Question 15.11: A single acting reciprocating pump has a plunger diameter 0.......
A single acting reciprocating pump has a plunger diameter 0.25 m and a stroke length 0.45 m. The suction pipe is 12.5 cm in diameter and 12 m long with 3 m suction lift. An air vessel is fitted to the suction pipe at a distance 1.5 m from the cylinder and 10.5 m from the sump level. If the barometer reads 10.3 m of water and separation takes at 2.5 m vacuum, find the speed at which the crank can operate without separation to occur. Take Darcy’s coefficient of friction f of the pipe to be 0.01.
Given: D = 0.25 m, d_{s} = 12.5 cm = 0.125 m, L = 0.45 m, r = 0.45/2 = 0.225 m, l_{s}= 12 m, f = 0.01, H_{atm} = 10.3 m of water, l_{s} = 10.5 m, l_{sv} = 1.5 m, H_{s}= 3 m, and H_{sep} = 2.5 m of water
A sketch of the example is shown in Figure 15.12, where different values of data are given.
Area of the plunger A={\frac{\pi}{4}}(0.25)^{2}=0.049\,\mathrm{m^{2}}
Area of the suction pipe a={\frac{\pi}{4}}(0.125)^{2}=0.01227\,\mathrm{m^{2}}
The mean velocity of suction pipe due to fitting of air vessel is:
{\overline{{V}}}={\frac{A}{a}}{\frac{r\omega}{\pi}}
={\frac{0.0419}{0.0127}}\times{\frac{0.225\times\omega}{\pi}}=0.286\omega\,\mathrm{m/s}
The head loss due to friction in the length 10.5 m is given as:
h_{f s}={\frac{f l{\bar{V}}^{2}}{2g d}}
={\frac{0.01\times10.5\times(0.286\omega)^{2}}{2\times9.81\times0.125}}=0.0035\omega^{2}
The acceleration pressure head in the length 1.5 m is:
h_{av} = \frac{l_{sv}}{g} \frac{A}{a} r ω^2 \\ = \frac{1.5}{9.81} \times \frac{0.049}{0.01227} \times 0.225 \times ω^2 = 0.1374ω^2Now, Separation head H_{\mathrm{sep}}=H_{\mathrm{atm}}-(h_{f s}+H_{s}+h_{av})
or 2.5 = 10.3 – (0.00352ω²+ 3 + 0.1374ω²)
or 0.1409 ω² = 4.8
or ω²= 34.06671398
or {{\omega}}=5.83666={\frac{2\pi N}{60}}\,\mathrm{rad/s}
Therefore, N = 55.736 rpm
