Calculate the adiabatic flame temperature of burning CH4 (g) at 1 atm when burned with 50%
excess air, and the reactants enter at 25oC.
Standard heat of formation:
Ch4: Hr = -49.963kj/mol
CO2: Hr = - 393.25 kJ/mol
H20(g) : Hr = - 241.835 kj/mol
Heat capacities (J/mol), T in oC:
CO2 (g) : 36.11 + 4.233×10-2 T - 2.887×10-5T2 + 7.464×10-9 T3
H2O (g) : 33.46 + 0.688×10-2 T + 0.7604×10-5T2 + 3.593×10-9 T3
O2 (g) : 29.1 + 1.158×10-2 T - 0.6076×10-5T2 + 1.311×10-9 T3
N2 (g) : 29 + 0.2199×10-2 T + 0.5723×10-5T2 - 2.787×10-9 T3
To calculate the adiabatic flame temperature of burning CH4 (g) with 50% excess air, we need to consider the heat transfer and energy balance during the combustion process. Adiabatic conditions mean that no heat is transferred to the surroundings.
Here's how you can calculate the adiabatic flame temperature:
1. Convert the given temperature from Celsius to Kelvin:
T = 25 + 273.15 = 298.15 K
2. Determine the reactants and products of the combustion reaction:
CH4 (g) + (oxygen from air) → CO2 (g) + H2O (g)
3. Calculate the heat of reaction (∆H) for the combustion reaction:
∆H = [moles of CO2 * ∆Hf(CO2)] + [moles of H2O * ∆Hf(H2O)] - [moles of CH4 * ∆Hf(CH4)]
Assuming that all the carbon in CH4 forms CO2 and all the hydrogen forms H2O, we get:
∆H = [1 * (-393.25 kJ/mol)] + [2 * (-241.835 kJ/mol)] - [1 * (-49.963 kJ/mol)]
∆H = -844.927 kJ/mol
4. Calculate the moles of CH4 required for the given conditions:
Assuming ideal gas behavior, we can use the ideal gas law and molar mass of CH4 to find the moles:
Moles of CH4 = (1 atm * V) / (R * T)
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- V is the volume in liters (for 1 mole of gas at STP, V is 22.414 L)
Let's assume 1 mole of CH4 for simplicity:
Moles of CH4 = (1 atm * 22.414 L) / (8.314 J/mol·K * 298.15 K) = 0.9451 moles
5. Determine the moles of air required:
Since we have a 50% excess air, we need twice the mole of oxygen as CH4:
Moles of O2 = 2 * 0.9451 moles = 1.8902 moles
Moles of N2 (from air) = 3.76 * Moles of O2 = 3.5727 moles
Total moles of air = Moles of O2 + Moles of N2 = 1.8902 moles + 3.5727 moles = 5.4629 moles
6. Calculate the heat capacity of the gas mixture at the inlet temperature (25°C):
Heat capacity = Heat capacity of CO2 * Moles of CO2 + Heat capacity of H2O * Moles of H2O + Heat capacity of O2 * Moles of O2 + Heat capacity of N2 * Moles of N2
Heat capacity = [36.11 + 4.233×10^(-2) * (25) - 2.887×10^(-5) * (25^2) + 7.464×10^(-9) * (25^3)] * 1 + [33.46 + 0.688×10^(-2) * (25) + 0.7604×10^(-5) * (25^2) + 3.593×10^(-9) * (25^3)] * 2 + [29.1 + 1.158×10^(-2) * (25) - 0.6076×10^(-5) * (25^2) + 1.311×10^(-9) * (25^3)] * 1 + [29 + 0.2199×10^(-2) * (25) + 0.5723×10^(-5) * (25^2) - 2.787×10^(-9) * (25^3)] * 3.5727
Heat capacity = 117.85 J/K
7. Calculate the heat released during combustion:
Heat released = ∆H * Moles of CH4 = -844.927 kJ/mol * 0.9451 mol = -798.849 kJ
8. Apply the energy balance equation:
Heat released = Heat capacity * (Tf - Ti)
- Tf is the adiabatic flame temperature we need to find.
- Ti is the initial temperature (298.15 K).
Rearrange the equation to solve for Tf:
Tf = (Heat released / Heat capacity) + Ti
Tf = (-798.849 kJ / 117.85 J/K) + 298.15 K
Tf = 1906.24 K
Therefore, the adiabatic flame temperature of burning CH4 with 50% excess air at 1 atm and an initial temperature of 25°C is approximately 1906.24 Kelvin.