Chapter 8.8, Problem 75P

a)

To determine

The final mass $\left(m_2\right)$ in the tank is $\underset{\_}{\text{3}\text{.90}\text{kg}}$.

Answer to Problem 75P

The final temperature of the helium is $\underset{\_}{321.7\text{ K}}$.

Explanation of Solution

Write the expression for the final mass $\left(m_2\right)$ stored in the tank.

$m_2=\frac{V}{v_2}$ (I)

Here, volume of the rigid tank is $V$.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of $800\text{kPa}$.

$\begin{array}{l}{v}_f=0.0008457{\text{m}}^3/\text{kg}\text{and}{v}_g=0.025645{\text{m}}^3/\text{kg}\\ u_f=94.80\text{kJ}/\text{kg}\text{and}{u}_g=246.82\text{kJ}/\text{kg}\\ s_f=0.35408\text{kJ}/\text{kg}\cdot \text{K}\text{and}{s}_g=0.91853\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, internal energy of saturated liquid and saturated vapor is $u_f\text{and}{u}_g$ respectively, specific volume of saturated liquid and saturated vapor is $v_f\text{and}{v}_g$ respectively, and specific entropy of saturated liquid and saturated vapor is $s_f\text{and}{s}_g$ respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of $800\text{kPa}$.

$\begin{array}{l}{h}_{e,h_f@800\text{kPa}}=95.48\text{kJ}/\text{kg}\\ s_{e,s_f@800\text{kPa}}=0.35408\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, enthalpy of saturated liquid is $h_f$.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of $800\text{kPa}$.

$\begin{array}{l}{v}_{2,v_g@800\text{kPa}}=0.025645{\text{m}}^3/\text{kg}\\ u_{2,u_g@800\text{kPa}}=246.82\text{kJ}/\text{kg}\\ s_{2,s_g@800\text{kPa}}=0.91853\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V$ and $0.025645{\text{m}}^3/\text{kg}$ for $v_2$ in equation (1).

$\begin{array}{c}{m}_2=\frac{\text{0}\text{.1}{\text{m}}^{\text{3}}}{0.025645{\text{m}}^3/\text{kg}}\\ =3.899\text{kg}\\ \cong \text{3}\text{.90}\text{kg}\end{array}$

Thus, the final mass $\left(m_2\right)$ in the tank is $\underset{\_}{\text{3}\text{.90}\text{kg}}$.

b)

To determine

The reversible work $\left(W_{\text{rev,out}}\right)$ associated with the process

Answer to Problem 75P

The reversible work $\left(W_{\text{rev,out}}\right)$ associated with the process is $\underset{\_}{\text{16}\text{.6}\text{kJ}}$.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

$\begin{array}{l}{m}_{\text{in}}-m_{\text{out}}=\Delta m_{\text{system}}\\ m_e=m_1-m_2\end{array}$ (II)

Here, mass of the refrigerant at the inlet is $m_{\text{in}}$, mass of the refrigerant at the exit is $m_{\text{out}}$, change in mass within the system is $\Delta m_{\text{system}}$, initial and final mass of the refrigerant is $m_1\text{and}{m}_2$ respectively, and mass at the exit is $m_e$.

Write the expression for the energy balance for the system.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ Q_{\text{in}}=m_e{h}_e+m_2{u}_2-m_1{u}_1\end{array}$ (III)

Here, net energy transfer in to the control volume is $E_{\text{in}}$, net energy transfer exit from the control volume is $E_{\text{out}}$, change in internal energy of system is $\Delta E_{\text{system}}$, amount of heat transfer to the tank from the source is $Q_{\text{in}}$, mass of refrigerant is $m$, initial internal energy is $u_1$ and final internal energy is $u_2$.

Write the expression for the initial mass $\left(m_1\right)$ stored in tank.

$\begin{array}{c}{m}_1=m_f+m_g\\ =\frac{0.3V_f}{v_f}+\frac{\left(1-0.3\right)V_g}{v_g}\\ =\frac{0.3V_f}{v_f}+\frac{0.7V_g}{v_g}\end{array}$ (IV)

Here, mass of the liquid refrigerant is $m_f$, mass of the vapor refrigerant is $m_g$, volume of the liquid refrigerant is $V_f$ and volume of the vapor refrigerant is $V_g$.

Write the expression for the initial internal energy $\left(U_1\right)$ stored in tank.

