-5.075X 10 kJ/kmole The exit state at the adiabatic flame temperature is specified by ∑n2Mh2=5075X10°kJ/ mole We find the adiabatic flame temperature in three ways, approximate solution using an average value of Cp, a more accurate one using the tabulated evolution of cp with temperature and a more precise solution using the tabulated values for gas enthalpy in Table A 8 of sB&v a) Approximate solution using"average"values of specific heat From Figure C-2 we can use the values at 500K as representative. These are (kJ/mole) CO? H2O 30 Using 4h=Cn,△T ∑n=48c)lo,+9()20+374)+18 where AT=Tfinal-25 C=Tfinal-298 K 2n42=74047 AT=682→T=980K b) Solution for adiabatic flame temperature using evolutions of specific heats with temperature Tables give the following evolutions of specific heats with temperature Evolution of cp/R with t(kJ/kmol) 2.40+8735.10-3xT-6607.102xT2+2.002.10°xT H2O 4.070-1.108.103xT+4.152.103x12-2.964.10°xT2+08071012xT O 3.626-1.878103xT+705106x12-6.76410°xT3+2156012xT4 3.675-1.208.103xT+2.324102x120.632.10°xT30.226.10-12xT Using Ah= c, (T) dT and the same equation as above, we obtain T899K c) Solution for adiabatic flame temperature using tabulated values for gas enthalpy M H20 Ah T=900K 28.041 21,92 19246 18. 221 KJ/mole T=1000K 33.405 25978 22,707 21. 460 kJ/mole 2C-9= -5.075 X 106 kJ/kmole. The exit state at the adiabatic flame temperature is specified by: ∑n h∆ = 5.075 X 106 kJ/kmole e e P We find the adiabatic flame temperature in three ways, approximate solution using an average value of cp , a more accurate one using the tabulated evolution of cp with temperature and a more precise solution using the tabulated values for gas enthalpy in Table A.8 of SB&VW. a) Approximate solution using “average” values of specific heat: From Figure C-2 we can use the values at 500K as representative. These are: Gas cp (kJ/kmole) CO2 45 H2O 35 O2 30 N2 30. Using ∆h c = p ∆T , "ave" ∑n h∆ c c c c e e = ∆T{8( ) p CO2 + 9( ) p H2O + 37.5( ) p O2 +188( ) p } P N2 o where ∆T T = final − 25 C = Tfinal − 298 K ∑n h∆ e = 7440∆T kJ e P ∆T = 682 ⇒ Tfinal = 980 K . b) Solution for adiabatic flame temperature using evolutions of specific heats with temperature Tables give the following evolutions of specific heats with temperature: Gas Evolution of cp/ R with T (kJ/kmol) CO2 2.401+8.735.10-3xT-6.607.10-6xT2 +2.002.10-9xT3 H2O 4.070-1.108.10-3xT+4.152.10-6xT2 -2.964.10-9xT3 +0.807.10-12xT4 O2 3.626-1.878.10-3xT+7.055.10-6xT2 -6.764.10-9xT3 +2.156.10-12xT4 N2 3.675-1.208.10-3xT+2.324.10-6xT2 -0.632.10-9xT3 -0.226.10-12xT4 Tf in K Using ∆h = ∫ c ( ) T .dT and the same equation as above, we obtain: p 298 K Tf= 899 K c) Solution for adiabatic flame temperature using tabulated values for gas enthalpy: ∆hCO2 ∆hH2O ∆hO2 ∆hN2 T= 900 K 28,041 21,924 19,246 18,221 kJ/kmole T=1000K 33,405 25,978 22,707 21,460 kJ/kmole 2C-9