Adiabatic Flame Temperature CH4 is pre-mixed with air at 298 K at an equivalence ratio of 1.0 and the combustion is complete. Determine the constant pressure AFT.

Question

Adiabatic Flame Temperature
[latex]CH_{4}[/latex] is pre-mixed with air at 298 K at an equivalence ratio of 1.0 and the combustion is complete. Determine the constant pressure AFT.

Accepted Answer

First of all, set up stoichiometric combustion reaction equation using the methods introduced in Sect. 3.4,
[latex]CH_{4} +2 (O_{2} + 3.76N_{2} ) → CO_{2} + 2H_{2}O + 7.52N_{2}[/latex]
For an adiabatic constant pressure system, with [latex]T_{R} = T_{0}[/latex] = 298 K, left hand side (LHS) of Eq. (3.65) is
[latex]\sum\limits_{R}{n_{i}h^{0}_{f,i}}=\sum\limits_{P}{n_{i}}\left[\left(a_{i}(T_{a}-298K)+\frac{b_{i}}{2}(T^{2}_{a}-298K)^{2} \right)+h^{0}_{f,i} \right][/latex] (3.65)

[latex]LHS = h^{0}_{f,CH_{4}}+2h^{0}_{f,O_{2}}+7.52h^{0}_{f,N_{2}}[/latex] (1)

right-hand side (RHS) of the equation is simplified as

[latex]RHS = \left[\left(a_{CO_{2}}(T_{a}-298K)+\frac{b_{CO_{2}}}{2}({T^{2}_{a}}-(298K)^{2} \right)+h^{0}_{f,CO_{2}} \right][/latex]
+ [latex]2\left[\left(a_{H_{2}O}(T_{a}-298K)+\frac{b_{H_{2}O}}{2}({T^{2}_{a}}-(298K)^{2} \right)+h^{0}_{f,H_{2}O} \right][/latex] (2)
+ [latex]7.52\left[\left(a_{N_{2}}(T_{a}-298K)+\frac{b_{N_{2}}}{2}({T^{2}_{a}}-(298K)^{2} \right)+h^{0}_{f,N_{2}} \right][/latex]

From Table A.4, we can get

Species	[latex]h^{o}_{f,i} (J/mol)[/latex]	[latex]C_{p}(T)[/latex] [J/mol K]
Species	[latex]h^{o}_{f,i} (J/mol)[/latex]	[latex]a_{i}[/latex]	[latex]b_{i}[/latex]
[latex]CO_{2}[/latex]	−394,088	44.3191	0.0073
[latex]H_{2}O[/latex]	−242,174	32.4766	0.00862
[latex]N_{2}[/latex]	0	29.2313	0.00307
[latex]O_{2}[/latex]	0	30.5041	0.00349
[latex]CH_{4}[/latex]	−74,980	44.2539	0.02273

With these values
LHS = -74,980(J/mol)

[latex]RHS = \left[\left(44.3191(T_{a}-298K)+\frac{0.0073}{2}({T^{2}_{a}}-(298K)^{2} \right) - 394.088 \right][/latex]
+ [latex]2\left[\left(32.4766(T_{a}-298K)+\frac{0.00862}{2}({T^{2}_{a}}-(298K)^{2} \right) - 242.174 \right][/latex]
+ [latex]2(3.76)\left[\left(29.2313(T_{a}-298K)+\frac{0.00307}{2}({T^{2}_{a}}-(298K)^{2} \right) + 0 \right][/latex]

Equating LHS and RHS we have
[latex]0.02387 T^{2}_{a} + 330.26 T_{a} - 903994 = 0[/latex]
Solving this equation we can get, [latex]T_{a}[/latex] = 2341 K; Another root [latex]T_{a}[/latex] = −16.177 K is ignored because it is physically wrong. So the estimated AFT is
[latex]T_{a}[/latex] = 2,341 K

Note that, with this high temperature, it is no longer reasonable to assume that nitrogen does not react with oxygen. We can proceed with the calculation for better accuracy by considering the knowledge to be introduced in Sect. 7.7.1.

Question 3.10: Adiabatic Flame Temperature CH4 is pre-mixed with air at 298...

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