Spirit evaporating from a bowl

[brotherbobby]

Homework Statement

Spirit in a bowl evaporates at the rate that is proportional to the surface area of the liquid. Initially, the height of the liquid in the bowl is $H_0$. It becomes $\dfrac{H_0}{2}$ in a time $t_0$. How much more time is needed for the height of the liquid to become $\dfrac{H_0}{4}$?
$\mathbf{\text{Answer =}}\;\; \boxed{\boldsymbol{t_0/2}}$

Relevant Equations

1. The area of the (circular) surface of radius $R(t)$ at an instant $t$: $A(t) = \pi R^2(t)$
2. [I do not know how to connect the height $H(t)$ of the water surface to the radius : $H(t) \stackrel{?}{=} R(t)$]

Problem statement : I draw the problem statement above. I hope I am correct in inferring that the bowl is hemispherical.

Attempt : I could not attempt to the solve the problem. We are given that the rate of change (decrease) in volume is proportional to the surface area : $\dfrac{dV}{dt}\propto A$. The surface area $A = \pi R^2$. The problem is, how to find an equation for the volume of the spirit as a function of radius $R$ and height $H$ : $V \stackrel{?}{=} V(R,H)$.

A hint or suggestion would be most welcome.

 

[Orodruin]

The shape of the bowl is irrelevant. Ask yourself how dh/dt depends on dV/dt.

 

[Delta2]

Well the easy stuff is to do it for a cylindrical bowl, indeed I get the same answer. I also did it for spherical bowl it is harder and I also get the same answer. But for some reason I can't prove that the answer is independent of the bowl shape.

NVM I see it now, I wonder why I couldn't see it myself and I had to read @Orodruin comment to realize it.

All I can do more is to give a small hint to the OP: You can express $dV$ in a very simple manner as a function of the current area $A(t)$ and of the infinitesimal change in height $dh$.

 

[Chenkel]

I did some math that you may find helpful.

Eq1$$ \frac {dV}{dt} = k_{1}A$$
Eq2$$ \frac {dH}{dt} = k_{2}k_{1}$$
Here I'm writing that the rate of change of volume is proportional to the surface area exposed. The dimensions of $k_{1}$ are cubic meters of volume of evaporated spirit per square meter of exposed surface area per second, the units cancel and you get meters per second for $k_{1}$. The constant $k_{2}$ has no dimension.

We can integrate both sides of Eq2

$$\int dH = k_{2}k{1}\int dt$$
$$H = k_{2}k_{1}t + C$$

We solve for $t=0$ and we find $C=H_{0}$

We solve for $t=t_{0}$ and height equal to $\frac {H_0} 2$ to find $k_{2}k_{1}=-\frac {H_{0}} {2t_0}$

We then plugin $\frac {H_{0}} 4$ for the height and solve for $t$. Finally subtract $t_0$ and we find the additional time we have to wait $t=\frac {t_0} 2$

Let me know if that helps!

 

[brotherbobby]

Thank you @Delta2, @Chenkel and @Orodruin for your inputs. I have made some progress, though I must admit that I still don't get how te shape of the vessel is irrelevant. Let me start by rewriting the problem.

Problem statement : Spirit in a bowl evaporates at the rate that is proportional to the surface area of the liquid. Initially, the height of the liquid in the bowl is $H_0$. It becomes $\frac{H_0}{2}$ in a time $t_0$. How much more time is needed for the height of the liquid to become $\frac{H_0}{4}$?

