Low Fault Current Results in High Arc Flash

This might work..

http://www.geindustrial.com/cwc/Dispatcher?REQUEST=PRODUCTS&pnlid=5&famid=40&catid=82&id=arcvaultPS

I you have a low fault current from the utility you should have a reduce fault on the secondary of the transformer. Is that your question?

One of the major contributing factors to the secondary side fault current is the size of the transformer feeding the system. You may have fairly low primary (utility) fault current, but if your transformer is pretty large, the transformer ratio is high, and/or it has a low % impedance, it can source a pretty hefty fault current at secondary voltage. A quick estimate for available secondary fault current (assuming infinite bus on primary), is to divide Full Load Amps (FLA) secondary by the impedance as a decimal. Try this with impedances of 5%, 8%, and 10% (.05, .08, .10), for example, and you will see the difference. Maybe it has to do with your choice of transformer. It is possible to specify higher % impedance from standard ANSI value to reduce fault currents (utilities do it all the time!), but you may pay some extra.

But, if the source fault level is low, that would surely reduce the secondary side fault current too. Isn't it?

The arc flash energy is greatly dependant on the speed of the protection.

In general, a low fault current will take longer to clear (by the protective device). While the arc energy will be lower, the exposure time will be longer. The combined effect can be that the cumulative emitted radiation is higher demanding a higher level of protection (high calories).

You can reduce the calorific level by selecting faster breakers and/or tightening the coordination levels.

Using protections that are fast and current limiting can also make a big difference.

At the extreme, try a semiconductor rated fuse (or a very fast one). The extreme speed will bring the calories down quickly. It might be difficult to coordinate with other protections though, but that is another problem...

Marcot has the explaination to the OP's question. If you choose and infinite bus to get your fault value you can get in trouble Example

Transformer 1500 kVA 3.5%z 208 VOLTS

Infinite Bus = 118 ka

5000 amps available primary = 85 ka

1000 amps available primary = 40 ka

Extreme examples but it shows the difference when you use available fault values.