Question 16.17: A three-stage centrifugal pump with an impeller in series ha......
A three-stage centrifugal pump with an impeller in series has outlet diameter 50 cm and width 2.5 cm. The vanes are curved back at outlet angle of 45°, and the circumferential area is reduced by 15%. The manometric efficiency is 80%, and the overall efficiency is 70%. Determine the head generated by the pump, when it is running at 900 rpm and discharging at 0.05 m³/s. Also find the shaft power.
Given: D_{1}= 50 cm = 0.5 m, B_{1}= 2.5 cm = 0.025 m, Φ = 45°, Reduction of circumferential area = 15%, \eta_{mano} = 80%, \eta_{o} = 70%, N= 900 rpm and Q= 0.05 m³/s
Now, Effective area of flow at outlet = \pi D_{1}B_{1}(1-0.15)
=\pi\times0.5\times0.025\times0.85=0.0338\ \mathrm{m^{2}}From the equation of discharge, we have
Q=\pi D_{1}B_{1}V_{f_{1}}or V_{f_{1}}={\frac{Q}{\pi D_{1}B_{1}}}={\frac{0.05}{0.0338}}=1.497\,{\mathrm{m/s}}
Again, u_{1}={\frac{\pi D_{1}N}{60}}={\frac{\pi\times0.5\times900}{60}}=23.562\,{\mathrm{m/s}}
From outlet velocity triangle of Figure 16.19, we have
\mathrm{tan}\,45^{\circ}={\frac{V_{f_{1}}}{u_{1}-V_{w_{1}}}}\qquad\mathrm{or}\qquad1={\frac{1.497}{23.562-V_{w_{1}}}}Thus, V_{w_{1}}=22.065\ {\mathrm{m/s}}
Now, \eta_{\mathrm{mano}}=\frac{g H_{m}}{V_{w_{1}}u_{1}}\qquad\mathrm{or}\qquad\mathrm{0.8}=\frac{9.81\times H_{m}}{22.065\times23.562}
Hence, H_{m}=42.397\,{\mathrm{m}}
Total head generated = nH_{m}= (3 × 42.397)m = 127.191 m
{\mathrm{Overall \ efficiency}} \eta_{o}={\frac{\mathrm{Power~oufput~of~pump}}{{\mathrm{Power~supplied~at~pump}}}}or \eta_{o}={\frac{\rho g Q\times127.191}{100\times{\mathrm{Shaft}}\;{\mathrm{Power}}\quad}}\mathrm{~or~}\qquad0.70={\frac{1000\times9.81\times0.0.65\times127.191}{100\times{\mathrm{Shaft~Power}}}}
Therefore, Shaft power = 89.124 kW
