Chapter 9.12, Problem 145P

To determine

The total exergy destruction for each process of an ideal dual cycle

Answer to Problem 145P

The total exergy destruction in an ideal dual cycle is $25.6\text{Btu}/\text{lbm}$.

Explanation of Solution

Draw the ideal dual cycle on $P-v$ diagram.

Consider, the pressure is $P_i$ , the specific volume is $v_i$, the temperature is $T_i$, the entropy is $s_i$, the enthalpy is $h_i$ corresponding to $i^{th}$ state.

Write the expression of temperature and volume relation for the isentropic compression process 1-2.

$\begin{array}{c}{T}_2=T_1{\left(\frac{v_1}{v_2}\right)}^{k-1}\\ =T_1{\left(r\right)}^{k-1}\end{array}$ (I)

Here, compression ratio is r and specific heat ratio is k.

Write the expression of pressure, and volume relation for the isentropic expansion process 1-2.

$\begin{array}{c}{P}_2=P_1{\left(\frac{v_1}{v_2}\right)}^k\\ =P_1{r}^k\end{array}$ (II)

Write the expression of pressure ratio relation.

$\begin{array}{c}{P}_x=P_3\\ =r_p{P}_2\end{array}$ (III)

Here, pressure ratio is $r_p$.

Write the expression of temperature, and pressure relation for the constant volume heat addition process $2-x$.

$T_x=T_2\left(\frac{P_x}{P_2}\right)$ (IV)

Write the expression of temperature, and volume relation for the constant pressure heat addition process $x-3$.

$\begin{array}{c}{T}_3=T_x\left(\frac{v_3}{v_x}\right)\\ =T_x{r}_c\end{array}$ (V)

Here, cutoff ratio is $r_c$.

Write the expression of temperature, and volume relation for the constant pressure heat addition process 3-4.

$\begin{array}{c}{T}_4=T_3{\left(\frac{v_3}{v_4}\right)}^{k-1}\\ =T_3{\left(\frac{r_c}{r}\right)}^{k-1}\end{array}$ (VI)

Write the expression to calculate the heat added to the cycle during process $4-1$.

$q_{out}=c_v\left(T_4-T_1\right)$ (VII)

Write the expression to calculate the heat added to the cycle during process $2-x$.

$q_{2-x}=c_v\left(T_x-T_2\right)$ (VIII)

Here, heat input to the process $2-x$ is $q_{2-x}$.

Write the expression to calculate the heat added to the cycle during process $x-3$.

$q_{x-3}=c_p\left(T_3-T_x\right)$ (IX)

Here, specific heat at constant pressure is $c_p$ and the work done by the process $x-3$ is $q_{x-3}$.

Write the expression of net heat addition to the cycle $\left(q_{\text{in,net}}\right)$.

$q_{\text{in,net}}=q_{2-x}+q_{x-3}$ (X).

Write the expression for exergy destruction during the process of the cycle.

$\begin{array}{c}{x}_{\text{dest}}=T_0{s}_{gen}\\ =T_0\left(\Delta s-\frac{q_{\text{in}}}{T_{\text{source}}}+\frac{q_{\text{out}}}{T_{\text{sink}}}\right)\end{array}$

Here, temperature of the surroundings is $T_0$, and the temperature of the heat source is $T_{\text{source}}$.

Write the expression of entropy change for the process $2-x$ of an ideal dual cycle $\left(s_x-s_2\right)$.

$s_x-s_2=c_v\ln \frac{T_x}{T_1}+R\ln \frac{v_x}{v_1}$ (XI)

Here, specific heat of air at constant volume is $c_v$.

Write the expression for the exergy loss for the isothermal process $2-x$$\left(x_{\text{dest,}2-x}\right)$.

$x_{\text{dest,}2-x}=T_0\left(s_x-s_2-\frac{q_{i\text{n},2-x}}{T_{\text{source}}}\right)$ (XII)

Write the expression of entropy change for the isothermal process $x-3$ of an ideal dual cycle $\left(s_3-s_x\right)$.

$s_3-s_x=c_p\ln \frac{T_3}{T_x}+R\ln \frac{P_3}{P_x}$ (XIII)

Write the expression for the exergy loss for the process $x-3$$\left(x_{\text{dest,}x-3}\right)$.

$x_{\text{dest,}x-3}=T_0\left(s_3-s_x-\frac{q_{i\text{n},x-3}}{T_{\text{source}}}\right)$ (XIV)

Write the expression of entropy change for the process $4-1$ of an ideal dual cycle $\left(s_1-s_4\right)$.

