In alt.energy.homepower BobG said:
In any case with your example how is a 100HP engine at 40% power
different
than a 40HP engine at peak power the whole time? I don't understand
how
a engine capable of 100HP is not capable of 100mpg as long as it isn't
used very often for its full 100HP output.
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ok, lets say a more aerodynamic car can go 60 all day with 30 HP.
(22.38KW). If the gas engine is 40% efficient, you need 55.95KW from
the fuel. If you do this for an hour and get 60 miles down the rd, you
have used .9325KWhr per mi. Assuming gas is 6.1 lb/gal and 15000
BTU/lb, you have used 191K BTU->12.7 lb->2 gal-> 30 mi/gal To get 100
mi/gal, you need to make something 3 times bigger or smaller, but it
wont be any of those constants that are in the physics books.
You still haven't answered my original question. My car has a ~130HP engine.
I can get ~39mpg at a steady 60mph. Obviously it takes a whole lot less
than 130HP to drive it at that speed. (Probably low 20s HP.) If I were to
drop a 30HP or a 230HP engine of the same design in the same car, I'd expect
the gas mileage wouldn't change significantly at a steady cruse. The mileage
has little to do with the maximum output of the engine, and is highly related
to the amount of power the engine needs to output to maintain 60mph (or speed
X). I'd would guess this is highly related to aerodynamics, rolling resistance,
and mechanical resistance.
I'm guessing the reason that a diesel S-10 with a ~100HP diesel (slightly
higher BTU/lb than gas) can't get 100mpg has little to do with the engine as
the first reply in this thread indicated. It has mostly to do with the
crappy aerodynamics of an S-10. No?