Question about water drop's critical radius

In summary, the conversation revolved around solving an exercise from a book on cloud physics. The exercise involved showing that the vapor pressure in equilibrium over a pure water drop decreases with temperature, given a specific condition. The equation used in the exercise was derived from the equation of vapor pressure of saturation and a plain surface. The conversation also delved into the values and dependencies of different variables in the equation. Ultimately, it was suggested to use the Clausius-Clapeyron equation to solve the problem.
  • #1
Frank Einstein
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1

Homework Statement


Hi everybody. I am currently trinyg to solve the first exercice of the fifth chapter of R.R. Rogers book about cloud physics.

"Show that the vapor pressure in equilibrium over a pure water drop of radius r decreases with T if r<2σ/LρL.

Homework Equations



es(r)=es(r=infinite)*(2σ/rRvρLT).

This equation relates the vapor pressure of saturation of a drop of radius r and the one from a plain surface.
es is the vapor pressure, r is radius, σ is surface tension, Rv is the gas constant, ρL is density and T is temperature.

3. The Attempt at a Solution

As far as I understand, I have to derivate es(r) respect to T and see if the result is similar to the crtical radius which value is 2σ/RvρLTlnS.

If I have understood correctly, none of the values in the expression of es depends on T, therefore d/dT=∂/∂T. If I derivate exp(a/T) I obtain -a*exp(a/T)/T2. This expression doesn't look like the one expressed in the upper paragraph.
Note: a=2σ/rRvρL

If someone could give me some guidance I would be extremley grateful.
 
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  • #2
Frank Einstein said:
r<2σ/LρL
What is L there (not the subscript, the other one)?
Frank Einstein said:
none of the values in the expression of es depends on T
Are you saying svp does not depend on temperature? That would be a controversial claim.
For the question to make sense es(∞) must depend on temperature. You need to get two terms from the differentiation, one positive, one negative. Which wins will depend on whether r is above or below a threshold.
 
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  • #3
haruspex said:
What is L there (not the subscript, the other one)?

Are you saying svp does not depend on temperature? That would be a controversial claim.
For the question to make sense es(∞) must depend on temperature. You need to get two terms from the differentiation, one positive, one negative. Which wins will depend on whether r is above or below a threshold.
haruspex said:
What is L there (not the subscript, the other one)?

Are you saying svp does not depend on temperature? That would be a controversial claim.
For the question to make sense es(∞) must depend on temperature. You need to get two terms from the differentiation, one positive, one negative. Which wins will depend on whether r is above or below a threshold.

L is the latent heat of evaporation.

In the book I can't find the value of es(r=infinite) so I just assumed that it didn't change.
 
  • #4
Frank Einstein said:
L is the latent heat of evaporation.

In the book I can't find the value of es(r=infinite) so I just assumed that it didn't change.
Well, it must change, but I don't understand this equation:
Frank Einstein said:
es(r)=es(r=infinite)*(2σ/rRvρLT).
Where did you get that from? It looks extremely unlikely to me. As r tends to infinity, the factor on the right after the asterisk should tend to 1.
 
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  • #5
haruspex said:
Well, it must change, but I don't understand this equation:

Where did you get that from? It looks extremely unlikely to me. As r tends to infinity, the factor on the right after the asterisk should tend to 1.

Is the equation 6.1 from the book A Short Course in Cloud Physics. It is in page 84. If you wish, you can find it here:
https://es.scribd.com/doc/293198948/A-Short-Course-in-Cloud-Physics-pdf
 
  • #6
Frank Einstein said:
Is the equation 6.1 from the book A Short Course in Cloud Physics. It is in page 84. If you wish, you can find it here:
https://es.scribd.com/doc/293198948/A-Short-Course-in-Cloud-Physics-pdf
That seems to be behind a paywall.
Can you just post an image of that page?
 
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  • #8
Frank Einstein said:
It seems I can't post it directly, I will post the link.
https://postimg.org/image/4rzbx74u5/[PLAIN]https://postimg.org/image/4rzbx74u5/[/QUOTE]
That makes more sense. You left out the exp. It says

es(r)=es(∞)e2σ/rRvρLT.
 
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  • #9
haruspex said:
That makes more sense. You left out the exp. It says

es(r)=es(∞)e2σ/rRvρLT.
I'm sorry. I didn't realize. I posted it in a hurry and I didn't realize. I apollogize for the inconvenience I have caused you.

But still I have the same problem that I started with. In the attempted solution, I derivate the propper expression and I don't find the anwser I seek.

Then If I have understood you correctly, I have to seek the expression of es(r=infinite) and try to add it to the derivate to try to obtain the expression of the crytical radius?
 
  • #10
Frank Einstein said:
I have to seek the expression of es(r=infinite) and try to add it to the derivate to try to obtain the expression of the crytical radius?
You would have to apply the product rule.
I tried working backwards from the thing to be proved and arrived at ##e_s(\infty)=Ae^{-\frac L{R_vT}}## for some constant A. I must say the curves I've seen for svp do not look much like that.

Edit: but it seems like it might follow by combining Clausius-Clapeyron with PV=nRT.
In the C-C equation, I guess you can take Δv as the V of the gas, since the liquid phase has almost no volume in comparison.
 
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  • #11
haruspex said:
You would have to apply the product rule.
I tried working backwards from the thing to be proved and arrived at ##e_s(\infty)=Ae^{-\frac L{R_vT}}## for some constant A. I must say the curves I've seen for svp do not look much like that.

Edit: but it seems like it might follow by combining Clausius-Clapeyron with PV=nRT.
In the C-C equation, I guess you can take Δv as the V of the gas, since the liquid phase has almost no volume in comparison.

After some more thoght I think I can use Classius Clapeyro equation to solve this problem.

Thak you very much for the help
 
  • #12
Frank Einstein said:
After some more thoght I think I can use Classius Clapeyro equation to solve this problem.

Thak you very much for the help
You are welcome.
I wrote earlier that the shape of y=e-1/x seemed wrong for the svp curve, but later I looked more carefully and realized it looks right in the small x region. The larger x regions are probably beyond the range of interest for engineers, so do not appear in published svp curves.
It is quite an interesting curve. Zero gradient at the origin, curving swiftly upwards, in a roughly circular arc, through a point of inflection, then levelling off to a horizontal asymptote.
 

Related to Question about water drop's critical radius

1. What is the critical radius of a water drop?

The critical radius of a water drop is the minimum size that a drop of water can be before it will spontaneously evaporate into a vapor. This varies depending on factors such as temperature, humidity, and atmospheric pressure.

2. How is the critical radius of a water drop calculated?

The critical radius of a water drop can be calculated using the Kelvin equation, which takes into account the surface tension of the water and the vapor pressure of the surrounding atmosphere. It can also be experimentally determined by measuring the evaporation rate of different sized drops.

3. What happens to a water drop when it reaches its critical radius?

When a water drop reaches its critical radius, it will begin to evaporate and shrink in size. Eventually, it will disappear completely as it turns into a vapor. This process is known as evaporation.

4. Can the critical radius of a water drop be changed?

Yes, the critical radius of a water drop can be changed by altering the surrounding conditions. For example, increasing the temperature or decreasing the humidity can decrease the critical radius, while decreasing the temperature or increasing the humidity can increase the critical radius.

5. Why is the critical radius of a water drop important?

The critical radius of a water drop is important because it determines the stability of the drop. Drops smaller than the critical radius will spontaneously evaporate, while drops larger than the critical radius will continue to grow through condensation. It also plays a role in atmospheric phenomena such as cloud formation and precipitation.

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