That's correct. My 280 ah cells have an internal resistance of 0.13 milliohms per cell. If you have 16 similar cells in series at .13 milliohms each, that would be about .002 ohms for the 16 cell bank. 48v/0.002 ohms = 24,000 amps at the battery terminals if you have perfect "no resistance" bus bar contact with the cell terminals. In reality every contact has some resistance so an estimate of .003 ohms for the bank is probably reasonable. That would reduce max fault current to about 16,000 amps. The point is, it is very easy to generate more than 6000 amps of fault current near the battery terminals with 280 ah cells and your main fault interrupting device needs to be rated to handle it. Also, the normal load rating of the device should be about 25% higher than your expected max normal load or charging current. If you plan to have a max load of 200 amps, a 250 amp fuse with an interrupting capability of 20,000 amps would be sufficient to protect your bank. NOTE: if you parallel the 48 v banks, each bank should have its own Class T fuse. A paralleled battery bank would double the fault current available at the battery terminals because your internal resistance is cut in half. A 48v, 560 ah battery could easily generate over 32,000 amps for a close in fault, if you assume a 0.0015 ohm bank resistance. Don't skimp on your main battery bank protection.Class T fuse at battery terminal sounds like good advice.
In the interest of learning am I correct to push 14,000A at 48V would require a total circuit resistance of under .00343 ohms? This would need to include the internal resistance of 16 cells in series, busbars, connections, cables and the wrench, screwdriver or whatever causing the short.