# How much heat is generated by a Li-ion phosphate battery during discharge

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?

• 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). – Transistor Apr 10 at 15:14
• I don't understand what you're saying here. Your units don't work out at all. – Hearth Apr 10 at 15:15
• 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,? – Russell McMahon Apr 11 at 11:19
• Maximum power transfer theorem shows that max power is Vbat/2 into 330 milliohm or about 1800 Watt. Your resistance figure MUST be wrong. – Russell McMahon Apr 11 at 11:22
• There must be a datasheet for the batteries you are considering! – Transistor Apr 11 at 17:47

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.