Question 15.7: The plunger in the reciprocating pump moves with simple harm......
The plunger in the reciprocating pump moves with simple harmonic motion. The diameter and the stroke length of plunger are 25 cm and 45 cm respectively. The suction pipe is 12.5 cm in diameter and 12 m long. The suction lift is 3 m. Calculate the speed at which the pump can operate without separation at the beginning of the stroke. Assume no air vessel in the suction side. The barometer reads 10.15 m of water and separation occurs at 2.5 m of water.
Given: D = 25 cm = 0.25 m, L = 45 cm = 0.45 m, d_{s} = 12.5 cm = 0.125 m, l_{s} = 12 m, H_{s} = 3m, H_{at} = 10.15 m, and H_{sep} = 2.5 m
Now, at the beginning of the suction stroke, the velocity is zero; therefore, the friction loss in suction pipe h_{fs} is zero. But the accelerating head is maximum.
∴ H_{\mathrm{at}}-(H_{s}+h_{\mathrm{asmax}})\leq H_{\mathrm{sep}}
Taking the limiting case, 10.15-3-\frac{l_{s}}{g}\,\frac{A}{a_{s}}r\omega^{2}=2.5
or {\frac{l_{s}}{g}}{\frac{A}{a_{s}}}r\omega^{2}=10.15-3-2.5=4.65
or {\frac{12}{9.81}}\times\left\lgroup{\frac{0.25}{0.125}}\right\rgroup ^{2}\times{\frac{0.45}{2}}\times\left\lgroup{\frac{2\pi N}{60}}\right\rgroup ^{2}=4.65
or 0.003018227N²= 4.65
∴ N = 39.25 rpm