# Rate of Evaporation of a Vase

The base of a vase is created by rotating the curve

$$f(x)=2.393794315((1.25916975182872)\left(\frac{x}{2.24581}\right)^4+(-4.29578020022745)(\frac{x}{2.24581})^3+(4.37766951188496)(\frac{x}{2.24581})^2+(-0.553610482668319)(\frac{x}{2.24581})+(-1.6343566433531))$$

...defined in the domain {$${0.224581\le x\le 2.919553}$$}, rotated $$2\pi$$ radians about the x-axis.

The units of the coordinate axes are in centimetres.

My goal is to find the time it takes when the base of the vase is completely filled with water to evaporate.

Through some basic evaporation equations from thermodynamics, I have discovered that:

$$\frac{dV}{dt}=-2.55\cdot Exposed\ Surface\ Area$$

As the curve above is being revolved around the x-axis, it is obvious that:

$$Exposed\ Surface\ Area=\pi \cdot (f(x))^2$$

However, this is as far as I've been able to go. In a worked example I saw, the rate of change in volume ($$\frac{dV}{dt}$$) was linked to the volume in the container remaining, and then the total volume of the vase was plugged in to solve the differential equation for time. Here it is:

SOLUTION

• that worked out example is very uselessly complicated, as the answer here has shown, time is height divided by rate of absorption, so $100 cm$ / $3 (cm/h)$ = $100/3$ hours. There is no need to go through complicated differential equations. Sep 26, 2016 at 10:05
• Which side is the top of the vase?
– Hrhm
Sep 30, 2016 at 12:32
• The left hand side Sep 30, 2016 at 12:33
• Now that I think about it, it doesn't matter which side the top is.
– Hrhm
Sep 30, 2016 at 12:59
• Also, Dan Robertson's solution below is correct.
– Hrhm
Sep 30, 2016 at 12:59

There are far more numbers in this question than I'd like so I'm going to try to remove everything not relevant to solving a slightly more general problem and then you should be able to finish off the problem.

Assume we have a cylindrical symmetrical vase and represent it by its radius $f$ as a function of the height above the ground $x$.

If the vase is filled water to a height $h$ then the exposed surface area of the water is $A(h)$ where we define: $$A(x)=\pi(f(x))^2.$$

The volume contained is the volume of revolution: $$V(h) = \int_0^h A\mathrm d x.$$

Now we have evaporation given by some rate and the exposed surface area: $$\frac{\mathrm d V}{\mathrm d t} = -k A(h),$$ where $h$ is the instantaneous water level and $k>0$ is some constant for the evaporation rate.

We want to know what the water level $h$ will be at various times so we want to consider $\frac{\mathrm d h}{\mathrm d t}.$

Ok. Now we're going to finish this off in a few lines. Look out for:

• The chain rule
• The fundamental theorem of calculus
• A bit of rearranging and substitution
• The assumption that $A\ne0$
• Solving a simple differential equation

Here we go: \begin{align} \frac{\mathrm d V}{\mathrm d t}&=\frac{\mathrm d h}{\mathrm d t}\frac{\mathrm d V}{\mathrm d h}\\ \frac{\mathrm d V}{\mathrm d h} &= \frac{\mathrm d }{\mathrm d h} \int_0^h A(x) \mathrm d x\\ &= A(h)\\ A(h) &= -\frac1k\frac{\mathrm d V}{\mathrm d t} \\ &=-\frac1k\frac{\mathrm d h}{\mathrm d t} A(h)\\ -k&=\frac{\mathrm d h}{\mathrm d t}\\ h&=h_0-k t \end{align}

Given the initial height $h_0$ you can solve the final equation for the time at which $h=0.$

Perhaps this result may be surprising but if you accept that it is reasonable to work out a volume of revolution by integrating surface area (summing up a bunch of lamina) then it seems reasonable that removing them at a constant rate by evaporation would reduce the water level at a constant rate.

Another way to think of this is by applying some dimensional analysis to the evaporation rate: The evaporation rate is some volume (length cubed) that evaporates per a given surface area (length squared) in a given time. Thus it has dimension $\frac{L^3}{L^2T}=LT^{-1}$ which is a velocity so it is quite natural to suppose that evaporation rate is the same as the rate that the depth of fluid decreases, and indeed it is.

To finish off your problem, you have $h_0=2.919553-0.224581=2.694972\text{ cm}$ and $k=2.55$ (unknown units; dimension $L^3T^{-1}L^{-2}=LT^{-1}$) so the time for the vase to empty would be $\frac{h_0}{k}\approx 1.06$ (unknown time units)

• I'm not sure that answers my question. As you can see, I have a worked solution for when $f(x)=x^2$. Thus, I understand the method. The problem arises when I try to apply it on my fourth order polynomial, which you completely ignored. Sep 26, 2016 at 8:56
• The point is that this proves that the time to evaporate depends only on the height of the vase and the type of fluid in it, not on the precise shape of the vase Sep 26, 2016 at 9:02
• You're saying that a cylindrical pool of a depth of $1$ meter with a constant surface area of $100$ $m^2$ would have the same time for evaporation as a tall cylindrical glass that holds water to a depth of $1$ meter that has a constant surface area of only $4$ $cm^2$ (yes, $cm^2$)? Sep 26, 2016 at 9:06
• I'm saying that under your model of evaporation whereby evaporation happens at a constant rate of volume per surface area, that would be true. Note that your model does not account for, say, the air above the surface becoming saturated. It surely seems more reasonable that it would take the same time to evaporate the thin film of water made from mopping your kitchen floor as the thin film of water made by mopping a gymnasium floor, so long as the films were the same depth. You are welcome to find a fault with your own model but I do not accept that my conclusion is wrong under the current one. Sep 26, 2016 at 9:12
• The worked example is an exercise in elementary calculus. The object is not to solve the problem simply but to demonstrate one's ability to abstractly manipulate some symbols. If only the information for the question and part (c) were kept, it could be solved as: $\frac{3\text{ cm}^3}{1\text{ cm}^21\text{ hour}}=3\frac{\text{cm}}{\text{hour}}$. 100cm (the height of the glass) at this rate is $\frac{100}{33}\text{ hours}$ Sep 26, 2016 at 12:04