Attempt (1) (The simplest case) : $\text{A rectangular bowl}$

We are given $dV/dt = kA$. After a time $t_0 (=\Delta t)$, the height drops to $H_0$, so the change in volume for the first time$\Delta V_1= AH_0/2$. Thus, for this first decline, we can divide by $t_0 (=\Delta t)$ on both sides to get $\frac{\Delta V_1}{\Delta t} = \frac{AH_0}{2t_0}\Rightarrow k = \frac{H_0}{2t_0}$. The second decline is by a height of $H_0/4$. Hence the change in the volume a second time is $\Delta V_2 = AH_0/4$. Thus again, $\frac{\Delta V_2}{\Delta t} = \frac{AH_0}{4\Delta t}$, and this is where we have to find $\Delta t = ?$. But the earlier (proportional) equation still holds, so we can say that $\frac{\Delta V_2}{\Delta t} = \frac{\cancel{A}H_0}{4\Delta t} = k\cancel{A}\Rightarrow \Delta t = \frac{H_0}{4k} = \frac{H_0}{4\frac{H_0}{2t_0}}=\boxed{\frac{t_0}{2}}$,

$\mathbf{\huge{\checkmark}}$matching the answer.

Attempt (2) : (Second simplest) $\text{A (conical) bucket}$

I draw an image for the problem. I do not want to show my calculations here, for there is plenty, but I will say that the rate of evaporation of the fluid, $\frac{dV}{dt}=kA$, leads to the value of $k$ which is not a constant. It depends on the initial height $H_0$ as well as its (time) derivative $\dot{H_0}$. Needless to say, I could not progress with my solution to where the volume first reduces to a half and then to a quarter.

I am willing to abandon this line of enquiry (I had a hemispherical bowl next in mind). I would be willing to try out the leads of both @Delta2 and @Chenkel above. I would be glad if you two could elaborate.

 

[Orodruin]

leads to the value of k which is not a constant.

This is wrong. The value of k is constant because you are given that dV/dt is proportional to the area. The constant k is the proportionality constant of this relationship. What is not constant is dV/dt because A changes with height.

I am willing to abandon this line of enquiry (I had a hemispherical bowl next in mind). I would be willing to try out the leads of both @Delta2 and @Chenkel above. I would be glad if you two could elaborate.

The shape of the bowl is irrelevant. Ask yourself how dh/dt depends on dV/dt.

Regardless of evaporation rate. How much does the height change if you change the volume by dV?

 

[Chenkel]

The k value is always constant, what changes is the area. The change in height with respect to time is constant, meaning during equal intervals of time, the height will decrease by a equal amounts, regardless of shape. We can see this is true for variable surface area A, for example dV = dH * A, we know dV/dt = kA so we can solve to find the equation dH = k * dt.

 

[Fred Wright]

We seek the time dependence of the depth of a liquid in a hemispherical shell due to evaporation assuming the evaporation rate (change in volume w.r.t time) is proportional to the surface area and assuming constant pressure, temperature, humidity, and no wind. That is,
$$
\frac{dV}{dt}=\gamma A
$$
where $V$ is the volume,$\gamma$ is the constant of proportionality, and $A$ is the surface area. We first compute the volume in terms of the depth of the liquid and construct the volume integral. From the diagram below,

