Chapter 8, Problem 182RQ

a)

To determine

The final temperature in each tank A and tank B.

Explanation of Solution

Given:

The volume of the tank A $\left(v_A\right)$ is $0.3{\text{ m}}^3$

The pressure of the tank A $\left(P_A\right)$ is $400\text{ kPa}$.

The quality of the tank A${\left(x\right)}_A$ is 60%.

The mass of the tank B $\left(m_B\right)$ is 2 kg.

The pressure of the tank $\left(P_B\right)$ is $200\text{ kPa}$.

The temperature of the tank $\left(T_B\right)$ is $250°\text{C}$.

The pressure of the tank A after the mixture is 200 kPa.

The heat transfer to surrounding is $300\text{ kJ}$.

The surrounding temperature $\left(T_{surr}\right)$ is $17°\text{C}$.

Calculation:

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties of water at initial pressure $\left(P_{1A}\right)$ of $400\text{ kPa}$ in tank A as follows:

$\begin{array}{l}{v}_f=0.001084{\text{m}}^3/\text{kg}\\ v_g=0.46242{\text{m}}^3/\text{kg}\\ u_f=604.22\text{kJ}/\text{kg}\end{array}$

$\begin{array}{l}{u}_{fg}=1948.9\text{kJ}/\text{kg}\\ s_f=1.7765\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=5.1191\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the specific volume $\left(v\right)$.

$v=v_f+x\left(v_g-v_f\right)$        (I)

$\begin{array}{l}{v}_{1,A}=0.001084{\text{m}}^3/\text{kg}+0.6\left(0.46242{\text{m}}^3/\text{kg}-0.001084{\text{m}}^3/\text{kg}\right)\\ v_{1,A}=0.27788{\text{m}}^3/\text{kg}\end{array}$

Calculate the specific internal energy of steam from tables $\left(u\right)$.

$u=u_f+x\left(u_{gf}\right)$        (II)

$\begin{array}{l}{u}_{1,A}=604.22\text{kJ}/\text{kg}+0.6\left(1948.9\text{kJ}/\text{kg}\right)\\ u_{1,A}=1773.6\text{kJ}/\text{kg}\end{array}$

Write the formula to calculate the specific entropy of steam from tables $\left(s\right)$.

$s=s_f+x\left(s_{gf}\right)$        (III)

$\begin{array}{l}{s}_{1,A}=1.7765\text{kJ}/\text{kg}\cdot \text{K}+0.6\left(5.1191\text{kJ}/\text{kg}\cdot \text{K}\right)\\ s_{1,A}=4.8479\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties of water at final pressure $\left(P_{2A}\right)$ of $200\text{ kPa}$ in tank A as follows:

$\begin{array}{l}{T}_{2A}=T_{sat}=120.2°\text{C}\\ v_f=0.001061{\text{m}}^3/\text{kg}\\ v_g=0.8858{\text{m}}^3/\text{kg}\\ u_f=504.50\text{kJ}/\text{kg}\end{array}$

$\begin{array}{l}{u}_{fg}=2024.6\text{kJ}/\text{kg}\\ s_f=1.5302\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=5.5968\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, final temperature of steam in tank A is $T_{2A}$.

The steam in tank A undergoes isentropic process, Thus, the final specific entropy of steam in tank A$\left(s_{2A}\right)$ is expressed as:

  $s_{2A}=s_{1A}$

Substitute $s_{2A}=4.8479\text{kJ}/\text{kg}\cdot \text{K}$, $s_f=1.5302\text{kJ}/\text{kg}\cdot \text{K}$, and $s_{fg}=5.5968\text{kJ}/\text{kg}\cdot \text{K}$ in Equation (III)  to obtain the final dryness fraction of steam in tank A$\left(x_{2,A}\right)$.