$s_1-s_4=c_v\ln \frac{T_1}{T_4}+R\ln \frac{v_1}{v_4}$ (XV)

Write the expression for the exergy loss for the process $4-1$ $\left(x_{\text{dest,}4-1}\right)$.

$x_{\text{dest,}4-1}=T_0\left(s_1-s_4+\frac{q_{\text{out}}}{T_{\text{sink}}}\right)$ (XVI)

Here, temperature of the sink is $T_{\text{sink}}$.

Write the expression to calculate the total in an ideal dual cycle.

$x_{\text{dest,total}}=x_{\text{dest,1-2}}+x_{\text{dest,}2-x}+x_{\text{dest,x}-3}+x_{\text{dest,3-4}}+x_{\text{dest,}4-1}$ (XVII)

Conclusion:

From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the following properties of air at room temperature.

$\begin{array}{l}k=1.4\\ c_v=0.171\text{Btu}/\text{lbm}\cdot \text{R}\\ c_p=0.240\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

From Table A-2Ea, “Ideal-gas specific heats of various common gases”, obtain the value for gas content $\left(R\right)$ as $0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for air at room temperature.

Substitute $530\text{R}$ for $T_1$, $20$ for r, and 1.4 for k in Equation (I).

$\begin{array}{c}{T}_2=\left(530\text{R}\right){\left(20\right)}^{1.4-1}\\ =1757\text{R}\end{array}$

Substitute $14\text{psia}$ for $P_1$, $20$ for r and 1.4 for kin Equation (II).

$\begin{array}{c}{P}_2=\left(14\text{psia}\right){\left(20\right)}^{1.4}\\ =928\text{psia}\end{array}$

Substitute $1.2$ for $r_p$ and $928\text{psia}$ for $P_2$ in Equation (III).

$\begin{array}{c}{P}_x=\left(1.2\right)\left(928\text{psia}\right)\\ =1114\text{psia}\end{array}$

Substitute $1757\text{R}$ for $T_2$, $1114\text{psia}$ for $P_x$, and $928\text{psia}$ for $P_2$ in Equation (IV).

$\begin{array}{c}{T}_x=\left(1757\text{R}\right)\left(\frac{1114\text{psia}}{928\text{psia}}\right)\\ =2109\text{R}\end{array}$

Substitute $2109\text{R}$ for $T_x$ and 1.4 for $r_c$ in Equation (V).

$\begin{array}{c}{T}_3=\left(2109\text{R}\right)\left(1.3\right)\\ =2742\text{R}\end{array}$

Substitute $2742\text{R}$ for $T_3$, $20$ for r and $1.3$ for $r_c$ in Equation (VI).

$\begin{array}{c}{T}_4=\left(2742\text{R}\right){\left(\frac{1.3}{20}\right)}^{1.4-1}\\ =918.8\text{R}\end{array}$

Substitute $0.171\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$, $1757\text{R}$ for $T_2$ and $2109\text{R}$ for $T_x$ in Equation (VIII).

$\begin{array}{c}{q}_{2-x}=\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2109\text{R}-1757\text{R}\right)\\ =\text{60}\text{.19}\text{Btu}/\text{lbm}\end{array}$

Substitute $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$, $2742\text{R}$ for $T_3$ and $2109\text{R}$ for $T_x$ in Equation (IX).

$\begin{array}{c}{q}_{x-3}=\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2742\text{R}-2109\text{R}\right)\\ =\text{157}\text{.9}\text{Btu}/\text{lbm}\end{array}$

Substitute $\text{60}\text{.19}\text{Btu}/\text{lbm}$ for $q_{2-x}$ and $\text{157}\text{.9}\text{Btu}/\text{lbm}$ for $q_{x-3}$ in Equation (X).

$\begin{array}{c}{q}_{\text{in,net}}=\text{60}\text{.19}\text{Btu}/\text{lbm}+\text{157}\text{.9}\text{Btu}/\text{lbm}\\ =212.1\text{Btu}/\text{lbm}\end{array}$

Substitute $0.171\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$ , $918.8\text{K}$ for $T_4$ and $530\text{R}$ for $T_1$ in

Equation (VII).

$\begin{array}{c}{q}_{out}=0.171\text{Btu}/\text{lbm}\cdot \text{R}\left(918.8\text{K}-530\text{R}\right)\\ =66.48\text{Btu}/\text{lbm}\end{array}$

The exergy loss for the isothermal process 1-2 $\left(x_{\text{dest,1-2}}\right)$ is equal to zero.

