Question 5.15: Pumping Water from a Lake to a Pool A submersible pump with ...
Pumping Water from a Lake to a Pool
A submersible pump with a shaft power of 5 kW and an efficiency of 72 percent is used to pump water from a lake to a pool through a constant diameter pipe (Fig. 5–62). The free surface of the pool is 25 m above the free surface of the lake. If the irreversible head loss in the piping system is 4 m, determine the discharge rate of water and the pressure difference across the pump.
Water from a lake is pumped to a pool at a given elevation. For a given head loss, the flow rate and the pressure difference across the pump are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Both the lake and pool are large enough that their surface elevations remain fixed.
Properties We take the density of water to be 1 kg/L = 1000 kg/m³.
Analysis The pump delivers 5 kW of shaft power and is 72 percent efficient. The useful mechanical power it imparts to the water is
\dot{W}_{pump u} = \eta _{pump} \dot{W}_{shaft} = (0.72)(5 kW) = 3.6 kW
We take point 1 at the free surface of the lake, which is also taken as the reference level (z_1 = 0), and point 2 at the free surface of the pool. Also, both points 1 and 2 are open to the atmosphere (P_1 = P_2 = P_{atm}), and the velocities are negligible there (V_1 ≅ V_2 ≅ 0). Then the energy equation for steady, incompressible flow through a control volume between these two surfaces that includes the pump is expressed as
\dot{m} \left(\frac{P_1}{\rho } + \alpha _1 \frac{V^2_1}{2} + gz_1 \right) + \dot{W}_{pump, u} = \dot{m} \left(\frac{P_2}{\rho } + \alpha _2\frac{V^2_2}{2} + gz_2 \right) + \dot{W}_{turbine, e} + \dot{E}_{mech loss, piping}
Under the stated assumptions, the energy equation reduces to
\dot{W}_{pump, u} = \dot{m}gz_2 + \dot{E}_{mech loss, piping}
Noting that \dot{E}_{mech loss, piping} = \dot{m}gh_L , the mass and volume flow rates of water become
\dot{m} = \frac{\dot{W}_{pump, u} }{gz_2 + gh_L} = \frac{\dot{W}_{pump, u} }{g(z_2 + h_L)} = \frac{3.6 kJ/s}{(9.81 m/s^2)(25 + 4 m)} \left(\frac{1000 m^2/s^2}{1 kJ} \right) = 12.7 kg/s
\dot{V} = \frac{\dot{m} }{\rho }= \frac{12.7 kg/s}{1000 kg/m^3} = 12.7 \times 10^{-3} m^3/s = 12.7 L/s
We now take the pump as the control volume. Assuming that the elevation difference and the kinetic energy change across the pump are negligible, the energy equation for this control volume yields
\Delta P = P_{out} – P_{in} = \frac{\dot{W}_{pump, u} }{\dot{V} } = \frac{3.6 kJ/s}{12.7 \times 10^{-3} m^3/s} \left(\frac{1 kN.m}{1 kJ} \right) \left(\frac{1 kPa}{1 kN/m^2} \right)
= 283 kPa
Discussion It can be shown that in the absence of head loss (h_L = 0) the flow rate of water would be 14.7 L/s, which is an increase of 16 percent. Therefore, frictional losses in pipes should be minimized since they always cause the flow rate to decrease.