Understanding the Science Behind Water Vapor and Cloud Formation

[Edison Bias]

At some temperature, some particles will have higher energy and some lower. As you increase the temperature, more molecules will have enough energy to escape the liquid phase and fewer molecules in the gas phase will get captured into the liquid. At any temperature, you can have an equilibrium where as many molecules are escaping the liquid into gas as there are being absorbed by the liquid from the gas. At equilibrium you are at a relative humidity of 100%. Since the temperature and pressure keep changing, the water is generally not in equilibrium. In areas where relative humidity is less than 100% you have evaporation, and in areas where relative humidity is greater than 100% the water vapor will condense into clouds, fog, and dew. Cloud droplets can eventually collect and become heavy enough to fall.

This is an extremely interesting reply, thanks! Let's think step by step. At Tav there are particles with higher and lower Tk. Increasing temperature Tk(max) gets so high that the particles escapes the liquid phase because Tk is now higher than T(liquid/gas)=Tboil. To me this sounds like what I think I now understand according to my former calculations. It evaporates because Tk>Tboil. I can't understand it in another way. The liquid is of course not boiling but some particles still reaches a temperature that is higher than boiling temperature (Tboil). Then you say that as some escapes the liquid phase, fewer gets recaptured. Why (relative humidity <100%)? Or is it perhaps due to the higher temperature where the matter approaches a gas, so to speak? The equliblium speak is interesting and obviously due to 100% humidity which I think can be called saturation. <100% humidity =evaporation, this I think I understand (room for more water excist in the air). But how can humidity >100%? You must mean that as a figure of speach because at the moment you hit 100% humidity everything must condense, right?

Edison

 

[Khashishi]

Close, but you should refer to the energy of a particle, not the temperature. The temperature refers to the whole distribution of all the energies of all the particles. Some molecules have more energy than the "work function" of the liquid. The work function is the energy it takes to escape the liquid (or solid). It is the work function, not the billing point, which you should be comparing to at the microscopic level. Humidity>100% means the air is supersaturated with vapor and will condense if there are enough nucleation sites. It doesn't immediately condense.

 

[sophiecentaur]

That article was just bla, bla, bla like most Wikipedia articles.

If you know something, why not simply tell it?

Edison

I have never said I don't like Wiki,

Which of your posts am I supposed to take seriously?
You may have committed a Trumpism.

 

[Edison Bias]

Close, but you should refer to the energy of a particle, not the temperature. The temperature refers to the whole distribution of all the energies of all the particles. Some molecules have more energy than the "work function" of the liquid. The work function is the energy it takes to escape the liquid (or solid). It is the work function, not the billing point, which you should be comparing to at the microscopic level. Humidity>100% means the air is supersaturated with vapor and will condense if there are enough nucleation sites. It doesn't immediately condense.

Another extremely interesting reply, thanks! I thought it was easier to see Ek as a temperature but obvioulsy I was wrong. The bold part is especially interesting for my misunderstanding. And I repeat for I find this profoundly educational: The temperature refers to the whole distribution of all the energies of all the particles.

It is also interesting that (I write this just because I love Tex):

$$n=\int_{-\infty} ^ {\infty} e^{-\frac{mv^2/2}{kT}}dv$$

Which, according to my Plasma Physics student litterature gives the particle density (note that the Maxwellian distribution came from that same book which I however did not consult, I remembered!).

Anyway, n is here the particle density (not the molar density as in p=nRT).

In my mind today I thought very much about this integral, to me it became known as a way of determining N (not n) and thus the number of particles. But n is almost the same, all you have to consider is that it is N/V and thus of more general use.

I really like the concept of "work function", it especially tells me how thermal emission in a tube is done, that is exceeding the work function by heat.

Finally, please explain the last bald sentence. Especially "nucleation sites" is hard to understand.

Edison

 

[houlahound]

Nucleation sites are typically impurities ie a physical scratch or some fur from string or dust that particles can "meet and chillax" and form an "island" and grow on.

