Dehumidification calculations

Question

I'm looking at options for DIY dehumidification and am hoping to get a sanity check on some calculations, and verification that I'm using the right factors in the right way.

As a reference point, I looked at a popular dehumidifier to see if my calculations made sense (which is why I question the results).

My calculations:

• Under DOE standards, dehumidifier capacity ratings are based on an environment at 65°F and 60% RH. To remove moisture, they need to cool the air to below the dew point (in this case 50°F), at which moisture in excess of saturation precipitates out.

• At 65°F and 60% RH, there is 0.0082 lb of water per pound of dry air according to Water Vapor in Air, so 1 pound of water is contained in 122 pounds of dry air.

• Given the range of temperature and humidity that the dehumidifier I'm using for reference can handle (and the associated dew points), I'm assuming it reduces the temperature of the air by 25°F.

• At 40°F, saturated air holds 0.0052 lb of water per lb of air, so 36.43% of the original water is removed. To extract 1 lb of water, you need to start with 335 lb of air (containing 2.74 lb of water).

• The specific heat of dry air at 65°F is 0.24 if I'm correctly reading Specific Heat at Constant Pressure and Varying Temperature.

• As indicated in the comments, I need to account for heat of condensation, which is 23.86 BTU/lb of water condensed.

• So to cool enough air to extract 1 pint of water (1.04125 lb) should calculate as follows:

  Temp reduction per lb:
  335 lb air * 0.24 + 2.74 lb water * 1 = 83.11 BTU/°F * 25°F = 2078 BTU
  Heat of condensation per lb:
  23.86 BTU
  Total per pint:
  2,102 BTU/lb * 1.04125 lb/pint = 2,188 BTU

• The reference dehumidifier is rated at 50 pints/day, or 2.083 pints/hr, which would require removing 4,559 BTU/hr. If the process was 100% efficient, this would convert to:
  4,559 BTU/hr * 0.29 = 1,322 watts.

• The nameplate data on the dehumidifier says it uses 897 watts. Arbitrarily allowing 72 watts for the fan and electronics (ballpark), leaves 825 watts for refrigeration.

Mass-market compressor-based refrigeration has a lot of losses, so I assume my estimated power requirement is at least double the actual, and obviously cannot be correct.

Am I doing the BTU calculations correctly, and do those properly represent the requirements for dehumidifying air?

Answer 1

you're on the right track (I didn't check your numbers but the method is sound), except:

1. your energy. Expect at least a 3 to 1 gain of a refrigeration system - you can move more energy than what you put in. I learned this as a coefficient of performance.

2. temps are off. You won't be able to cool air to 40 degrees - that would require the coils to be too cold and they would start to freeze. Expect 50 degree outlet temps. Most HVAC systems expect a room output near 75°F, 50% relative humidity because that generally matches the moisture content (usu expressed in grains in the US) of the conditioned air. Look up psychrometric tables and learn how they work - it will help you.

3. a strict dehumidifier puts all its energy into the room, so you then have to remove that heat.