Question 15.8: A single acting reciprocating pump has plunger diameter 200 ......
A single acting reciprocating pump has plunger diameter 200 mm and stroke length 300 mm. The suction pipe is 100 mm in diameter and 8 m long. The water surface from which the pump draws water is 4 m below the pump cylinder axis. If the pump is working at 30 rpm, find the pressure head in the cylinder at the beginning, middle and end of suction stroke. Take friction factor f = 0.04.
Given: D = 200 m = 0.2 m, d_{s}, = 100 mm = 0.1 m, L = 300 mm = 0.3 m, r = 0.3/2 = 0.15 m, l_{s} = 8 m, H_{s}= 4 m, f = 0.04 and N = 30 rpm
Let us take atmospheric pressure head H_{at} = 10.3 m of water
Again, Angular velocity of the crank {{\omega}}={\frac{2\pi N}{60}}={\frac{2\pi\times30}{60}}=\pi
At the beginning of the suction stroke, h_{fs} = 0.
Hence, Pressure head = H_{at}-(H_{s}+h_{as})
or =10.3-\left\lgroup4+{\frac{l_{s}}{g}}\,{\frac{A_{s}}{a_{s}}}\,r\omega^{2}\right\rgroup
or =\,10.3-\left\lgroup4+{\frac{8}{9.81}}\left\lgroup{\frac{0.2}{0.1}}\right\rgroup ^{2}\times0.15\times\pi^{2}\right\rgroup
or = 1.4708 m of water absolute
At the middle of the suction stroke, h_{as} is zero. But loss due to friction is maximum given by Eq. (15.23).
h_{f\operatorname*{max}}={\frac{f l}{2g d}}\left\lgroup{\frac{A}{a}}r\omega^{2}\right\rgroup (15.23)
Therefore,
Pressure head in the middle of the stroke = [H_{a t}-(H_{s}+h_{f{s}})]
=\;10.3-4-\frac{{fl}_{s}}{2g d_{s}}\left\lgroup\frac{A}{a_{s}}r\omega\right\rgroup ^{2}
=\,6.3-\frac{0.04\times8}{2\times9.81\times0.1}\left\lgroup\frac{(\pi/4)\times(0.2)^{2}}{(\pi/4)\times(0.1)^{2}}\times0.15\times\pi\right\rgroup ^{2}
= 6.3 – 0.5795 = 5.7205 m
= 5.7205 m of water absolute
At the end of the stroke, h_{fs} is zero and pressure head H_{a t}-(H_{s}-h_{a s})
=10.3-\left\lgroup4-{\frac{l_{s}}{g}}\,{\frac{A_{s}}{a_{s}}}\,r\omega^{2}\right\rgroup
=10.3-\left\lgroup4-{\frac{8}{9.81}}\left\lgroup{\frac{0.2}{0.1}}\right\rgroup ^{2}\times0.15\times\pi^{2}\right\rgroup
= [10.3 – {4 – 4.8292}]
= 11.129 m of water absolute