Question 4.7: What is the maximum temperature that can be reached by the c...
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.
The reaction is CH_{4} + 2O_{2} → CO_{2} + 2H_{2} O(g) for which,
ΔH^{°}_{298}= −393,509 + (2)(−241,818) − (−74,520) = −802,625 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 W_{s} = 0, 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 O_{2} required = 2.0
M oles excess O_{2} = (0.2)( 2.0) = 0.4
Moles N_{2} entering = (2.4)(79/21) = 9.03
The mole numbers ni of the gases in the product stream leaving the burner are 1 mol CO_{2}, 2 mol H_{2}O(g), 0.4 mol O_{2}, and 9.03 mol N_{2} . Because the enthalpy change must be independent of path,
\Delta H^{º}_{298} + \Delta H^{º}_{P} = \Delta H =0 (A)
where all enthalpies are on the basis of 1 mol CH_{4} burned. The enthalpy change of the products as they are heated from 298.15 K to T is:
\Delta H^{º}_{P} = \left\langle C^{º}_{P}\right\rangle_{H} \left(T − 298.15)\right) (B)
where we define \left\langle C^{º}_{P}\right\rangle_{H} as the mean heat capacity for the total product stream:
\left\langle C^{º}_{P}\right\rangle_{H} = \sum\limits_{i}{ni\left\langle C^{º}_{P_{i}}\right\rangle_{H} }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:
\left\langle C^{º}_{P}\right\rangle_{H} = \sum\limits_{i}{ni\left\langle C^{º}_{P_{i}}\right\rangle_{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} } \right]
Data from Table C.1 are combined as follows:
A =\sum\limits_{i}{ n_{i}A_{i} } = (1)(5.457) + (2)(3.470) + (0.4)(3.639) + (9.03)(3.280) = 43.471Similarly,[/latex]B =\sum\limits_{i}{ n_{i}B_{i} } = 9.502 \times 10^{-3} and D = \sum\limits_{i}{ n_{i}D_{i} }=− 0.645 \times 10^{-5}[/latex]
For the product stream \left\langle C^{o}_{P} \right\rangle _{H}/R is therefore represented by:
MCPH(298.15, T; 43.471, 9.502\times 10^{-3} , 0.0, − 0.645 \times 10^{5} )
Equations (A) and (B) are combined and solved for T:
T = 298.15 − \frac{\Delta H^{o}_{298} }{\left\langle C^{o}_{P} \right\rangle _{H}}Because the mean heat capacities depend on T, one first evaluates \left\langle C^{o}_{P} \right\rangle _{H} 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 \left\langle C^{o}_{P} \right\rangle _{H} 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.