739) / 6
  • Total Cable Reactance =0.0018475 Ω
  • Total Cable Impedance (Zc5)=√(RXR)+(XxX)
  • Total Cable Impedance (Zc4)=0.003587 Ω——–(16)
  • U Impedance at Motor Control Panel (Z Pu)= (Zc5 X Base KVA) / (Base KV x Base KVx1000)
  • U Impedance at Motor Control Panel (Z Pu)= (0.003587 X2.5×1000)/(0.415×0.415X1000)
  • P P.U Impedance at Motor Control Panel (Z Pu)= 05207 Ω ——(17)
  • Total Impedance up to Motor Control Panel (Z Pu-f)= (Zc5)+ (Z Pu-e) =(17)+(15)
  • Total Impedance up to Motor Control Panel (Z Pu-e)= 0.13879 +0.2211
  • Total Impedance up to Motor Control Panel (Z Pu-e)=0.27317 Ω.——(15)
  • Fault MVA at Motor Control Panel = Base MVA / Z Pu-a.
  • Fault MVA at Motor Control Panel = 2.5 / 0.27317
  • Fault MVA at Motor Control Panel = 9 MVA
  • Fault Current = Fault MVA / Base KV
  • Fault Current = 9 / (1.732X0.415)
  • Fault Current at Motor Control Panel = 13 KA

 Summary of Calculation:

Sr.NoFault LocationFault MVAFault Current (KA)
1
At HT Circuit Breaker
332
50
2
At Primary Side of Transformer
290
44
3
At Secondary Side of Transformer
31
43
4
At Main LT Panel
30
42
5
At Sub Main Panel
11
16
6
At Motor Control Panel
9
13
Source

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