Join Date: Feb 2010
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Re: water flow GPM per HP of an engine?
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.
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