Question 14.8: The pump shown in Fig. 14–17 is used to transfer 20°C sewage......
The pump shown in Fig. 14–17 is used to transfer 20°C sewage water from the wet well to the sewage treatment plant. If the flow through the 75-mm-diameter pipe is to be \frac{1}{60}\rm m^3/s, determine if cavitation occurs when the pump in Fig. 14–16 is selected. The pump turns off just after the water reaches its lowest level of h = 4 m. Take the friction factor for the pipe to be f = 0.02, and neglect minor losses.
Fluid Description. We will assume steady flow of an incompressible fluid. From table in Appendix A, \rho_w = 998.3 kg/m³ for water at T = 20°C.
Inlet Pressure. We can determine the available suction head at the pump’s impeller inlet by applying the energy equation. The control volume consists of the water in the vertical pipe and the well, Fig. 14–17. The largest suction at B occurs when h = 4 m. Here we will work with absolute pressures since the vapor pressure is generally reported as an absolute pressure. At A the atmospheric pressure is p_A = 101.3 kPa. Since
then
\frac{p_{A}}{\gamma}+\frac{V_{A}^{2}}{2g}+z_{A}+h_{\mathrm{pump}}=\frac{p_{B}}{\gamma}+\frac{V_{B}^{2}}{2g}+z_{B}+h_{\mathrm{turb}}+f\frac{L}{D}\frac{V^{2}}{2g}+\Sigma K_{L}\frac{V^{2}}{2g}\\\frac{101.3(10^{3})\;\mathrm{N/m^{3}}}{(998.3\;\mathrm{kg/m^{3}})(9.81\;\mathrm{m/s^{2}})}+0-4\;\mathrm{m}\;+\;0=\\\frac{p_{B}}{(998.3\,\mathrm{kg/m}^{3})\,(9.81\,\mathrm{m/s}^{2})}+\frac{(3.7726\,\mathrm{m/s})^{2}}{2(9.81\,\mathrm{m/s}^{2})}+0+0+0.02\left(\frac{4\,\mathrm{m}}{0.075\,\mathrm{m}}\right)\left[\frac{(3.7726\,\mathrm{m/s})^{2}}{2(9.81\,\mathrm{m/s}^{2})}\right]+0\\p_{B}=47.445(10^{3})\,\mathrm{Pa}The available suction head at the pump inlet is therefore
{\frac{p_{B}}{\gamma}}+{\frac{V_{B}^{2}}{2g}}={\frac{47.445(10^{3})\,\mathrm{N/m^{2}}}{(998.3\,\mathrm{kg/m^{3}})\,(9.81\,\mathrm{m/s^{2}})}}+{\frac{(3.7726\,\mathrm{m/s})}{2(9.81\,\mathrm{m/s^{2}})}}=5.570\,\mathrm{m}From Appendix A, at 20°C, the (absolute) vapor pressure for water is 2.34 kPa. Therefore, the available NPSH is
({\mathrm{NPSH}})_{\mathrm{avail}}=5.570\,{\mathrm{m}}-{\frac{2.34(10^{3})\,{\mathrm{N/m}}^{2}}{(998.3\,{\mathrm{kg/m}}^{3})\,{(9.81\,{\mathrm{m}}/{\mathrm{s}}^{2})}}}=5.331\,{\mathrm{m}}The flow in liters per minute is
Q=\left({\frac{\frac{1}{60}\mathrm{m}^{3}}{\mathrm{s}}}\right)\left(\frac{1000\;\mathrm{liters}}{1\;\mathrm{m}^{3}}\right)\left({\frac{60\,\mathrm{s}}{1\,\mathrm{min}}}\right)=1000\;\mathrm{liters}/\mathrm{min}From Fig. 14–16, at Q = 1000 liters/min, (NPSH)_{req‘d} = 2.33 m (point D), and since (NPSH)_{avail} > (NPSH)_{req‘d}, cavitation will not occur in the pump.
Also note from Fig. 14–16 that if a 175-mm-diameter impeller is used in this pump (point C), then the pump will have an efficiency of about 77% and will have a shaft or brake power of about 13.5 kW.