Differential equation related to water clock

Question

This is the question:-

A clepsydra, or water clock, is a container with a small hole in the bottom through which water can flow. Water clocks were used by ancient Egyptians, Greeks, Romans and Chinese to measure time by observing the change in the height of water in the container. The device is calibrated for measuring time by placing markings on the container corresponding to water levels at equally spaced times.

Suppose the container of a water clock is obtained by revolving around the y-axis the graph of a continuous function $x = f(y)$ defined on an interval $[0, b]$, where $b$ is some positive real number and $f(0) > 0$. Let $V$ denote the volume of water in the container and $h$ the height of the water level at time $t$.

(a) Find an expression for the amount of water $V$ in the container as a function of the water level $h$.

(b) Find a formula for the function $f(y)$ that would result in a water clock with equally spaced markings on the container, i.e. with markings that are an equal vertical distance from each other.

You may use Torricelli's Law, which states that the speed of fluid flowing out of a small hole of a container is related to the fluid level in the container. Specifically, Torricelli's Law says that the rate of change of the volume of water $V$ in a container with a (small) hole of area $A$ at the bottom of the container is proportional to the square root of the water height $h$,

$$\frac{dV}{dt} = -A\sqrt{2gh}$$

where $g$ is the acceleration due to gravity.

Here is what I have tried:-

$$V = \pi\int_0^h{f^2(y)dh}$$

$$\frac{dV}{dt} = \pi f^2(h)\frac{dh}{dt}$$

$$-A\sqrt{2gh} = \pi f^2(h)\frac{dh}{dt}$$

I don’t really know where to go from here Also I think that,

$$A = \pi f^2(0)$$

Answer 1

A method for you to use.

First, note that $\frac{dV}{\text dh}= \pi x^2=\pi f(h)^2$.

Note also that $\frac{dV}{\text dt}=k\sqrt h $ for some constant $k$.

For the markings to be equally spaced, $\frac{\text dh}{\text dt}$ must be constant.

Then use the chain rule.

Answer 2

You have already solved the problem. Point (b) tells you that $ dh/dt $ is a (negative) constant, therefore you have an equation which gives you $f(h)$, i.e. $f(h)\propto h^{1/4}$.

The only formal "inconsistency" of the model is for $h\rightarrow 0^+$, because the solution implies $f(h)=0$ but $f(0)$ cannot be equal to zero (because the orifice must have a non zero area $A=\pi f(0)^2$). This happens because in the Bernoulli equation you have assumed $\pi f(h)^2 \gg A$ (i.e. $f(h)^2\gg f(0)^2$) and therefore neglected the velocity of the surface with respect to the discharge velocity at the orifice (by using continuity equation, you find that their ratio is $f(0)^2/f(h)^2\ll 1$ and, in the end, this implies $|h'(t)|\ll \sqrt{2gh}$). Without such approximation that small inconsistency would have not been present in the result, which would have been $$ f(h) = f(0)\left(1+\frac{2gh}{h'^2}\right)^{1/4} $$ Of course, since $|h'(t)|\ll \sqrt{2gh}$ except in the final stage, the correction is small everywhere except for $h\rightarrow 0^+$.