In a steady-state flow process carried out at atmospheric pressure, 1 mol.s^−1 of air at 600 K is continuously mixed with 2 mol.s^−1 of air at 450 K. The product stream is at 400 K and 1 atm. A schematic representation of the process is shown in Fig. 5.4. Determine the rate of heat transfer

Question

In a steady-state flow process carried out at atmospheric pressure,[latex] 1 mol⋅s^−1[/latex] of air at 600 K is continuously mixed with 2 mol⋅s−1 of air at 450 K. The product stream is at 400 K and 1 atm. A schematic representation of the process is shown in Fig. 5.4. Determine the ideal-gas state for air with [latex]C_{P}^{ig} = (7/ 2)R[/latex], that the surroundings are at 300 K, and that kinetic- and potential-energy changes of the streams are negligible.

Accepted Answer

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),

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

with [latex] dot{m} replaced by dot{n} , and then replacing \dot{n} ext { with } \dot{n}_A+\dot{n}_B [/latex]

[latex] \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} ight)+\dot{n}_B\left(H^{i g} - H_B^{i g} ight) [/latex]

[latex] \dot{Q}=\dot{n}_A C_P^{i g}\left(T - T_A ight)+\dot{n}_B C_P^{i g}\left(T - T_B ight)=C_P^{i g}\left[\dot{n}_A\left(T - T_A ight)+\dot{n}_B\left(T - T_B ight) ight] [/latex]

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

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

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

[latex] \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} ight)+\dot{n}_B\left(s^{i g}-S_B^{i g} ight)-\frac{\dot{Q}}{T_\sigma} [/latex]

[latex] =\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} ight)-\frac{\dot{Q}}{T_\sigma} [/latex]

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

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

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

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