I took the thermodynamic inefficiency of fuel oxidation into account. The main thermodynamic inefficiency for which I didn't account that of the hydrolysis, because my CRC handbook is at work.
Of course, it is true that if you add enough batteries, you can achieve the 4 m.p.g. increase. No matter how you slice it though, you need something like 3.3 million joules to go the extra four miles. You're getting it from a battery or from batteries.
Let's say you use 200 amp hour/12 volt batteries. These can deliver 2400 watts for 3600 seconds, or 6,240,000 joules. However this is, at best, a 20 hour capacity, so it will not be able to deliver this much energy at a rate which discharges it in an hour. For your 90 minute commute, assuming Peukert's constant for these batteries is 1.2, such a battery can deliver about 80 amps at 12 volts. This is 960 watts for 5400 seconds or 5,184,000 joules. The battery makes HHO (sigh) which is then burned in cylinders. If 25% efficiency is achieved in the overall process (battery to generator to HHO to H2O), about 1,300,000 joules go to move the car. So something like 3 of these batteries might do the trick.
From here on, it's purely estimation. If your 90 minute commute is at 35 m.p.h. on average, you'd take 2.57 hours. You'd be using electrical energy at the rate of 80*12*3 watts, or 2880 watts. 2.88 kilowatts for 1.5 hours is 4.32 kilowatt hours. Depending on where and how you purchase electricity, that could be something like $0.50 worth of electricity. You'd be saving something like a third of a gallon of gas, worth maybe $0.80.
Just as a reality check, 2.88 kilowatts is about 3.86 horsepower that the batteries would be saving your gasoline from having to deliver. That's certainly in the ballpark for a 16.7% savings.
Last edited by PA32R; 11-09-2008 at 10:08 PM..
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