Drain times with Toricelli's law

 It's easy to calculate the drain time of water in a cylindrical tank using Toricelli's law, $v=\sqrt{2gh}$. In fact, using a simple differential equation, we can see that

$\displaystyle \sqrt{\frac{h}{H}}=1-\frac{t}{t_H}$

where $H$ is the initial height and $t_H$ is the time it takes for the tank to drain a water column of height $H$.

I got curious about the case when there is a constant inflow of water. Then, we have the differential equation,

$a\sqrt{2gh_0} \,dt-a\sqrt{2gh}\,dt=A\,dh$

where $a$ is the area of discharge, $A$ the cylindrical area and $h_0$ the measure of constant inflow. Solving this gives,

$\displaystyle \frac{t}{t_0}=\sqrt{H/h_0}-\sqrt{h/h_0}+\ln\left(\frac{1-\sqrt{H/h_0}}{1-\sqrt{h/h_0}}\right)$

where $t_0$ is the time taken to drain a water column of height $h_0$.

Plotting this shows that irrespective of the height of the water column we start with, we always end up with a steady state water column of height $h_0$.

This brings us to the next case. Suppose we have a tank with an initial water column of height $H$ which drains water into a identical tank. We know that the first tank takes time $t_H$ to drain completely and therefore, will be restricting our analysis to this timeframe.

In this case, the differential equation (using the first equation of this post) becomes

$\displaystyle a\sqrt{2gH}\left(1-\frac{t}{t_H}\right)\,dt-a\sqrt{2gh}\,dt=A\,dh$

This can be simplified to

$\displaystyle 1 - \frac{t}{t_H}-\sqrt{\frac{h}{H}}=\frac{t_H}{2H}\frac{dh}{dt}$

We now makes substitutions, $x=1 - t/t_H$ and $y=\sqrt{h/H}$ so that the above equation becomes

$\displaystyle y-x=y\frac{dy}{dx}$

Solving this homogenous differential equation, we get

$\displaystyle\ln(y^2-yx+x^2)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\frac{y}{x}-\frac{1}{\sqrt{3}}\right)=C$

Let's see some special cases of height of the water column in the second tank when the first tank runs out of water.

If the second tank is empty to start with, we have $h=0,t=0 \implies y=0,x=1$ as the initial conditions. With this we get,

$\displaystyle\ln(y^2-yx+x^2)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\frac{y}{x}-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{3\sqrt{3}}$

Putting $x=0$ in the above (to find the water column at $t=t_H$), we see that $\displaystyle h(t_H) =\exp\left(-\frac{4\pi}{3\sqrt{3}}\right)H$

Similarly, if the two tanks start with the same level from the start, we use the initial conditions $h=H,t=0 \implies y=1,x=1$. This time we get,

$\displaystyle\ln(y^2-yx+x^2)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\frac{y}{x}-\frac{1}{\sqrt{3}}\right)=\frac{\pi}{3\sqrt{3}}$

Now, if we put $x=0$ in the above we get, $\displaystyle h(t_H) =\exp\left(-\frac{2\pi}{3\sqrt{3}}\right)H$

Finally, note that if the initial water level in the second tank is lower than that of the first tank, it's water level rises initially, reaches a maximum and then falls off.

To find the maximum height it reaches, we need $dh=0 \implies dy=0 \implies y=x$.

If we let $y_{\text{max}}$ to denote the maximum $y$ and $y_1$ to denote the value of $y$ at $x=1$, we see that

$\displaystyle\ln(y^2-yx+x^2)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\frac{y}{x}-\frac{1}{\sqrt{3}}\right)=\ln y_1^2+\frac{\pi}{\sqrt{3}}$

Putting $y=x$ in the above, we have

$\displaystyle \ln y_{\text{max}}^2+\frac{\pi}{3\sqrt{3}}=\ln y_1^2+\frac{\pi}{\sqrt{3}}$

This then shows that

$\displaystyle h_{\text{max}}=\exp\left(\frac{2\pi}{3\sqrt{3}}\right)h_1$

While it was already surprising that both $\pi$ and $e$ made an appearance in this problem, this result that the max. height and final height are in a constant ratio made it all the more satisfying for me.

Until then

Yours Aye

Me