Chapter 5.5, Problem 178RP

A liquid R-134a bottle has an internal volume of 0.0015 m³. Initially it contains 0.55 kg of R-134a (saturated mixture) at 26°C. A valve is opened and R-134a vapor only (no liquid) is allowed to escape slowly such that temperature remains constant until the mass of R-134a remaining is 0.15 kg. Find the heat transfer with the surroundings that is needed to maintain the temperature and pressure of the R-134a constant.

To determine

The heat transfer with the surrounding that is needed to maintain the temperature and pressure of R-134a constant.

Answer to Problem 178RP

The heat transfer with the surrounding that is needed to maintain the temperature and pressure of R-134a constant is $72.8\text{kJ}$.

Explanation of Solution

Write the equation of mass balance.

$m_{\text{in}}-m_{\text{e}}=\Delta m_{\text{system}}$ (I)

Here, the inlet mass is $m_{\text{in}}$, the exit mass is $m_{\text{e}}$ and the change in mass of the system is $\Delta m_{\text{system}}$.

The change in mass of the system for the control volume is expressed as,

$\Delta m_{\text{system}}={\left(m_2-m_1\right)}_{\text{cv}}$

Here, the subscripts 1 and 2 indicates the initial and final states of the system.

Consider the given rigid container as the control volume.

Initially the container is filled with liquid refrigerant and the valve is in closed position, further no other mass is allowed to enter the container. Hence, the inlet mass is neglected i.e. $m_{\text{in}}=0$.

Rewrite the Equation (I) as follows.

$\begin{array}{l}0-m_{\text{e}}={\left(m_2-m_1\right)}_{\text{cv}}\\ m_{\text{e}}=m_1-m_2\end{array}$ (II)

Write the formula for initial specific volume $\left(v_1\right)$.

$v_1=\frac{\nu }{m_1}$ (III)

Write the formula for final specific volume $\left(v_2\right)$.

$v_2=\frac{\nu }{m_2}$ (IV)

Here, the volume is $\nu$, the specific volume is $v$, the mass is $m$, and the subscript 1 indicates the initial state, the subscript 2 indicates the final state.

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{e}}+W_{\text{e}}+m_{\text{e}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{e}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (V)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The process is maintained at isothermal at the open condition of valve, there is no heat transfer while the mass leaves the container .i.e. $Q_{\text{e}}=0$. There is no work transfer, i.e. $\left(W_{\text{in}}=W_{\text{e}}=0\right)$. Neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$. There is no mass inlet for the rigid container, the inlet mass is neglected i.e. $\left(m_{\text{in}}=0\right)$.

The Equation (V) reduced as follows.

$\begin{array}{l}{Q}_{\text{in}}-m_{\text{e}}{h}_{\text{e}}=m_2{u}_2-m_1{u}_1\\ Q_{\text{in}}=m_2{u}_2-m_1{u}_1+m_{\text{e}}{h}_{\text{e}}\end{array}$ (VI)

At the initial state 1:

The rigid container consist of saturated mixture refrigerant at $26°\text{C}$.

Refer Table A-11, “Saturated refrigerant-134a-Temperature table”.

Obtain the following corresponding to the temperature of $26°\text{C}$.

$\begin{array}{l}{v}_{f,1}=0.0008312{\text{m}}^3\text{/kg}\\ v_{g,1}=0.030008{\text{m}}^3\text{/kg}\\ u_{f,1}=87.26\text{kJ/kg}\\ u_{fg,1}=156.89\text{kJ/kg}\end{array}$

The quality of the refrigerant at state 1 is expressed as follows.

$x_1=\frac{v_1-v_{f,1}}{v_{g,1}-v_{f,1}}$ (VII)

The internal energy of the refrigerant at state 1 is expressed as follows.

$u_1=u_{f,1}+x_1{u}_{fg,1}$ (VIII)

At the final state (2):

When the valve is opened, the vapor refrigerant only allowed to escape and the temperature is kept constant.

The final temperature of the refrigerant is also $26°\text{C}$.

Refer Table A-12, “Saturated refrigerant-134a-Temperature table”.

Obtain the following corresponding to the temperature of $26°\text{C}$.

$\begin{array}{l}{v}_{f,2}=0.0008312{\text{m}}^3\text{/kg}\\ v_{g,2}=0.030008{\text{m}}^3\text{/kg}\\ u_{f,2}=87.26\text{kJ/kg}\\ u_{fg,2}=156.89\text{kJ/kg}\end{array}$

The quality of the refrigerant at state 2 is expressed as follows.

$x_2=\frac{v_2-v_{f,2}}{v_{g,2}-v_{f,2}}$ (IX)

The internal energy of the refrigerant at state 2 is expressed as follows.

