Thermodynamics

84 | Thermodynamicshturbine = 0.75 hgenerator= 0.97TurbineGeneratorhturbine–gen= hturbinehgenerator= 0.75 × 0.97= 0.73FIGURE 2–59The overall efficiency of aturbine–generator is the product of theefficiency of the turbine and theefficiency of the generator, andrepresents the fraction of themechanical energy of the fluidconverted to electric energy.andElectric power outputGenerator: h generator (2–48)Mechanical power input W# elect,outW # shaft,inA pump is usually packaged together with its motor, and a turbine with itsgenerator. Therefore, we are usually interested in the combined or overallefficiency of pump–motor and turbine–generator combinations (Fig. 2–59),which are defined asandh pumpmotor h pump h motor W# pump,uW # ¢E# mech,fluidelect,in W # elect,inW# elect,out(2–49)h turbinegen h turbine h generator W# elect,out(2–50)W # turbine,e 0 ¢E # mech,fluidAll the efficiencies just defined range between 0 and 100 percent. Thelower limit of 0 percent corresponds to the conversion of the entire mechanicalor electric energy input to thermal energy, and the device in this casefunctions like a resistance heater. The upper limit of 100 percent correspondsto the case of perfect conversion with no friction or other irreversibilities,and thus no conversion of mechanical or electric energy to thermal energy.Lakeh = 50 mTurbinem · = 5000 kg/sFIGURE 2–60Schematic for Example 2–16.h generator = 0.951862 kWGeneratorEXAMPLE 2–16Performance of a Hydraulic Turbine–GeneratorThe water in a large lake is to be used to generate electricity by the installationof a hydraulic turbine–generator at a location where the depth of thewater is 50 m (Fig. 2–60). Water is to be supplied at a rate of 5000 kg/s. Ifthe electric power generated is measured to be 1862 kW and the generatorefficiency is 95 percent, determine (a) the overall efficiency of the turbine–generator, (b) the mechanical efficiency of the turbine, and (c) the shaftpower supplied by the turbine to the generator.Solution A hydraulic turbine–generator is to generate electricity from thewater of a lake. The overall efficiency, the turbine efficiency, and the turbineshaft power are to be determined.Assumptions 1 The elevation of the lake remains constant. 2 The mechanicalenergy of water at the turbine exit is negligible.Properties The density of water can be taken to be r 1000 kg/m 3 .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 thechange in its mechanical energy per unit mass becomese mech,in e mech,out P r 0 gh 19.81 m>s2 2150 m2a1 kJ>kg1000 m 2 >s b 2 0.491 kJ>kg