Chapter 8.8, Problem 73P

A 0.6-m³ rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Determine (a) the amount of heat transfer and (b) the reversible work and exergy destruction for this process. Assume the surroundings to be at 25°C and 100 kPa.

To determine

The amount of heat transfer during the process.

Answer to Problem 73P

The amount of heat transfer during the process is $1115\text{kJ}$.

Explanation of Solution

Write the expression for the mass balance for a tank which acts as a system.

$m_{\text{in}}-m_{\text{out}}=\Delta m_{\text{system}}$

$m_e=m_1-m_2$ (I)

Here, mass of water entering the system is $m_{\text{in}}$, mass of water leaving the system is $m_{\text{out}}$, change in mass of the system is $\Delta m_{\text{system}}$, initial mass in the tank is $m_1$, final mass in the tank is $m_2$ and mass at the exit is $m_e$.

Write the expression for the energy balance equation for a closed system.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (II)

Here, net energy transfer in to the control volume is $E_{\text{in}}$, net energy transfer exit from the control volume is $E_{\text{out}}$ and change in internal energy of system is $\Delta E_{\text{system}}$.

For steady state, the change in internal energy of system is zero.

$\Delta E_{\text{system}}=0$

Substitute 0 for $\Delta E_{system}$ in Equation (II).

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=0\\ E_{\text{in}}=E_{\text{out}}\\ Q_{\text{in}}=m_e{h}_e+m_2{u}_2-m_1{u}_1\end{array}$ (III)

Here, amount of heat transfer during the process is $Q_{\text{in}}$, specific enthalpy at exit is $h_e$, internal energy at exit is $u_e$, initial internal energy is $u_1$ and final internal energy is $u_2$.

Write the formula to calculate initial mass in the tank $\left(m_1\right)$.

$m_1=\frac{V}{v_1}$ (IV)

Here, volume of the tank is $V$ and initial specific volume of the water is $v_1$.

Write the relation between the initial and final mass of the tank.

$m_2=\frac{1}{2}{m}_1$                                                                                                             (V)

Write the formula to calculate final mass in the tank $\left(m_2\right)$.

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

Here, final specific volume of the water is $v_2$.

Write the expression to calculate the quality at state 2 $\left(x_2\right)$.

$v_2=v_f+x_2\left(v_g-v_f\right)$ (VII)

Here, specific volume of saturated liquid is $v_f$ and specific volume of saturated vapor is $v_f$.

Write the expression to calculate final internal energy $\left(u_2\right)$.

$u_2=u_f+x_2{u}_{fg}$ (VIII)

Here, internal energy of saturated liquid is $u_f$, and internal energy saturated mixture is $u_{fg}$.

Write the expression to calculate the final specific entropy $\left(s_2\right)$.

$s_2=s_f+x_2{s}_{fg}$ (IX)

Here, specific entropy of saturated liquid is $s_f$ and specific entropy saturated mixture is $s_{fg}$.

Conclusion:

Refer Table A-4, “Saturated water-Temperature table”, obtain the following properties of water at the temperature of $135°\text{C}$.

$\begin{array}{l}{v}_1=v_f=\text{0}\text{.001705}{\text{m}}^3/\text{kg}\\ v_g=\text{0}\text{.58179}{\text{m}}^3/\text{kg}\\ u_1=u_f=\text{567}\text{.41}\text{kJ}/\text{kg}\\ u_{fg}=1977.3\text{kJ}/\text{kg}\\ s_1=s_f=\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-4, “Saturated water-Temperature table”, obtain the following properties of water at the temperature of $135°\text{C}$.

$\begin{array}{l}{h}_e=h_f=567.75\text{kJ}/\text{kg}\\ s_e=s_f=\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=\text{5}\text{.2901}\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.6{\text{ m}}^3$ for $V$ and $\text{0}\text{.001705}{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (IV).

$\begin{array}{c}{m}_1=\frac{0.6{\text{ m}}^3}{\text{0}\text{.001705}{\text{m}}^3/\text{kg}}\\ =558.32\text{ kg}\end{array}$

Substitute $558.32\text{ kg}$ for $m_1$ in Equation (V).

$\begin{array}{c}{m}_2=\frac{1}{2}\left(558.32\text{ kg}\right)\\ =276.16\text{ kg}\end{array}$

Substitute $558.32\text{ kg}$ for $m_1$ and $276.16\text{ kg}$ for $m_2$ in Equation (I).

$\begin{array}{c}{m}_e=558.32\text{ kg}-276.16\text{ kg}\\ =276.16\text{ kg}\end{array}$

Substitute $276.16\text{ kg}$ for $m_2$ and $0.6{\text{ m}}^3$ for $V$ in Equation (VI).

$\begin{array}{l}276.16\text{ kg}=\frac{0.6{\text{ m}}^3}{v_2}\\ v_2=0.002149{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $0.002149{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$,$\text{0}\text{.001705}{\text{m}}^3/\text{kg}$ for $v_f$, $\text{0}\text{.58179}{\text{m}}^3/\text{kg}$ for $v_g$ in Equation (VII).

