Thermodynamics

238 | ThermodynamicsAnalysis We take the turbine as the system. This is a control volume sincemass crosses the system boundary during the process. We observe that thereis only one inlet and one exit and thus ṁ 1 ṁ 2 ṁ. Also, work is done bythe system. The inlet and exit velocities and elevations are given, and thusthe kinetic and potential energies are to be considered.(a) At the inlet, steam is in a superheated vapor state, and its enthalpy isAt the turbine exit, we obviously have a saturated liquid–vapor mixture at15-kPa pressure. The enthalpy at this state isThen¢ke V 2 2 V 2 12(b) The energy balance for this steady-flow system can be expressed in therate form asRate of net energy transferby heat, work, and massm # a h 1 V 1 2P 1 2 MPaT 1 400C f h 1 3248.4 kJ/kg (Table A–6)h 2 h f x 2 h fg [225.94 (0.9)(2372.3)] kJ/kg 2361.01 kJ/kg¢h h 2 h 1 12361.01 3248.42 kJ>kg 887.39 kJ/kg¢pe g 1z 2 z 1 2 19.81 m>s 2 2316 102 m4a1 kJ>kg1000 m 2 >s b 0.04 kJ /kg2E # in E # out dE system >dt 0⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭ 1180 m>s22 150 m>s2 22E # in E # outRate of change in internal, kinetic,potential, etc., energies2 gz 1 b W # out m # a h 2 V 2 2Dividing by the mass flow rate ṁ and substituting, the work done by the turbineper unit mass of the steam is determined to bew out c1h 2 h 1 2 V 2 2 2 V 1 g 1z22 z 1 2d 1¢h ¢ke ¢pe2[887.39 14.95 0.04] kJ/kg 872.48 kJ/kg(c) The required mass flow rate for a 5-MW power output ism # W# out 5000 kJ/sw out 872.48 kJ/kg 5.73 kg /sa1 kJ>kg1000 m 2 >s b 14.95 kJ /kg2¡ 0 (steady)2 gz 2 b1since Q # 02Discussion Two observations can be made from these results. First, thechange in potential energy is insignificant in comparison to the changes inenthalpy and kinetic energy. This is typical for most engineering devices.Second, as a result of low pressure and thus high specific volume, the steamvelocity at the turbine exit can be very high. Yet the change in kinetic energyis a small fraction of the change in enthalpy (less than 2 percent in ourcase) and is therefore often neglected.