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 $89.4\%$.

The exergy destruction during the expansion process is $79\text{kJ}/\text{kg}$.

Explanation of Solution

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

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

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

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

$N_{{\text{CO}}_{\text{2}}}=\frac{m_{{\text{CO}}_{\text{2}}}}{M_{{\text{CO}}_{\text{2}}}}$ (II)

Here, molar mass of ${\text{CO}}_2$ is $M_{{\text{CO}}_{\text{2}}}$ and mass of ${\text{CO}}_2$ is $m_{{\text{CO}}_{\text{2}}}$.

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

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

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 mass fraction of ${\text{O}}_{\text{2}}$ $\left({\text{mf}}_{{\text{O}}_{\text{2}}}\right)$.

${\text{mf}}_{{\text{O}}_{\text{2}}}=\frac{m_{{\text{O}}_{\text{2}}}}{m_m}$ (IV)

Write the expression to obtain the mass fraction of ${\text{CO}}_2$ $\left({\text{mf}}_{{\text{CO}}_{\text{2}}}\right)$.

${\text{mf}}_{{\text{CO}}_{\text{2}}}=\frac{m_{{\text{CO}}_{\text{2}}}}{m_m}$ (V)

Write the expression to obtain the mass fraction of $\text{He}$ $\left({\text{mf}}_{\text{He}}\right)$.

${\text{mf}}_{\text{He}}=\frac{m_{\text{He}}}{m_m}$ (VI)

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

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

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

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

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

$c_p={\text{mf}}_{{\text{O}}_{\text{2}}}{c}_{p,{\text{O}}_{\text{2}}}+{\text{mf}}_{{\text{CO}}_{\text{2}}}{c}_{p,{\text{CO}}_{\text{2}}}+{\text{mf}}_{\text{He}}{c}_{p,\text{He}}$ (IX)

Here, constant pressure specific heat of ${\text{O}}_{\text{2}}{\text{, He, and CO}}_{\text{2}}$ are $c_{p,{\text{O}}_{\text{2}}}$, $c_{p,\text{He}}$, and $c_{p,{\text{CO}}_{\text{2}}}$ respectively.

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

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

Here, the universal gas constant is $R_u$.

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

$c_v=c_p-R$ (XI)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Conclusion:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of ${\text{O}}_2$, ${\text{CO}}_{\text{2}}$ and $\text{He}$ as $32\text{kg}/\text{kmol}$, $44\text{kg}/\text{kmol}$ and $4\text{kg}/\text{kmol}$ respectively.

Substitute $0.1\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{0.1\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.003125\text{kmol}\end{array}$

Substitute $1\text{kg}$ for $m_{{\text{CO}}_2}$ and $44\text{kg}/\text{kmol}$ for $M_{{\text{CO}}_2}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{CO}}_2}=\frac{1\text{kg}}{\left(44\text{kg}/\text{kmol}\right)}\\ =0.02273\text{kmol}\end{array}$

Substitute $0.5\text{kg}$ for $m_{\text{He}}$ and $4\text{kg}/\text{kmol}$ for $M_{\text{He}}$ in Equation (III).

$\begin{array}{c}{N}_{\text{He}}=\frac{0.5\text{lbm}}{\left(4\text{kg}/\text{kmol}\right)}\\ =0.125\text{kmol}\end{array}$

Substitute $0.003125\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.02273\text{kmol}$ for $N_{{\text{CO}}_2}$ and $0.125\text{kmol}$ for $N_{\text{He}}$ in Equation (VII).

$\begin{array}{c}{N}_m=0.003125\text{kmol}+0.02273\text{kmol}+0.125\text{kmol}\\ =0.15086\text{kmol}\end{array}$

Substitute $1.6\text{kg}$ for $m_m$ and $0.15086\text{kmol}$ for $N_m$ in Equation (X).

$\begin{array}{c}{M}_m=\frac{1.6\text{kg}}{0.15086\text{kmol}}\\ =10.61\text{kg}/\text{kmol}\end{array}$

Substitute $8.314\text{kJ}/\text{Kmol}\cdot \text{K}$ for $R_u$ and $10.61\text{kg}/\text{kmol}$ for $M_m$ in Equation (IX).

$\begin{array}{c}R=\frac{8.314\text{kJ}/\text{Kmol}\cdot \text{K}}{10.61\text{kg}/\text{kmol}}\\ =0.7836\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.1\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $1.6\text{kg}$ for $m_m$ in Equation (IV).

$\begin{array}{c}{\text{mf}}_{{\text{O}}_{\text{2}}}=\frac{0.1\text{kg}}{1.6\text{kg}}\\ =0.0625\end{array}$

Substitute $1\text{kg}$ for $m_{{\text{CO}}_{\text{2}}}$ and $1.6\text{kg}$ for $m_m$ in Equation (V).

$\begin{array}{c}{\text{mf}}_{{\text{CO}}_{\text{2}}}=\frac{1\text{kg}}{1.6\text{kg}}\\ =0.625\end{array}$

Substitute $0.5\text{kg}$ for $m_{\text{He}}$ and $1.6\text{kg}$ for $m_m$ in Equation (VI).

