Chapter 8.8, Problem 47P

A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the work done, (b) the heat transfer, (c) the exergy destroyed, and (d) the second-law efficiency of this process.

FIGURE P8–48

To determine

The work done.

Answer to Problem 47P

The work done is $\text{0}\text{.497}\text{kJ}$.

Explanation of Solution

Express the boundary work done.

$W_{\text{b,out}}=m{P}_2\left({\nu }_2-{\nu }_1\right)$ (I)

Here, mass is $m$, final pressure is $P_2$ and initial and final velocity is ${\nu }_1\text{and}{\nu }_2$ respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from $\text{kPa}\text{to}\text{MPa}$.

$\begin{array}{c}{P}_1=100\text{kPa}\\ =100\text{kPa}\left[\frac{\text{MPa}}{\text{1000}\text{kPa}}\right]\\ =0.1\text{MPa}\end{array}$

$\begin{array}{c}{P}_2=120\text{kPa}\\ =120\text{kPa}\left[\frac{\text{MPa}}{\text{1000}\text{kPa}}\right]\\ =0.12\text{MPa}\end{array}$

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of $\text{0}\text{.1}\text{MPa}$ and initial temperature $\left(T_1\right)$ of $\text{20°C}$.

$\begin{array}{l}{v}_1=0.23373{\text{m}}^{\text{3}}/\text{kg}\\ u_1=248.81\text{kJ}/\text{kg}\\ s_1=1.0919\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, initial specific volume, internal energy and entropy is $v_1,u_1\text{and}{s}_1$ respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of $\text{0}\text{.12}\text{MPa}$ and final temperature $\left(T_2\right)$ of $\text{80°C}$ using interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at $\text{0}\text{.10}\text{MPa}\text{and}\text{0}\text{.14}\text{MPa}$ as in Table (1).

Final pressure $P_2\left(\text{MPa}\right)$	Final specific volume $v_2\left({\text{m}}^{\text{3}}/\text{kg}\right)$
0.10 $\left(x_1\right)$	0.28465 $\left(y_1\right)$
0.12 $\left(x_2\right)$	$\left(y_2=?\right)$
0.14 $\left(x_3\right)$	0.20242 $\left(y_3\right)$

Substitute $\text{0}\text{.10}\text{MPa,}\text{0}\text{.12}\text{MPa}\text{and}\text{0}\text{.14}\text{MPa}$ for $x_1,x_2\text{and}{x}_3$ respectively, $0.28465{\text{m}}^3/\text{kg}$ for $y_1$ and $0.20242{\text{m}}^3/\text{kg}$ for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(\text{0}\text{.12}-\text{0}\text{.10}\right)\text{MPa}\left(0.20242-0.28465\right){\text{m}}^3/\text{kg}}{\left(\text{0}\text{.14}-\text{0}\text{.10}\right)\text{MPa}}+0.28465{\text{m}}^3/\text{kg}\\ =0.23669{\text{m}}^3/\text{kg}\\ =v_2\end{array}$

Show the final specific internal energy at $\text{0}\text{.10}\text{MPa}\text{and}\text{0}\text{.14}\text{MPa}$ as in Table (2).

Final pressure $P_2\left(\text{MPa}\right)$	Final specific internal energy $u_2\left(\text{kJ}/\text{kg}\right)$
0.10 $\left(x_1\right)$	297.10 $\left(y_1\right)$
0.12 $\left(x_2\right)$	$\left(y_2=?\right)$
0.14 $\left(x_3\right)$	296.77 $\left(y_3\right)$

Substitute $\text{0}\text{.10}\text{MPa,}\text{0}\text{.12}\text{MPa}\text{and}\text{0}\text{.14}\text{MPa}$ for $x_1,x_2\text{and}{x}_3$ respectively, $\text{297}\text{.10}\text{kJ}/\text{kg}$ for $y_1$ and $\text{296}\text{.77}\text{kJ}/\text{kg}$ for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(\text{0}\text{.12}-\text{0}\text{.10}\right)\text{MPa}\left(296.77-297.10\right)\text{kJ}/\text{kg}}{\left(\text{0}\text{.14}-\text{0}\text{.10}\right)\text{MPa}}+\text{297}\text{.10}\text{kJ}/\text{kg}\\ =296.94\text{kJ}/\text{kg}\\ =u_2\end{array}$

Show the final specific entropy at $\text{0}\text{.10}\text{MPa}\text{and}\text{0}\text{.14}\text{MPa}$ as in Table (3).

