Falling Raindrop - Evaporation and - Diff Eq project

[Mobilizer]

I'm working on a project where I'm trying to find the differential equation of a falling raindrop. The drop I'm considering will be under 10mm to eliminate the drops breaking apart and I've found that for that size, the drops will be spherical and traveling 6-8m/s. I've determined my air resistance to be F = =bv$$^{2}$$ because of the size of raindrops being molded.
So far my equation is simple with ma + bv$$^{2}$$ = mg

Next I was trying to figure out the part for evaporation. The raindrop will be decreasing in size due to evaporation but and also it's velocity will be changing too due to a smaller size; and here's where I got stuck.

What would be the best way to go about this?

 

[gato_]

the problem will be harder the more detail you get into, but roughly, both mass and friction coefficients will vary in time because of the change in volume/geometry of the drop. You need to model too the process of evaporation. The rate is constant in time? does it depend on velocity? (on account of the drop heating), and add a new equation for the rate of volume/friction area loss rate.
The simplest approximation I can think about (may be pretty unaccurate) is: consider the friction to be given, over the rate of velocities considered, by
$$F_{D}=1/2\rho A C_{D}v^{2}$$
where A is the cross area of the drop, and Cd a drag coefficient. Let us consider it constant, and the drop, to keep a spherical shape (again, unaccurate) of volume $$4/3\pi r(t)^{3}$$ and cross section $$A=4\pi r(t)^{2}$$. You still need to specify the rate at which r changes. Let's imagine it loses mass at a constant rate q:
$$\dot{m}=\rho 4/3\pi \dot{r^3}=q$$
you would now have two equations for variables (r(t),v(t)).

 

[Mute]

Also note that the net force starting point should be

$$\frac{dp}{dt} = \frac{d(m(t)v(t))}{dt} = \sum F$$

as the mass is changing. (Or alternatively you could still say the net force is ma, but there is now an additional force on the drop due to the changing mass).

 

[gato_]

True. Also, I think it is better to consider evaporation rate to be proportional to the surface area rather than constant, the reason being only molecules in the interface can evaporate. That results in a linearly varying radius with time.