Specifically if the cathode and anode are known materials how do you calculate the theoretical capacity and energy density of the full cell? For example if you have a Lithium Iron Phosphate cathode and graphite anode.
Grab a bunch of cells of that make, weigh them, find a typical number for AH per gram. For A123 I get 0.035 AH/Gram for their 20AH pouch cells, 0.033 for their cylinder cell. IMO, A123 is top of the line, so generic LiFePo might be a bit lower. So say 30mAh/g typical.
Compare that to a computed 'theoretical max' from these sources:
mAh charge capacity of LiFePo on Wikipedia of 170mAh/g
Check that Wiki number:
Weight of 1 Mole of LiFePO4: 158g
Coulombs in 1 Mole (one charge per Li):9.65E4
Coulombs in 1 mAh: 3.6
mAh per mole of charge: 9.65E4/3.6 = 2.68E4
mAh per gram of LiFePO4: 2.68E4/158 = 170 mAh/g. Ha! Spot on.
mAh charge capacity of graphite sheet 372 mAh/g
Convert the two numbers to grams per Ah:
LiFePO4: 5.9 g/Ah
Graphite: 2.7 g/Ah
add, invert, to get 116 mAh/g of graphite and LiFePO
That is too high, of course. How are you going to get the current out? With copper and aluminium sheets. They weigh maybe as much as the anode and cathode, so divide by two to get 58mAh/g. That's closer to the real world.
A little more reading: I -know- you can not get all the Li out of LiFePO. Mr Borong Wu et all say you can only get about 0.6. So that means 102mAh/g for LiFePO, not 170. With that correction, I get 80 mAh/g for the materials alone. Add in the weight of the anode and cathode as before, I get 40 mAh/g. Pretty darn close!
protected by Dave Tweed♦ Aug 23 '15 at 14:14
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