EXAMPLE 3.12
For a boiler, 200 lb‐mol/h of air at STP will be required for complete combustion of methane. What is the actual flow rate per minute at stack condition at 800 °F at 1 atm.
SOLUTION
Corrections for Percent O2
Corrections for percent O2 are often needed for referencing combustion source emissions. For boilers, comparisons are made at 3% oxygen, consistent with the near stoichiometric conditions in boiler combustors. For combustion turbines, comparisons are made at 15% oxygen, consistent with the high excess air flow in a combustion turbine. To correct for concentrations at other oxygen levels, the following equation is used (Eq. 3.16):
where m% O2 is the nonstandard oxygen concentration. To calculate the concentration at 15% O2, substitute 15.0 for 3.0 in Eq. (3.18).
Boiler Flue Gas Concentrations Are Usually Corrected to 3% Oxygen
EXAMPLE 3.13
The particulates in a stack gas are measured as 0.045 gr/dry standard cubic foot (dscf) with a stack gas oxygen concentration of 7% by volume. What is the particulate concentration at 3% oxygen?
SOLUTION
Air‐to‐Fuel Ratio and Stoichiometric Ratio
In combustion of a hydrocarbon fuel, it is important to accurately calculate the mass of air‐to‐fuel ratio (AFR). Then, the stoichiometric ratio (SR) can be defined as the actual AFR divided by the stoichiometric AFR, as illustrated by the following equation:
(3.19)
The complete combustion of hydrocarbons produces CO2, the desirable aspect of rapid combustion, although CO2 is also an agent of global warming. The following equation shows the reaction:
(3.20)
From a practical standpoint, discounting the disastrous effect of CO2 on global warming, Eq. (3.7) is the desirable reaction. This reaction liberates all the heats of reaction of the hydrocarbon.
The study of combustion frequently uses the terms percent excess air, SR, and equivalence ratio. Percent excess air is the difference between the actual mass of air used and the theoretical divided by the theoretical times 100. The SR is the ratio of the theoretical fuel/air mass ratio to the actual fuel/air mass ratio. The equivalence ratio is the reciprocal of the SR. Air is 21% O2 and 79% N2.
EXAMPLE 3.14
A solid‐waste‐to‐energy recovery plant fires solid waste represented by C50H100O40N. The plant burns 100 T/day of solid waste, producing 20 T/day of residue containing 5% C. (a) If the solid waste burns under conditions of insufficient air, how much CO is produced per day? (b) If the solid waste burns at 20% excess, how much CO2 is produced per day? (c) What are the stoichiometric and equivalence ratios? Assume that fuel nitrogen converts to NO.
SOLUTION
- Since the ratio of H to O in H2O is 2 : 1, the H in H2O of the solid waste is 2(40) = 80, from the formula of C50H100O40N. Therefore, the H of the hydrocarbon portion of the solid waste is 100 – 80 = 20. Hence, CxHy = C50H2O.
The oxidation state of O has been changed from 0 to −2, involving a total of 30(2)(2) = 120 mol of electrons. Hence, the number of reference species = 120 mol of electrons.
Carbon burned to gases = 100(0.44) − 20(0.05) = 43 T/day (i.e. the carbon burned to gases in the C50 of C50H20).
Thus,
- C50H20 + 55O2 → 50CO2 + 10H2OReferences species = 55(2)(2) = 220 mol of electrons
- C50H100O40N +
(=55.5)O2 → 50CO2 + 50H2O + NO2
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