Question 2.73E: A water pump delivers 3 hp of shaft power when operating. If......

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the pump is measured to be 1.2 psi when the flow rate is 8 ft³/s and the changes in velocity and elevation are negligible, determine the mechanical efficiency of this pump.

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A water pump raises the pressure of water by a specified amount at a specified flow rate while consuming a known amount of electric power. The mechanical efficiency of the pump is to be determined.

Assumptions 1 The pump operates steadily. 2 The changes in velocity and elevation across the pump are negligible. 3 Water is incompressible.

Analysis To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is

\displaystyle \begin{aligned} \Delta \dot{E}_{\text {mech,fluid }}& =\dot{m}\left(e_{\text {mech,out }}-e_{\text {mech,in }}\right)=\dot{m}\left[(P \boldsymbol{\nu})_2-(P \boldsymbol{\nu})_1\right]=\dot{m}\left(P_2-P_1\right) \boldsymbol{\nu} \\[10pt] & =\dot{\boldsymbol{V}}\left(P_2-P_1\right)=\left(8 \mathrm{~ft}^3 / \mathrm{s}\right)(1.2 \mathrm{~psi})\left(\frac{1 \mathrm{~Btu}}{5.404 \mathrm{~psi} \cdot \mathrm{ft}^3}\right)=1.776 \mathrm{~Btu} / \mathrm{s}=2.51 \mathrm{~hp} \end{aligned}

since 1 \mathrm{~hp}=0.7068 \mathrm{~Btu} / \mathrm{s}, \dot{m}=\rho \dot{\boldsymbol{V}}=\dot{\boldsymbol{V}} / \boldsymbol{v}, and there is no change in kinetic and potential energies of the fluid. Then the mechanical efficiency of the pump becomes

\displaystyle \eta_{\text {pump }}=\frac{\Delta \dot{E}_{\text {mech,fluid }}}{\dot{W}_{\text {pump, shaft }}}=\frac{2.51 \mathrm{~hp}}{3 \mathrm{~hp}}=0.838 \text { or } \mathbf{8 3 . 8 \%}

Discussion The overall efficiency of this pump will be lower than 83.8 \% because of the inefficiency of the electric motor that drives the pump.

2-73e
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