How can evaporative cooling air conditioning work?

[russ_watters]

please feel free to add some goo to the marbles in the example

I honestly don't see a way to make this analogy work.

 

[genekuli]

i need this simplified,
would this be correct:
the hot (vibrating) marbles that escape the set lose energy to breaking the goo bond on the way.
they then have low vibration upon mixing with the air marbles and so do not increase the air marbles' vibration.
is that it?

 

[Dale]

the explanation given in the video has the hottest water molecules going into the air

Hmm, that doesn’t make sense. The water is one temperature, so there aren’t hotter and colder water molecules. At a given temperature the molecules have a distribution of kinetic energy, but that doesn’t make some hotter than others.

 

[russ_watters]

Hmm, that doesn’t make sense. The water is one temperature, so there aren’t hotter and colder water molecules. At a given temperature the molecules have a distribution of kinetic energy, but that doesn’t make some hotter than others.

The argument being made is that if molecules that are higher on that bell curve leave, the bell curve shifts in the opposite direction. If the temperature is somewhat a function of the average kinetic energy, then the shifting of the curve down is a drop in temperature.

I'm pretty sure I learned something like this in high school, but had forgotten about it. It makes sense in a self-contained way, but to me it lacks quantification. I don't see how you would calculate the temperature drop of the water based on this description. And it seems incomplete; it implies the energy of the bond (enthalpy of vaporization) is irrelevant to the temperature change in the water as it evaporates.

 

[russ_watters]

i need this simplified,
would this be correct:
the hot (vibrating) marbles that escape the set lose energy to breaking the goo bond on the way.
they then have low vibration upon mixing with the air marbles and so do not increase the air marbles' vibration.
is that it?

That really makes no sense to me. It doesn't capture the chemical energy involved (vague reference to "goo bond" is not really meaningful). To me, the entire problem you had from the beginning was treating temperature as the only component of this problem, and ignoring the chemical bond energy. Here you have an analogy that does the same thing. Maybe someone else can craft a useful analogy from that, but I'm not seeing it.

 

[Dale]

The argument being made is that if molecules that are higher on that bell curve leave, the bell curve shifts in the opposite direction. If the temperature is somewhat a function of the average kinetic energy, then the shifting of the curve down is a drop in temperature.

The argument is fine, but the terminology is the problem. At a single temperature there is a distribution of energy. To call the molecules at the high end of the curve “hot” means that at a single temperature there is a distribution of temperatures!

This is exactly the problem the OP is having. He has been falsely told that there are hot molecules and so his conclusion that the air should get hotter is reasonable given that misinformation.

But the molecules are all at a single temperature, and molecules at a single temperature have a distribution of energies. The high energy molecules can leave, but that does not increase temperature because they are not hotter. I think it is exactly this that is causing the OP’s confusion.

 

[jartsa]

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?

Let's say a water bucket has been standing in a room for a long time with a lid on.

Now we open the lid.

The first small puff of steam coming out is at the same temperature as the water and the room.

The second small puff of steam coming out is at the same temperature as the water, which has cooled very slightly, because of a few of the faster molecules leaving.

The faster molecules become average molecules in the process of leaving the liquid.

 

[genekuli]

The argument is fine, but the terminology is the problem. At a single temperature there is a distribution of energy. To call the molecules at the high end of the curve “hot” means that at a single temperature there is a distribution of temperatures!

This is exactly the problem the OP is having. He has been falsely told that there are hot molecules and so his conclusion that the air should get hotter is reasonable given that misinformation.

But the molecules are all at a single temperature, and molecules at a single temperature have a distribution of energies. The high energy molecules can leave, but that does not increase temperature because they are not hotter. I think it is exactly this that is causing the OP’s confusion.

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

 

[russ_watters]

The argument is fine, but the terminology is the problem. At a single temperature there is a distribution of energy. To call the molecules at the high end of the curve “hot” means that at a single temperature there is a distribution of temperatures!

I don't think that's the OP's confusion -- he said he understands why the water gets cooler, just not why the air gets cooler.

And to be honest the whole "one molecule can't have a temperature" thing has always confused me too. It begs the question of how many it takes before we can call it an "average energy". But we can easily avoid this by discussing bundles of molecules instead of individual molecules. Then, I see nothing wrong with dividing a distribution into two distributions. If you have 1 kg of water at 20 C and 1 kg of water at 30 C and you mix them together to get 2 kg of water at 25 C, the reverse should be possible -- if the distribution is wide enough and you can select the molecules.

I just don't think that's what's happening.

This is exactly the problem the OP is having. He has been falsely told that there are hot molecules and so his conclusion that the air should get hotter is reasonable given that misinformation.

But the molecules are all at a single temperature, and molecules at a single temperature have a distribution of energies. The high energy molecules can leave, but that does not increase temperature because they are not hotter. I think it is exactly this that is causing the OP’s confusion.

"hot molecules" or "higher energy molecules"? Because the molecules that leave are higher energy and do increase the average energy of the air. Just not in the way he thinks (via latent heat, not sensible heat).

Similar to above, if I take 1 kg of air at 20C and 1 kg of water vapor at 30C I get a mixture at about 23C. That's what the OP thinks is happening, but it isn't because the water vapor temperature isn't above the air temperature.

 

[russ_watters]

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

I have tried to tell you several times that there is more than one kind of heat/energy -- and even, there's more than one kind of temperature. It's what you've been missing since the beginning, I've said it several times, but it isn't getting through.

No, having more energy doesn't necessarily mean being at a higher temperature.

At this point, I think you simply don't want to believe it.

But if it makes you feel better about it, I don't think I mentioned the name of that other temperature before: it's called Dew Point Temperature. And it does rise in this situation.

