Question 21.13: A single-acting reciprocating pump operating at 60 rpm has a......

Given data:

Diameter of piston                                   D = 20 cm = 0.2 m

Stroke length                                   L= 30 cm = 0.30 m

Speed of pump                                N= 60 rpm

Diameter of suction pipe          d_{s}=10\;{\mathrm{cm}}=0.10\;{\mathrm{m}}

Length of suction pipe                l_{s}=7\;{\mathrm{m}}

Separation pressure head          h_{s e p}=2.5\;\mathrm{m\;of~water\;(abs)}

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

Crank radius is given by

r={\frac{L}{2}}={\frac{0.3}{2}}=0.15\,\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 suction pipe is given by

a_{s}=\frac{\pi}{4}d_{s}^{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\times40}{60}}=4.189\,\mathrm{rad/s}

Pressure head at the beginning of the suction stroke is given by

=h_{a t m}-(h_{s}+h_{a s})

where h_{s} is the suction lift at the beginning of the suction stroke and h_{a s} is the acceleration head.

Since the absolute pressure head during suction stroke is minimum at the beginning of the stroke, separation can take place at the beginning of the stroke. When the suction lift is equal to the maximum permissible value, the pressure head at the beginning of the suction stroke will be same as the separation pressure head.

Thus, one can write

h_{s e p}=h_{a t m}-(h_{s}+h_{a s})

or                                        h_{s e p}=h_{a t m}-h_{s}-{\frac{l_{s}}{g}}\times{\frac{A}{a_{s}}}\omega^{2}r              [\because \theta=0^{\circ}]

Substituting the values, we have

2.5=10.3-h_{s}-\frac{5}{9.81}\times\frac{0.03\,14}{0.00785}\times4.189^{2}\times0.15

or                    2.5=10.3-h_{s}-5.366

or                    h_{s}=10.3-2.5-5.366=2.434\,\mathrm{m}