When water is cooled in a direct contact cooling tower of air and water, some of the heat is removed by the sensible heat of air in contact with the water. Most of the heat is removed by evaporation of a portion of the circulating water. This mass transfer of water occurs normally from the water stream to the air stream. However, this transfer will be in the opposite direction if the entering water temperature is lower than the entering air wet bulb temperature. In the usual cooling tower operation the water evaporation rate is essentially fixed by the rate of removal of sensible heat from the water, and the evaporation loss can be roughly estimated as 0.1% of the circulating water flow for each degree F of cooling range.

Sensible heat transfer involves an increase in the dry bulb temperature of mixture but evaporation heat transfer involves a change in the humidity ratio of the mixture. Thus, a sensible heat transfer from water to air inside a cooling tower involves a horizontal change on the psychrometric chart while evaporative transfer involves a vertical movement as is illustrated in psychrometric curve. Sensible heat transfer refers to heat transferred by virtue of a temperature difference between the water and air. Evaporative heat removal refers to the energy removal from the water as latent heat of evaporation; this heat removal is the result of the evaporation of water into air during the direct-contact cooling process. In a wet cooling tower, where the temperature of water is greater than the ambient wet bulb temperature, the air humidity always increases as the air passes through the tower. Sensible heat transfer may be either positive or negative. When the temperature is less than the ambient dry bulb temperature, the sensible heat transfer may be negative and the air dry bulb temperature will be lowered as the air passes through the tower; under these circumstance, the air as well as the water is cooled by evaporative transfer in the cooling tower.

In normal cooling tower operation the amount of heat removal by the evaporation is about 60 to 95% to the total heat, and it varies upon the cooling range, air flow rate, relative humidity, and dry bulb temperature, etc.

Example 12-1. Determine the evaporation loss in a percentage for the previous example 6-1.

(Solution)
Evaporation Loss Rate = (Absolute Humidity @ Tower Exit - Absolute Humidity @ Tower Inlet) x 1/(L/G) x 100

Absolute Humidity @ Tower Exit (97^oF WBT) = 0.039166
Absolute Humidity @ Tower Inlet (85.24^oF DBT & 80% RH )= 0.021117
Evaporation Loss Rate = (0.039166 - 0.021117) x 1 / 1.4760 x 100 = 1.22%

Download the example file (exe12_1.zip)

The above calculation is based on a value of L/G , which was obtained from a result of ignoring the term of evaporation loss in the heat balance. In case of considering the loss of water due to the evaporation, L/G must be computed again as follows;

L₂/G = {(ha₂ - ha₁) - (tw₁ -32) x (w₂ - w₁)} / (tw₂ - tw₁)        (tw₂ - tw₁ = Actual Range)
Air Enthalpy at Exit (97^o F) = 66.5773 Btu/lb
Air Enthalpy at Inlet (80^oF) = 43.6907 Btu/lb
Then, L₂/G = {(66.5773 - 43.6907) - (89 - 32) x (0.039166 - 0.021117)} / 15.507 = 1.4096
Evaporation Loss Rate = (Absolute Humidity @ Tower Exit - Absolute Humidity @ Tower Inlet) x 1/ (L₂/G) x 100 = 1.28%

Download the example file (exe12_1A.zip)

Example 12-2.  Determine the heat removal in the percentage by the evaporation for the example 6-1.

(Solution)
Evaporation Rate = (w₂ - w₁) x Latent Heat of Water / (Enthalpy @ Exit - Enthalpy @ Inlet) 
Latent Heat of Water: About 1,040 BTU/Lb of Water 
(Note: For each pound of water that a cooling tower evaporates, it removes somewhere near 1,040 BTU from water. Evaporative heat removal refers to the energy removal from water as latent heat of evaporation. This heat removal is the result of the evaporation of water into air stream during the direct contact cooling process.)

Evaporation Rate = (0.039166 - 0.021117) x 1040 / (66.5773 - 43.6907) x 100 (%) = 82.02%

Example 12-3. Determine the rate of heat removal by to the evaporation under the assumption that the L/G ratio was changed to 1.600 for the initial conditions of example 6-1.

(Solution)
First, let's calculate the enthalpy of exit air.

Enthalpy of Exit Air = Enthalpy of Inlet Air + L/G x Actual Range = 43.6907 + 1.6 x 15.506
= 68.5019 BTU/lb

Exit Air Temperature = 98.14^oF

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Download the example file (exe12_3.zip)

Absolute Humidity @ Tower Exit = 0.040639
Absolute Humidity @ Tower Inlet = 0.021117
Therefore, evaporation rate = (0.040639 - 0.021117) x 1040 / (68.5008 - 43.6907) x 100 (%) = 81.83%

Download the example file (exe12_3A.zip)

Through above two examples the heat removal rate by the evaporation varies with the ratio of water and air mass flow rate. Under the same water flow rate, the higher L/G the smaller evaporation rate.

Example 12-4. Determine the rate of heat removal due to the evaporation under the assumption that RH was changed to 60% from 80% for the example 6-1.

(Solution)
First, calculate the dry bulb temperature of inlet air and find the humidity ratio with the dry bulb temperature & relative humidity.

Absolute Humidity @ Tower Exit = 0.039167
Absolute Humidity @ Tower Inlet = 0.019563
Therefore, evaporation rate = (0.039167 - 0.019563) x 1040 / (66.5780 - 43.6907) x 100 (%) = 89.08%

Download the example file (exe12_4.zip)

Note that the evaporation rate of heat removal is being highly effected by the change of relative humidity. Sensible heat transfer involves an increase in the dry bulb temperature of the mixture but evaporative heat transfer involves a change in the humidity ratio of the mixture. Therefore, a sensible heat transfer from water to air inside a cooling tower involves a horizontal change on the psychometric chart while evaporative transfer involves a vertical movement on the psychometric chart. In a wet cooling tower, which the inlet water temperature is greater than the ambient wet bulb temperature, the air humidity always increase as the air passes through the tower. However, Sensible heat transfer may be either positive or negative. When the inlet water temperature is less than the ambient air dry bulb temperature, the sensible heat transfer may be negative and air dry bulb temperature will be lowered as the air passes through the tower. Under these circumstances, the air as well as the water is cooled by evaporative transfer in the cooling tower.