Chapter 4.5, Problem 29P

A 0.5-m³ rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality. Heat is now transferred to the refrigerant until the pressure reaches 700 kPa. Determine (a) the mass of the refrigerant in the tank and (b) the amount of heat transferred. Also, show the process on a P-V diagram with respect to saturation lines.

To determine

The mass of the refrigerant.

Answer to Problem 29P

The mass of the refrigerant is $\underset{\_}{10.03\text{kg}}$.

Explanation of Solution

Write the expression for the energy balance equation.

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

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

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

$\begin{array}{c}{Q}_{\text{in}}=\Delta U=m\left(u_2-u_1\right)\\ =m\left(u_2-u_1\right)\end{array}$ (II)

Here, the mass is $m$, the initial specific internal energy is $u_1$, and the final specific internal energy is $u_2$.

Calculate the specific volume of the refrigerant.

$\begin{array}{c}{v}_1=v_f+x_1\times v_{fg}\\ =v_f+x_1\times \left(v_g-v_f\right)\end{array}$ (III)

Here, the specific volume of saturated liquid is $v_f$, specific volume of saturated vapour is $v_g$, and the dryness fraction is $x_1$.

Calculate the specific internal energy of the refrigerant.

$u_1=u_f+x_1\times u_{fg}$ (IV)

Here, the specific internal energy of saturated liquid is $u_f$, the specific internal energy change upon vaporization is $u_{fg}$, and the dryness fraction is $x_1$.

Write the expression for mass of the refrigerant.

$m=\frac{V_1}{v_1}$ (V)

Here, the initial specific volume is $v_1$ and the volume of tank is $V_1$.

Conclusion:

From the Table A-12, “Saturated refrigerant-134a” to obtain the value of specific volume of saturated liquid, the specific volume of saturated vapour, the specific internal energy of saturated liquid, and the specific internal energy change upon vaporization of the saturated refrigerant 134a at 160 kPa of pressure as $0.0007435{\text{m}}^{\text{3}}/\text{kg}$, $0.12355{\text{m}}^{\text{3}}/\text{kg}$, $31.06\text{kJ}/\text{kg}$, and $190.31\text{kJ}/\text{kg}$.

Substitute $0.0007435{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.12355{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$, and 0.4 for $x_1$ in Equation (III).

$\begin{array}{c}{v}_1=\left(0.0007435{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.4\right)\times \left(0.12355{\text{m}}^{\text{3}}/\text{kg}-0.0007435{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(0.0007435{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.4\right)\times \left(0.122807{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.049866{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $31.06\text{kJ}/\text{kg}$ for $u_f$, $190.31\text{kJ}/\text{kg}$ for $u_{fg}$, and 0.4 for $x_1$ in Equation (IV).

$\begin{array}{c}{u}_1=\left(31.06\text{kJ}/\text{kg}\right)+\left(0.4\right)\times \left(190.31\text{kJ}/\text{kg}\right)\\ =107.184\text{kJ}/\text{kg}\end{array}$

Refer to Table A-13, “Superheated refrigerant-134a”, obtain the below properties at the final specific volume $0.049866{\text{m}}^{\text{3}}/\text{kg}$ using interpolation method of two variables.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y are specific volume and internal energy.

Show the specific volume at $0.047306{\text{m}}^{\text{3}}/\text{kg}$ and $0.048597{\text{m}}^{\text{3}}/\text{kg}$ as in Table (1).

S. No	specific volume, ${\text{m}}^{\text{3}}/\text{kg}$$\left(x\right)$	Specific internal energy , $\text{kJ}/\text{kg}$ $\left(y\right)$
1	$0.047306$	$357.42$
2	$0.049866$	$y_2=?$
3	$0.048597$	367.31

Calculate final specific internal energy of refrigerant at the final specific volume $0.049866{\text{m}}^{\text{3}}/\text{kg}$ for liquid phase using interpolation method.

Substitute $0.047306$ for $x_1$, $0.049866$ for $x_2$, $0.048597$ for $x_3$, $357.42$ for $y_1$, and 367.31 for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left(0.049866{\text{m}}^{\text{3}}/\text{kg}-0.047306{\text{m}}^{\text{3}}/\text{kg}\right)\left(367.31\text{kJ}/\text{kg}-357.42\text{kJ}/\text{kg}\right)}{\left(0.048597{\text{m}}^{\text{3}}/\text{kg}-0.047306{\text{m}}^{\text{3}}/\text{kg}\right)}+\left(357.42\text{kJ}/\text{kg}\right)\\ =377.0315\text{kJ}/\text{kg}\\ \cong 377.032\text{kJ}/\text{kg}\end{array}$

From above calculation the final specific internal energy of refrigerant at the final specific volume $0.049866{\text{m}}^{\text{3}}/\text{kg}$ is $377.032\text{kJ}/\text{kg}$.

Substitute $0.5{\text{m}}^{\text{3}}$ for ${\nu }_1$ and $0.049866{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (V)

$\begin{array}{c}m=\frac{0.5{\text{m}}^{\text{3}}}{0.049866{\text{m}}^{\text{3}}/\text{kg}}\\ =10.0268\text{kg}\\ \cong \text{10}\text{.03}\text{kg}\end{array}$

Thus, the mass of the refrigerant is $\underset{\_}{10.03\text{kg}}$.

To determine

The amount heat transfer to a rigid tank.

Answer to Problem 29P

The amount heat transfer to a rigid tank is $\underset{\_}{2707\text{kJ}}$.

Explanation of Solution

Substitute $377.032\text{kJ}/\text{kg}$ for $u_2$, $107.184\text{kJ}/\text{kg}$ for $u_1$, and $10.03\text{kg}$ for $m$ in Equation (II).

$\begin{array}{c}{Q}_{\text{in}}=\left(10.03\text{kg}\right)\times \left(377.032-107.184\right)\text{kJ}/\text{kg}\\ =2706.57\text{kJ}\\ \cong 2707\text{kJ}\end{array}$

Thus, the amount heat transfer to a rigid tank is $\underset{\_}{2707\text{kJ}}$.