3.5 Combustion reactions

Combustion reactions are crucial in chemical engineering, involving the burning of fuels with oxygen. They're key to energy production and industrial processes. Understanding these reactions helps engineers optimize fuel efficiency and minimize environmental impact.

Material balances in combustion reactions are essential for analyzing reactant and product flows. By applying conservation of mass principles, engineers can calculate air requirements, determine combustion product compositions, and solve complex combustion problems in various applications.

Combustion Reactions for Fuels

Balanced Combustion Reactions

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• Combustion is a chemical reaction between a fuel and an oxidant that produces heat and light
  • The most common oxidant is oxygen in air
• Common fuels include hydrocarbons (natural gas, propane, butane), alcohols (methanol, ethanol), and biomass (wood, agricultural waste)
• A balanced combustion reaction has the correct stoichiometric coefficients for the reactants and products, ensuring conservation of mass
  • For example, the balanced combustion reaction for methane (CH4) is: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

Complete and Incomplete Combustion

• Complete combustion occurs when a fuel reacts with sufficient oxygen to produce only carbon dioxide and water as products
  • Example: Complete combustion of propane (C3H8): $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
• Incomplete combustion occurs when there is insufficient oxygen, resulting in the formation of carbon monoxide and other byproducts
  • Example: Incomplete combustion of ethanol (C2H5OH): $2C_2H_5OH + 3O_2 \rightarrow 2CO + 4H_2O + 2CH_4$

Air Requirements for Combustion

Theoretical and Excess Air

• Theoretical (stoichiometric) air is the minimum amount of air required for complete combustion of a fuel, based on the balanced chemical reaction
  • For example, the theoretical air required for complete combustion of methane is 2 moles of oxygen per mole of methane
• Excess air is the additional air supplied beyond the theoretical requirement to ensure complete combustion and improve efficiency
  • Example: If 20% excess air is used in the combustion of methane, the actual air supplied is 1.2 times the theoretical air

Air-Fuel Ratio and Percent Excess Air

• The air-fuel ratio (AFR) is the mass ratio of air to fuel in a combustion process
  • Stoichiometric AFR corresponds to theoretical air, while actual AFR accounts for excess air
• Percent excess air is the amount of air supplied in excess of the theoretical requirement, expressed as a percentage
  • For example, if the actual air supplied is 1.5 times the theoretical air, the percent excess air is 50%
• Calculating theoretical and excess air requirements involves using the balanced combustion reaction and the desired percent excess air

Composition of Combustion Products

Complete and Incomplete Combustion Products

• Combustion products include the compounds formed during the reaction, such as carbon dioxide, water vapor, and nitrogen (from air)
  • Complete combustion of hydrocarbons produces carbon dioxide and water vapor
  • Incomplete combustion also yields carbon monoxide, hydrogen, and other byproducts
• Flue gases are the gaseous mixtures that exit the combustion chamber, consisting of combustion products and any unused air

Calculating Composition of Combustion Products

• The composition of combustion products and flue gases depends on the fuel composition, air-fuel ratio, and combustion efficiency
• Calculating the composition of combustion products and flue gases involves applying the balanced combustion reaction, stoichiometry, and accounting for excess air
  • For example, to calculate the mole fraction of CO2 in the flue gas from the complete combustion of methane with 20% excess air:
    • Write the balanced combustion reaction: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
    • Calculate the actual moles of air supplied: $2 \times 1.2 = 2.4$ moles of O2 per mole of CH4
    • Calculate the moles of CO2 produced: 1 mole of CO2 per mole of CH4
    • Calculate the total moles of flue gas: $1 + 2 + (2.4 - 2) \times 3.76 = 4.504$ moles (assuming air is 21% O2 and 79% N2)
    • Mole fraction of CO2: $1 / 4.504 = 0.222$

Material Balances in Combustion

Conservation of Mass in Combustion Reactions

• Material balances are based on the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction
• Combustion material balance problems involve analyzing the flow of reactants (fuel and air) and products (combustion products and flue gases) in a system

Solving Combustion Material Balance Problems

• Key steps in solving combustion material balance problems include:
  • Writing the balanced combustion reaction
  • Identifying given information (fuel composition, air-fuel ratio, percent excess air)
  • Calculating the theoretical air requirement
  • Determining the actual air supplied based on percent excess air
  • Calculating the composition and flow rates of combustion products and flue gases
• Material balance problems may involve analyzing the effect of variables such as fuel composition, air-fuel ratio, and percent excess air on combustion efficiency and emissions
  • For example, a problem may ask to determine the mass flow rate of flue gases produced from the combustion of 100 kg/h of propane (C3H8) with 30% excess air
    • Write the balanced combustion reaction: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
    • Calculate the theoretical air requirement: $5 \times (32 \text{ kg O}_2 / \text{kmol O}_2) = 160 \text{ kg air} / \text{kmol C}_3\text{H}_8$
    • Determine the actual air supplied: $160 \times 1.3 = 208 \text{ kg air} / \text{kmol C}_3\text{H}_8$
    • Calculate the molar flow rate of propane: $100 \text{ kg/h} / (44 \text{ kg/kmol}) = 2.27 \text{ kmol/h}$
    • Calculate the mass flow rate of flue gases: $2.27 \text{ kmol/h} \times (3 \times 44 + 4 \times 18 + 208) = 442.7 \text{ kg/h}$