I have used all of the above to promote nucleation in crystal growth .

 

[houlahound]

There is a lot of deep physics around nucleation, one can assume I feel the net result is it takes less energy to be together than free.

All I know is dipping some string in some copper sulphate solution or use a dirty beaker causes crystals to start.

Edison when you get bored with Boltzmann/Maxwell move onto Bose and prepare for a good mind blowing.

 

[Khashishi]

In the sky, dust particles act as nucleation sites.

$$n=\int_{-\infty} ^ {\infty} e^{-\frac{mv^2/2}{kT}}dv$$

Something is missing in your equation. I think it should look something like
$n=\frac{n_0}{\sqrt{2\pi kT/m}} \int_{-\infty}^{\infty} e^{-\frac{mv^2}{2kT}}dv$

<Moderator's note: fixed the LaTeX>

 

[houlahound]

Something went horribly wrong in the last post or my cell phone has a virus.

 

[DrClaude]

That integral should definitely not start from -∞

 

[Edison Bias]

In the sky, dust particles act as nucleation sites.

Something is missing in your equation. I think it should look something like
$n=\frac{n_0}{\sqrt{2\pi kT/m}} \int_{-\infty}^{\infty} e^{-\frac{mv^2}{2kT}}dv$

<Moderator's note: fixed the LaTeX>

Thank you for correcting me! I missed a constant in the Maxwellian distribution (MD) to be determined by my integral above, let's say this constant is A then MD' becomes

$$f(v)=Ae^{-\frac{mv^2/2}{kT}}$$

With the use of this along with my integral above the result for A is tedious but

$$A=\frac{n}{\sqrt{2\pi kT/m}}$$

which makes MD' equal to

$$f(v)=\frac{n}{\sqrt{2\pi kT/m}}e^{-\frac{mv^2/2}{kT}}$$

Integrating this with respect for v gives your result. However, n_o is kind of new to me, what does it mean? Is it perhaps the density at Ek_av? Because this distribution just tells that the density lessens for both higher and lower speed where I think speed should be vritten v=v'-v_o where v_o is related to Ek_av(?) I'm sorry, I still don't understand other than perhaps this: while there are fewer particles with a higer/lower speed in the same volume, desity has to be smaller for them, right?

Edison

 

[Edison Bias]

There is a lot of deep physics around nucleation, one can assume I feel the net result is it takes less energy to be together than free.

All I know is dipping some string in some copper sulphate solution or use a dirty beaker causes crystals to start.

Edison when you get bored with Boltzmann/Maxwell move onto Bose and prepare for a good mind blowing.

This is interesting, I will however not ask why due to deep physics, as you say :)

Edison

 

[Edison Bias]

Dust is also required for rain to fall. Additionally, there was a paper from PRL two years back that said that small vortices within clouds literally spin the saturated dust particles into rain drops. I can't find the original publication, but here's an the story about it:

http://physics.aps.org/story/v7/st14

Hi PQ!

I have now read your article, very interesting, short and easy to read too :)

And as Khashishi talks about "nucleation sites" for supersaturated air (humidity>100%) to condense this with dust for rain to condense/fall kind of feel related. Especially when houlahound talks about nucleation sites as impurities. I probably oversimplify too much but all these statements seem related and are very interesting.

Edison

 

[Edison Bias]

Let's consider a related natural phenomena:

If we have a puddle of water and the temperature outside is 300K and the pressure is 1atm, how long will it take for that puddle to dry up i.e vaporize totally if the amount is 1L and humidity is 50%?

Does evaporation depend on shape of water, i.e does it differ in time if the water is inside a tube or in a puddle?

I don't think so, it seems like there's a mass evaporation, regardless of shape, right?

So it seems like:

$$-\frac{dm}{dt}$$

is constant given a fixed temperature and pressure, right?

Let's rewrite that

$$-n\frac{dV}{dt}$$

or better

$$-V\frac{dn}{dt}$$

where n is the density, and the expression is constant.

