Spirit evaporating from a bowl

[jbriggs444]

Indeed, and this is true regardless of the shape of the container.

Barring pathological shapes, of course.

 

[brotherbobby]

I respond not only to @Orodruin but others in this thread too. While the problem is solved, a small curiosity has brought something else to light. Let me begin by refreshing your mind with what exactly is going on.

Problem Statement :

Additional question : How much time is needed for the bowl to become empty of the spirit?

Attempt : I repeat my argument as in post# 29 above. An infinitesimal volume of the spirit, $dV = A(z) dz$, where $z$ is the (vertical) height of the fluid at that instant (clearly $z=z(t)$). The expression for

the elementary volume implies that whatever be the shape of the container, it is locally a cylinder (or a prism if it has straight edges). Thus $\dfrac{dV}{dt}=A(z)\dfrac{dz}{dt}$, but since $\dfrac{dV}{dt} = -k A(z)\; (k>0)$ from the problem statement, we have $\dfrac{dz}{dt} = -k\mathbf{\;\cdots (1)}$, a constant.

This led to the answer, but also a curiousity.

If the shape is that of a cone which I have shown to the right, we have at an instant $t$, $V(t) = \frac{1}{3}\pi r^2(z) z$, where $r(z)$ is the radius of the surface of the spirit at that instance. Then, $\dfrac{dV}{dt} = \dfrac{1}{3}\cancel{\pi}\left( 2r \dfrac{dr}{dt}z+r^2\dfrac{dz}{dt} \right) = -k\cancel{\pi}r^2\Rightarrow 2r\dot rz- r^2k=-3kr^2$, since $\dot z= -k$. This simplifies to $\dot r z=-kr\Rightarrow \dot r=-k\dfrac{r}{z}=-k\tan\phi$, where $\phi$ is the semi-vertical angle of the cone.
Thus $\dfrac{dr}{dt} = -k\tan\phi\; \mathbf{\cdots (2)}$
Hence we find that just as $\dfrac{dh}{dt} (=-k)$ was a constant, so is $\dfrac{dr}{dt}$, for a cone.
(It remains to be seen, and I intend to settle the matter presently, whether $\dfrac{dr}{dt}$ would continue to be constant for, let's say, a hemispherical shape also.)

So, to answer the question above, how much time is required by the evaporating spirit to drain?

We can find that from either of the equations (1) or (2) above, in blue. Integrating (1) $\dfrac{dz}{dt} = -k\Rightarrow z(t) = -kt+H_0$, where $H_0 = z(0)$. If $T_D$ is the total time to drain, then $z(T_D)=0\Rightarrow kT_D=H_0\Rightarrow \boxed{T_D = \dfrac{H_0}{k}}$.

The same can be checked integrating (2). We find $r(t)= -k\tan\phi t+R_0$, where $R_0$ is the radius of the spirit at time $t=0$, or $r(0) = R_0$. When the liquid has drained, $r(T_D) = 0\Rightarrow k\tan\phi T_D = R_0$. This leads to $T_D = \dfrac{R_0}{k\tan\phi} = \dfrac{R_0}{k\frac{R_0}{H_0}}=\boxed{\dfrac{H_0}{k}}$.

I believe I am right as the answers match. However, I am open to comments regarding the constancy of $\dfrac{dr}{dt}$.

To summarise, when liquid evaporates from a shape proportional to the surface area of the liquid, , irrespective of what shape it is, $\boldsymbol{\dfrac{dz}{dt} = \;\text{constant}}$, i.e. the height decreases at a constant rate.. Additionally, if the shape is conical, $\boldsymbol{\dfrac{dr}{dt} = \;\text{constant}}$, i.e., the radius (and surface area) decrease at constant rates too.

Thank you for your interest.

 

[hutchphd]

However, I am open to comments regarding the constancy of drdt.

This is true but (if I may be blunt) neither requested nor relevant to the question. I would not include it in a formal solution to the problem as stated.
If doing each shape is useful practice for you then I recommend it. But the solution is for the general case and therefore is complete.

 

[Orodruin]

That radius decreases linearly for a cone is a trivial result since the radius is directly proportional to the height and the height decreases linearly. This will no longer be true for other shapes. The general statement is just the chain rule
$$
\frac{dr}{dt} = \frac{dr}{dz} \frac{dz}{dt}
= -k r’(z).
$$
Thus, $dr/dt$ is constant only if $dr/dz$ is, ie, a cone.