Question

A pump draws water (density \(=1000\ \mathrm{kg\,m^{-3}}\)) at a steady rate of \(10\ \mathrm{kg\,s^{-1}}\). The pressures at the suction and discharge sides of the pump are \(-20\ \mathrm{kPa}\) (gauge) and \(350\ \mathrm{kPa}\) (gauge), respectively. The pipe diameters at the suction and discharge side are \(70\ \mathrm{mm}\) and \(50\ \mathrm{mm}\), respectively. The suction and discharge lines are at the same elevation, and the pump operates at an efficiency of \(80%\). Neglecting frictional losses in the system, the power (in kW) required to drive the pump is __________ (rounded off to two decimal places).

Hint:

For pump power with negligible losses and equal elevation: \[ P_{\text{shaft}}=\frac{\dot m}{\eta}\left(\frac{\Delta p}{\rho}+\frac{V_d^{2}-V_s^{2}}{2}\right). \]
Use gauge pressures directly for \(\Delta p\) when both are gauge values.

Answer 1

Correct Answer: 4.7

Step 1: Convert mass flow to volumetric flow and compute velocities
\[ \dot m = 10\ \mathrm{kg\,s^{-1}},\quad \rho=1000\ \mathrm{kg\,m^{-3}} \Rightarrow Q=\frac{\dot m}{\rho}=0.01\ \mathrm{m^3\,s^{-1}}. \] Areas: \[ A_s=\frac{\pi (0.07)^2}{4}=3.848\times10^{-3}\ \mathrm{m^2},\quad A_d=\frac{\pi (0.05)^2}{4}=1.963\times10^{-3}\ \mathrm{m^2}. \] Velocities: \[ V_s=\frac{Q}{A_s}=2.598\ \mathrm{m\,s^{-1}},\quad V_d=\frac{Q}{A_d}=5.093\ \mathrm{m\,s^{-1}}. \] Step 2: Specific energy added to the fluid (Bernoulli with pump)
Elevation change \(=0\). Pressure rise: \[ \Delta p = 350-(-20)=370\ \mathrm{kPa}=3.70\times10^{5}\ \mathrm{Pa}. \] Specific energy added: \[ e=\frac{\Delta p}{\rho}+\frac{V_d^{2}-V_s^{2}}{2} =\frac{3.70\times10^{5}}{1000}+\frac{5.093^{2}-2.598^{2}}{2} =379.59\ \mathrm{J\,kg^{-1}}. \] Step 3: Hydraulic power and shaft power
\[ P_{\text{hyd}}=\dot m\,e=10\times 379.59=3795.93\ \mathrm{W}. \] With efficiency \(\eta=0.80\), \[ P_{\text{shaft}}=\frac{P_{\text{hyd}}}{\eta} =\frac{3795.93}{0.80}=4744.91\ \mathrm{W} =4.74\ \mathrm{kW}\ (\text{to two decimals}). \]