Question 3 of 13: Power for a Submersible Garden Fountain

Question 3 — Power for a Submersible Garden Fountain (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The power required to run the fountain, P_in.

Approach. Torricelli's relation gives the exit velocity needed to reach the target height; continuity gives the flow rate; the hydraulic (useful) power equals the kinetic power of the jet, and dividing by the overall efficiency gives the required input power.

1. Required exit velocity. Neglecting air drag, all kinetic energy converts to potential energy at the apex: $$V = \sqrt{2gh} = \sqrt{2(9.81)(2)} = \boxed{6.264\text{ m/s}}$$
2. Flow rate. $$Q = V\left(\frac{\pi d^2}{4}\right) = 6.264\times \frac{\pi(0.020)^2}{4} = \boxed{1.968\times10^{-3}\text{ m}^3\text{/s} = 1.968\text{ L/s}}$$
3. Ideal (hydraulic) power. $$P_{ideal} = \rho g Q h = 1000\times 9.81\times 1.968\times10^{-3}\times 2 = \boxed{38.61\text{ W}}$$
4. Power required from the pump. Substituting the overall efficiency: $$P_{in} = \frac{P_{ideal}}{\eta} = \frac{38.61}{0.65} = \boxed{59.4\text{ W}}$$