So, you can increase the enthalpy of the vapor mixture by increasing its sensible/dry bulb temperature or by increasing its dew point temperature.

So, does that help?

 

[genekuli]

I have tried to tell you several times that there is more than one kind of heat/energy -- and even, there's more than one kind of temperature. It's what you've been missing since the beginning, I've said it several times, but it isn't getting through.

No, having more energy doesn't necessarily mean being at a higher temperature.

At this point, I think you simply don't want to believe it.

But if it makes you feel better about it, I don't think I mentioned the name of that other temperature before: it's called Dew Point Temperature. And it does rise in this situation.

So, you can increase the enthalpy of the vapor mixture by increasing its sensible/dry bulb temperature or by increasing its dew point temperature.

So, does that help?

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?

 

[256bits]

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

You pick a reference point to describe the energy content of a mass of molecules.
For water, we pick 0.01 C, where the enthapy is designated as being 0.

To gain any other HIGHER temperature, or to vapourize, one has to add energy, so that is where the talk about having more energy come from.
https://www.engineeringtoolbox.com/water-properties-d_1508.html

You will experience a more severe burn from steam than water, even though they may be at the same temperature, due to the higher energy content at the vapourized state. The steam will condense on your skin releasing heat 'of condensation' to the liquid state, while the liquid water will only have conduction available to transfer heat.

 

[genekuli]

You pick a reference point to describe the energy content of a mass of molecules.
For water, we pick 0.01 C, where the enthapy is designated as being 0.

To gain any other HIGHER temperature, or to vapourize, one has to add energy, so that is where the talk about having more energy come from.
https://www.engineeringtoolbox.com/water-properties-d_1508.html

and that energy would be called heat energy, no?

 

[256bits]

and that energy would be called heat energy, no?

I edited , and added more .

 

[russ_watters]

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?

It is "heat energy". That's still not specific enough though. Specifically, it is latent heat energy. Because, again, there is more than one kind.

Your mistake is still thinking heat = temperature. It does not.

Every post, I'm saying the same thing over and over in different ways.

Here's another description of the process:

Evaporative cooling takes place along lines of constant wet bulb temperature or enthalpy. This is because there is no change in the amount of energy in the air. The energy is merely converted from sensible energy to latent energy. The moisture content of the air increases as the water is evaporated which results in an increase in relative humidity along a line of constant wet bulb temperature.

http://www.coolbreeze.co.za/psyevap.htm

 

[genekuli]

right, so the energy of the heat of the water, breaks the H bond, and that kinetic heat energy (sensible energy) changes to potential energy (latent energy) and therefore does not transfer the energy to the air, as the E stays in the H bond potential. is THIS right?

 

[jbriggs444]

i need this simplified,
would this be correct:
the hot (vibrating) marbles that escape the set lose energy to breaking the goo bond on the way.
they then have low vibration upon mixing with the air marbles and so do not increase the air marbles' vibration.
is that it?

That matches the way I think of it. The molecules that leave the liquid phase expend energy breaking free of the liquid and, so, no longer have the high kinetic energy they started with.

 

[sophiecentaur]

the video said the water molecule "is" hot, that is why it escapes the H bonds. when it leaves the sweat, it decreases the total heat in the sweat.
it didn't say: The act of evaporating lowers the temperature of the escaping water vapor.
those are two different explanations/phenomenons right?

Your second paragraph is more like the correct message.
Firstly, it's not a good idea to talk about 'hot' and 'cold' molecules; they are fast or slow. Temperature is a bulk quantity.
The description of the system seems to be failing to stress the total dependence of the effect on change of state. The escaping molecules of water are not going as fast as the temperature of the liquid would imply; they are only just managing to escape the Potential Energy of the surface, having lost most of their Kinetic Energy. ````Only by blowing them away with a fan can you manage to remove that energy permanently from the liquid (cooling it down).

This can't go on forever in a closed room because, eventually, the air can become saturated with water vapour. Neither does the system work at all well when outside air is used but when it's already very humid. You end up with cold wet air in the room which actually feels disgusting. In hot dry locations, the system is fine and cheap to buy but not in UK.

 

[Dale]

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

This indicates a misunderstanding about thermal energy. Thermal energy is not something that even makes sense for individual molecules. Thermal energy is only something that a large ensemble of molecules have. Thermal energy is energy distributed randomly in internal degrees of freedom.

What these individual molecules have is kinetic energy. Inside a large group of molecules, all at a single temperature, there are molecules with a distribution of kinetic energy. When you are looking at the details at the scale of the individual degrees of freedom then what is distributed is no longer thermal energy, it is kinetic energy or whatever other degree of freedom you are looking at.

They are not hotter because they are in thermal equilibrium with the rest of the water. They have more kinetic energy. Kinetic energy and thermal energy are not the same thing.

 

[Dale]

he said he understands why the water gets cooler, just not why the air gets cooler

Right, and that misunderstanding is due to the fact that he incorrectly believes the air is getting hot molecules.

Then, I see nothing wrong with dividing a distribution into two distributions.

Yes, but it is a distribution of kinetic energies at a fixed temperature. It is not a distribution of temperature. The faster molecules are not hotter, they are in thermal equilibrium with the rest.

"hot molecules" or "higher energy molecules"? Because the molecules that leave are higher energy and do increase the average energy of the air.

Yes, energy and temperature are not the same thing. They are high energy molecules, they are not hot molecules.

 

[Dale]

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.

 

[JT Smith]

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.

 

[sophiecentaur]

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.

 

[JT Smith]

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.

 

[sophiecentaur]

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.

 

[JT Smith]

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?

 

[Herman Trivilino]

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?

 

[256bits]

Who cares what happens to the air?

An engineer who design evaporative cooling systems would be one guess.

 

[256bits]

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.

 

[russ_watters]

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).