  $\begin{array}{l}{x}_{2,A}=\frac{4.8479\text{kJ}/\text{kg}\cdot \text{K}-1.5302\text{kJ}/\text{kg}\cdot \text{K}}{5.5968\text{kJ}/\text{kg}\cdot \text{K}}\\ x_{2,A}=0.5928\end{array}$

Substitute $v_f=0.001061{\text{m}}^3/\text{kg}$, $v_g=0.8858{\text{m}}^3/\text{kg}$, and $x_{2,A}=0.5928$ in Equation (I) to obtain the final specific volume of steam in tank A$\left(v_{2,A}\right)$.

  $\begin{array}{l}{v}_{2,A}=0.001061{\text{m}}^3/\text{kg}+0.5928\left(0.8858{\text{m}}^3/\text{kg}\right)\\ v_{2,A}=0.52552{\text{m}}^3/\text{kg}\end{array}$

Substitute $u_f=504.50\text{kJ}/\text{kg}$, $u_{fg}=2024.6\text{kJ}/\text{kg}$, and $x_{2,A}=0.5928$ in Equation (II) to obtain the final specific internal energy of steam in tank A$\left(u_{2,A}\right)$.

  $\begin{array}{l}{u}_{2,A}=504.50\text{kJ}/\text{kg}+0.5928\left(2024.6\text{kJ}/\text{kg}\right)\\ u_{2,A}=1704.7\text{kJ}/\text{kg}\end{array}$

Refer Table A-6, “Superheated water”, note the properties for steam in tank B initially at the pressure of $200\text{ kPa}$ and temperature of $250°\text{C}$ as follows:

  $\begin{array}{l}{v}_{1,B}=1.1989{\text{m}}^3/\text{kg}\\ u_{1,B}=2731.4\text{kJ}/\text{kg}\\ s_{1,B}=7.7100\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the mass of the steam $\left(m\right)$.

  $m=\frac{\nu }{v}$        (IV)

  $\begin{array}{l}{m}_{1,A}=\frac{0.3{\text{ m}}^3}{0.27788{\text{m}}^3/\text{kg}}\\ m_{1,A}=1.08\text{ kg}\end{array}$

Calculate the final mass of steam in tank A$\left(m_{2,A}\right)$ using equation (IV).

  $\begin{array}{l}{m}_{2,A}=\frac{0.3{\text{ m}}^3}{0.52552{\text{m}}^3/\text{kg}}\\ m_{2,A}=0.5709\text{ kg}\end{array}$

Write the expression for the mass balance.

  $\Delta m=m_{in}-m_{out}$        (V)

Here, mass of the water entering into the system is $m_{in}$, the mass of water leaving out of the system is $m_{out}$ and the change in mass is $\Delta m$.

Substitute $m_{1,A}=1.080\text{ kg}$ and $m_{2,A}=0.5709\text{ kg}$ using Equation (V) to calculate the mass of steam flows into tank B from tank A$\left(\Delta m_A\right)$.

$\begin{array}{l}\Delta m_A=1.08\text{ kg}-0.5709\text{ kg}\\ \Delta m_A=0.5091\text{ kg}\end{array}$

Rewrite the Equation (V) to calculate the final total mass of steam in tank B$\left(m_{2,B}\right)$.

$m_{2,B}=m_{1,B}+\Delta m_A$        (VI)

$\begin{array}{l}{m}_{2,B}=2\text{ kg}+0.5091\text{ kg}\\ m_{2,B}=2.5091\text{ kg}\end{array}$

Substitute $m_{1,B}=2\text{ kg}$ and $v_{1,B}=1.1989{\text{m}}^3/\text{kg}$ in Equation (IV) to obtain the volume of steam in tank B$\left({\nu }_B\right)$.

  $\begin{array}{l}{\nu }_B=2\text{ kg}\left(1.1989{\text{m}}^3/\text{kg}\right)\\ {\nu }_B=2.3978{\text{ m}}^3\end{array}$

Substitute ${\nu }_B=2.3978{\text{ m}}^3$ and $m_{2,B}=2.5091\text{ kg}$ in Equation (IV) to calculate the final specific volume of steam in tank B$\left(v_{2,B}\right)$.

