Chapter 3.8, Problem 30P

A piston–cylinder device contains 0.85 kg of refrigerant-134a at −10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

FIGURE P3–30

To determine

The final pressure of the refrigerant R-134a.

Answer to Problem 30P

The final pressure of the refrigerant R-134a is $\underset{\_}{\text{90}\text{.4}\text{kPa}}$

Explanation of Solution

The final pressure is equal to the initial pressure of the refrigerant R-134a.

$\begin{array}{c}{P}_2=P_1\\ =P_{\text{atm}}+\frac{m_{p}g}{\left(\frac{\pi D^2}{4}\right)}\end{array}$ (I)

Here, atmospheric pressure is $P_{\text{atm}}$, mass of piston-cylinder device is $m_p$, acceleration due to gravity is g, and diameter of a piston is $D$.

Conclusion:

Substitute 88 kPa for $P_{\text{atm}}$, 12 kg for $m_p$, $\text{9}\text{.81}{\text{m/s}}^{\text{2}}$ for g, and 25 cm for D in Equation (I).

$\begin{array}{c}{P}_2=88\text{kPa}+\frac{\left(12\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)}{\left(\frac{\pi {\left(\text{25}\text{cm}\right)}^2}{4}\right)}\\ =88\text{kPa}+\frac{\left(12\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)}{\left(\frac{\pi {\left(\text{25}\text{cm}\right)}^2}{4}\right)}\left(\frac{1\text{kN}}{1000\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =90.4\text{kPa}\end{array}$

Thus, the final pressure of the refrigerant R-134a is $\underset{\_}{\text{90}\text{.4}\text{kPa}}$.

To determine

The change in the volume of the cylinder.

Answer to Problem 30P

The change in the volume of the cylinder is $\underset{\_}{0.02142{\text{m}}^{\text{3}}}$.

Explanation of Solution

Convert the unit of initial pressure from kPa to MPa.

$\begin{array}{c}{P}_1=90.4\text{kPa}\\ =90.4\text{kPa}\left(\frac{1\text{MPa}}{1000\text{kPa}}\right)\\ =0.0904\text{MPa}\end{array}$

Write the formula of interpolation method of two variables at $-10°\text{C}$.

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

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

Calculate the initial volume of cylinder.

${\nu }_1=m{v}_1$ (III)

Here, the initial state specific volume is $v_1$.

Calculate the final volume of cylinder.

${\nu }_2=m{v}_2$ (IV)

Here, the final state specific volume is $v_2$.

Calculate the change in the volume of cylinder.

$\Delta \nu ={\nu }_2-{\nu }_1$ (V)

Conclusion:

Refer to Table A-13, obtain the values of below variables as in Table (I) at $-10°\text{C}$.

Pressure, MPa $\left(x\right)$	Specific volume, ${\text{m}}^{\text{3}}/\text{kg}$ $\left(y\right)$
0.06	0.35048
0.0904	?
0.10	0.20743

Substitute 0.06 for $x_1$, 0.0904 for $x_2$, 0.10 for $x_3$, 0.35048 for $y_1$, and 0.20743 for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(0.0904-0.06\right)\left(0.20743-0.35048\right)}{\left(0.10-0.06\right)}+0.35048\\ =0.2418\end{array}$

Thus, the specific volume of refrigerant R-134a at the initial state of 90.4 kPa and $-10°\text{C}$ is $0.2418{\text{m}}^{\text{3}}/\text{kg}$.

Refer to Table A-13, obtain the values of below variables as in Table (II) at $-10°\text{C}$.

Pressure, MPa $\left(x\right)$	Enthalpy, $\text{kJ}/\text{kg}$ $\left(y\right)$
0.06	248.60
0.0904	?
0.10	247.51

Substitute 0.06 for $x_1$, 0.0904 for $x_2$, 0.10 for $x_3$, 248.60 for $y_1$, and 247.51 for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(0.0904-0.06\right)\left(247.51-248.60\right)}{\left(0.10-0.06\right)}+248.60\\ =247.77\end{array}$

Thus, the enthalpy of refrigerant R-134a at the initial state of 90.4 kPa and $-10°\text{C}$ is $247.77\text{kJ}/\text{kg}$.

