Second-Law Analysis of Adiabatic Combustion Methane (CH4) gas enters a steady-flow adiabatic combustion chamber at 25°C and 1 atm. It is burned with 50 percent excess air that also enters at 25°C and 1 atm, as shown in Fig. 15-33. Assuming complete combustion, determine (a) the temperature

Question

Second-Law Analysis of Adiabatic Combustion
Methane ([latex]\mathrm{CH}_{4}[/latex]) gas enters a steady-flow adiabatic combustion chamber at 25°C and 1 atm. It is burned with 50 percent excess air that also enters at 25°C and 1 atm, as shown in Fig. 15-33. Assuming complete combustion, determine (a) the temperature of the products, (b) the entropy generation, and (c) the reversible work and exergy destruction. Assume that [latex]T_{0}[/latex] = 298 K and the products leave the combustion chamber at 1 atm pressure.

Accepted Answer

Methane is burned with excess air in a steady-flow combustion chamber. The product temperature, entropy generated, reversible work, and exergy destroyed are to be determined.

Assumptions 1 Steady-flow conditions exist during combustion. 2 Air and the combustion gases are ideal gases. 3 Changes in kinetic and potential energies are negligible. 4 The combustion chamber is adiabatic and thus there is no heat transfer. 5 Combustion is complete.

Analysis (a) The balanced equation for the complete combustion process with 50 percent excess air is

[latex]\mathrm{CH}_{4}(g)+3\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+11.28 \mathrm{~N}_{2}[/latex]

Under steady-flow conditions, the adiabatic flame temperature is determined from [latex]H_{ ext {prod }}=H_{ ext {react, }}[/latex] which reduces to

[latex]\sum N_{p}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ} ight)_{p}=\sum N_{r} \bar{h}_{f, r}^{\circ}=\left(N \bar{h}_{f}^{\circ} ight)_{\mathrm{CH}_{4}}[/latex]

since all the reactants are at the standard reference state and [latex]\bar{h}_{f}^{\circ}=0[/latex] for [latex]\mathrm{O}_{2}[/latex] and [latex]\mathrm{N}_{2}[/latex]. Assuming ideal-gas behavior for air and for the products, the [latex]\bar{h}_{f}^{\circ}[/latex] and h values of various components at 298 K can be listed as

Substance	[latex]\bar{h}_{f}^{\circ}[/latex] kJ/kmol	[latex]\bar{h}_{298 \mathrm{K}}[/latex] kJ/kmol
[latex]\mathrm{CH}_{4}(g)[/latex]	-74,850	—
[latex]\mathrm{O}_{2}[/latex]	0	8682
[latex]\mathrm{N}_{2}[/latex]	0	8669
[latex]\mathrm{H}_{2} \mathrm{O}(g)[/latex]	-241,820	9904.0
[latex]\mathrm{C} \mathrm{O}_{2}(\ell)[/latex]	-393,520	9364

Substituting, we have

[latex]\left(1 \mathrm{kmol} \quad\mathrm{CO}_{2} ight)\left[\left(-393,520+\bar{h}_{\mathrm{CO}_{2}}-9364 ight) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{CO}_{2} ight][/latex]

[latex]+\left(2 \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O} ight)\left[\left(-241,820+\bar{h}_{\mathrm{H}_{2} \mathrm{O}}-9904 ight) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O} ight][/latex]

[latex]+\left(11.28 \mathrm{kmol} \quad\mathrm{N}_{2} ight)\left[\left(0+\bar{h}_{\mathrm{N}_{2}}-8669 ight) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{N}_{2} ight][/latex]

[latex]+\left(1 \mathrm{kmol} \quad\mathrm{O}_{2} ight)\left[\left(0+\bar{h}_{\mathrm{O}_{2}}-8682 ight) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{O}_{2} ight][/latex]

