Question 4 of 4: Pump Power to Water

Problem 4: Pump Power to Water (25 points)

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. A pump draws water from an open reservoir through a $D=12$ cm suction line rising 2 m above the free surface to the pump, then discharges through a $D_e=5$ cm nozzle to atmosphere at the same elevation as the pump.

Find. The pump power delivered to the water, in kW.

Approach. Apply the steady-flow energy equation between the reservoir free surface (1) and the nozzle exit (2), both at atmospheric pressure with $V_1\approx0$; solve for the pump head $h_p$, then convert to power via $\dot W_p=\rho g Q h_p$.

1. Nozzle exit velocity from continuity. $$A_e = \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2,\qquad V_2 = \frac{Q}{A_e} = \frac{0.0611}{1.9635\times10^{-3}} = \boxed{V_2\approx31.12\ \text{m/s}}.$$
2. Energy equation, free surface (1) to nozzle exit (2). With $p_1=p_2=0$ (gauge) and $V_1\approx0$, $$h_p = (z_2-z_1)+\frac{V_2^2}{2g}+h_f = 2+\frac{31.12^2}{2(9.81)}+5 = \boxed{h_p\approx56.4\ \text{m}}.$$ The nozzle's kinetic-energy term dominates the pump head, as expected for a jet discharging at over 31 m/s.
3. Pump power delivered to the water. $$\dot W_p = \rho g Q h_p = (998)(9.81)(0.0611)(56.4) = \boxed{\dot W_p\approx33.7\ \text{kW}}.$$