Chapter 13.3, Problem 97RP

To determine

The second law efficiency and the exergy destruction during the expansion process.

Answer to Problem 97RP

The second law efficiency is $83.7\%$.

The exergy destruction during the expansion process is $33.7\text{Btu}/\text{lbm}$.

Explanation of Solution

Write the expression to obtain the mole number of ${\text{N}}_{\text{2}}$ $\left(N_{{\text{N}}_{\text{2}}}\right)$.

$N_{{\text{N}}_{\text{2}}}=\frac{m_{{\text{N}}_{\text{2}}}}{M_{{\text{N}}_{\text{2}}}}$ (I)

Here, molar mass of ${\text{N}}_{\text{2}}$ is $M_{{\text{N}}_{\text{2}}}$ and mass of ${\text{N}}_{\text{2}}$ is $m_{{\text{N}}_{\text{2}}}$.

Write the expression to obtain the mole number of $\text{He}$ $\left(N_{{\text{CH}}_4}\right)$.

$N_{\text{He}}=\frac{m_{\text{He}}}{M_{\text{He}}}$ (II)

Here, molar mass of $\text{He}$ is $M_{\text{He}}$ and mass of $\text{He}$ is $m_{\text{He}}$.

Write the expression to obtain the mole number of ${\text{CH}}_4$ $\left(N_{{\text{CH}}_4}\right)$.

$N_{{\text{CH}}_4}=\frac{m_{{\text{CH}}_4}}{M_{{\text{CH}}_4}}$ (III)

Here, molar mass of ${\text{CH}}_4$ is $M_{{\text{CH}}_4}$ and mass of ${\text{CH}}_4$ is $m_{{\text{CH}}_4}$.

Write the expression to obtain the mole number of ${\text{C}}_2{\text{H}}_6$ $\left(N_{{\text{C}}_2{\text{H}}_6}\right)$.

$N_{{\text{C}}_2{\text{H}}_6}=\frac{m_{{\text{C}}_2{\text{H}}_6}}{M_{{\text{C}}_2{\text{H}}_6}}$ (IV)

Here, molar mass of ${\text{C}}_2{\text{H}}_6$ is $M_{{\text{C}}_2{\text{H}}_6}$ and mass of ${\text{C}}_2{\text{H}}_6$ is $m_{{\text{C}}_2{\text{H}}_6}$.

Write the expression to obtain the equation to calculate the mole number of the mixture $\left(N_m\right)$.

$N_m=N_{{\text{N}}_{\text{2}}}+N_{\text{He}}+N_{{\text{CH}}_{\text{4}}}+N_{{\text{C}}_2{\text{H}}_6}$ (V)

Write the expression to obtain the molar mass of the gas mixture $\left(M_m\right)$.

$M_m=\frac{m_m}{N_m}$ (VI)

Write the expression to obtain the equation to calculate the constant-pressure specific heat of the mixture $\left(c_p\right)$.

$c_p={\text{mf}}_{{\text{N}}_{\text{2}}}{c}_{p,{\text{N}}_{\text{2}}}+{\text{mf}}_{\text{He}}{c}_{p,\text{He}}+{\text{mf}}_{{\text{CH}}_{\text{4}}}{c}_{p,{\text{CH}}_{\text{4}}}+{\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{c}_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ (VII)

Here, mass fraction of ${\text{N}}_{\text{2}}{\text{, He, CH}}_{\text{4}}\text{, and}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ are ${\text{mf}}_{{\text{N}}_{\text{2}}}$, ${\text{mf}}_{\text{He}}$, ${\text{mf}}_{{\text{CH}}_{\text{4}}}$, and ${\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ respectively and constant pressure specific heat of ${\text{N}}_{\text{2}}{\text{, He, CH}}_{\text{4}}\text{, and}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ are $c_{p,{\text{N}}_{\text{2}}}$, $c_{p,\text{He}}$, $c_{p,{\text{CH}}_{\text{4}}}$, and $c_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ respectively.

Write the expression to obtain the gas constant of the mixture $\left(R\right)$.

$R=\frac{R_u}{M_m}$ (VIII)

Here, the universal gas constant is $R_u$.

Write the expression to obtain the constant volume specific heat $\left(c_v\right)$.

$c_v=c_p-R$ (IX)

Write the expression to obtain the specific heat ratio $\left(k\right)$.

$k=\frac{c_p}{c_v}$ (X)

Write the expression to obtain the temperature at the end of the expansion for the isentropic process $\left(T_{2s}\right)$.

$T_{2s}=T_1{\left(\frac{P_2}{P_1}\right)}^{\left(k-1\right)/k}$ (XI)

Write the expression to obtain the actual outlet temperature $\left(T_2\right)$.

$T_2=T_1-{\eta }_{\text{turb}}\left(T_1-T_{2s}\right)$ (XII)

Here, efficiency of the turbine is ${\eta }_{\text{turb}}$.

