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On a battery manufacturer specification (https://www.scib.jp/en/product/cell.htm, "High power 10Ah cell") it says: Output power 1800W* (SOC:50%, 10s, 25°C), *This value is calculated from the internal resistance.

How is this power rating calculation from the battery's internal resistance carried out exactly?

How does the above calculated output power relate to the heat generation rate (W) of the battery that is also calculated from the internal resistance?

And how does this theoretical output power (typically) relate to the actual battery output power available to the user in practice?

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  • \$\begingroup\$ Does the product even exist? "Do you sell SCiB™ cells and modules to individual customers? answer No, we cannot sell SCiB™ cells or modules to individual customers. Please understand they are intended only for corporate customers." \$\endgroup\$ – Bruce Abbott Nov 28 '19 at 9:28
  • \$\begingroup\$ Excellent question, given the vagueness of the specification. But it does - my university has purchased 12 modules with these cells in from a local vendor. \$\endgroup\$ – AzizG Nov 28 '19 at 10:22
  • \$\begingroup\$ (and we have received the modules) \$\endgroup\$ – AzizG Nov 28 '19 at 10:32
  • \$\begingroup\$ Disappointing that they don't supply internal resistance figures. However since you have the batteries you can measure it. For 1800W output it needs to be 0.8 milliohms or less (assuming open circuit voltage is 2.4V). \$\endgroup\$ – Bruce Abbott Nov 28 '19 at 20:10
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How is this power rating calculation from the battery's internal resistance carried out exactly?

My guess is that it is based on the maximum power transfer theorem, hence the maximum power you can deliver to a load depends on the internal resistance of the battery.

How does the above calculated output power relate to the heat generation rate (W) of the battery that is also calculated from the internal resistance?

If the above guess is correct, then the power delivered is the same as the power dissipated in the battery itself.

And how does this theoretical output power (typically) relate to the actual battery output power available to the user in practice?

At maximum power transfer, the output voltage of the battery cell would be half of the no-load voltage. In addition, the efficiency would be 50% and hence a lot of the energy is lost in internal dissipation. For a practical case, the allowable voltage drop and criteria for energy efficiency would limit the power delivery. I have no industry experience on battery systems, but to increase the energy efficiency to above 80% or 90%, the power must be decreased less than 64% or 36% of the stated maximum output power value (again using the same theorem). However, the internal resistance will vary with several operating parameters, so my numbers are for an ideal case.

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  • \$\begingroup\$ Would this imply that I could then work out an estimate of the battery's internal resistance backwards from P=U^2/R, using the specified output power (P = 1800 W) and the nominal battery voltage (U = 2.4V)? That would yield 3.2 mOhm, which is probably in the ballpark. \$\endgroup\$ – AzizG Nov 28 '19 at 10:31
  • \$\begingroup\$ @AzizG This source from 2009 [1] quotes 6 mOhm for a 10Ah Li-ion battery, so 3.2 mOhm seems like a probable value. \$\endgroup\$ – ocspro Nov 28 '19 at 12:02
  • \$\begingroup\$ Note that in that case the battery would also be dissipating 1800W, so have fun holding it at 25°C. It must be a peak power rating. \$\endgroup\$ – user253751 Nov 28 '19 at 12:25
  • \$\begingroup\$ Yeah, cooling system design for a battery with these cells is specifically our goal. We have (more or less) been able to identify the current demand profile from the battery in typical usage conditions; fortunately, it remains far below the peak. But in the absence of proper impedance tests, for the time being, the above internal resistance value would still yield our first guesstimate for heat generation. \$\endgroup\$ – AzizG Nov 28 '19 at 13:21

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