# Homework Help: Sherwood number of evaporating spherical droplet

1. Nov 23, 2014

### Maylis

1. The problem statement, all variables and given/known data
From Fick's Law of Diffusion, $N_{A}C_{B} - N_{B}C_{A} = -DC \frac {dC_{A}}{dz}$

Prove that Sh=2 for a volatile spherical droplet evaporating in a quiescent atmosphere. (Sherwood number for a sphere is given by $Sh = \frac {k_{A}d}{D}$, where d = initial droplet diameter). At large distances from the sphere the concentration of the volatile component is zero. The saturated vapor pressure is negligibly small compared to the atmospheric pressure.
2. Relevant equations

3. The attempt at a solution
So I know $P_{atm} \gg P^{*}$, and $C_{A,\infty} = 0$. But how the heck can I use the equation to prove this relationship? I don't understand what the equation means. Is $C_{B}$ the concentration of air molecules in the spherical droplet? I would think the first term is just approximated as zero.

2. Nov 23, 2014

### Staff: Mentor

Would you believe that this is the same problem as the heat transfer problem you just solved? In the heat transfer problem, h = k/a= 2k/d. So the Nussult number is Nu = hd/k = 2.

Chet

3. Nov 24, 2014

### Maylis

Wow, I would have never believed it. I did the derivation pretty much identical to that heat transfer problem, and got it.

Here is a follow up question

A volatile sphere has radius 0.01 m, molar density 5.0 kmol.m-3, and its saturated
vapor pressure at 20 C is 30 Pa. The time for it to evaporate at 20 C and 1 bar
pressure in still air is 32 days. Find the diffusivity of the vapor through the stagnant
air.

So, I know 30 Pa << 1 atm, so this should be basically a continuation of the previous problem. So
$Sh = 2 = \frac {k_{A} d}{D}$, and I am looking for D. However, how can I go about calculating $k_{A}$, since it is an unknown?

I have $V \frac {dC_{A}}{dt} = -k_{A}AC_{A}^{*}$.
$$\frac {4}{3} \pi r^3 \frac {dC_{A}}{dt} = -k_{A} 4 \pi r^2 C_{A}^{*}$$
$$\frac {r}{3} \frac {dC_{A}}{dt} = -k_{A}C_{A}^{*}$$
So I will say $\frac {dC_{A}}{dt} = \frac {5 - 0}{32 \times 24 \times 3600} = 1.81 \times 10^{-6}$, since it starts at 5 kmol/m^3, but after it evaporates it will have zero concentration. I guess this process is so slow that it is assumed to be steady state. so I calculate $k_{A} = 2 \times 10^{-10}$, then plug back in and get $D = 2 \times 10^{-12} \frac {kmol}{m^{2} s}$. Although, I still have that negative sign in here that shouldn't be there...

Last edited: Nov 24, 2014
4. Nov 24, 2014

### Staff: Mentor

Regarding the negative sign, I guess the equation for dC/dt should have 0-5, rather than 5-0. The units on the diffusion coefficient look wrong. They should be m2/s. I haven't checked your arithmetic, but the diffusion coefficient looks way low (if it were in the correct units).

Chet

5. Nov 24, 2014

### Maylis

That is a good point, so then the units of $k_{A} [=] \frac {kmol \hspace{0.05 in} s}{m \hspace{0.05 in} kg}$, but then multiplying by the diameter means D has units of $\frac {kmol \hspace{0.05 in} s}{kg}$, so I can't get rid of this mole term, and without a specified species it shouldn't have anything to do with the molar mass.

6. Nov 24, 2014

### Staff: Mentor

Maylis,

I don't think you set up this problem correctly. The left side of the equation should read CdV/dt, where C is the molar density of the drop, which is not changing. The units of kA should be m/s. I assume you calculated C* correctly. You are solving for the radius of the drop as a function of time.

Chet

7. Nov 24, 2014

### Maylis

Thanks a ton, I now get on the order of 10^-5, which sounds much better