I am doing the calculations for a system composed of around 200 batteries, linked in series. The anticipated current is expected to be a little over 2500 amps. I got the internal battery resistance value and calculated the bolted fault to be around 80 kA. I looked at the specifications for the DC breaker and was concerned the Interrupt Rating was only 10 kA. My question is: Is the 10 kA Interrupt Rating of the DC Breaker for the battery string going to be a problem? We have Instantaneous Trip modules in the breaker that should trip well before 3000 amps with a clearing time of 33 mS.

The thing with short circuit calculations is that they don't take rise time into account, which appeas to be your argument. You need to calculate inductance and capacitance and apply the LCR equation if you want to figure out rise time and whether or not the relay/breaker can trip before you reach short circuit current.

In general this is always a problem. Breakers for DC don't have much in the way of interrupting rating simply because you have to break the arc by brute force only...pulling the arc apart. With AC you can take advantage of the current zero that occurs and the requirements for the breaker are much more mild...simply preventing restrikes in most cases.

This is a good example of where fuses, especially intermodule fuses, or else an alternative arrangement (series-parallel) might make more sense to get the short circuit current down to something reasonable. Your description by the way is a little strange. You said "batteries" and I'm assuming then that this means you have multi-cell jars or simply 200 cells? The battery is the whole thing...each container contains one or more cells.

It is just a LOT of extremely large 6 volt batteries in series.

Unfortunately, they don't want to put fuses between the cells. The other thing is that I am not even sure if in some conditions the breaker would even see the fault. If something fell across the cell, the cell would see the short, but the rest of the system would only see a 6 volt drop.

And you're right - that 80 kA is not going to be there very long at all since the cell voltage will be plummetting

I am a bit confused. You stated that the batteries are connected in series. The available fault current from the battery bank connected in series is the anticipated fault current from one battery. Contact the manufacturer, they should be able to tell you what the maximum available fault current from the battery will be. 80 kA sounds extremely high.

In summary, if I have a battery bank of 12 batteries and each battery is capable of providing 5, 000 amperes, then when connected in series, the battery bank will have only 5,000 amperes of fault current. If I had two of the (12) battery banks connected in parallel, then there would be 10,000 amperes available.

Hope this helps!

The batteries are connected in series. I got the internal resistance from the manufacturer. I got the Bolted Fault Current from Ohms Law. I = E /R - which equals a little over 80kA

This still sounds way too high to me. I suggest contacting the battery manufacturer and verify the available short circuit current.

The question really wasn't about the battery, but about the breaker interrupt time, but as I said in an earlier post I'm not even sure in some cases the breaker would even see the bolted fault. I've seen the battery and the Manufacturer's specifications

I agree with the other posts as far as the idea that you can't possibly have a huge bolted fault current with a huge number of cells in SERIES. The bolted fault current would be the same as a single cell. So something isn't right here.

Large arrays of power devices of ANY KIND frequently have intermediate fuses specifically due to large short circuit currents and the inability to deal with them. Off the top of my head that includes large inverters, large solar arrays, cap banks, and batteries. Due to cost and space more often than not these are always semiconductor fuses and are strictly designed to interrupt bolted faults and similar high current faults...the downstream devices handle overloads.

As an example of where this was violated, Siemens built and commissioned a new system called UDD for 6 large electric powered excavators in coal mines in Australia. Normally you'd do one at a time but because it was a CSIRO government sponsored project, they had to all be upgraded at one time. Since there were several IGBT's in parallel and because Siemens believed that their controls were so fast, they believed that they could detect and shut down the inverters before anything bad would happen and opted to skip using fuses as a cost saving measure. Due to some misunderstandings about the application, the inverters were undersized by approximately a factor of 2. About 18 months later (Arrhenius was right), the IGBT's in the drives started to fail. But they didn't just emit a small amount of noxious smoke...they actually exploded and shot shrapnel through the drive cabinets.

So if 80 kA is indeed correct, consider this sobering example of why intermediate fuses are not just a good idea...they are an absolute must.

And on a final note, there are a few AC circuit breakers out there that hit the 35-65 kA range. Above this you can still find a few 100 kA rated AC switchgear circuit breakers. But keep this in mind...most of them are not true "100 kA" circuit breakers. They are actually 35-65 kA circuit breakers but they have backup fuses and the fuses are rated to 100 kA. As long as the fault is below the range that the circuit breaker operates in, everything works as expected. But above the circuit breaker's interrupting rating, the fuse does the heavy lifting.

This is in the much easier AC interrupting category. You're trying to do this with DC...you need fuses almost no matter what. Even a TCL (triggered current limiter) with the exception of the very expensive superconducting kind are actually essentially glorified fuses.

Using the internal resistance that was measured under load conditions to extrapolate to fault current level assumes that resistance remains constant over this range. I is equal E/R, but R may not necessarily be independent and constant. I'd look at the manufacturer's fault current data.

Stevenal - I did not get the resistance value by measuring it. It was provided by the manufacturer (<.08 milliOhms for the battery)

Paul,
I share your concern regarding the 80kA bolted fault current. These are large batteries (Each one weighs over 100 pounds). I expressed my concern regarding the lack of fusing.

This particular system is just batteries in series, nothing in parallel. My biggest concern is a single battery shorting out

I am aware of Grid Tie Inverter applications. As a matter of fact, I wrote the first manuals for our solar grid tie inverters, and on those inverters each PV input is fused and paralleled in the Inverter enclosure. And we have done Battery powered Grid Tie inverters as well, where individually fused parallel strings of batteries are combined in the Grid Tie Inverter.

I believe what Stevenal was referring to was to obtain the fault current from the manufacturer not the resistance. The fault current for some manufacturers is found on the data sheet for the battery.

I would think for the resistance you would also have to take the intercell connectors into consideration.

I have the manufacturer data sheet

100 lb. jars is not that unusual.

Look at the batteries in any emergency services truck such as fire trucks. Quite often these use a pretty standard nominal 12 VDC battery called an "8D" that weighs about 150 lbs. But that's nothing compared to substation and power plant batteries. It's not unusual to see jars weighing in at 100-500 lbs. and having strings of 30-120 jars in a single string, often with two parallel strings for redundancy reasons.

The only trick with these things is that you have to plan on having an engine jack or some other kind of hoist system and using a spreader bar with hooks to get the batteries on and off the rack from the shipping pallets, and a fork truck to run the pallets around. That's the reason the NECA posts 3.5 hours of labor expected per jar.

Then you should have the short circuit current from the manufacturer.