Chapter 8.8, Problem 32P

A well-insulated rigid tank contains 6 lbm of a saturated liquid–vapor mixture of water at 35 psia. Initially, three-quarters of the mass is in the liquid phase. An electric resistance heater placed in the tank is turned on and kept on until all the liquid in the tank is vaporized. Assuming the surroundings to be at 75°F and 14.7 psia, determine (a) the exergy destruction and (b) the second-law efficiency for this process.

To determine

The exergy destruction.

Answer to Problem 32P

The exergy destruction is $\text{2766}\text{Btu}$.

Explanation of Solution

Express the entropy balance for the two constant pressure devices.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ S_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=m\left(s_2-s_1\right)\end{array}$ (I)

Here, net entropy transfer by heat and mass is $S_{\text{in}}-S_{\text{out}}$, entropy generation is $S_{\text{gen}}$, change in entropy is $\Delta S_{\text{system}}$, mass is $m$ and initial and final specific entropy is $s_1\text{and}{s}_2$.

Substitute $0$ for $S_{\text{in}}-S_{\text{out}}$ in Equation (I)

$\begin{array}{l}0+S_{\text{gen}}=m\left(s_2-s_1\right)\\ S_{\text{gen}}=m\left(s_2-s_1\right)\end{array}$ (II)

Express the exergy destruction.

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

Here, surrounding temperature is $T_0$.

Here, final specific volume is $v_2$ and specific volume at saturated liquid and vapor is $v_f\text{and}{v}_g$ respectively.

Express initial specific volume.

$v_1=v_f+x_1\left(v_g-v_f\right)$ (IV)

Here, initial quality is $x_1$, specific volume at saturated vapor and liquid is $v_g\text{and}{v}_f$ respectively.

Express initial specific internal energy.

$u_1=u_f+x_1{u}_{fg}$ (V)

Here, specific internal energy at saturated liquid and evaporation is $u_f\text{and}{u}_{fg}$ respectively.

Express initial specific entropy.

$s_1=s_f+x_1{s}_{fg}$ (VI)

Here, specific entropy at saturated liquid and evaporation is $s_f\text{and}{s}_{fg}$ respectively.

Conclusion:

Express initial quality.

$\begin{array}{c}{x}_1=1-\frac{3}{4}\\ =1-0.75\\ =0.25\end{array}$

Refer Table A-5E, “saturated water-pressure table” and write the properties corresponding to initial pressure $\left(P_1\right)$ of $\text{35}\text{psia}$ and initial quality of $0.25$.

$\begin{array}{l}{v}_f=0.01708{\text{ft}}^{\text{3}}/\text{lbm}\\ v_g=11.901{\text{ft}}^{\text{3}}/\text{lbm}\\ u_f=227.92\text{Btu}/\text{lbm}\\ u_{fg}=862.19\text{Btu}/\text{lbm}\end{array}$

$\begin{array}{l}{s}_f=0.38093\text{Btu}/\text{lbm}\cdot \text{R}\\ s_{fg}=1.30632\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $0.01708{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_f$, $11.901{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_g$ and $0.25$ for $x_1$ in Equation (IV).

$\begin{array}{c}{v}_1=\left(0.01708{\text{ft}}^{\text{3}}/\text{lbm}\right)+0.25\left[\left(11.901-0.01708\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\\ =2.9880{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $0.25$ for $x_1$, $227.92\text{Btu}/\text{lbm}$ for $u_f$ and $862.19\text{Btu}/\text{lbm}$ for $u_{fg}$ in Equation (V).

$\begin{array}{c}{u}_1=227.92\text{Btu}/\text{lbm}+\left(0.25\right)\left(862.19\text{Btu}/\text{lbm}\right)\\ =443.47\text{Btu}/\text{lbm}\end{array}$

Substitute $0.25$ for $x_1$, $0.38093\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_f$ and $1.30632\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_{fg}$ in Equation (VI).

