Thermodynamics

Chapter 9 | 529EXAMPLE 9–10Second-Law Analysis of an Otto CycleDetermine the exergy destruction associated with the Otto cycle (all fourprocesses as well as the cycle) discussed in Example 9–2, assuming thatheat is transferred to the working fluid from a source at 1700 K and heat isrejected to the surroundings at 290 K. Also, determine the exergy of theexhaust gases when they are purged.Solution The Otto cycle analyzed in Example 9–2 is reconsidered. For specifiedsource and sink temperatures, the exergy destruction associated with thecycle and the exergy purged with the exhaust gases are to be determined.Analysis In Example 9–2, various quantities of interest were given or determinedto beProcesses 1-2 and 3-4 are isentropic (s 1 s 2 , s 3 s 4 ) and therefore donot involve any internal or external irreversibilities; that is, X dest,12 0 andX dest,34 0.Processes 2-3 and 4-1 are constant-volume heat-addition and heat-rejectionprocesses, respectively, and are internally reversible. However, the heat transferbetween the working fluid and the source or the sink takes place through afinite temperature difference, rendering both processes irreversible. Theexergy destruction associated with each process is determined from Eq. 9–32.However, first we need to determine the entropy change of air during theseprocesses:s 3 s 2 s° 3 s° 2 R ln P 3P 2Also,Thus, 0.7540 kJ>kg # Kr 8 P 2 1.7997 MPaT 0 290 K P 3 4.345 MPaT 1 290 K q in 800 kJ>kgT 2 652.4 K q out 381.83 kJ>kgT 3 1575.1 K w net 418.17 kJ>kg 13.5045 2.49752 kJ>kg # K 10.287 kJ>kg # K2 ln4.345 MPaq in 800 kJ>kgandT source 1700 Kx dest,23 T 0 c1s 3 s 2 2 sys 1290 K2c0.7540 kJ>kg # K 800 kJ>kg1700 K d 82.2 kJ>kgq inT sourcedFor process 4-1, s 1 s 4 s 2 s 3 0.7540 kJ/kg · K, q R,41 q out 381.83 kJ/kg, and T sink 290 K. Thus,x dest,41 T 0 c1s 1 s 4 2 sys q outT sinkd1.7997 MPa