Question 21.14: The piston diameter and stroke length of a single-acting rec......
The piston diameter and stroke length of a single-acting reciprocating pump are 150 mm and 350 mm, respectively. It delivers water to a tank 15 m above the pump through a delivery pipe of 100 mm in diameter and 20 m long. If the pump is running at 35 rpm, determine the pressure head in the cylinder at the beginning, middle and the end of the delivery stroke. Take atmospheric pressure as 10.3 m of water and Darcy’s friction factor f as 0.015.
Given data:
Diameter of piston D =150 mm = 0.15 m
Stroke length L= 350 cm = 0.35 m
Delivery head h_{d}=15\;\mathrm{m}
Speed of pump N = 35 rpm
Diameter of delivery pipe d_{d}=100\,\mathrm{mm}=0.10\,\mathrm{m}
Length of delivery pipe l_{d}=20\ m
Atmospheric pressure head h_{a t m}=10.3\ \mathrm{m~of~water}
Darcy’s friction factor f = 0.015 .
Crank radius is given by
\displaystyle r={\frac{L}{2}}={\frac{0.35}{2}}=0.175\,\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}}
Area of delivery pipe is
a_{s}=a_{d}=\frac{\pi}{4}d_{s}^{2}=\frac{\pi}{4}(0.1)^{2} =0.00785\,{\mathrm{m}}^{2}
Angular speed is given by
\omega={\frac{2\pi N}{60}}=\frac{2\pi\times35}{60}=3.665\,\,\mathrm{rad/s}
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{20}{9.81}}\times{\frac{0.0177}{0.00785}}\times3.665^{2}\times0.175\times\cos\theta=10.80\cos\theta
The loss of head due to friction in delivery pipe is computed from Eq. (21.24) as
h_{f d}=\frac{f l_{d}}{d_{d}\times2g}\biggl(\frac{A}{d_{d}}\omega r\;\sin\theta\biggr)^{2}
=\frac{0.015\times20}{0.1\times2\times9.81}{\left(\frac{0.0177}{0.00785}\times3.665\times0.175\times\sin\theta\right)^{2}}
={0.32\sin^{2}\theta}
At the beginning of delivery stroke At the beginning of delivery stroke, \theta=0^{\circ}. Thus,
h_{a d}=10.80\;\mathrm{cos}\;0^{\circ}=10.80\;\mathrm{m\;and}\;h_{f d}=0
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
= 15 + 10.80 = 25.80 m above the atmospheric pressure head
= Atmospheric pressure head + 25.80 m
= 10.3 + 25.80 = 36.1 m of water (abs)
At the middle of delivery stroke At the middle of delivery strokes, \theta=90^{\circ}. Thus,
h_{a d}=0\;\mathrm{and}\;h_{f d}=0.32\;\mathrm{sin}^{2}\;90^{\circ}=0.32
The pressure head in the cylinder at the middle of the delivery stroke is (refer to Fig. 21.7)
= (h_{d}+h_{f d}) above the atmospheric pressure head
= 15 + 0.32 = 15.32 m above the atmospheric pressure head
= Atmospheric pressure head + 15.32 m
= 10.3 + 15.32 = 25.62 m of water (abs)
At the end of delivery stroke At the beginning of suction and delivery strokes, \theta = 180°. Thus
h_{a d}=10.80\;\mathrm{cos}\;180^{\circ}=-10.80\operatorname{m and}h_{f d}=0
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
= 15 – 10.80 = 4.2 m above the atmospheric pressure head
= Atmospheric pressure head + 4.2 m
= 10.3 + 4.2 = 14.5 m of water (abs)