Now, n is according to PF:

$$n=\frac{n_0}{\sqrt{2\pi kT/m}}\int_{-\infty}^{\infty}e^{-\frac{mv^2/2}{kT}}dv$$

But the temperature is constant so n does not vary even though it is interesting :)

So what makes -dV/dt?

To be continued...

Edison
PS
The only thing I have going for me is the Maxwellian Distribution and I have already calcylated that some 27% reach 400K (Tk) at 300K (Tav) but I kind of understand that that way of thinking is wrong.

Edit: Added humidity=50%, struck me as important before Khashishi told me below but it is because of Khashishi that I understand the importance of this.

 

[Khashishi]

Of course it depends on shape, and on humidity. Evaporation will be faster if the exposed surface area of the water is higher.

 

[Edison Bias]

Here I feel stupid to not understand this because intuition tells me you are right. But why is it shape-dependent? Do I dare to guess? It isn't more photons per area because water evaporates regardless of sunlight (even though sunlight heats the surface and speed up the process but we are talking 300K here). Maybe simply more air-molecules to "connect with" if the area is large? Thus more water molecules evaporates, right?

I have corrected for humidity above.

Edison

 

[Edison Bias]

I am thinking that there should be a change to the density if the fast particles dissappear into evaporation all the time so maybe this relationship holds:

$$A=-\frac{dm}{dt}=-V\frac{dn}{dt} \propto -\frac{dn}{dt}$$

which we could write

$$-A'dt=dn$$

or

$$-A'\int_{0}^{t}dt =\int_{n}^{0}dn$$

where t is the time it takes to evaporate all the water and

$$n=\frac{n_0}{\sqrt{2\pi kT/m}}\int_{-\infty}^{\infty}e^{-\frac{mv^2/2}{kT}}dv$$

which differetiated becomes

$$dn=\frac{n_0}{\sqrt{2\pi kT/m}} \frac{mv}{kT}e^{-\frac{mv^2/2}{kT}}dv$$

or

$$dn=\frac{n_0}{\sqrt{2\pi (kT/m)^3}}e^{-\frac{mv^2/2}{kT}}vdv$$

Sorry for taking up your time

Edison

 

[jbriggs444]

Here I feel stupid to not understand this because intuition tells me you are right. But why is it shape-dependent? Do I dare to guess? It isn't more photons per area because water evaporates regardless of sunlight (even though sunlight heats the surface and speed up the process but we are talking 300K here). Maybe simply more air-molecules to "connect with" if the area is large? Thus more water molecules evaporates, right?

Air molecules are not the answer. Water evaporates in a vacuum too.

The evaporation of water at the surface of a puddle is limited by at least three things: The rate at which water vapor is carried away from the surface (typically by diffusion or convection), the rate at which heat can be supplied to the water surface (typically by conduction or convection) and the rate at which water molecules sufficiently near the surface can randomly attain sufficient kinetic energy to escape.

All three factors increase with increasing surface area.

 

[houlahound]

When you guys say escape I thought vapour pressure was an equilibrium thing where escaping and condensing molecules are on average fixed at constant value all other factors being constant?? That's why vapour pressure is quoted as a fixed value at a specific temp, pressure etc??

 

[Edison Bias]

Air molecules are not the answer. Water evaporates in a vacuum too.

The evaporation of water at the surface of a puddle is limited by at least three things: The rate at which water vapor is carried away from the surface (typically by diffusion or convection), the rate at which heat can be supplied to the water surface (typically by conduction or convection) and the rate at which water molecules sufficiently near the surface can randomly attain sufficient kinetic energy to escape.

All three factors increase with increasing surface area.

I find this answer very interesting but maybe not so educational. I mean that in a nice way such as me not really understanding neither of these three factors. They sound accurate and I believe you but let's look at the three factors:

1) The rate at which water vapor is carried away from the surface (typically by diffusion or convection)
2) The rate at which heat can be supplied to the water surface (typically by conduction or convection)
3) The rate at which water molecules sufficiently near the surface can randomly attain sufficient kinetic energy to escape.

