Question 5.4: Methane gas at 550 K and 5 bar undergoes a reversible adiaba...
Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming ideal-gas-state methane at these conditions, find its final temperature.
For this process ΔS^{ig} = 0, and Eq. (5.14) becomes:
\frac{\Delta S^{i g}}{R} = \frac{\left\langle C_P^{i g}\right\rangle_S}{R} \ln \frac{T}{T_0}-\ln \frac{P}{P_0} (5.14)
\frac{\left\langle C_P^{i g}\right\rangle_S}{R} \ln \frac{T_2}{T_1}=\ln \frac{P_2}{P_1}=\ln \frac{1}{5}=-1.6094
Because \left\langle C_P^{i g}\right\rangle_S depends on T_{2} , we rearrange this equation for iterative solution:
\ln \frac{T_2}{T_1}=\frac{-1.6094}{\left\langle C_P^{i g}\right\rangle_S / R} \quad \text { or } \quad T_2=T_1 \exp \left(\frac{-1.6094}{\left\langle C_P^{i g}\right\rangle_S / R}\right)With constants from Table C.1 of App. C, \left\langle C_P^{i g}\right\rangle_S / R is evaluated by Eq. (5.13) written in its functional form:
\frac{\left\langle C_P^{i g}\right\rangle_S}{R}=A+\left[B+\left(C+\frac{D}{T_0^2 T^2}\right)\left(\frac{T+T_0}{2}\right)\right] \frac{T-T_0}{\ln \left(T / T_0\right)} (5.13)
\frac{\left\langle C_P^{i g}\right\rangle_S}{R}=\operatorname{MCPS}\left(550, T_2 ; 1.702,9.081 \times 10^{-3},-2.164 \times 10^{-6}, 0.0\right)For an initial value of T_2 < 550 , compute a value of \frac{\left\langle C_P^{i g}\right\rangle_S}{R} for substitution into the equation for T_2 . This new value of T_2 allows recalculation of \frac{\left\langle C_P^{i g}\right\rangle_S}{R} , and the process continues to convergence on a final value of T_2 = 411.34 K.
As with Ex. 4.3, a trial procedure is an alternative approach, with Microsoft Excel’s Goal Seek providing a prototypical automated version.