Short Circuit Current Estimate (Transformer Terminals and Cable End)

Why calculate the short circuit current?

The highest current that can flow during a fault sets the breaking capacity of the breakers, the withstand rating of busbars and cables and the protection scheme. This tool gives a rough first estimate of the three phase fault current from the transformer nameplate.

Current at the transformer terminals

If the network is taken as infinite, only the impedance of the transformer limits the current:

I_k = I_n / u_k, where I_n = S_n / (√3 × U) is the rated current of the transformer.

The lower the impedance voltage, the higher the fault current. At 4 percent it is 25 times the rated current, at 6 percent 16.7 times.

Example

A 1000 kVA, 400 V transformer with 6 percent impedance voltage has a rated current of 1,000,000 / (1.732 × 400) = 1443 A. The fault current is 1443 / 0.06 = 24,056 A, roughly 24 kA. The transformer impedance is 0.06 × 400² / 1,000,000 = 9.6 mΩ.

Effect of the network and the cable

In reality the network has impedance too. A 500 MVA fault level adds 400² / 500,000,000 = 0.32 mΩ and brings the current down to 23.3 kA. If the fault is some distance away, the cable resistance and reactance are added as well: Z_k = √(R² + X²), I_k = U / (√3 × Z_k). The current falls quickly as the cable gets longer or thinner.

Where do I get the cable resistance?

You need the one way resistance of one phase conductor. The conductor resistance output of the cable voltage drop calculator can be used. Take the cold conductor resistance for the maximum current. Without manufacturer data, about 0.08 mΩ per metre is used for the reactance.

Limits

This is a preliminary estimate and does not replace an IEC 60909 study. Motor contribution, peak current, single phase to earth faults, minimum fault current and parallel transformers are not considered. Choosing the breaking capacity and setting the protection is the job of a qualified electrical engineer.