[mackerel]

I had a couple more thoughts.

If the machinery was outside the conditioned space, unlike a standard dehumidifier, you won't have the waste heat inside.

Also, from a simple thermodynamics point of view, you should be able to calculate how much water you're condensing out of the air.

Say you have 85° air with 100% humidity, and you hope to have 85° air with 50% humidity. You're removing 90 grains of water per pound of dry air. You're going from 180 grains total to 90 grains total. You're bringing the air down to 65°, where 100%RH means 90 grains of moisture as vapor, and the rest condneses, and then you're warming this air back to room temperature by cooling the air coming into the dehumidifier system.

A grain is .000143 pounds. A pound of air is about 13 cubic feet. Say your house is 2000 square feet, with 8' ceilings, and gets the recommended .35 air changes per hour natural ventilation. That's 5600 cubic feet. That's 430 pounds of air. That's .062 pounds of water to condense. The latent heat of evaporation is 970 BTU/lb for water, so we're talking about ~60 BTU per hour, which is virtually nothing.

EDIT: I found my math error: 5600 ft3 -> 430 #air -> 38800 grains h2o -> 5.54 #h2o -> 5400 BTU -> 1.6 kWh. Something like a 20 amp air conditioner, including equipment inefficiencies.

There will also be efficiency losses, and the air will be cooled in the process. If we can keep the cooling to 5°F, at .0182 BTU/(ft3*°F), that's another 500 BTU, and the room is down to 80°F, but up to about 57% relative humidity.

Somebody (STILL) check my math. If I want to be certain, I use a spreadsheet and a second person to check things. What I calculated is all air. If there are solid surface (through the insulation) heat gains, those will have to be dealt with separately. I also didn't account for the moisture you've added to the air from normal human activity, like breathing and doing dishes.