Adiabatic Flame Temperature

Calculate the theoretical flame temperature of a gas having 20% CO and 80% N₂ when burnt with 150% excess air. Both the reactants are at 25^oC.

Data:

ΔH^o_f of CO₂ = -94052 cal/gmol

ΔH^o_f of CO = -26412 cal/gmol

Mean heat capacities in cal/gmol.^oC are CO₂ = 12, O₂ = 7.9, N₂ = 7.55

Calculations:

Basis : 1 mole of gas (having 20% CO and 80% N₂)

Combustion reaction:

CO + 1/2 O₂ → CO₂

Heat of reaction at 25^oC for the above reaction is calculated as:

Heat of reaction = heat of formation of products - heat of formation of reactants

= -94052 - (-26412) = -94052 + 26412 = -67640 cal/gmol of CO converted

This represents exothermic heat of reaction.

Total heat liberated = 67640 x 0.2 = 13528 cal.

If a base temperature of 25^oC is assumed, enthalpy of reactants at 25^oC = 0 cal

Total enthalpy of product stream = enthalpy of reactants + heat added by reaction = 0 + 13528 = 13528 cal.

This enthalpy rise of product stream with respect to feed is accomplished by a temperature increase, if the reaction is in adiabatic conditions.

Constituents of product stream are estimated as follows:

CO₂ in the product stream = 0.2 gmol

O₂ theoretically needed = 0.2 x 0.5 = 0.1 gmol.

O₂ actually entering (150% excess) = 0.1 x 2.5 = 0.25 gmol

N₂ entering along with O₂ in the air = 0.25 x 79/21 = 0.9405 gmol

O₂ in the leaving gases (assuming complete combustion of CO to CO₂) = 0.25 - 0.1 = 0.15 gmol

N₂ in the leaving gases = 0.9405 + 0.8 = 1.7405 gmol.

Adiabatic temperature (T) is calculated as follows:

13528 = 0.2 x 12 x (T - 25) + 0.15 x 7.9 x (T - 25) + 1.7405 x 7.55 x (T - 25)

i.e., 13528 = 16.726 (T - 25)

808.8 = T - 25

T = 808.8 + 25 = 833.8^oC

Theoretical or Adiabatic flame temperature = 833.8^oC