I am designing an enclosed container with 10 kWh 50 V battery in it which requires me to actively remove the heat. The battery feeds an 8.8 kVA inverter. Assume I have a load of 8 kW, which would draw 160 amp at 50 V.

One of the sources https://batteryuniversity.com/learn/archive/how_does_internal_resistance_affect_performance referred to small Li-ion and gave internal resistance of about 320 mΩ to 340 mΩ. Other sources were more academic and incomprehensible. Most addressed charging - nothing on large batteries.

If this is the case the internal heat generated would be I2 × R = 1602 × 320/1000 = 8192 W, an impossible result.

Either the internal resistance is wrong or I am using the wrong logic. How much heat would the battery create during discharge?

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    \$\begingroup\$ A battery would normally be rated in Ah or Wh but never VAh as it is not AC so there is no reactive component (which is where VA is used). Can you edit your question to fix this and quote the discharge in current (A) or power (W). \$\endgroup\$ – Transistor Apr 10 at 15:14
  • \$\begingroup\$ I don't understand what you're saying here. Your units don't work out at all. \$\endgroup\$ – Hearth Apr 10 at 15:15
  • \$\begingroup\$ Can you quote a source for battery resistance. I'd that for a cell or your actual battery and what cells are used in what configuration,? \$\endgroup\$ – Russell McMahon Apr 11 at 11:19
  • \$\begingroup\$ Maximum power transfer theorem shows that max power is Vbat/2 into 330 milliohm or about 1800 Watt. Your resistance figure MUST be wrong. \$\endgroup\$ – Russell McMahon Apr 11 at 11:22
  • \$\begingroup\$ There must be a datasheet for the batteries you are considering! \$\endgroup\$ – Transistor Apr 11 at 17:47

Either your battery is 10 kWh or 10 kAh but not normally referred to as 10 kVAh (a term we might use in AC circuits due to power-factor).

If your battery's internal resistance is 320 mΩ then the maximum current you could draw into a dead short (not recommended) would be \$ I = \frac V R = \frac {50}{0.33} = 150 \ \text A \$ but you would have zero volts at the terminals so no power.

I suggest that your internal resistance calculation is wrong or your battery can only supply about 15 A with a 10% voltage drop.

  • \$\begingroup\$ Thanks. Rephrased question. You are correct in that batteries are rated in kWh. With DC, PF = 1 then kVah is also correct but unconventional.. \$\endgroup\$ – IanO Apr 11 at 11:08
  • \$\begingroup\$ Edit again to show how you figured 320 mΩ. That's got to be wrong. \$\endgroup\$ – Transistor Apr 11 at 11:20

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