Chapter 8.8, Problem 101RP

(a)

To determine

The final temperature of the steam in tank A.

The final temperature of the steam in tank B.

Answer to Problem 101RP

The final temperature of the steam in tank A is $133.52°\text{C}$.

The final temperature of the steam in tank B is $110.1°\text{C}$.

Explanation of Solution

Write the expression to calculate the specific volume of saturated water $\left(v\right)$.

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

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

Write the expression to calculate the specific internal energy of saturated water $\left(u\right)$.

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

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

Write the expression to calculate the specific entropy of saturated water $\left(s\right)$.

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

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

Write the expression to calculate the mass from the specific volume.

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

Write the mass balance equation for the fluid flow process.

$m_{\text{in}}=m_{\text{out}}$ (V)

Here, mass of the steam entered is $m_{\text{in}}$ and the mass of the steam exit out from system is $m_{\text{out}}$.

Write the energy balance equation for the entire system considering it as a stationary closed system.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ -Q_{\text{out}}=\Delta U\\ -Q_{\text{out}}={\left(\Delta U\right)}_A+{\left(\Delta U\right)}_B\end{array}$

$-Q_{\text{out}}={\left(m_2{u}_2-m_1{u}_1\right)}_A+{\left(m_2{u}_2-m_1{u}_1\right)}_B$ (VI)

Here, net energy input to the system is $E_{\text{in}}$, net energy output to the system is $E_{\text{out}}$, the change in net energy is $\Delta E_{\text{system}}$, the amount of heat lost is $Q_{\text{out}}$, change in internal energy of steam in tank A is ${\left(\Delta U\right)}_A$, change in internal energy of steam in tank B is $\left(\Delta U_B\right)$, final mass of steam is $m_2$, initial mass of steam is $m_1$, final internal energy is $u_2$, and the initial internal energy is $u_1$.

Conclusion:

For Tank A:

Refer the Table A-5 of “Saturated water: Pressure”, obtain the properties of steam at the pressure $\left(P_1\right)$ of $400\text{kPa}$ as follows:

$\begin{array}{l}{v}_f=0.001084{\text{m}}^3/\text{kg}\\ v_g=0.46242{\text{m}}^3/\text{kg}\\ u_f=604.22\text{kJ}/\text{kg}\\ u_{fg}=1948.9\text{kJ}/\text{kg}\\ s_f=1.7765\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=5.1191\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.001084{\text{m}}^3/\text{kg}$ for $v_f$, $0.46242{\text{m}}^3/\text{kg}$ for $v_g$, and 0.8 for $x_1$ in Equation (I).

$\begin{array}{c}{v}_{1,A}=0.001084{\text{m}}^3/\text{kg}+0.8\left(0.46242{\text{m}}^3/\text{kg}-0.001084{\text{m}}^3/\text{kg}\right)\\ =0.37015{\text{m}}^3/\text{kg}\end{array}$

Substitute $604.22\text{kJ}/\text{kg}$ for $u_f$, $1948.9\text{kJ}/\text{kg}$ for $u_{fg}$, and 0.8 for $x_1$ in Equation (II).

$\begin{array}{c}{u}_{1,A}=604.22\text{kJ}/\text{kg}+0.8\left(1948.9\text{kJ}/\text{kg}\right)\\ =2163.3\text{kJ}/\text{kg}\end{array}$

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 0.8 for $x_1$ in Equation (III).

$\begin{array}{c}{s}_{1,A}=1.7765\text{kJ}/\text{kg}\cdot \text{K}+0.8\left(5.1191\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =5.8717\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer the Table A-5 of “Saturated water: Pressure”, obtain the properties of steam at the pressure $\left(P_2\right)$ of $300\text{kPa}$ as follows:

$\begin{array}{l}{v}_f=0.001073{\text{m}}^3/\text{kg}\\ v_g=0.60582{\text{m}}^3/\text{kg}\\ u_f=561.11\text{kJ}/\text{kg}\\ u_{fg}=1982.1\text{kJ}/\text{kg}\\ s_f=1.6717\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=5.320\text{kJ}/\text{kg}\cdot \text{K}\\ T_{2,A}=T_{\text{sat}}=133.52°\text{C}\end{array}$

Thus, the final temperature of the steam in tank A is $133.52°\text{C}$.

