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I have a battery pack consisting of 720 cells. I want to calculate the heat generated by it. The current of the pack is 345Ah and the pack voltage is 44.4Volts. Each cell has a voltage of 3.7V and current of 5.75Ah.

The pack provides power to a motor which in turn drives the wheels of an EV. I wanted to design the cooling system for the battery pack, so wanted to know the heat generated by the battery pack.

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  • \$\begingroup\$ Impossible to tell based on your data. Will need info on internal resistance of cells. \$\endgroup\$
    – Marcus Müller
    Commented Aug 22, 2018 at 7:18
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    \$\begingroup\$ Heat out of pack is a simple P=RI^2 equation. You know the current out of each cell, and you know (or should be able to find out) the internal resistance of each cell. So you know the power, which then just needs to be removed for the pack. \$\endgroup\$
    – Puffafish
    Commented Aug 22, 2018 at 7:34
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    \$\begingroup\$ Ah is not the unit of current but the unit of charge (current multiplied by time). For a battery this is usually called capacity. \$\endgroup\$
    – Arsenal
    Commented Aug 22, 2018 at 7:43

2 Answers 2

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The battery heat is generated in the internal resistance of each cell and all the connections (i.e. terminal welding spots, metal foils, wires, connectors, etc.). You'll need an estimation of these, in order to calculate the total battery power to be dissipated (P=R*I^2).

Considering your data to make an example, with a 1C discharge current (5.75A per cell) and estimating, let's say, a resistance of 50mOhm per cell, each cell is contrubuting 1.65W of dissipated power (Pcell=0.05*5.75*5.75), and the total dissipated power for the battery is Ptot=1.65*720=1190W.

As the real battery current, you can consider the mean discharge current during normal cruising on flat road, which is usually constant speed and quite low current. But take also into account the peak current the battery needs to deliver, for example during acceleration or regenerative braking or climbing. In particular, during the two latter cases, the current value can be high for a prolonged amout of time.

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So first of all there are two ways the battery can produce heat.

  • Due to Internal resistance (Ohmic Loss)
  • Due to chemical loss

Your battery configuration is 12S60P, which means 60 cells are combined in a parallel configuration and there are 12 such parallel packs connected in series to provide 44.4V and 345AH.

Now if the cell datasheet says the Internal resistance of the cell is R mΩ. then,

The internal resistance of battery pack = 0.2R mΩ

For ohmic loss, take the worst case. If your battery dies in let's say in T time(Hours), The load current is 345/T Amps, where T in hours.

So, Total Ohmic power loss will be,

Ohmic Loss = (345x345)/(TxT)x(0.2R/1000) Watts

Ignore Chemical losses as it is hard to calculate from the formula, you can check it by Power flow method.

So use a cooling system that can remove atleast the power we calculated as Ohmic Loss.

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