A gaseous fuel has the following composition by volume – [latex]CH_4 = 4.0\%; CO = 26\%; CO_2 = 10\%; H_2 = 10\%; N_2= 50\%[/latex]. If 20% excess air is supplied calculate the volume of air supplied and the % composition of dry flue gases.

Minimum volume of air required [latex]= 0.260 × \frac{100}{21} = 1.238 m^3[/latex] Excess air [latex]= 1.238 × \frac{20}{100} = 0.2476 m^3[/latex] Total air supplied = 1.238 m³ + 0.2476 m³ = 1.4856 m³ Calculation of volume of dry flue gases The flue gas contains [latex]CO_2, O_2[/latex] from excess air and [latex]N_2[/latex] of fuel + [latex]N_2[/latex] of total air supplied. [latex]CO_2 = 0.40 m^3\\ \\ \\ O_2 = 0.2476 × \frac{21} {100} = 0.052 m^3 \\ \\ \\ N_2 = 0.50 m^3 \text{(of fuel gas)} +1.4856 × \frac{79}{100} \text{(from total air supplied)}\\ \\ \\ N_2 = 0.50 + 1.1174 = 1.674 m^3.[/latex] Total volume of dry flue gases = (0.40 + 0.052 + 1.674)m³ = 2.126 m³ [latex]\% CO_2 =\frac{0.40}{2.126} × 100 = 18.81\% \\ \\ \\ \%O_2 = \frac{0.052} {2.126} × 100 = 2.446\%\\ \\ \\ \% N_2 = \frac{1.674}{ 2.126} × 100 = 78.74\%[/latex]

Question 1.2.4: A gaseous fuel has the following composition by volume – CH4......

Engineering Chemistry: Fundamentals and Applications [EXP-169979]

A gaseous fuel has the following composition by volume – $CH_4 = 4.0\%; CO = 26\%; CO_2 = 10\%; H_2 = 10\%; N_2= 50\%$ . If 20% excess air is supplied calculate the volume of air supplied and the % composition of dry flue gases.

Step-by-Step

The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

Minimum volume of air required $= 0.260 × \frac{100}{21} = 1.238 m^3$

Excess air $= 1.238 × \frac{20}{100} = 0.2476 m^3$

Total air supplied = 1.238 m³ + 0.2476 m³ = 1.4856 m³

Calculation of volume of dry flue gases

The flue gas contains $CO_2, O_2$ from excess air and $N_2$ of fuel + $N_2$ of total air supplied.

CO_2 = 0.40  m^3\\ \\ \\ O_2 = 0.2476 × \frac{21} {100}  = 0.052  m^3 \\ \\ \\ N_2 = 0.50  m^3  \text{(of fuel gas)} +1.4856 × \frac{79}{100}  \text{(from total air supplied)}\\ \\ \\ N_2 = 0.50 + 1.1174 = 1.674  m^3.

Total volume of dry flue gases = (0.40 + 0.052 + 1.674)m³ = 2.126 m³

\%  CO_2 =\frac{0.40}{2.126}  × 100 = 18.81\% \\ \\ \\ \%O_2 = \frac{0.052} {2.126} × 100 = 2.446\%\\ \\ \\ \%  N_2 = \frac{1.674}{ 2.126}  × 100 = 78.74\%

Constituents	Volume in 1 m³ of gas	Combustion Reaction	$\bf O_2$ required (in m³)	Volume of dry flue gases(m³)
$CH_4$	0.04 m³	$CH_4 +2 O_2 \rightarrow CO_2 +2H_2O$	0.04 × 2 = 0.08 m³	$CO_2 =0.04 \times1 =0.04 m³$
$CO$	0.26 m³	$CO +\frac{1}{2} O_2 \rightarrow CO_2$	$0.26 × \frac{1}{2}= 0.13 m³$	$CO_2 =0.26 \times1 =0.26 m³$
$H_2$	0.10 m³	$H_2 + \frac{1}{2} O_2 \rightarrow H_2O$	$0.10 × \frac{1}{2}= 0.05 m³$
$CO_2$	0.10 m³	–	–	$CO_2 =0.10 m³$
$N_2$	0.50 m³	–	–	$0.50 m^3 + N_2 \text{in total air supplied}$
			Volume of $O_2$ required = 0.260 m³	Total $CO_2$ in flue gas = 0.40 m³