Question 5.7: What is the maximum work that can be obtained in a steady-st...
What is the maximum work that can be obtained in a steady-state flow process from 1 mol of nitrogen in its ideal-gas state at 800 K and 50 bar? Take the temperature and pressure of the surroundings as 300 K and 1.0133 bar
The maximum possible work is obtained from any completely reversible process that reduces the nitrogen to the temperature and pressure of the surroundings, i.e., to 300 K and 1.0133 bar. Required here is the calculation of W_{ideal} by Eq. (5.22),
W_{\text {ideal }}=\Delta H – T_\sigma \Delta S (5.22)
in which ΔS and ΔH are the molar entropy and enthalpy changes of the nitrogen as its state changes from 800 K and 50 bar to 300 K and 1.0133 bar. For the ideal-gas state, enthalpy is independent of pressure, and its change is given by:
\Delta H^{i g}=\int_{T_1}^{T_2} C_P^{i g} d T
The value of this integral is found from Eq. (4.8), and is represented by:
\int_{T_0}^T \frac{C_P}{R} d T=A\left(T – T_0\right)+\frac{B}{2}\left(T^2 – T_0^2\right)+\frac{C}{3}\left(T^3 – T_0^3\right)+D\left(\frac{T – T_0}{T T_0}\right) (4.8)
8.314 \times ICPH \left(800,300 ; 3.280,0.593 \times 10^{-3}, 0.0,0.040 \times 10^{-5}\right)=-15,060 J \cdot mol ^{-1}
where the parameters for nitrogen come from Table C.1 of App. C.
Similarly, the entropy change is found from Eq. (5.10), here written:
\frac{\Delta S^{i g}}{R}=\int_{T_0}^T \frac{C_P^{i g}}{R} \frac{d T}{T}-\ln \frac{P}{P_0} (5.10)
The value of the integral, found from Eq. (5.11), is represented by:
x_i \gamma_i^l=z_i \gamma_i^s \psi_i (all i ) (15.11)
8.314 \times \operatorname{ICPS}\left(800,300 ; 3.280,0.593 \times 10^{-3}, 0.0,0.040 \times 10^{-5}\right)=-29.373 J \cdot mol ^{-1} \cdot K ^{-1}
Whence, \Delta S^{i g}=-29.373 – 8.314 \ln \frac{1.0133}{50}=3.042 J \cdot mol ^{-1} \cdot K ^{-1}
With these values of \Delta H^{i g} \text { and } \Delta S^{i g} \text {, } , Eq. (5.22) becomes:
W_{\text {ideal }}=-15,060 – (300)(3.042)=-15,973 J \cdot mol ^{-1}