A tall cylindrical tank stands on the ground…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A tall cylindrical tank stands on the ground. Its cross-section is $0.50\ \text{m}^2$ , and it has a small hole of area $5.0\ \text{cm}^2$ in its flat bottom. Starting with the tank empty, a pipe pours water into it at a steady $2.0\ \text{litres}$ per second, as shown in the figure. Water leaves through the hole at the speed $\sqrt{2gh}$ , where $h$ is the depth of the water at that moment. Take $g = 10\ \text{m/s}^2$ , and ignore the contraction of the jet.

A tall cylindrical tank of cross-section 0.50 square m, partly filled with water of depth h. A pipe above it pours in 2.0 litres per second, and water leaves through a small hole of area 5.0 square cm in the bottom.

The water level rises towards a steady depth. Find that depth.
How long after the start does the water reach a depth of $0.60\ \text{m}$ ?
Once the depth has (very nearly) reached its steady value, the pipe is shut off. How long does the tank then take to empty?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Write how the volume in the tank changes: in through the pipe, out through the hole. Then integrate.

Step 1. The steady depth (a). Out through the hole: $a\sqrt{2gh}$ . Steady when this equals $Q$ :

\displaystyle h_e = \frac{Q^2}{2ga^2} = \frac{(2.0 \times 10^{-3}\ \text{m}^3/\text{s})^2}{2 \times 10\ \text{m/s}^2 \times (5.0 \times 10^{-4}\ \text{m}^2)^2} = 0.80\ \text{m}

Step 2. Filling (b). With $k = a\sqrt{2g} = 2.2361 \times 10^{-3}\ \text{m}^{5/2}/\text{s}$ :

\displaystyle A\,\frac{dh}{dt} = Q - k\sqrt{h}

Put $h = u^2$ , so $dh = 2u\,du$ , and separate the variables:

\displaystyle t = \int_0^{U} \frac{2Au\,du}{Q - ku} = \frac{2A}{k^2}\left[Q\ln\frac{Q}{Q - kU} - kU\right], \qquad U = \sqrt{0.60\ \text{m}}

Step 3. Numbers. $kU = 1.7321 \times 10^{-3}\ \text{m}^3/\text{s}$ , which is $0.866$ of $Q$ . Then $\dfrac{2A}{k^2} = 200000\ \text{s}^2/\text{m}^3$ and the bracket is $2.0 \times 10^{-3} \times 2.0101 - 1.7321 \times 10^{-3} = 2.2882 \times 10^{-3}\ \text{m}^3/\text{s}$ , so

\displaystyle t = 458\ \text{s}

Step 4. Emptying (c). Now $A\,\dfrac{dh}{dt} = -k\sqrt{h}$ , so $\dfrac{dh}{\sqrt h} = -\dfrac{k}{A}\,dt$ and $2\sqrt{h_e} = \dfrac{k}{A}\,t$ :

\displaystyle t = \frac{A}{a}\sqrt{\frac{2h_e}{g}} = 1000 \times \sqrt{\frac{2 \times 0.80}{10}}\ \text{s} = 400\ \text{s}

Answer: (a) $0.80\ \text{m}$ . (b) $458\ \text{s}$ . (c) $400\ \text{s}$ .

Check. If the hole were plugged, the pipe would fill $0.60\ \text{m}$ in $\dfrac{0.50 \times 0.60}{2.0 \times 10^{-3}}\ \text{s} = 150\ \text{s}$ . With the hole open it must take longer, and it does. The steady depth itself is never quite reached: the closer the level gets, the smaller the net inflow.

Watch out. Filling and emptying are not mirror images. Emptying from the steady depth takes a finite time, $400\ \text{s}$ , but filling up to the steady depth would take forever.

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