Page 102 - Air pollution and greenhouse gases from basic concepts to engineering applications for air emission control

76 3 Basics of Gas Combustion

At a macroscopic scale, it seems that the combustion process is finished, but at a
microscopic scale, there may be chemical equilibrium established among different
products in the form of, say,

1 1
A $ B þ C and 2C þ D $ 3X þ Y
2 2
These chemical equilibrium reactions determine the actual quantities of the
products, numerically, a; b; c...z. With the known equilibrium constants of these
reactions, we can derive the amount of the gas products.
Let us use an example to explain how it works for a fuel rich combustion.
Example 3.8: Fuel rich combustion and chemical equilibrium
Combustion products from the reactant mixture of 1,000 mol of CO 2 , 500 mol of
O 2 and 500 mol of N 2 consist of CO 2 , CO, O 2 ,N 2 and NO at 3,000 K and 1 atm.
Determine the equilibrium composition of the combustion product.

Solution
Since the reactant mixture and the product compounds are already known, the
combustion stoichiometry can be described as

1 1
CO 2 þ O 2 þ N 2 ! aCO + bNO + cCO 2 þdO 2 þeN 2
2 2

with a, b, c, d, e as coefficients to be determined.
From the mass balances for C, O, and N we can set up three equations:

Carbon balance : a þ c þ 1 ð1Þ
Oxygen balance : a þ b þ 2c þ 2d ¼ 3 ð2Þ

Nitrogen balance : b þ 2e ¼ 1 ð3Þ

We need two more equations, which can be obtained from two equilibrium
reactions:
1
2
CO 2 $ CO þ = O 2 ð R 1 Þ
1
1 = O 2 þ = N 2 $ NO ð R 2 Þ
2
2
The corresponding chemical equilibrium constants can be found in Table 3.2,
which gives,

K P1 ð3,000 KÞ¼ expð 1:111Þ¼ 0:3273
K P2 ð3,000 KÞ¼ expð 2:114Þ¼ 0:1222