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Combustion Reaction
FUEL + AIR (or PURE O2) CO2, CO, H2O, SOX.
Combustion Reaction Combustion reactions are rapid oxidation reactions. One usually burns a fuel (typically coal, oil or natural gas) in air (or rarely pure oxygen). Types of Fuel
Solid: Coal.Gaseous: Natural gas.Liquid: Liquefied petroleum gas.
The principal atomic species in combustion reactions are carbon, hydrogen, sulfur and oxygen.
Why ImportantReleases tremendous amount of energy, which for example can be used to boil water to produce steam; the steam is then used to drive turbines to produce electricity.
Combustion Reaction Complete combustionIf all of the fuel is completely oxidized, we have complete combustion. For the principal species the combustion products are CO2, SO2, and H2O. Partial or incomplete combustionIf not all of the fuel is completely oxidized, or if some is not oxidized at all, we have partial or incomplete combustion.
A typical product of incomplete combustion is CO.
Terminology Theoretical oxygen: Amount of oxygen needed for complete
combustion – all carbon in the fuel is oxidized to CO2, and – all the hydrogen is oxidized to H2O
Theoretical air: Theoretical air is the amount of air containing the stoichiometric oxygen needed for complete combustion.
Air(theo)= 4.76 x O2(theo)
Excess air: The amount by which the air fed to reactor exceeds the theoretical air.
Percent Excess Air
Burning in excess air does not guarantee complete combustion.
Test Yourself
Example 5.1-1 Determination of a Solution Density Composition of Flue or Stack Gas Flue Gas: The product gas that leaves a combustion furnace is referred to as the stack gas or flue gas.
Wet basis composition: Wet basis includes the water vapor in the flue gas composition.
Dry basis (Orsat analysis) Composition: Dry basis does not include water vapor in the flue gas composition. One apparatus for measuring flue gas composition was named after Orsat. It reports compositions on a dry basis.
S
Example 5.1-1 Determination of a Solution Density Composition of Flue or Stack Gas
Wet Basis = => Dry Basis - Basis: 1 mole wet gas - basic idea, subtract the water and express the rest in %
Dry Basis = => Wet Basis - Basis: 1 mole wet gas - need one extra information: How much water is there in one mole of wet gas? (Say, y mole fraction out of 1 mole wet gas)
- subtract water from one mole of wet gas and get the mole of dry gas. (dry gas=1-y)
- Now, dry gas fraction is (1-y)*yi, where yi is mole fraction of ith dry gas components
Example 4.8-1Composition on Wet and Dry Bases
Problem 1: Wet Basis Dry Basis.
Example 4.8-1Problem 2: Dry Basis Wet Basis.An Orsat analysis (a technique for stack analysis) yields the following dry basis composition:
A humidity measurement shows that the mole fraction of H20 in the stack gas is 0.0700. Calculate the stack gas composition on a wet basis.
Solution 2: Dry Basis Wet Basis.
Also, try the alternative approach shown in class
From example shown in book
Exercise 4.64Question: A gas contains 80.0 wt% propane, 15.0 wt% n-butane, and the balance water.(a) Calculate the molar composition of this gas on both a wet and a dry
basis and the ratio (mol H2O/ mol dry gas).(b) If 100 Kg/h of this fuel is to be burned with 30% excess air, what is
the required air feed rate (kmol/h)? How would the answer change if the combustion were only 75% complete?
Answer:(a)
Exercise 4.64If 100 Kg/h of this fuel is to be burned with 30% excess air, what is
the required air feed rate (kmol/h)? How would the answer change if the combustion were only 75% complete?
Exercise 4.71
Exercise 4.71
