How does a cooling tower work and how much water does it use?

The evaporative cooler is a very simple device, which serves to cool water by direct exchange with air.

There are different types of evaporative towers:

induced, with suction fans
forced, using forced ventilation
counter flow, with countercurrent water and air flow
cross flow, that uses a crossing flow of water and air

in any case all of these machines work using the same concept.

The counter-flow type is the most widely used, whether it be forced or induced, because it presents greater functioning efficiency.

EVAPORATION…EVAPORATIVE, why?

Compared to a conventional radiator it is more efficient because the evaporation effect of the water is exploited as a result of the direct exchange, thereby taking advantage of the latent heat evaporation.

The reference parameter therefore isn’t the ambient temperature, it is the temperature of the moist bulb, or rather the temperature that we can determine using the psychrometric chart, based on ambient temperature and relative humidity.

The tower’s ambient temperature is usually 4 ° C to 6 ° C (or even higher) less than the ambient temperature. As a result the cooling tower is able to cool the water at a temperature below ambient temperature.

Furthermore, by exploiting the latent heat evaporation efficiency is considerably increased. In fact, on average for every kilogram of evaporated water there is a dispersion of 600 kcal/h (average value). This fact implies that a cooling tower, by its very nature, consumes water, therefore we cannot consider the cooling tower circuit as a closed circuit. In fact there is a constant consumption of water, which is linked to the potential dissipation.

WATER CONSUMPTION

The following formula enables us to evaluate the average water consumption of an evaporative tower:

Qc = P / 600

where:

Qc is the amount of water used expressed in Kg/h
P is the tower’s thermal power expressed in Kcal/h
600 is the amount of heat extracted from each kilo of evaporated water (KCal / Kg) on average…the correct value should be taken from the steam diagram based on actual operating conditions. On the first approximation however, this value is rather accurate, as we will see in the explanation below.

The result of this simple formula gives us the value of only the water that is consumed by evaporation.

We must add to this value:

the amount of water loss due to drag
the amount of wastewater purge

The losses due to drag, using current drift eliminators, are almost irrelevant. The amounts in question are losses of roughly 0.1 to .05% of the total value of the recirculating flow.

The amount of wastewater purge however depends on several factors:

the quality of the water that is re-integrated
the type of conditioning applied to the water
concentration cycles

Essentially these three parameters are closely linked to each other and depend mainly on the quality of the water that is re-integrated.

In any case due to the continuous evaporation of water, a gradual concentration phenomenon takes place, which will change clean water to a state that is unacceptable for cooling circuits. This phenomenon forces you to re-integrate, clean and condition the water.

Rather than going into a detailed discussion of these aspects we can say that at a first approximation, particularly if we’re making some considerations regarding the consumption of water by a cooling tower, we can assume that the flow of the wastewater purge is equal to the water lost due to evaporation. In brief we can say that the average consumption of water in an evaporative tower is the result of the sum of water losses due to drag in addition to double the evaporation losses.