A simple sublimation problem

[Yoruichi-san]

You have a spherical mothball which sublimes at a rate of k cm³/s per cm² of exposed solid surface (assuming constant T and P in the room).

When will it be at half its original radius? When will it have completely sublimed?

[jim]

You would need to know the initial radius R. Simplifying your rate you get k cm/s as the rate at which the radius decreases so it would take 0.5R/k to get to half the radius and R/k to sublime completely.

[Yoruichi-san]

It should be apparent that a dimensional analysis is not sufficient. k is dependent on the material the object is made of, and T and P, not geometry. However, the sublimation rate is highly dependent on geometry; geometries with more surface area/volume ratio sublime at a higher rate.

Try a right circular cylinder with height = 2R, made of the same material at the same T and P.

Yes, you can assume initial conditions of R₀, t₀, etc.

[bushindo]

Try a right circular cylinder with height = 2R, made of the same material at the same T and P.

Yes, you can assume initial conditions of R₀, t₀, etc.

I get the same results with a right cylinder as with the sphere

We first assume that the right cylinder holds its cylindrical shape as it sublimates. The Area and Volume of the cylinder are

A = 6 pi r²

V = 2 pi r³

The given differential equation is

dV/dt = k * A = k * ( 6 pi r² )

We can compute dV/dr as

dV/dr = 6 pi r²

We convert dV/dt into dr/dt using the above, and we get

dr/dt = k

So the change in radius with respect to time is a constant k, which implies that the half-sublimated and fully-sublimated times are (1/2)* R₀/k and R₀/k, respectively.

[Yoruichi-san]

I get the same results with a right cylinder as with the sphere

We first assume that the right cylinder holds its cylindrical shape as it sublimates. The Area and Volume of the cylinder are

A = 6 pi r²

V = 2 pi r³

The given differential equation is

dV/dt = k * A = k * ( 6 pi r² )

We can compute dV/dr as

dV/dr = 6 pi r²

We convert dV/dt into dr/dt using the above, and we get

dr/dt = k

So the change in radius with respect to time is a constant k, which implies that the half-sublimated and fully-sublimated times are (1/2)* R₀/k and R₀/k, respectively.

That was the derivation I was looking for.

What happens when we increase the height of the cylinder to H=3R or decreasing to H=R, or general nR, or assume the cylinder is sitting on a base so that it is not exposed?

Edited August 20, 2012 by Yoruichi-san

[jim]

You don't need calculus. We just need to know the area of the part of the solid at least t / k cm away from an exposed surface. The radius decreases at the rate of t / k cm per second. The upper surface drops at the same rate, and so does the bottom surface except in the case where the bottom is not an exposed surface. As long as the height is at least 2R(or R if the bottom is not exposed) then the radius decreases linearly down to zero. Otherwise the height goes to zero while the radius approaches a positive limit.

In your first point the limit is a point. With height 3R it approaches a limit that is a line of height R. If the height is R it takes half the time to disappear, and the limit is a 2D disc of radius R, etc.

[Yoruichi-san]

Um...that is calculus ;P.

Calculus is more than just equations with "d/dt" or the integral of something. It's the concepts of rate of change and limits etc. It's from these concepts that the equations were based on, but you don't need to use the equations to be using calculus, just like, say, you're using algebra when you try to figure out how many cobs of 25 cent corn you can buy for $2 even if you don't write the equation for it.

[beppe]

to find the solutions requires some calculus, albeit basic one:

for the sphere to melt down to half its radius requires t = (7/6)*(pi/k)*r³ seconds

for the sphere to melt altogether requires t = (4/3)*(pi/k)*r³ seconds

[mmiguel]

You have a spherical mothball which sublimes at a rate of k cm³/s per cm² of exposed solid surface (assuming constant T and P in the room).

When will it be at half its original radius? When will it have completely sublimed?

Radius at time t in cm = r(t)

dV/dt = -k*A(t)

A(t) = 4*pi * r(t)^2

V(t) = 4/3*pi*r(t)^3

dV/dt = 4*pi*r(t)^2 *dr/dt = A(t)*dr/dt = -k*A(t) = -k*4*pi*r(t)^2

dr/dt = -k

r(t) = -kt + r(0)

Half sublimed is

r(0)/2 = r(0) - kt

t = r(0)/(2k)

Fully sublimed is

0 = r(0) - kt

t = r(0)/k