Evaporation rates for waste electropolishing solutions
Hello All: I am looking for information on evaporation rates for a waste electropolishing solution. 1. Would 2 to 3 kWh be a typical amount of energy needed to evaporate a gallon of waste electropolishing solution from an evaporator? 2. If only the blower (heater element not operational) was running on the evaporator (Unfortunately I do not have the model type/number of the evaporator handy, but will post it later) how long would it take to evaporate one gallon of water? Any help would be greatly appreciated Thanks in advance.Mike Ellenbecker
Wisconsin DNR - Sturtevant, Wisconsin
As a starting point, the latent heat of evaporation of water is about 1050 BTUs per pound. So, to evaporate a gallon of water, it should take 8760 BTUs, which equals 2.57 kW-hr. Your numbers seem to be right on.
Without the heater on, evaporation would never take place in a closed system because it cannot happen until those 8760 BTUs are supplied. So in this case the requisite BTUs must be supplied by the atmosphere, conductance from the floor, etc., which is quite difficult to model, and obviously impossible to calculate without a lot of data about surface area, temperature, humidity, etc. But the short answer is it would take nearly forever.
If your interest is in concentrating a waste solution for disposal, it sounds like a mess in the first place, and is totally impractical without heat input. But if you are trying to evaporate a gallon of water from an operating electropolishing tank to make headroom available for the return of rinse water, that's a different story because you'll find that the rectifiers (and possibly heating coils) in the electropolishing bath provide plenty of input heat. The most important piece of data in that case is the operating temperature of the electropolishing bath.
Ted Mooney, P.E. RET
Pine Beach, New Jersey
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