From the diagram we see,
$$
h=\rho - H
$$
$$
\alpha = \sin^{-1}(\frac{h}{\rho})= \sin^{-1}(1 -\frac{H}{\rho})\\
$$
where $H$ is the liquid depth. The volume integral, in spherical coordinates, becomes
$$
V=\int_0^{2\pi} \int_{\frac{\pi}{2}+\sin^{-1}(1 -\frac{H}{\rho})}^{\pi}\int_{\rho -H}^{\rho}\rho^{'2}\sin(\theta^{'})d\rho{'}d\theta^{'}d\phi^{'}
$$
Evaluating the integral we find
$$
V=\frac{2\pi}{3}\rho^{3}y^2(3 -3y +y^2)
$$
$$
y=\frac{H}{\rho}
$$
We take the derivative of $V$ w.r.t time;
$$
\frac{dV}{dt}=\frac{2\pi}{3}\rho^{3}y(6-9y+4y^2)\frac{dy}{dt}
$$
The surface area is $\pi R^2$ with $R^2=\rho^2-H^2$ we have
$$
\frac{dV}{dt}=\gamma \pi \rho^2 y(2-y)
$$
$$
\frac{2\pi \rho^3}{3}y(6-9y+4y^2)\frac{dy}{dt}=\gamma \pi \rho^2 y(2-y)
$$
We integrate both sides of the equation
$$
\frac{2\rho}{3\gamma}\int \frac{(6-9y+4y^2)}{(2-y)}dy=\int dt
$$
These integrations evaluate to
$$
-y^2+\frac{y}{2}-\log((y-2)^2)+C=\frac{3\gamma}{4\rho}t
$$
where $C$ is the constant of integration. To determine $C$ we assume at time $t=0$, $y=1$, i.e. the hemisphere is filled to the brim. We find
$$
C=\frac{1}{2}
$$
To evaluate $\gamma$ take the measurement of depth ($H_0$) at time $t_0$
$$
-(\frac{H_0}{\rho})^2 + \frac{H_0}{\rho}+\frac{1}{2}-\log((\frac{H_0}{\rho}-2)^2)=\frac{3\gamma}{4\rho}t_0
$$
and we find for the equation for normalized depth vs. time
$$
-y^2 + y + \frac{1}{2}-\log((y-2)^2)=\frac{t}{t_0}(-(\frac{H_0}{\rho})^2 +\frac{H_0}{\rho}+\frac{1}{2}-\log((\frac{H_0}{\rho}-2)^2))
$$
The equation of $y$ as a function of time can't be expressed in terms of elementary functions and must be evaluated by numerical techniques. I wrote a C++ routine to find $y(t)$ with accuracy ~one part in $10^{-10}$. Taking $\frac{H_0}{\rho}=\frac{1}{2}$ and $t_0=500$ I obtained the following plot

 

[Orodruin]

Taking $\frac{H_0}{\rho}=\frac{1}{2}$ and $t_0=500$ I obtained the following plot
View attachment 303568

Anything that is not a straight line is wrong.

 

[Fred Wright]

Anything that is not a straight line is wrong.

Is that a religious doctrine or do you have mathematical proof of your assertion?

 

[Delta2]

Simply put it is $$dV(t)=A(t)dH(t)\Rightarrow \frac{dV}{dt}=A(t)\frac{dH}{dt}$$ do you have any objections to that?

 

[Orodruin]

Is that a religious doctrine or do you have mathematical proof of your assertion?

See the hints earlier in the thread. It is a mathematical fact.

Since posting full solutions in the homework forums before OP has solved the problem is against PF rules I will not say more than that.

Suffice to say, it is true regardless of the shape of the container.

 

[Fred Wright]

Simply put it is $$dV(t)=A(t)dH(t)\Rightarrow \frac{dV}{dt}=A(t)\frac{dH}{dt}$$ do you have any objections to that?

I don't have any objections and you may attack the problem as you wish, but I think calculating the answer in terms of $\frac{H}{\rho}$ results in the remarkable beauty of the answer being universal, i.e. independent of the radius of the hemispherical shell.

 

[hutchphd]

Is that a religious doctrine or do you have mathematical proof of your assertion?

I don't have any objections and you may attack the problem as you wish, but I think calculating the answer in terms of Hρ results in the remarkable beauty of the answer being universal, i.e. independent of the radius of the hemispherical shell.

There is no "beauty" in taking a four line solution and expanding it to fill several pages, getting the wrong result, and insulting folks all along the way.

 

[Delta2]

I don't have any objections and you may attack the problem as you wish, but I think calculating the answer in terms of $\frac{H}{\rho}$ results in the remarkable beauty of the answer being universal, i.e. independent of the radius of the hemispherical shell.

