Thermodynamics

460 | Thermodynamicscold stream, provided that the cold stream is not at a lower temperature thanthe surroundings. Then the second-law efficiency of the heat exchangerbecomesh II,HX m# cold 1c 4 c 3 2m # hot 1c 1 c 2 2 orh II,HX 1 T 0 S # genm # hot 1c 1 c 2 2(8–55)where Ṡ gen ṁ hot (s 2 s 1 ) + ṁ cold (s 4 s 3 ). Perhaps you are wondering whathappens if the heat exchanger is not adiabatic; that is, it is losing some heatto its surroundings at T 0 . If the temperature of the boundary (the outer surfaceof the heat exchanger) T b is equal T 0 , the definition above still holds(except the entropy generation term needs to be modified if the second definitionis used). However, if T b T 0 , then the exergy of the lost heat at theboundary should be included in the recovered exergy. Although no attempt ismade in practice to utilize this exergy and it is allowed to be destroyed, theheat exchanger should not be held responsible for this destruction, whichoccurs outside its boundaries. If we are interested in the exergy destroyedduring the process, not just within the boundaries of the device, then itmakes sense to consider an extended system that includes the immediate surroundingsof the device such that the boundaries of the new enlarged systemare at T 0 . The second-law efficiency of the extended system reflects theeffects of the irreversibilities that occur within and just outside the device.An interesting situation arises when the temperature of the cold streamremains below the temperature of the surroundings at all times. In that casethe exergy of the cold stream actually decreases instead of increasing. Insuch cases it is better to define the second-law efficiency as the ratio of thesum of the exergies of the outgoing streams to the sum of the exergies of theincoming streams.For an adiabatic mixing chamber where a hot stream 1 is mixed with a coldstream 2, forming a mixture 3, the exergy supplied is the sum of the exergiesof the hot and cold streams, and the exergy recovered is the exergy of themixture. Then the second-law efficiency of the mixing chamber becomesh II,mix m # 3c 3m # 1c 1 m # 2c 2orh II,mix 1 where ṁ 3 ṁ 1 + ṁ 2 and S . gen ṁ 3 s 3 ṁ 2 s 2 ṁ 1 s 1 .T 0 S # genm # 1c 1 m # 2c 2(8–56)3 MPa450°C300 kWSTEAMTURBINET 0 = 25°CP 0 = 100 kPaFIGURE 8–450.2 MPa150°CWSchematic for Example 8–15.EXAMPLE 8–15Second-Law Analysis of a Steam TurbineSteam enters a turbine steadily at 3 MPa and 450°C at a rate of 8 kg/s andexits at 0.2 MPa and 150°C, (Fig. 8–45). The steam is losing heat to the surroundingair at 100 kPa and 25°C at a rate of 300 kW, and the kinetic andpotential energy changes are negligible. Determine (a) the actual power output,(b) the maximum possible power output, (c) the second-law efficiency, (d) theexergy destroyed, and (e) the exergy of the steam at the inlet conditions.Solution A steam turbine operating steadily between specified inlet and exitstates is considered. The actual and maximum power outputs, the second-lawefficiency, the exergy destroyed, and the inlet exergy are to be determined.