$\begin{array}{l}0.002149{\text{m}}^{\text{3}}/\text{kg}=\text{0}\text{.001705}{\text{m}}^3/\text{kg}+x_2\left(\text{0}\text{.58179}{\text{m}}^3/\text{kg}-\text{0}\text{.001705}{\text{m}}^3/\text{kg}\right)\\ x_2=\text{0}\text{.001851}\end{array}$

Substitute $\text{567}\text{.41}\text{kJ}/\text{kg}$ for $u_f$, $\text{0}\text{.001851}$ for $x_2$, $1977.3\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (VIII).

$\begin{array}{c}{u}_2=\text{567}\text{.41}\text{kJ}/\text{kg}+\left(\text{0}\text{.001851}\right)\left(1977.3\text{kJ}/\text{kg}\right)\\ =571.01\text{kJ}/\text{kg}\end{array}$

Substitute $\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $\text{0}\text{.001851}$ for $x_2$, $\text{5}\text{.2901}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (IX).

$\begin{array}{c}{s}_2=\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}+\left(\text{0}\text{.001851}\right)\left(\text{5}\text{.2901}\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =1.6970\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $276.16\text{ kg}$ for $m_e$, $567.75\text{kJ}/\text{kg}$ for $h_e$, $276.16\text{ kg}$ for $m_2$, $571.01\text{kJ}/\text{kg}$ for $u_2$, $\text{558}\text{.32 kg}$ for $m_1$ and $\text{567}\text{.41}\text{kJ}/\text{kg}$ for $u_1$ in Equation (III).

$\begin{array}{c}{Q}_{\text{in}}=\left[\begin{array}{l}\left(276.16\text{ kg}\right)\left(567.75\text{kJ}/\text{kg}\right)+\left(276.16\text{ kg}\right)\left(571.01\text{kJ}/\text{kg}\right)\\ -\left(\text{558}\text{.32 kg}\right)\left(\text{567}\text{.41}\text{kJ}/\text{kg}\right)\end{array}\right]\\ =1115\text{ kJ}\end{array}$

Thus, the amount of heat transfer during the process is $1115\text{kJ}$.

To determine

The reversible work and exergy destruction during the process.

Answer to Problem 73P

The exergy destruction during the process is $126\text{kJ}$.

The reversible work for the process is $126\text{kJ}$.

Explanation of Solution

Write the expression for the entropy balance on the extended system.

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

Here, entropy generation is $S_{\text{gen}}$, change in entropy of the extended system is $\Delta S_{\text{tank}}$, net entropy at inlet condition is $S_{\text{in}}$, net entropy at outlet condition is $S_{\text{out}}$ and entropy leaving the tank is $s_e$ and base temperature is $T_{b,in}$

Write the expression to calculate exergy destruction during the process $\left(X_{\text{destroyed}}\right)$.

$X_{\text{destroyed}}=T_0{S}_{\text{gen}}$ (XI)

Here, dead state temperature is $T_0$.

For the given process, the actual work is zero, then the reversible work will be equal to exergy destruction.

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

$W_{\text{rev,out}}=X_{\text{destroyed}}$ (XII)

Conclusion:

Substitute $276.16\text{ kg}$ for $m_2$,$\text{1}\text{.6970}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, $\text{558}\text{.32}\text{kg}$ for $m_1$,$\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $276.16\text{ kg}$ for $m_e$, $\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_e$, $\text{1115 kJ}$ for $Q_{\text{in}}$, and $\text{210}°\text{C}$ for $T_{\text{b,in}}$ in Equation (X).

$\begin{array}{c}{S}_{\text{gen}}=\left[\begin{array}{c}\left(276.16\text{ kg}\right)\left(\text{1}\text{.6970}\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(\text{558}\text{.32}\text{kg}\right)\left(\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +\left(276.16\text{ kg}\right)\left(\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{1115 kJ}}{\text{210}°\text{C}}\end{array}\right]\\ \left[\begin{array}{c}\left(276.16\text{ kg}\right)\left(\text{1}\text{.6970}\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(\text{558}\text{.32}\text{kg}\right)\left(\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +\left(276.16\text{ kg}\right)\left(\text{1}\text{.6872}\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{\text{1115 kJ}}{\left(\text{210}+273\right)\text{ K}}\end{array}\right]\\ =0.42282\text{kJ}/\text{K}\end{array}$

Substitute $25°\text{C}$ for $T_0$ and $0.42282\text{kJ}/\text{K}$ for $S_{\text{gen}}$ in Equation (V).

$\begin{array}{c}{X}_{\text{destroyed}}=\left(25°\text{C}\right)\left(0.42282\text{kJ}/\text{K}\right)\\ =\left[\left(25+273\right)\text{K}\right]\left(0.42282\text{kJ}/\text{K}\right)\\ =126\text{ kJ}\end{array}$

Thus, the exergy destruction during the process is $126\text{kJ}$.

Substitute $126\text{kJ}$ for $X_{\text{destroyed}}$ in Equation (XII).

$W_{\text{rev,out}}=126\text{kJ}$

Thus, the reversible work during the process is $126\text{kJ}$.