Martinm210

I think understand the time part, you can convert the equation to give you something in terms of flow rate, you would just want to figure out the volume per second and apply that to the 4.18 Joules factor.

This link supports that 4.18 factor:
Specific Heat
4.18 joules/grams*C
and if a joule is a watt* second, if you figured it all out in one second of time, that would be:
4.18 watts/grams*C

Something like

Watts = ((Flow in milliters/second ) * (Change in temperature in degrees Celcius))/4.18

Or for GPM

Watts = (63.09(Flow GPM) * (Change in Temperature in degrees Celcius))/4.18

OR

Watts = 15.0933 * (Flow in GPM) * (Change in temperautre in degrees Celcius)


But does that make sense?? if we had 2 gpm and a 2 C rise, that would give you 64 watts? Does this seem like it would be in the ballpark??

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