The overall efficiency of a centrifugal pump when head is 25 m, discharge = 0.04 m³/s and output power p = 16 kW (take g = 10 m/s²? and ρ = 1000) is

Explanation:

Overall Efficiency (η): It is defined as a ratio of the power output of the pump to the power input to the pump.

The overall efficiency of the pump will be given as,

\({{\rm{\eta }}_{\rm{}}} = \frac{{{\rm{water\;power}}}}{{{\rm{shaft\;power\;}}}} = \frac{{{\rm{\omega QH}}}}{{\rm{P}}}\)

\(P = \frac{{{\bf{\omega QH}}\;}}{{{\eta }}}\)

Calculation:

\(\eta = \frac{{{\bf{\rho g QH}}\;}}{{{P }}} = \frac{{{\bf{1000\times10\times0.04\times25}}\;}}{{{16000}}}=0.625\)

Additional Information

Manometric Efficiency (ηman): It is the ratio of the manometric head to head imparted by the impeller to the water.

\({\eta _{man}} = \frac{{{H_m}}}{{\frac{{{V_{w2}}{u_2}}}{g}}} = \frac{{g{H_m}}}{{{V_{w2}}{u_2}}}\)

Mechanical Efficiency (ηm): It is the ratio of the power available at the impeller to the power at the shaft of the centrifugal pump.

\({\eta _m} = \frac{{{\rm{Power\;at\;the\;impeller}}}}{{{\rm{Power\;at\;the\;shaft}}}} = \frac{{\frac{W}{g}\left( {\frac{{{V_{w2}}{u_2}}}{{1000}}} \right)}}{{{\rm{SP}}}}\)