MAE204_F11_HW3_Extra_Solutions - 2-68 A hydraulic...

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Unformatted text preview: 2-68 A hydraulic turbine-generator is to generate electricity from the water of a lake. The overall efficiency, the turbine efficiency, and the shaft power are to be determined. Assumptions 1 The elevation of the lake and that of the discharge site remains constant. 2 Irreversible losses in the pipes are negligible. Properties The density of water can be taken to be p= 1000 kg/m3. The gravitational acceleration is g = 9.81 m/sz. Analysis (a) We take the bottom of the lake as the reference level for convenience. Then kinetic and potential energies of water are zero, and the mechanical energy of water consists of pressure energy only which is P emecth _ emech.oul : — = p =( .81 m/s2 )(50 m)(L/kzgzj 1000 m /s = 0.491 kJ/kg Then the rate at which mechanical energy of fluid supplied to the turbine and the overall efficiency become | AEmech’fluid |= rh(emech’in — emech‘in ) = (5000 kg/s)(0.49l kJ/kg) = 2455 kW _ M l AEmcchfluid l novemll =77turbine»gen _ = 0.760 (b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from nturbinc-gcn 77turbine-gen = Ururbinengenerator _) 77turbine : 2 0 95 = Ugenerator ' (c) The shaft power output is determined from the definition of mechanical efficiency, Wshaft.out = 77mm lmmechfluid |= (0.800)(2455 kW) = 1964 kW z 1960 kW Therefore, the lake supplies 2455 kW of mechanical energy to the turbine, which converts 1964 kW of it to shaft work that drives the generator, which generates 1862 kW of electric power. 2-69 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed. _ Wind Properties The density of air is given to be p = 1.25 kg/m3. 6 Analysis Kinetic energy is the only form of mechanical energy 7 INS E the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and ran / 2 for a given mass flow rate: A 2 2 emech -ke — V — (7 W5) [ “(J/1‘23 2 ]— 0.0245 kJ/kg 2 2 lOOOm /S 7:192 2 m = pVA = pV T = (1.25 kg/m3 )(7 11mm = 43,982 kg/s Wmax = Em, = mm = (43,982 kg/s)(0.0245 kJ/kg) =1078 kW The actual electric power generation is determined by multiplying the power generation potential by the efficiency, W = nwind turbineWmax = (0.30)(1078 kW) 2 323 kW elect Therefore, 323 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions. 2-76 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible. Properties The density of oil is given to be p = 860 kg/m3. Analysis Then the total mechanical energy of a fluid is the sum of the potential, flow, and kinetic energies, and is expressed per unit mass as emech = gh + Pv+ V2 /2. 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 _ . . V22 V12 . V22 _ Vlz AEmechfluid : m(emech,out _emech,in ) = m (PV)2 +7_(Pv)l _7 2 V (P2 _Pl)+pT since n'1 = pV : V / u , and there is no change in the potential energy of the fluid. Also, . . 3 Vl=l= l: z—O‘lm/ZS =19.9m/s A. no, /4 7r(0.08m) /4 . . 3 V2 V V 0.1m /s 28.84m/S —A—2—7rD22/4_7r(0.12m)2/4 Substituting, the useful pumping power is determined to be W = AEmechfluid pump.u 2 _ 2 norms/s) 40OkN/mz +(860kg/ms)w 2 lOOOkg6m/s2 lkN'm/s = 26.3 kW Then the shaft power and the mechanical efficiency of the pump become Wpump.shaft : Umotor Welectric : (0‘90)(35 kw) : 31‘5 kw Wpump,u _ 26.3 kW Wpump. Shaft 31.5 kW 77me = = 0.836 = 83.6% Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9x0.836 = 0.75. ...
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This document was uploaded on 10/27/2011 for the course MAE 204 at SUNY Buffalo.

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MAE204_F11_HW3_Extra_Solutions - 2-68 A hydraulic...

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