// Given:- // The combustion equation is // CH4 + 2O2 + 7.52N2 --- CO2 + 2H2O + 7.52N2 // Part(a) // With enthalpy of formation values from Table A-25 hfbarCO2 = -393520 // in kj/kmol hfbarH2O = -285830 // in kj/kmol hfbarCH4 = -74850 // in kj/kmol M = 16.04 // molar mass of CH4 in kg/kmol // Calculations hRPbar = hfbarCO2 + 2*hfbarH2O - hfbarCH4 // in kj/kmol hRP = hRPbar/M // in kj/kg // Result printf( ' Part(a)the enthalpy of combustion of gaseous methane, fuel is: %f kJ/kg.',hRP) // Part(b) hfbarCO2 = -393520 // in kj/kmol hfbarH2O = -241820 // in kj/kmol hfbarCH4 = -74850 // in kj/kmol // Calculations hRPbar = hfbarCO2 + 2*hfbarH2O - hfbarCH4 // in kj/kmol hRP = hRPbar/M // in kj/kg // Result printf( ' Part(b)the enthalpy of combustion of gaseous methane, fuel is: %f kJ/kg',hRP); // Part(c) // From table A-23 deltahbarO2 = 31389-8682 // in kj/kmol deltahbarH2O = 35882-9904 // in kj/kmol deltahbarCO2 = 42769-9364 // in kj/kmol // Using table A-21 // Calculations // function cpbar = f(T) T=298 // in kelvin function T = cpbar(T) T = (3.826 - (3.979e-3)*T + 24.558e-6*T**2 - 22.733e-9*T**3 + 6.963e-12*T**4)*8.314 endfunction deltahbarCH4 = intg(298,1000,cpbar) var = deltahbarCH4(1) hRPbar = hRPbar + (deltahbarCO2 + 2*deltahbarH2O - var -2*deltahbarO2) hRP = hRPbar/M // in kj/kg // Result printf( ' Part(c)the enthalpy of combustion of gaseous methane, per kg of fuel is %.f kJ/kg',hRP);