Question 4 of 4: Pump Power for Gasoline Delivery

Problem 4: Pump Power for Gasoline Delivery (equal value)

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.

Flow rate $Q=12\ \text{m}^3\text{/h}$; specific weight $\gamma=\rho g=6671\ \text{N/m}^3$; motor efficiency $\eta=75\%$.

Find. The motor power required to drive the pump.

Approach. Apply the steady-flow energy equation per unit weight (Bernoulli with a pump-head term) between inlet and outlet to find the pump head $h_p$, convert to hydraulic (fluid) power via $\dot{W}_{\text{fluid}}=\gamma Q h_p$, then divide by motor efficiency to get the power the motor must supply.

1. Pump head from the energy balance. $$h_p=\frac{p_2-p_1}{\gamma}+\frac{V_2^2-V_1^2}{2g}+(z_2-z_1)$$ $$h_p=\frac{(500-100)\times10^3}{6671}+\frac{3^2-2^2}{2(9.81)}+(4-1)=59.97+0.255+3=\boxed{63.2\ \text{m}}$$
2. Hydraulic (fluid) power. $Q=12/3600=3.333\times10^{-3}\ \text{m}^3\text{/s}$: $$\dot{W}_{\text{fluid}}=\gamma Q h_p=(6671)(3.333\times10^{-3})(63.2)=\boxed{1.41\ \text{kW}}$$
3. Motor power required. The motor must overcome its own 75% efficiency to deliver that hydraulic power: $$\dot{W}_{\text{motor}}=\frac{\dot{W}_{\text{fluid}}}{\eta}=\frac{1.41}{0.75}=\boxed{1.87\ \text{kW}}\ (\approx2.51\ \text{hp})$$