A siphon empties a large open tank of water…

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

A siphon empties a large open tank of water. It is a tube of cross-section $4.0\ \text{cm}^2$ , with a nozzle of cross-section $2.0\ \text{cm}^2$ fitted at its outlet. The highest point of the tube is $1.0\ \text{m}$ above the water surface in the tank, and the nozzle is $2.2\ \text{m}$ below that surface, as shown in the figure. Treat the water as an ideal fluid in steady flow, take the tank to be so large that its surface hardly moves, and take the air's pressure as $1.0 \times 10^{5}\ \text{Pa}$ , $g = 10\ \text{m/s}^2$ and the density of water as $1000\ \text{kg/m}^3$ .

A large open tank of water. A siphon tube of cross-section 4.0 square cm rises from under the water, over the rim of the tank, to a highest point 1.0 m above the water surface, then runs down outside the tank to a nozzle of cross-section 2.0 square cm, 2.2 m below the water surface, from which a jet of water comes out.

How fast does the water leave the nozzle?
How many litres of water per second does the siphon carry?
What is the pressure in the water at the highest point of the tube?
The siphon stops working if the pressure at the highest point would fall below zero. Keeping the nozzle where it is, how high above the water surface could the highest point of the tube be at most?

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. Apply Bernoulli's equation from the tank's surface to the places we want, and continuity between the tube and the nozzle.

Step 1. The jet (a). At the tank's surface and in the jet the pressure is the air's, and the tank's surface is almost still. So $\tfrac12\rho v_2^2 = \rho g h_2$ :

\displaystyle v_2 = \sqrt{2gh_2} = \sqrt{2 \times 10 \times 2.2}\ \text{m/s} = 6.63\ \text{m/s}

Step 2. Flow rate (b). $Q = A_2 v_2 = 2.0 \times 10^{-4}\ \text{m}^2 \times 6.633\ \text{m/s} = 1.327 \times 10^{-3}\ \text{m}^3/\text{s}$ , or $1.33\ \text{L/s}$ .

Step 3. Speed in the tube. By continuity, $v_1 = v_2\,\dfrac{A_2}{A_1} = 6.633 \times \dfrac{2.0}{4.0}\ \text{m/s} = 3.317\ \text{m/s}$ .

Step 4. The highest point (c). From the tank's surface to the top of the bend:

\displaystyle P_0 = P_{\text{top}} + \tfrac12\rho v_1^2 + \rho g h_1

\displaystyle P_{\text{top}} = 100\ \text{kPa} - 5.5\ \text{kPa} - 10\ \text{kPa} = 84.5\ \text{kPa}

This is below the air's pressure, which is what lifts the water over the bend.

Step 5. The limit (d). Raising the bend does not change the flow, which depends only on $h_2$ . So $P_{\text{top}} = P_0 - \tfrac12\rho v_1^2 - \rho g h_1$ reaches zero at

\displaystyle h_1 = \frac{P_0 - \tfrac12\rho v_1^2}{\rho g} = \frac{100\ \text{kPa} - 5.5\ \text{kPa}}{10\ \text{kPa/m}} = 9.45\ \text{m}

Answer: (a) $6.63\ \text{m/s}$ . (b) $1.33\ \text{L/s}$ . (c) $84.5\ \text{kPa}$ . (d) $9.45\ \text{m}$ .

Check. Bernoulli from the top of the bend to the jet: $P_{\text{top}} + \tfrac12\rho v_1^2 + \rho g(h_1 + h_2) = P_0 + \tfrac12\rho v_2^2$ . In kilopascals the left side is $84.5 + 5.5 + 32 = 122.0$ , and the right side is $100 + 22 = 122.0$ . They agree.

Watch out. The nozzle helps the siphon. Without it, the water in the tube would move at the full jet speed, and the pressure at the top would be only $68\ \text{kPa}$ .

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