Never really thought about it, but my guess would be the sum of:
1. NumberOfCells * IR(20mOhm?) * CurrentPerCell^2, so using floydR's values: 112 Watts of heat
2. All wires and fuses that will heat up under high loads, using same formula ( Power = Resistance * Current^2 )
Only current matters, not voltage. Thus you'll want to maximize voltage and minimize current to reduce heat losses.
Ah I see so using that formula for a 13s15p pack, I get the following:
(assuming the max discharge current is 20A; some sites say 20A but others say 15A, but this is all theoretical so let's assume 20A for now)
P = 195*0.02*20A^2 = 1560 watts of heat?
when you say current per cell, does this mean how much current each cell will discharge for the 200A load, so:
200A load / # of parallel cells = 200 / 15 = 13.33A, then
P = 195*0.02*13.33^2 = 693 watts of heat?