How Is Virga Raindrop Evaporation Time Calculated Using Differential Equations?

[pociteh]

Hi,


I was given several applied differential equation problems (which are ungraded), and I have trouble solving 4 of them. This time I'm going to ask about 1 first though (and after that I'll try solving the other 3 again on my own). Help (even hint) is very much appreciated. Here it is:

In very dry regions, the phenomenon called Virga is very important because it can endanger aeroplanes. [See http://en.wikipedia.org/wiki/Virga" ]

Virga is rain in air that is so dry that the raindrops evaporate before they can reach the ground. Suppose that the volume of a raindrop is proportional to the 3/2 power of its surface area. [Why is this reasonable? Note: raindrops are not spherical, but let's assume that they always have the same shape, no mater what their size may be.]

Suppose that the rate of reduction of the volume of a raindrop is proportional to its surface area. [Why is this reasonable?]

Find a formula for the amoung of time it takes for a virga raindrop to evaporate completely, expressed in terms of the constants you introduced and the initial surface area of a raindrop. Check that the units of your formula are correct. Suppose somebody suggests that the rate of reduction of the volume of a raindrop is prpoportional to the square of the surface area. Argue that this cannot be correct.

What I've come up with so far is just:

V(t) = k_{1} A(t)^{3/2}

where V is the volume and A is the surface area of the raindrop, and

\frac{dV}{dt} = -k_{2} A(t)


Please help.. thanks!

 

[HallsofIvy]

In our first equation, surely you meant
\frac{dV}{dt}= -k_1A(t)^{3/2}

Since, for any sphere, the volume is given by V= (4/3)\pi r^3 and the surface area by A= 4\pi r^2, A^{3/2}= (4 \pi)^{3/2} r^3 and so "proportional to surface area to the 3/2 power" is the same as "proportional to the volume" which is, at least, "reasonable" since many things decay that way: for example, radioactive substance decay that way.

Volume decreasing in proportion to surface area is also reasonable (perhaps more reasonable) since raindrops decrease by water evaporating into the air which can only happen at the surface.

You have
\frac{dV(t)}{dt}= -k_1 A(t)^{3/2}
and
\frac{d(t)}{dt}= -k_2 A(t)[/itex]<br /> Now replace V(t) by (4/3)\pi r(t)^3 and A by 4\pi r(t)^2 in each of those so you have equations for r as a function of t.

 

[Rainbow Child]

@ HallsofIvy I think that the 1st equation pociteh wrote is correct since

...that the volume of a raindrop is proportional to the 3/2 power of its surface area...

or am I missing something?

Thus he can substitude V(t) from the 1st equation to the 2nd and solve the resulting ODE for A(t)