What is the maximum temperature that can be reached by the combustion of methane with 20% excess air? Both the methane and the air enter the burner at 25°C.

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Accepted Answer

The reaction is [latex]CH_{4} + 2O_{2} → CO_{2} + 2H_{2} O(g)[/latex] for which,

[latex]ΔH^{°}_{298}= −393,509 + (2)(−241,818) − (−74,520) = −802,625[/latex] J

Because the maximum attainable temperature (called the theoretical flame temperature) is sought, assume that the combustion reaction goes to completion adiabatically (Q = 0). If the kinetic- and potential-energy changes are negligible and if [latex]W_{s} = 0[/latex], the overall energy balance for the process reduces to ΔH = 0. For purposes of calculating the final temperature, any convenient path between the initial and final states may be used. The path chosen is indicated in the diagram. J

When one mole of methane burned is the basis for all calculations, the following quantities of oxygen and nitrogen are supplied by the entering air:

Moles [latex]O_{2}[/latex] required = 2.0
M oles excess [latex]O_{2}[/latex] = (0.2)( 2.0) = 0.4
Moles [latex]N_{2}[/latex] entering = (2.4)(79/21) = 9.03

The mole numbers ni of the gases in the product stream leaving the burner are [latex]1 mol CO_{2}, 2 mol H_{2}O(g), 0.4 mol O_{2}, and 9.03 mol N_{2}[/latex] . Because the enthalpy change must be independent of path,

[latex]\Delta H^{º}_{298} + \Delta H^{º}_{P} = \Delta H =0[/latex] (A)

where all enthalpies are on the basis of 1 mol [latex] CH_{4}[/latex] burned. The enthalpy change of the products as they are heated from 298.15 K to T is:

[latex] \Delta H^{º}_{P} = \left\langle C^{º}_{P} ight angle_{H} \left(T − 298.15) ight)[/latex] (B)

where we define [latex]\left\langle C^{º}_{P} ight angle_{H}[/latex] as the mean heat capacity for the total product stream:

[latex]\left\langle C^{º}_{P} ight angle_{H} = \sum\limits_{i}{ni\left\langle C^{º}_{P_{i}} ight angle_{H} }[/latex]

The simplest procedure here is to sum the mean-heat-capacity equations for the products, each multiplied by its appropriate mole number. Because C = 0 for each product gas (Table C.1), Eq. (4.9) yields:

[latex]\left\langle C^{º}_{P} ight angle_{H} = \sum\limits_{i}{ni\left\langle C^{º}_{P_{i}} ight angle_{H} } = R \left[\sum\limits_{i}{n_{i} A_{i} + \frac{\sum{}_{i}n_{i} B_{i} }{2} }(T - T_{0} ) + \frac{\sum {}_{i} n_{i} D_{i}}{TT_{0} } ight][/latex]

Data from Table C.1 are combined as follows:

[latex]A =\sum\limits_{i}{ n_{i}A_{i} }  = (1)(5.457) + (2)(3.470) + (0.4)(3.639) + (9.03)(3.280) = 43.471[/latex]

Similarly,[/latex]B =\sum\limits_{i}{ n_{i}B_{i} } = 9.502  imes  10^{-3} and D = \sum\limits_{i}{ n_{i}D_{i} }=− 0.645 imes 10^{-5}[/latex]

For the product stream [latex]\left\langle C^{o}_{P} ight angle _{H}/R[/latex] is therefore represented by:

[latex]MCPH(298.15, T; 43.471, 9.502 imes 10^{-3} , 0.0, − 0.645  imes  10^{5} )[/latex]

Equations (A) and (B) are combined and solved for T:

[latex]T = 298.15 −  \frac{\Delta H^{o}_{298} }{\left\langle C^{o}_{P} ight angle _{H}}[/latex]

Because the mean heat capacities depend on T, one first evaluates [latex]\left\langle C^{o}_{P} ight angle _{H}[/latex] for an assumed value of T > 298.15, then substitutes the result in the preceding equation. This yields a new value of T for which [latex]\left\langle C^{o}_{P} ight angle _{H}[/latex] is reevaluated. The procedure continues to convergence on the final value,

T = 2066 K or 1793° C

Again, solution can be easily automated with the Goal Seek or Solver function in a spreadsheet or similar solve routines in other software packages.

Question 4.7: What is the maximum temperature that can be reached by the c...

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