Methane gas is burned with 30 percent excess air. This fuel...

Methane gas is burned with 30 percent excess air. This fuel enters a steady flow combustor at $101 \mathrm{kPa}$ and $25^{\circ} \mathrm{C}$, and is mixed with the air. The products of combustion leave this reactor at $101 \mathrm{kPa}$ and $1600 \mathrm{K}$. Determine the equilibrium composition of the products of combustion, and the amount of heat released by this combustion, in $\mathrm{kJ} / \mathrm{kmol}$ methane.

Key Concepts

Air Composition

Air is primarily composed of oxygen and nitrogen, with oxygen participating in the combustion reaction while nitrogen remains largely inert. Understanding the standard composition of air is essential for accurately modeling combustion reactions as it influences the reactant mix and the resulting product distribution, especially when considering excess air conditions.

Thermodynamic Analysis and Energy Balance

This concept involves calculating the energy released during combustion by considering the enthalpy changes associated with the reactants and products. It utilizes principles such as conservation of energy and the heats of formation of species. In combustion analysis, an energy balance allows the determination of the heat release (often expressed per unit fuel, e.g., kJ/kmol methane) and provides insights into the efficiency and thermal performance of the combustion process.

Chemical Equilibrium

In combustion processes, chemical equilibrium refers to the state where the forward reaction (fuel oxidation) and the reverse reaction occur at equal rates. Determining the equilibrium composition of combustion products at specified temperature and pressure conditions involves applying thermodynamic principles to identify the most stable mixture of species, thereby predicting the final concentrations of products such as CO2, H2O, O2, and N2.

Excess Air

Excess air is the additional air provided beyond the theoretical requirement for complete combustion. This concept is crucial because it ensures that the fuel burns completely under practical conditions. Accounting for excess air adjusts the reactant ratios, influencing the composition of combustion products (notably leaving excess oxygen) and affecting the thermal and emission characteristics of the process.

Combustion Stoichiometry

This concept involves balancing the chemical equation for the combustion reaction so that the reactants and products are correctly quantified. In combustion calculations, stoichiometry helps determine the minimum amount of oxygen needed for complete combustion of the fuel, establishes the molar ratios of reactants and products, and forms the basis for further equilibrium and energy calculations.

Transcript

00:01 All right guys, first the reaction of the burning of the methane with excess air is considered.

00:09 That is, ch4 plus 1 .3a .o2 plus 3 .76n2, bco2 plus c .h2 plus c .h2 plus 0, plus 0 .3ao2 plus 1 .3a into 3 .76 m2.

00:48 An analysis of the reaction gives us the following values for the coefficients, that is a is equal to 2, b is equal to 1, and c is equal to 2.

01:03 Now the nitrogen monoxide is considered and the reaction then looks like this.

01:10 With coefficient b and c express through a so c h4 plus 2 .6 or 2 plus 3 .76 n2 co2 plus 2h2 plus a .o plus a .o plus c and 2 plus cn2 where b is equal to 0 .6 minus 0 .5 a and c and c is equal to 9 .7 76 minus 0 .5a.

01:50 So the stoichiomatic reaction for this example along with this coefficient is n2 plus o2, 2no, whereas vn2 is equal to 1, vo2 is equal to 1, vno is equal to 2.

02:23 The reaction coefficients are now determined from the equilibrium coefficient relation.

02:28 The value of the equilibrium coefficient for this reaction is taken from table a28 for the given temperature, that is e, raised to power 2, natural log kp, is equal to nno raised to power vno, divided by no2 raised to power vo2 into nn2, raised to power vo2, into nn2, raised to power vo2, n2, into p over n raised to power v n o minus v o two minus v n two so guys e raised to power minus 2 .2 into 5 .294 is equal to a square divided by 0 .6 minus 0 .5 a into 9 .776 minus 0 .5a.

03:38 So if we have 0 .99 -99934a square plus 1 .31 into 10 raise to power minus 4a minus 1 .48 into 10 raised to power minus 4 is equal to 0.

03:59 So guys, we've got a is equal to a...

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