How can evaporative cooling air conditioning work?

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  • #51
Dale
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you said: "having more energy doesn't necessarily mean being at a higher temperature." so please tell me specifically what this energy is called if it is not called heat energy?
For a liquid and its vapor it is potential energy. The vapor has a higher potential energy, not a higher temperature. For water this potential energy is primarily in the hydrogen bonds that are formed in the liquid phase and broken in the vapor phase.
 
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The average temperature of the escaping water molecules is really beside the point. The idea behind a swamp cooler is to cool the air. And that happens because heat is required to evaporate the water. That heat has to come from the environment, from the air.
 
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  • #53
sophiecentaur
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The average temperature of the escaping water molecules is really beside the point.
Molecules don't have a temperature. They have Energy and the average energy is the temperature (crudely). The molecules with most energy in the water are the ones that leave the surface. It is the low temperature of the water vapour that cools the room. The water that evaporated 'left' a chunk of its Energy in the form of Potential energy of its bonding to the water it left behind. Blown vapour and the remaining water both end up with lower temperature.
 
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Yes, of course, I was being sloppy in using the word temperature instead of energy.

But I'm not convinced that the vapor is the primary way the room is cooled. I think the air is cooled by contact with the liquid water. Heat must transfer from air to liquid for evaporation to take place.
 
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  • #55
sophiecentaur
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Heat must transfer from (incoming) air to liquid for evaporation to take place.
That's true. I guess there's a logic that says it has to be the incoming air that prevents the water from cooling so much that evaporation stops. But it's got to be true that it's the Latent heat of vaporisation of the water that's the source of 'minus' energy and that has to be supplied by the air flow. So it's perhaps not a case of one mechanism or the other that cools the room.
I guess some numbers would be appropriate.
 
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I guess I'm thinking of standard kitchen refrigerators where the vapor is contained. The air is cooled because heat is required to evaporate the refrigerant. Compression aside, is an open evaporative cooler fundamentally different?

I've never lived where a swamp cooler made sense. But don't they have some sort of substrate that is kept wet that the air blows through? It seems like the air would give up heat there. And I guess you're right, there would be cooler water vapor that comes out as well.

How would you measure the relative contributions?
 
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you said: "it [the video] doesn't say that the air gets hotter.". but the explanation given in the video has the hottest water molecules going into the air; so surely that must make the air warmer, right?

No. What if the air is already warmer than the hottest air molecules? Then the process will tend to cool the air. But the air is such a large heat sink in this scenario that it doesn't really matter.

I mean it made the skin cooler by leaving and now is part of the air, making the air warmer, right?

Making the skin cooler is all that really matters when you're trying to explain this evaporative cooling effect. Who cares what happens to the air?
 
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Who cares what happens to the air?
An engineer who design evaporative cooling systems would be one guess.
 
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Yes, of course, I was being sloppy in using the word temperature instead of energy.

But I'm not convinced that the vapor is the primary way the room is cooled. I think the air is cooled by contact with the liquid water. Heat must transfer from air to liquid for evaporation to take place.
With a mist system, such as an ultrasonic humidifier, where all the liquid water is evaporated, it is more easy to see where all the water evaporates.
Air is readily in contact with the small air droplets, which evaporate, and with the process being endothermic, the heat transfer is from the air to the water droplets., resulting in the air temperature decrease.
The specific humidity is in the order of grams of water vapour to kg of dry air, so little bit of liquid water turning into a gas will cool the air due to the high heat of vapourization.

Its not so readily seen what's going on for something such as the surface of a lake, where evaporation keeps the lake cooler than the surrounding air temperature. The air temperature above the lake would be cooler also, which one can notice on calm days.
 
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But I'm not convinced that the vapor is the primary way the room is cooled. I think the air is cooled by contact with the liquid water. Heat must transfer from air to liquid for evaporation to take place.
For a real swamp cooler it doesn't matter. Maximum cooling effectiveness is achieved by maximizing mixing. So it doesn't matter where the heat comes from/goes to - ultimately the entire mixture is at a uniform temperature.