$u_2=u_{f,2}+x_2{u}_{fg,2}$ (X)

Conclusion:

Substitute $0.55\text{kg}$ for $m_1$ and $0.15\text{kg}$ for $m_2$ in Equation (II).

$\begin{array}{c}{m}_{\text{e}}=0.55\text{kg}-0.15\text{kg}\\ =0.40\text{kg}\end{array}$

Substitute $0.0015{\text{m}}^3$ for $\nu$ and $0.55\text{kg}$ for $m_1$ in Equation (III).

$\begin{array}{c}{v}_1=\frac{0.0015{\text{m}}^3}{0.55\text{kg}}\\ =0.002727{\text{m}}^3\text{/kg}\end{array}$

Substitute $0.0015{\text{m}}^3$ for $\nu$ and $0.15\text{kg}$ for $m_2$ in Equation (IV).

$\begin{array}{c}{v}_2=\frac{0.0015{\text{m}}^3}{0.15\text{kg}}\\ =0.01{\text{m}}^3\text{/kg}\end{array}$

Substitute $0.002727{\text{m}}^3\text{/kg}$ for $v_1$, $0.0008312{\text{m}}^3\text{/kg}$ for $v_{f,1}$, and $0.030008{\text{m}}^3\text{/kg}$ for $v_{g,1}$ in Equation (VII).

$\begin{array}{c}{x}_1=\frac{0.002727{\text{m}}^3\text{/kg}-0.0008312{\text{m}}^3\text{/kg}}{0.030008{\text{m}}^3\text{/kg}-0.0008312{\text{m}}^3\text{/kg}}\\ =\frac{0.0018958}{0.0291768}\\ =0.064976\end{array}$

Substitute $0.064976$ for $x_1$, $87.26\text{kJ/kg}$ for $u_{f,1}$, and $156.89\text{kJ/kg}$ for $u_{fg,1}$ in

Equation (VIII).

$\begin{array}{c}{u}_1=87.26\text{kJ/kg}+\left(0.064976\right)\left(156.89\text{kJ/kg}\right)\\ =87.26\text{kJ/kg}+10.194085\text{kJ/kg}\\ =97.4541\text{kJ/kg}\\ \simeq 97.45\text{kJ/kg}\end{array}$

Substitute $0.01{\text{m}}^3\text{/kg}$ for $v_2$, $0.0008312{\text{m}}^3\text{/kg}$ for $v_{f,2}$, and $0.030008{\text{m}}^3\text{/kg}$ for $v_{g,2}$ in Equation (IX).

$\begin{array}{c}{x}_2=\frac{0.01{\text{m}}^3\text{/kg}-0.0008312{\text{m}}^3\text{/kg}}{0.030008{\text{m}}^3\text{/kg}-0.0008312{\text{m}}^3\text{/kg}}\\ =\frac{0.0091688}{0.0291768}\\ =0.31425\end{array}$

Substitute $0.31425$ for $x_2$, $87.26\text{kJ/kg}$ for $u_{f,2}$, and $156.89\text{kJ/kg}$ for $u_{fg,2}$ in

Equation (X).

$\begin{array}{c}{u}_2=87.26\text{kJ/kg}+\left(0.31425\right)\left(156.89\text{kJ/kg}\right)\\ =87.26\text{kJ/kg}+49.3027\text{kJ/kg}\\ =136.5627\text{kJ/kg}\\ \simeq 136.56\text{kJ/kg}\end{array}$

Here, the temperature is kept constant until the final state and the vapor only exits the tank. Hence the exit enthalpy is expressed as follows.

$h_e=h_{g@26°\text{C}}$

Refer Table A-12, “Saturated refrigerant-134a-Temperature table”.

The exit enthalpy $\left(h_e\right)$ corresponding to the temperature of $26°\text{C}$ is $264.73\text{kJ/kg}$.

Substitute $0.15\text{kg}$ for $m_2$, $136.56\text{kJ/kg}$ for $u_2$, $0.55\text{kg}$ for $m_1$, $97.45\text{kJ/kg}$ for $u_1$, $0.40\text{kg}$ for $m_{\text{e}}$, and $264.73\text{kJ/kg}$ for $h_{\text{e}}$ in Equation (VI).

$\begin{array}{c}{Q}_{\text{in}}=\left[\begin{array}{l}\left(0.15\text{kg}\right)\left(136.56\text{kJ/kg}\right)-\left(0.55\text{kg}\right)\left(97.45\text{kJ/kg}\right)\\ +\left(0.40\text{kg}\right)\left(264.73\text{kJ/kg}\right)\end{array}\right]\\ =\left(20.484-53.5975+105.892\right)\text{kJ}\\ =72.7785\text{kJ}\\ \simeq 72.8\text{kJ}\end{array}$

Thus, the heat transfer with the surrounding that is needed to maintain the temperature and pressure of R-134a constant is $72.8\text{kJ}$.