Why Does Misting Fan Cool Air?

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SUMMARY

The discussion centers on the principles of evaporative cooling as applied to misting fans. It establishes that when water evaporates, it absorbs energy from the surrounding air, leading to a decrease in air temperature. The conversation clarifies that while the energy levels of liquid water and water vapor may be similar, the kinetic energy of surrounding air decreases due to the potential energy increase of water molecules transitioning from liquid to vapor. This results in a cooling effect experienced by individuals near the misting fan.

PREREQUISITES
  • Understanding of evaporative cooling principles
  • Basic knowledge of thermodynamics, specifically kinetic and potential energy
  • Familiarity with the properties of water in different states (liquid, vapor)
  • Concept of entropy in thermodynamic systems
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  • Research the physics of evaporative cooling mechanisms
  • Explore the relationship between kinetic energy and temperature in gases
  • Learn about the thermodynamic properties of water and its phase transitions
  • Investigate the design and efficiency of misting fans in various environments
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Individuals interested in thermodynamics, HVAC professionals, engineers designing cooling systems, and anyone seeking to understand the science behind misting fans and evaporative cooling.

hacillunation
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So, as far as I know evaporative cooling is when the top layer of a material (lets say water) evaporates- taking energy with it leaving the rest of the material cooler. Also leaving the air, which transported some heat to it, cooler.

But in a misting fan, the mist evaporates, takes energy from the air, and then- it just remains there.
So the overall energy level of the system should be the same, no? Same entlapy, same entropy.

Then why when this mixture of air and evaporated water droplets blows- even that you suposedly get a system with the same energy level as before, it cools you down?

Thx.
 
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hacillunation said:
So, as far as I know evaporative cooling is when the top layer of a material (lets say water) evaporates- taking energy with it leaving the rest of the material cooler. Also leaving the air, which transported some heat to it, cooler.

But in a misting fan, the mist evaporates, takes energy from the air, and then- it just remains there.
So the overall energy level of the system should be the same, no? Same entlapy, same entropy.

Then why when this mixture of air and evaporated water droplets blows- even that you suposedly get a system with the same energy level as before, it cools you down?

Thx.
The energy liquid water may have the same as the energy of the water vapour but the temperature has decreased. The water mixing with the air results in an increase in potential energy of the water molecules. This potential energy comes at the expense of the kinetic energy of the surrounding air. So the temperature of the surrounding air decreases.

The potential energy of water molecules in vapour comes from the fact that water molecules are polar - in liquid and solid form they stick together and it takes energy to separate them.

AM
 
Thank you for the succint answer.

That raises another question then...
So, youre saying that water vapor is cooler than the water it escapes from?
But when water evaporates, doen't it increase its particle speed?
 
The mist is liquid not vapor. That's why it is visible.

Also when you said "same entropy", the statement is wrong.
 
hacillunation said:
Thank you for the succint answer.

That raises another question then...
So, youre saying that water vapor is cooler than the water it escapes from?
But when water evaporates, doen't it increase its particle speed?
It is a bit complicated because the water molecules in liquid form are stuck together and have vibrational kinetic energy. When they separate into individual molecules (vaporization) they no longer vibrate against each other but have translational kinetic energy. The average translational kinetic energy of those water molecules can be less than the original liquid water but because they are not stuck to other water molecules they stay in vapour form. The energy required to separate the molecules is at the expense of the kinetic energy of the surrounding air molecules. So the temperature of the air decreases as more water is vaporized.

AM
 

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