At 01:22 PM 4/25/2010, Jake de Villiers wrote:
>When approaching the long (15 - 30 minute) second and third gear
>grades through Eastern Washington and Idaho you'd turn on the fan
>before the rad got too hot.

Ok, I see the idea.  But do the numbers make sense?

Lessee... at 75 horsepower output and 1/3 efficiency the engine's
going to be putting out something like 200 kw of waste heat, or 3.4
kWh per minute.

18 quarts of 50/50 coolant weighs around 36 lb and has a specific
heat capacity of roughly 0.85 BTU per lb per deg F at working
temperature.  So changing the entire coolant by 20F would get you
0.85 x 20 x 36 = ~600 BTU, about what a sedentary human throws off in an hour.

600 BTU = ~0.18 kWh.  3.4 kWh x 0.18 x 60 = ~32

So a twenty-degree change in the entire coolant would absorb the
waste heat of the engine for about thirty two seconds.  I think this
is conservative since the coolant inside the engine won't be
participating in this, and because 75 hp may be conservative for
climbing a steep hill.

Somebody check my numbers.  If this number is anywhere near correct I
don't think it's worth the hassle.

Yours,
David