Originally Posted by powerhouse
Hmm... I don't understand how the 10ah battery pack with more internal resistance has the same voltage sag, thus power, as the 8ah battery pack ?
I wouldn't place too much faith into calculations of what the sag *should* be based on these Chinese cell manufacturer's datasheets. To paraphrase Mark Twain, there are lies, damn lies and battery datasheets.
From the PM you sent to me, you wish to construct an 8p112s pack (note the order of p and s - that helps to clarify that you are first paralleling the cells then putting them in series) and the claimed internal resistance - Ri or R[int] - is 6 milliohms per cell.
The Ri in milliohms of a group of 8 cells in parallel would be:
1/((1/6mΩ)*8) = 0.75mΩ
And 112 of those in series would be 112 * 0.75mΩ = 84mΩ
To calculate the voltage drop due to Ri you simply multiply the battery current and the total resistance of the pack while to calculate power *lost* to Ri you multiply by current again:
1000A * 84mΩ = 84V; 1000A * 84V = 84kW
2000A * 84mΩ = 168V; 2000A * 168V = 336kW
The maximum power occurs when the load impedance (or resistance) matches the source impedance (or Ri here) because at that point half of the voltage is lost to Ri and half to the load. In other words, when voltage sag from Ri is half the nominal pack voltage (assume 3.2V per LFP cell, so 358.4V total) then trying to draw more current results in less actual power being delivered. In this case we want to find the amount of current that results the pack dropping to half of 358.4V, or 179.2V:
179.2V / 84mΩ = 2133.3A
179.2V * 2133.3A = 382.3kW
If you need more convincing this is true, the voltage drop and total power at 3000A and 4000A:
3000A * 84mΩ = 252V; (358.4V - 252V) * 3000A = 319.2kW (63.1kW less)
4000A * 84mΩ = 336V; (358.4V - 336V) * 4000A = 89.6kW (292.7kW less)
But all of this is the *theoretical* performance of your battery pack. You won't actually know what the voltage sag of your pack is until you have built it and measured the actual difference in voltage at two non-zero currents (as I did previously in this thread).