Heat Transfer Problem with a Project...

Well, wouldn't the efficiency be indicated by the rate of evaporation? Could you get away with just weighing the water before and after. I think it takes something like 540 degrees C to evaporate a gram; not sure if that applies at sub boiling temperatures or not.

Well well, yes, that would be the mmm... what could we name it, water efficiency? But not the thermal efficiency. You see, in a heat exchanger, since there is no actual work done (W), its just the ratio in the temperatures of the fluid that at the beginning and in the end.

Something like that, I'm not really good in this subject...

Isn't roger on to something, though? At 100% efficiency the amount of heat removed from whatever it is you're trying to cool would be the same as the amount of heat required to make the amount of water lost evaporate. So a simple efficiency equation could be...

(heat lost)/(amount of heat required to evaporate water lost)



Or are we both missing something?


(Is this a closed system where the evaporated water gets recycled?)

Ok ok, for those heat lovers, heres a step by step resolve:

(Its an open system, since the water adds moisture to the air that flows out of the cooler, creating the fresh sensation )

1. Measure the pump current and calc W (A).
2. Measure the inlet air dry and wet bulb temps.
3. Measure the outlet air dry and wet bulb temps.
4. Use a psychrometric chart to determine enthalpy at each of the inlet and outlet.
5. Measure inflow air velocity. Calculate mass airflow.
6. Multiply the mass flow by the difference in enthalpies to get cooling effect, W.
7. Divide to get efficiency.

Any opinions?

Crikey.