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 25oC.
ΔHof of CO2 = -94052 cal/gmol
ΔHof of CO = -26412 cal/gmol
Mean heat capacities in cal/gmol.oC are CO2 = 12, O2 = 7.9, N2 = 7.55
Basis : 1 mole of gas (having 20% CO and 80% N2)
CO + 1/2 O2 → CO2
Heat of reaction at 25oC 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 25oC is assumed, enthalpy of reactants at 25oC = 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 = 833.8oC
Theoretical or Adiabatic flame temperature = 833.8oC
Last Modified on: 04-Feb-2022
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