High School Maximum efficiency of a dehumidifier

[SlowThinker]

There's a youtube channel where someone nicknamed Thunderf00t is making fun of a water source that is supposed to be a wind-powered dehumidifier. He makes some arguments that don't look entirely valid.

He argues that the energy needed is the evaporation heat of water, or some 2000 kJ/kg. In this video he says the energy is retrieved and needs to be dissipated, in another video he says this energy has to be supplied. Clearly he's somewhat confused, and so am I.

On the one hand, energy should be gained by turning vater vapor into liquid. On the other hand, water evaporates spontaneously, so putting an ideal dehumidifier into a closed loop would create a perpetuum mobile of the 2nd kind.

So I was wondering, how much energy is REALLY needed to take water vapor out of the air, using an ideal heat engine, ideal heat exchangers to pre-cool the input air, etc.? Should venting be accounted for?

 

[willem2]

To condense water out of the air you have to dissipate energy and you need to cool the air down to the dew point. To produce the 37 litres of water a day, you need to remove about 1 kW of heat, using only conduction to the ground, and only using the temperature difference between the ground and the dew point.
That's only the enthalpy of evaporation, the air itself needs to be cooled as well.
In most places, the ground won't be cold enough to cool any air to the dew point for half of the year.
If the air is dry, the dew point will be lower and this will make it even harder to produce water. Look at the relative humidity in New Mexico (9:30 in the video). The device will never work, because the ground will never be as cold as the dew point.

 

[SlowThinker]

Yes this device won't work, but another could. It can have a refrigerator inside. Dehumidification would happen inside. Outgoing air would cool the incoming air, so that very little actual power would be needed.
Or is this the answer? You have to supply energy and have to generate heat? I was hoping there could be a way to use the energy gained from condensation to power the device.
Even then, should not $$\eta=\frac{T_{air}-T_{dew}}{T_{dew}}$$ play some role in the result? After all, if the air temperature is at dew point, it should be somewhat easier to extract water out of it, than if the air is very hot.