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I'm trying to gauge the efficiency of a night-cooled stone to condense water from the humid air at seashores at noon. I'm assuming the stone is thermally insulated once it's in thermal equilibrium with the lowest possible temperature at night. Then at noon, the humid air is passed through the stone and the (assume thermally isolated) system now consists of hot, humid air and the cool stone.

I understand that at lower temperatures, air holds lesser vapor, so if hot humid air is condensed, some vapor will condense.

I'm using the specific heat capacity of vapor, air and stone to try and come up with the final temperature, so that I can see how much vapor will air condense at that temperature by subtracting the after-cooling holding capacity per meter-cubed from the initial one.

The problem here that I must use the latent heat of vaporisation for the energy lost to the stone for the water condensed, but then the vapor is at a temperature lower than 100 Celsius so should I use the specific heat capacity of vapor or latent heat or specific heat capacity of water?! I just got confused.

As $q_{vapor}+q_{air}=-q_{stone}$, I came up with this $$m_{vapor}c_{Pvapor}(T_{1}-T_{F})+m_{air}c_{Pair}(T_{1}-T_{F})=-m_{stone}c_{Pstone}(T_{1}-T_{F})$$

Something just seems wrong to me here because of the vapor temperature being below boiling point of water. I know that temperature is AVERAGE internal energy of a mass of substance and all. However it still confuses me. Someone clear it up for me.

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You confused me too, lol. Lets back up.

Water will only condense when moist air is cooled below its dew point. The amount of water the air can store is not important to us in this case; just when it is at the dew point.

This means that in order for this system to produce any water, the night temperature must be less than the noon dew point. The final temp will be the noon dew point because any warmer and water will not condense, it will evaporate. As such, if it continues past that point it will evaporate all the water it gained.

To calculate how much water will be produced you will use something like:

Rock_Specific_Heat * Rock Mass * ΔT(night_temp to noon_dewpoint) =

      Water_Latent_Heat_of_Vaporization * Water Mass 
                   + Air_Specific_Heat(with humidity) * Air Mass * ΔT(noon_temp to noon_dewpoint)

Also note that the latent heat equation for condensation in atmospheric conditions is different than pure steam.

Then you will have to add in the humidity. Use a Psychometric Chart convert relative humidity to a humidity mass ratio kg/kg.

Still half baked, but I think that gets you on the right track. Good Luck!

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    $\begingroup$ Your sincerity is leagues ahead of everyone here! Thank you ericnutsch! $\endgroup$ – El Flea May 4 at 13:59

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