How can I build a dehumidifier?

  • #1
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I would like to build a dehumidifier. I don't even know where to start. I read that refridgerative coils are use in mechanical dehumidifiers. How should I go about doing this, or is it inviable?
 
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  • #2
http://home.howstuffworks.com/question76.htm

For a very simple solution, I might try using a thermoelectric cooler in conjunction with a pair of heatsinks. Blow air over the cold side of the TEC to remove humidity (water condenses on the cold heat sink, and can drip down into a collection tray) and then blow the air over the hot side of the TEC to heat it back up. While TEC's are ineffiecient, they are simple and cheap, compared to trying to build your own refrigeration system.

It would be useful to know what the dew point is of the air your dehumidifying; as this information will determine how cold the TEC has to get. You would be able to get the necessary information off of a psychometric chart.

Also, how much humidity do you want to remove?
 
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  • #3
How much dehumidification do you want, how much do you want to spend, and how much "building" are we talking about and why? A store-bought dehumidifier is basically the same device as a store-bought window air conditioner, just in a different package and with a tank.
 
  • #4
How much water could be extrated from the air... could a large enough dehumidifier be used for irrigation purposes?
 
  • #5
smackdammer said:
How much water could be extrated from the air... could a large enough dehumidifier be used for irrigation purposes?

It all depends on how much humidity is in the air to start with, and what you're trying to irrigate, and how big your dehumidifier is. On the coast it might be possible to get a few gallons per day with a relatively small unit (but probably not more easily or cheaply than desalination). Inland in arid environments it's doubtful there would be enough water in the air to do anything useful, even with a very large flow rate.
 
  • #6
An easy way to find properties of air and water is a psychrometric chart (http://en.wikipedia.org/wiki/Psychrometrics). Take for example this one:

PsychrometricChart-SeaLevel-SI.jpg


So if you know a few values of the air, such as the temperature and relative humidity, you can easily look up the amount of water in the air in terms of grams water per gram air. Knowing the specific mass of water in the air, and how much water you're wanting to produce, you can calculate what the flow rate of air through your dehumidifier would have to be. For a humid environment it could be within the realm of possibility; for an arid climate I doubt it will be possible without very large flow rates and a lot of power.

Hope this helps.
 
  • #7
Example: A typical 100% outside air AC unit in Pennsylvania will be sized for a peak of 92F at a 68F dewpoint and can cool the air to 55/55, for a capacity of 15.2 BTU/lb of air at an input power of 1 kW per ton of cooling (12,000 btu/hr). With that 15.2 BTU/lb of air, you remove 0.015 lb of water from the air or 11.8 lb of water per hour per ton of cooling.

From the first google link on the subject, I find that irrigating 1 acre of land requires on the order of 8,000 gallons of water per day. 8000 gallons is 67,000 lb per day or 2800 lb per hour. And that's 237 tons or 237 kW. At Pennsylvania residential costs, that would be about $850 per acre per day. That's an awful lot of money/energy.
 

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