[latex]=\left(1 \mathrm{kmol} \quad\mathrm{CH}_{4} ight)\left(-74,850 \mathrm{~kJ} / \mathrm{kmol} \quad\mathrm{CH}_{4} ight)[/latex]

which yields

[latex]\bar{h}_{\mathrm{CO}_{2}}+2 \bar{h}_{\mathrm{H}_{2} \mathrm{O}}+\bar{h}_{\mathrm{O}_{2}}+11.28 \bar{h}_{\mathrm{N}_{2}}=937,950 \mathrm{~kJ}[/latex]

By trial and error, the temperature of the products is found to be

[latex]T_{ ext {prod }}=1789 \mathrm{~K}[/latex]

(b) Noting that combustion is adiabatic, the entropy generation during this process is determined from Eq. 15-20:

[latex]S_{ ext {gen }}=S_{ ext {prod }}-S_{ ext {react }}=\sum N_{p} \bar{s}_{p,}-\sum N_{r} \bar{s}_{r}[/latex]

The [latex]\mathrm{CH}_{4}[/latex] is at [latex]25^{\circ} \mathrm{C}[/latex] and 1 atm, and thus its absolute entropy is [latex]\bar{s}_{\mathrm{CH}_{4}}=186.16 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K}[/latex] (Table A-26). The entropy values listed in the ideal-gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components, which is equal to [latex]P_{i}=y_{i} P_{ ext {total }}[/latex], where [latex]y_{i}[/latex] is the mole fraction of component i. From Eq. 15-22:

[latex]S_{i}=N_{i} \bar{s}_{i}\left(T, P_{i} ight)=N_{i}\left[\bar{s}_{i}^{o}\left(T, P_{0} ight)-R_{u} \ln y_{i} P_{m} ight][/latex]

The entropy calculations can be represented in tabular form as follows:

	[latex]\mathrm{N}_{i}[/latex]	[latex]\mathrm{y}_{i}[/latex]	[latex]\bar{S}_i^{\circ}(T, 1 ext { atm) }[/latex]	[latex]-R_{u} \ln y_{i} P_{m}[/latex]	[latex]N_{i} \bar{S}_{i}^{\circ}[/latex]
[latex]\mathrm{C} \mathrm{H}_{4}[/latex]	1	1.00	186.16	—	186.16
[latex]\mathrm{O}_{2}[/latex]	3	0.21	205.04	12.98	654.06
[latex]\mathrm{N}_{2}[/latex]	11.28	0.79	191.61	1.96	2183.47
[latex]S_{ ext {react }}=3023.69[/latex]
[latex]\mathrm{CO}_{2}[/latex]	1	0.0654	302.517	22.674	325.19
[latex]\mathrm{H}_2 \mathrm{O}[/latex]	2	0.1309	258.957	16.905	551.72
[latex]\mathrm{O}_{2}[/latex]	1	0.0654	264.471	22.674	287.15
[latex]\mathrm{N}_{2}[/latex]	11.28	0.7382	247.977	2.524	2825.65
[latex]S_{ ext {prod }}=3989.71[/latex]

Thus,

[latex]S_{ ext {gen }} =S_{ ext {prod }}-S_{ ext {react }}=(3989.71-3023.69) \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K} \quad\mathrm{CH}_{4} [/latex]

[latex]=966.0 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K}[/latex]

(c) The exergy destruction or irreversibility associated with this process is determined from Eq. 15-23,

[latex]X_{ ext {destroyed }} =T_{0} S_{ ext {gen }}=(298 \mathrm{~K})(966.0 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K}) [/latex]

[latex]=288 \mathrm{M} \mathrm{J} / \mathrm{kmol} \quad\mathrm{CH}_{4}[/latex]

That is, 288 MJ of work potential is wasted during this combustion process for each kmol of methane burned. This example shows that even complete combustion processes are highly irreversible.

This process involves no actual work. Therefore, the reversible work and exergy destroyed are identical:

[latex]W_{ ext {rev }}=288 \mathrm{MJ} / \mathrm{kmol} \mathrm{CH}_{4}[/latex]

That is, 288 MJ of work could be done during this process but is not. Instead, the entire work potential is wasted.

Question 15.10: Second-Law Analysis of Adiabatic Combustion Methane (CH4) ga...

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