// Given:- y1 = 0.8 // mole fraction of CO2 y2 = 0.2 // mole fraction of O2 T1 = 700.0 // in kelvin p1 = 5.0 // in bars V1 = 3.0 // in m/s p2 = 1.0 // in bars // Part(a) // From table A-23 sO2barT1 = 231.358 sCO2barT1 = 250.663 // Calculations RHS = y2*sO2barT1 + y1*sCO2barT1 + 8.314*log(p2/p1) // Using table A-23 LHSat510K = y2*221.206 + y1*235.7 LHSat520K = y2*221.812 + y1*236.575 // Using linear interpolation, T2 = 510 +((520-510)/(LHSat520K-LHSat510K))*(RHS-LHSat510K) // Result printf( 'The temperature at the nozzle exit in K is: %f',T2); // Part(b) // From table A-23 sbarO2T2 = 221.667 // in kj/kmol.K sbarO2T1 = 231.358 // in kj/kmol.K sbarCO2T2 = 236.365 // in kj/kmol.K sbarCO2T1 = 250.663 // in kj/kmol.K // Calculations deltasbarO2 = sbarO2T2-sbarO2T1-8.314*log(p2/p1) // in kj/kmol.K deltasbarCO2 = sbarCO2T2-sbarCO2T1-8.314*log(p2/p1) // in kj/kmol.K // Results printf( 'The entropy changes of the CO2 from inlet to exit, in KJ/Kmol.K is: %f',deltasbarCO2) printf( 'The entropy change of the O2 from inlet to the exit in kj/kmol.k is: %f',deltasbarO2) // Part(c) // From table A-23, the molar specific enthalpies of O2 and CO2 are h1barO2 = 21184.0 h2barO2 = 15320.0 h1barCO2 = 27125.0 h2barCO2 = 18468.0 // Calculations M = y1*44.0 + y2*32.0 // apparent molecular weight of the mixture in kg/kmol deltah = (1.0/M)*(y2*(h1barO2-h2barO2) + y1*(h1barCO2-h2barCO2)) V2 = sqrt(V1**2+ 2*deltah*10**3) // Result printf( 'The exit velocity in m/s is: %f',V2)