$\begin{array}{c}{U}_1=m_1{u}_1\\ =m_f{u}_f+m_g{u}_g\\ =\frac{0.3V_f}{v_f}\left(u_f\right)+\frac{0.7V_g}{v_g}\left(u_g\right)\end{array}$ (V)

Write the expression for the initial entropy $\left(S_1\right)$ stored in tank.

$\begin{array}{c}{S}_1=m_1{s}_1\\ =m_f{s}_f+m_g{s}_g\\ =\frac{0.3V_f}{v_f}\left(s_f\right)+\frac{0.7V_g}{v_g}\left(s_g\right)\end{array}$ (VI)

Write the expression for the entropy balance for refrigerant.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ \frac{Q_{\text{in}}}{T_{\text{source}}}-m_e{s}_e+S_{\text{gen}}=m_2{s}_2-m_1{s}_1\\ S_{\text{gen}}=m_2{s}_2-m_1{s}_1+m_e{s}_e-\frac{Q_{\text{in}}}{T_{\text{source}}}\end{array}$ (VI)

Here, entropy generation is $S_{\text{gen}}$, change of entropy is $\Delta S_{\text{system}}$, entropy at inlet condition is $S_{\text{in}}$, entropy at outlet condition is $S_{\text{out}}$, and source temperature is $T_{\text{source}}$.

Write the expression for the reversible work $\left(W_{\text{rev,out}}\right)$ associated with the process.

$W_{\text{rev,out}}=X_{\text{destroyed}}+W_{\text{act,out}}$

Here, exergy destroyed is $X_{\text{destroyed}}$ and actual work associated with the process is $W_{\text{act,out}}$.

Since, the process does not involve any actual work, substitute 0 for $W_{\text{act,out}}$.

$\begin{array}{c}{W}_{\text{rev,out}}=X_{\text{destroyed}}+0\\ W_{\text{rev,out}}=X_{\text{destroyed}}\\ =T_0{S}_{\text{gen}}\end{array}$ (VII)

Here, dead state temperature is $T_0$.

Conclusion:

Substitute $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V_f$, $0.0008457{\text{m}}^3/\text{kg}$ for $v_f$, $0.025645{\text{m}}^3/\text{kg}$ for $v_g$ and $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V_g$ in Equation (IV).

$\begin{array}{c}{m}_1=\frac{0.3\left(\text{0}\text{.1}{\text{m}}^{\text{3}}\right)}{0.0008457{\text{m}}^3/\text{kg}}+\frac{0.7\left(\text{0}\text{.1}{\text{m}}^{\text{3}}\right)}{0.025645{\text{m}}^3/\text{kg}}\\ =\frac{\text{0}\text{.03}{\text{m}}^{\text{3}}}{0.0008457{\text{m}}^3/\text{kg}}+\frac{\text{0}\text{.07}{\text{m}}^{\text{3}}}{0.025645{\text{m}}^3/\text{kg}}\\ =38.202\text{kg}\end{array}$

Substitute $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V_f$, $0.0008457{\text{m}}^3/\text{kg}$ for $v_f$, $0.025645{\text{m}}^3/\text{kg}$ for $v_g$, $94.80\text{kJ}/\text{kg}$ for $u_f$, $246.82\text{kJ}/\text{kg}$ for $u_g$ and $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V_g$ in Equation (V)

$\begin{array}{c}{U}_1=m_1{u}_1\\ =\frac{0.3\left(\text{0}\text{.1}{\text{m}}^{\text{3}}\right)}{0.0008457{\text{m}}^3/\text{kg}}\left(94.80\text{kJ}/\text{kg}\right)+\frac{0.7\left(\text{0}\text{.1}{\text{m}}^{\text{3}}\right)}{0.025645{\text{m}}^3/\text{kg}}\left(246.82\text{kJ}/\text{kg}\right)\\ =3,362.73\text{kJ}+673.81\text{kJ}\\ =4,036.4\text{kJ}\end{array}$

Substitute $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V_f$, $0.0008457{\text{m}}^3/\text{kg}$ for $v_f$, $0.025645{\text{m}}^3/\text{kg}$ for $v_g$, $0.35408\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $0.91853\text{kJ}/\text{kg}\cdot \text{K}$ for $s_g$, $\text{0}\text{.1}{\text{m}}^{\text{3}}$ for $V_g$ in Equation (VI)

