Design "Reversed Cooling tower"

[henj]

Hi
I am going to design a “scrubber” with packed filling to remove the heat from a process stream. So the air stream is cooled down and water in the air is condensing out and the circulating water is heated up, say the reversed process of traditional Cooling towers where water evaporates from the water.

Could I assume that the transfer of heat and condensation of water are the same as for traditional Cooling towers (Evaporate cooling), for which I have some figures for the design say Merkel equation for “Ka V / L”. I have not found this mentioned anywhere in the literature. Any sources?

I also tried to simulate this process in ChemCad, but how to get enough condensation of the water in the air cooled down due to psychometric? I get say 0,0657 kg H2O/kg dry air at 40 C at top of the column which is higher than saturated air around 0,050 kg H2O/kg dry air when using 78% Nitrogen and 21% Oxygen and settings as K:UNIF H:LATE. Any suggestions on doing psychometric in ChemCad?

[mkhurram79]

Dear, Check out this book.

Cooling tower fundamentals.

http://ifile.it/i81rbdw/cooling_tower_fundamentals.pdf

[henj]

Thanks for the link!
Another useful source is the Cooling Tower Thermal Design Manual at:

[link Point to another website Only the registered members can access]
But these does not cover my reversed operation where I want to heat the water by condensation the humility in the air stream?

[mgms]

Merkel ecuation do not work for what you descrive.

[henj]

Merkel ecuation do not work for what you descrive.

Thanks for your reply, Mgms.

Could you please tell a bit more why the Merkel equation should work on this reverse mass transfer?

Merkel showed that:

Q = K x S x (hw - ha)
where,

* Q = total heat transfer Btu/h
* K = overall enthalpy transfer coefficient lb/hr.ft2
* S = heat transfer surface ft2. S equals to a x V, which "a" means area of transfer surface per unit of tower volume. (ft2/ft3), and V means an effective tower volume (ft3).
* hw = enthalpy of air-water vapor mixture at the bulk water temperature, Btu/Lb dry air
* ha = enthalpy of air-water vapor mixture at the wet bulb temperature, Btu/Lb dry air

By condensing in stead of evaporation of the water in the air is to me the same as absorption and stripping in common filled towers, where the equilibrium and operating lines changes places, but still gives a driving force in the enthalpy difference - or am I total wrong in this matter?

[kwy1970]

thanks