Determine the maximum useful shaft work that can be obtained from natural gas that we will model as methane, which is available at 25ºC and 1 bar. The methane is to be burned in air that contains nitrogen and oxygen in the mole ratio of 3.76 to 1.
[a] Use the approximate availability equation, and
[b] Use the exact availability equation.
The stoichiometry of the reaction is
CH _{4}+2 O _{2}= CO _{2}+2 H _{2} O
Since methane is available at ambient conditions Eq. (14.5-7) above neglecting solution nonidealities and the RTlnx term reduces to
W_{S, \max }=\sum_{\text {species i }} N_{2, i } 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 } \underline{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) \\=& \Delta \underline{G}_{ f , CO _{2}}\left(25^{\circ} C , 1 bar \right)+2 \Delta \underline{G}_{ f , H _{2} O }\left(25^{\circ} C , 1 bar \right)-\Delta \underline{G}_{ f , CH _{4}}\left(25^{\circ} C , 1 bar \right) \\&-2 \Delta \underline{G}_{ f , O _{2}}\left(25^{\circ} C , 1 bar \right) \\=&-394.4-2 \times 228.6-(-50.5+0)=-801.1 kJ / mol \text { methane }\end{aligned}
Thus, the maximum amount of shaft work that can be produced per mol of methane is 801.1 kJ.
[b] Now the stoichiometry of the reaction is
CH _{4}+2 O _{2}+2 \times 3.76 N _{2}= CO _{2}+2 H _{2} O +2 \times 3.76 N _{2}
The stoichiometric table for the inlet conditions is
| Species | Initial Number of Moles | Mole fraction | N_{ i } \ln y_{ i } |
| CH _{4} | 1.0 | 0.0947 | −2.3570 |
| O _{2} | 2.0 | 0.1894 | −3.3278 |
| N _{2} | 7.56 | 0.7159 | −2.5267 |
| H _{2} | 0 | 0 | 0 |
| CO _{2} | 0 | 0 | 0 |
| Total | 10.56 | −8.2115 |
The stoichiometric table for the outlet conditions is
| Species | Final Number of Moles | Mole fraction | N_{ i } \ln y_{ i } |
| CH _{4} | 0 | 0 | 0 |
| O _{2} | 0 | 0 | 0 |
| N _{2} | 7.56 | 0.7159 | −2.5267 |
| H _{2} | 2 | 0.1894 | −3.3278 |
| CO _{2} | 1.0 | 0.0947 | −2.3570 |
| Total | 10.56 | −8.2115 |
Comments
Note that for combustion reactions such as the one above, the reaction generally goes to completion, that is all the fuel burns, so we do not have to solve the chemical equilibrium problem. Also, in computation the Gibbs energy of formation at 25oC has been used for the molar Gibbs energy. At other than 25oC, these values would have to be corrected for temperature.
Note also by the fortuitous of this reaction and that a stoichiometric amount of air has been used, the sum of the N_{ i } \ln y_{ i } terms cancel out, and the approximate and exact equations give the same result, This would not be the case if the stoichiometry were different or excess air was used to insure complete combustion (Problems 14.23 to 27) was used to insure complete combustion.