Evaporative cooling: Who takes the call air or water?
What happens when air embraces water in a cooler?
Air in terms of its internal energy with two non-polar gases N2 and O2 is much weaker than a strongly polar water. Therefore, air has practically very small energy compared to water. Air's energy is its sensible heat. Air has no stored energy except the bond energies in N2 and O2 which are not available in physical process like cooling tower heat and mass transfer.
Air’s energy which it shares in HX like CT is essentially a tiny amount of latent heat in the water it carries. At 50RH at ambient temp, this is < 0.01% moisture in air. That is air's latent heat to share in HX like CT. Therefore, air acts as a sink in CT where water dumps its energy while cooling. Since, energy transfer takes place always in the direction from high energy to low energy water transfers its energy to air as it cools. Except sensible heat transfer between air and water which maybe just a few degc, air does not contribute energy for water evaporation. When water is sprayed in CT some molecules which have high kinetic energy escape from water to the surrounding air. Unsaturated air is unstable until it gets saturated. Air like a magnet pulls those water into its stomach. This process goes on until air has a capacity at a given temp to absorb water. This is the driving force for water evaporation in CT. Since air pulls water vapor, water meets this air’s requirement until the air reaches wet bulb temp by supplying energy from its internal energy and cools. Water has no other option. The point at which air has saturated, Gibbs free energy change =0.
In other words, when water and air are in intimate contact in a properly designed packed mass transfer column it is like one energy donor and one acceptor of energy are in contact. Both become unstable and like to come to stable state. Water loses energy and air gains energy in the form transfer of water molecules from one to another.
Typical calculation
Imagine 1 lb air enters a cooler at DBT / RH = 80 F / 50 RH and cools 1 lb water cools by 15 deg F adiabatically by taking its own internal energy.
Internal energy consumption by water
= [1[lb mass] *1[Btu/lb-F specific heat]*15 F [?t] ] * 1000/1000 = 15 Btu [ Water’s latent heat = 1000 Btu/lb]
Water transfers enthalpy of 15 Btu to 1 lb air near isothermally since there is no change in air temperature in cooler.
Enthalpy change ?H of air inside CT= [32+15] = 47 Btu [ Air’s initial enthalpy = 32 Btu/lb from psychrometric chart]
Enthalpy of air post absorption of moisture from water
?H = 32 + 15 = 47 Btu
The enthalpy received by air from water goes as work energy when water vapor pushes air molecules to make space for their inclusion in air [ Very important concept] Water vapor has to do work to push air to find its place. Water while does this work on air, water adds 15 Btu energy to air. The total enthalpy of air thus becomes 47 Btu. This value ties up with psychrometric chart air out enthalpy.
Air leaves the cooler at 100 RH [ assumed fully saturated] with 47 Btu enthalpy into ambient condition 50 RH / 80F. Air from cooler can’t hold 47 Btu enthalpy once it is out of cooling tower. It goes back to outside air enthalpy of 32 Btu releasing 15 Btu heat isothermally, this is work energy, air releases as air ejects out water vapor and which expands. 15 Btu heat loss by 1 lb air = 15/1000 = 0.015 lb water vapor. This is the amount of suspended water in the plume/lb dry air.
Finally, it is air which takes the call in evaporating cooling.