Available fault current calculation
Infinite Bus Method
1. Fault Current At Transformer
Inputs:
- Transformer Kva
- Secondary Voltage of Transformer (Line-Line)
- Secondary Voltage of Transformer (Line-Neutral)
- Transformer Impedance (Zt)
- Transformer Secondary Full Load Current I (Line-Line)
- Transformer Secondary Full Load Current I (Line-Neutral)
- KVA Short Circuit At Primary Side
- Source Impedance (Zu)
- Total Impedance (Z =Zu+Zt)
- Short Circuit Current RMS Symmetrical
- Fault Current At Transformer Secondary ( Isc(L-L)=I (L-L)/Total Impedance)
- Fault Current At Transformer Secondary ( Isc(L-N)=I (L-N)/Total Impedance)
2. Fault Current At Main Panel
Inputs:
- Cable Length from Transformer to Main Panel
- Type of Cable
- No of Conductor/Phase
- Size of Phase Conductor
- Size of Neutral Conductor
- Type of Race Way
- Phase Conductor Constant (C1)
- Neutral Conductor Constant (C2)
- F1(L-L) [F(L-L)=(1.732 X Length X Isc)/(C1 X No of Conductor X Volt)]
- F2(L-N) [F(L-L)=(1.732 X Length X Isc)/(C2 X No of Conductor X Volt)]
- M1(L-L) [M=(1/(1+F1)]
- M2(L-N) [M=(1/(1+F2)]
- Fault Current at Main Panel Disconnection (L-L) [Isc X M1]
- Fault Current at Main Panel Disconnection (L-N) [Isc X M2]
3. Fault Current At Sub Panel
Inputs:
- Cable Length from Main Panel to Sub Panel
- Type of Cable
- No of Conductor/Phase
- Size of Phase Conductor
- Size of Neutral Conductor
- Type of Race Way
- Phase Conductor Constant (C1)
- Neutral Conductor Constant (C2)
- F1(L-L) [F(L-L)=(1.732 X Length X Isc)/(C1 X No of Conductor X Volt)]
- F2(L-N) [F(L-L)=(1.732 X Length X Isc)/(C2 X No of Conductor X Volt)]
- M1(L-L) [M=(1/(1+F1)]
- M2(L-N) [M=(1/(1+F2)]
- Fault Current at Sub Panel Disconnection (L-L) [Isc X M1]
- Fault Current at Sub Panel Disconnection (L-N) [Isc X M2]
4. Fault Current At Distribution Board
Inputs:
- Cable Length from Sub Panel to Distribution Board
- Type of Cable
- No of Conductor/Phase
- Size of Phase Conductor
- Size of Neutral Conductor
- Type of Race Way
- Phase Conductor Constant (C1)
- Neutral Conductor Constant (C2)
- F1(L-L) [F(L-L)=(1.732 X Length X Isc)/(C1 X No of Conductor X Volt)]
- F2(L-N) [F(L-L)=(1.732 X Length X Isc)/(C2 X No of Conductor X Volt)]
- M1(L-L) [M=(1/(1+F1)]
- M2(L-N) [M=(1/(1+F2)]
- Fault Current at Distribution Board Disconnection (L-L) [Isc X M1]
- Fault Current at Distribution Board Disconnection (L-N) [Isc X M2]
| Software: | Short Circuit Current Calculation at Various Point of Electrical Circuits (Isc) |
| Version: | – |
| Developer: | Jignesh Parmar |
| Size: | 85 Kb |
| Price: | Free |
| Download: | Right here | Video Courses | Membership | Download Updates |
Is it possible that in the base MVA method tab there is an error in the calculation of fault current at Main LT Panel?
I think that there is an error because the impedance of the transformer is not added to the impedance used for the calculation and I’m getting results like 2 000 KA -> 2 000 000 A
Please confirm
PD.: Can I have the password to modify the spreadsheets?
Good day Edward could you pls send me a copy of short ckt current calculation at various point of electrical circuits in excel..thank you
Un saludo
please send me
[email protected]
regards
All: Where the system peak fault current and the cable diameter are known, the formula featured in the below Article, excerpted from the international standard IEC 61914, can be used to calculate the forces between two conductors in the event of a 3 phase fault in order to specify the correct type of cable cleats.
IEC 61914 – Cable Cleats & Short Circuit Protection Calculations
https://www.powerandcables.com/iec-61914-cable-cleats-short-circuit-protection/
Chris Dodds
Thorne & Derrick http://www.powerandcables.com