$\begin{array}{c}{s}_1=0.38093\text{Btu}/\text{lbm}\cdot \text{R}+\left(0.25\right)\left(1.30632\text{Btu}/\text{lbm}\cdot \text{R}\right)\\ =0.70751\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

As the specific volumes are constant, take initial and final specific volume as equal.

$\begin{array}{c}{v}_1=v_2\\ =2.9880{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Here, specific final volume is $v_2$.

Refer Table A-5E, “saturated water-pressure table”, obtain the below properties at specific volume at saturated vapor of $\text{2}\text{.9880}{\text{ft}}^{\text{3}}/\text{lbm}$ using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (VII)

Here, the variables denote by x and y are specific volume at saturated vapor and specific internal energy at saturated vapor respectively.

Show final specific internal energy at specific volume of $2.9880{\text{ft}}^{\text{3}}/\text{lbm}$ as in Table (1).

$v_g\left({\text{ft}}^{\text{3}}/\text{lbm}\right)$	$u_2\left(\text{Btu}/\text{lbm}\right)$
	Saturated vapor, $u_g$
3.2202 $\left(x_1\right)$	1109.9 $\left(y_1\right)$
2.9880 $\left(x_2\right)$	$\left(y_2=?\right)$
3.0150 $\left(x_3\right)$	1110.8 $\left(y_3\right)$

Express final specific internal energy using interpolation method.

Substitute $\text{3}\text{.2202}{\text{ft}}^{\text{3}}/\text{lbm}$ for $x_1$, $2.9880{\text{ft}}^{\text{3}}/\text{lbm}$ for $x_2$, $3.0150{\text{ft}}^{\text{3}}/\text{lbm}$ for $x_3$, $\text{1109}\text{.9}\text{Btu}/\text{lbm}$ for $y_1$ and $\text{1110}\text{.8}\text{Btu}/\text{lbm}$ for $y_3$ in Equation (VII).

$\begin{array}{c}{y}_2=\frac{\left[\left(2.9880-\text{3}\text{.2202}\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\left[\left(\text{1110}\text{.8}-\text{1109}\text{.9}\right)\text{Btu}/\text{lbm}\right]}{\left(3.0150-\text{3}\text{.2202}\right){\text{ft}}^{\text{3}}/\text{lbm}}+\text{1109}\text{.9}\text{Btu}/\text{lbm}\\ =1110.9\text{Btu}/\text{lbm}\\ =u_2\end{array}$

From above calculation the final specific internal energy is $1110.9\text{Btu}/\text{lbm}$.

Refer Table A-5E, “saturated water-pressure table”, obtain the below properties at specific volume at saturated vapor of $\text{2}\text{.9880}{\text{ft}}^{\text{3}}/\text{lbm}$ using interpolation method of two variables.

Show final specific entropy at specific volume of $2.9880{\text{ft}}^{\text{3}}/\text{lbm}$ as in Table (2).

$v_g\left({\text{ft}}^{\text{3}}/\text{lbm}\right)$	$s_2\left(\text{Btu}/\text{lbm}\cdot \text{R}\right)$
	Saturated vapor, $s_g$
3.2202 $\left(x_1\right)$	1.5757$\left(y_1\right)$
2.9880 $\left(x_2\right)$	$\left(y_2=?\right)$
3.0150 $\left(x_3\right)$	1.5700$\left(y_3\right)$

Substitute $\text{3}\text{.2202}{\text{ft}}^{\text{3}}/\text{lbm}$ for $x_1$, $2.9880{\text{ft}}^{\text{3}}/\text{lbm}$ for $x_2$, $3.0150{\text{ft}}^{\text{3}}/\text{lbm}$ for $x_3$, $\text{1}\text{.5757}\text{Btu}/\text{lbm}\cdot \text{R}$ for $y_1$ and $\text{1}\text{.5700}\text{Btu}/\text{lbm}\cdot \text{R}$ for $y_3$ in Equation (VII).

