Question

Please calculate the adiabatic flame temperature for methanol combustion for the following case : 12% CH3OH (g) in air at 1 atm when the maximum possible conversion of methanol is achieved. It is believed that CO is not formed during the combustion.

Answer 1

Basis : 1 mole of mixture of Methanol and air.

Moles of methanol= 0.12, moles of air =0.88, Air contains 21% O2 and 79% N2, mole of oxygen =0.88*0.21=0.1848 and moles of N2= 0.79*0.88=0.6952 moles

The combustion of methanol is CH3OH+ 1.5O2----->CO2 + 2H2O

1 mole of methanol requires 1.5 moles of oxygen

0.12 moles require 0.12*1.5= 0.18 moles of oxygen. Moles of oxygen available =0.1848, So all the CH3OH gets reacted. Products contains O2= 0.1848-0.18=0.0048 moles Oxygen, N2= 0.6952, CO2= 0.18 and H2O(g)= 2*0.1848= 0.3696 moles

Standard heat of reaction for combustion of methanol is = -675.6 Kj/mole

for 0.12 moles, standard heat of reaction = -675.6*0.12 Kj=-81.072 Kj/mol

Let the final temperature = T K

since the temperature envisaged is adiabatic flame temperature and the reactants are at 25 deg.c, all the enthalpy change of formation is transferred to products

Hence 61.072*1000 Joules = moles of N2* Molas mas* specif heat* ( T-298.15)+ Moles of CO2* molar mass* speocific heat* (T-298.15)+ moles of water vapor* molar mass* specific heat*(T-298.15) + moles of O2* molas mass specific heat* temperature difference

specific heat CpN2= 1.278J/g.deg.c Cp H2O= 2.798 J/g.deg.c and Cp CO2= 1.3.64 j/g.deg.c, Cp O2= 1.173 J/g.deg.c

Products contains O2= 0.1848-0.18=0.0048 moles Oxygen, N2= 0.6952, CO2= 0.18 and H2O(g)= 2*0.1848= 0.3696 moles

81.072*1000 =( 0.0048*32*1.173 +0.6952*28*1.278+0.18*44*1.364+0.3696*18*2.798)*(T-298.15)=50.127*(T-298.15)

T- 298.15 = 81.072*1000/54.47 =1488.379

T= 1786 K ( the specific heat is assumed at 1800 K which is close to the adiabatic flame temperature)

T=1915.708 K