Question 21.18: The piston diameter and stroke length of a single-acting rec......

Given data:

Diameter of piston                             D = 20 cm = 0.20 m

Stroke length                                    L= 35 cm = 0.35 m

Speed of pump                                 N = 40 rpm

Delivery head                                  h_{d}=15\;{\mathrm{m}}

Diameter of delivery pipe              d_{d}=10  cm=0.10\,\mathrm{m}

Length of delivery pipe              l = 25 m

Distance between air vessel and centre of the pump l_{d}^{\prime}=1.5\;\mathrm{m}

Length of delivery pipe above the air vessel is l_{d}=l-l_{d}^{\prime}=25-1.5=23.5\,{\mathrm{m}}

Atmospheric pressure head            h_{a t m}=10.3\;\mathrm{m~of~water}

Darcy’s friction factor                 f = 0.03

Crank radius is

r={\frac{L}{2}}={\frac{0.35}{2}}=0.175\;{\mathrm{m}}

Area of piston is given by

A={\frac{\pi}{4}}D^{2}={\frac{\pi}{4}}(0.2)^{2}=0.0314\,\mathrm{m}^{2}

Area of delivery pipe is given by

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

Angular speed is given by

{\boldsymbol{\omega}}={\frac{2\pi N}{60}}={\frac{2\pi\times40}{60}}=4.189\,\mathrm{rad/s}

The pressure head in the cylinder at the beginning of the delivery stroke is given by Eq. (21.40) as

h=h_{a m}+\Biggl[h_{d}+\frac{l_{d}^{\prime}}{g}\times\frac{A}{a_{d}}\omega^{2}r+\frac{{f}l_d}{d_{d}\times2g}\biggl(\frac{A}{a_{d}}\times\frac{\omega r}{\pi}\biggr)^{2}+\frac{1}{2g}\biggl(\frac{A}{a_{d}}\times\frac{\omega r}{\pi}\biggr)^{2}\Biggr]

=10.3+15+{\frac{1.5}{9.81}}\times{\frac{0.0314}{0.00785}}\times4.189^{2}\times0.175

+\,\frac{0.03\times25}{0.1\times2\times 9.81}{\bigg(}\frac{0.0314}{0.00785}\times\frac{4.189\times0.175}{\pi}{\bigg)}^{2}+\,\frac{1}{2\times9.81}\biggl(\frac{0.0314}{0.00785}\times\frac{4.189\times0.175}{\pi}\biggr)^{2}

= 10.3 + 15 + 1.878 + 0.333 + 0.044 = 27.555 m of water (abs)          \ast

The pressure head in the cylinder at the middle of the delivery stroke is given by Eq. (21.41) as

h=h_{a t m}+ \left[h_{d}+{\frac{\mathcal{fl}_{d}}{d_{d}\times2g}}\right. \left({\frac{A}{a_{d}}}\times{\frac{\omega r}{\pi}}\right]^{2} +\,\frac{fl_{d}^{\prime}}{d_{d}\times2g}\biggl(\frac{A}{a_{d}}\omega r\biggr)^{2}+\,{\frac{1}{2g}}\bigg({\frac{A}{a_{d}}}\times{\frac{\omega r}{\pi}}\bigg)^{2}\,\bigg]

=10.3+15+\frac{0.03\times25}{0.1\times2\times9.81}\left({\frac{0.0314}{0.00785}}\times{\frac{4.189\times0.175}{\pi}}\right)^{2}

+\,\frac{0.03\times1.5}{0.1\times2\times9.81}\!\left(\frac{0.03\,14}{0.00785}\times4.189\times0.175\right)^{2}+\frac{1}{2\times9.81}\!\left(\frac{0.03\,14}{0.00785}\times\frac{4.189\times0.175}{\pi}\right)^{2}

= 10.3 + 15 + 0.333 + 0.197 + 0.044 = 25.874 m of water (abs)

The pressure head in the cylinder at the end of the delivery stroke is given by Eq. (21.43) as

h=h_{a t m} + \biggl[h_{d}-\frac{l_{d}^{\prime}}{g}\times {\frac{\mathcal A}{a_{d}}}\omega^{2}r +\frac{f l_{d}}{d_{d}\times2g} \left({\frac{A}{a_{d}}}\times{\frac{\omega r}{\pi}}\right)^{2} \displaystyle\left.+\,\frac{1}{2\,g}\bigg(\frac{A}{\,a_{d}}\times\frac{\omega_{^T}}{\pi}\bigg)^{2}\right]

=10.3+1.5-{\frac{1.5}{9.81}}\times{\frac{0.03\,14}{0.00785}}\times4.189^{2}\times0.175

+\frac{0.03\times25}{0.1\times2\times9.81}\biggl(\frac{0.03\,14}{0.00785}\times\frac{4.189\times0.175}{\pi}\biggr)^{2}\, \displaystyle{}+\frac1{2\times9.81}\bigg(\frac{0.0314}{0.00785}\times\frac{4.189\times0.175}{\pi}\bigg)^{2}

= 10.3 + 15 – 1.878 + 0.333 + 0.044 = 23.799 m of water (abs)