18 A - Sample Problem on Combustion of Liquid Fuels (Case 1)

Lawrence Belo
1 Dec 202214:21

Summary

TLDRIn this video, the process of solving sample problems related to the combustion of liquid fuels is explored, using the octane number as a key fuel characteristic. The problem focuses on an automobile that uses unleaded gasoline with an octane number of 85 and analyzes the dry exhaust gases produced. The script walks through a detailed calculation involving fuel composition, air excess, and the molar ratios of carbon dioxide to carbon monoxide and hydrogen to carbon monoxide in the exhaust gases. By applying stoichiometric principles, the video provides a comprehensive guide on calculating the composition of exhaust gases in a combustion system.

Takeaways

  • 😀 The problem involves solving combustion problems for liquid fuels based on the octane number and fuel characteristics.
  • 😀 The fuel mixture consists of 85% iso-octane (C₈H₁₈) and 15% heptane (C₇H₁₆) by volume.
  • 😀 The octane number represents the percentage by volume of iso-octane in a mixture, indicating its knocking tendency.
  • 😀 The process begins with the calculation of the fuel's mass using the densities of iso-octane (0.6918 g/mL) and heptane (0.684 g/mL).
  • 😀 The molar masses of iso-octane (114 g/mol) and heptane (100 g/mol) are used to calculate the number of moles of each fuel component.
  • 😀 The total number of moles of carbon (from both iso-octane and heptane) is calculated to be 4.846 moles, while the total moles of hydrogen are 10.919 moles.
  • 😀 The theoretical oxygen needed for combustion is 7.5761 moles, but an excess of 30% air increases the oxygen supply to 9.8489 moles.
  • 😀 The nitrogen in the air is calculated based on the ratio of nitrogen to oxygen in air, resulting in 37.0507 moles of nitrogen.
  • 😀 The combustion products include CO₂, CO, H₂, O₂, and N₂, with the mole ratios of CO₂ to CO being 5:2, and H₂ to CO being 1:1.
  • 😀 The final analysis of the dry exhaust gases shows that CO₂ accounts for 7.37%, CO for 2.95%, H₂ for 2.95%, O₂ for 7.79%, and N₂ for 78.94%.

Q & A

  • What is the octane number and what does it represent in this problem?

    -The octane number is the percentage by volume of iso-octane in a mixture of iso-octane and heptane, representing the fuel's resistance to knocking. In this problem, the fuel has an octane number of 85, meaning it is 85% iso-octane and 15% heptane by volume.

  • What does the term 'excess air' mean in the context of this combustion problem?

    -Excess air refers to the additional amount of air supplied beyond the theoretical amount required for complete combustion. In this case, 30% excess air means the oxygen supplied is 30% more than what is theoretically necessary for the fuel's combustion.

  • How does the mole ratio of CO2 to CO (5:2) help in determining the composition of the exhaust gases?

    -The mole ratio of CO₂ to CO helps determine how the carbon atoms from the fuel are distributed in the exhaust gases. Since 5 moles of CO₂ are produced for every 2 moles of CO, knowing the total amount of carbon in the fuel allows us to calculate the exact number of moles of CO₂ and CO in the exhaust gases.

  • What is the significance of the mole ratio of H₂ to CO (1:1) in this problem?

    -The mole ratio of H₂ to CO (1:1) indicates that for every mole of CO in the exhaust, there is an equal number of moles of H₂. This relationship is used to calculate the amount of hydrogen present in the exhaust gases, which is important for the overall gas analysis.

  • Why is it important to calculate the moles of carbon and hydrogen atoms in the fuel?

    -Calculating the moles of carbon and hydrogen atoms is essential to determine how much oxygen is required for complete combustion, as well as to calculate the resulting moles of carbon dioxide, carbon monoxide, and hydrogen in the exhaust gases.

  • How is the total oxygen supply calculated when 30% excess air is provided?

    -The total oxygen supply is calculated by adding the theoretical oxygen required for combustion (based on the carbon and hydrogen content of the fuel) and the oxygen from the excess air, which is 30% more than the theoretical oxygen.

  • What is the method for calculating the nitrogen content in the exhaust gases?

    -The nitrogen content is calculated based on the air supplied. Since air is composed of about 79% nitrogen, the moles of nitrogen are determined by multiplying the moles of oxygen supplied by the ratio of nitrogen in air (79:21).

  • What is the significance of the fuel composition (iso-octane and heptane) in determining the exhaust gas composition?

    -The fuel composition directly influences the types and quantities of gases produced during combustion. Iso-octane and heptane have different carbon and hydrogen atom counts, which affects the amount of carbon dioxide, carbon monoxide, and hydrogen in the exhaust gases.

  • How do you calculate the free oxygen in the exhaust gases?

    -Free oxygen in the exhaust gases is calculated by adding the excess oxygen and the oxygen that has reacted with carbon monoxide and hydrogen to form CO₂ and water. The oxygen needed to react with carbon monoxide and hydrogen is calculated by considering the 1:2 and 1:2 stoichiometries for CO and H₂.

  • What is the final step in determining the exhaust gas analysis?

    -The final step is to calculate the total moles of all the gases (CO₂, CO, H₂, O₂, N₂) and then convert these mole values into percentages to determine the composition of the dry exhaust gases.

Outlines

plate

此内容仅限付费用户访问。 请升级后访问。

立即升级

Mindmap

plate

此内容仅限付费用户访问。 请升级后访问。

立即升级

Keywords

plate

此内容仅限付费用户访问。 请升级后访问。

立即升级

Highlights

plate

此内容仅限付费用户访问。 请升级后访问。

立即升级

Transcripts

plate

此内容仅限付费用户访问。 请升级后访问。

立即升级
Rate This

5.0 / 5 (0 votes)

相关标签
Combustion ChemistryFuel AnalysisOctane NumberExhaust GasesGasoline CombustionChemical EngineeringAir ExcessAutomobile FuelsEnergy SystemsFuel CharacteristicsMolar Ratios
您是否需要英文摘要?