A fuel is found to contain C = 90%; H = 6.0%; S = 2.5%; O = 1.0% and ash = 0.5%. Calculate the amount of air required for complete combustion of 1 kg of fuel. If 25% excess air is used for combustion, calculate the percentage composition of the dry products of combustion.

Question

Accepted Answer

[latex]\begin{aligned} O_2 \text{required from air} &= \text{Total} O_2 \text{required} – O_2 \text{in fuel}\ & = 2.905 - 0.01 = 2.895 ~kg\end{aligned}[/latex]

[latex]\text{Minimum weight of air required for combustion} \quad \quad 2.895 × \frac{100}{ 23} =12.587 ~kg[/latex]

According to the question 25% excess air is supplied

[latex]\text{Excess air} = 12.587 × \frac{25}{100}= 3.147 ~kg\[/latex]

[latex]\begin{aligned}\text{Total air supplied} &= 12.587 ~kg + 3.147 ~kg \&= 15.734 ~kg\end{aligned}[/latex]

(i) Calculation of dry products in the flue gases

The flue gases contains [latex]CO_2, SO_2, O_2 \text{from excess air and} N_2[/latex] from the total air supplied.

Weight of [latex]CO_2[/latex] = 3.3 kg

Weight of [latex]SO_2[/latex] = 0.05 kg

Weight of [latex]O_2 = 3.147 × \frac{23} {100} = 0.724 ~kg[/latex]

Weight of [latex]N_2 = 15.734 × \frac{77} {100} = 12.115 ~kg[/latex]

[latex]\begin{aligned} \text{Total weight of flue gases} &= 3.3 + 0.05 + 0.724 + 12.115 \&= 16.189 ~kg\end{aligned}[/latex]

Percentage composition of dry flue gases

[latex]\% CO_2 = \frac{3.3}{16.189} × 100 = 20.38\% \ \ \% SO_2 = \frac{0.05} {16.189} × 100 = 0.309\% \ \ \% O_2 = \frac{0.724}{16.189} × 100 = 4.47\%\ \ \%N_2 = \frac{12.115}{16.189} × 100 = 74.835\%[/latex]

Constituent	Amount per kg of the coal sample	Combustion reaction	Weight of $\bf O_2$ required	Weight of dry flue gas
C	0.90 kg	$C +O_2 \rightarrow CO_2$	$0.90 × \frac{32}{12} = 2.4 ~kg$	$0.90 × \frac{44}{12} = 3.3 ~kg$
H	0.06 kg	$H_2 +\frac{1}{2} O_2 \rightarrow H_2O$	$0.06 × \frac{16}{2} = 0.48 ~kg$	$H_2O$ does not constitute dry flue gas
S	0.025 kg	$S + O_2 \rightarrow SO_2$	$0.025 × \frac{32}{32} = 0.025 ~kg$	$0.025 × \frac{64}{32} = 0.05 ~kg$
O	0.01 kg	–	–	–
Ash	0.005 kg	–	–	–
		–	$\text{Total} O_2 \text{required} = 2.4+\\0.48+0.025 ~kg=2.905 ~kg$

Question 1.1.4: A fuel is found to contain C = 90%; H = 6.0%; S = 2.5%; O = ......

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