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Through-Fault Current Calculator

A Resource for Selecting a Substation Transformer Protective Device

One criterion in selecting a distribution substation transformer protective device is its ability to protect the transformer from secondary-side limited faults . . . or “through-faults.” These faults are difficult to detect by the overcurrent relay of the line-terminal circuit breaker, because the magnitude of the fault current is relatively low — being limited by the impedance of the transformer. These faults are a challenge to clear as well, because of their high transient recovery voltage.

That’s why a stand-alone protective device such as an S&C Circuit-Switcher is typically applied at each transformer. S&C has tested each of its distribution transformer protective devices — Series 2000, Mark V, and Mark VI Circuit-Switchers, and Trans-Rupter II® Transformer Protector — specifically for this duty, and assigned a secondary-fault interrupting rating to each device. 

This formula can be used to determine the secondary-fault interrupting rating required for the substation transformer protective device, to properly protect a particular size and impedance of transformer:

I = (57.8P) / [(%Z)E]
The formula assumes an infinite (zero-impedance) source.

 

Through-Fault Current Calculator

Conditions of Use

S&C does not accept responsibility for, nor warrant the proper application of non-S&C protective devices. Moreover, S&C is not responsible for any errors, mistakes, or miscalculations that may occur through use of the Through-Fault Current Calculator application. Results should be reviewed and approved by a consultant or end user who is familiar with the principles of determining the secondary-fault interrupting rating required for the substation transformer protective devices, to properly protect a particular size and impedance of a transformer.

This calculator is the property of S&C Electric Company, Chicago, IL and may not be sold nor be used such as to produce a commercial, marketable product.

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Where:

I =   Inherent secondary-fault current, amperes

P =   Transformer self-cooled three-phase base rating, kVA

E =   System phase-to-phase voltage, kV

%Z =   Percent transformer primary-to-secondary impedance, referred

A protective device is appropriate for the application if its secondary-fault interrupting rating is equal to or greater than the value for “I” calculated above. 

EXAMPLE:
The inherent secondary-fault current for a 20/40/50-MVA, 115-kV, 8%-impedance transformer would be:

I = [(57.8) (20 000 kVA)] / [(8) (115 kV)] = 1257 A