Question 15.P.9: The dimensionless specific speed of a centrifugal pump is 0.......
The dimensionless specific speed of a centrifugal pump is 0.06. Static head is 30 m. Flow rate is 50 l/s. The suction and delivery pipes are each of 15 cm diameter. The friction factor is 0.02. Total length is 55 m other losses equal 4 times the velocity head in the pipe. The vanes are forward curved at 120°. The width is one tenth of the diameter. There is a 6% reduction in flow area due to the blade thickness. The manometric efficiency is 80%. Determine the impeller diameter. Inlet is radial.
Frictional head is calculated first. Velocity in the pipe
={\frac{0.05\times4}{\pi\times0.15^{2}}}=2.83\ \mathrm{m/s}
Total loss of head ={\frac{f l\ V^{2}}{2g\ D}}+{\frac{4\ V^{2}}{2g}}
={\frac{0.02\times55\times2.83^{2}}{2\times9.81\times0.15}}+{\frac{4\times2.83^{2}}{2\times9.81}}=4.63\;\mathrm{m}
Total head against which pump operates = 34.63 m
Speed is calculated from specific speed N_{s}=N\sqrt{{ Q}}\ /(g H)^{3/4}
N\,=\,\frac{0.06\times(9.81\times34.63)^{3/4}}{0.05^{1/2}}\,=21.23\,\,\mathrm{rps}
Flow velocity is determined :
Flow area =\pi\times D\times{\frac{D}{10}}\times0.94=0.2953\ D^{2}
V_{f2}=\frac{0.05}{0.2953\,D^{2}}=0.1693/D^{2}\qquad\qquad\qquad…(1) \\ u_{2}=\pi\,D N=21.23\times\pi\times D=66.7\,D\qquad…(2) \\ \eta_{m}=0.8=\frac{9.81\times34.63}{66.7\,D\times V_{u2}}
∴ \qquad V_{u2}=\frac{6.367}{D} \qquad\qquad\qquad\qquad\qquad\qquad …(3)From velocity diagram,
\tan\,60={\frac{V_{f2}}{V_{u2}-u_{2}}}={\frac{0.1693}{D^{2}}}\cdot{\frac{1}{\left({\frac{6.367}{D}} – 66.7\;D\right)}}
Rearranging, 115.52 D³ – 11.028 D + 0.1693 = 0
Solving, D = 0.3 m.
