Pump Power to Water: 22-Agric-A4 Fluid Flow, December 2017, Q4

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.

The 6-m submerged intake depth shown in the figure does not affect the energy balance since the calculation starts at the free surface itself.

Given. A pump draws water from an open reservoir (intake 6 m below the free surface) through a $D=12$ cm suction line rising 2 m above the free surface to the pump, then discharges horizontally through a $D_e=5$ cm nozzle to atmosphere at the same elevation as the pump.

Quantity	Value
Flow rate, $Q$	220 m³/hr $=0.0611$ m³/s
Nozzle exit diameter, $D_e$	5 cm
Elevation rise, pump/nozzle above free surface	2 m
Total friction head loss, $h_f$	5 m
Water, 20°C	$\rho=998\ \text{kg/m}^3$

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

Figure 4 — Pump draws water from the open reservoir (1) via a $D=12$ cm suction line (intake 6 m below the free surface) rising 2 m above the surface, then discharges through a $D_e=5$ cm nozzle to atmosphere at (2).

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 $\dot W_p=\rho g Q h_p$. The 6-m submerged intake depth only describes where the suction pipe draws from within the reservoir — it does not enter the energy balance since station (1) is the free surface itself.

Nozzle exit velocity from continuity. $$A_e = \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2,\qquad V_2 = \frac{Q}{A_e} = \frac{0.0611}{1.9635\times10^{-3}} = \boxed{V_2\approx31.12\ \text{m/s}}.$$
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.
Pump power delivered to the water. $$\dot W_p = \rho g Q h_p = (998)(9.81)(0.0611)(56.4) = \boxed{\dot W_p\approx33.7\ \text{kW}}.$$

Quantity	Result
Nozzle exit velocity, $V_2$	31.12 m/s
Pump head, $h_p$	56.4 m
Pump power delivered to water	33.7 kW

Concept & Theory

Topic. Pump sizing via the steady-flow (extended Bernoulli) energy equation with a head-loss and pump-head term.

Key relations. Energy equation $p_1/\rho g+V_1^2/2g+z_1+h_p = p_2/\rho g+V_2^2/2g+z_2+h_f$; power delivered to the fluid $\dot W_p=\rho g Q h_p=\gamma Q h_p$.

Why this works. Choosing the reservoir free surface and the nozzle exit as the two energy-equation stations eliminates the unknown internal pipe pressures entirely — both stations are at atmospheric pressure, and the reservoir surface velocity is negligible for a large tank, leaving only the elevation rise, the nozzle kinetic energy, and the stated friction loss to balance against the pump head.

Common pitfall. Forgetting that "pump power delivered to the water" is $\rho g Q h_p$, the hydraulic (water) power — not the shaft input power, which would additionally divide by the pump efficiency (not given here, so not applied).

Source. White, Fluid Mechanics, Ch. 3 (Integral Relations) / Ch. 11 (Turbomachinery) — Energy Equation with Pumps and Turbines.

Question 4 of 4: Pump Power to Water

Problem 4: Pump Power to Water (25 points)