The following details refer to working of a single-acting reciprocating pump with an air vessel fitted on the delivery pipe very close to the cylinder.
Piston diameter = 20 cm
Crank radius = 15 cm
Speed of the pump = 50 rpm.
Diameter of the delivery pipe = 10 cm
Length of the delivery pipe = 25 m
Darcy’s friction factor = 0.02
Atmospheric pressure head = 10.3 m of water
Find the power saved in overcoming the friction by fitting the air vessel.
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