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2inmech,outmech,fluidmech, VV PP V Pv V PvmeemE \u3c1V&&&& since , and there is no change in the potential energy of the fluid. Also, vVV /&&& == \u3c1m 1 2 Motor 35 kW Pump inlet PUMP m/s 9.19 4/m) (0.08 /sm 1.0 4/ 2 3 2 11 1 ==== \u3c0\u3c0DAV VV && m/s 84.8 4/m) (0.12 /sm 1.0 4/ 2 3 2 22 2 ==== \u3c0\u3c0DAV VV && Substituting, the useful pumping power is determined to be kW3.26 m/s kN1 kW1 m/s kg1000 kN1 2 m/s) (19.9)m/s 84.8( ) kg/m860( kN/m400/s)m (0.1 2 22 323 fluidmech,upump, = \uf8f7\uf8f8 \uf8f6\uf8ec\uf8ed \uf8eb \u22c5\uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8ec\uf8ec\uf8ed \uf8eb \uf8f7\uf8f7\uf8f8 \uf8f6 \uf8ec\uf8ec\uf8ed \uf8eb \u22c5 \u2212+= \u2206= EW && Then the shaft power and the mechanical efficiency of the pump become kW 5.31kW) 35)(90.0(electricmotorshaftpump, === WW && \u3b7 83.6%==== 836.0 kW31.5 kW3.26 shaft pump, upump, pump W W & & \u3b7 Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9×0.836 = 0.75. 2-36 2-78E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The mechanical power used to overcome frictional effects is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the free surface of the pool is constant. 3 The average flow velocity is constant since pipe diameter is constant. Properties We take the density of water to be \u3c1 = 62.4 lbm/ft3. Analysis The useful mechanical pumping power delivered to water is hp76.8 hp)12)(73.0(pumppumpupump, === WW && \u3b7 The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and for a given mass flow rate. That is, gzm& 2 1 35 ft Pump Lake Pool zgzgmpememE \u2206=\u2206=\u2206=\u2206=\u2206 V&&&&& \u3c1mechmech Substituting, the rate of change of mechanical energy of water becomes hp 76.4 ft/slbf 550 hp 1 ft/slbm 32.2 lbf 1)ft 35)(ft/s /s)(32.2ft )(1.2lbm/ft (62.4 2 233 mech =\uf8f7\uf8f8 \uf8f6\uf8ec\uf8ed \uf8eb \u22c5\uf8f7\uf8f8 \uf8f6\uf8ec\uf8ed \uf8eb \u22c5=\u2206E & Then the mechanical power lost in piping because of frictional effects becomes hp 4.0 =\u2212=\u2206\u2212= hp 76.476.8mechu pump,frict EWW &&& Discussion Note that the pump must supply to the water an additional useful mechanical power of 4.0 hp to overcome the frictional losses in pipes. Forms of Energy Energy Transfer by Heat and Work Mechanical Forms of Work The First Law of Thermodynamics Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on. Assumptions 1 The house is well-sealed, so no air enters or heaves the house. 2 All the lights and appliances are kept on. 3 The house temperature remains constant. Energy Conversion Efficiencies