Final pressure $P_2\left(\text{MPa}\right)$	Final specific entropy $s_2\left(\text{kJ}/\text{kg}\cdot \text{K}\right)$
0.10 $\left(x_1\right)$	1.2573 $\left(y_1\right)$
0.12 $\left(x_2\right)$	$\left(y_2=?\right)$
0.14 $\left(x_3\right)$	1.2289 $\left(y_3\right)$

Substitute $\text{0}\text{.10}\text{MPa,}\text{0}\text{.12}\text{MPa}\text{and}\text{0}\text{.14}\text{MPa}$ for $x_1,x_2\text{and}{x}_3$ respectively, $\text{1}\text{.2573}\text{kJ}/\text{kg}\cdot \text{K}$ for $y_1$ and $\text{1}\text{.2289}\text{kJ}/\text{kg}\cdot \text{K}$ for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(\text{0}\text{.12}-\text{0}\text{.10}\right)\text{MPa}\left(\text{1}\text{.2289}-\text{1}\text{.2573}\right)\text{kJ}/\text{kg}\cdot \text{K}}{\left(\text{0}\text{.14}-\text{0}\text{.10}\right)\text{MPa}}+\text{1}\text{.2573}\text{kJ}/\text{kg}\cdot \text{K}\\ =1.2419\text{kJ}/\text{kg}\cdot \text{K}\\ =s_2\end{array}$

Thus, write the values obtained from interpolation method:

$\begin{array}{l}{v}_2=0.23669{\text{m}}^{\text{3}}/\text{kg}\\ u_2=296.94\text{kJ}/\text{kg}\\ s_2=1.2419\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $\text{1}\text{.4}\text{kg}$ for $m$, $\text{120}\text{kPa}$ for $P_2$, $0.23669{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$ and $0.23373{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (I).

$\begin{array}{c}{W}_{\text{b,out}}=\left(\text{1}\text{.4}\text{kg}\right)\left(\text{120}\text{kPa}\right)\left(0.23669-0.23373\right){\text{m}}^{\text{3}}/\text{kg}\\ =0.497\text{kJ}\end{array}$

Hence, the work done is $\text{0}\text{.497}\text{kJ}$.

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is $\text{67}\text{.9}\text{kJ}$.

Explanation of Solution

Express heat transfer.

$Q_{\text{in}}=m\left(u_2-u_1\right)+W_{\text{b,out}}$ (III)

Conclusion:

Substitute  for , $\text{0}\text{.497}\text{kJ}$ for $W_{\text{b,out}}$, $296.94\text{kJ}/\text{kg}$ for $u_2$ and $248.81\text{kJ}/\text{kg}$ for $u_1$ in Equation (III).

$\begin{array}{c}{Q}_{\text{in}}=\left(\text{1}\text{.4}\text{kg}\right)\left(296.94\text{kJ}/\text{kg}-248.81\text{kJ}/\text{kg}\right)+\text{0}\text{.497}\text{kJ}\\ =67.9\text{kJ}\end{array}$

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is $\text{14}\text{.8}\text{kJ}$.

Explanation of Solution

Express the exergy destruction.

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

Here, entropy generation is $S_{\text{gen}}$ and surrounding temperature is $T_0$.

Express the entropy generation by taking entropy balance on an extended system.

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

Here, net entropy transfer by heat and mass is $S_{\text{in}}-S_{\text{out}}$, entropy generation is $S_{\text{gen}}$, change in entropy of system is $\Delta S_{\text{system}}$ and source temperature is $T_{\text{source}}$.

