Question 21.20: A single-acting reciprocating pump has a bore of 15 cm and s......

Given data:

Diameter of piston                          D =15 cm = 0.15 m

Stroke length                                    L = 40 cm = 0.40 m

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

Length of suction pipe                    l_{s}=8\;\mathrm{m}

Speed of pump                                  N = 30 rpm

Crank radius is given by

r={\frac{L}{2}}={\frac{0.4}{2}}=0.2~{\mathrm{m}}

Area of piston is given by

A={\frac{\pi}{4}}D^{2}= \frac{\pi}{4}(0.15)^{2}=0.0177\;\mathrm{m^{2}}

Angular speed is given by

\omega={\frac{2\pi N}{60}}= {\frac{2\pi\times90}{60}}=9.425\operatorname{rad/s}

For single-acting reciprocating pump, the rate of flow to or from air vessel is given by Eq. (21.51) as

Q_{air  vessel}=A\omega r\biggl(\sin\theta-\frac{1}{\pi}\biggr)      (21.51)

=A\omega r\biggl(\sin\theta-\frac{1}{\pi}\biggr)

=0.0177\times9.425\times0.2\Biggl({\sin\theta-{\frac{1}{\pi}}}\Biggr)

=0.0336(\sin\theta-0.3183)

Since the air vessel is fitted on suction pipe, positive sign of the above equation implies that the flow will take place from the air vessel. Similarly, negative sign of the above equation implies that the flow will take place into the air vessel

For θ = 45°

Flow = 0.03336 (sin 45° – 0.3183)

= 0.03336 (0.707 – 0.3183) = 0.01297 m³/s

Positive sign implies that the flow is taking place from the air vessel.
For \theta=90^{\circ}

Flow = 0.03336 (sin 90° – 0.3183)

= 0.03336 (1 – 0.3183) = 0.02274 m³/s

Positive sign implies that the flow is taking place from the air vessel

Flow = 0.03336 (sin 170° – 0.3183)

= 0.03336(0.1736 – 0.3183) = -0.00483 m³/s

Negative sign implies that the flow is taking place into the air vessel