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Question 1.2.3: A gaseous fuel has the following composition by volume – CH4......

A gaseous fuel has the following composition by volume – CH_4 = 20\%; CO = 10\%;  CO_2 = 5\%; O_2 = 2\%; C_2H_4 = 5\%; C_3H_8 = 8\%,  \text{rest}  N_2.Calculate the volume of air supplied per m³ of fuel and the % composition of dry flue gases.

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Volume of air required for the combustion of 1 m³ of fuel = 0.98 × \frac{100}{21} = 4.67  m^3

Calculation of dry flue gases

CO_2 = (0.20+0.10+0.05+0.10+0.24)  m^3 = 0.69  m^3 \\ \\ N_2 = N_2  \text{(fuel)} + N_2  \text{(from air)} = 4.19  m^3

Total volume of dry products = 0.69 + 4.19 = 4.88 m³

\%  CO_2 = \frac{0.69}{4.88}  × 100 = 14.139\% \\ \\ \\ \%  N_2 = \frac{4.19}{4.88} × 100 = 85.861\%
Constituent Volume in 1 m³ of gas Combustion Reaction
\bf O_2 required (in m³)
 Volume of  dry  flue  gases(m³)
CH_4 0.20 m³ CH_4 +2  O_2 \rightarrow CO_2 +2H_2O 0.20 × 2 = 0.40 m³ CO_2 = 0.20×1 = 0.20  m^3
CO 0.10 m³ CO +\frac{1}{2}  O_2 \rightarrow CO_2 0.10 × \frac{1}{2} = 0.05  m³ CO_2 = 0.10 × 1 = 0.10  m^3
CO_2 0.05 m³ –
–
 CO_2 = 0.05  m^3
O_2 0.02 m³ – –
C_2H_4 0.05 m³
C_2H_4 +3 O_2 \rightarrow 2CO_2 +2H_2O
 0.05 × 3 = 0.15 m³ CO_2 = 0.05 × 2 = 0.10  m^3
C_3H_8 0.08 m³
C_3H_8 +5O_2 \rightarrow 3CO_2 +4H_2O
 0.08 × 5 = 0.40 m³ CO_2 = 0.08 × 3 = 0.24  m^3
N_2
 0.50 – – N_2 = 0.50 + \frac{79}{100} × 4.67 = 4.19  m³
Total O_2 reqd =1.0 m³
O_2 in fuel =0.020 m³
Net O_2 reqd. = 0.98 m³

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