Question 21.11: The piston diameter and stroke length of a single-acting rec......

Given data:

Diameter of piston                               D=100 mm = 0.10 m

Stroke length                                        L = 250 mm = 0.25 m

Diameter of suction pipe                  {d}_{s} = 60 mm = 0.06 m

Length of suction pipe                      l_{s} = 7 m

Suction lift                                          h_{s}=4\;{\mathrm{m}}

Atmospheric pressure head            h_{a tm} = 10.3 m of water

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

Crank radius is given by

r={\frac{L}{2}}={\frac{0.25}{2}}=0.125\operatorname{m}

Area of piston is given by

A={\frac{\pi}{4}}D^{2}={\frac{\pi}{4}}(0.10)^{2}=0.00785\,{\mathrm{m}}^{2}

Area of suction pipe is given by

a_{s}=\frac{\pi}4d_{s}^{2}=\frac{\pi}4(0.06)^{2}=0.00283\,\mathrm{m}^{2}

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

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

where h_{a s}  is the acceleration head and is given by Eq. (21.17) as

h_{a s}=\frac{l_{s}}{g}\times\frac{A}{a_{s}}\omega^{2}r\cos 0^{\circ}                                          \left[\because \theta=0^{\circ}\right]

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

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. For that the pressure head at the beginning of the stroke is equal to 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{d}{a_{s}}}\omega^{2}r

Substituting the values, we have

2.5=10.3-4-{\frac{7}{9.81}}\times{\frac{0.00785}{0.00283}}\times\omega^{2}\times0.125

or                  {\frac{7}{9.81}}\times{\frac{0.00785}{0.00283}}\times\omega^{2}\times0.125=10.3-4-2.5=3.8

or                  0.2474\omega^{2}=3.8

or                \omega^{2}{=}\frac{3.8}{0.2474}=15.3597

or                \omega={\sqrt{15.3597}}=3.919\,\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\times3.919}{2\pi}}=37.42\;\mathrm{rpm}