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Combustion analysis and stoichiometry are key to understanding how fuels burn and release energy. These concepts help us balance chemical equations, calculate air-fuel ratios, and determine the composition of combustion products.

By mastering these principles, we can optimize combustion efficiency and reduce harmful emissions. This knowledge is crucial for designing better engines, power plants, and industrial processes that use combustion to generate energy.

Balancing combustion reactions

Reactants and products in combustion

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  • Combustion reactions involve the rapid oxidation of a fuel (typically a hydrocarbon) with oxygen gas, releasing heat and light energy
  • The reactants are the fuel and oxygen, while the products are carbon dioxide, water vapor, and heat
  • Nitrogen in the air can also react with oxygen at high temperatures to form nitrogen oxides (NOx), which are that contribute to smog and acid rain
    • The formation of NOx depends on factors such as temperature, pressure, and residence time in the combustion chamber

Balancing combustion reaction equations

  • The general form of a hydrocarbon combustion reaction is: CxHy+(x+y/4)O2xCO2+(y/2)H2O+heatCxHy + (x + y/4)O2 → xCO2 + (y/2)H2O + heat
    • Carbon and hydrogen in the fuel react with oxygen to form carbon dioxide and water vapor, respectively
  • Balancing combustion reactions requires adjusting the coefficients of the reactants and products to ensure that the number of atoms of each element is equal on both sides of the equation, following the
  • occurs when there is sufficient oxygen for all the carbon in the fuel to be converted to carbon dioxide and all the hydrogen to be converted to water vapor
    • Example: CH4+2O2CO2+2H2O+heatCH4 + 2O2 → CO2 + 2H2O + heat (methane combustion)
  • occurs when there is insufficient oxygen, resulting in the formation of carbon monoxide and/or unburned hydrocarbons
    • Example: 2C2H6+5O23CO2+CO+6H2O+heat2C2H6 + 5O2 → 3CO2 + CO + 6H2O + heat (incomplete ethane combustion)

Air-fuel ratios for combustion

Theoretical and actual air-fuel ratios

  • The air-fuel ratio (AFR) is the mass ratio of air to fuel in a combustion process
  • The theoretical (or stoichiometric) AFR is the minimum amount of air required for complete combustion of the fuel, assuming that all the oxygen in the air reacts with the fuel
    • To calculate the theoretical AFR, determine the molar quantities of fuel and oxygen required for complete combustion using the balanced reaction equation, then convert the molar quantities to mass using the molecular weights of the fuel and air (assuming air is 21% oxygen by volume)
  • The actual AFR is the ratio of the mass of air supplied to the mass of fuel consumed in a real combustion process
    • It is typically higher than the theoretical AFR to ensure complete combustion and to control the combustion temperature

Equivalence ratio and excess air

  • The equivalence ratio (φ) is the ratio of the actual AFR to the theoretical AFR
    • An equivalence ratio greater than 1 indicates a fuel-rich mixture (excess fuel), while a ratio less than 1 indicates a fuel-lean mixture (excess air)
  • The percent excess air is another way to express the amount of air supplied beyond the theoretical requirement
    • It is calculated as: %ExcessAir=(ActualAFRTheoreticalAFR)/TheoreticalAFR×100%\%Excess Air = (Actual AFR - Theoretical AFR) / Theoretical AFR × 100\%
    • Example: If the actual AFR is 18 and the theoretical AFR is 15, the percent excess air is (1815)/15×100%=20%(18 - 15) / 15 × 100\% = 20\%

Combustion product composition

Determining product composition using stoichiometry

  • The composition of combustion products can be determined using the balanced combustion reaction equation and stoichiometric relationships
    • The molar quantities of the products are directly proportional to the coefficients in the balanced equation
  • For complete combustion of hydrocarbons, the products are carbon dioxide, water vapor, and nitrogen (from the air)
    • The molar quantities of CO2 and H2O can be determined from the coefficients of the balanced equation, while the molar quantity of N2 is calculated based on the composition of air (79% nitrogen by volume)
  • In the case of incomplete combustion, additional products such as carbon monoxide (CO) and unburned hydrocarbons may be present
    • The molar quantities of these products can be determined by measuring their concentrations in the exhaust gases and using the balanced equation for incomplete combustion