Conclusion:

Substitute $\text{1}\text{.4}\text{kg}$ for $m$, $\text{1}\text{.2419}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $\text{1}\text{.0919}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $\text{67}\text{.9}\text{kJ}$ for $Q_{\text{in}}$ and $\text{150°C}$ for $T_{\text{source}}$ in Equation (V).

$\begin{array}{c}{S}_{\text{gen}}=\left(\text{1}\text{.4}\text{kg}\right)\left(\text{1}\text{.2419}-1.0919\right)\text{kJ}/\text{kg}\cdot \text{K}-\frac{\text{67}\text{.9}\text{kJ}}{\text{150°C}}\\ =\left(\text{1}\text{.4}\text{kg}\right)\left(0.15\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{67}\text{.9}\text{kJ}}{\left(\text{150}+273\right)\text{K}}\\ =\left(\text{1}\text{.4}\text{kg}\right)\left(0.15\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{67}\text{.9}\text{kJ}}{423\text{K}}\end{array}$

Substitute $\left(\text{1}\text{.4}\text{kg}\right)\left(0.15\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{67}\text{.9}\text{kJ}}{423\text{K}}$ for $S_{\text{gen}}$ and $\text{25°C}$ for $T_0$ in Equation (IV).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(\text{25°C}\right)\left[\left(\text{1}\text{.4}\text{kg}\right)\left(0.15\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{67}\text{.9}\text{kJ}}{423\text{K}}\right]\\ =\left(25+273\right)\text{K}\left[\left(\text{1}\text{.4}\text{kg}\right)\left(0.15\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{67}\text{.9}\text{kJ}}{423\text{K}}\right]\\ =\left(298\text{K}\right)\left[\left(\text{1}\text{.4}\text{kg}\right)\left(0.15\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{67}\text{.9}\text{kJ}}{423\text{K}}\right]\\ =14.8\text{kJ}\end{array}$

Hence, the exergy destroyed is $\text{14}\text{.8}\text{kJ}$.

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is $\text{26}\text{.2\%}$.

Explanation of Solution

Express exergy expended.

$\begin{array}{c}{X}_{\text{expended}}={\eta }_{\text{th,rev}}{Q}_{\text{in}}\\ =\left[1-\frac{T_L}{T_H}\right]Q_{\text{in}}\end{array}$ (VI)

Here, efficiency for reversible cycle is ${\eta }_{\text{th,rev}}$, temperature of low source is $T_L$ and temperature of high source is $T_H$.

Express the second law efficiency.

${\eta }_{\text{II}}=1-\frac{X_{\text{destroyed}}}{X_{\text{expended}}}\times 100\%$ (VII)

Conclusion:

Substitute $\text{25°C}$ for $T_L$ and $\text{150°C}$ for $T_H$ and $\text{67}\text{.9}\text{kJ}$ for $Q_{\text{in}}$ in Equation (VI).

$\begin{array}{c}{X}_{\text{expended}}=\left[1-\frac{\text{25°C}}{\text{150°C}}\right]\left(67.9\text{kJ}\right)\\ =\left[1-\frac{\left(\text{25}+273\right)\text{K}}{\left(\text{150}+273\right)\text{K}}\right]\left(67.9\text{kJ}\right)\\ =\left[1-\frac{298\text{K}}{423\text{K}}\right]\left(67.9\text{kJ}\right)\\ =20.06\text{kJ}\end{array}$

Substitute $20.06\text{kJ}$ for $X_{\text{expended}}$ and $14.8\text{kJ}$ for $X_{\text{destroyed}}$ in Equation (VII).

$\begin{array}{c}{\eta }_{\text{II}}=1-\frac{14.8\text{kJ}}{20.06\text{kJ}}\times 100\%\\ =0.262\times 100\%\\ =26.2\%\end{array}$

Hence, the second law efficiency of the process is $\text{26}\text{.2\%}$.