How does a cooling tower function?

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  • #1
dRic2
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So I'm having trouble understanding the physics behind evaporative cooling. This is what I know: I want to cool some water so I nebulize it and I let an air flow (coming from the outside) pass through this mist of water. Now some water has to evaporate. Here I am stuck because I don't understand how water evaporates: since it evaporates it must take energy (sensible heat) from somewhere and convert it into latent heat. Does it take the heat both from the liquid phase and the air ?
 

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  • #2
darth boozer
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A simple search returns several articles about this, on being
https://en.wikipedia.org/wiki/Cooling_tower
Although many people look down on Wikipedia, it has improved vastly over the last few years, especially for scientific subjects, as stated by Richard Dawkins in his book "Science in the soul" (2017).
 
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  • #3
dRic2
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I did read the article but I still do not get it very much, that's why I asked.

The article says that air gives sensible heat to the water (so the temperature of air drops) and the water evaporates (thus converting the sensible heat in latent heat). The vapor then goes into the gas phase bringing back the latent heat so, over all, the entalphy of the gas phase does not change (isoentaphic process).

This is fine by me, but then I don't understand how water itself is cooled if the energy is taken from the air and not from the liquid water.

Plus I do not understand the process at the atomic scale. The article mainly uses classical equilibrium thermodynamics and it is completely fine, but I can't picture it. For instance, the air is colder then the liquid, right ?
I know that heat does not flow from colder to hotter so how exactly does the air make the water evaporate? How does the air supply the necessary heat? I suspect the answer lies in diffusion of water molecules from one phase for the other and maybe the surface tension of the droplets plays a role, but I can't figure out an explanation that satisfies me.
 
  • #4
Asymptotic
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the air is colder then the liquid, right ?
Not necessarily. If this were so, a cooling tower would be useless during summer months.

I don't understand how water itself is cooled if the energy is taken from the air and not from the liquid water.
An open loop tower removes heat by evaporating some of the incoming water (it takes energy to convert liquid water to vapor). What isn't evaporated becomes cooler, and returned to the system. Most of what a cooling tower does is to increase evaporation by hiking air flow rate, and increasing water surface area by distributing it over a "fill" material.

970 BTU is removed per pound of water evaporated (2 MJ/kg), and while leakage and "drift" (small droplets of unevaporated water blown out into the airstream) factor in, it is possible to get an idea of how much cooling a tower provides by monitoring make-up water demand. For example, if 290 gallons per hour is consumed, then (with water at 8.34 lb/gallon, and ignoring losses) approximately 2.35 million BTU/hr (about 195 tons of cooling; 1 cooling ton = 12,000 BTU) of heat is removed.

SPX and other cooling tower manufacturers are a valuable resource. This is a good start.
https://spxcooling.com/library/detail/cooling-tower-fundamentals
 
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There is a combination of heat- and mass transfer occurring. The mass-transfer part goes like this: the partial pressure of the water vapor in the air at the water-air interface is the equilibrium vapor pressure of water at the water temperature. This partial pressure is higher than in the bulk of the flowing air, so there is driving force for water molecules to diffuse away from the interface, through the air in the convective mass transfer boundary layer. This water mass transfer must be satisfied by water being caused to evaporate. The heat for this comes from the water (and partly from the air). For a more detailed and quantitative discussion of all this, see Treybel, Mass Transfer Operations.
 
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  • #6
essenmein
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Reading that paragraph probably saves you a chapter or two in a textbook.
 
  • #7
dRic2
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@Chestermiller Thank you. In the book you cited, can I find a way to calculate the ratio
$$\frac { \text{ heat from water } }{ \text{ heat from air} } $$
?
That's should tell me which cooling effect is prevalent.
 
  • #8
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@Chestermiller Thank you. In the book you cited, can I find a way to calculate the ratio
$$\frac { \text{ heat from water } }{ \text{ heat from air} } $$
?
That's should tell me which cooling effect is prevalent.
Sure. That would be determined from the overall enthalpy balance on the cooling tower.
 
  • #9
dRic2
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Thanks. As soon as I have enough free time I'll check the book. Maybe I'll be back for help :D
 

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