Question 21.10: A single-acting reciprocating pump running at 30 rpm has plu......

Given data:

Diameter of plunger                          D = 12 cm = 0.12 m

Stroke length                                     L= 25 cm

Diameter of delivery pipe              d_{\mathrm{{e}}}=7.5\;{\mathrm{cm}}=0.075\;{\mathrm{m}}

Length of delivery pipe                    l_{d}=25{\mathrm{~m}}

Delivery head                                    h_{d}=18\;\mathrm{m}

Speed of pump                                  N= 30 rpm

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

Area of plunger is given by

A={\frac{\pi}{4}}D^{2}={\frac{\pi}{4}}(0.12)^{2}=0.0113\ \mathrm{m^{2}}

   Area of delivery pipe is given by

a_{d}=\frac{\pi}{4}\,d_{d}^{2}=\frac{\pi}{4}(0.075)^{2}=0.00442\;\mathrm{m}^{2}

Crank radius r={\frac{L}{2}}={\frac{25}{2}}=12.5\,\mathrm{cm}=0.125\,\mathrm{m}

Angular speed \omega={\frac{2\pi N}{60}}={\frac{2\pi\times30}{60}}=3.142\;{\mathrm{rad/s}}

The acceleration head in the delivery pipe is given by Eq. (21.18) as

h_{a d}=\frac{l_{d}}{g}\times\frac{d}{a_{d}}\,\omega^{2}r\cos\theta

At the beginning of the delivery stroke (\theta=0^{\mathrm{o}}), the pressure head due to acceleration is

h_{a d}=\frac{25}{9.81}\times\frac{0.0113}{0.00442}\times3.142^{2}\times0.125\times\cos0^{\circ}=7.065\,\mathrm{m}

At the middle of the delivery stroke (\theta=90^{\circ}), the pressure head due to acceleration is

h_{a d}={\frac{25}{9.81}}\times{\frac{0.0113}{0.00442}}\times3.142^{2}\times0.125\times\cos 90^{\circ}=0

At the end of the delivery stroke (\theta=180^{\circ}), the pressure head due to acceleration is

h_{a d}={\frac{25}{9.81}}\times{\frac{0.0113}{0.00442}}\times3.142^{2}\times0.125\times\cos180^{\circ}=-7.065\,{\mathrm{m}}

The pressure head in the cylinder at the beginning of the delivery stroke is (refer to Fig. 21.5)

= (h_{d}+h_{a d}) above the atmospheric pressure head
=18 + 7.065 = 25.065 m above the atmospheric pressure head
= Atmospheric pressure head + 25.065 m
=10.3 + 25.065 = 35.365 m of water (abs)

The pressure head in the cylinder at the middle of the delivery stroke is (refer to Fig. 21.5)

= h_{d} above the atmospheric pressure head
=18 m above the atmospheric pressure head
= Atmospheric pressure head + 18 m
=10.3 +18 =18.3 m of water (abs)

The pressure head in the cylinder at the end of the delivery stroke is (refer to Fig. 21.5)

= (h_{d}-h_{a d}) above the atmospheric pressure head
=18 – 7.065 =10.935 m above the atmospheric pressure head
= Atmospheric pressure head + 10.935 m
=10.3 +10.935 = 21.235 m of water (abs)