Combustion of a Gaseous Fuel with Moist Air
A certain natural gas has the following volumetric analysis: 72 percent \mathrm{CH}_{4}, 9 percent \mathrm{H}_{2}, 14 percent \mathrm{N}_{2}, 2 percent \mathrm{O}_{2}, and 3 percent \mathrm{CO}_{2}. 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.
Question 15.3: Combustion of a Gaseous Fuel with Moist Air A certain natura...
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 \mathrm{CO}_{2} and all the hydrogen to \mathrm{H}_{2} \mathrm{O} . 2 The fuel is burned with the stoichiometric amount of air and thus there is no free \mathrm{O}_{2} 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 \mathrm{H}_{2} \mathrm{O} 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,
\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}\right)}^{\text { fuel }} \quad+\overbrace{a_{\mathrm{th}} {\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right)}}^{\text { dryair }} \quad\longrightarrow
x \mathrm{CO}_{2}+y \mathrm{H}_{2} \mathrm{O}+z \mathrm{~N}_{2}
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
\mathrm{O}_{2}: 0.02+0.03+a_{\text {th }}=x+\frac{y}{2} \rightarrow a_{\text {th }}=1.465
\mathrm{N}_{2}: 0.14+3.76 a_{\text {th }}=z \quad \rightarrow \quad z=5.648
Next we determine the amount of moisture that accompanies 4.76 a_\text{th} = (4.76 × 1.465) = 6.97 kmol of dry air. The partial pressure of the moisture in the air is
P_{\text {v,air }}=\phi_{\text {air }} P_{\text {sat } @ 20^{\circ} \mathrm{C}}=(0.80)(2.3392 \mathrm{kPa})=1.871 \mathrm{kPa}
Assuming ideal-gas behavior, the number of moles of the moisture in the air is
N_{\text {v,air }}=\left(\frac{P_{v, a i r}}{P_{\text {total }}}\right) N_{\text {total }}=\left(\frac{1.871 \mathrm{kPa}}{101.325 \mathrm{kPa}}\right)\left(6.97+N_{v, \text { air }}\right)
which yields
N_{v, \mathrm{air}}=0.131 \mathrm{kmol}
The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.131 kmol of \mathrm{H}_{2} \mathrm{O} to both sides of the equation:
\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}\right)}^{\text { fuel }} \quad+\overbrace{1.465 {\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right)}}^{\text { dryair }} \quad
+\quad \overbrace{0.131 \mathrm{H}_{2} \mathrm{O}}^{\text {moisture }}\rightarrow0.75\mathrm{CO}_2 + \overbrace{1.661 \mathrm{H}_{2} \mathrm{O}}^{\text {includes moisture }} +5.648 \mathrm{N}_2
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
P_{v, \text { prod }}=\left(\frac{N_{v, \text { prod }}}{N_{\text {prod }}}\right) P_{\text {prod }}=\left(\frac{1.661 \mathrm{kmol}}{8.059 \mathrm{kmol}}\right)(101.325 \mathrm{kPa})=20.88 \mathrm{kPa}
Thus,
T_{\mathrm{dp}}=T_{\text {sat } @ 20.88 \mathrm{kPa}}=60.9^{\circ} \mathrm{C}
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.
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