Bolted Fault Current on a DC System
Bolted Fault Current on a DC System
(OP)
I am looking at an arc flash calculation for a battery bank. It will be done using Annex D of NFPA 70E. I have read that, read Chapter 12 in Jim Phillips book and reviewed the manufacturer of the battery's material.
In order to begin I need to determine the bolted fault current at the battery terminals. In order to do this, I need the internal resistance of the battery bank. Lacking any testing, how does one determine that?
Thanks in advance.
In order to begin I need to determine the bolted fault current at the battery terminals. In order to do this, I need the internal resistance of the battery bank. Lacking any testing, how does one determine that?
Thanks in advance.






RE: Bolted Fault Current on a DC System
RE: Bolted Fault Current on a DC System
RE: Bolted Fault Current on a DC System
RE: Bolted Fault Current on a DC System
RE: Bolted Fault Current on a DC System
RE: Bolted Fault Current on a DC System
http://www.enersysreservepower.com/documents/US-SB...
If you wanted to be especially conservative you could just use this figure for you study, but in real life you would be way over the top.
As an example I recently had to do a calculation on a telco centre battery rack which consisted of 7 strings of 4 x SBS 190F batteries. From the link above the short circuit rating of an SBS 190F is 3800 amps (3.3 x 10-3 ohm), so with 7 strings the fault current was 7 x 3800 = 26.6 kA.
The battery resistance can also be worked out from the nominal voltage (12 volts) divided by the short circuit current (3800 amps): 12/3800 = 3.16 x 10-3 ohms - very close to the given figures in the PDF.
Each individual string providing 3800 amps.
Now take into account the connecting devices.
Each battery link, 3 per string = 1.5 x 10-3 ohm.
Each cable from battery to circuit breaker, 2 per string = 3.5 x 10-3 ohm.
Circuit breaker resistance = 1.2 x 10-3 ohm.
Add that all up and you get about 2.5 x 10-2 ohms.
Now use the nominal string voltage = 48 volts divided by the worked out resistance = 48/2.5 x 10-2 = about 1900 amps.
So with 7 stings in parallel we have 7 x 1900 = 13.3 kA.
Just by adding in all the link elements, the short circuit current from this battery bank was brought down from the conservative 26.6 kA to a figure of 13.3 kA.