Originally Posted by Tesseract
There's one bad assumption in there - the *theoretical* (Carnot) efficiency of an internal combustion engine is 37% at 25C using ordinary materials. If you were to build the engine out of refractory metals like Inconel, Titanium, etc., then you *might* be able to push that a few more points higher.
The typical internal combustion engine in a car gets a best case efficiency of around 18%, or half the theoretical ideal.
Skipping over lots of math, the 33kWh of energy in a gallon of gas is equal to about 7.1kWh of battery pack (100% DoD). Divide by whatever fraction of DoD you want to use to get the needed real world battery pack capacity; e.g. - 8.9kWh if DoD is limited to 80%.
1 gal. of gas weighs 6.lbs (2.76kg)... and takes up 3785cc. 7-9kWh of battery pack weighs... umm... well, let's see. A Winston (neč Thundersky) 200Ah cell takes up 5115cc and weighs 7.3kg but only stores ~0.64kWh, so you'd need 11-14 of those to equal one gallon of gas.
All of this was obtained via Wikipedia with the exception of the Winston cell data, btw...
Ok, so to distill all that down:
- 8kWh of battery =~ (approx) 1 gallon of gas for propulsive purposes.
- That's approximately 12 LiIon batteries in 2011 (I'll be optimistic), each weighing about 7.3Kg.
- 12 batteries * 7.3Kg = 87.6Kg = 192lbs of batteries = 1 gallon of gas
- 192 lbs / 6lbs = 32 times the weight of a gallon of gas.
Is that math correct? Gawd, no wonder we're not all driving EVs...
By this formula, it would take 7,800 lbs of batteries to replace 42 gallons (252 lbs) of gas in my 1,500 lb airplane...