Question 14.2: A system like the one shown in Figure 14.7 is to be assemble......
A system like the one shown in Figure 14.7 is to be assembled to pump water. The inlet pipe to the centrifugal pump is 12 cm in diameter and the desired flow rate is 0.025 m³/s. At this flow rate, the specifications for this pump show a value of NPSH of 4.2 m. The minor loss coefficient for the system may be taken as K=12. Water properties are to be evaluated at 300 K. Determine the maximum value of y, the distance between pump inlet and reservoir level.
The quantity desired, y, is given by
y={\frac{P_{\mathrm{atm}}-P_{v}}{\rho g}}-\Sigma h_{L}-\mathrm{NPSH} (14-14)
Water properties required, at 300 K, are
\begin{array}{c}{{\rho=997\,\mathrm{kg/m}^{3}}}\\ {{P_{v}=3598\,\mathrm{Pa}}}\end{array}
and we have
v=\frac{\dot{\mathrm{V}}}{A}=\frac{0.025\,{\mathrm{m}}^{2}/{\mathrm{s}}}{\frac{\mathrm{\pi }}{4} (0.12\ m)^{2}}=2.21\,{\mathrm{m}}/{\mathrm{s}}
\Sigma h_{L}=K_{L}{\frac{v^{2}}{2g}}={\frac{12(2.21\,{\mathrm{m/s}}^{2})^{2}}{2(9.81\,{\mathrm{m/s}}^{2})}}=2.99\,{\mathrm{m}}
We can now complete the solution
y\;=\frac{(101,1360-3598)Pa}{(997\ \mathrm{kg/m^{3}})(9.81\ \mathrm{m/s^{2}})}-2.99\mathrm{~m-4.2~m}
\quad =2.805\ \mathrm{m}\ \ \ (9.2\,\mathrm{ft})