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LiFePo4 or NMC discharge heatup calculation?

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I would like to determine the heat generated by a battery under load.
You can calculate heat generated over time, if you know the internal resistance.
Internal resistance is often specified (in a range). However that doesn't tell the whole story.
Because I am sure the internal resistance in discharge varies over time, because of different power output (higher and lower C rating)
as well as temperature.
What I would like to know is how I determine that. And that in a theoretical way, because I don't have the batteries to do any testing.
This is also because I still need to make the choice which batteries I use.

So for example, we take a typical LiFePo4 cell of around 300Ah (could be EVE or CATL). It is at ambient temperature of 10 degrees Celcius.
Then put a 2C load on it, so 600A. Lets assume the internal resistance is 0.25mOhm because that is often listed.
P = I^2 * R so 600A^2 * 0.00025Ohm = 90W dissipation in the cell right?

Lets say the cell weighs 5.5 kilogram.
And specific heat of LiFePo4 is listed at 1130 J/kg.K (but this number varies over temperature right?)

For 1 minute load of 600A means 90W for 60 seconds is 1.5 Watthour = 5400 Joules

Then heat capacity of LiFePo4 at 1130 J/kg.K
5.5 kg * 1130 J/kg.K means that it needs 6215 Joules for the cell to heat up 1 degree. Right?
So at 60 seconds it will have almost raised 1 degree, actually 0.87 degrees to be exact.

Does the above sound about right or did I make big errors?
I expect the internal resistance to change based on temperature and amount of current drawn.
The same goes for the specific heat, but it might not have a huge impact on the calculations.
And it seems that internal resistance will also become larger as the state of charge drops.

Of course, it leaves out any heat up of busbars and other connections, or inverter/motor etc. It is just about the batteries.

Background: I am checking if I should use LiFePo4, which I could keep at atleast 10 degrees Celcius (which is better when you draw some current)
in an insulated box. Of course, under load (2C will be maximum motor power draw) it will heat up, and due to the isolation not so easy to lose that heat.
However, if the rise in heat is acceptable, than I think it would be a great setup. Skipping a battery cooling system for simplicity.
But in general I am trying to understand the theory, to later verify it in practical test.

Edit: For some NMC EV cells you can find a bit more info, like discharge internal resistance for several conditions.
See the example below of a Samsung SDI 94Ah cell (as used in BMW i3).
At 94Ah you need three in parallel to be comparable in energy density (ok NMC has higher voltage but just for simplicity)
So internal resistance of 0.7mOhm becomes 0.23mOhm and thus very comparable to LFP heat up profiles?
Only specific heat is a bit less per C/kg.K so they heat up slightly faster?
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· Registered
1,950 Posts
This is an interesting topic.

i look at the cell as a chemical reaction transfer of ions and electrons thru the electrolyte; the concept of "internal resistance" is not a real resistor, but a calculated equivalent value of an impedance that might characterize the chemical reaction at the time, temperature and conditions of the measurement.

i consider "IR" like the reactance of a capacitor, the impedance related to the frequency of the applied signal.

i think IR would be a non-linear value subject to temperature and SOC such as you have shown in the chart.

The heating of a cell would be related to parameters of the chemical reaction, e.g. rate, temperature, Gibb's free energy, etc.
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