  $\begin{array}{l}{v}_{2,B}=\frac{2.3978{\text{ m}}^3}{2.5091\text{ kg}}\\ v_{2,B}=0.9558{\text{m}}^3/\text{kg}\end{array}$

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

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

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

From first law of thermodynamics, Re-write the Equation (VII) for heat transfer $\left(Q\right)$ as follows:

$\begin{array}{l}Q-W=\Delta U\\ Q-W={\left(\Delta U\right)}_A+{\left(\Delta U\right)}_B\end{array}$

$Q-W=\left(m_{2,A}{u}_{2,A}-m_{1,A}{u}_{1,A}\right)+\left(m_{2,B}{u}_{2,B}-m_{1,B}{u}_{1,B}\right)$

  $\begin{array}{l}-300\text{ kJ}-0=\left[\begin{array}{l}\left(0.5709\text{ kg}\times 1704.7\text{kJ}/\text{kg}-1.08\text{ kg}\times 1773.6\text{kJ}/\text{kg}\right)\\ +\left(2.5091\text{ kg}\left(u_{2,B}\right)-2\text{ kg}\times 2731.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ u_{2,B}=2433.3\text{kJ}/\text{kg}\end{array}$

Refer Table A-5, “Saturated water-Temperature table”, obtain the following properties of water at $u_{2,B}$ of $2433.3\text{kJ}/\text{kg}$ and $v_{2,B}$ of $0.9558{\text{m}}^3/\text{kg}$ as follows:

  $\begin{array}{l}{T}_{2,B}=116.1°\text{C}\\ s_{2,B}=6.9156\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, the temperature of the steam in tank at final state is $T_{2,B}$, and the specific entropy of steam in tank B finally is $s_{2,B}$.

Thus, the final temperature of steam in tank A is $120.2°\text{C}$, and final temperature of steam in tank B is $116.1°\text{C}$.

b)

To determine

The entropy generated during the process.

Explanation of Solution

Write the expression for the entropy balance Equation of the system.

  ${\dot{S}}_{in}-{\dot{S}}_{out}+{\dot{S}}_{gen}=\Delta {\dot{S}}_{system}$        (VIII)

Here, rate of net entropy in is ${\dot{S}}_{in}$, rate of net entropy out is ${\dot{S}}_{out}$, rate of entropy generation is ${\dot{S}}_{gen}$ and change of entropy of the system is $\Delta {\dot{S}}_{system}$

Re-write the Equation (VIII) to obtain the entropy generated $\left(S_{\text{gen}}\right)$ during the process.

$\begin{array}{l}\frac{Q}{T_{\text{surr}}}+S_{\text{gen}}=\Delta S_{\text{A}}+\Delta S_{\text{B}}\\ S_{\text{gen}}=\Delta S_{\text{A}}+\Delta S_{\text{B}}-\frac{Q}{T_{\text{surr}}}\end{array}$

$S_{\text{gen}}=\left(m_{2,A}{s}_{2,A}-m_{1,A}{s}_{1,A}\right)+\left(m_{2,B}{s}_{2,B}-m_{1,B}{s}_{1,B}\right)-\frac{Q}{T_{\text{surr}}}$

$\begin{array}{c}{S}_{\text{gen}}=\left\{\begin{array}{l}\left[\left(0.5709\text{ kg}\times 4.8479\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(1.08\text{ kg}\times 4.8479\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ +\left[\left(2.5091\text{ kg}\times 6.9156\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(2\text{ kg}\times 7.7100\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ -\frac{\left(-300\text{ kJ}\right)}{17°\text{C}}\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\left(0.5709\text{ kg}\times 4.8479\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(1.08\text{ kg}\times 4.8479\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ +\left[\left(2.5091\text{ kg}\times 6.9156\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(2\text{ kg}\times 7.7100\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ +\frac{300\text{ kJ}}{\left(17+273\right)\text{K}}\end{array}\right\}\\ =0.498\text{kJ}/\text{K}\end{array}$

Thus, the entropy generated during this process is $0.498\text{kJ}/\text{K}$.