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post #12 of Old 05-05-2012
JimsCAL
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Re: water flow GPM per HP of an engine?

Quote:
Originally Posted by SlowButSteady View Post
Here's a very rough estimate:

Twenty hp is about 15 kW. If we assume about 30% efficiency, the engine would need about a 50 kW input. The difference, 35 kW, ends up as heat (one way or another). Converting 35 kWHr to kcal we get about 30,000 kcal. It takes 1 kcal to raise the temperature of a liter of water 1˚C. If we set the intake water to 20˚C and the output water to 40˚C, each liter of water will conduct 20 kcal away from the system. So, getting rid of 30,000 kcal/hr will require 1500 l/hr = 25 l/min = 6.6 gal/min. Of course, the efficiency of the engine may be somewhat higher or lower; the water temps may be higher or lower; and we haven't accounted for the heat transfer from the engine to the air surrounding it (remember, your engine compartment heats up quite a bit, particularly with the engine running at or near full load). But that's my rough estimate.
More importantly, you forgot the rather significant heat in the exhaust. Basically the heat rejected is roughly split equally between the cooling water and the exhaust. So your cooling water estimate is about double what's needed.
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