Chapter 3.8, Problem 42P

10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

FIGURE P3–42

To determine

The temperature and total enthalpy at initial and final states in the container.

The temperature and total enthalpy at initial and final states when the heating is completed.

Answer to Problem 42P

The temperature and total enthalpy at initial and final states in the container are $\underset{\_}{0.61°\text{C}}$ and $\underset{\_}{\text{545}\text{.6}\text{kJ}}$ respectively.

The temperature and total enthalpy at initial and final states when the heating is completed are are $\underset{\_}{21.55°\text{C}}$ and $\underset{\_}{846.4\text{kJ}}$ respectively.

Explanation of Solution

Since the process is a constant volume, calculate the specific volume.

$\begin{array}{c}{v}_1=v_2\\ =\frac{\nu }{m}\end{array}$ (I)

Here, specific volume at states 1 and 2 are $v_1\text{and}{v}_2$ respectively, volume of a rigid container is $\nu$, and mass of a refrigerant R-134a is m.

The initial state represents the mixture, so temperature to be considered as the saturation temperature at given pressure.

Calculate the dryness fraction at state 1.

$x_1=\frac{v_1-v_f}{v_g-v_f}$ (II)

Here, specific volume at state 1 is $v_1$, specific volume at saturated water is $v_f$, and specific volume at saturated vapor is $v_g$.

Calculate the specific enthalpy at state 1.

$h_1=h_f+x_1\left(h_{fg}\right)$ (III)

Here, specific enthalpy at saturated liquid is $h_f$ and specific enthalpy at evaporation is $h_{fg}$.

Calculate the total enthalpy at state 1.

$H_1=m{h}_1$ (IV)

Here, mass of the refrigerant R-134a is m.

Similarly, calculate for $x_2$, $h_2$, and $H_2$ respectively.

$x_2=\frac{v_2-v_f}{v_g-v_f}$ (V)

Here, dryness fraction at state 2 is $x_2$ and specific volume at state 2 is $v_2$.

Calculate the specific enthalpy at state 2.

$h_2=h_f+x_2\left(h_{fg}\right)$ (VI)

Calculate the total enthalpy at state 2.

$H_2=m{h}_2$ (VII)

Conclusion:

Substitute 14 L for $\nu$ and 10 kg for m in Equation (I).

$\begin{array}{c}{v}_1=\frac{14\text{L}}{10\text{kg}}\\ =\frac{14\text{L}}{10\text{kg}}\left(\frac{1{\text{m}}^{\text{3}}}{1000\text{L}}\right)\\ =0.0014{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y are pressure and temperature.

Refer to Table A-12, obtain the values of below variables as in Table (I).

Pressure, kPa $\left(x\right)$	Temperature, $°\text{C}$ $\left(y\right)$
280	–1.25
300	?
320	2.46

Substitute 280 for $x_1$, 300 for $x_2$, 320 for $x_3$, –1.25 for $y_1$, and 2.46 for $y_3$ in Equation (VIII).

$\begin{array}{c}{y}_2=\frac{\left(300-280\right)\left(2.46-\left(-1.25\right)\right)}{\left(320-280\right)}+\left(-1.25\right)\\ =0.605\end{array}$

Thus, the value obtained for saturation temperature at the given pressure is $0.605°\text{C}$.

$\begin{array}{c}{T}_1=0.605°\text{C}\\ \simeq \text{0}\text{.65}°\text{C}\end{array}$

Similarly, calculate the values of $v_f$, $v_g$, $h_f$, and $h_{fg}$ using interpolation method at pressure of 300 kPa as $0.0007735{\text{m}}^{\text{3}}/\text{kg}$, $0.067776{\text{m}}^{\text{3}}/\text{kg}$, $52.71\text{kJ/kg}$, and 198.17 kJ/kg respectively.

Substitute $0.0014{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$, $0.0007735{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, and $0.067776{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (II).

$\begin{array}{c}{x}_1=\frac{0.0014{\text{m}}^{\text{3}}/\text{kg}-0.0007735{\text{m}}^{\text{3}}/\text{kg}}{0.067776{\text{m}}^{\text{3}}/\text{kg}-0.0007735{\text{m}}^{\text{3}}/\text{kg}}\\ =0.009351\end{array}$

Substitute $52.71\text{kJ/kg}$ for $h_f$, 0.009351 for $x_1$, and 198.17 kJ/kg for $h_{fg}$ in Equation (III).

$\begin{array}{c}{h}_1=52.71\text{kJ/kg}+\left(0.009351\right)\left(198.17\text{kJ/kg}\right)\\ =54.56\text{kJ/kg}\end{array}$

Substitute 10 kg for m and 54.56 kJ/kg for $h_1$ in equation (IV).

$\begin{array}{c}{H}_1=10\text{kg}\left(54.56\text{kJ/kg}\right)\\ =545.6\text{kJ}\end{array}$

Thus, the temperature and total enthalpy at initial and final states in the container are $\underset{\_}{0.61°\text{C}}$ and $\underset{\_}{\text{545}\text{.6}\text{kJ}}$ respectively.

Similarly, calculate the values of $v_f$, $v_g$, $h_f$, and $h_{fg}$ using interpolation method at pressure of 300 kPa as $0.0008198{\text{m}}^{\text{3}}/\text{kg}$, $0.034335{\text{m}}^{\text{3}}/\text{kg}$, $81.50\text{kJ/kg}$, and $180.95\text{kJ/kg}$ respectively.

Repeat the above steps for $T_2,x_2,h_2,\text{and}{H}_2$ by substituting the values of $v_f$, $v_g$, $h_f$, and $h_{fg}$ in equations (V), (VI), and (VII).

$\begin{array}{l}{T}_2=21.55°\text{C}\\ x_2=0.01731\\ h_2=84.64\text{kJ/kg}\\ H_2=846.4\text{kJ}\end{array}$

Thus, the temperature and total enthalpy at initial and final states in the container are $\underset{\_}{21.55°\text{C}}$ and $\underset{\_}{846.4\text{kJ}}$ respectively.