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Chapter 8 | 461Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0, E CV 0, and X CV 0. 2 Thekinetic and potential energies are negligible.Analysis We take the turbine as the system. This is a control volume sincemass crosses the system boundary during the process. We note that there isonly one inlet and one exit and thus ṁ 1 ṁ 2 ṁ. Also, heat is lost to thesurrounding air and work is done by the system.The properties of the steam at the inlet and exit states and the state ofthe environment areInlet state:Exit state:Dead state:P 1 3 MPaT 1 450°C fh 1 3344.9 kJ>kgs 1 7.0856 kJ>kg # KP 2 0.2 MPaT 2 150°Cfh 2 2769.1 kJ>kgs 2 7.2810 kJ>kg # K(Table A–6)P 0 100 kPaf h 0 h f @ 25°C 104.83 kJ>kgT 0 25°C s 0 s f @ 25°C 0.3672 kJ>kg # K(Table A–6)(Table A–4)(a) The actual power output of the turbine is determined from the rate formof the energy balance,S0 (steady)E # in E # out dE system /dt 0Rate of net energy transferby heat, work, and mass⎫⎪⎬⎪⎭⎫⎪⎪⎪⎬⎪⎪⎪⎭Rate of change in internal, kinetic,potential, etc., energiesE # in E # outm # h 1 W # out Q # out m # h 2 1since ke pe 02W # out m # 1h 1 h 2 2 Q # out 18 kg>s2313344.9 2769.12 kJ>kg4 300 kW 4306 kW(b) The maximum power output (reversible power) is determined from therate form of the exergy balance applied on the extended system (system +immediate surroundings), whose boundary is at the environment temperatureof T 0 , and by setting the exergy destruction term equal to zero,S 0 (reversible)S 0 (steady)X # in X # out X # destroyed dX system >dt 0⎫ ⎪⎬⎪⎭⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭Rate of net exergy transfer Rate of exergy Rate of changeby heat, work, and mass destruction in exergyX # in X # outm # c 1 W # rev,out X # 0 Qheat m # c 2W # rev,out m # 1c 1 c 2 2 m # 31h 1 h 2 2 T 0 1s 1 s 2 2 ¢ke Q0 ¢pe Q0 4Note that exergy transfer with heat is zero when the temperature at the pointof transfer is the environment temperature T 0 . Substituting,W # rev,out 18 kg>s2313344.9 2769.12 kJ>kg 1298 K2 17.0856 7.28102kJ>kg # K4 4665 kW

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