Write the expression to obtain the entropy change of the gas mixture $\left(s_2-s_1\right)$.

$s_2-s_1=c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$ (XIII)

Write the expression to obtain the actual work output $\left(w_{\text{out}}\right)$.

$w_{\text{out}}=c_p\left(T_1-T_2\right)$ (XIV)

Write the expression to obtain the reversible work output $\left(w_{\text{rev,out}}\right)$.

$w_{\text{rev,out}}=w_{\text{out}}-T_0\left(s_1-s_2\right)$ (XV)

Write the expression to obtain the second law efficiency $\left({\eta }_{II}\right)$.

${\eta }_{II}=\frac{w_{\text{out}}}{w_{\text{rev,out}}}$ (XVI)

Write the expression to obtain the exergy destruction $\left(x_{\text{dest}}\right)$.

$x_{\text{dest}}=w_{\text{rev,out}}-w_{\text{out}}$ (XVII)

Conclusion:

Refer Table A-1E, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of ${\text{N}}_2$, $\text{He}$, ${\text{CH}}_{\text{4}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ as $28.0\text{lbm}/\text{lbmol}$, $4.0\text{lbm}/\text{lbmol}$, $16.0\text{lbm}/\text{lbmol}$ and $30.0\text{lbm}/\text{lbmol}$.

Substitute 0.15 lbm for $m_{{\text{N}}_{\text{2}}}$ and $28\text{lbm/lbmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{0.15\text{lbm}}{28\text{lbm/lbmol}}\\ =0.005357\text{lbmol}\end{array}$

Substitute 0.5 lbm for $m_{\text{He}}$ and $4\text{lbm/lbmol}$ for $M_{\text{He}}$ in Equation (II).

$\begin{array}{c}{N}_{\text{He}}=\frac{0.5\text{lbm}}{\left(4\text{lbm/lbmol}\right)}\\ =0.0125\text{lbmol}\end{array}$

Substitute 0.6 lbm for $m_{{\text{CH}}_{\text{4}}}$ and $16\text{lbm/lbmol}$ for $M_{{\text{CH}}_{\text{4}}}$ in Equation (III).

$\begin{array}{c}{N}_{{\text{CH}}_{\text{4}}}=\frac{0.6\text{lbm}}{\left(16\text{lbm/lbmol}\right)}\\ =0.0375\text{lbmol}\end{array}$

Substitute 0.2 lbm for $m_{{\text{C}}_2{\text{H}}_6}$ and $30\text{lbm/lbmol}$ for $M_{{\text{C}}_2{\text{H}}_6}$ in Equation (IV).

$\begin{array}{c}{N}_{{\text{C}}_2{\text{H}}_6}=\frac{0.2\text{lbm}}{\left(30\text{lbm/lbmol}\right)}\\ =0.006667\text{lbmol}\end{array}$

Substitute $0.005357\text{lbmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.0125\text{lbmol}$ for $N_{\text{He}}$, $0.0375\text{lbmol}$ for $N_{{\text{CH}}_{\text{4}}}$, and $0.006667\text{lbmol}$ for $N_{{\text{C}}_2{\text{H}}_6}$ in Equation (V).

$\begin{array}{c}{N}_m=0.005357\text{lbmol}+0.00125\text{lbmol}+0.00375\text{lbmol}+0.006667\text{lbmol}\\ =0.06202\text{lbmol}\end{array}$

Substitute 1 lbm for $m_m$ and $0.06202\text{lbmol}$ for $N_m$ in Equation (VI).

$\begin{array}{c}{M}_m=\frac{1\text{lbm}}{0.06202\text{lbmol}}\\ =16.12\text{lbm/lbmol}\end{array}$

Refer Table A-2Ea, “Ideal gas specific heats of various common gases”, obtain the constant pressure specific heats of ${\text{N}}_2$, $\text{He}$, ${\text{CH}}_{\text{4}}$, and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ as $0.248\text{Btu}/\text{lbm}\cdot \text{R}$, $1.25\text{Btu}/\text{lbm}\cdot \text{R}$, $0.532\text{Btu}/\text{lbm}\cdot \text{R}$ and $0.427\text{Btu}/\text{lbm}\cdot \text{R}$.

Substitute 0.15 for ${\text{mf}}_{{\text{N}}_{\text{2}}}$, 0.05 for ${\text{mf}}_{\text{He}}$, 0.60 for ${\text{mf}}_{{\text{CH}}_{\text{4}}}$, 0.20 for ${\text{mf}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$, $0.248\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_{p,{\text{N}}_{\text{2}}}$, $1.25\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_{p,\text{He}}$, $0.532\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_{p,{\text{CH}}_{\text{4}}}$, and $0.427\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_{p,{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}$ in Equation (VII).

