B Understanding the Science Behind Water Vapor and Cloud Formation

[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

 

[jbriggs444]

Let's say

V=V_oe^{jwt}

That assumes that the incremental volumetric decay rate of a puddle scales linearly with current puddle volume. But that's patently false.

 

[Khashishi]

You need to stop making up equations willy nilly and start using your intuition to form the equations. Where are you getting these equations from? How can make up an equation and not know what the variables mean? Anyways, you can find the right answer if you do some research. You could write down random equations all day and not get anywhere if you don't have a physical understanding of what's going on.

 

[Edison Bias]

I got it now!

From the beginning while I think it is important to actually see the derivation of equations, this makes up for lots of reading if you understand the differential equation and thereby the formula and the way it was created.

So seriously now, let's begin all over:

-BS=\frac{dm}{dt}

thus the decay of mass is proportional to the surface (S) times a constant (B).

Putting Vn=m gives

-BS=n\frac{dV}{dt}+V\frac{dn}{dt}

Solving for dV/dt gives

-\frac{dV}{dt}=\frac{BS}{n}+\frac{V}{n}\frac{dn}{dt}

Now to the fun part, multiplying both sides with dt gives

-\int_{V}^{0}dV=\int_{0}^{t}\frac{BS}{n}dt+\int_{n}^{n_0}V\frac{dn}{n}

where n_0 came to me due to the fact that ln(0) simply don't exist, so I thought for a while then it hit me, 0 should be n_0 as in the ambient particle densirty, eureka!

Solving the eqaution gives:

V=\frac{BSt}{n}-V(ln(n)-ln(n_0))=\frac{BSt}{n}-Vln(\frac{n}{n_0})

so t is then

t=\frac{nV}{BS}(1+ln(\frac{n}{n_0}))

here it is tempting to say that V/S=h while S disappears from the formula

t=\frac{nh}{B}(1+ln(\frac{n}{n_0}))

But B still sits there and B is related to S so S has not fully disappeared. Sadly though, B needs to be determined to get a proper t.

If any part of this formula is correct, the time for my puddle to completelly dry up is proportional to particle density (n, seems right), hight (h, also seems right) and reversely proportional to B (higher S, lower B I think for a constant dm/dt anyway, so this also seems right) and then logaritmically proportional to particle density (n) over ambient density (n_0) which we in this case also could see as particle pressure P=nkT devided by ambient pressure P_0=n_0kT due to the actual division, does not seem all that wrong, or is it?

In other words my solution may be rewritten as:

t=\frac{Ph}{kTB}(1+ln(\frac{P}{P_0}))

Am I somewhere in the vicinity of the truth?

Edison

 

[jbriggs444]

So seriously now, let's begin all over:

-BS=\frac{dm}{dt}

thus the decay of mass is proportional to the surface (S) times a constant (B).

Putting Vn=m gives

-BS=n\frac{dV}{dt}+V\frac{dn}{dt}

Seriously? You are going to approach the evaporation of a puddle of water by accounting for the changing density of the liquid water in the puddle? [n here is, I believe, the density of the fluid expressed, for instance, in kilograms per liter].

 

[Edison Bias]

You need to stop making up equations willy nilly and start using your intuition to form the equations. Where are you getting these equations from? How can make up an equation and not know what the variables mean? Anyways, you can find the right answer if you do some research. You could write down random equations all day and not get anywhere if you don't have a physical understanding of what's going on.

I am starting to lose my respect for you. I don't say that to be mean but "making up equations willy nilly", what do you mean by that? I am struggling with both basic physics which I know I can derive from the back of my head as well as kind of advanced math that I at least some 20 years ago knew very well. I have made several stupid calculations in this thread but that's just me getting back into the game. I don't need to study anymore, I'm a Master of Science and it's hightime I use that knowledge!

Edison
PS
I thought up the initial differential equation for the mass decay rate almost by my own (with some tip from you skilled guys, which is why I'm here).

 

[sophiecentaur]

I am starting to lose my respect for you. I don't say that to be mean but "making up equations willy nilly", what do you mean by that? I am struggling with both basic physics which I know I can derive from the back of my head as well as kind of advanced math that I at least some 20 years ago knew very well. I have made several stupid calculations in this thread but that's just me getting back into the game. I don't need to study anymore, I'm a Master of Science and it's hightime I use that knowledge!

Edison
PS
I thought up the initial differential equation for the mass decay rate almost by my own (with some tip from you skilled guys, which is why I'm here).

The process that you are describing here - the way you are trying to get to grips with this - is risking losing the 'respect' of PF. You are 'thinking aloud' and 'washing your dirty linen in public' (to use common phrases). What you should be doing is working on your own and reading stuff, with the occasional visit to PF for incremental help with concrete ideas. How can we know which of your random jottings are really though out and which ones are just 'straws in the wind'?
It matters little what your qualifications are. We respond to what you write. (And many of us are in our waning years, too.)

 

[Edison Bias]

Seriously? You are going to approach the evaporation of a puddle of water by accounting for the changing density of the liquid water in the puddle? [n here is, I believe, the density of the fluid expressed, for instance, in kilograms per liter].

Thank you for your answer!

Well it sounds strange that the density of water really changes but as I see it, high speed water molecules over a certain Tk evaporates and this means that there are less molecules per unit volume left but this goes on all the time and perhaps the density change is "digital" i.e there is a certain number of molecules that evaporate but in that "cell" there is always the same amount of molecules. Because why should a small puddle evaporate differently from a large? So there is a constant decay rate of mass (as I have stated). I thereby think that there is a density change but that that change is constant and proportional to Tk/T_av.

Even though you kind guys have taught me that Tk really is not correct, I like the concept because it makes more sense. Tk for me is the "equivalent" kinetic temperature derived from mv^2/2=kT. With respect to the Maxwellian Distribution this means 1/e=37% of available velocities. Tk then is used for lower probabilities but higher velocities in particular "boiling temperature".

Edison
PS
It is very ok if you direct me to a usable solution.

 

[Doc Al]

I think it's time to close this thread.