$\begin{array}{c}{S}_1=m_1{s}_1\\ =\frac{0.3\left(\text{0}\text{.1}{\text{m}}^{\text{3}}\right)}{0.0008457{\text{m}}^3/\text{kg}}\left(0.35408\text{kJ}/\text{kg}\cdot \text{K}\right)+\frac{0.7\left(\text{0}\text{.1}{\text{m}}^{\text{3}}\right)}{0.025645{\text{m}}^3/\text{kg}}\left(0.91853\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =\frac{0.03{\text{m}}^{\text{3}}}{0.0008457{\text{m}}^3/\text{kg}}\left(0.35408\text{kJ}/\text{kg}\cdot \text{K}\right)+\frac{0.07{\text{m}}^{\text{3}}}{0.025645{\text{m}}^3/\text{kg}}\left(0.91853\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =12.55\text{kJ}/\text{K}+2.507\text{kJ}/\text{K}\end{array}$

$=15.067\text{kJ}/\text{K}$

Substitute $38.202\text{kg}$ for $m_1$ and $\text{3}\text{.90}\text{kg}$ for $m_2$ in Equation (II).

$\begin{array}{c}{m}_e=m_1-m_2\\ =38.202\text{kg}-\text{3}\text{.90}\text{kg}\\ =34.30\text{kg}\end{array}$

Substitute $\text{34}\text{.30}\text{kg}$ for $m_e$, $95.48\text{kJ}/\text{kg}$ for $h_e$, $38.202\text{kg}$ for $m_1$, $\text{3}\text{.90}\text{kg}$ for $m_2$, $246.82\text{kJ}/\text{kg}$ for $u_2$ and $4036.4\text{kJ}$ for $m_1{u}_1$ in Equation (III) .

$\begin{array}{c}{Q}_{\text{in}}=m_e{h}_e+m_2{u}_2-m_1{u}_1\\ =\left[\left(\text{34}\text{.30}\text{kg}\right)\left(95.48\text{kJ}/\text{kg}\right)\right]+\left[\left(\text{3}\text{.90}\text{kg}\right)\left(246.82\text{kJ}/\text{kg}\right)\right]-4,036.4\text{kJ}\\ =3,274.96\text{kJ}+962.598\text{kJ}-4,036.4\text{kJ}\\ =201\text{kJ}\end{array}$

Substitute $\text{3}\text{.899}\text{kg}$ for $m_2$, $15.067\text{kJ}/\text{K}$ for $m_1{s}_1$, $34.30\text{kg}$ for $m_e$, $0.35408\text{kJ}/\text{kg}\cdot \text{K}$ for $s_e$, $0.91853\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $201\text{kJ}$ for $Q_{\text{in}}$ and $\text{60°C}$ for $T_{\text{source}}$ in Equation (VI).

$\begin{array}{c}{S}_{\text{gen}}=\left\{\begin{array}{l}\left[\left(\text{3}\text{.899}\text{kg}\right)\left(0.91853\text{kJ}/\text{kg}\cdot \text{K}\right)\right]-15.067\text{kJ}/\text{K}+\\ \left[\left(34.30\text{kg}\right)\left(0.35408\text{kJ}/\text{kg}\cdot \text{K}\right)\right]-\frac{201\text{kJ}}{\text{60°C}}\end{array}\right\}\\ =3.5813\text{kJ}/\text{K}-15.067\text{kJ}/\text{K}+12.1449\text{kJ}/\text{K}-\frac{201\text{kJ}}{\left(\text{60}+\text{273}\right)\text{K}}\\ =0.6592\text{kJ}/\text{K}-\frac{201\text{kJ}}{333\text{K}}\\ =0.6592\text{kJ}/\text{K}-0.6036\text{kJ}/\text{K}\end{array}$

$=0.0556\text{kJ}/\text{K}$

Substitute $\text{25°C}$ for $T_0$ and $0.0556\text{kJ}/\text{K}$ for $S_{\text{gen}}$ in Equation (VII).

$\begin{array}{c}{W}_{\text{rev,out}}=\left(\text{25°C}\right)\left(0.0556\text{kJ}/\text{K}\right)\\ =\left[\left(25+273\right)\text{K}\right]\left(0.0556\text{kJ}/\text{K}\right)\\ =\left(298\text{K}\right)\left(0.0556\text{kJ}/\text{K}\right)\\ =16.6\text{kJ}\end{array}$

Thus, the reversible work $\left(W_{\text{rev,out}}\right)$ associated with the process is $\underset{\_}{\text{16}\text{.6}\text{kJ}}$.