Write the final specific entropy of steam in tank A from isentropic relation.

$\begin{array}{c}{s}_{2,A}=s_{1,A}\\ =5.8717\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $5.8717\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2,A}$, $1.6717\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $5.320\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (III).

$\begin{array}{l}5.8717\text{kJ}/\text{kg}\cdot \text{K}=1.6717\text{kJ}/\text{kg}\cdot \text{K}+x_{2,A}\left(5.320\text{kJ}/\text{kg}\cdot \text{K}\right)\\ x_{2,A}=0.7895\end{array}$

Substitute $0.001073{\text{m}}^3/\text{kg}$ for $v_f$, $0.60582{\text{m}}^3/\text{kg}$ for $v_g$, and 0.7895 for $x_{2,A}$ in Equation (I).

$\begin{array}{c}{v}_{2,A}=0.001073{\text{m}}^3/\text{kg}+0.7895\left(0.60582{\text{m}}^3/\text{kg}-0.001073{\text{m}}^3/\text{kg}\right)\\ =0.4785{\text{m}}^3/\text{kg}\end{array}$

Substitute $561.11\text{kJ}/\text{kg}$ for $u_f$, $1982.1\text{kJ}/\text{kg}$ for $u_{fg}$, and 0.7895 for $x_{2,A}$ in Equation (II).

$\begin{array}{c}{u}_{2,A}=561.11\text{kJ}/\text{kg}+0.7895\left(1982.1\text{kJ}/\text{kg}\right)\\ =2125.9\text{kJ}/\text{kg}\end{array}$

For Tank B:

Refer to Table A-6 of “Superheated water”, obtain the properties of steam for pressure $\left(P_1\right)$ of $200\text{kPa}$ and temperature $\left(T_1\right)$ of $250°\text{C}$ as

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

Substitute $0.2{\text{m}}^3$ for $V_A$ and $0.37015{\text{m}}^3/\text{kg}$ for $v_{1,A}$ in Equation (IV).

$\begin{array}{c}{m}_{1,A}=\frac{0.2{\text{m}}^3}{0.37015{\text{m}}^3/\text{kg}}\\ =0.5403\text{kg}\end{array}$

Substitute $0.2{\text{m}}^3$ for $V_A$ and $0.47850{\text{m}}^3/\text{kg}$ for $v_{2,A}$ in Equation (IV).

$\begin{array}{c}{m}_{2,A}=\frac{0.2{\text{m}}^3}{0.47850{\text{m}}^3/\text{kg}}\\ =0.418\text{kg}\end{array}$

Write the expression to calculate the mass flowing into tank B $\left(m_e\right)$ from Equation (V).

$m_e=m_{1,A}-m_{2,A}$ (VII)

Substitute $0.5403\text{kg}$ for $m_{1,A}$ and $0.418\text{kg}$ for $m_{2,A}$ in Equation (VII).

$\begin{array}{c}{m}_e=0.5403\text{kg}-0.418\text{kg}\\ =0.122\text{kg}\end{array}$

Calculate the final mass of steam in tank B $\left(m_{2,B}\right)$ from Equation (V).

$m_{2,B}=m_{1,B}+m_e$ (VIII)

Substitute $3\text{kg}$ for $m_{1,B}$ and $0.122\text{kg}$ for $m_e$ in Equation (VIII).

$\begin{array}{c}{m}_{2,B}=3\text{kg}+0.122\text{kg}\\ =3.122\text{kg}\end{array}$

Write the expression to calculate the final specific volume of steam in tank B from Equation (IV).

