Some calculations related to battery temperature and cooling
An IC member reported a 92% cycle efficiency while charging and discharging an IMA battery. This applies below the threshold of overcharging, where charging efficiency drops to zero. This prompted me to do a simple calculation:
If you model the IMA battery as 22kg of nickel, it has a specific heat of 0.445kJ/(Kg*°C) = 10 kJ/°C. Adding 1kWh of charge at 96% efficency dissipates 0.04kWh, or 144kJ. If the battery is insulated (or has no fan and the charging happens quickly), its temperature rises 14°C. If the entire pack is uniform, no cells peak early, and the charge terminates at that time, you won't have a problem, and your fan never needs to kick on.
Suppose heat can be conducted very quickly throughout an entire stick of six cells. If one cell peaks early, and you're passing 350mA through it, it's dissipating 0.5W. The temperature of the stick will rise by 3.6°C per hour. Once all six cells in the stick peak, it's 22°C per hour. Obviously, this spells a serious problem if the fan is not running.
What this means: If the pack doesn't cool while you're charging it, you'll be able to start with a pack at >9°C on a -5°C day. It's not terribly impressive, but it's better than nothing.
Also: once the pack peaks, you will begin to do major damage to the cells after about four hours if you're not running a fan.
|
