A classical mechanics problem by Mr X

Relevant wiki: Fluid Mechanics

Let $H$ be the fixed height of container, $h$ be the current height of water, $A$ be the cross-sectional area of the cylinder, and $a$ be the cross-sectional area of the pipe at the bottom.

For the $300$ second case, assuming a constant rate of water flow, $\dfrac{dh}{dt} = \dfrac{H}{300}$

For the $600$ second case, by Torricelli's Law, $v=\sqrt{2gh}$ $\dfrac{dV}{dt} = A\dfrac{dh}{dt} = -a\sqrt{2gh}$ $\dfrac{dh}{\sqrt{h}} = -\dfrac{a\sqrt{2g}}{A}dt$ $\int_H^0 \dfrac{dh}{\sqrt{h}} = -\dfrac{a\sqrt{2g}}{A} \int_0^{600} dt$ $-2\sqrt{H} = -\dfrac{600a\sqrt{2g}}{A}$ $H=\dfrac{2g(300)^2a^2}{A^2}$

Thus combining both the water flow from the tap and out of the pipe, $\dfrac{dh}{dt} = \dfrac{H}{300} - \dfrac{a\sqrt{2gh}}{A}$ $dt=\dfrac{dh}{\dfrac{H}{300} - \dfrac{a\sqrt{2gh}}{A}}$ $\dfrac{a\sqrt{2g}}{A}dt = \dfrac{dh}{\dfrac{AH}{300a\sqrt{2g}}-\sqrt{h}}$

Note that $\dfrac{AH}{300a\sqrt{2g}}=\dfrac{A}{300a\sqrt{2g}}\dfrac{2g(300)^2a^2}{A^2}$ $=\dfrac{300a\sqrt{2g}}{A}=\sqrt{H}$

Hence $\dfrac{a\sqrt{2g}}{A}T = \int_0^{H} \dfrac{dh}{\sqrt{H}-\sqrt{h}}$

which approaches $+\infty$ , thus the container can never fill up fully.

Let $\tau$ be the time it takes to empty the tank when full; and $\tau/K$ the time it takes to fill the tank. In this problem, $K = 2$ .

If the tank fills to its height $h$ in time $\tau/K$ , then the change of water level due to inflow is $\frac{dy}{dt} = \frac {hK}\tau.$ For the emptying of the tank, consider that the kinetic energy (proportional to $(dy/dt)^2$ ) of the water leaving is equal to the potential energy (proportional to $y$ ) of the water disappearing from the top. Therefore $dy/dt = -c\sqrt y$ for some constant $c$ . Combine this with the fact that $y = h$ at $t = 0$ , and $t = 0$ at $t = \tau$ , we find $y = h\left(1 - \frac t \tau\right)^2,$ and $\frac{dy}{dt} = -\frac{2\sqrt h}{\tau}\:\sqrt y.$

Qualitative analysis

To find a quick answer to the question, consider under what circumstances the outflow rate would be equal to inflow rate-- in that case, the water level will no longer increase. We solve $\frac{hK}\tau = \frac{2\sqrt{h}}{\tau}\:\sqrt y\ \ \ \ \ \ \ \therefore\ \ \ \ \ \ \ y = \left(\frac K 2\right)^2 h.$ In our situation, $K = 2$ so that the outflow and inflow are matched at $y = h$ . While this does not quite answer the questions, note

• In this case, the tank may fill up all the way but will never overflow.

• If the inflow is reduced ( $K$ increases), $y < h$ , which means that an equilibrium is reached before the tank is full.

Detailed analysis

For the situation of filling with open pipe, we combine the two differential equations: $\frac{dy}{dt} = \frac {hK}\tau - \frac{2\sqrt{h}}{\tau}\:\sqrt y,$ which may be separated as $\frac{dy}{hK - 2\sqrt{hy}} = \frac {dt}{\tau}.$ The solution is $-\frac K 2 \ln (hK - 2\sqrt{hy}) - \sqrt{\frac y h} = \frac t{\tau} + C;$ Since $y = 0$ when $t = 0$ , we must have $C = -\tfrac K 2 \ln hK$ ; incorporating this in the logarithm on the left, we obtain $-\frac K 2 \ln \left(1 - \frac 2K\sqrt{\frac y h}\right) - \sqrt{\frac y h} = \frac t{\tau}.$ Now we wish to know when the tank will be full, i.e. $y = h$ and therefore $\sqrt{y/h} = 1$ . The equation simplifies as $-\frac K 2 \ln \left(1 - \frac 2K\right) - 1 = \frac t{\tau},$ with solution $t = \left(-\frac K 2 \ln \left(1 - \frac 2 K \right) - 1\right)\:\tau.$ For $K = 2$ , we obtain a logarithm of zero, making $t$ infinitely large; the tank will never be filled!

Torricelli theorem (for a reservoir that receives no inflow) predicts that both the velocity and the discharge (drainage) rate are constant with time whereas only the relationship of the residual water height in the reservoir with time is nonlinear (it is represented by a concave curve, i.e. the residual water column height will decrease with time.)

Now, the problem is that the reservoir receives certain recharge rate with Recharge = reservoir volume / filling time when the bottom orifice is closed.

This particular case is a modified Torricelli problem (where the relationship of the residual water column head in the draining reservoir with time is not anymore a linear relationship whereas such relationship is linear for Torricelli reservoir that receives no recharge.)

Everybody will give an answer and defend its correctness whereas we have no model to solve it. Moreover, the initial conditions to solve it could be that the reservoir has been filled-up before opening the bottom orifice, or the reservoir was just empty at time zero. This complicates the issue dealt with a bit further. For sure the input (inflow rate) is assumed constant (and this is again a special case of a changing inflow rate that will generalized the case). When we have a modified Torricelli and settle the initial conditions (filled or empty reservoir when the bottom orifice gets opened) we can solve this problem, otherwise it cannot be solved. Who has courage to rum an experiment to learn more using a set of actual observations?

Disregarding the above-mentioned remarks on the imposed problem, a simplistic approximate (but acceptable) solution will say that during a time step of one minute the inflow is 1/5th of the reservoir's volume, while the discharge will be 1/10th of the reservoir's volume. Accordingly, the reservoir will seemingly get fully filled-up during 10 min. But this is just an approximation.

Apparently, without using calculus, the reservoir will fill close to its full volume within the approximate 10 min as mentioned above, but it will not fill-up quite equal to its full volume since there is a small - and changing - increase in the discharge rate due to the changing increase in drainage velocity by the change of the residual head in the reservoir (same fact as applying a changing pneumatic pressure on the top pf the residual water column in the reservoir.)

Nonetheless, the 10 min approximation is good, and acceptable, for all practical purposes.