2nd-Law Analysis of Otto CyclesConsider an engine operating on the ideal Otto cycle with a compression ratio of8. At the beginning of the compression process, air is at 100 kPa and 17°C. Duringthe constant-volume heat-addition process, 800 kJ/kg of heat is transferred to airfrom a source at 1700 K and waste heat is rejected to the surroundings at 290 K.Accounting for the variation of specific heats of air with temperature, determine(a) the exergy destruction associated with each of the four processes and thecycle and (b) the second-law efficiency of this cycle.P. kPaIsentropicinJoutIsentropic10038V = V =V = V4

given:compression ratio in an otto cycle, rk=8P1=100 kPaT1=17oC=290.15KHeat addition=800 kJ/kgEngine is operating between heat source at TH temperature and surrounding at To TemperatureAir behaves like an ideal gas also specific heat of air is varying with temperature so we will calculate enthalpy and entropy at different temperature with the help of table of properties of ideal gas Process 1-2: isentropic compressionfor adiabatic processat T1=290 K , enthalpy = h1=206.91 kJ/kg (from data table for air)also specific volume of air vr1=676.1 m3/kgv2v1=1rk =vr2vr1, vr2=84.51 m3/kg for this value we get from tableT2=652.4 K , h2=475.11 kJ/kgP2v2T2=P1v1T1 , we get P2=1799.7 kPaprocess 2-3: Qin=enthalpy change of air =h3-h2800 =h3-475.11h3=1275.11 kPa, for this value T3=1575.1 K and vr3=6.108 m3/kgP3v3T3=P2v2T2 , v3 and v2 are sameP3v31575.1=1799.7×v2652.4 , P3=4.345 MPaprocess 3-4: vr4vr3=v4v3=rkvr4=rkvr3=8×6.108=48.86 m3/kgfrom tables h4=588.74 kJ/kgprocess 4-1 :Qout=h4-h1=588.74-206.91 =381.83 kJ/kgWnet=Qinput-Qout=418.17 kJ/kg process 1-2 and 3-4 are isentropic so there is no irreversibilitiesNow for process 2-3:entropy change, s3-s2=s3'-s2'-RlnP3P2 , m=1kgs3'=3.5045 kJ/kg-K and s2'=2.4975 kJ/kg-K , data taken from properties of air table s3-s2=3.5045-2.4975-0.287ln4.8451.799s3-s2=0.754 kJ/kg-KQinput=800 kJ/kg and TH=1700 K sourceEdistruction 2-3=Tss3-s2system-QinputTsourceEdistruction 2-3=290(0.754-8001700)=82.2 kJ/kgFor process (4-1),Edistruction 4-1=Tss1-s4system+QoutTsinks1-s4=s3-s2=-0.754Qout=381.83 kJ/kg , Tsink=290 K after putting valuesEdistruction 4-1=163.2 kJ/kgTotal energy distruction for a cycle=82.2+163.2=245.4 kJ/kgPart b: second law efficiencyη2=1-Edistruction totalEexpendedEexpended=1-TsTHQinput = (1-2901700)800=663.5 kJ/kgη2=1-245.4663.5 =0.6307=63.07%