My suspicion though is that it's all about the gas, since you should be able to achieve the effect by evaporating all of a mass of water. But modeling a real-world situation can be complicated. For a sweaty person on a sunny 90F day, you have all four forms of heat transfer, some in each direction, and complexities that cause them to interfere with each other (such as varying temperature and humidity with distance from the skin).
 
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For a real swamp cooler it doesn't matter. Maximum cooling effectiveness is achieved by maximizing mixing. So it doesn't matter where the heat comes from/goes to - ultimately the entire mixture is at a uniform temperature.

My suspicion though is that it's all about the gas, since you should be able to achieve the effect by evaporating all of a mass of water. But modeling a real-world situation can be complicated. For a sweaty person on a sunny 90F day, you have all four forms of heat transfer, some in each direction, and complexities that cause them to interfere with each other (such as varying temperature and humidity with distance from the skin).
I think part of the problem in understanding, and explaining, is that it is not an equilibrium process, as that which would be found in a container of water and water vapour ( with no air, or it could be ). The vapour and liquid water will achieve the same temperature. With open systems difficulties arise.
 
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sophiecentaur
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The vapour and liquid water will achieve the same temperature.
Assume an arrangement where there is no air in the system and that the pressure of the vapour is the same as the partial pressure is in the room. Not difficult to achieve in a lab with a large bell jar with two chambers, a vacuum pump and a fan. One chamber would be large (the world) and the other would be small (the room).

You could achieve a swamp cooler operation without any air. Use the fan to blow vapour across the surface of the water and the high energy molecules will be removed from the water and reduce the temperature of the water. Incoming vapour (from the world) would have a proportion of high energy molecules which would condense into the cooler water join the liquid water and warm it up.

The Energy for the overall cooling would presumably come from the electricity supply to the fan?? It would maintain a temperature difference between world and room until the heat leaking into the room balances the ('negative') heat that's supplied by the molecules emerging from the cooler. I think the room would end up warmer than the world!!
it is not an equilibrium process
This is very true. Except that the room will reach an equilibrium temperature eventually - but the walls will be streaming with condensation. The condensation process will release heat into the room.
I've never lived where a swamp cooler made sense.
Nor have I but I was in a Lab where they tried (and failed). The rise in humidity in the room was very noticeable and we ended up in a room that was a bit chilly but it still felt sweaty. Disgusting
 
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Use the fan to blow vapour across the surface of the water and the high energy molecules will be removed from the water and reduce the temperature of the water. Incoming vapour (from the world) would have a proportion of high energy molecules which would condense into the cooler water join the liquid water and warm it up.
You can look up adiabatic saturation temperature, which is the coolest temperature air can be cooled to by this process. I think it does involve some expansion work, as water molecules have to also overcome the pressure of the air ( the added water molecules take up space so they push the air molecules aside ).
One can test for AST by passing air over water in a long tunnel.
 
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sophiecentaur
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You can look up adiabatic saturation temperature, which is the coolest temperature air can be cooled to by this process. I think it does involve some expansion work, as water molecules have to also overcome the pressure of the air ( the added water molecules take up space so they push the air molecules aside ).
One can test for AST by passing air over water in a long tunnel.
My point was that the Air is not actually necessary for discussing basic principles here. The vapour pressure just adds to the air pressure (Dalton's law of partial pressures) so it's valid to work with a model with no air, at the same partial pressure(s) of Water as exist in the World and the Room.
The 'usual' description of air as being like a sponge that can hold water is not a good one. It only works because the behaviour of a mixture of gases and vapours makes it appear that way.
Your AST applies in practice, of course but water and water vapour don't only exist in the presence of air.
You could imagine a refrigeration machine that worked with water only ```(instead of a CFC). It would probably need to operate between different sink temperatures but you could have a phase change, a pump and two heat exchangers. (The term Heat Pipe comes to mind whilst we're on that topic)
 
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The 'usual' description of air as being like a sponge that can hold water is not a good one. It only works because the behaviour of a mixture of gases and vapours makes it appear that way.
We might not have as much weather if the atmosphere was only water vapour.
Humidified air is less dense that dry air.
 
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sophiecentaur
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We might not have as much weather if the atmosphere was only water vapour.
Humidified air is less dense that dry air.

