Chapter 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

Explanation of Solution

Given:

The initial pressure $\left(P_1\right)$ is 200 kPa.

The initial quality of the steam is $40\%$.

The final pressure $\left(P_2\right)$ is $400\text{ kPa}$.

The volume of the tank $\left(\nu \right)$ is $0.5{\text{m}}^3$.

The source temperature $\left(T_{source}\right)$ is $35°\text{C}$.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of $\text{200}\text{kPa}$.

  $\begin{array}{l}{u}_f=38.26\text{kJ}/\text{kg}\\ u_{fg}=186.25\text{kJ}/\text{kg}\end{array}$

  $\begin{array}{l}{s}_f=0.15449\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=0.78339\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

  $\begin{array}{l}{v}_f=\text{0}\text{.0007532}{\text{m}}^3/\text{kg}\\ v_g=\text{0}\text{.099951}{\text{m}}^3/\text{kg}\end{array}$

Calculate the initial internal energy of the refrigerant $\left(u_1\right)$.

  $u_1=u_f+x_1{u}_{fg}$

  $\begin{array}{c}{u}_1=38.26\text{kJ}/\text{kg}+\left(0.4\right)\left(186.25\text{kJ}/\text{kg}\right)\\ =112.76\text{kJ}/\text{kg}\end{array}$

Calculate the initial entropy of the refrigerant $\left(s_1\right)$ .

  $s_1=s_f+x_1{s}_{fg}$

  $\begin{array}{c}{s}_1=0.15449\text{kJ}/\text{kg}\cdot \text{K}+\left(0.4\right)\left(0.78339\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.4678\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the initial specific volume of the refrigerant $\left(v_1\right)$.

  $v_1=v_f+x_1\left(v_g-v_f\right)$

  $\begin{array}{c}{v}_1=\text{0}\text{.0007532}{\text{m}}^3/\text{kg}+\left(0.4\right)\left(\text{0}\text{.099951}{\text{m}}^3/\text{kg}-\text{0}\text{.0007532}{\text{m}}^3/\text{kg}\right)\\ =0.04043{\text{m}}^3/\text{kg}\end{array}$

Specific volume remains constant for a rigid tank $\left(v_2=v_1\right)$.

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of $\text{400}\text{kPa}$.

  $\begin{array}{l}{v}_f=0.0007905{\text{m}}^3/\text{kg}\\ v_g=0.051266{\text{m}}^3/\text{kg}\end{array}$

  $\begin{array}{l}{u}_f=63.61\text{kJ}/\text{kg}\\ u_{fg}=171.49\text{kJ}/\text{kg}\end{array}$

  $\begin{array}{l}{s}_f=0.24757\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=0.67954\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the final quality of the refrigerant using the specific volume.

  $v_2=v_f+x_2\left(v_g-v_f\right)$

  $\begin{array}{l}0.04043{\text{m}}^3/\text{kg}=\left[\begin{array}{l}0.0007905{\text{m}}^3/\text{kg}\\ +x_2\left(0.051266{\text{m}}^3/\text{kg}-0.0007905{\text{m}}^3/\text{kg}\right)\end{array}\right]\\ x_2=\frac{0.04043-0.0007905}{0.051266-0.0007905}\\ =0.7853\end{array}$

Calculate the final internal energy of the refrigerant $\left(u_2\right)$ .

  $\begin{array}{c}{u}_2=u_f+x_2{u}_{fg}\\ u_2=63.61\text{kJ}/\text{kg}+\left(0.7853\right)\left(171.49\text{kJ}/\text{kg}\right)\\ =198.29\text{kJ}/\text{kg}\end{array}$

Calculate the final entropy of the refrigerant $\left(s_2\right)$.

  $\begin{array}{c}{s}_2=s_f+x_2{s}_{fg}\\ s_2=0.24757\text{kJ}/\text{kg}\cdot \text{K}+\left(0.7853\right)\left(0.67954\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =0.7813\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the mass of the refrigerant $\left(m\right)$.

  $\begin{array}{c}m=\frac{\nu }{v_1}\\ m=\frac{0.5{\text{m}}^3}{0.04043{\text{m}}^3/\text{kg}}\\ =12.37\text{kg}\end{array}$

Calculate the expression for the entropy change of the refrigerant $\left(\Delta S_{system}\right)$.

  $\begin{array}{c}\Delta S_{system}=m\left(s_2-s_1\right)\\ \Delta S_{system}=12.37kg\left(0.7813\text{kJ}/\text{kg}\cdot \text{K}-0.4678\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =3.878\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the refrigerant is $3.878\text{kJ}/\text{K}$.

b)

To determine

The entropy change of the heat source

Explanation of Solution

Write the expression for the energy balance equation for closed system.

  $E_{in}-E_{out}=\Delta E_{system}$        (I)

Here, energy transfer into the control volume is $E_{in}$, energy transfer exit from the control volume is $E_{out}$ and change in internal energy of system is $\Delta E_{system}$.

Substitute $E_{in}=Q_{in}$, $E_{out}=0$ and $\Delta E_{system}=m\left(u_2-u_1\right)$ in Equation (I).

  $\begin{array}{l}{Q}_{in}-0=m\left(u_2-u_1\right)\\ Q_{in}=m\left(u_2-u_1\right)\end{array}$

  $\begin{array}{c}{Q}_{in}=\left(12.37\text{kg}\right)\left(198.29\text{kJ}/\text{kg}-112.76\text{kJ}/\text{kg}\right)\\ =1058\text{kJ}\end{array}$

Heat transfer for the source $\left(Q_{source}\right)$ is equal to heat transfer input $\left(Q_{in}\right)$. The heat transfer for the source $\left(Q_{source}\right)$ is equal in magnitude but opposite in direction.

  $Q_{source}=-Q_{in}=-1058\text{kJ}$

Calculate the entropy change of the heat source.

  $\Delta S_{source}=\frac{Q_{source}}{T_{source}}$        (X)

  $\begin{array}{c}\Delta S_{source}=\frac{-1058\text{kJ}}{35°\text{C}}\\ =\frac{-1058\text{kJ}}{\left(35+273\right)\text{K}}\\ =-3.435\text{kJ}/\text{K}\end{array}$

The entropy change of the heat source is $-3.435\text{kJ}/\text{K}$.

c)

To determine

The total entropy change during the process.

Explanation of Solution

Calculate the total entropy change during the process.

  $\begin{array}{c}\Delta S_{total}=\Delta S_{system}+\Delta S_{source}\\ \Delta S_{total}=3.876\text{kJ}/\text{K}-3.434\text{kJ}/\text{K}\\ =0.442\text{kJ}/\text{K}\end{array}$

Thus, the total entropy change during the process is $0.442\text{kJ}/\text{K}$.