Mole and mass fractions of combustion products

  • The mole fractions of the combustion products can be calculated by dividing the molar quantity of each product by the total number of moles of the products
    • Example: For the combustion reaction CH4+2O2+7.52N2CO2+2H2O+7.52N2CH4 + 2O2 + 7.52N2 → CO2 + 2H2O + 7.52N2, the mole fraction of CO2 is 1/(1+2+7.52)=0.0951 / (1 + 2 + 7.52) = 0.095
  • The mass fractions can be determined by multiplying the mole fractions by the respective molecular weights and dividing by the total mass of the products
    • Example: For the same reaction, the mass fraction of CO2 is (0.095×44)/(44+36+210.56)=0.144(0.095 × 44) / (44 + 36 + 210.56) = 0.144

Adiabatic flame temperature

  • The adiabatic flame temperature is the maximum temperature that can be achieved in a combustion process, assuming no heat loss to the surroundings
    • It can be calculated using the enthalpy balance equation, considering the enthalpies of formation and specific heats of the reactants and products
    • Example: For the combustion of methane at standard conditions, the adiabatic flame temperature is approximately 2200 K

Combustion efficiency analysis

Factors affecting combustion efficiency

  • Combustion efficiency is a measure of how effectively the chemical energy in the fuel is converted to heat energy
  • Excess air is necessary to ensure complete combustion and to control the combustion temperature
    • However, too much excess air can reduce the combustion efficiency by cooling the combustion products and increasing the heat losses in the exhaust gases
  • Incomplete combustion occurs when there is insufficient oxygen or poor mixing of the reactants, resulting in the formation of carbon monoxide and unburned hydrocarbons
    • These products represent a loss of potential heat energy and can also be harmful pollutants

Calculating combustion efficiency

  • The combustion efficiency can be calculated based on the actual and theoretical air-fuel ratios: ηcomb=(TheoreticalAFR/ActualAFR)×100%ηcomb = (Theoretical AFR / Actual AFR) × 100\%
    • A higher efficiency indicates better utilization of the fuel's energy content
    • Example: If the theoretical AFR is 14.7 and the actual AFR is 16, the combustion efficiency is (14.7/16)×100%=91.9%(14.7 / 16) × 100\% = 91.9\%

Exhaust gas analysis for incomplete combustion

  • The exhaust gas analysis can be used to determine the extent of incomplete combustion
    • The presence of carbon monoxide and unburned hydrocarbons in the exhaust indicates incomplete combustion and reduced efficiency
  • The carbon monoxide concentration is typically expressed as parts per million (ppm) or as a percentage of the total exhaust volume
    • Higher CO levels indicate more incomplete combustion and lower efficiency
    • Example: An exhaust gas with 1000 ppm of CO indicates more incomplete combustion than one with 100 ppm
  • The unburned hydrocarbon concentration is usually measured in ppm of carbon (ppmC) and represents the amount of fuel that did not react completely
    • Higher hydrocarbon levels also indicate lower combustion efficiency
    • Example: An exhaust gas with 500 ppmC of unburned hydrocarbons indicates more incomplete combustion than one with 50 ppmC

Optimizing combustion efficiency

  • Optimizing the air-fuel ratio, improving mixing, and maintaining proper combustion temperatures can help minimize incomplete combustion and increase the overall efficiency of the combustion process
    • Example: Using a fuel injector that atomizes the fuel into fine droplets can improve mixing with the air and promote more complete combustion
    • Example: Installing a turbulator in the combustion chamber can create turbulence and enhance the mixing of the reactants, leading to higher efficiency
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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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