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.
The stoichiometry of the reaction is
C _{8} H _{18}+12.5 \times xO _{2}+12.5 \times 3.76 N _{2}=8 CO _{2}+9 H _{2} O +12.5 \times 3.76 N _{2}
[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
W_{S, \max }=\sum_{\text {species i }} N_{2, i } \underline{G}_{ i }\left(T_{ amb }, P_{ amb }\right)-\sum_{\text {species i }} N_{1, i} B _{ i }\left(T_{1}, P_{1}\right) (14.5-7)
\begin{aligned}W_{S, \max }=& \sum_{\text {species i }} N_{2, i} \underline{G}_{ i }\left(25^{\circ} C , 1 bar \right)-\sum_{\text {species i }} N_{1, i } G_{ i }\left(25^{\circ} C , 1 bar \right) \\=& \sum_{\text {species i }} N_{2, i} \Delta \underline{G}_{ f , i }\left(25^{\circ} C , 1 bar \right)-\sum_{\text {species i }} N_{1, i } \Delta \underline{G}_{ f , i }\left(25^{\circ} C , 1 bar \right) \\=& 8 \Delta \underline{G}_{ f , CO _{2}}\left(25^{\circ} C , 1 bar \right)+9 \Delta \underline{G}_{ f , H _{2} O }\left(25^{\circ} C , 1 bar \right)-\Delta \underline{G}_{ f , C _{ H } H _{18}}\left(25^{\circ} C , 1 bar \right) \\&-2 \Delta \underline{G}_{ f , O _{2}}\left(25^{\circ} C , 1 bar \right) \\=&-8 \times 394.4-9 \times 228.6-(-16.3)=-4565.3 kJ / mol \text { octane }\end{aligned}
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 | N_{ i } \ln y_{ i } |
| C _{8} H _{18} | (1.0) | 0 | 0 |
| O _{2} | 12.5 | 0.2101 | −19.503 |
| N _{2} | 47 | 0.7899 | −11.0843 |
| H _{2} | 0 | 0 | 0 |
| CO _{2} | 0 | 0 | 0 |
| H _{2} O | 0 | 0 | 0 |
| Total | 59.5 | −30.587 |
The stoichiometric table for the outlet conditions is
| Species | Final Number of Moles | Mole fraction | N_{ i } \ln y_{ i } |
| C _{8} H _{18} | 0 | 0 | 0 |
| O _{2} | 0 | 0 | 0 |
| N _{2} | 47 | 0.6144 | −22.896 |
| H _{2} | 9 | 0.1177 | −3.3278 |
| CO _{2} | 1.0 | 0.0947 | −19.603 |
| H _{2} O | 9 | 0.1177 | −19.261 |
| Total | 76.5 | −83.608 |
Comments
So the maximum shaft work available is
\begin{aligned}W_{S, \max } &=-4565.26+R T_{ amb }(83.608-30.587) / 1000=\\&=-4565.26+8.314 \times 298.15 \times 53.021 / 1000=-4565.26+131.42 J / mol \\&=-4433.8 kJ / mol\end{aligned}
which is not insignificant, and the error in using the approximate equation is about 2.9% in this case.