$\begin{array}{c}{c}_p=\left[\begin{array}{l}\left(0.15\right)\left(0.248\text{Btu}/\text{lbm}\cdot \text{R}\right)+\left(0.05\right)\left(1.25\text{Btu}/\text{lbm}\cdot \text{R}\right)+\\ \left(0.60\right)\left(0.532\text{Btu}/\text{lbm}\cdot \text{R}\right)+\left(\text{0}\text{.20}\right)\left(0.427\text{Btu}/\text{lbm}\cdot \text{R}\right)\end{array}\right]\\ =0.5043\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $1.9858\text{lbm}/\text{lbmol}\cdot \text{R}$ for $R_u$ and $16.12\text{lbm/lbmol}$ for $M_m$ in Equation (VIII).

$\begin{array}{c}R=\frac{1.9858\text{lbm}/\text{lbmol}\cdot \text{R}}{16.12\text{lbm/lbmol}}\\ =0.1232\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $0.1232\text{Btu}/\text{lbm}\cdot \text{R}$ for R and $0.5043\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ in Equation (IX).

$\begin{array}{c}{c}_v=0.5043\text{Btu}/\text{lbm}\cdot \text{R}-0.1232\text{Btu}/\text{lbm}\cdot \text{R}\\ =0.3811\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $0.5043\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ and $0.3811\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$ in Equation (X).

$\begin{array}{c}k=\frac{0.5043\text{Btu}/\text{lbm}\cdot \text{R}}{0.3811\text{Btu}/\text{lbm}\cdot \text{R}}\\ =1.323\end{array}$

Substitute 1.323 for k, $400°\text{F}$ for $T_1$, 15 psia for $P_2$, and 200 psia for $P_1$ in Equation (XI).

$\begin{array}{c}{T}_{2s}=\left(400°\text{F}+460\right)\text{R}{\left(\frac{15\text{psia}}{200\text{ psia}}\right)}^{\left(1.323-1\right)/1.323}\\ =456.8\text{ R}\end{array}$

Substitute $400°\text{F}$ for $T_1$, 456.8 for $T_{2s}$ and 0.85 for ${\eta }_{\text{turb}}$ in Equation (XII).

$\begin{array}{c}{T}_2=\left(400°\text{F}+460\right)\text{R}-0.85\left(\left(400°\text{F}+460\right)\text{R}-456.8\text{ R}\right)\\ =517.3\text{ R}\end{array}$

Substitute $0.5043\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $517.3\text{ R}$ for $T_2$, $400°\text{F}$ for $T_1$, $0.1232\text{Btu}/\text{lbm}\cdot \text{R}$ for R, 15 psia for $P_2$, and 200 psia for $P_1$ in Equation (XIII).

$\begin{array}{c}{s}_2-s_1=\left\{\begin{array}{c}\left(0.5043\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \left(\frac{517.3\text{ R}}{\left(400\text{°F}+460\right)\text{R}}\right)\\ -\left(0.1232\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \left(\frac{15\text{ psia}}{200\text{ psia}}\right)\end{array}\right\}\\ =0.06270\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $0.5043\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $517.3\text{ R}$ for $T_2$ and $400°\text{F}$ for $T_1$ in Equation (XIV).

$\begin{array}{c}{w}_{\text{out}}=0.5043\text{Btu}/\text{lbm}\cdot \text{R}\left(\left(400°\text{F}+460\right)\text{R}-517.3\text{ R}\right)\\ =172.8\text{Btu}/\text{lbm}\end{array}$

Substitute $172.8\text{Btu}/\text{lbm}$ for $w_{\text{out}}$, $77°\text{F}$ for $T_0$ and $\left(-0.06270\text{Btu}/\text{lbm}\cdot \text{R}\right)$ for $\left(s_1-s_2\right)$ in Equation (XV).

$\begin{array}{c}{w}_{\text{rev,out}}=172.8\text{Btu}/\text{lbm}-\left(77°\text{F}+460\right)\text{R}\left(-0.06270\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ =206.5\text{Btu}/\text{lbm}\end{array}$

Substitute $172.8\text{Btu}/\text{lbm}$ for $w_{\text{out}}$ and $206.5\text{Btu}/\text{lbm}$ for $w_{\text{rev,out}}$ in Equation (XVI).

$\begin{array}{c}{\eta }_{II}=\frac{172.8\text{Btu}/\text{lbm}}{206.5\text{Btu}/\text{lbm}}\\ =0.837\\ =83.7\%\end{array}$

Thus, the second law efficiency is $83.7\%$.

Substitute $172.8\text{Btu}/\text{lbm}$ for $w_{\text{out}}$ and $206.5\text{Btu}/\text{lbm}$ for $w_{\text{rev,out}}$ in Equation (XVII).

$\begin{array}{c}{x}_{\text{dest}}=206.5\text{Btu}/\text{lbm}-172.8\text{Btu}/\text{lbm}\\ =33.7\text{Btu}/\text{lbm}\end{array}$

Thus, the exergy destruction during the expansion process is $33.7\text{Btu}/\text{lbm}$.