Question

The gas of the following composition is burned with 70% excess air in the furnace.
CH4 60%, C2H6 20%, CO 5%, O2 5%, N210%.
It is known that the conversion of CH4 to form CO2 is 95%, forming 5% CO. Conversion of C2H6 forms 90% CO2, 5% CO, and does not burn 5%
Count it
a. Theoretical air requirements
b. The amount of air that enters the furnace
c. The remaining gas leaving the combustion chamber on a dry basis
d. Analyze the gas leaving the furnace on a wet basis

Answer 1

for the combustion of methane,

$CH_4 + a(O_2 + 3.76N_2) \to bCO_2 + cH_2O + dN_2$

where;

C: b = 1

H: 2c = 4

O: 2b + c = 2a

N: d = 3.76a

Solving for each

The balanced equation becomes;

$CH_4 + 2(O_2 + 3.76N_2) \to CO_2 + 2H_2O + 7.52N_2$

The amount of combustion air is 2 moles of oxygen plus 2 x 3.76 moles of nitrogen,

giving a total of 9.52 moles of air per mole of fuel. The theoretical air requirements for the above combustion is therefore;

Air requirements = $\dfrac{9.52×28.97}{16.04} = 17.19$ kg

Therefore, during the combustion of methane, 1kg of methane requires 17.19kg of air..

For your question,

$CH_4 + aC_2H_6 + bCO + c (O_2 + 3.76N_2) \to dCO_2 + eCO + fH_2O + gN_2$

for C, 2a + b + 1 = d + e

for N, 3.76c = g

for H, 4 + 6a = 2f

for O, b + 2c = 2d + e + f

after solving for all of them, use the formula above to get the theoretical air requirements.

The amount of air that enters the system is 75% of the theoretical air requirements.