I have a coworker who said that if an infinite source is applied to the primary of a transformer, that the fault current would be infinite.
I know this is wrong because the maximum fault current is set by the impedance (%Z) of the transformer, but couldn't explain why in the moment. I'm pretty sure my coworker is incorrect. Or am i incorrect? Is it possible to get higher fault current than the theoretical max based on the %Z impedance?
If you graph the secondary fault current as a function of the primary fault current, you will notice that it is an increasing function, but that it asymptotically approaches a limiting "cruising altitude", as the primary fault current gets large. You will also notice that the cruising altitude of this function also doesn't depend on the primary voltage. This gives us an upper bound on the fault current, to calculate it on the secondary, regardless of either of the primary side parameters.
Here is the formulas behind the calculation.
Given:
Iscp = primary side fault current
Vp = primary side voltage, phase-to-phase
Vs = secondary side voltage, phase-to-phase
Z = Percent impedance, translated to a decimal
VA = VA rating of the transformer. 1000*the KVA rating of the transformer, or 1 million * the MVA rating
phi = a factor relating to whether it is single phase or three phase. For single phase, phi = 1, for three phase, phi = sqrt(3).
Start by calculating the f-factor:
f=phi*Iscp*Vp*Z/VA
Then calculate the M-multiplier, from the f-factor
M = 1/(1+f)
From the M-multiplier,you calculate the secondary fault current. Multiply the primary fault current by the voltage ratio, and then by the M-multiplier.
Iscs = M*Iscp * Vp/Vs
M and f are placeholder variables to make use of the formula a lot easier. If you combine the steps to make a single equation, you get the following equation:
Iscs = 1/(1 + phi*Iscp*Vp*Z/VA) * Iscp*Vp/Vs
Define K and x as the following, to simplify the equation:
K = phi*Z/VA
x = Iscp*Vp
K is a constant, consisting of the impedance, 3-phase factor if applicable, and KVA. None of these depend on Iscp or Vp. Since Iscp and Vp always appear together, we can define these terms as x. The equation becomes:
Iscs = x/((1 + K*x)*Vs)
Take the limit as x approaches infinity. This provides an upper limit of what the fault current on the secondary could be, regardless of how large Iscp or Vp may be. The 1 in 1 + K*x becomes insignficant. Once eliminating the 1+, we can cancel the x. We get:
Iscs_infbus = lim as x-> inf of Iscs =
x/((1+K*x)*Vs) =
x/(K*x*Vs) =
1/(K*Vs)
Reconstruct the formula, recalling the definition of K:
Iscs_infbus = VA/(phi*Z*Vs)
And you see that it equals amps associated with KVA rating, divided by impedance as a decimal. This is the upper boundary on how large the fault current can get. The transformer impedance is ultimately what limits it.
