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Principle of Cooling< HOME 

Principle of Cooling

 

 
   
 


Why can we feel to be cooled in front of fan motor in the summer? The fan motor is not a device to generate the cool air, but just a device to move the air. The reason is that a part of sweat extracted through the sweat glands is evaporated by the air stream and extract a heat of skin of body . The cooling tower is a device to utilize the natural principle which extracts the heat from the water during a part of water is evaporated.

The cooling tower encloses the fan which induces the atmospheric air into the cooling tower for increasing the evaporation phenomenon, the water distribution system which makes the bulk water with the optimum size of water droplets for more effectively being evaporated, the fill which extends the surface area and the contact time of water and air stream. (The fill is shaped to properly disturb the flow of water and rolls to maximize the evaporation of water as increasing the contact surface area.)

The heat of vaporization to be required to evaporate 1 kilogram of water is 597㎉. This means that the heat of 597㎉ from the water is extracted when the 1 kilogram of water is evaporated. For example, the temperature of water is lowered as much as 5.97℃ if the 1 percentage of 1 kilogram of water is evaporated. The process which the heat of water is transferred to the air could be illustrated as below:

 

The air which is induced into the cooling tower is saturated at the temperature of water droplet and is formed as an air film surrounding the water droplet. This is named as "Air Film" and the heat and water is transferred to the air from the water through the air film until the state of air film equilibriums with the state of bulk air surrounding the air film. (The state of equilibrium means a condition of which the state of air equals to the state of water after the heat and water from the water is moved to the air stream. The heat and water is not transferred to the air after reaches to the state of equilibrium. That is, if the air is completely saturated at the temperature of water droplet, there is no movement in the heat and water from water.)

Is it possible to cool the 1000㎏ of water from 35℃ to 23℃ with the 1000㎏ of air which the temperature is 25℃ and the relative humidity is 60%? The air temperature to cool the water is higher than the water temperature to be cooled…. In general, an air cooled heat exchanger can cool the water to 28~30℃ in the temperature, but it is impossible to cool the water to 23℃. However, the wet cooling tower can cool the water to such temperature. This is why to use the cooling tower.

The previous example is a case which the L/G ratio is 1.0 and the cooling range is 12℃ (= 35-23). The enthalpy of air leaving the cooling tower that is calculated from the correlation of Enthalpy of Exit Air = Enthalpy of Inlet Air + L/G x Cooling Range is 25.2982㎉/㎏'. (The temperature of air corresponding to this enthalpy is about 31.08℃.) That is, the enthalpy of air was increased from 13.2982㎉/㎏' to 25.2982㎉/㎏' by 12.0㎉/㎏' and the temperature of air was increased from 25℃ to 31.08℃ by 6.08℃.

The water vapor amount in the 1000㎏ of 25℃ & RH 60% air is 11.94㎏ and the water vapor amount of 31.08℃ & RH 100% air, which the heat was exchanged to the air from the water, is 29.11㎏. Therefore, the amount of water vapor transferred from water to air is 17.17㎏ (= 29.11 - 11.94). The heat of vaporization to be required to evaporate the 17.17㎏ of water is 10,250.5㎉. This means that 85.4 percentage of total removal heat (12,000㎉ = 1,000㎏ x 12℃ of cooling range) was extracted by the evaporation. (Of course, the state of air leaving the tower is being changed according to the value of L/G ratio.)

The previous example shows the fact that the cooling tower can cool the water below the dry bulb temperature of air. This is not a magic, but is a natural principle. The actual temperature of air being contacted with the water is not a dry bulb temperature, but a wet bulb temperature. The wet bulb temperature corresponding to the 25℃ and RH 60% of air is 19.42℃. That is, the 35℃ of water was cooled to 23℃ with the 19.42℃ of air. While the variation of temperature in the water is 12℃, the variation of temperature in the air is 11.66℃.