Chapter 13.3, Problem 73P

a)

To determine

The total entropy change and exergy destruction by treating the mixture as an ideal gas.

Answer to Problem 73P

The entropy generated is $8.496\text{kJ}/\text{K}$.

The energy destroyed is $2489\text{kJ}$.

Explanation of Solution

Write the entropy balance equation to obtain the expression of entropy generation in terms of ${\text{H}}_{\text{2}}\text{and}{\text{N}}_2$.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ \frac{Q_{\text{in}}}{T_{b,\text{surr}}}+S_{\text{gen}}=m\left(s_2-s_1\right)\\ S_{\text{gen}}=m\left(s_2-s_1\right)+\frac{Q_{\text{in}}}{T_{\text{surr}}}\\ S_{\text{gen}}=m_{{\text{H}}_{\text{2}}}{\left(s_2-s_1\right)}_{{\text{H}}_{\text{2}}}+m_{{\text{N}}_{\text{2}}}{\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}-\frac{Q_{\text{in}}}{T_{\text{surr}}}\\ S_{\text{gen}}=\Delta S_{{\text{H}}_{\text{2}}}+\Delta S_{{\text{N}}_{\text{2}}}-\frac{Q_{\text{in}}}{T_{\text{surr}}}\end{array}$ (I)

Here, mass of ${\text{H}}_{\text{2}}$ is $m_{{\text{H}}_{\text{2}}}$, mass of ${\text{N}}_{\text{2}}$ is $m_{{\text{N}}_{\text{2}}}$, boundary temperature is $T_{\text{b,surr}}$, surrounding temperature is $T_{\text{surr}}$, entropy at state 1 and 2 is $s_1\text{and}{s}_2$, entropy generation is $S_{\text{gen}}$, heat input is $Q_{\text{in}}$, outlet entropy is $S_{\text{out}}$, and inlet entropy is $S_{\text{in}}$.

Write the expression to obtain the energy destroyed during a process $\left(X_{\text{destroyed}}\right)$.

$X_{\text{destroyed}}=T_0{S}_{\text{gen}}$ (II)

Here, initial temperature is $T_0$.

Conclusion:

Refer Table A-2b, “Ideal gas specific heats of various common gases”, obtain the specific heat at constant pressure of ${\text{H}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$ at $200\text{K}$ and $160\text{K}$ as, $13.6\text{kJ}/\text{kg}\cdot \text{K}$ and $1.039\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

From Equation (I) obtain the value of ${\left(s_2-s_1\right)}_{{\text{H}}_{\text{2}}}$.

${\left(s_2-s_1\right)}_{{\text{H}}_{\text{2}}}=m_{{\text{H}}_{\text{2}}}{\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2{}^{↵0}}{P_1}\right)}_{{\text{H}}_{\text{2}}}$ (III)

The partial pressure of ${\text{H}}_{\text{2}}$ gas remains constant in the total mixture pressure, thus calculate ${\left(s_2-s_1\right)}_{{\text{H}}_{\text{2}}}$ from Equation (III).

$\begin{array}{l}{\left(s_2-s_1\right)}_{{\text{H}}_{\text{2}}}=m_{{\text{H}}_{\text{2}}}{\left(c_p\ln \frac{T_2}{T_1}\right)}_{{\text{H}}_{\text{2}}}\\ \Delta S_{{\text{H}}_{\text{2}}}=m_{{\text{H}}_{\text{2}}}{\left(c_p\ln \frac{T_2}{T_1}\right)}_{{\text{H}}_{\text{2}}}\end{array}$ (IV)

Here, constant pressure specific heat is $c_p$, initial temperature and pressure is $T_1\text{and}{P}_1$, and apparent gas constant is R,

Substitute 6 kg for $m_{{\text{H}}_{\text{2}}}$, $13.60\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 200 K for $T_2$, and 160 K for $T_1$ in Equation (IV).

$\begin{array}{c}\Delta S_{{\text{H}}_{\text{2}}}=6\text{kg}\left(\left(13.60\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{200\text{K}}{160\text{K}}\right)\\ =18.21\text{kJ}/\text{K}\end{array}$

From Equation (I) obtain the value of ${\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}$.

