E3.4 Combustion Efficiency - Environmental Concerns

Example 3.4: A stack gas from a liquid injection incinerator contains 7% oxygen by volume on wet basis at standard conditions. Toluene is burnt at a rate of 184 lb/h with air. What percent excess air is required? What is the combustion efficiency if the CO content of the flue gas is 500 ppm. How much excess air would be required, if the stack gas measurements were on a dry basis. Molecular weight of toluene, M= 92

Solution

Mass flow rate, W = 184 lb/h

Molar flow rate, F = W/M = 2 lbmol/h

The combustion reaction of toluene is given as

Wet basis: Oxygen analysis in the products is given as 7% or 0.07 mole fraction.

Mole fraction of oxygen, y_O2

y_O2 = (X - 18)/(4.764X + 4) = 0.07

X = 27.39 lbmol

Stoichiometric amount of oxygen requirement = 18 moles

Excess O₂ = (27.39 - 18)/18 = 52.2%

Theoretical mole fraction of CO₂, y_CO2 = 14/(4.764(27.39) + 4) = 0.104

Mole fraction of CO, y_CO = 500 ppm = 5 ´ 10^-4

Combustion efficiency (Equation 3.12):

(y_CO2 - y_CO)/y_CO2 = (0.104 - 5 ´ 10^-4)/0.104 = 99.52%

Dry basis: If the results are given on dry basis, then

y_O2 = (X - 18)/(4.764X - 4) = 0.07

X = 26.56 lbmol

Excess O₂ = (26.56 - 18)/18 = 47.5%

Theoretical mole fraction of CO₂, y_CO2 = 14/(4.764(26.56) - 4) = 0.114

Combustion efficiency (Equation 3.12)_:

(y_CO2 - y_CO)/y_CO2 = (0.114 - 5 ´ 10^-4)/0.114 = 99.56%