Tautochronous tanks

Toricelli's law states that the velocity of a discharge, say $v$, located at the bottom of a tank filled to a height $h$ is $v=\sqrt{2gh}$.

Assuming the shape of the tank to be a right prism (with its axis along the $y$-axis), we can use the conservation of mass (or volume in this context) to get

$A \,dy=-a \cdot \sqrt{2gy}\,dt$

where $A$ is the cross sectional area of the prism and $a$ is the area of discharge. Note that the negative sign indicates that $y$ (the height of the water column) decreases as time $t$ increases.

Rearranging and integrating, we have

$\displaystyle \int_0^T \,dt=-\frac{A}{a\sqrt{2g}}\int_h^0 \frac{1}{\sqrt{y}}\,dy \implies T=\frac{A}{a}\sqrt{\frac{2h}{g}}$

where $T$ is the total time taken to drain the tank which is proportional to the square root of the height of the liquid column.

But can we design a tank such that the time of discharge is directly proportional to height? If we assume that the shape of the tank is a solid of revolution such that the area of the cross section at $y$ is given by $A(y)$, then by the similar reasoning as above, we have

$\displaystyle T=\frac{1}{a\sqrt{2g}}\int_0^h \frac{A(y)}{\sqrt{y}}\,dy$

Therefore, for $A(y)=\pi x^2 \propto \sqrt{y}$ (or) $x \propto \sqrt[4]{y}$, we get that $T(h)$ is directly proportional to $h$.

This clearly shows that the draining time for a tank formed by the solid of revolution quartic curve $y=x^4$ is directly proportional to height of the liquid column.

This naturally leads to the second question that we want to address in this post. Can we design a tank such that the draining time is independent of the height of the liquid column.

A direct attack (for example, using Abel's solution of the Tautochrone curve) quickly suggests that there is no such curve.

This might seem like a dead end and it sort of is. But, Let's consider an almost similar question: Can we design a tank, initially filled to the brim, such that the time of discharge is independent of the height at which the discharging hole is located?

For ease of what follows, we orient the coordinate system such that gravity points in the positive $y$ direction. Then for a tank whose shape is a solid of revolution, using the conservation of volume, we have,

$\displaystyle A(y)\,dy=a\cdot \sqrt{2g(y_0-y)}\,dt \implies T(y_0)=\frac{1}{a\sqrt{2g}}\int_0^{y_0}\frac{A(y)}{\sqrt{y_0-y}}\,dy$

where $y_0$ is the height at which the discharge is located.

Now, if we use Abel's idea, we can see that

$\displaystyle A(y)=T_0\frac{a\sqrt{2g}}{\pi}\frac{1}{\sqrt{y}}$

And there we have it! The inverse quartic equation $y=x^{-4}$ is tautochronous such that the time taken to empty the tank is independent of the 'depth' of the water column.

Because the $x$-axis is asymptotic to the inverse quartic curve, we should also ensure that the tank holds a finite volume of water of water for a given height of discharge. But it's an easy exercise to check that the volume is indeed finite.

Hope you enjoyed this. See ya soon.

Until then, Yours Aye

Me