${\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}=m_{{\text{N}}_{\text{2}}}{\left(c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2{}^{↵0}}{P_1}\right)}_{{\text{N}}_{\text{2}}}$ (V)

The partial pressure of ${\text{N}}_{\text{2}}$ gas remains constant in the total mixture pressure, thus calculate ${\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}$ from Equation (V).

$\begin{array}{l}{\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}=m_{{\text{N}}_{\text{2}}}{\left(c_p\ln \frac{T_2}{T_1}\right)}_{{\text{N}}_{\text{2}}}\\ \Delta S_{{\text{N}}_{\text{2}}}=m_{{\text{N}}_{\text{2}}}{\left(c_p\ln \frac{T_2}{T_1}\right)}_{{\text{N}}_{\text{2}}}\end{array}$ (VI)

Substitute $\text{21}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$, $1.039\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 200 K for $T_2$, 160 K for $T_1$ in Equation (VI).

$\begin{array}{c}\Delta S_{{\text{N}}_{\text{2}}}=\text{21}\text{kg}\left(\left(1.039\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{200\text{K}}{160\text{K}}\right)\\ =4.87\text{kJ}/\text{K}\end{array}$

Substitute $18.21\text{kJ}/\text{K}$ for $\Delta S_{{\text{H}}_{\text{2}}}$, $4.87\text{kJ}/\text{K}$ for $\Delta S_{{\text{N}}_{\text{2}}}$, 4273 kJ for $Q_{\text{in}}$ and 293 K for $T_{\text{surr}}$ in Equation (I).

$\begin{array}{c}{S}_{\text{gen}}=\left(18.21\text{kJ}/\text{K}\right)+\left(4.87\text{kJ}/\text{K}\right)-\frac{4273\text{kJ}}{293\text{K}}\\ =8.496\text{kJ}/\text{K}\end{array}$

Thus, the entropy generated is $8.496\text{kJ}/\text{K}$.

Substitute 293 K for $T_0$ and $8.496\text{kJ}/\text{K}$ for $S_{\text{gen}}$ in Equation (II).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(\text{293}\text{K}\right)\left(8.496\text{kJ}/\text{K}\right)\\ =2489\text{kJ}\end{array}$

Thus, the energy destroyed is $2489\text{kJ}$.

b)

To determine

The total entropy change and exergy destruction by treating the mixture as a non ideal gas using Amagat’s law.

Answer to Problem 73P

The entropy generation is $9.390\text{kJ}/\text{K}$.

The energy destroyed is $2,751\text{kJ}$.

Explanation of Solution

Write the expression to obtain the initial reduced temperature of ${\text{H}}_{\text{2}}$ $\left(T_{R1{\text{,H}}_{\text{2}}}\right)$.

$T_{R_1{\text{,H}}_{\text{2}}}=\frac{T_{m,1}}{T_{\text{cr},{\text{H}}_{\text{2}}}}$ (VII)

Here, critical temperature of ${\text{H}}_{\text{2}}$ is $T_{\text{cr},{\text{H}}_{\text{2}}}$.

Write the expression to obtain the initial and final reduced pressure of ${\text{H}}_{\text{2}}$ $\left(P_{R_1{\text{,H}}_{\text{2}}}\right)$.

$\begin{array}{c}{P}_{R_1{\text{,H}}_{\text{2}}}=P_{R_2,H_2}\\ =\frac{P_m}{P_{\text{cr},{\text{H}}_{\text{2}}}}\end{array}$ (VIII)

Here, critical temperature of ${\text{H}}_{\text{2}}$ is $P_{\text{cr},{\text{H}}_{\text{2}}}$.

Write the expression to obtain the final reduced temperature of ${\text{H}}_{\text{2}}$ $\left(T_{R_2{\text{,H}}_{\text{2}}}\right)$.

