Question 4 of 13: Electrical Power for a Gasoline Transfer Pump

Question 4: Electrical Power for a Gasoline Transfer Pump (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 electrical power drawn by the pump motor.

Approach. Apply the steady-flow energy equation between the suction and discharge gauge points, with the pump head h_p as the only unknown; convert the resulting fluid power to electrical power through the stated efficiency.

1. Pipe velocities from continuity (A = πD²/4): $$V_s=\frac{Q}{A_s}=\frac{0.035}{\frac{\pi}{4}(0.15)^2}=1.98\ \text{m/s}, \qquad V_d=\frac{Q}{A_d}=\frac{0.035}{\frac{\pi}{4}(0.10)^2}=4.46\ \text{m/s}$$
2. Energy equation, suction → discharge, pump head h_p added between the gauges: $$h_p=\frac{p_d-p_s}{\rho g}+\frac{V_d^2-V_s^2}{2g}+(z_d-z_s)$$
3. Substitute (ρ = 750 kg/m³; p_d−p_s = 150−(−30) = 180 kPa; z_d−z_s = 1.5−(−0.5) = 2.0 m): $$h_p=\frac{180\,000}{750(9.81)}+\frac{4.46^2-1.98^2}{2(9.81)}+2.0 = 24.47+0.81+2.0=\boxed{27.28\ \text{m}}$$
4. Fluid (hydraulic) power delivered to the gasoline: $$P_{fluid}=\rho g Q h_p = 750(9.81)(0.035)(27.28) = 7024\ \text{W} = 7.02\ \text{kW}$$
5. Electrical power (fluid power divided by pump efficiency): $$P_{elec}=\frac{P_{fluid}}{\eta}=\frac{7.02}{0.75}=\boxed{9.37\ \text{kW}}$$