Balancing the Combustion Equation One kmol of octane (C8H18) is burned with air that contains 20 kmol of O2, as shown in Fig. 15-7. Assuming the products contain only CO2, H2O, O2, and N2, determine the mole number of each gas in the products and the air-fuel ratio for this combustion process.

Question

Balancing the Combustion Equation
One kmol of octane [latex]\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)[/latex] is burned with air that contains [latex]20 \mathrm{kmol}[/latex] of [latex]\mathrm{O}_{2}[/latex], as shown in Fig. 15-7. Assuming the products contain only [latex]\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{O}_{2}[/latex], and [latex]\mathrm{N}_{2}[/latex], determine the mole number of each gas in the products and the air-fuel ratio for this combustion process.

Accepted Answer

The amount of fuel and the amount of oxygen in the air are given. The amount of the products and the AF are to be determined.

Assumptions The combustion products contain [latex]\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{O}_{2}[/latex], and [latex]\mathrm{N}_{2}[/latex] only.

Properties The molar mass of air is [latex]M_{ ext {air }}=28.97 \mathrm{~kg} / \mathrm{kmol} \cong 29.0 \mathrm{~kg} / \mathrm{kmol}[/latex] (Table A-1).

Analysis The chemical equation for this combustion process can be written as

[latex]\mathrm{C}_{8} \mathrm{H}_{18}+20\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow x \mathrm{CO}_{2}+y \mathrm{H}_{2} \mathrm{O}+z \mathrm{O}_{2}+w \mathrm{~N}_{2}[/latex]

where the terms in the parentheses represent the composition of dry air that contains 1 kmol of [latex]\mathrm{O}_{2}[/latex] and x, y, z, and w represent the unknown mole numbers of the gases in the products. These unknowns are determined by applying the mass balance to each of the elements-that is, by requiring that the total mass or mole number of each element in the reactants be equal to that in the products:

C: 8 = x → x = 8

H: 18 = 2y → y = 9

O: 20 × 2 = 2x + y + 2z → z = 7.5

[latex]\mathrm{N}_{2}[/latex]: (20) (3.76) = w → w = 75.2

Substituting yields

[latex]\mathrm{C}_{8} \mathrm{H}_{18}+20\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}+7.5 \mathrm{O}_{2}+75.2 \mathrm{~N}_{2}[/latex]

Note that the coefficient 20 in the balanced equation above represents the number of moles of oxygen, not the number of moles of air. The latter is obtained by adding 20 × 3.76=75.2 moles of nitrogen to the 20 moles of oxygen, giving a total of 95.2 moles of air. The air-fuel ratio (AF) is determined from Eq. 15-3 by taking the ratio of the mass of the air and the mass of the fuel,

[latex]\mathrm{AF} =\frac{m_{ ext {air }}}{m_{ ext {fuel }}}=\frac{(N M)_{ ext {air }}}{(N M)_{\mathrm{C}}+(N M)_{\mathrm{H}_{2}}} [/latex]

[latex]=\frac{(20 imes 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{(8 \mathrm{kmol})(12 \mathrm{~kg} / \mathrm{kmol})+(9 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})} [/latex]

[latex]=24.2 \mathrm{~kg} \mathrm{air} / \mathrm{kg} ext { fuel }[/latex]

That is, 24.2 kg of air is used to burn each kilogram of fuel during this combustion process.

Question 15.1: Balancing the Combustion Equation One kmol of octane (C8H18)...

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