Chapter 13.3, Problem 91RP

a)

To determine

The total entropy change and exergy destruction associated with the process using ideal gas approximation.

Answer to Problem 91RP

The entropy generated is $1.61\text{kJ}/\text{kmol}\cdot \text{K}$.

The energy destroyed is $488\text{kJ}/\text{kmol}$.

Explanation of Solution

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

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

Here, mass of ${\text{O}}_{\text{2}}$ is $m_{{\text{O}}_{\text{2}}}$, 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 ${\text{O}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$ at $210\text{K}$ and $180\text{K}$ as $0.918\text{kJ}/\text{kg}\cdot \text{K}$, and $1.039\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Obtain the value of ${\left(s_2-s_1\right)}_{{\text{O}}_{\text{2}}}$ from Equation (I).

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

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.

The partial pressure of ${\text{O}}_2$ gas remains constant in the total mixture pressure, thus calculate ${\left(s_2-s_1\right)}_{{\text{O}}_2}$ from Equation (III).

${\left(s_2-s_1\right)}_{{\text{O}}_2}={\left(c_p\ln \frac{T_2}{T_1}\right)}_{{\text{O}}_2}$ (IV)

Substitute $0.918\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 210 K for $T_2$, and 180 K for $T_1$ in Equation (IV).

$\begin{array}{c}{\left(s_2-s_1\right)}_{{\text{O}}_2}=\left(\left(0.918\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{210\text{K}}{180\text{K}}\right)\\ =0.141\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

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

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

${\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}={\left(c_p\ln \frac{T_2}{T_1}\right)}_{{\text{N}}_{\text{2}}}$ (VI)

Substitute $1.039\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 210 K for $T_2$, 180 K for $T_1$ in Equation (VI).

$\begin{array}{c}{\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}=\left(\left(1.039\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{210\text{K}}{180\text{K}}\right)\\ =0.160\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $32\text{kg}/\text{kmol}$ for $m_{{\text{O}}_{\text{2}}}$, $28\text{kg}/\text{kmol}$ for $m_{{\text{N}}_{\text{2}}}$, $0.141\text{kJ}/\text{kg}\cdot \text{K}$ for ${\left(s_2-s_1\right)}_{{\text{H}}_{\text{2}}}$, $0.160\text{kJ}/\text{kg}\cdot \text{K}$ for ${\left(s_2-s_1\right)}_{{\text{N}}_{\text{2}}}$, 872 kJ for $Q_{\text{in}}$, and 303 K for $T_{\text{surr}}$ in Equation (I).

$\begin{array}{c}{S}_{\text{gen}}=\left[\begin{array}{c}\left(32\text{kg}/\text{kmol}\right)\left(0.141\text{kJ}/\text{kg}\cdot \text{K}\right)+\\ \left(28\text{kg}/\text{kmol}\right)\left(0.160\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{872\text{kJ}}{303\text{K}}\end{array}\right]\\ =1.61\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Thus, the entropy generated is $1.61\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute 303 K for $T_0$ and $1.61\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_{\text{gen}}$ in Equation (II).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(303\text{K}\right)\left(1.61\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =488\text{kJ}/\text{kmol}\end{array}$

Thus, the energy destroyed is $488\text{kJ}/\text{kmol}$.

b)

To determine

The total entropy change and exergy destruction associated with the process using Kay’s rule.

Answer to Problem 91RP

The entropy generation is $3.41\text{kJ}/\text{kmol}\cdot \text{K}$.

The energy destroyed is $1,034\text{kJ}/\text{kmol}$.

Explanation of Solution

Write the expression to obtain the pseudo-critical temperature of the mixture $\left({T^{\prime }}_{\text{cr},m}\right)$.

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{T}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{T}_{{\text{cr,O}}_{\text{2}}}+y_{{\text{N}}_{\text{2}}}{T}_{{\text{cr,N}}_{\text{2}}}\end{array}$ (VII)

Here, mole fraction of $O_2,\text{and}{\text{N}}_{\text{2}}$ is $y_{{\text{O}}_{\text{2}}}\text{and}{y}_{{\text{N}}_{\text{2}}}$ and critical temperature of $O_2,\text{and}{\text{N}}_{\text{2}}$ are $T_{{\text{cr,O}}_2},and{T}_{{\text{cr,N}}_2}$.

Write the expression to obtain the pseudo-critical pressure of the mixture $\left({P^{\prime }}_{\text{cr},m}\right)$.

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{P}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{P}_{{\text{cr,O}}_{\text{2}}}+y_{{\text{N}}_{\text{2}}}{P}_{{\text{cr,N}}_{\text{2}}}\end{array}$ (VIII)

Here, critical pressure of $O_2,\text{and}{\text{N}}_{\text{2}}$ are $P_{{\text{cr,O}}_2},and{P}_{{\text{cr,N}}_2}$.

Write the expression to obtain the initial reduced temperature $\left(T_{R,1}\right)$.

$T_{R\text{,1}}=\frac{T_{m.1}}{{T^{\prime }}_{\text{cr},m}}$ (IX)

Here, mixing critical temperature is ${T^{\prime }}_{\text{cr},m}$.

