# Exiting air temperature in direct evaporative cooler vs cooling tower

In a direct evaporative cooler, dry air is passed by a water reservoir to evaporate water into the air, causing the latent heat of vaporization to be absorbed from both the air and the water; this results in the air leaving the cooler at a lower temperature, as desired. An image for this process is below:

In the type of recirculating cooling tower typically connected to the condenser of a thermoelectric power plant (image below), dry air is passed by hot water coming from the power plant's condenser. This causes water to evaporate into the air, absorbing the latent heat of evaporation from the water and the air, resulting in a lower water temperature, as desired.

The thing that is confusing me is the fact that evaporation of water into air in the direct evaporative cooler (first image) causes the air to leave the cooler at a lower temperature, but in the recirculating cooling tower, the air leaves at a higher temperature than it came in at. My questions are as follows:

1) What is the main reason for the difference in air temperature leaving each of the respective cooling systems?

2) Is the higher temperature of the air leaving the cooling tower due to sensible heat transfer (from warm water to cooler air) in excess of absorption of latent heat from the air?

• You have to consider the temperature of the input water in both systems. Evap cooling works because the water is at ambient temperature. The air blowing over it cools off by evaporating some water. Any water that does not evaporate, may cool off but we don't care. In the cooling tower case, the water is HOT compared to the ambient air. The air blowing in warms up because it is taking heat from the water. There is a benefit from the evaporation of the water but the biggie is heat transfer. The little droplets of water have a lot of surface area to allow them to cool off. Nov 28, 2018 at 2:58

Mixing two media (air and water) just averages the temperature between them (proportionally to their respective heat capacities and total volumes) - through classic conduction.

Additionally, evaporation absorbs a large amount of heat, reducing the resulting temperature (of both media) on exit. Rapid flow of large amounts of air speeds up evaporation considerably.

Air has a very low heat capacity comparing to water, so mixing the two will definitely shift the average towards the temperature of water - unless you push volume of air massive enough to make a difference; also, such volume will result in more evaporation - so even if the air is rather warm, the water temperature rise will be low - or even there will be a temperature drop, evaporative cooling outpacing heating from air.

So, in the first case a certain, moderate volume of moderately warm air is passed through cool water; the evaporative cooling is quite enough to keep the water - and the exiting air - cool.

In case of the cooling tower, there's a massive amount of cool air being pumped - enough to absorb (through both evaporation and plain conduction) most of the heat from the hot water. (and make no mistake, the "Hot saturated discharge air" is a very similar temperature as "Cooled water out"; maybe a little warmer as it meets hotter water later, further up the tower.

• The answer is still a bit unclear to me because the latent heat of vaporization of water is very large (2256 kJ/kg) and I would, therefore, expect that absorption of latent heat of vaporization from air (in cooling tower) would reduce its temperature greatly, considering that air has such a low specific heat capacity (1 kJ/kg/K). Or is latent heat almost exclusively absorbed from the part of the water that isn't evaporated? If so, then why? Perhaps because the temperature of the air is lower than the temperature of the unevaporated water? Nov 29, 2018 at 23:47
• @ryanDavid: " Or is latent heat almost exclusively absorbed from the part of the water that isn't evaporated? If so, then why?" - Yes, because specific heat capacity of water is so much higher than of air. 4.18 kJ/kg/K but that kg is 1dm^3 (of water), not almost a m^3 (of air) - so by mass you're getting much more water than air. The temperature of both drops as the evaporation happens on the interface between the two and both cool by the same amount, but since there's so much more water by mass, then it gives out most of the heat.
– SF.
Nov 30, 2018 at 0:11
• Also, you get 2256 kJ by evaporating the entire kilogram of water. In these devices only tiny amount of water from the exposed surfaces is evaporated. Below boiling point air can hold about 22g of water per m^3, or about 1.2kg of air. So if completely dry air leaves at 100% humidity, at best ~50kJ per m^3 of air will be absorbed. through evaporation. Water specific heat is 4.186 kJ/kg/K so a m^3 of air can drop the temperature of 1dm^3 of water (and all that air) by about 12 degrees - at best.
– SF.
Nov 30, 2018 at 0:13