Question 5.5: In a steady-state flow process carried out at atmospheric pr...

We start by applying an energy balance to determine the rate of heat transfer, which we must know to compute the rate of entropy generation. Writing the energy balance, Eq. (2.29),

\Delta\left[\left(H+\frac{1}{2} u^2+z g\right) \dot{m}\right]_{ fs }=\dot{Q}+\dot{W}_s      (2.29)

with dot{m}  replaced  by  dot{n}  ,   and  then  replacing  \dot{n}  \text { with }  \dot{n}_A+\dot{n}_B

\dot{Q}=\dot{n} H^{i g}  –  \dot{n}_A H_A^{i g}  –  \dot{n}_B H_B^{i g}=\dot{n}_A\left(H^{i g}-H_A^{i g}\right)+\dot{n}_B\left(H^{i g}  –  H_B^{i g}\right)

= (7 / 2)(8.314)[(1)(400  −  600) + (2)(400  −  450)] = − 8729.7 J⋅ s ^{−1}

The steady-state entropy balance, Eq. (5.17), with m∙ again replaced by \dot{n} , can similarly be written as

\Delta(\dot{S} m)_{ fs }-\sum_i \frac{\dot{Q}_j}{T_{\sigma, j}}=\dot{S}_G \geq 0       (5.17)

\dot{S}_G=\dot{n} S^{i g}-\dot{n}_A S_A^{i g}-\dot{n}_B S_B^{i g}-\frac{\dot{Q}}{T_\sigma}=\dot{n}_A\left(S^{i g}-S_A^{i g}\right)+\dot{n}_B\left(s^{i g}-S_B^{i g}\right)-\frac{\dot{Q}}{T_\sigma}

=\dot{n}_A C_P^{i g} \ln \frac{T}{T_A}+\dot{n}_B C_P^{i g} \ln \frac{T}{T_B} \frac{\dot{Q}}{T_\sigma}=C_P^{i g}\left(\dot{n}_A \ln \frac{T}{T_A}+\dot{n}_B \ln \frac{T}{T_B}\right)-\frac{\dot{Q}}{T_\sigma}

=(7 / 2)(8.314)\left[(1) \ln \frac{400}{600}+(2) \ln \frac{400}{450}\right] \frac{8729.7}{300}=10.446 J \cdot K ^{-1} \cdot s ^{-1}

The rate of entropy generation is positive, as it must be for any real process.