Question 6 of 13: Power Required for a Submersible Garden Fountain
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2013-Dec. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.
Air density is taken from the paper's own Constants table at the temperature each question states: 1.19 kg/m³ at 20°C (Q5's wind, Q7's inlet air).
Q1's touching-rod array is modelled as a repeating square unit cell of four mutually tangent rods (pitch = rod diameter, per the question's own "closely packed" wording), giving a curvilinear-square pore whose perimeter/area ratio drives the capillary rise.
Q8's Moody diagram and Q9's drag-coefficient diagram are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 and the Morrison (2013) curve-fit for sphere drag vs. Reynolds number for Q9.
Question 6: Power Required for a Submersible Garden Fountain (5 marks)
Given. Jet diameter Djet = 20 mm; required jet height h = 2 m; overall (pump + hydraulic) efficiency η = 65%; nozzle discharges essentially at the pond's own free surface.
Find. Shaft (electrical/mechanical input) power required to run the fountain.
Approach. Use the free-jet (Torricelli) relation to find the exit velocity needed to reach the target height, get the flow rate from the jet area, then convert the ideal hydraulic power to actual shaft power via the stated efficiency.
Required jet velocity. Neglecting air drag, all kinetic energy at the nozzle converts to potential energy at the peak: $V_{jet}=\sqrt{2gh}=\sqrt{2(9.81)(2)}=\boxed{6.264\ \text{m/s}}$.
Flow rate. $A_{jet}=\tfrac{\pi}{4}(0.020)^2=3.1416\times10^{-4}\ \text{m}^2$, so $Q=V_{jet}A_{jet}=6.264\times3.1416\times10^{-4}=1.968\times10^{-3}\ \text{m}^3\text{/s}$.
Ideal hydraulic power. The pump must add exactly the kinetic energy leaving in the jet (equivalently, $\rho g Q h$, since $\tfrac12 V_{jet}^2=gh$):
$$\dot{P}_{ideal} = \rho g Q h = 1000\times9.81\times1.968\times10^{-3}\times2 = \boxed{38.6\ \text{W}}$$
Actual (shaft) power. Accounting for the 65% overall efficiency,
$$\dot{P}_{actual} = \frac{\dot{P}_{ideal}}{\eta} = \frac{38.6}{0.65} = \boxed{59.4\ \text{W}}$$
