Second-Law Analysis of Isothermal Combustion Methane (CH4) gas enters a steady-flow combustion chamber at 25°C and 1 atm and is burned with 50 percent excess air, which also enters at 25°C and 1 atm, as shown in Fig. 15-34. After combustion, the products are allowed to cool to 25°C. Assuming

Question

Second-Law Analysis of Isothermal Combustion

Methane ([latex]\mathrm{CH}_{4}[/latex]) gas enters a steady-flow combustion chamber at 25°C and 1 atm and is burned with 50 percent excess air, which also enters at 25°C and 1 atm, as shown in Fig. 15-34. After combustion, the products are allowed to cool to 25°C. Assuming complete combustion, determine (a) the heat transfer per kmol of [latex]\mathrm{CH}_{4}[/latex], (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

This is the same combustion process we discussed in Example 15-10, except that the combustion products are brought to the state of the surroundings by transferring heat from them. Thus the combustion equation remains the same:

[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]

At 25°C, part of the water will condense. The amount of water vapor that remains in the products is determined from (see Example 15-3)

[latex]\frac{N_{v}}{N_{ ext {gas }}}=\frac{P_{v}}{P_{ ext {total }}}=\frac{3.1698 \mathrm{kPa}}{101.325 \mathrm{kPa}}=0.03128[/latex]

and

[latex]N_{v}=\left(\frac{P_{v}}{P_{ ext {total }}} ight) N_{\mathrm{gas}}=(0.03128)\left(13.28+N_{v} ight) ightarrow N_{v}=0.43 \mathrm{kmol}[/latex]

Therefore, 1.57 kmol of the [latex]\mathrm{H}_{2} \mathrm{O}[/latex] formed is in the liquid form, which is removed at 25°C and 1 atm. When one is evaluating the partial pressures of the components in the product gases, the only water molecules that need to be considered are those that are in the vapor phase. As before, all the gaseous reactants and products are treated as ideal gases.

(a) Heat transfer during this steady-flow combustion process is determined from the steady-flow energy balance [latex]E_{ ext {out }}=E_{ ext {in }}[/latex] on the combustion chamber,

[latex]Q_{ ext {out }}+\sum N_{p} \bar{h}_{j, p}^{o}=\sum N_{r} \bar{h}_{f, r}^{\circ}[/latex]

since all the reactants and products are at the standard reference of 25°C and the enthalpy of ideal gases depends on temperature only. Solving for [latex]Q_{ ext {out }}[/latex] and substituting the [latex]\bar{h}_{f}^{\circ}[/latex] values, we have

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

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

[latex]-\left[0.43 \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O}(g) ight]\left[-241,820 \mathrm{~kJ} / \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O}(g) ight] [/latex]

[latex]-\left[1.57 \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O}(\ell) ight]\left[-285.830 \mathrm{~kJ} / \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O}(\ell) ight] [/latex]

[latex]= 871,400 \mathrm{~kJ} / \mathrm{kmol} \quad\mathrm{CH}_{4}[/latex]

(b) The entropy of the reactants was evaluated in Example 15-10 and was determined to be [latex]S_{ ext {react }}[/latex] = 3023.69 kJ/kmol.K [latex]\mathrm{CH}_{4}[/latex]. By following a similar approach, the entropy of the products is determined to be

	[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}[/latex]
[latex]\mathrm{H}_{2} \mathrm{O}(\ell)[/latex]	1.57	1.0000	69.92	--	109.77
[latex]\mathrm{H}_{2} \mathrm{O}[/latex]	0.43	0.0314	188.83	28.77	93.57
[latex]\mathrm{CO}_{2}[/latex]	1	0.0729	213.80	21.77	235.57
[latex]\mathrm{O}_{2}[/latex]	1	0.0729	205.04	21.77	226.81
[latex]\mathrm{N}_{2}[/latex]	11.28	0.8228	191.61	1.62	2179.63
[latex]S_{ ext {prod }}=2845.35[/latex]

Then the total entropy generation during this process is determined from an entropy balance applied on an extended system that includes the immediate surroundings of the combustion chamber

[latex]S_{ ext {gen }} =S_{ ext {prod }}-S_{ ext {react }}+\frac{Q_{ ext {out }}}{T_{ ext {surr }}} [/latex]

[latex]=(2845.35-3023.69) \mathrm{kJ} / \mathrm{kmol}+\frac{871,400 \mathrm{~kJ} / \mathrm{kmol}}{298 \mathrm{~K}} [/latex]

[latex]=2746 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K} \quad\mathrm{CH}_{4}[/latex]

(c) The exergy destruction and reversible work associated with this process are determined from

[latex]X_{ ext {destroyed }}=T_{0} S_{ ext {gen }}=(298 \mathrm{~K})(2746 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K}) [/latex]

[latex]=818 \mathrm{MJ} / \mathrm{kmol} \mathrm{CH}_{4}[/latex]

and

[latex]W_{ ext {rev }}=X_{ ext {destroyed }}=818 \mathrm{MJ} / \mathrm{kmol} \mathrm{CH}_{4}[/latex]

since this process involves no actual work. Therefore, 818 MJ of work could be done during this process but is not. Instead, the entire work potential is wasted. The reversible work in this case represents the exergy of the reactants before the reaction starts since the products are in equilibrium with the surroundings, that is, they are at the dead state.

Discussion Note that, for simplicity, we calculated the entropy of the product gases before they actually entered the atmosphere and mixed with the atmospheric gases. A more complete analysis would consider the composition of the atmosphere and the mixing of the product gases with the gases in the atmosphere, forming a homogeneous mixture. There is additional entropy generation during this mixing process, and thus additional wasted work potential.

Question 15.11: Second-Law Analysis of Isothermal Combustion Methane (CH4) g...

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