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
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
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:
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℃
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℃.