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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?

fixer1234
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  • You've forgotten about the energy required to condense the water (heat of condensation). – Drew Aug 19 '21 at 03:20
  • @Drew, I've been away from this stuff forever. Atmospheric water is below the boiling point. Doesn't it exist as tiny particles of water suspended in the air rather than as a gas? If that's the case, it seems like heat of condensation wouldn't apply. – fixer1234 Aug 19 '21 at 03:42
  • @fixer1234 even below boiling point there is a part of the water that becomes vapor. The air at a temperature can hold a specific amount of vapor (see RH). – NMech Aug 19 '21 at 04:06
  • Added heat of condensation to the calculation (which made the result even farther from actual). There's obviously a fundamentally wrong assumption in how I'm trying to calculate this. – fixer1234 Aug 19 '21 at 05:16

1 Answers1

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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.

J. Ari
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Tiger Guy
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  • Thanks for weighing in. I tried plugging 50°F into the calculation. Air saturation at that temperature removes only 6% of the water, so the process would need to start with over 16x the air. The result was 4,700W, which is way farther from actual. I may be basically on the right track, but there has to be a fundamental error in the method. – fixer1234 Aug 19 '21 at 17:11
  • @fixer1234, how dry do you need the air to end up? – Tiger Guy Aug 19 '21 at 17:24
  • The target is 50% RH or less (starting with 60%-65% at 65°F to 75°F in my application). BTW, good point about moving energy. My calculations would put energy moved at 8x energy used, which seems unreasonable. I'm actually looking at making something based on a "swamp cooler" (circulating ice water), so energy use will be different, but the water extraction will be the same principle. I'm trying to validate the calculations by comparing to a dehumidifier to see if they're in the ballpark, but they aren't. – fixer1234 Aug 19 '21 at 18:14
  • Just a quick update based on more calculations. Dehumidifier specs must be based on 65°F/60%RH, so 50°F output would put it right at the dew point, with very little water removal. It looks like 278 lb of air contains the water that must be removed per hr for 50 pts/day. The fan on the dehumidifiers can move ~900 lb of air per hour. So the unit must remove ~31% of the water in the air passing through it. That requires a temperature drop of about 25°F (40°F output). The units are supposed to operate down to about 41°F, and often use a defrost cycle, so the coil must be below 32°F. – fixer1234 Aug 20 '21 at 11:00
  • @fixer1234 , 50% rh at what final temperature? That is the moisture content your exiting air needs to be at. – Tiger Guy Aug 20 '21 at 13:26
  • The humidity range is for mold control, so temp isn't critical. With a normal dehumidifier, the heat it adds to the room is small relative to the space. Raising the temperature helps reduce the RH. In my case, the heat will go into melting ice and raising the temp of insulated water, which will be colder than the room. The outlet air will cool the room, offsetting some RH reduction. If the effect is too much, I can add heat to the room. So I was looking only at removing water. But, it looks like just putting a heater there and raising the room temp 5°F would reduce RH by 10 points. – fixer1234 Aug 20 '21 at 19:38
  • @fixer1234, unless you are discharging the condenser heat outside, you will get zero cooling effect from a dehumidifier. The is similar to the classic question about trying to cool a room by leaving a refrigerator door open. Heat gain for the room will be equal to the power consumed. – Tiger Guy Aug 20 '21 at 22:41
  • Right, a normal dehumidifier will add heat. What I'm investigating extracts heat from the air and uses it to melt ice and warm ice water. The heat is eventually removed from the space by dumping the water containing the heat (so discharged outside). The only energy added is a few watts for a tiny circulation pump. The cooled outlet air will cool the room air it mixes with. If you just put a big block of ice in a room and collect/remove the water that melts, you cool the room without introducing additional energy there. – fixer1234 Aug 20 '21 at 23:40