Write the expression to obtain the initial reduced pressure $\left(P_{R,1}\right)$.

$\begin{array}{c}{P}_{R\text{,1}}=P_{R\text{,2}}\\ =\frac{P_m}{{P^{\prime }}_{\text{cr},m}}\end{array}$ (X)

Here, mixing critical pressure is ${P^{\prime }}_{\text{cr},m}$.

Write the expression to obtain the final reduced temperature $\left(T_{R,2}\right)$.

$T_{R\text{,2}}=\frac{T_{m,2}}{{T^{\prime }}_{\text{cr},m}}$ (XI)

Write the expression to obtain the specific change in entropy of a system $\left(\Delta {\overline{s}}_{\text{sys}}\right)$.

$\Delta {\overline{s}}_{\text{sys}}=R_u\left(Z_{s_1}-Z_{s_2}\right)+\left(\Delta {\overline{s}}_{\text{sys,ideal}}\right)$ (XII)

Here, universal gas constant is $R_u$.

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

$S_{\text{gen}}=\Delta {\overline{s}}_{\text{sys}}-\frac{Q_{\text{in}}}{T_{\text{surr}}}$ (XIII)

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

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

Here, initial temperature is $T_0$.

Conclusion:

From the Table of critical properties, obtain the critical temperature and pressure for ${\text{O}}_2$ as 154.8 K and 5.08 MPa and critical temperature and pressure for ${\text{N}}_{\text{2}}$ as 126.2 K and 3.39 MPa.

Substitute 0.25 for $y_{{\text{O}}_{\text{2}}}$, 154.8 K for $T_{{\text{cr,O}}_2}$, 0.75 for $y_{{\text{N}}_{\text{2}}}$, and 126.2 K for $T_{{\text{cr,N}}_2}$ in Equation (VII).

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}=0.25\left(154.8\text{K}\right)+0.75\left(126.2\text{K}\right)\\ =133.4\text{K}\end{array}$

Substitute 0.25 for $y_{{\text{O}}_{\text{2}}}$, 5.08 MPa for $P_{{\text{cr,O}}_2}$, 0.75 for $y_{{\text{N}}_{\text{2}}}$, and 3.39 MPa for $P_{{\text{cr,N}}_2}$ in Equation (VIII).

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}=0.25\left(5.08\text{MPa}\right)+0.75\left(3.39\text{MPa}\right)\\ =3.81\text{MPa}\end{array}$

Substitute 180 K for $T_{m,1}$ and 133.4 K for ${T^{\prime }}_{\text{cr},m}$ in Equation (IX).

$\begin{array}{c}{T}_{R,1}=\frac{180\text{K}}{133.4\text{K}}\\ =1.349\end{array}$

Substitute 8 MPa for $P_m$ and 3.81 MPa for ${P^{\prime }}_{\text{cr},m}$ in Equation (X).

$\begin{array}{c}{P}_{R,1}=\frac{8\text{MPa}}{3.81\text{}\text{MPa}}\\ =2.1\end{array}$

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of initial entropy departure $\left(Z_{s_2}\right)$ as 0.8 by taking $P_{R\text{,1}}$ as 2.1 and $T_{R,1}$ as 1.349.

Substitute 210 K for $T_{m,2}$ and 133.4 K for ${T^{\prime }}_{\text{cr},m}$ in Equation (XI).

$\begin{array}{c}{T}_{R,2}=\frac{210\text{K}}{133.4\text{K}}\\ =1.574\end{array}$

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of final entropy departure $\left(Z_{s_2}\right)$ as 0.45 by taking $P_{R\text{,2}}$ as 2.1 and $T_{R,2}$ as 1.574.

Substitute $8.314\text{kJ}/\text{kmol}\cdot \text{K}$ for $R_u$, 0.8 for $Z_{s_1}$, 0.45 for $Z_{s_2}$, and $4.49\text{kJ}/\text{kmol}\cdot \text{K}$ for $\Delta {\overline{s}}_{\text{sys,ideal}}$ in Equation (XII).

$\begin{array}{c}\Delta {\overline{s}}_{\text{sys}}=8.314\text{kJ}/\text{kmol}\cdot \text{K}\left(0.8-0.45\right)+4.49\text{kJ}/\text{kmol}\cdot \text{K}\\ =7.40\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $7.40\text{kJ}/\text{kmol}\cdot \text{K}$ for $\Delta {\overline{s}}_{\text{sys}}$, $1,204.7\text{kJ}/\text{kmol}$ for $Q_{\text{in}}$, and $303\text{K}$ for $T_{\text{surr}}$ in Equation (XIII).

$\begin{array}{c}{S}_{\text{gen}}=7.40\text{kJ}/\text{kmol}\cdot \text{K}-\frac{1,204.7\text{kJ}/\text{kmol}}{303\text{K}}\\ =3.41\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Thus, the entropy generation is $3.41\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute 303 K for $T_0$ and $3.41\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_{\text{gen}}$ in Equation (XIV).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(303\text{K}\right)\left(3.41\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =1,034\text{kJ}/\text{kmol}\end{array}$

Thus, the energy destroyed is $1,034\text{kJ}/\text{kmol}$.