$\begin{array}{c}{\text{mf}}_{\text{He}}=\frac{0.5\text{kg}}{1.6\text{kg}}\\ =0.3125\end{array}$

Refer Table A-2a, “Ideal gas specific heats of various common gases”, obtain the constant pressure specific heats of ${\text{O}}_2$, $\text{He}$ and ${\text{CO}}_2$ as $0.918\text{kJ}/\text{kg}\cdot \text{K}$, $5.1926\text{kJ}/\text{kg}\cdot \text{K}$,  and $0.846\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $0.0625$ for ${\text{mf}}_{{\text{O}}_{\text{2}}}$, $0.625$ for ${\text{mf}}_{{\text{CO}}_{\text{2}}}$, $0.3125$ for ${\text{mf}}_{\text{He}}$, $0.918\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,{\text{O}}_{\text{2}}}$, $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,\text{He}}$, and $0.846\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,{\text{CO}}_{\text{2}}}$ in Equation (IX).

$\begin{array}{c}{c}_p=\left[\begin{array}{l}\left(0.0625\right)\left(0.918\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.625\right)\left(0.846\text{kJ}/\text{kg}\cdot \text{K}\right)+\\ +\left(0.3125\right)\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =2.209\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.7836\text{kJ}/\text{kg}\cdot \text{K}$ for R and $2.209\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (XI).

$\begin{array}{c}{c}_v=2.209\text{kJ}/\text{kg}\cdot \text{K}-0.7836\text{kJ}/\text{kg}\cdot \text{K}\\ =1.425\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $2.209\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ and $1.425\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$ in Equation (XII).

$\begin{array}{c}k=\frac{2.209\text{kJ}/\text{kg}\cdot \text{K}}{1.425\text{kJ}/\text{kg}\cdot \text{K}}\\ =1.550\end{array}$

Substitute $1.550$ for k, $327°\text{C}$ for $T_1$, $100\text{kPa}$ for $P_2$, and $1000\text{kPa}$ for $P_1$ in Equation (XIII).

$\begin{array}{c}{T}_{2s}=\left(327°\text{C}\right){\left(\frac{100\text{kPa}}{1000\text{kPa}}\right)}^{\left(1.550-1\right)/1.550}\\ =\left(327+273\right)\text{K}{\left(\frac{100\text{kPa}}{1000\text{kPa}}\right)}^{\left(1.550-1\right)/1.550}\\ =265\text{K}\end{array}$

Substitute $327°\text{C}$ for $T_1$, $265\text{K}$ for $T_{2s}$ and $0.90$ for ${\eta }_{\text{turb}}$ in Equation (XIV).

$\begin{array}{c}{T}_2=327°\text{C}-0.90\left(327°\text{C}-265\text{K}\right)\\ =\left(327+273\right)\text{K}-0.90\left(\left(327+273\right)\text{K}-265\text{K}\right)\\ =299\text{K}\end{array}$

Substitute $2.209\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $299\text{K}$ for $T_2$, $327°\text{C}$ for $T_1$, $0.7836\text{kJ}/\text{kg}\cdot \text{K}$ for R,  $100\text{kPa}$ for $P_2$, and $1000\text{kPa}$ for $P_1$ in Equation (XV).

$\begin{array}{c}{s}_2-s_1=\left\{\begin{array}{c}\left(2.209\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{299\text{K}}{327°\text{C}}\right)\\ -\left(0.7836\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{100\text{kPa}}{1000\text{kPa}}\right)\end{array}\right\}\\ =\left\{\begin{array}{c}\left(2.209\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{299\text{K}}{\left(327+273\right)\text{K}}\right)\\ -\left(0.7836\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{100\text{kPa}}{1000\text{kPa}}\right)\end{array}\right\}\\ =0.2658\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $2.209\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $299\text{K}$ for $T_2$ and $327°\text{C}$ for $T_1$ in Equation (XVI).

$\begin{array}{c}{w}_{\text{out}}=2.209\text{kJ}/\text{kg}\cdot \text{K}\left(327°\text{C}-299\text{K}\right)\\ =2.209\text{kJ}/\text{kg}\cdot \text{K}\left(\left(327+273\right)\text{K}-299\text{K}\right)\\ =665\text{kJ}/\text{kg}\end{array}$

Substitute $665\text{kJ}/\text{kg}$ for $w_{\text{out}}$, $25°\text{C}$ for $T_0$ and $\left(-0.2658\text{kJ}/\text{kg}\cdot \text{K}\right)$ for $\left(s_1-s_2\right)$ in Equation (XVII).

$\begin{array}{c}{w}_{\text{rev,out}}=665\text{kJ}/\text{kg}-25°\text{C}\left(0.2658\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =665\text{kJ}/\text{kg}-\left(25+273\right)\text{K}\times \left(-0.2658\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =744\text{kJ}/\text{kg}\end{array}$

Substitute $665\text{kJ}/\text{kg}$ for $w_{\text{out}}$ and $744\text{kJ}/\text{kg}$ for $w_{\text{rev,out}}$ in Equation (XVIII).

$\begin{array}{c}{\eta }_{II}=\frac{665\text{kJ}/\text{kg}}{744\text{kJ}/\text{kg}}\\ =0.894\times 100\%\\ =89.4\%\end{array}$

Thus, the second law efficiency is $89.4\%$.

Substitute $665\text{kJ}/\text{kg}$ for $w_{\text{out}}$ and $744\text{kJ}/\text{kg}$ for $w_{\text{rev,out}}$ in Equation (XIX).

$\begin{array}{c}{x}_{\text{dest}}=744\text{kJ}/\text{kg}-665\text{kJ}/\text{kg}\\ =79\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction during the expansion process is $79\text{kJ}/\text{kg}$.