Apply spreadsheet and solve the final state specific volume at $15°\text{C}$ and 0.0904 MPa using interpolation method.

Refer to Table A-13, obtain the values of below variables as in Table (III) at $-15°\text{C}$ and 0.06 MPa.

Temperature, $°\text{C}$ $\left(x\right)$	Specific volume, ${\text{m}}^{\text{3}}/\text{kg}$ $\left(y\right)$
10	0.37893
15	?
20	0.39302

Substitute 10 for $x_1$, 15 for $x_2$, 20 for $x_3$, 0.37893 for $y_1$, and 0.39302 for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(15-10\right)\left(0.39302-0.37893\right)}{\left(20-10\right)}+0.37893\\ =0.386\end{array}$

Similarly, solve final state specific volume at $-15°\text{C}$ and 0.10 MPa using interpolation method as $0.2294{\text{m}}^{\text{3}}/\text{kg}$.

Now use interpolation method again to solve the final state specific volume at $15°\text{C}$. Refer to Table A-13, obtain the values of below variables as in Table (IV) at $-15°\text{C}$ and 0.0904 MPa.

Pressure, MPa $\left(x\right)$	Specific volume, ${\text{m}}^{\text{3}}/\text{kg}$ $\left(y\right)$
0.06	0.386
0.0904	?
0.10	0.2294

Substitute 0.06 for $x_1$, 0.0904 for $x_2$, 0.10 for $x_3$, 0.386 for $y_1$, and 0.2294 for $y_3$ in Equation (II).

$\begin{array}{c}{y}_2=\frac{\left(0.0904-0.06\right)\left(0.2294-0.386\right)}{\left(0.10-0.06\right)}+0.386\\ =0.267\end{array}$

Thus, the final state specific volume at $15°\text{C}$ and 90.4 kPa is $0.267{\text{m}}^{\text{3}}/\text{kg}$.

Apply the above steps to calculate the enthalpy at $15°\text{C}$ and 90.4 kPa using interpolation method as $268.19\text{kJ/kg}$.

Substitute 0.85 kg for m and $0.2418{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (III).

$\begin{array}{c}{\nu }_1=0.85\text{kg}\left(0.2418{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.20553{\text{m}}^{\text{3}}\end{array}$

Substitute 0.85 kg for m and $0.267{\text{m}}^{\text{3}}/\text{kg}$ for ${\nu }_2$ in Equation (IV).

$\begin{array}{c}{\nu }_2=0.85\text{kg}\left(0.267{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.22695{\text{m}}^{\text{3}}\end{array}$

Substitute $0.22695{\text{m}}^{\text{3}}$ for ${\nu }_2$ and $0.20553{\text{m}}^{\text{3}}$ for ${\nu }_1$ in Equation (V).

$\begin{array}{c}\Delta \nu =0.22695{\text{m}}^{\text{3}}-0.20553{\text{m}}^{\text{3}}\\ =0.02142{\text{m}}^{\text{3}}\end{array}$

Thus, the change in the volume of the cylinder is $\underset{\_}{0.02142{\text{m}}^{\text{3}}}$.

To determine

The change in the enthalpy of the refrigerant R-134a.

Answer to Problem 30P

The change in the enthalpy of the refrigerant R-134a is $\underset{\_}{17.35\text{kJ/kg}}$.

Explanation of Solution

Calculate the total enthalpy change of refrigerant R-134a.

$\Delta H=m\left(h_2-h_1\right)$ (VI)

Here, enthalpy at initial state and final state are $h_1\text{and}{h}_2$ respectively.

Conclusion:

Substitute 0.85 kg for m, $\text{247}\text{.77}\text{kJ/kg}$ for $h_1$ and $\text{268}\text{.19}\text{kJ/kg}$ for $h_2$ in equation (VI).

$\begin{array}{c}\Delta H=0.85\text{kg}\left(268.19\text{kJ/kg}-247.77\text{kJ/kg}\right)\\ =17.35\text{kJ/kg}\end{array}$

Thus, the change in the enthalpy of the refrigerant R-134a is $\underset{\_}{17.35\text{kJ/kg}}$.