First I obviously need to study diffusion, convection and heat conduction.

But let's look at 1). Can we somehow know the rate? For my puddle, either (not both) diffusion or convection takes place, right? Let's say it's convection, warm fluid then goes up and gives space for colder fluid that goes down and the circle is closed until there's no more fluid. Diffusion is more abstract and a phenomena that might more happen inside the puddle, that is for the high-Ek molecules to actually reach the surface, right?

If we look at 2) heat needs to be supplied to the surface. Once again I think convection is the answer. Feels like 1&2 actually might be the same. Because if both uses convection there is both a carrying away of vapor and a supply of heat.

If we look at 3), this might be related to my precious Maxwellian Distribution but how may that give an answer?

In other words, how may I involve these three things to accuratelly determine the time it takes for my puddle to dry up? (We may suppose that the puddle has a constant area too).

Thank you for your reply and be aware that I really don't know so much about anything!

Best Regards, Edison

 

[jbriggs444]

But let's look at 1). Can we somehow know the rate? For my puddle, either (not both) diffusion or convection takes place, right?

Both take place. Just because the wind is blowing, that does not mean that diffusion ceases to act.

Let's say it's convection, warm fluid then goes up and gives space for colder fluid that goes down and the circle is closed until there's no more fluid.

Convection is more than just warm air rising and cold air sinking. A fan blowing air on your face is an example of convection. A spoon stirring a pot of water on the stove is an example of convection.

Diffusion is more abstract and a phenomena that might more happen inside the puddle, that is for the high-Ek molecules to actually reach the surface, right?

What I had in mind was for the water vapor in an area that is at a saturated partial pressure to diffuse to an area that is at the same total pressure but a smaller vapor pressure fraction.

In other words, how may I involve these three things to accuratelly determine the time it takes for my puddle to dry up? (We may suppose that the puddle has a constant area too).

Measure it or look it up. Do not try to figure it out from first principles.

 

[Edison Bias]

I'm a sucker for Tex, it's kind of hard to code but it sure is fun how nice the formulas look like :)

We have for a constant S

$$B=-S\frac{dm}{dt}=-S\frac{d(Vn)}{dt}=-S(\frac{ndV}{dt}+\frac{Vdn}{dt})=-S(n\frac{dV}{dt}+V\frac{dn}{dt})$$

Let's evaluate:

$$-B/S=(n\frac{dV}{dt}+V\frac{dn}{dt})$$

$$\frac{dV}{dt}=-(B/S+V\frac{dn}{dt})/n$$

which may be written as:

$$\frac{dV}{dt}=-\frac{B}{Sn}-\frac{V}{n}\frac{dn}{dt}$$

that is

$$dV=-\frac{Bt}{Sn}-V\frac{dn}{n}$$

$$t=-\frac{Sn}{B}*(dV+V\frac{dn}{n})$$

I'm struggling with both math and physics here, been 20 years since I did advanced math...

Edison

 

[Edison Bias]

Both take place. Just because the wind is blowing, that does not mean that diffusion ceases to act.

I think I understand this one (my referense of understanding diffusion is what happens inside a semiconductor).

Convection is more than just warm air rising and cold air sinking. A fan blowing air on your face is an example of convection. A spoon stirring a pot of water on the stove is an example of convection.

Convection for me is how an electronic tube is cooled. I have difficulties with the fact that a fan can be an example of convection but if you think about it you are blowing air with a certain temperature (sometimes heated, sometimes cooled) and this makes the molecules just in front of your face move away at the expence of the new molecules from the fan hitting your face, so ok maybe I recognize this but the last example, what is that? Is it the gradient of heat? Because if you stir it, the heat gradient becomes less because you manually move hotter areas together with cooler areas, evening up the temperature and it is this "flow" that may be considered convection? Very interesting examples, thanks!

What I had in mind was for the water vapor in an area that is at a saturated partial pressure to diffuse to an area that is at the same total pressure but a smaller vapor pressure fraction.

Interesting. The bold part is however greek to me.

