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jerry_nj

Dehumidification heat gain due to condensation

jerry_nj
13 years ago

I have read, for reference, that the latent heat of condensation for a gallon of water (condensed by cooling coils, e.g., an A/C or a dehumidifier) is 1065 BTU/Pound.

So, it seems one could measure the rate of condensation in a dehumidifier and thus compute the heat gain due to the condensation of that moisture.

I have a 70 pint per day dehumidifier that removes about 2 pints per hour at an ambient of about 70 degrees and 60% RH. If I use the number (rounded to 1 KBUTU/Pound) quoted I can computer:

HeatOfCondensation = 32 oz/Hr x 0.0625 pounds/oz x 1,000 KBTU/pound, or ~ 2KBTU/Hr. going for watt-hours, 2K/3.41, I get about 600 watts.

From this I conclude the dehumidifier is adding the heating of 600 watts plus the power consumed by its compressor and fan, about another 600 watts. Or, dehumidification is adding the equivalent heat of approximately a 1.2 KW heater or 4K BTUs per Hour.

I am not trying to get high accuracy, but ask if the exercise about approximately correct? The only thing I think I explicitly ignored is that some of the energy into the dehumidifier that is converted to moving air, via the fan. I'd guess that is less than a small fraction of the 80 watts or so the fan is consuming.

Thinking in another way, in weather where one needs to both heat and dehumidify the dehumidifier may be more efficient as a heater than a pure resistive heater.

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