# Question 21.15: A single-acting reciprocating pump having a bore and stroke ......

A single-acting reciprocating pump having a bore and stroke of 20 cm and 30 cm respectively runs at 30 rpm. The suction and delivery pipes are both of 10 cm in diameter. The lengths of the suction and delivery pipes are 5 m and 15 m respectively. The suction and delivery heads are 3.2 m and 10 m, respectively. Calculate the pressure head in the cylinder at the start, middle and end of each stroke. Assume that the piston has a simple harmonic motion. Take atmospheric pressure as 10.3 m of water and Darcy’s friction factor f as 0.005.

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Given data:

Diameter of piston                      D = 20 cm = 0.2 m

Stroke length                                L = 30 cm = 0.30 m

Speed of pump                            N = 30 rpm

Suction head                                $\quad h_{_{S}}=3.2\;\operatorname{m}$

Delivery head                              $h_{d}=10\,\mathrm{m}$

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

Length of suction pipe            $l_{s}=5\;{\mathrm{m}}$

Diameter of delivery pipe      $d_{d}=10\;\mathrm{cm}=0.10\;\mathrm{m}$

Length of delivery pipe            $l_{d}=15~\mathrm{m}$

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

Darcy’s friction factor              f = 0.005

$r={\frac{L}{2}}={\frac{0.3}{2}}=0.15\;\mathrm{m}$

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 and delivery pipes are given by

$a_{s}=a_{d}=\frac{\pi}{4}d_{s}^{2}=\frac{\pi}{4}(0.1)^{2}=0.00785\;{\mathrm{m}}^{2}$

Angular speed is

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

The pressure head due to acceleration on suction pipe is obtained from Eq. (21.17) as

$h_{a s}=\frac{l_{s}}{g}\times\frac{A}{a_{s}}\,\omega^{2}r\,\mathrm{cos}\,\theta$

$=\frac{5}{9.81}\times\frac{0.03\,14}{0.00785}\times3.142^{2}\times 0.15\times\cos\theta=3.02\;\cos\theta$

The pressure head due to acceleration on delivery pipe is found from Eq. (21.18) as

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

$\,=\frac{15}{9.81}\times\frac{0.03\,14}{0.00785}\times3.142^{2}\times 0.15\times\cos\theta=9.05\;\mathrm{cos}\;\theta$

The loss of head due to friction in suction pipe is computed from Eq. (21.23) as

$h_{f s}=\frac{\mathcal{fl}_{s}}{d_{s}\times2g}\!\left(\frac{A}{a_{s}}\omega r\;\mathrm{sin}\;\theta\right)^{2}$

$={\frac{0.005\times5}{0.1\times2\times9.81}}\left({\frac{0.03\,14}{0.00785}}\times3.142\times0.15\times\mathrm{sin}\;\theta\right)^{2}$

$=0.045\ \mathrm{sin}^{2}\,\theta\,$

The loss of head due to friction in delivery pipe is computed from Eq. (21.24) as

$\mathcal{h}_{f d}\ \equiv\frac{\mathcal{fl}_{d}}{\mathcal{d}_{d}\times2\mathcal{g}}\left({\frac{A}{a_{d}}}\omega r\sin\theta\right)^2$

$={\frac{0.005\times15}{0.1\times2\times9.81}}{\left({\frac{0.0314}{0.00785}}\times3.142\times0.15\times\sin\theta\right)}^{2}$

$=0.136\,\sin^{2}\theta$

At the beginning of suction and delivery strokes         At the beginning of suction and delivery strokes, $\theta=0^{\circ}. \mathrm{Thus},$

$h_{a s}=3.02\cos{0}^{\circ}=3.02\,{\mathrm{m}}$

$h_{a d}=9.05\cos0^{\circ}=9.05\ m\ {\mathrm{and}}$

$h_{f s}=h_{f d}=0$

The pressure head in the cylinder at the beginning of the suction stroke is (refer to Fig. 21.7)

= $(h_{s}+h_{a s})$ below the atmospheric pressure head
= 3.2 + 3.02 = 6.22 m below the atmospheric pressure head
= Atmospheric pressure head – 6.22 m
= 10.3 – 6.22 = 4.08 m of water (abs)

The pressure head in the cylinder at the beginning of the delivery stroke is (refer to Fig. 21.7)

= $(h_{d} + h_{a d})$ above the atmospheric pressure head
= 10 + 9.05 = 19.05 m above the atmospheric pressure head
= Atmospheric pressure head +19.05 m
= 10.3 + 19.05 = 29.35 m of water (abs)

At the middle of suction and delivery strokes            At the middle of suction and delivery strokes, $\theta$ = 90°. Thus,

$h_{a s}=h_{a d}=0\quad.$

$h_{f s}=0.045\;\mathrm{sin}^{2}\;90^{\circ}=0.045$

$h_{fd}=0.136\sin^{2}90^{\circ}=0.136$

The pressure head in the cylinder at the middle of the suction stroke is (refer to Fig. 21.7)

= $(h_{s}+h_{f s})$ below the atmospheric pressure head
= 3.2 + 0.045 = 3.245 m below the atmospheric pressure head
= Atmospheric pressure head – 3.245 m
= 10.3 – 3.245 = 7.055 m of water (abs)

The pressure head in the cylinder at the middle of the delivery stroke is (refer to Fig. 21.7)

= $(h_{d\lnot}h_{f d})$ above the atmospheric pressure head
= 10 + 0.136 =10.136 m above the atmospheric pressure head
= Atmospheric pressure head +10.136 m
= 10.3 + 10.136 = 20.436 m of water (abs)

At the end of suction and delivery strokes          At the end of suction and delivery strokes, ${\boldsymbol{\theta}}=$ 180°. Thus,

$h_{a s}=3.02\cos180^{\circ}=-3.02\operatorname{m}$

$h_{_{\alpha d}}=9.05\cos180^{\circ}=-9.05\mathrm{~m},\operatorname{and}$

$h_{f s}=h_{f d}=0$

The pressure head in the cylinder at the end of the suction stroke is (refer to Fig. 21.7)

= $(h_{s}+h_{a s})|$ below the atmospheric pressure head
= 3.2 – 3.02 = 0.18 m below the atmospheric pressure head
= Atmospheric pressure head – 0.18 m
= 10.3 – 0.18 = 10.12 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
= 10 – 9.05 = 0.95 m above the atmospheric pressure head
= Atmospheric pressure head + 0.95 m
= 10.3 + 0.95 =11.25 m of water (abs)

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