Question 21.19: The following details refer to working of a single-acting re......

Diameter of piston                  D = 20 cm = 0.20 m

Crank radius                              r = 15 cm= 0.15 m

Speed of pump                          N = 50 rpm

Diameter of delivery pipe      $d_{{d}}$  =10 cm = 0.10 m

Length of delivery pipe         $l_{d}=25{\mathrm{~m}}$

Darcy’s friction factor            f = 0.02

Atmospheric pressure head  $h_{a t m}$ = 10.3 m of water

Area of piston is given by

$A=\frac{\pi}{4}D^{2}=$ $\frac{\pi}{4}(0.15)^{2}=0.0177\,{\mathrm{m}}^{2}$

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

$\omega={\frac{2\pi N}{60}}=$${\frac{2\pi\times50}{60}}=5.236\,\mathrm{rad/}{s}$

Without air vessel: The loss of head due to friction in delivery pipe is computed from Eq. (21.24) as

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

$={\frac{0.02\times25}{0.1\times2\times9.81}}$$\left({\frac{0.0177}{0.00785}}\times5.236\times0.15\right)^{2}$ = 0.7984 m

Power required in overcoming the friction is found to be

$P_{w i t h o u t}=\rho g Q$ $\times{\frac{2}{3}}h_{f d,m a x}$ $={\frac{\rho g A L N}{60}}\times{\frac{2}{3}}h_{f d,m a x}$           $\left[\because Q = \frac{ALN}{60} \right]$

$={\frac{1000\times9.81\times0.0177\times0.3\times40}{60}}$ $\times{\frac{2}{3}}\times0.7984=18.484{\mathrm{~W~}}$

With air vessel: The loss of head due to friction in delivery pipe by fitting an air vessel is computed as

$h_{f d}=\frac{fl_{d}\nu_{a\nu}^{2}}{d_{d}\times2g}=$ ${\frac{{\mathcal{fl}}_{d}}{d_{d}\times2g}}\times\left({\frac{A}{a_{d}}}\times{\frac{\omega r}{\pi}}\right)^{2}$

$={\frac{0.02\times25}{0.1\times2\times9.81}}\times$ $\left({\frac{0.0177}{0.00785}}\times{\frac{5.236\times0.15}{\pi}}\right)^{2}=0.081\,{\mathrm{m}}$

Power required in overcoming the friction is found to be

$P_{w i th}=\rho g Q\times h_{f d}$ $=\frac{\rho g A L N}{60}\times h_{f d}$

$={\frac{1000\times9.8\ 1\times0.0177\times0.3\times40}{60}}$ $\times0.081=2.183\,\,\mathrm{W}$

The power saved in overcoming the friction by fitting the air vessel is

$P_{w i t h o u t}-P_{w i t h}=$ 18.484 – 2.183 = 16.301 W