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We are going to assume the battery pack has a linear charge / discharge curve.

There are 4320 cells in the battery. The battery consists of 96 blocks wired in series, with each block comprised of 45 cells in parallel.

I'm not concerned with the cells, just the 96 blocks.

The battery has a total energy of 73.5 kWh.
100% State of charge per cell is 4.15v.
0% State of charge per cell is 3.00v

Total battery voltage at 100% is 398.4v. At 0% it is 288v.

Each cell is 4.8 Ah.

I want to know the rate of change in the battery voltage with a 1kW load.

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    \$\begingroup\$ The only way to get the information you want is to measure it or get data from someone else who has already measured it. However, you only have to measure one cell. \$\endgroup\$
    – user57037
    Commented Feb 20, 2022 at 18:22

2 Answers 2

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The question is not that clear to me however based on assumptions of linear voltage drop, stored energy, and 1kW load then the discharge time can be determined in seconds and the linear voltage drop per second would be the equation of a line:

73.5 kWh x 3600 s/h = 264,600 kWs

264,600{kWs}/1{kW} = 264,600 s

Vdrop = (4.15 - 3.00)/264,600 V/s

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  • \$\begingroup\$ linear voltage drop is an extremely poor assumption for two reasons: 1) at constant current discharge, the drop is not linear, it is sort of "S" shaped. 2) because the OP mentioned constant power discharge as opposed to constant current discharge The "S" shape will be exacerbated. \$\endgroup\$
    – user57037
    Commented Feb 20, 2022 at 18:24
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We are going to assume the battery pack has a linear charge / discharge curve.

Unrealistic assumption and therefore not worth considering.

Instead, let's assume reality: a Li-ion battery pack has a OCV vs SoC curve that looks somewhat like this (assuming LCO cells):

LCO cell discharge curves

As the discharge current is only 0.013 C, we can use the OCV (Open Circuit Voltage) instead of the terminal voltage.

Note how the rate of voltage drop as the cell is being discharged varies radically from 0.05 V/% to 0.225 V/% as the SoC varies from 0 to 100 %, with an average of 0.01 V/%.

Therefore, the answer to your question is that, on the average, the total battery voltage (350 V nominal) will drop by 0.9 V for every 1% drop in SOC, but will range widely, from 0.45 to 21 V for every 1% drop in SOC.

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    \$\begingroup\$ A note for the OP: This graph shows open-circuit voltage. While being discharged, the open circuit voltage is not known. Even after the load is removed, the cell will take some time to return to its open-circuit voltage. The time constant of the recovery is measured in minutes, not seconds. So the loaded voltage can't really be used for high precision capacity estimates, nor for precision cutoff at end of discharge unless a lot more work is done. \$\endgroup\$
    – user57037
    Commented Feb 20, 2022 at 18:30
  • \$\begingroup\$ On the contrary, because in this case the current is so low (0.013 C), the OCV can be used. But, you're right: in most cases, batteries are discharged faster and therefore the terminal voltage is lower than the OCV. \$\endgroup\$
    – Davide Andrea
    Commented Feb 20, 2022 at 19:21
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    \$\begingroup\$ Oh, yeah, good point. I guess my comment to the OP only applies if the discharge rate is somewhat higher. But when discharging at 0.013 C, the OP should probably terminate at 3.5 V per cell or something like that. Taking OCV down to 3.0 V seems kind of harsh. \$\endgroup\$
    – user57037
    Commented Feb 20, 2022 at 19:58
  • \$\begingroup\$ The purpose of obtaining an answer to my question was to allow me to populate my battery "simulator" with something that approximates the charge/discharge rate, as seen by the naked eyeball. I'll add further detail if anyone is interested, but the actual resultant number isn't critical, nor is it necessary to be entirely accurate. I understand that the notion of a linear charge/discharge "curve" is preposterous, but I didn't want to have to calculate/lookup the rate of charge/discharge for this particular battery. In the future the "simulator" may need this, - if so, I'll be back. \$\endgroup\$
    – Dave Robinson
    Commented Feb 22, 2022 at 2:18

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