Bolted Fault Current on a DC System

[wbd]

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.

 

[ScottyUK]

Battery manufacturers publish the internal resistance of their cells. Adding more cells to a battery does not increase the bolted fault current because that calculation method assumes that the shorting element has zero resistance: doubling the number of cells also doubles the resistances in series so the bolted fault remains the same. In reality the interconnecting link(s) add a bit of resistance so the theoretical short circuit value is not achieved. Whose cells are they?

 

[wbd]

Thank you ScottyUK. I could not find any published internal resistance for SBS batteries.

 

[wbd]

I contacted the manufacturer and they provided the values for the specific cells at time of manufacture.

 

[pwrtran]

Be aware of the terminals of a battery bank normally is upstream of the main DC breaker. In other words, there will be no interrupting device to break the arc if a bolted short occurs. Our practice is to select the battery with terminal covers, and/or plastic shielding.

 

[mikekilroy]

since cell resistance increases as batt wears, my 350vdc batt bank of cells on EV has a battery managment circuit that constantly measures each cell's internal resistance & voltage and reports it via the gui. I believe they simply are watching the cell voltage drop @ current and doing ohms law r=v/i: could you not do same? just read nominal voltage, put small resistive load across the fully charged cell and measure the drop?

 

[sibeen]

Battery manufacturers will give tables showing the short circuit current and the internal resistance, for example see:

http://www.enersysreservepower.com/documents/US-SBSF-RS-003_0610.pdf

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.