Combustion of a Gaseous Fuel with Moist Air A certain natural gas has the following volumetric analysis: 72 percent CH4, 9 percent H2, 14 percent N2, 2 percent O2, and 3 percent CO2. This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 20°C, 1 atm,

Question

Combustion of a Gaseous Fuel with Moist Air
A certain natural gas has the following volumetric analysis: 72 percent [latex]\mathrm{CH}_{4}[/latex], 9 percent [latex]\mathrm{H}_{2}[/latex], 14 percent [latex]\mathrm{N}_{2}[/latex], 2 percent [latex]\mathrm{O}_{2}[/latex], and 3 percent [latex]\mathrm{CO}_{2}[/latex]. This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 20°C, 1 atm, and 80 percent relative humidity, as shown in Fig. 15-12. Assuming complete combustion and a total pressure of 1 atm, determine the dew-point temperature of the products.

Accepted Answer

A gaseous fuel is burned with the stoichiometric amount of moist air. The dew point temperature of the products is to be determined.

Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to [latex]\mathrm{CO}_{2}[/latex] and all the hydrogen to [latex]\mathrm{H}_{2} \mathrm{O} [/latex]. 2 The fuel is burned with the stoichiometric amount of air and thus there is no free [latex]\mathrm{O}_{2}[/latex] in the product gases. 3 Combustion gases are ideal gases.

Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).

Analysis We note that the moisture in the air does not react with anything; it simply shows up as additional [latex]\mathrm{H}_{2} \mathrm{O}[/latex] in the products. Therefore, for simplicity,

we balance the combustion equation by using dry air and then add the moisture later to both sides of the equation.

Considering 1 kmol of fuel,

[latex]\overbrace{\left(0.72 \mathrm{CH}_{4}+0.09 \mathrm{H}_{2}+0.14 \mathrm{~N}_{2}+0.02 \mathrm{O}_{2}+0.03 \mathrm{CO}_{2} ight)}^{ ext { fuel }} \quad+\overbrace{a_{\mathrm{th}} {\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight)}}^{ ext { dryair }} \quad\longrightarrow[/latex]

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

The unknown coefficients in the above equation are determined from mass balances on various elements,

C: 0.72 + 0.03 = x → x = 0.75

H: 0.72 × 4 + 0.09 × 2 = 2y → y = 1.53

[latex]\mathrm{O}_{2}[/latex]: [latex]0.02+0.03+a_{ ext {th }}=x+\frac{y}{2} ightarrow a_{ ext {th }}=1.465[/latex]

[latex]\mathrm{N}_{2}[/latex]: [latex]0.14+3.76 a_{ ext {th }}=z \quad ightarrow \quad z=5.648[/latex]

Next we determine the amount of moisture that accompanies 4.76 a[latex]_ ext{th}[/latex] = (4.76 × 1.465) = 6.97 kmol of dry air. The partial pressure of the moisture in the air is

[latex]P_{ ext {v,air }}=\phi_{ ext {air }} P_{ ext {sat } @ 20^{\circ} \mathrm{C}}=(0.80)(2.3392 \mathrm{kPa})=1.871 \mathrm{kPa}[/latex]

Assuming ideal-gas behavior, the number of moles of the moisture in the air is

[latex]N_{ ext {v,air }}=\left(\frac{P_{v, a i r}}{P_{ ext {total }}} ight) N_{ ext {total }}=\left(\frac{1.871 \mathrm{kPa}}{101.325 \mathrm{kPa}} ight)\left(6.97+N_{v, ext { air }} ight)[/latex]

which yields

[latex]N_{v, \mathrm{air}}=0.131 \mathrm{kmol}[/latex]

The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.131 kmol of [latex]\mathrm{H}_{2} \mathrm{O}[/latex] to both sides of the equation:

[latex]\overbrace{\left(0.72 \mathrm{CH}_{4}+0.09 \mathrm{H}_{2}+0.14 \mathrm{~N}_{2}+0.02 \mathrm{O}_{2}+0.03 \mathrm{CO}_{2} ight)}^{ ext { fuel }} \quad+\overbrace{1.465 {\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight)}}^{ ext { dryair }} \quad[/latex]

[latex]+\quad \overbrace{0.131 \mathrm{H}_{2} \mathrm{O}}^{ ext {moisture }} ightarrow0.75\mathrm{CO}_2 + \overbrace{1.661 \mathrm{H}_{2} \mathrm{O}}^{ ext {includes moisture }} +5.648 \mathrm{N}_2[/latex]

The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled. Again, assuming ideal-gas behavior, the partial pressure of the water vapor in the combustion gases is

[latex]P_{v, ext { prod }}=\left(\frac{N_{v, ext { prod }}}{N_{ ext {prod }}} ight) P_{ ext {prod }}=\left(\frac{1.661 \mathrm{kmol}}{8.059 \mathrm{kmol}} ight)(101.325 \mathrm{kPa})=20.88 \mathrm{kPa}[/latex]

Thus,

[latex]T_{\mathrm{dp}}=T_{ ext {sat } @ 20.88 \mathrm{kPa}}=60.9^{\circ} \mathrm{C}[/latex]

Discussion If the combustion process were achieved with dry air instead of moist air, the products would contain less moisture, and the dew-point temperature in this case would be 59.5°C.

Question 15.3: Combustion of a Gaseous Fuel with Moist Air A certain natura...

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