Fault current calculations

[BRYAN AMBROGIO]

I have been able to find the fault current equation through transformers all over the internet but only using the infinite bus calculation method. I cannot find the equation using the finite bus calculation. Can anyone please give me this equation. Also, could you please verify I am doing the infinite bus calculation correctly? Please see below.


Infinite bus : 112.5kVA : 480P : 120/208S : impedance (z) = 1.5.

112500 / 360 = 312 ------> 312 / 1.5% = 20818A AFC @ Secondary side of the transformer.


Finite Bus : Same transformer with 9500A AFC on primary side of transformer.

What will be the AFC @ Secondary side of the transformer?

 

[rlundsrud]

Your infinite bus Calc is correct, however I can't help on the finite question. I believe you would need info about the power supply system to answer that.

 

[ron]

ohmic or per unit should do the trick
http://www1.cooperbussmann.com/library/docs/EDP-1.pdf

 

[Phil Corso]

Bryan,

Here is my Gi.Fi.E.S. method (pronounced Jiffy?s) for problem solution, where:
Gi represents "Given"; Fi for "Find"; E for "Equation"; and S for "Solution!" Th following examples will solve your problem:

Example 1
Given:
A 3-phase transformer having the following parameters; design or rated capacity:
o 112.5kVA; kVpri = 0.48; kVsec = 0.208; Imp Z=1.5%; connected to an infinite source.
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Infinite means no impedance between the transformer primary and the source voltage, then the short-circuit current magnitude is the transformer's rated kVA, divided by its impedance in per-unit,
o kVAsc = kVA / (Z/100), and then,
o Isc = kVAsc / (1.732 * kVsec)
Solution:
o kVAsec = 112.5 / 0.015 = 7,500 kVA.
o Isc = 7,500 / (1.732 * 0.208) = 20,820A = 20.8 kA

Example 2
Given:
Parameters above except that SC fault-current level at xfmr primary is given in Amperes = 9,500 A
Then, primary SC Duty, kVApri = 1.732 * Isc * kVpri ~ 7,900 kVA
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Because only reactance?s are involved, that is, resistances are ignored, I will use a short-cut procedure called the kVA Value Method.
Solution:
Combine the pri and sec fault MVA?s together as if they were two resistances in parallel, that is,
o kVAequiv = (kVApri * kVAsec) / (kVApri + kVAsec).
Solve equations as required to find equivalent SC-Duty on 208V bus:
o kVAequiv = (7,900*7,500) /(7,900+7,500) = 3,850 kVA.
o Isc = 3,850/ (1.732 * 0.208) = 10.7 kA

Please let me know if additional detail is required.

Ps: Are you sure that SC-Duty at the Xfmr?s primary is only 9,500A

Regards, Phil Corso

 

[BRYAN AMBROGIO]

Thank You

Thank You

Thank you Phil,

I spent about half of a day looking for this equation, its almost impossible to find online.
Online examples typically don't define all the variables.
You broke it apart and defined all the variables, this makes it very easy to follow.

To answer your question about 9500A on the primary, I just used this as a default number for calculation purposes.
I am waiting for more information about the existing system so I can calculate this number accurately.

Bryan