$\begin{array}{c}{y}_2=\left\{\begin{array}{c}\frac{\left[\left(2.9880-\text{3}\text{.2202}\right){\text{ft}}^{\text{3}}/\text{lbm}\right]\left[\left(\text{1}\text{.5700}-\text{1}\text{.5757}\right)\text{Btu}/\text{lbm}\cdot \text{R}\right]}{\left(3.0150-\text{3}\text{.2202}\right){\text{ft}}^{\text{3}}/\text{lbm}}\\ +\text{1}\text{.5757}\text{Btu}/\text{lbm}\cdot \text{R}\end{array}\right\}\\ =1.5692\text{Btu}/\text{lbm}\cdot \text{R}\\ =s_2\end{array}$

From above calculation the final specific entropy is $1.5692\text{Btu}/\text{lbm}\cdot \text{R}$.

Substitute $\text{6}\text{lbm}$ for $m$, $1.5692\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_2$ and $0.70751\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_1$ in Equation (II).

$S_{\text{gen}}=\left(\text{6}\text{lbm}\right)\left(1.5692-0.70751\right)\text{Btu}/\text{lbm}\cdot \text{R}$

Substitute $\text{75°F}$ for $T_0$ and $\left(\text{6}\text{lbm}\right)\left(1.5692-0.70751\right)\text{Btu}/\text{lbm}\cdot \text{R}$ for $S_{\text{gen}}$ in Equation (III).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(\text{75°F}\right)\left[\left(\text{6}\text{lbm}\right)\left(1.5692-0.70751\right)\text{Btu}/\text{lbm}\cdot \text{R}\right]\\ =\left(75+460\right)\text{R}\left[\left(\text{6}\text{lbm}\right)\left(1.5692-0.70751\right)\text{Btu}/\text{lbm}\cdot \text{R}\right]\\ =\left(535\text{R}\right)\left[\left(\text{6}\text{lbm}\right)\left(1.5692-0.70751\right)\text{Btu}/\text{lbm}\cdot \text{R}\right]\\ =2766\text{Btu}\end{array}$

Hence, the exergy destruction is $\text{2766}\text{Btu}$.

To determine

The second law efficiency for the process.

Answer to Problem 32P

The second law efficiency for the process is $30.9\%$.

Explanation of Solution

Write the expression for the energy balance equation.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ W_{\text{u,in}}=\Delta U\\ =m\left(u_2-u_1\right)\end{array}$ (VIII)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, the change in the total energy of the system is $\Delta E_{\text{system}}$ and useful work input is $W_{\text{u,in}}$.

Express the reversible work during process.

$W_{\text{rev,in}}=W_{\text{u,in}}-X_{\text{destroyed}}$ (IX)

Express the second law efficiency for the process.

${\eta }_{\text{II}}=\frac{W_{\text{rev,in}}}{W_{\text{u,in}}}\times 100\%$ (X)

Conclusion:

Substitute $\text{6}\text{lbm}$ for $m$, $\text{1110}\text{.9}\text{Btu}/\text{lbm}$ for $u_2$ and $\text{443}\text{.47}\text{Btu}/\text{lbm}$ for $u_1$ in Equation (VIII).

$\begin{array}{c}{W}_{\text{u,in}}=\left(\text{6}\text{lbm}\right)\left(\text{1110}\text{.9}-\text{443}\text{.47}\right)\text{Btu}/\text{lbm}\\ =4005\text{Btu}\end{array}$

Substitute $4005\text{Btu}$ for $W_{\text{u,in}}$ and $2766\text{Btu}$ for $X_{\text{destroyed}}$ in Equation (IX).

$\begin{array}{c}{W}_{\text{rev,in}}=4005\text{Btu}-2766\text{Btu}\\ =\text{1239}\text{Btu}\end{array}$

Substitute $\text{1239}\text{Btu}$ for $W_{\text{rev,in}}$ and $4005\text{Btu}$ for $W_{\text{u,in}}$ in Equation (X).

$\begin{array}{c}{\eta }_{\text{II}}=\frac{\text{1239}\text{Btu}}{\text{4005}\text{Btu}}\times 100\%\\ =0.309\times 100\%\\ =30.9\%\end{array}$

Hence, the second law efficiency for the process is $30.9\%$.