Combustion of Coal with Theoretical Air Coal from Pennsylvania which has an ultimate analysis (by mass) as 84.36 percent C, 1.89 percent H2, 4.40 percent O2, 0.63 percent N2, 0.89 percent S, and 7.83 percent ash (non-combustibles) is burned with theoretical amount of air (Fig. 15-11).

Question

Combustion of Coal with Theoretical Air
Coal from Pennsylvania which has an ultimate analysis (by mass) as 84.36 percent C, 1.89 percent [latex]\mathrm{H}_2[/latex], 4.40 percent [latex]\mathrm{O}_2[/latex], 0.63 percent [latex]\mathrm{N}_2[/latex], 0.89 percent S, and 7.83 percent ash (non-combustibles) is burned with theoretical amount of air (Fig. 15-11). Disregarding the ash content, determine the mole fractions of the products and the apparent molar mass of the product gases. Also determine the air-fuel ratio required for this combustion process.

Accepted Answer

Coal with known mass analysis is burned with theoretical amount of air. The mole fractions of the product gases, their apparent molar mass, and the air-fuel ratio are to be determined.

Assumptions 1 Combustion is stoichiometric and thus complete. 2 Combustion products contain [latex]\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{SO}_{2}[/latex], and [latex]\mathrm{O}[/latex] only (ash disregarded). 3 Combustion gases are ideal gases.

Analysis The molar masses of [latex]\mathrm{C}, \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{~S},[/latex] and air are 12, 2, 32, 32, and 29 kg/kmol, respectively (Table A-1). We now consider 100 kg of coal for simplicity. Noting that the mass percentages in this case correspond to the masses of the constituents, the mole numbers of the constituent of the coal are determined to be

[latex]N_{\mathrm{C}}=\frac{m_{C}}{M_{C}}=\frac{84.36 \mathrm{~kg}}{12 \mathrm{~kg} / \mathrm{kmol}}=7.030 \mathrm{kmol} [/latex]

[latex]N_{\mathrm{H} 2}=\frac{m_{\mathrm{H}_2}}{M_{\mathrm{H}_{2}}}=\frac{1.89 \mathrm{~kg}}{2 \mathrm{~kg} / \mathrm{kmol}}=0.9450 \mathrm{kmol} [/latex]

[latex]N_{\mathrm{O} 2}=\frac{m_{\mathrm{O}_{2}}}{M_{\mathrm{O}_{2}}}=\frac{4.40 \mathrm{~kg}}{32 \mathrm{~kg} / \mathrm{kmol}}=0.1375 \mathrm{kmol} [/latex]

[latex]N_{\mathrm{N} 2}=\frac{m_{\mathrm{N}_{2}}}{M_{\mathrm{N}_{2}}}=\frac{0.63 \mathrm{~kg}}{28 \mathrm{~kg} / \mathrm{kmol}}=0.0225 \mathrm{kmol} [/latex]

[latex]N_{\mathrm{S}}=\frac{m_{\mathrm{S}}}{M_{\mathrm{S}}}=\frac{0.89 \mathrm{~kg}}{32 \mathrm{~kg} / \mathrm{kmol}}=0.0278 \mathrm{kmol}[/latex]

Ash consists of the non-combustible matter in coal. Therefore, the mass of ash content that enters the combustion chamber is equal to the mass content that leaves. Disregarding this non-reacting component for simplicity, the combustion equation may be written as

[latex]7.03 \mathrm{C}+0.945 \mathrm{H}_{2}+0.1375 \mathrm{O}_{2}+0.0225 \mathrm{~N}_{2}+0.0278 \mathrm{~S}+a_{\mathrm{th}}\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) [/latex]

[latex] ightarrow x \mathrm{CO}_{2}+y \mathrm{H}_{2} \mathrm{O}+z \mathrm{SO}_{2}+w \mathrm{~N}_{2}[/latex]

Performing mass balances for the constituents gives

C balance: x = 7.03
[latex]\mathrm{H}_{2}[/latex] balance: y = 0.945
S balance: z = 0.0278

[latex]\mathrm{O}_{2}[/latex] balance: 0.1375 + [latex]a_ ext{th}[/latex] = x + 0.5y + z → [latex]a_ ext{th}[/latex] = 7.393

[latex]\mathrm{N}_{2}[/latex] balance: w = 0.0225 + 3.76 [latex]a_ ext{th}[/latex] = 0.0225 + 3.76 × 7.393 = 27.82

Substituting, the balanced combustion equation without the ash becomes

[latex]7.03 \mathrm{C}+0.945 \mathrm{H}_{2}+0.1375 \mathrm{O}_{2}+0.0225 \mathrm{~N}_{2}+0.0278 \mathrm{~S}+7.393\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) [/latex]

[latex] ightarrow 7.03 \mathrm{CO}_{2}+0.945 \mathrm{H}_{2} \mathrm{O}+0.0278 \mathrm{SO}_{2}+27.82 \mathrm{~N}_{2}[/latex]

The mole fractions of the product gases are determined as follows.

[latex]N_{ ext {prod }} =7.03+0.945+0.0278+27.82=35.82 \mathrm{kmol} [/latex]

[latex]y_{\mathrm{CO}_2}= \frac{N_{\mathrm{CO}_{2}}}{N_{ ext {prod }}}=\frac{7.03 \mathrm{kmol}}{35.82 \mathrm{kmol}}=0.1963 [/latex]

[latex]y_{\mathrm{H}_{2} \mathrm{O}}=\frac{N_{\mathrm{H}_{2} \mathrm{O}}}{N_{ ext {prod }}}=\frac{0.945 \mathrm{kmol}}{35.82 \mathrm{kmol}}=0.02638 [/latex]

[latex]y_{\mathrm{SO}_{2}}= \frac{N_{\mathrm{SO}_{2}}}{N_{ ext {prod }}}=\frac{0.0278 \mathrm{kmol}}{35.82 \mathrm{kmol}}=0.000776 [/latex]

[latex]y_{\mathrm{N}_{2}}= \frac{N_{\mathrm{N}_{2}}}{N_{ ext {prod }}}=\frac{27.82 \mathrm{kmol}}{35.82 \mathrm{kmol}}=0.7767[/latex]

Then, the apparent molar mass of product gases becomes

[latex]M_{ ext {prod }}=\frac{m_{ ext {prod }}}{N_{ ext {prod }}} =\frac{(7.03 imes 44+0.945 imes 18+0.0278 imes 64+27.82 imes 28) \mathrm{kg}}{35.82 \mathrm{kmol}} [/latex]

[latex]=30.9 \mathrm{~kg} / \mathrm{kmol}[/latex]

Finally, the air-fuel mass ratio is determined from its definition to be

[latex]\mathrm{AF}=\frac{m_{ ext {air }}}{m_{ ext {fuel }}}=\frac{(7.393 imes 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{100 \mathrm{~kg}}=10.2 ext{ kg air}/ \mathrm{kg}[/latex] fuel

That is, 10.2 kg of air is supplied for each kg of coal in the furnace.

Discussion We could also solve this problem by considering just 1 kg of coal, and still obtain the same results. But we would have to deal with very small fractions in calculations in this case.

Question 15.2: Combustion of Coal with Theoretical Air Coal from Pennsylvan...

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