Question 21.17: The piston diameter and stroke length of a single-acting rec......
The piston diameter and stroke length of a single-acting reciprocating pump are 250 mm and 300 mm, respectively. The centre of the pump is 4 m above the water level in the sump. The diameter and length of the suction pipe are 100 mm and 7 m, respectively. It delivers water to a tank 12 m above the pump through a delivery pipe of 100 mm in diameter and 18 m long. Separation occurs at 75 kN/m² below the atmospheric pressure. Find the maximum speed at which pump can run without separation. Take atmospheric pressure as 10.3 m of water.
Given data:
Diameter of piston D = 250 mm = 0.25 m
Stroke length L = 300 cm = 0.30 m
Suction head h_{s}=4\;\mathrm{m}
Delivery head h_{d}=12\;\mathrm{m}
Diameter of suction pipe d_{s}=100\,\mathrm{mm}=0.10\,\mathrm{m}
Length of suction pipe l_{s}=7\,{\mathrm{m}}
Diameter of delivery pipe d_{d}=100\,\mathrm{mm}=0.10\,\mathrm{m}
Length of delivery pipe l_{d}=18\;\mathrm{m}
Atmospheric pressure head h_{a t m}=10.3\;\mathrm{m~of~water}
Separation pressure p_{s e p}=75\operatorname{kN/m^{2}}=75\times10^{3}\operatorname{N/m^{2}}\ below atmospheric
Separation pressure head h_{s e p}={\frac{75\times10^{3}}{10^{3}\times9.81}}=7.645\;\mathrm{m} of water below atmospheric
h_{s e p}=10.3-7.645=2.655\,\mathrm{m~of~water~(abs)}
Crank radius is given by
r={\frac{L}{2}}={\frac{0.3}{2}}=0.15\,{\mathrm{m}}
Area of piston is
A={\frac{\pi}{4}}D^{2}={\frac{\pi}{4}}(0.25)^{2}=0.049\,\mathrm{m^{2}\quad\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}}
First of all we will find the speed of the pump without separation during suction and delivery strokes separately.
Maximum speed of the pump without separation during suction stroke: The maximum speed of the pump without separation during suction stroke is computed from Eq. (21.33) as
h_{s e p}=h_{a t m}-h_{s}-{\frac{l_{s}}{g}}\times{\frac{A}{a_{s}}}\omega^{2}r
Substituting the values, we have
2.655=10.3-4-{\frac{7}{9.81}}\times\frac{0.049}{0.00785}\times\omega^{2}\times0.15
or {\frac{7}{9.81}}\times{\frac{0.049}{0.00785}}\times{\omega}^{2}\times{\ 0}.15=10.3-4-2.655=3.645
or 0.6681\omega^{2}=3.645
or \omega^{2}={\frac{3.645}{0.6681}}=5.4558
or \omega={\sqrt{5.4558}}=2.336\,{\mathrm{rad/s}}
Speed of the pump without separation during delivery stroke: The maximum speed of the pump without separation during delivery stroke is computed from Eq. (21.35) as
h_{s e p}=h_{a t m}+h_{d}-{\frac{l_{d}}{g}}\times{\frac{A}{a_{d}}}\omega^{2}r
Substituting the values, we have
\begin{array}{c l c r}{{{2.655=10.3+12-\frac{18}{9.81}}\times\frac{0.049}{0.00785}\times{\omega}^{2}\times0.15}}\end{array}
or \frac{18}{9.81}\times\frac{0.049}{0.00785}\times{\omega}^{2}\times0.15=10.3+12-2.655=19.645
or 1.718\omega^{2}=19.645
or {\boldsymbol{\omega}}^{2}={\frac{19.645}{1.718}}=11.4348
or {\boldsymbol{\omega}}={\sqrt{11.4348}}=3.382\ \mathrm{rad/s}
The maximum speed at which the pump may be operated without separation is the minimum of the above two speeds, i.e., 2.336 rad/s.
Rotational speed is given by
N={\frac{60\omega}{2\pi}} \left[\because \omega={\frac{2\pi N}{60}}\right]
={\frac{60\times2.236}{2\pi}}=21.35\,{\mathrm{rpm}}