Question 21.16: Water is flowing through a single-acting reciprocating pump ......

Given data:
Diameter of piston                          D = 250 mm = 0.25 m

Crank radius                                    r = 200 cm = 0.20 m

Suction head                                    h_{s}=3\;{\textrm{m}}

Delivery head                                  h_{d}=14\;\mathrm{m}

Diameter of suction pipe              d_{s}=100\;\mathrm{mm}=0.10\;\mathrm{m}

Length of suction pipe                l_{s}=5  {\textrm{m}}

Diameter of delivery pipe          d_{d}=100\,\mathrm{mm}=0.10\,\mathrm{m}

Length of delivery pipe            l_{d}=18\operatorname{m}

Speed of pump                            N = 30 rpm

Darcy’s friction factor                f = 0.01

Area of piston is given by

A={\frac{\pi}{4}}D^{2}={\frac{\pi}{4}}(0.25)^{2}=0.049\,\mathrm{m^{2}\quad}

Area of suction and delivery pipe are given by

 a_{s}=a_{d}=\frac{\pi}{4}d_{s}^{2}=\frac{\pi}{4}(0.1)^{2}=0.00785\,\mathrm{m}^{2}

Angular speed is

\omega={\frac{2\pi N}{60}}={\frac{2\pi\times30}{60}}=3.142\;{\mathrm{rad/s}}

The maximum loss of head due to friction in suction pipe is computed from Eq. (21.23) as

h_{f s,m a x}={\frac{\mathcal{fl}_{s}}{d_{s}\times2g}}{\left({\frac{A}{a_{s}}}\omega r\right)}^{2}

={\frac{0.01\times5}{0.1\times2\times9.81}}\left({\frac{0.049}{0.00785}}\times3.142\times0.2\right)^{2}=0.392\,\mathrm{m}

The maximum loss of head due to friction in delivery pipe is computed from Eq. (21.24) as

h_{fd,m a x}=\frac{\mathcal{fl}_{d}}{d_{d}\times2g}{\left(\frac{A}{a_{d}}\omega r\right)^{2}}

={\frac{0.01\times18}{0.1\times2\times9.81}}\left({\frac{0.049}{0.00785}}\times3.142\times0.2\right)^{2}=1.411\,\mathrm{m}

Power required to drive the pump is given by Eq. (21.30) as