Calculate the theoretical air fuel ratio for the combustion of octane and ethanol. If 150% theoretical air and octane are burned to complete combustion, specify the equivalence ratio for the reaction.
The Correct Answer and Explanation is :
To calculate the theoretical air-fuel ratio (AFR) for the combustion of octane (C₈H₁₈) and ethanol (C₂H₅OH), we first need to write the balanced combustion reactions for both fuels.
Combustion of Octane (C₈H₁₈):
The balanced combustion reaction for octane is:
[
2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O
]
From this, we can calculate the theoretical air-fuel ratio. The molecular weight of octane is 114 g/mol, and the molecular weight of oxygen (O₂) is 32 g/mol.
For 2 moles of octane (2 × 114 = 228 g), 25 moles of oxygen are required (25 × 32 = 800 g). Since air contains approximately 21% oxygen by mass, the mass of air required is calculated as:
[
\text{Mass of air} = \frac{800}{0.21} \approx 3809.52 \, \text{g}
]
Therefore, the theoretical air-fuel ratio for octane is:
[
\text{AFR (octane)} = \frac{3809.52}{228} \approx 16.7 \, \text{(mass of air per unit mass of fuel)}
]
Combustion of Ethanol (C₂H₅OH):
The balanced combustion reaction for ethanol is:
[
C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O
]
Using the same method as for octane, the molecular weight of ethanol is 46 g/mol, and oxygen is 32 g/mol. For 1 mole of ethanol (46 g), 3 moles of oxygen are required (3 × 32 = 96 g). The mass of air required is:
[
\text{Mass of air} = \frac{96}{0.21} \approx 457.14 \, \text{g}
]
Therefore, the theoretical air-fuel ratio for ethanol is:
[
\text{AFR (ethanol)} = \frac{457.14}{46} \approx 9.95 \, \text{(mass of air per unit mass of fuel)}
]
Equivalence Ratio:
The equivalence ratio (( \phi )) is the ratio of the actual fuel-to-air ratio to the stoichiometric fuel-to-air ratio. It can be defined as:
[
\phi = \frac{\text{Actual fuel-to-air ratio}}{\text{Theoretical fuel-to-air ratio}}
]
If 150% theoretical air is used (meaning 1.5 times the theoretical amount of air), the equivalence ratio can be calculated as:
For octane:
[
\phi_{\text{octane}} = \frac{1}{1.5} \approx 0.67
]
Thus, the equivalence ratio for octane is approximately 0.67, which indicates a lean mixture (more air than necessary for complete combustion).
In summary:
- AFR for octane: 16.7
- AFR for ethanol: 9.95
- Equivalence ratio for octane (with 150% theoretical air): 0.67
This demonstrates the difference in air-fuel ratios for the combustion of different fuels and how the equivalence ratio is affected by the air supply.