Question 5.12: Pumping Power and Frictional Heating in a Pump The pump of a...

The pressures across a pump are measured. The mechanical efficiency of the pump and the temperature rise of water are to be determined.
Assumptions   1 The flow is steady and incompressible. 2  The pump is driven by an external motor so that the heat generated by the motor is dissipated to the atmosphere. 3  The elevation difference between the inlet and outlet of the pump is negligible, z_1 ≅ z_2. 4  The inlet and outlet diameters are the same and thus the average inlet and outlet velocities are equal, V_1 = V_2. 5  The kinetic energy correction factors are equal, \alpha_1 = \alpha_2.
Properties   We take the density of water to be 1 kg/L = 1000 kg/m³ and its specific heat to be 4.18 kJ/kg·°C.
Analysis   (a) The mass flow rate of water through the pump is

\dot{m} = \rho \dot{V} = (1  kg/L)(50  L/s) = 50  kg/s

The motor draws 15 kW of power and is 90 percent efficient. Thus the mechanical (shaft) power it delivers to the pump is

\dot{W}_{pump,  shaft} = \eta _{motor} \dot{W}_{electric} = (0.90)(15  kW) = 13.5  kW

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

\Delta \dot{E}_{mech,  fluid} = \dot{E}_{mech,  out}  –  \dot{E}_{mech,  in} = \dot{m}\left(\frac{P_2}{\rho } + \alpha _2 \frac{V^2_2}{2} + gz_2 \right) –  \dot{m}\left(\frac{P_1}{\rho } + \alpha _1\frac{V^2_1}{2} + gz_1 \right)

Simplifying it for this case and substituting the given values,

\Delta \dot{E}_{mech,  fluid} = \dot{m} \left(\frac{P_2  –  P_1}{\rho } \right) = (50  kg/s)\left(\frac{(300  –  100)  kPa}{1000  kg/m^3} \right)\left(\frac{1  kJ}{1  kPa  .  m^3} \right) = 10.0  kW

Then the mechanical efficiency of the pump becomes

\eta _{pump} = \frac{\dot{W}_{pump,  u} }{\dot{W}_{pump,  shaft} } = \frac{\Delta \dot{E}_{mech, fluid} }{\dot{W}_{pump, shaft} } = \frac{10.0  kW}{13.5  kW} = 0.741       or      74.1%

(b) Of the 13.5-kW mechanical power supplied by the pump, only 10.0 kW is imparted to the fluid as mechanical energy. The remaining 3.5 kW is converted to thermal energy due to frictional effects, and this “lost” mechanical energy manifests itself as a heating effect in the fluid,

\dot{E}_{mech,  loss} = \dot{W}_{pump,shaft}  –  \Delta \dot{E}_{mech,  fluid} = 13.5  –  10.0 = 3.5  kW

The temperature rise of water due to this mechanical inefficiency is determined from the thermal energy balance, \dot{E}_{mech,  loss} = \dot{m}(u_2  –  u_1) = \dot{m}c\Delta T . Solving for ΔT,

\Delta T = \frac{\dot{E}_{mech,  loss} }{\dot{m}c } = \frac{3.5  kW}{(50  kg/s)(4.18  kJ/kg.°C)} = 0.017°C

Therefore, the water experiences a temperature rise of 0.017°C which is very small, due to mechanical inefficiency, as it flows through the pump.
Discussion   In an actual application, the temperature rise of water would probably be less since part of the heat generated would be transferred to the casing of the pump and from the casing to the surrounding air. If the entire pump and motor were submerged in water, then the 1.5 kW dissipated due to motor inefficiency would also be transferred to the surrounding water as heat.