Chapter 7.13, Problem 201RP

a)

To determine

The final temperature in each tank A and tank B.

Answer to Problem 201RP

The final temperature in tank A is $120.2°\text{C}$.

The final temperature in tank B is $116.1°\text{C}$.

Explanation of Solution

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

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

Here, specific volume of saturated liquid is $v_f$, specific volume of saturated vapor is $v_g$, and dryness fraction of water is $x$.

Write the formula to calculate the specific internal energy of steam from tables $\left(u\right)$.

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

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

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

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

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

Write the formula to calculate the mass of the steam $\left(m\right)$.

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

Here, volume of the steam is $\nu$.

Write the expression for the mass balance.

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

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

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

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 $\Delta E_{system}$

Conclusion:

From 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:

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

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

Substitute $0.001084{\text{m}}^3/\text{kg}$ for $v_f$, $0.46242{\text{m}}^3/\text{kg}$ for $v_g$, and $0.6$ for $x$ in Equation (I) for initial specific volume of steam in tank A $\left(v_{1,A}\right)$.

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

Substitute $604.22\text{kJ}/\text{kg}$ for $u_f$, $1948.9\text{kJ}/\text{kg}$ for $u_{fg}$, and $0.6$ for $x$ in Equation (II) for initial specific volume of steam in tank A $\left(u_{1,A}\right)$.

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

Substitute $1.7765\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $5.1191\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$, and $0.6$ for $x$ in Equation (III) for initial specific entropy of steam in tank A $\left(s_{1,A}\right)$.

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

From 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 final specific entropy of steam in tank A $\left(s_{2A}\right)$ is expressed as:

$s_{2A}=s_{1A}$

Substitute $4.8479\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2A}$, $1.5302\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $5.5968\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (III) for 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 $0.001061{\text{m}}^3/\text{kg}$ for $v_f$, $0.8858{\text{m}}^3/\text{kg}$ for $v_g$, and $0.5928$ for $x_{2,A}$ in Equation (I) to calculate 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 $504.50\text{kJ}/\text{kg}$ for $u_f$, $2024.6\text{kJ}/\text{kg}$ for $u_{fg}$, and $0.5928$ for $x_{2,A}$ in Equation (II) to calculate the final specific internal energy of steam in tank A $\left(u_{2,A}\right)$.

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

From 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:

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

Substitute $0.3{\text{ m}}^3$ for ${\nu }_A$ and $0.27788{\text{m}}^3/\text{kg}$ for $v_{1,A}$ in Equation (IV) to calculate the initial mass of steam in tank A $\left(m_{1,A}\right)$.

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

Substitute $0.3{\text{ m}}^3$ for ${\nu }_A$ and $0.27788{\text{m}}^3/\text{kg}$ for $v_{1,A}$ in Equation (IV) to calculate the final mass of steam in tank A $\left(m_{2,A}\right)$.

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

Substitute $1.080\text{ kg}$ for $m_{1,A}$ and $0.5709\text{ kg}$ for $m_{2,A}$ 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 $\left(m_{2,B}\right)$.

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

Here, initial mass of steam in tank B is $m_{1,B}$.

Substitute $2\text{ kg}$ for $m_{1,B}$ and $0.5091\text{ kg}$ for $\Delta m_A$ in Equation (VII).

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

Substitute $2\text{ kg}$ for $m_{1,B}$ and $1.1989{\text{m}}^3/\text{kg}$ for $v_{1,B}$ in Equation (IV) to calculate 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 $2.3978{\text{ m}}^3$ for ${\nu }_B$ and $2.5091\text{ kg}$ for $m_{2,B}$ in Equation (IV) to calculate the final specific volume of steam in tank B $\left(v_{2,B}\right)$.

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

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

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

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

Here, work done is $W$, total change in internal energy is $\Delta U$, change in internal in tank A is ${\left(\Delta U\right)}_A$, the change in internal energy in tank B is ${\left(\Delta U\right)}_B$, final specific internal energy of steam in tank B is $u_{2,B}$.

Substitute $-300\text{ kJ}$ for $Q$, $0$ for $W$, $0.5709\text{ kg}$ for $m_{2,A}$, $1704.7\text{kJ}/\text{kg}$ for $u_{2,A}$, $1.08\text{ kg}$ for $m_{1,A}$, $1773.6\text{kJ}/\text{kg}$ for $u_{1,A}$, $2.5091\text{ kg}$ for $m_{2,B}$, $2\text{ kg}$ for $m_{1,B}$, and $2731.4\text{kJ}/\text{kg}$ for $u_{1,B}$ in Equation (VIII).

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

From 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.

Answer to Problem 201RP

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

Explanation of Solution

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

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

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 $\Delta {\dot{S}}_{system}$

Conclusion:

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

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

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

Here, temperature of the surroundings is $T_{\text{surr}}$, entropy change for steam in tank A is $\Delta S_{\text{A}}$, and the entropy change for the steam in tank B is $\Delta S_{\text{B}}$.

Substitute $-300\text{ kJ}$ for $Q$, $0.5709\text{ kg}$ for $m_{2,A}$, $1.08\text{ kg}$ for $m_{1,A}$, $2.5091\text{ kg}$ for $m_{2,B}$, $2\text{ kg}$ for $m_{1,B}$, $4.8479\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2A}$, $4.8479\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{1A}$, $6.9156\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2,B}$, $7.7100\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{1,B}$, and $17°\text{C}$ for $T_{\text{surr}}$ in Equation (X).

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

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