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Old 10-17-2012
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Re: Optimal propulsion system

Ok I wanted to run some numbers and see how things looked in more detail. Hopefully I don't make a mistake.

Diesel fuel is listed at 45MJ/kilogram which is 20.45MJ/lb or 36MJ/liter.

A Group 27 from Deka weighs 53 lbs and has 90Ah. This is 90Ah*12V=1080Wh.
A joule is a Watt-Second so 1080*3600 (seconds per hour) is 3.888MJ.
So per pound this is 3.888MJ/53lbs = 73kJ /lb
or by volume it's 3.888MJ/13 liters = 0.3MJ/l

This puts diesel at 280X as dense as the lead acid by weight or 120X as dense by volume. Or another way, one Group 27 is 0.0285306 gallons..

So to compare this to the example from earlier 2200lbs gets you 42 batteries which nets about 1.2 gallons...

As pointed out earler this gets better (by about 75%) when factoring in conversion efficiency (electric is more efficient than diesel) but then worse (by a factor of about 2) when you figure usable capacity of a bank.

Man that is pretty bleak for electric. Lithium is several times more dense than lead acid, at least about 5x according to some quick numbers but that's still not too good.

My only last thought is that the 45% number for the desel engine seems a bit high, perhaps that number is a train engine or a power plant operating at optimum RPM? What's a small boat engine get when running at 1/2 or 2/3 throttle?
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