Chapter 13.3, Problem 68P

To determine

The thermal efficiency of the cycle.

Answer to Problem 68P

The thermal efficiency of the cycle is $\underset{\_}{0.468}$.

Explanation of Solution

Refer to Table A-1E, Obtain the molar masses of ${\text{N}}_{\text{2}},{\text{O}}_2\text{,}{\text{H}}_{\text{2}}\text{O},\text{and}{\text{CO}}_2$ as below:

$\begin{array}{l}{M}_{{\text{N}}_{\text{2}}}=28.0\text{lbm/lbmol}\\ M_{{\text{O}}_{\text{2}}}=32.0\text{lbm/lbmol}\\ M_{{\text{H}}_{\text{2}}\text{O}}=18.0\text{lbm/lbmol}\\ M_{{\text{CO}}_{\text{2}}}=44.0\text{lbm/lbmol}\end{array}$

Refer to Table A-2Ea, obtain the constant-pressure specific heats of the gases at room temperature.

$\begin{array}{l}{c}_{p,{\text{N}}_{\text{2}}}=0.248\text{Btu/lbm}\cdot \text{R}\\ c_{p,{\text{O}}_{\text{2}}}=0.219\text{Btu/lbm}\cdot \text{R}\\ c_{p,{\text{H}}_{\text{2}}\text{O}}=0.445\text{Btu/lbm}\cdot \text{R}\\ c_{p,{\text{CO}}_2}=0.203\text{Btu/lbm}\cdot \text{R}\end{array}$

Write the mass of ${\text{N}}_{\text{2}}$.

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

Here, the mole number of ${\text{N}}_{\text{2}}$ is $N_{{\text{N}}_{\text{2}}}$.

Write the mass of ${\text{O}}_{\text{2}}$.

$m_{{\text{O}}_{\text{2}}}=N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}$ (II)

Here, the mole number of ${\text{O}}_{\text{2}}$ is $N_{{\text{O}}_{\text{2}}}$.

Write the mass of ${\text{H}}_{\text{2}}\text{O}$.

$m_{{\text{H}}_{\text{2}}\text{O}}=N_{{\text{H}}_{\text{2}}\text{O}}{M}_{{\text{H}}_{\text{2}}\text{O}}$ (III)

Here, the mole number of ${\text{H}}_{\text{2}}\text{O}$ is $N_{{\text{H}}_{\text{2}}\text{O}}$.

Write the mass of ${\text{CO}}_2$.

$m_{{\text{CO}}_2}=N_{{\text{CO}}_2}{M}_{{\text{CO}}_2}$ (IV)

Here, the mole number of ${\text{CO}}_2$ is $N_{{\text{CO}}_2}$.

Write the equation to calculate the total mass of the mixture.

$m_m=m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}+m_{{\text{H}}_{\text{2}}\text{O}}+m_{{\text{CO}}_{\text{2}}}$ (V)

Calculate the mass fraction of ${\text{N}}_{\text{2}}$.

${\text{mf}}_{{\text{N}}_{\text{2}}}=\frac{m_{{\text{N}}_{\text{2}}}}{m_m}$ (VI)

Calculate the mass fraction of ${\text{O}}_{\text{2}}$.

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

Calculate the mass fraction of ${\text{H}}_{\text{2}}\text{O}$.

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

Calculate the mass fraction of ${\text{CO}}_2$.

${\text{mf}}_{{\text{CO}}_2}=\frac{m_{{\text{CO}}_2}}{m_m}$ (IX)

Calculate the molar mass of the gas mixture.

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

Write the equation to calculate the constant-pressure specific heat of the mixture.

$c_p={\text{mf}}_{{\text{N}}_{\text{2}}}{c}_{p,{\text{N}}_{\text{2}}}+{\text{mf}}_{{\text{O}}_{\text{2}}}{c}_{p,{\text{O}}_{\text{2}}}+{\text{mf}}_{{\text{H}}_{\text{2}}\text{O}}{c}_{p,{\text{H}}_{\text{2}}\text{O}}+{\text{mf}}_{{\text{CO}}_2}{c}_{p,{\text{CO}}_2}$ (XI)

Here, mass fraction of ${\text{O}}_2\text{,}{\text{N}}_{\text{2}}\text{,}{\text{CO}}_{\text{2}}\text{,}\text{and}{\text{H}}_{\text{2}}\text{O}$ are ${\text{mf}}_{{\text{O}}_{\text{2}}}$, ${\text{mf}}_{{\text{N}}_{\text{2}}}$, ${\text{mf}}_{{\text{CO}}_{\text{2}}}$, and ${\text{mf}}_{{\text{H}}_{\text{2}}\text{O}}$, respectively and constant pressure specific heat of ${\text{O}}_2\text{,}{\text{N}}_{\text{2}}\text{,}{\text{CO}}_{\text{2}}\text{,}\text{and}{\text{H}}_{\text{2}}\text{O}$ are $c_{p,{\text{O}}_{\text{2}}}$, $c_{p,{\text{N}}_{\text{2}}}$, $c_{p,{\text{CO}}_2}$, and $c_{p,{\text{H}}_{\text{2}}\text{O}}$ respectively.

