Chapter 4, Problem 41P

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.

Explanation of Solution

Given:

The mass of a refrigerant R-134a $\left(m\right)$ is 10 kg.

The volume of a rigid container $\left(\nu \right)$ is 14 L.

The final pressure of the container is $600\text{ kPa}\text{.}$

Calculation:

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

  $v_1=v_2$

Here, specific volume at states 1 and 2 are $v_1\text{and}{v}_2$ respectively.

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

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

Write the formula of interpolation method of two variables.

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

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

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

Thus, the initial temperature $\left(T_1\right)$ is $\underset{\_}{0.61°\text{C}\text{.}}$

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

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.

Calculate the dryness fraction at state 1.

  $x_1=\frac{v_1-v_f}{v_g-v_f}$

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

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 .

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

Calculate the specific enthalpy at state 1.

  $h_1=h_f+x_1\left(h_{fg}\right)$

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

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

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

Calculate the total enthalpy at state 1.

  $H_1=m{h}_1$

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

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, total enthalpy final states in the container is $\underset{\_}{\text{545}\text{.6}\text{kJ}}$.

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

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

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.

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

Calculate the total enthalpy at state 2.

  $H_2=m{h}_2$        (III)

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 (I), (II), and (III).

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

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.