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.” S&C Electric Company features an online through-fault calculator on its web site at http://www.sandc.com/webzine/032707_1.asp.
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.
| 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 to transformer self-cooled three-phase kVA rating |
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
