Chapter 7.13, Problem 43P

A 0.5-m³ rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 35°C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.

To determine

The entropy change of the refrigerant.

Answer to Problem 43P

The entropy change of the refrigerant is $3.876\text{kJ}/\text{K}$.

Explanation of Solution

Write the expression to calculate initial specific volume of the refrigerant.

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

Here, initial specific volume is $v_1$, saturated liquid specific volume is $v_f$, initial vapor quality is $x_1$ and saturated vapor specific volume is $v_g$.

Write the expression to calculate the initial internal energy of the refrigerant.

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

Here, initial internal energy is $u_1$, saturated liquid internal energy is $u_f$, initial vapor quality is $x_1$  and evaporated internal energy is $u_{fg}$.

Write the expression to calculate the initial entropy of the refrigerant.

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

Here, initial entropy is $s_1$, saturated liquid entropy is $s_f$, initial vapor quality is $x_1$ and evaporated entropy is $s_{fg}$.

Write the expression to calculate the final specific volume of the refrigerant.

$v_2=v_f+x_2\left(v_g-v_f\right)$ (IV)

Here, final specific volume is $v_2$, saturated liquid specific volume is $v_f$, final vapor quality is $x_2$ and saturated vapor specific volume is $v_g$.

Write the expression to calculate the final internal energy of the refrigerant.

$u_2=u_f+x_2{u}_{fg}$ (V)

Here, final internal energy is $u_2$, saturated liquid internal energy is $u_f$, final vapor quality is $x_2$ and evaporated internal energy is $u_{fg}$.

Write the expression to calculate the final entropy of the refrigerant.

$s_2=s_f+x_2{s}_{fg}$ (VI)

Here, final entropy is $s_2$, saturated liquid entropy is $s_f$, final vapor quality is $x_2$ and evaporated entropy is $s_{fg}$.

Write the expression to calculate the mass of the refrigerant.

$m=\frac{\nu }{v_1}$ (VII)

Here, mass of the refrigerant is m, volume of the tank is $\nu$ and initial specific volume is $v_1$.

Write the expression to calculate the expression for the entropy change of the refrigerant.

$\Delta S_{system}=m\left(s_2-s_1\right)$ (VIII)

Here, entropy change of the refrigerant is $\Delta S_{system}$, mass of the refrigerant is m, initial entropy is $s_1$ and final entropy is $s_2$.

Conclusion:

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

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

Substitute $38.26\text{kJ}/\text{kg}$ for $u_f$, 0.4 for $x_1$ and $186.25\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (II).

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

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

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

Substitute $0.15449\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, 0.4 for $x_1$ and $0.78339\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (III).

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

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

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

Substitute $\text{0}\text{.0007532}{\text{m}}^3/\text{kg}$ for $v_f$, 0.4 for $x_1$ and $\text{0}\text{.099951}{\text{m}}^3/\text{kg}$ for $v_g$ in Equation (I).

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

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

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

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

Substitute $0.04043{\text{m}}^3/\text{kg}$ for $v_2$, $0.0007905{\text{m}}^3/\text{kg}$ for $v_f$ and $0.051266{\text{m}}^3/\text{kg}$ for $v_g$ in Equation (IV).

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

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

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

Substitute $63.61\text{kJ}/\text{kg}$ for $u_f$, 0.7853 for $x_2$ and $171.49\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (V).

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

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

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

Substitute $0.24757\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, 0.7853 for $x_2$ and $0.67954\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (VI).

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

Substitute $0.5{\text{m}}^3$ for $\nu$ and $0.04043{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (VII).

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

Substitute 12.37 kg for m, $0.7813\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $0.4678\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (VIII).

$\begin{array}{c}\Delta S_{system}=12.37kg\left(0.7813\text{kJ}/\text{kg}\cdot \text{K}-0.4678\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =3.876\text{kJ}/\text{K}\end{array}$

Thus, the entropy change of the refrigerant is $3.876\text{kJ}/\text{K}$.

To determine

The entropy change of the heat source

Answer to Problem 43P

The entropy change of the heat source is $-3.434\text{kJ}/\text{K}$.

Explanation of Solution

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

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

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 $\Delta E_{system}$.

Write the expression to calculate the entropy change of the heat source.

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

Here, source temperature is $T_{source}$.

Conclusion:

Substitute $Q_{in}$ for $E_{in}$, 0 for $E_{out}$ and $m\left(u_2-u_1\right)$ in Equation (IX).

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

Here, heat transfer input is $Q_{in}$, mass of the refrigerant is m, final internal energy is $u_2$, and initial internal energy is $u_1$.

Substitute 12.37 kg for m, $198.29\text{kJ}/\text{kg}$ for $u_2$ and $112.76\text{kJ}/\text{kg}$ for $u_1$ in Equation (XI).

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

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.

$Q_{source}=-Q_{in}=-1058\text{kJ}$

Substitute –1058 kJ for $Q_{source}$ and $35°\text{C}$ for $T_{source}$ in Equation (X).

$\begin{array}{c}\Delta S_{source}=\frac{-1058\text{kJ}}{35°\text{C}}\\ =\frac{-1058\text{kJ}}{\left(35+273\right)\text{K}}\\ =-3.434\text{kJ}/\text{K}\end{array}$

The entropy change of the heat source is $-3.434\text{kJ}/\text{K}$.

To determine

The total entropy change during the process.

Answer to Problem 43P

The total entropy change during the process is $0.441\text{kJ}/\text{K}$.

Explanation of Solution

Write the expression for the total entropy change during the process.

$\Delta S_{total}=\Delta S_{system}+\Delta S_{source}$ (XII)

Here, total entropy change during the process is $\Delta S_{total}$.

Conclusion:

Substitute $3.876\text{kJ}/\text{K}$ for $\Delta S_{system}$, and $-3.434\text{kJ}/\text{K}$ for $\Delta S_{source}$ in Equation (XII).

$\begin{array}{c}\Delta S_{total}=3.876\text{kJ}/\text{K}-3.434\text{kJ}/\text{K}\\ =0.441\text{kJ}/\text{K}\end{array}$

Thus, the total entropy change during the process is $0.441\text{kJ}/\text{K}$.