Question 15.P.14: A centrifugal pump of impeller diameter 0.4 m runs at 1450 r......
A centrifugal pump of impeller diameter 0.4 m runs at 1450 rpm. The blades are curved back at 30° to the tangent at the outlet. The velocity of flow is 3 m per second. Determine the theoretical maximum lift if the outlet velocity is reduced by the diffuser by 50%.
Inlet whirl is assumed to be zero
u_{2}={\frac{\pi\times0.4\times1450}{60}}=30.37{\mathrm{~m/s}} \\ V_{u2}=30.37-\frac{3}{\tan30}=25.17{\mathrm{~m/s}} \\ V_{2}=(25.17^{2}+3^{2})^{0.5}=25.35~\mathrm{m}
Head imparted ={\frac{30.37\times25.17}{9.81}}=77.92\ \mathrm{m}
Static head =77.92-{\frac{25.35^{2}}{2\times9.81}}=45.17\operatorname{m}
Without diffuser the pump can pump to a head of 45.17 m theoretically.
If velocity is reduced to 50% of the value
New velocity = 12.675 m/s
∴ Head recovered ={\frac{25.35^{2}-12.675^{2}}{2\times9.81}}=24.57\,\,{\mathrm{m}}
∴ Theoretical maximum lift
= 45.17 + 24.57 = 69.74 m