The density thing can't be a surprise when you think that water molecules are a lot lighter than O2 or N2 or CO2. Avagadro tells us that there are the same total number of molecules in a given volume so hence, the density.
It's interesting to consider that the air that forms clouds could well be more saturated than the air around it - so it's probably due to go up in any case. (No one should treat that too seriously; I realise that weather is more complicated than that)
 
  • #67
The average temperature of the escaping water molecules is really beside the point. The idea behind a swamp cooler is to cool the air. And that happens because heat is required to evaporate the water. That heat has to come from the environment, from the air.
Energy is conserved. The total energy in a system remains the same. The system we are considering is the water in the evaporative cooler and the air drawn through by the fan. Temperature is a measure of the average kinetic energy of particles passing through a unit of area. Heat is the total kinetic energy of these particles. Looking at the individual molecules of water, these are moving relatively slowly in the liquid but many more pass through a given area than in a gas at the same temperature. They contain much more heat than a similar volume of air.
Think of baseballs rattling around in a box. If one of the molecules of water at the surface of the water is struck by a molecule of air that is moving at the speed of sound (the average velocity of molecules of a gas) the water molecule gains energy from the gas molecule and becomes part of the gas and is now moving basically at the speed of sound. That change in energy came from the gas molecule which is now moving slower (at a slower temperature). Looking back at our baseballs consider one bouncing above the box (a water molecule) and you strike it with your bat (the air molecule) the baseball's velocity is greatly increased and the bat's velocity is reduced conserving energy. Energy gained by the water molecule is lost by the air. Energy gained by the baseball is lost by the bat.
 
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Summary:: how does evaporative cooling air conditioning work when the hottest water molecules that have escaped the water now should make the air hotter because the water became cooler?

If evaporative cooling (Such as sweating) is due to the escape of the hottest molecules into the air thereby lowering the total average temperature of the water then that means that that hot water molecule has gone into the air and has made the air hotter, so then how does evaporative cooling air conditioning work when these hot water molecules that have escaped the water now make the air hotter because the water became cooler?
As an analogy, ask yourself why an ice cube cools a glass of water. The only water molecules leaving the ice cube are ones that are the high energy liquid ones at 0-degrees-Celsius.

You say the air must get hotter, but it is not boiling steam you are adding, it is water molecules from room temperature water. A high energy liquid water molecule is a low energy gaseous water molecule. (Or else it would all be gone already).

A swamp cooler works best with low humidity. If you have an isolated box with air at 0% humidity, and you drop in a bit of water, the air temperature drops a bit and the water turns into a gas. In that isolated box, the pressure goes up, and the total energy is the same. But the temperature drops. If the air was already at 100% humidity, then the water you drop in just stays water, at equilibrium with the humidity.

A swamp cooler uses moving air across water. The high surface area allows the humidity of the moving air to increase. The pressure of the open system is not noticeably affected. The air temperature drops as energy was used to evaporate water into the air.

Also, sweating is ineffective in high humidity.
 
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Dear questioner, try this: Given that the evaporation predominates control over the resulting air temperature [making the air temperature lower], all other factors become trivial and irrelevant to whether the resulting air temperature will be higher or lower because the other factors will have little to no effect and do not cause the resulting air temperature to rise. The worthy endeavor of discussing and understanding the other factors are a separate endeavor than seeking the answer to your question now that you know that the evaporation has the predominant effect on the change in air temperature.

An example of how evaporation predominates is in air conditioning systems, which use evaporation to create the colder air even though the liquid being evaporated is hotter than the air. Since analogies help you understand, try this analogy with your desired analogy friends, the marbles, except instead of reducing temperature, we will be reducing noise while demonstrating that the change of state is the predominant factor determining the change in noise level, not the temperature of components: The Question: "Why isn't the total noise of marbles contacting each other reduced when we remove more marbles?" The Answer: "The change of state that occurs when the marbles are removed by smashing them with a hammer causes much more noise than the reduction in noise caused by there being fewer marbles contacting each other." In other words, the state of change predominates control over the results regardless of other factors that may have a relatively trivial effect; therefore, the endeavor of asking why the other factors don't determine the ultimate air temperature difference is pointless.
 
  • #70
We might not have as much weather if the atmosphere was only water vapour.
Humidified air is less dense that dry air.
We wouldn't have much Life either!
 

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