Carbon Monoxide and Combustion Efficiency:

Carbon dioxide monitors have been added to many incinerators to assess combustion efficiency (C.E). The combustion efficiency is calculated as follows:

Where CO₂ = volume concentration (dry) of CO₂ (parts per million, volume, ppmv)

CO = volume concentration (dry) of CO (ppmv)

Combustion efficiency can also calculated as equal to (y_CO2 - y_CO)/y_CO2.

Example 3.4: Dry and wet basis of reporting analyses: 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

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

C The combustion reaction of toluene is given as

	Feed	Rx	Product
C6H5CH3	2	-2	0
O2	X	-18	X - 18
CO2	0	14	14
H2O	0	8	8
N2	3.764X	0	3.764X
Total	Wet-basis		4.764X + 4
Total	Dry-basis		4.764X - 4

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

C 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

C Excess O₂ = (27.39 - 18)/18 = 52.2%

C 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

C 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

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

X = 26.56 lbmol

C Excess O₂ = (26.56 - 18)/18 = 47.5%

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

C Combustion efficiency (Equation 3.12)_:

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