I think your approach is not correct. I think you have a mistake either in calculating the triple integral or calculating the area of the current surface of the liquid. Because for me at least it is clear that ($A(h)$ is the surface of the bowl at height $h$)$$V=\int A(h) dh\Rightarrow \frac{dV}{dh}=A(h)\Rightarrow \frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=A(h)\frac{dh}{dt}$$ and because it is given that $$\frac{dV}{dt}=kA(h)$$ we can conclude that $$\frac{dh}{dt}=k$$ but you find a not so simple ODE for $\frac{dy}{dt}$...

 

[Orodruin]

I think you have a mistake either in calculating the triple integral or calculating the area of the current surface of the liquid.

The error seems to be in the setup of the triple integral. The integration boundary for the inner integral should depend on $\theta’$.

The point is that it is trivial to find the differential equation for $h$ in a single line and therefore concluding that $h$ is a linear function of $t$ regardless of the shape of the container (not only independent of $H/\rho$). You don’t even need to find $V$ or $A$ as functions of $h$, reducing the risk of errors in finding them.

 

[Orodruin]

As to the correct volume of the liquid in the bowl, the easiest way is to not use spherical coordinates but instead use the same kind of integral as given in #15. We have
$$
A(H) = \pi r(H)^2 = \pi[ R^2 - (R-H)^2]
= \pi [R^2 - R^2 + 2RH - H^2] = \pi (2RH -H^2)
$$
and therefore
$$
V(H) = \pi \int_0^H (2RH’ - H’^2) dH’
= \pi [RH^2 - H^3/3] = \pi R^3 y^2(1-y/3)
$$
The sanity check of $V = 4\pi R^3/3$ when $y = 2$ works out nicely.

 

[Delta2]

Ehm, sorry @Orodruin you seem to replace $H^2$ by $R^2-(R-H)^2$ is that valid?

 

[Orodruin]

Ehm, sorry @Orodruin you seem to replace $H^2$ by $R^2-(R-H)^2$ is that valid?

No, I am using the function $r(H) = \sqrt{R^2 - (R-H)^2}$ and square it $r(H)^2 = \ldots$

$r(H)$ being the radius of the surface at height $H$.

 

[Delta2]

Must be my lack of sleep, somehow I read $\pi H^2$ instead of $\pi (r(H))^2$.

 

[brotherbobby]

Sorry for coming in late; I am the OP. I respond for now to @Orodruin only.

Regardless of evaporation rate. How much does the height change if you change the volume by dV?

At a given instant $t$, infinitesimally we have $dV(t) = A(t) dh(t)$, where the symbols have their usual meanings.

It is given that the spirit evaporates at a rate proportional to the surface area $A$. Can we say from above that $\dfrac{dV}{dt} = A(t) \dfrac{dh}{dt}$? What about the derivative $\dfrac{dA}{dt}$?

 

[Orodruin]

Can we say from above that $\dfrac{dV}{dt} = A(t) \dfrac{dh}{dt}$?

Yes, and therefore… (using the given evaporation rate proportional to $A$)

What about the derivative $\dfrac{dA}{dt}$?

Not really relevant to the problem.

 

[brotherbobby]

Yes, and therefore… (using the given evaporation rate proportional to A)

Therefore, since it is given that $\dfrac{dV}{dt} = kA$ and we have $\dfrac{dV}{dt} = A\dfrac{dh}{dt}$, we can say that $\dfrac{dh}{dt} = k$. So it means that if the volume decreases (on increases) with time proportional to the surface area, the rate of decrease (or increase) of the height of the liquid is a constant?

 

[Orodruin]

Therefore, since it is given that $\dfrac{dV}{dt} = kA$ and we have $\dfrac{dV}{dt} = A\dfrac{dh}{dt}$, we can say that $\dfrac{dh}{dt} = k$. So it means that if the volume decreases (on increases) with time proportional to the surface area, the rate of decrease (or increase) of the height of the liquid is a constant?

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

Since we are talking evaporation, $k$ would typically be negative here.