Calculate the gas constant of the mixture.

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

Here, the universal gas constant is $R_u$.

Calculate the constant volume specific heat.

$c_v=c_p-R$ (XIII)

Calculate the specific heat ratio.

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

Refer to Table A-2Ea, obtain the air properties at room temperature.

$\begin{array}{l}{c}_{p,air}=0.240\text{Btu/lbm}\cdot \text{R}\\ c_{v,air}=0.171\text{Btu/lbm}\cdot \text{R}\\ k_{air}=1.4\end{array}$

Calculate the constant pressure and constant velocity specific heat on an average.

$c_{p,avg}=\frac{c_p+c_{p,\text{air}}}{2}$ (XV)

$c_{v,avg}=\frac{c_v+c_{v,air}}{2}$ (XVI)

Calculate the average specific heat ratio.

$k_{\text{avg}}=\frac{k+k_{air}}{2}$ (XVII)

Calculate the final temperature during the compression process.

$T_2=T_1{r}^{k_{air}-1}$ (XVIII)

Here, the compression ratio is r.

Express the heat addition process.

$q_{\text{in}}=c_{v,\text{avg}}\left(T_3-T_2\right)$ (XIX)

Here, specific heat at constant volume on an average is $c_{v,\text{avg}}$.

Write the equation to calculate the temperature during the expansion process.

$T_4=T_3{\left(\frac{1}{r}\right)}^{k-1}$ (XX)

Express the heat rejection process.

$q_{\text{out}}=c_{v,\text{avg}}\left(T_4-T_1\right)$ (XXI)

Calculate the thermal efficiency of the cycle.

${\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}$ (XXII)

Conclusion:

Consider 100 lbmol of this mixture.

$\begin{array}{l}{N}_{{\text{N}}_{\text{2}}}=25\text{lbmol}\\ N_{{\text{O}}_{\text{2}}}=7\text{lbmol}\\ N_{{\text{H}}_{\text{2}}\text{O}}=28\text{lbmol}\\ N_{{\text{CO}}_{\text{2}}}=40\text{lbmol}\end{array}$

Substitute 25 lbmol for $N_{{\text{N}}_2}$, 28 lbm/lbmol for $M_{{\text{N}}_2}$, 7 lbmol for $N_{{\text{O}}_{\text{2}}}$, 32 lbm/lbmol for $M_{{\text{O}}_{\text{2}}}$, 28 lbmol for $N_{{\text{H}}_{\text{2}}\text{O}}$, 18 lbm/lbmol for $M_{{\text{H}}_{\text{2}}\text{O}}$, 40 lbmol for $N_{{\text{CO}}_{\text{2}}}$, and 44 lbm/lbmol for $M_{{\text{CO}}_{\text{2}}}$ in Equations (I) to (IV).

$\begin{array}{l}{m}_{{\text{N}}_2}=700\text{lbm}\\ m_{{\text{O}}_{\text{2}}}=224\text{lbm}\\ m_{{\text{H}}_{\text{2}}\text{O}}=504\text{lbm}\\ m_{{\text{CO}}_{\text{2}}}=1760\text{lbm}\end{array}$

Substitute 224 lbm for $m_{{\text{O}}_{\text{2}}}$, 700 lbm for $m_{{\text{N}}_2}$, 1760 lbm for $m_{{\text{CO}}_{\text{2}}}$, and 504 lbm for $m_{{\text{H}}_{\text{2}}\text{O}}$ in Equation (V).

$\begin{array}{c}{m}_m=700\text{lbm}+224\text{lbm}+504\text{lbm}+1760\text{lbm}\\ =3188\text{lbm}\end{array}$

Substitute 224 lbm for $m_{{\text{O}}_{\text{2}}}$, 700 lbm for $m_{{\text{N}}_2}$, 1760 lbm for $m_{{\text{CO}}_{\text{2}}}$, and 504 lbm for $m_{{\text{H}}_{\text{2}}\text{O}}$, and 3188 lbm for $m_m$ in Equations (VI) to (IX).

$\begin{array}{l}{\text{mf}}_{{\text{N}}_{\text{2}}}=0.2196\\ {\text{mf}}_{{\text{O}}_{\text{2}}}=0.07026\\ {\text{mf}}_{{\text{H}}_{\text{2}}\text{O}}=0.1581\\ {\text{mf}}_{{\text{CO}}_{\text{2}}}=0.5521\end{array}$

Substitute 3188 lbm for $m_m$ and $100\text{lbmol}$ for $N_m$ in Equation (X).

$\begin{array}{c}{M}_m=\frac{3188\text{lbm}}{100\text{lbmol}}\\ =31.88\text{lbm/lbmol}\end{array}$

Substitute 0.07026 for ${\text{mf}}_{{\text{O}}_{\text{2}}}$, 0.2196 for ${\text{mf}}_{{\text{N}}_2}$, 0.5521 for ${\text{mf}}_{{\text{CO}}_{\text{2}}}$, 0.1581 for ${\text{mf}}_{{\text{H}}_{\text{2}}\text{O}}$, $0.219\text{Btu/lbm}\cdot \text{R}$ for $c_{p,{\text{O}}_{\text{2}}}$, $0.248\text{Btu/lbm}\cdot \text{R}$ for $c_{p,{\text{N}}_2}$, $0.203\text{Btu/lbm}\cdot \text{R}$ for $c_{p,{\text{CO}}_{\text{2}}}$, and $0.445\text{Btu/lbm}\cdot \text{R}$ for $c_{p,{\text{H}}_{\text{2}}\text{O}}$ in Equation (XI).

