Determine the maximum useful shaft work that can be obtained from n-octane, a model for gasoline, which is available at 25ºC and 1 bar. The octane is to be burned in air that contains nitrogen and oxygen in the mole ratio of 3.76 to 1, and that n-octane liquid at ambient conditions

[a] Use the approximate availability equation, and

[b] Use the exact availability equation.

Question

Determine the maximum useful shaft work that can be obtained from n-octane, a model for gasoline, which is available at 25ºC and 1 bar. The octane is to be burned in air that contains nitrogen and oxygen in the mole ratio of 3.76 to 1, and that n-octane liquid at ambient conditions

[a] Use the approximate availability equation, and

[b] Use the exact availability equation.

[a] Use the approximate availability equation, and

[b] Use the exact availability equation.

Accepted Answer

The stoichiometry of the reaction is

[latex]C _{8} H _{18}+12.5 imes xO _{2}+12.5 imes 3.76 N _{2}=8 CO _{2}+9 H _{2} O +12.5 imes 3.76 N _{2}[/latex]

[a]

Since n-octane is available at ambient conditions Eq. (14.5-7) at which it is a liquid, neglecting solution nonidealities and the RTlny term reduces to

[latex]W_{S, \max }=\sum_{ ext {species i }} N_{2, i } \underline{G}_{ i }\left(T_{ amb }, P_{ amb } ight)-\sum_{ ext {species i }} N_{1, i} B _{ i }\left(T_{1}, P_{1} ight)[/latex] (14.5-7)

[latex]\begin{aligned}W_{S, \max }=& \sum_{ ext {species i }} N_{2, i} \underline{G}_{ i }\left(25^{\circ} C , 1 bar ight)-\sum_{ ext {species i }} N_{1, i } G_{ i }\left(25^{\circ} C , 1 bar ight) \=& \sum_{ ext {species i }} N_{2, i} \Delta \underline{G}_{ f , i }\left(25^{\circ} C , 1 bar ight)-\sum_{ ext {species i }} N_{1, i } \Delta \underline{G}_{ f , i }\left(25^{\circ} C , 1 bar ight) \=& 8 \Delta \underline{G}_{ f , CO _{2}}\left(25^{\circ} C , 1 bar ight)+9 \Delta \underline{G}_{ f , H _{2} O }\left(25^{\circ} C , 1 bar ight)-\Delta \underline{G}_{ f , C _{ H } H _{18}}\left(25^{\circ} C , 1 bar ight) \&-2 \Delta \underline{G}_{ f , O _{2}}\left(25^{\circ} C , 1 bar ight) \=&-8 imes 394.4-9 imes 228.6-(-16.3)=-4565.3 kJ / mol ext { octane }\end{aligned}[/latex]

Thus, the maximum amount of shaft work that can be produced per mol of methane is 4565.3 kJ/mol octane.

[b] The stoichiometric table for the inlet conditions is, noting that n-octane is a liquid at ambient conditions,

Species	Initial Number of Moles	Mole fraction	[latex]N_{ i } \ln y_{ i }[/latex]
[latex]C _{8} H _{18}[/latex]	(1.0)	0	0
[latex]O _{2}[/latex]	12.5	0.2101	−19.503
[latex]N _{2}[/latex]	47	0.7899	−11.0843
[latex]H _{2}[/latex]	0	0	0
[latex]CO _{2}[/latex]	0	0	0
[latex]H _{2} O[/latex]	0	0	0
Total	59.5		−30.587

The stoichiometric table for the outlet conditions is

Species	Final Number of Moles	Mole fraction	[latex]N_{ i } \ln y_{ i }[/latex]
[latex]C _{8} H _{18}[/latex]	0	0	0
[latex]O _{2}[/latex]	0	0	0
[latex]N _{2}[/latex]	47	0.6144	−22.896
[latex]H _{2}[/latex]	9	0.1177	−3.3278
[latex]CO _{2}[/latex]	1.0	0.0947	−19.603
[latex]H _{2} O[/latex]	9	0.1177	−19.261
Total	76.5		−83.608

Comments

So the maximum shaft work available is

[latex]\begin{aligned}W_{S, \max } &=-4565.26+R T_{ amb }(83.608-30.587) / 1000=\&=-4565.26+8.314 imes 298.15 imes 53.021 / 1000=-4565.26+131.42 J / mol \&=-4433.8 kJ / mol\end{aligned}[/latex]

which is not insignificant, and the error in using the approximate equation is about 2.9% in this case.

Question 14.5.2: Determine the maximum useful shaft work that can be obtained...

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