Edison

 

[jbriggs444]

What I had in mind was for the water vapor in an area that is at a saturated partial pressure to diffuse to an area that is at the same total pressure but a smaller vapor pressure fraction.

Interesting. The bold part is however greek to me.

Chester Miller could do a better job of explaining this, but I'll give it a whack.

You understand that in a mixture of gasses, the pressure of the mixture can be thought of as the sum of the partial pressures of each of the component gasses? The air we breathe has Nitrogen with a partial pressure of 12 pounds per square inch, Oxygen with a partial pressure of 3 pounds per square inch and additional contributions from water vapor, CO₂, Argon and other trace elements. The sum is about 15 pounds per square inch. On a hot and humid day in July the fraction of that pressure from water vapor in the atmosphere is, of course, much larger than the fraction on a cold dry day in December.

A useful point about partial pressures is that (to a good approximation and for most purposes), each gas in the mixture acts as if the other gasses are not there. If a container of water is placed in a chamber from which all the air has been removed, maintained at a fixed temperature and allowed to evaporate into that chamber the water will evaporate only until a certain water vapor pressure is reached. When the water vapor in the chamber is at that pressure, the rate of condensation of water vapor into the water is equal to the rate of evaporation of water into the vapor. The net is a rate of zero. If you perform the same experiment without removing the air first, the same thing happens. The water evaporates only until the partial pressure of the water vapor in the air reaches that same level. When the partial pressure of the water vapor reaches this level, we say that the air in the chamber is "saturated". It has as much water vapor as it can hold. We could also say that it is at 100% relative humidity.

Imagine the puddle on a day with no wind. Water is evaporating from the puddle. If nothing else were happening, the air very near the surface of the water would be getting more and more nearly saturated with water vapor. The atmosphere is very large. The air far from the puddle would remain largely unaffected. The partial pressure of water vapor in the air near the puddle would be high and the partial pressure of water vapor far from the puddle would be low.

The total atmospheric pressure is the same both near the puddle and far away. Remember that we assumed no wind. If there were any pressure imbalance, the air would quickly move to even it out.

So we have a high concentration of water vapor in the air over here. And a low concentration of water vapor over there. What is the process by which concentration gradients even themselves out? Diffusion.

 

[Edison Bias]

I thank you very much for this long and tedious explanation for me!

Very interesting and educational!

I didn't know that the partial pressure was different between Nitrogen and Oxygen in air, very interesting (at the same time, what is pressure then? P=nkT, with different n, or?)

And I now think I understand some of what diffusion is, given no wind the concentration gradients still will even out, right? But why/how?

Now I "just" need to know how to calculate with it.

I love calculations, math is the universal language and lessens the number of words needed to explain, if you understand math, that is (which I kind of did once upon a time).

I dream of a Wikipedia 2.0 where there are almost only math formulas becase now there's just too much to read and too many fancy words, it doesn't really add to once understandig to hear or learn what a certain thing is called. The only benefit from that is ease of human communication. But while there is LaTex, we can communicate much more efficiently and at the same time understand beyond words.

Edison

 

[jbriggs444]

I didn't know that the partial pressure was different between Nitrogen and Oxygen in air, very interesting (at the same time, what is pressure then? P=nkT, with different n, or?)

Just measure n as moles per liter. It doesn't matter what it's a mole of. Oxygen molecules, Nitrogen molecules, Helium atoms, Hydrogen molecules, water molecules, whatever. Just add them all up to get n as a total moles of whatever per liter. total P = total n times kT. Or don't add them up. Figure out P = nkT for each component gas and add up all the computed partial pressures. That's the magic of the distributive property of multiplication over addition.

I may have been a bit off on the 80/20 split for Nitrogen and Oxygen in the atmosphere. That's the approximate percentage breakdown by mass. The pressure ratio would be tilted a bit more in favor of Nitrogen since Nitrogen molecules are lighter than Oxygen by a ratio of 28:32. More nitrogen molecules per unit mass means more nitrogen molecules. More nitrogen molecules means more pressure. P = nkT after all.

