Question

Methane combusts with air in a furnace as \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \). The heat of reaction \( \Delta H_\text{r} = -880 \, \text{kJ/mol CH}_4 \), and is assumed to be constant. The furnace is well-insulated, and no other side reactions occur. All components behave as ideal gases with a constant molar heat capacity \( c_p = 40 \, \text{J mol}^{-1} \, \text{°C}^{-1} \). Air may be considered as 20 mol\% \( \text{O}_2 \) and 80 mol\% \( \text{N}_2 \). The air-fuel mixture enters the furnace at \( 50 \, \text{°C} \). The methane conversion \( X \) varies with the air-to-methane mole ratio, \( r \), as: \[ X = 1 - 0.1e^{-2(r-r_s)}, \, \text{with} \, 0.9r_s \leq r \leq 1.3r_s, \] where \( r_s \) is the stoichiometric air-to-methane mole ratio. For \( r = 1.05r_s \), the exit flue gas temperature in \( \, \text{°C} \), rounded off to 1 decimal place, is:

Hint:

For combustion problems, always calculate the stoichiometric ratio and use energy balances carefully to account for heat released.

Answer 1

Step 1: Calculate the stoichiometric air-to-methane ratio \( r_s \). The combustion reaction requires 2 moles of \( \text{O}_2 \) per mole of \( \text{CH}_4 \). Since air contains 20\% \( \text{O}_2 \): \[ r_s = \frac{2 \, \text{mol O}_2}{0.2 \, \text{mol O}_2/\text{mol air}} = 10. \] Step 2: Determine methane conversion \( X \). Given \( r = 1.05r_s \): \[ r = 1.05 \cdot 10 = 10.5. \] Substitute \( r \) into the conversion equation: \[ X = 1 - 0.1e^{-2(10.5 - 10)} = 1 - 0.1e^{-2(0.5)} = 1 - 0.1e^{-1}. \] \[ e^{-1} \approx 0.3679, \quad \text{so } X = 1 - 0.1 \cdot 0.3679 = 1 - 0.03679 = 0.9632. \] Step 3: Energy balance for the flue gas. The total heat released by combustion is: \[ Q = X \cdot \Delta H_r = 0.9632 \cdot (-880) = -847.6 \, \text{kJ/mol}. \] The molar heat capacity of the flue gases is given as \( c_p = 40 \, \text{J/mol °C} \). The temperature rise of the flue gas is: \[ \Delta T = \frac{-Q}{c_p}. \] Substitute \( Q = -847.6 \, \text{kJ/mol} = -847600 \, \text{J/mol} \) and \( c_p = 44 \, \text{J/mol °C} \): \[ \Delta T = \frac{847600}{44} = 21190 \, \text{°C}. \] The exit flue gas temperature is: \[ T_\text{exit} = T_\text{inlet} + \Delta T = \approx 1727 \, \text{°C}. \] Step 4: Conclusion. The exit flue gas temperature is \( 1727 \, \text{°C} \).