Question 21.12: A single-acting reciprocating pump has a piston diameter 15 ......

Given data:

Diameter of piston                         D = 15 cm = 0.15 m

Crank radius                                   r = 20 cm = 0.20 m

Diameter of delivery pipe            d_{d}=8\;{\mathrm{cm}}=0.08\;{\mathrm{m}}

Length of delivery pipe              l_{d}=22\;\mathrm{m}

Delivery head                                h_{d}=16\;\mathrm{m}

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

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

Area of piston is given by

\ A={\frac{\pi}{4}}D^{2}={\frac{\pi}{4}}(0.15)^{2}=0.01767\,{\mathrm{m}}^{2}

Area of delivery pipe is given by

a_{d}=\frac{\pi}{4}d_{d}^{2}=\frac{\pi}{4}(0.08)^{2}=0.00503\,{\mathrm{m}}^{2}

Pressure head at the end of the delivery stroke is given by

=h_{a t m}+h_{d}-h_{a d}

where h_{a d} is the acceleration head and is given by Eq. (21.18) as

h_{a d}=\frac{l_{d}}{g}\times\frac{A}{a_{d}}\,\omega^{2}r\cos\theta              [\because \theta=180^{\circ}]

={\frac{l_{d}}{g}}\times{\frac{A}{a_{d}}}\omega^{2}r

Since the absolute pressure head during delivery stroke is minimum at the end of the stroke, separation can take place at the end of the stroke. For that the pressure head at the end of the stroke is equal to the separation pressure head.

Thus, one can write

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

or                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

2.5=10.3+16-\frac{22}{9.81}\times\frac{0.01767}{0.00503}\times\omega^{2}\times0.20

or                  {\frac{22}{9.81}}\times{\frac{0.01767}{0.00503}}\times\omega^{2}\times 0.20=10.3+16-2.5

or                  1.5756\omega^{2}=23.8

or                {\boldsymbol{\omega}}^{2}={\frac{23.8}{1.5756}}=15.105

or                \omega={\sqrt{15.105}}=3.886\,\,{\mathrm{rad/s}}

Rotational speed is given by

N=\frac{60\omega}{2\pi}\               \left[\because \omega={\frac{2\pi N}{60}}\right]

={\frac{60\times3.886}{2\pi}}=37.11\,{\mathrm{rpm}}