 

[Edison Bias]

Interesting but you must mean "Just add them all up to get n as a total moles of whatever per liter", and not k, right? :)

Edison

 

[Khashishi]

Can you define your variables in your equations?

 

[Edison Bias]

I must understand you wrongly, a skilled guy like you can't be interested in my equations?

I really do not know what I'm doing :)

Repeating for convenience:

$$t=-\frac{Sn}{B}*(dV+V\frac{dn}{n})$$

where:

S=Surface of puddle
B=A constant depending on mass-escape rate and S
V=Volume of puddle
n=Density of puddle (which in my ignorant world should be constant)
t=Integrated time constant for the puddle to evaporate totally.

Best reagards, Edison

 

[jbriggs444]

Sanity check: The greater the surface area, the longer the puddle takes to evaporate? That does not ring true.

 

[Khashishi]

You can't have differential expressions like dV and dn without an integral or derivative.
Or, is d a variable?

 

[Edison Bias]

Sanity check: The greater the surface area, the longer the puddle takes to evaporate? That does not ring true.

I saw that too :D

Edison

 

[Edison Bias]

You can't have differential expressions like dV and dn without an integral or derivative.
Or, is d a variable?

Let's move up a notch from my final and faulty equation above:

We then have:

$$\frac{dV}{dt}=-(\frac{B}{Sn}+\frac{V}{n}\frac{dn}{dt})$$

Then I just thought that I multiply with dt from the left and from the right and integrate B/(Sn) which gives the above equation.

But looking at this equation is kind of interesting to me.

The rate of volume change is partly proportional to B and inversely proportional to the area (S) as well as the density.

Viewing this again we may say that a large surface gives a slow volume change (strange) and a high density gives a slow volume change (more particles to escape per area, perhaps?)

Looking at the other part we have that the rate of volume change is high if original volume is high (strange) or the density is low (few particles, perhaps?) and the rate of density change is high (no clue here).

I'm sorry for wasting your time!

Edison

 

[Khashishi]

My intuition tells me that the evaporation rate is proportional to surface area and the difference between the vapor pressure and partial pressure of vapor in the air.
$\frac{dN}{dt} = (P_0-P)CS$
where $N$ is the amount, $P_0$ is the vapor pressure, $P$ is the partial pressure, $C$ is some function of temperature and flow conditions, and $S$ is the surface area.

Of course, evaporation also depends on airflow. You avoid wikipedia because it's too detailed, but then you are getting into that level of detail now, so you should read the wikipedia page.

 

[Edison Bias]

I hear you, but I prefere trying to understand on my own with your kind help because forums like this is fun and gives my life meaning (studying by my own without a mentor to ask questions is far from that joyful and meaningful).

It struck me on the bus to get some more beer that my equation should read:

$$\frac{dV}{dt}=-\frac{1}{n}(BS+V\frac{dn}{dt})$$

due to

$$BS=-\frac{dm}{dt}$$

I stated that formula wrong.

Then we have that the rate of volume change is high when S is high (correct, right?), the rate is also high when n is low (sounds right due to few particles, right?), the rate is also high when the volume is high (strange) and the rate of density change is high (strange).

Edison
PS
Thank you Khashishi for that formula. And I'm learning as I go, which I like very much. Refreshing my old mathematical skills and perhaps even understanding some physics more than I thought I understood.

 

[Edison Bias]

Let's say

$$V=V_oe^{jwt}$$

and

$$n=n_0e^{jwt}$$

then my equation becomes

$$jwV=-(\frac{BS}{n}+Vjwn)$$

which gives

$$jwV(1+n)=-\frac{BS}{n}$$

taking the imaginary part gives

$$w=-\frac{BS}{nV(1+n)}$$

or

$$w=-\frac{BS}{V(n+n^2)}$$

The question now is what w is :D

This didn't work, but I promise you it solves many differential equations where you have an oscillation or gyration very easily!

Berst regards, Edison