Cooling tower: Two-phase Heat and Mass transfer

In evaporative cooling air is the master and the water is the follower. At every step, the air is the decision-maker. Tracking air is very important.

The image is a very familiar set of observations of a typical cooling tower temperature / RH profile. A simple heat balance on the basis of the data on the image is given below. 

Waterside

Heat lost by water = (1lb [ mass of water] x 1 [specific heat of water in Btu/lb-F] x [104-77] F [ Cooling of water])

= 27 Btu/lb

Airside

Heat gained by water = 1 [mass of air] x 0.24 [ specific heat of air in Btu/lb-F] x [88-68] F [ Heating air]

= 0.24 x 20

= 4.8 Btu/lb

Un-accounted heat missing in air = [27-4.8] = 22.2 Btu/lb

Question is how 22 Btu heat disappeared?

The answer is as follows:

Waterside

Water evaporates because there is someone to accept the heat and mass it releases when it evaporates. That someone is unsaturated air. If air says I am full and cannot accommodate more water, evaporative cooling ends there.

Heat lost by water while it evaporates = Mass of water x Specific heat x dT, dT is cooling. --- [1]

There is no other source of energy that water can use for cooling other than its latent heat. So, water uses its latent heat to evaporate. An unsaturated air and water cannot stay together. This is the most unstable situation. So, the water is forced by air to evaporate and get the energy of water transferred into the air. Energy flows from high energy water to low energy air.

Airside

Airside equation is Mass of air x [Enthalpy of out air - Enthalpy of in air] --- [2]

At a steady-state,

Equation [1] = Equation [2]

On the airside water vapor with its latent heat goes into the pockets of air. This is called enthalpy transfer to air. Enthalpy of vapor = Mass x Latent ht. Air needs space to accommodate this vapor. Air expands by using the energy it gets from water. Air does mechanical work to expand. Enthalpy H = U [internal energy + PdV [work]. Enthalpy is energy. The energy of air increases by gaining work energy and the energy of water reduces by losing its [internal] thermal energy, -latent heat.

22 Btu /lb missing heat was used by air to do mechanical work to expand to accommodate water vapor. 

Ideally for perfect heat and mass transfer in a cooling tower DBT = WBT at the exit if there is no sensible heat transfer from water to air. Any excess exit air temperature over the air- in temperature is an indication of only sensible heat transfer without mass transfer.