 

[brotherbobby]

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

Since nowhere have we referred to the shape of the container to conclude that $\dfrac{dh}{dt} = k$ from $\dfrac{dV}{dt} = kA$, that seems the case.
However, it is still a bit hard to understand that for most shapes, $V \ne AH$, except for "area maintaining" shapes like cuboids and cylinders, members of the family of prisms. However, for all shapes, prisms or otherwise, we can maintain that $dV = AdH$. Is this because all shapes are locally like prisms?

 

[Orodruin]

Since nowhere have we referred to the shape of the container to conclude that $\dfrac{dh}{dt} = k$ from $\dfrac{dV}{dt} = kA$, that seems the case.
However, it is still a bit hard to understand that for most shapes, $V \ne AH$, except for "area maintaining" shapes like cuboids and cylinders, members of the family of prisms. However, for all shapes, prisms or otherwise, we can maintain that $dV = AdH$. Is this because all shapes are locally like prisms?

If you slice the shape up in very thin horizontal slices of thickness $dh$, then the volume of the slice at $h$ is $A(h) dh$ as long as the area does not change appreciably within each slice (and since there is formally a limit of the slice thickness going to zero - that is going to be the case)

 

[brotherbobby]

Ok so locally, meaning for vanishing heights, a shape resembles a prism.
If we move into the domain of extended heights, i.e. $h>0$, can we write $dV = hdA+Adh$?

 

[Orodruin]

Ok so locally, meaning for vanishing heights, a shape resembles a prism.
If we move into the domain of extended heights, i.e. $h>0$, can we write $dV = hdA+Adh$?

No. $dV = A \, dh$ is a differential statement based on the forementioned slicing. The extended height statement is summing up all of those small volumes, ie, integrating, in the limit of vanishing slice thickness:
$$
V(H) = \int_0^H A(h) dh.
$$

 

[brotherbobby]

No. $dV = A \, dh$ is a differential statement based on the forementioned slicing. The extended height statement is summing up all of those small volumes, ie, integrating, in the limit of vanishing slice thickness:
$$
V(H) = \int_0^H A(h) dh.
$$

Thank you @Orodruin and sorry for my query in post#27 above which read:

If we move into the domain of extended heights, i.e. h>0, can we write dV=hdA+Adh?

If $dV=Adh$ then my query above makes no sense. I had in mind something like this : $\Delta V = A\Delta h+h\Delta A$. But that would make no sense either. Which $A$ and $H$ to take?

However, I think I should do my job and bring the thread to an end from my side by writing down the solution to the given problem, which was :

Solution :

Yes indeed the height decreases at a constant rate : $\dfrac{dh}{dt}= k$.
Now, according to the problem, the height becomes $\dfrac{H_0}{2}$ in a time $t_0$. Hence, it will reduce by a (further) amount of $\dfrac{H_0}{4}$ in half the time as above, that is $\boxed{\dfrac{t_0}{2}}$.

Insight : I think the crucial insight from the problem is that an incremental volume of a shape can be written as $dV=A dh$. This is true for all shapes, as all shapes are infinitesimally "regular" - i.e. prismatic, when the heights are vanishingly small.

 

[Orodruin]

Insight : I think the crucial insight from the problem is that an incremental volume of a shape can be written as $dV=A dh$. This is true for all shapes, as all shapes are infinitesimally "regular" - i.e. prismatic, when the heights are vanishingly small.

If one is uncomfortable with the arguments above, one can always start from the volume element $dV = dx\, dy\, dz$. Integrating over the full volume of fluid gives the total volume
$$
V = \iiint dV = \iiint dx\, dy\, dz.
$$
For a fixed $z$, the cross sectional area $A(z)$ is given by integrating $dx\, dy$ over the cross section
$$
A(z_0) = \iint_{z = z_0} dx\, dy.
$$
By performing the $x$ and $y$ integrals in the volume integral first, we therefore obtain
$$
V = \int A(z) dz
$$
as expected.

 

[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.