$\begin{array}{c}{c}_p=\left[\begin{array}{l}\left(0.2196\right)\left(0.248\text{kJ/kg}\cdot \text{K}\right)+\left(0.07026\right)\left(0.219\text{kJ/kg}\cdot \text{K}\right)+\\ \left(0.1581\right)\left(0.445\text{kJ/kg}\cdot \text{K}\right)+\left(0.5521\right)\left(0.203\text{kJ/kg}\cdot \text{K}\right)\end{array}\right]\\ =0.2523\text{Btu/lbm}\cdot \text{R}\end{array}$

Substitute $1.9858\text{Btu/lbmol}\cdot \text{R}$ for $R_u$ and 31.88 lbm/lbmol for $M_m$ in Equation (XII).

$\begin{array}{c}R=\frac{1.9858\text{Btu/lbmol}\cdot \text{R}}{31.88\text{lbm/lbmol}}\\ =0.06229\text{Btu/lbm}\cdot \text{R}\end{array}$

Substitute $0.06229\text{Btu/lbm}\cdot \text{R}$ for R and $0.2523\text{Btu/lbm}\cdot \text{R}$ for $c_p$ in Equation (XIII).

$\begin{array}{c}{c}_v=0.2523\text{Btu/lbm}\cdot \text{R}-0.06229\text{Btu/lbm}\cdot \text{R}\\ =0.1900\text{Btu/lbm}\cdot \text{R}\end{array}$

Substitute $0.2523\text{Btu/lbm}\cdot \text{R}$ for $c_p$ and $0.1900\text{Btu/lbm}\cdot \text{R}$ for $c_v$ in Equation (XIV).

$\begin{array}{c}k=\frac{0.2523\text{Btu/lbm}\cdot \text{R}}{0.1900\text{Btu/lbm}\cdot \text{R}}\\ =1.328\end{array}$

Substitute $0.2523\text{Btu/lbm}\cdot \text{R}$ for $c_p$, $0.2523\text{Btu/lbm}\cdot \text{R}$ for $c_{p,\text{air}}$, $0.1900\text{Btu/lbm}\cdot \text{R}$ for $c_v$, $0.171\text{Btu/lbm}\cdot \text{R}$ for $c_{v,\text{air}}$ in Equations (XV) and (XVI).

$\begin{array}{l}{c}_{p,avg}=0.2462\text{Btu/lbm}\cdot \text{R}\\ c_{v,avg}=0.1805\text{Btu/lbm}\cdot \text{R}\end{array}$

Substitute 1.328 for k and 1.4 for $k_{air}$ in Equation (XVII).

$\begin{array}{c}{k}_{\text{avg}}=\frac{1.328+1.4}{2}\\ =1.364\end{array}$

Substitute 1.4 for $k_{\text{air}}$, 7 for r, and 515 R for $T_1$ in Equation (XVIII).

$\begin{array}{c}{T}_2=\left(\text{515}\text{R}\right){\left(7\right)}^{\frac{1.4-1}{1.4}}\\ =1122\text{R}\end{array}$

Substitute $0.1805\text{Btu/lbm}\cdot \text{R}$ for $c_{v,avg}$, 1122 R for $T_2$, and 2060 R for $T_3$ in Equation (XIX).

$\begin{array}{c}{q}_{\text{in}}=0.1805\text{Btu/lbm}\cdot \text{R}\left(2060\text{R}-1122\text{R}\right)\\ =169.3\text{Btu/lbm}\end{array}$

Substitute 1.328 for $k$, 7 for r, and 2060 R for $T_3$ in Equation (XX).

$\begin{array}{c}{T}_4=\left(2060\text{R}\right){\left(\frac{1}{7}\right)}^{\frac{1.328-1}{1.328}}\\ =1014\text{R}\end{array}$

Substitute $0.1805\text{Btu/lbm}\cdot \text{R}$ for $c_{v,avg}$, 1014 R for $T_4$, and 515 R for $T_1$ in Equation (XXI).

$\begin{array}{c}{q}_{\text{out}}=0.1805\text{Btu/lbm}\cdot \text{R}\left(1014\text{R}-515\text{R}\right)\\ =90.1\text{Btu/lbm}\end{array}$

Substitute $90.1\text{Btu/lbm}$ for $q_{\text{out}}$ and $169.3\text{Btu/lbm}$ for $q_{\text{in}}$ in Equation (XXII).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{90.1\text{Btu/lbm}}{169.3\text{Btu/lbm}}\\ =0.468\end{array}$

Thus, the thermal efficiency of the cycle is $\underset{\_}{0.468}$.