Waste heat in battery pack - large pack vs small pack?

Question

Are larger battery packs more efficient than smaller battery packs due to "resistance in parallel" rule, assuming eveything else being the same?

As an example let's consider two 48V battery packs - 13S2P and 13S6P (let's assume that individual 18650 cell can handle 30Amp power draw). And electric bicycle that has a motor that consumes 2.88KW at cruise speed.

2.88KW at 48V would imply need to draw 60 Amps.

I found on Internet that Panasonic 18650 cells have 0.05 Ohms inner resistance. So, for 13 cells in series that would be 0.65 Ohms resistance.

With 13S2P battery pack the 60 Amps would be distributed over two parallel branches. Hence 30 Amps per branch. The waste loss due to inner battery pack resistance over both branches summed together would be P=2*(R*I^2)=2*(0.65*30^2)=1170 Watts.

With 13S6P battery pack the 60 Amps would be distributed over six parallel branches. Hence 10 Amps per branch. The waste loss due to inner resistance in battery pack over 6 branches summed together would be P=6*(R*I^2)=6*(0.65*10^2)=390 Watts.

Is there really 780W difference in battery pack waste heat? Does inner battery cell resistance stay constant over different current draws and temperatures?

Answer 1

Is there really 780W difference in battery pack waste heat?

Yes, your sums are correct. There is a large difference in internal heating between a 13S2P and a 13S6P, assuming the same load and the same cells.

Look at it in terms of just the battery internal resistance. The 2P pack has about 0.33 ohms, the 6P pack is 0.11 ohms.

At 60A, that gives you an \$I^2R\$ heating of 1188 Watts or 396 Watts.

If you reduce the internal resistance by whatever means (better cells, or more of them in parallel), the battery internal dissipation will fall.

Does inner battery cell resistance stay constant over different current draws and temperatures?

To some extent.

We often measure a voltage drop, and a current, take the ratio and call it resistance. This is only useful if that ratio stays fairly constant.

In a cell, there are several contributions to the excess voltage drop, at any given current. Amongst these are several reasonably constant terms like electrode resistance (increases slightly with temperature), less constant things like ion mobility in the electrolyte (changes with temperature, probably reduces), and things you've have no right to expect to be constant like polarisation potentials (the effect of a redistribution of the cell's active chemicals due to current flow, so it has a time history as well).

Whether any of these terms dominates depends on what chemistry and what quality of cell you are studying. The only way to measure the change in terminal voltage as the current changes (the definition of effective internal resistance) is to load the cell with a varying current, and measure the voltage.

When a cell manufacturer makes a claim about the maximum current the cell can deliver, that's an implicit claim about the maximum internal resistance you're likely to see during its operation. You would certainly expect it to deliver that current into a short circuit. You might hope it would deliver that current with only 25%, or maybe only 10%, terminal voltage drop due to internal resistance.

In all cases you need to read the data sheet, or do your own measurements.

Answer 2

Your question reduces to something like this: How does the efficiency compare between a cell operating at 1C versus one operating at 0.33C?

Any data sheet or description is going to tell you that at higher charge/discharge rates, the efficiency is lower.

As you are looking specifically at the losses due to heat (via the so-called internal resistance), consider that the cell operating at 1C supplies 3 times the current as the 0.33C cell. But power is I squared times R, so the 1C cell dissipates nine time the power.