$T_{R_2{\text{,H}}_{\text{2}}}=\frac{T_{m,2}}{T_{\text{cr},{\text{H}}_{\text{2}}}}$ (IX)

Here, critical temperature of ${\text{H}}_{\text{2}}$ is $T_{\text{cr},{\text{H}}_{\text{2}}}$.

Write the expression to obtain the initial reduced temperature of ${\text{N}}_2$ $\left(T_{R1{\text{,N}}_2}\right)$.

$T_{R_1{\text{,N}}_2}=\frac{T_{m,1}}{T_{\text{cr},{\text{N}}_2}}$ (X)

Here, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$.

Write the expression to obtain the initial and final reduced pressure of ${\text{N}}_2$ $\left(P_{R_1{\text{,N}}_2}\right)$.

$\begin{array}{c}{P}_{R_1{\text{,N}}_2}=P_{R_2,{\text{N}}_2}\\ =\frac{P_m}{P_{\text{cr},{\text{N}}_2}}\end{array}$ (XI)

Here, critical temperature of ${\text{N}}_2$ is $P_{\text{cr},{\text{N}}_2}$.

Write the expression to obtain the final reduced temperature of ${\text{N}}_2$ $\left(T_{R_2{\text{,N}}_2}\right)$.

$T_{R_2{\text{,N}}_2}=\frac{T_{m,2}}{T_{\text{cr},{\text{N}}_2}}$ (XII)

Here, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$.

Write the expression to obtain the entropy change for ${\text{H}}_{\text{2}}$ $\left(\Delta S_{{\text{H}}_{\text{2}}}\right)$.

$\Delta S_{{\text{H}}_{\text{2}}}=\Delta S_{{\text{H}}_{\text{2}},\text{ideal}}$ (XIII)

Write the expression to obtain the entropy change for ${\text{N}}_2$ $\left(\Delta S_{{\text{N}}_2}\right)$.

$\Delta S_{{\text{N}}_{\text{2}}}=N_{{\text{N}}_{\text{2}}}{R}_u\left(Z_{s_1}-Z_{s_2}\right)+\Delta S_{{\text{N}}_{\text{2}},\text{ideal}}$ (XIV)

Here, number of moles of ${\text{N}}_{\text{2}}$ is $N_{{\text{N}}_{\text{2}}}$, and universal gas constant is $R_u$.

Write the expression to obtain the surrounding entropy change $\left(\Delta S_{\text{surr}}\right)$.

$\Delta S_{\text{surr}}=\frac{Q_{\text{surr}}}{T_0}$ (XV)

Here, surrounding heat is $Q_{\text{surr}}$.

Write the expression to obtain the entropy generation $\left(S_{\text{gen}}\right)$.

$S_{\text{gen}}=\Delta S_{{\text{H}}_{\text{2}}}+\Delta S_{{\text{N}}_{\text{2}}}+\Delta S_{\text{surr}}$ (XVI)

Write the expression to obtain the energy destroyed during a process $\left(X_{\text{destroyed}}\right)$.

$X_{\text{destroyed}}=T_0{S}_{\text{gen}}$ (XVII)

Here, initial temperature is $T_0$.

Conclusion:

Substitute 160 K for $T_{m,1}$ and 33.3 K for $T_{\text{cr},{\text{H}}_{\text{2}}}$ in Equation (VII).

$\begin{array}{c}{T}_{R_1{\text{,H}}_{\text{2}}}=\frac{160\text{}\text{K}}{33.3\text{K}}\\ =4.805\end{array}$

Substitute 5 MPa for $P_m$ and 1.30 MPa for $P_{\text{cr},{\text{H}}_{\text{2}}}$ in Equation (VIII).

$\begin{array}{c}{P}_{R{\text{,H}}_{\text{2}}}=\frac{5\text{MPa}}{1.30\text{MPa}}\\ =3.846\end{array}$

Substitute 200 K for $T_{m,2}$ and 33.3 K for $T_{\text{cr},{\text{H}}_{\text{2}}}$ in Equation (IX).

