Reverse Combustion Analysis Octane (C8H18) is burned with dry air. The volumetric analysis of the products on a dry basis is (Fig. 15-13) CO2: 10.02 percent O2: 5.62 percent CO: 0.88 percent N2: 83.48 percent Determine (a) the air-fuel ratio, (b) the percentage of theoretical air used,

Question

Reverse Combustion Analysis Octane

Octane [latex]\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)[/latex] is burned with dry air. The volumetric analysis of the products on a dry basis is (Fig. 15-13)

[latex]\mathrm{CO}_{2}[/latex]: 10.02 percent

[latex] \mathrm{O}_{2}[/latex]: 5.62 percent

[latex]\mathrm{CO}[/latex]: 0.88 percent

[latex]\mathrm{N}_{2}[/latex]: 83.48 percent

Determine (a) the air-fuel ratio, (b) the percentage of theoretical air used, and (c) the amount of [latex]\mathrm{H}_{2} \mathrm{O}[/latex] that condenses as the products are cooled to [latex]25^{\circ} \mathrm{C}[/latex] at 100 kPa.

Accepted Answer

Combustion products whose composition is given are cooled to 25°C. The AF, the percent theoretical air used, and the fraction of water vapor that condenses are to be determined.

Assumptions Combustion gases are ideal gases.

Properties The saturation pressure of water at 25°C is 3.1698 kPa (Table A-4).

Analysis Note that we know the relative composition of the products, but we do not know how much fuel or air is used during the combustion process.

However, they can be determined from mass balances. The [latex]\mathrm{H}_{2} \mathrm{O}[/latex] in the combustion gases will start condensing when the temperature drops to the dewpoint temperature.

For ideal gases, the volume fractions are equivalent to the mole fractions. Considering 100 kmol of dry products for convenience, the combustion equation can be written as

[latex]x \mathrm{C}_{8} \mathrm{H}_{18}+a\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 10.02 \mathrm{CO}_{2}+0.88 \mathrm{CO}+5.62 \mathrm{O}_{2}+83.48 \mathrm{~N}_{2}+b \mathrm{H}_{2} \mathrm{O}[/latex]

The unknown coefficients x, a, and b are determined from mass balances,

N[latex]_{2}[/latex]: [latex]3.76 a=83.48 \quad ightarrow \quad a=22.20[/latex]

C: [latex]8 x=10.02+0.88 \quad ightarrow \quad x=1.36[/latex]

H: [latex]18 x=2 b \quad ightarrow \quad b=12.24[/latex]

O[latex]_{2}[/latex]: [latex]a=10.02+0.44+5.62+\frac{b}{2} ightarrow 22.20=22.20[/latex]

The [latex]\mathrm{O}_{2}[/latex] balance is not necessary, but it can be used to check the values obtained from the other mass balances, as we did previously. Substituting, we get

[latex]1.36 \mathrm{C}_{8} \mathrm{H}_{18}+22.2\left(\mathrm{O}_{2}+ ight.\left.3.76 \mathrm{~N}_{2} ight) ightarrow [/latex]

[latex]10.02 \mathrm{CO}_{2}+0.88 \mathrm{CO}+5.62 \mathrm{O}_{2}+83.48 \mathrm{~N}_{2}+12.24 \mathrm{H}_{2} \mathrm{O}[/latex]

The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 1.36,

[latex]\mathrm{C}_{8} \mathrm{H}_{18}+16.32\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow [/latex]

[latex]7.37 \mathrm{CO}_{2}+0.65 \mathrm{CO}+4.13 \mathrm{O}_{2}+61.38 \mathrm{~N}_{2}+9 \mathrm{H}_{2} \mathrm{O}[/latex]

[a) The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel (Eq. 15-3),

[latex]\mathrm{AF}=\frac{m_{ ext {air }}}{m_{ ext {fuel }}} =\frac{(16.32 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]=19.76 \mathrm{~kg} \mathrm{air} / \mathrm{kg} ext { fuel }[/latex]

(b) To find the percentage of theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of the fuel,

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

[latex]a_{\mathrm{th}}=8+4.5 ightarrow a_{\mathrm{th}}=12.5[/latex]

Then,

Percentage of theoretical air [latex]=\frac{m_{ ext {air,act }}}{m_{ ext {air,th }}}=\frac{N_{ ext {air,act }}}{N_{ ext {air,th }}}[/latex]

[latex]=\frac{(16.32)(4.76) \mathrm{kmol}}{(12.50)(4.76) \mathrm{kmol}} [/latex]

[latex]=131 \%[/latex]

That is, 31 percent excess air was used during this combustion process. Notice that some carbon formed carbon monoxide even though there was considerably more oxygen than needed for complete combustion.

(c) For each kmol of fuel burned, 7.37+0.65+4.13+61.38+9= 82.53 kmol of products are formed, including 9 kmol of [latex]\mathrm{H}_{2} \mathrm{O}[/latex]. Assuming that the dew-point temperature of the products is above 25°C, some of the water vapor will condense as the products are cooled to 25°C. If [latex]N_{w} \mathrm{kmol}[/latex] of [latex]\mathrm{H}_{2} \mathrm{O}[/latex] condenses, there will be [latex]\left(9-N_{w} ight) \mathrm{kmol}[/latex] of water vapor left in the products. The mole number of the products in the gas phase will also decrease to 82.53-N[latex]_{w}[/latex] as a result. By treating the product gases (including the remaining water vapor) as ideal gases, [latex]N_{w}[/latex] is determined by equating the mole fraction of the water vapor to its pressure fraction,

[latex]\frac{N_{v}}{N_{ ext {prod,gas }}} =\frac{P_{v}}{P_{ ext {prod }}} [/latex]

[latex]\frac{9-N_{w}}{82.53-N_{w}} =\frac{3.1698 \mathrm{kPa}}{100 \mathrm{kPa}} [/latex]

[latex]N_{w} =6.59 \mathrm{kmol}[/latex]

Therefore, the majority of the water vapor in the products (73 percent of it) condenses as the product gases are cooled to 25°C.

Question 15.4: Reverse Combustion Analysis Octane (C8H18) is burned with dr...

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