$x_{\text{dest,1-2}}=0$.

Here $v_x=v_2$

Substitute $v_x$ for $v_2$ , $0.171\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$ , $1757\text{R}$ for $T_2$ and $2109\text{R}$ for $T_x$ in Equation (XI).

$\begin{array}{c}{s}_x-s_2=\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{2109\text{R}}{1757\text{R}}+R\ln \frac{v_x}{v_x}\\ =\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{2109\text{R}}{1757\text{R}}+R\ln \left(1\right)\\ =0.03123\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $530\text{R}$ for $T_0$ , $0.03123\text{Btu}/\text{lbm}\cdot \text{R}$ for $\left(s_x-s_2\right)$ , $60.19\text{Btu}/\text{lbm}$ for $q_{i\text{n},2-x}$ and $2742\text{R}$ for $T_{\text{source}}$ in Equation (XII).

$\begin{array}{c}{x}_{\text{dest,}2-x}=530\text{R}\left(\left(0.03123\text{Btu}/\text{lbm}\cdot \text{R}\right)-\frac{60.19\text{Btu}/\text{lbm}}{2742\text{R}}\right)\\ =4.917\text{Btu}/\text{lbm}\end{array}$

Substitute $P_3$ for $P_x$ , $0.240\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ , $2742\text{R}$ for $T_3$ and $2109\text{R}$ for $T_x$ in Equation (XIII).

$\begin{array}{c}{s}_3-s_x=\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{2742\text{R}}{2109\text{R}}+R\ln \frac{P_3}{P_3}\\ =0.06299\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $530\text{R}$ for $T_0$ , $0.06299\text{Btu}/\text{lbm}\cdot \text{R}$ for $\left(s_3-s_x\right)$ , $151.9\text{Btu}/\text{lbm}$ for $q_{i\text{n},x-3}$ and $2742\text{R}$ for $T_{\text{source}}$ in Equation (XIV).

$\begin{array}{c}{x}_{\text{dest,x}-3}=530\text{R}\left(\left(0.06299\text{Btu}/\text{lbm}\cdot \text{R}\right)-\frac{151.9\text{Btu}/\text{lbm}}{2742\text{R}}\right)\\ =4.024\text{Btu}/\text{lbm}\end{array}$

The exergy loss for the isothermal process 3-4 $\left(x_{\text{dest,3-4}}\right)$ is equal to zero.

$x_{\text{dest,3-4}}=0$

Here, $v_1=v_4$

Substitute $v_1$ for $v_4$ , $0.171\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_v$ , $530\text{R}$ for $T_1$ and $918.8\text{R}$ for $T_4$ in Equation (XV).

$\begin{array}{c}{s}_1-s_4=\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{530\text{R}}{918.8\text{R}}+R\ln \frac{v_1}{v_1}\\ =\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{530\text{R}}{918.8\text{R}}+R\ln \left(1\right)\\ =-0.09408\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $530\text{R}$ for $T_0$ , $-0.09408\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_1-s_4$, $66.48\text{Btu}/\text{lbm}$ for $q_{out}$ and $530\text{R}$ for $T_{\text{sink}}$ in Equation (XVI).

$\begin{array}{c}{x}_{\text{dest,}4-1}=530\text{R}\left(\left(-0.09408\text{Btu}/\text{lbm}\cdot \text{R}\right)+\frac{66.48\text{Btu}/\text{lbm}}{530\text{R}}\right)\\ =16.62\text{Btu}/\text{lbm}\end{array}$

During heat rejection process the largest exergy destruction in an ideal dual cycle occur.

Substitute $0$ for $x_{\text{dest,1-2}}$ , $0$ for $x_{\text{dest,3-4}}$, $4.917\text{Btu}/\text{lbm}$ for $x_{\text{dest,}2-x}$, $4.024\text{Btu}/\text{lbm}$ for $x_{\text{dest,x}-3}$ and $16.62\text{Btu}/\text{lbm}$ for $x_{\text{dest,}4-1}$ in Equation (XVII).

$\begin{array}{c}{x}_{\text{dest,total}}=0+4.917\text{Btu}/\text{lbm}+4.024\text{Btu}/\text{lbm}+0+16.62\text{Btu}/\text{lbm}\\ =25.6\text{Btu}/\text{lbm}\end{array}$

Thus, the total exergy destruction in an ideal dual cycle is $25.6\text{Btu}/\text{lbm}$.