$\begin{array}{c}{T}_{R_2{\text{,H}}_{\text{2}}}=\frac{200\text{}\text{K}}{33.3\text{K}}\\ =6.006\end{array}$

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of $Z_{s_1}$ and $Z_{s_2}$ for ${\text{H}}_{\text{2}}$ as 1 and 1 by taking $T_{R_1{\text{,H}}_{\text{2}}}$ as 4.805, $P_{R_1{\text{,H}}_{\text{2}}}$ as 3.846, and $T_{R_2{\text{,H}}_{\text{2}}}$ as 6.006.

Substitute 160 K for $T_{m,1}$ and 126.2 K for $T_{\text{cr},{\text{N}}_2}$ in Equation (X).

$\begin{array}{c}{T}_{R_1{\text{,N}}_2}=\frac{160\text{}\text{K}}{126.2\text{K}}\\ =1.268\end{array}$

Substitute 5 MPa for $P_m$ and 3.39 MPa for $P_{\text{cr},{\text{N}}_2}$ in Equation (XI).

$\begin{array}{c}{P}_{R{\text{,N}}_2}=\frac{5\text{MPa}}{3.39\text{MPa}}\\ =1.475\end{array}$

Substitute 200 K for $T_{m,2}$ and 126.2 K for $T_{\text{cr},{\text{N}}_2}$ in Equation (XII).

$\begin{array}{c}{T}_{R_2{\text{,N}}_2}=\frac{200\text{}\text{K}}{126.2\text{K}}\\ =1.585\end{array}$

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of $Z_{s_1}$ and $Z_{s_2}$ for ${\text{N}}_2$ as 0.8 and 0.4 by taking $T_{R_1{\text{,N}}_2}$ as 1.268, $P_{R_1{\text{,N}}_2}$ as 1.475 and $T_{R_2{\text{,N}}_2}$ as 1.585.

Substitute $18.21\text{kJ}/\text{K}$ for $\Delta S_{{\text{H}}_{\text{2}},\text{ideal}}$ in Equation (XIII).

$\Delta S_{{\text{H}}_{\text{2}}}=18.21\text{kJ}/\text{K}$

Substitute 0.75 kmol for $N_{{\text{N}}_{\text{2}}}$, $8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$ for $R_u$, 0.8 for $Z_{s_1}$, 0.4 for $Z_{s_2}$, and $4.87\text{kJ}/\text{K}$ for $\Delta S_{{\text{N}}_{\text{2}},\text{ideal}}$ in Equation (XIV).

$\begin{array}{c}\Delta S_{{\text{N}}_{\text{2}}}=0.75\text{kmol}\left(8.314\text{kPa}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\right)\left(0.8-0.4\right)+4.87\text{kJ}/\text{K}\\ =7.37\text{kJ}/\text{K}\end{array}$

Substitute –4,745 kJ for $Q_{\text{surr}}$ and 293 K for $T_0$ in Equation (XV).

$\begin{array}{c}\Delta S_{\text{surr}}=\frac{-4,745\text{kJ}}{293\text{K}}\\ =-16.19\text{kJ}/\text{K}\end{array}$

Substitute $7.37\text{kJ}/\text{K}$ for $\Delta S_{{\text{N}}_{\text{2}}}$, $18.21\text{kJ}/\text{K}$ for $\Delta S_{{\text{H}}_{\text{2}}}$, and $-16.19\text{kJ}/\text{K}$ for $\Delta S_{\text{surr}}$ in Equation (XVI).

$\begin{array}{c}{S}_{\text{gen}}=18.21\text{kJ}/\text{K}+7.37\text{kJ}/\text{K}-16.19\text{kJ}/\text{K}\\ =9.390\text{kJ}/\text{K}\end{array}$

Thus, the entropy generation is $9.390\text{kJ}/\text{K}$.

Substitute 293 K for $T_0$ and $9.390\text{kJ}/\text{K}$ for $S_{\text{gen}}$ in Equation (XVII).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(\text{293}\text{K}\right)\left(9.390\text{kJ}/\text{K}\right)\\ =2751\text{kJ}\end{array}$

Thus, the energy destroyed is $2751\text{kJ}$.