B Understanding the Science Behind Water Vapor and Cloud Formation

[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.