$v_{2,B}=\frac{m_{1,B}{v}_{1,B}}{m_{2,B}}$ (IX)

Substitute $3\text{kg}$ for $m_{1,B}$, $1.1989{\text{m}}^3/\text{kg}$ for $v_{1,B}$, and $3.122\text{kg}$ for $m_{2,B}$ in Equation (IX).

$\begin{array}{c}{v}_{2,B}=\frac{3\text{kg}\times 1.1989{\text{m}}^3/\text{kg}}{3.122\text{kg}}\\ =1.152{\text{m}}^3/\text{kg}\end{array}$

Substitute $900\text{kJ}$ for $Q_{\text{out}}$, $0.5403\text{kg}$ for $m_{1,A}$, $0.418\text{kg}$ for $m_{2,A}$, $2125.9\text{kJ}/\text{kg}$ for $u_{2,A}$, $2163.3\text{kJ}/\text{kg}$ for $u_{1,A}$, $3\text{kg}$ for $m_{1,B}$, $3.122\text{kg}$ for $m_{2,B}$, and $2731.4\text{kJ}/\text{kg}$ for $u_{1,B}$ in Equation (VI).

$\begin{array}{l}-900\text{kJ}=\left(0.418\times 2125.9-0.5403\times 2163.3\right)+\left[3.122\left(u_{2,B}\right)-3\times 2731.4\right]\\ u_{2,B}=2425.9\text{kJ}/\text{kg}\end{array}$

From the Table A-4 of “Saturated water: Temperature”, obtain the properties of water in tank B at specific volume of $1.152{\text{m}}^3/\text{kg}$ and specific internal energy of $2425.9\text{kJ}/\text{kg}$ as

$\begin{array}{l}{T}_{2,B}=110.1°\text{C}\\ s_{2,B}=6.9772\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Thus, the final temperature of the steam in tank B is $110.1°\text{C}$.

(b)

To determine

The amount of work potential wasted during the process.

Answer to Problem 101RP

The amount of work potential wasted during the process is $337\text{kJ}$.

Explanation of Solution

Write the entropy generation $\left(S_{\text{gen}}\right)$ equation for the process from entropy balance.

$S_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}$

$S_{\text{gen}}=\Delta S_A+\Delta S_B+\frac{Q_{\text{out}}}{T_{\text{b,surr}}}$

$S_{\text{gen}}={\left(m_2{s}_2-m_1{s}_1\right)}_A+{\left(m_2{s}_2-m_1{s}_1\right)}_B+\frac{Q_{\text{out}}}{T_{\text{b,surr}}}$ (X)

Here, entropy input to the system is $S_{\text{in}}$, entropy exiting out is $S_{\text{out}}$, change in the entropy system is $\Delta S_{\text{system}}$, and the surrounding’s temperature is $T_{\text{b,surr}}$.

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

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

Here, the surrounding’s temperature is $T_{\text{0}}$.

Conclusion:

Substitute $900\text{kJ}$ for $Q_{\text{out}}$, $0.5403\text{kg}$ for $m_{1,A}$, $0.418\text{kg}$ for $m_{2,A}$, $5.8717\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2,A}$, $5.8717\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{1,A}$, $3\text{kg}$ for $m_{1,B}$, $3.122\text{kg}$ for $m_{2,B}$, $6.9772\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2,B}$, $7.7100\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{1,B}$, and $273\text{K}$ for $T_{\text{b,surr}}$ in Equation (X).

$\begin{array}{c}{S}_{\text{gen}}=\left[\begin{array}{l}\left(0.418\times 5.8717-0.5403\times 5.8717\right)\\ +\left(3.122\times 6.9772-3\times 7.7100\right)+\frac{900\text{kJ}}{273\text{K}}\end{array}\right]\\ =1.234\text{kJ}/\text{K}\end{array}$

Substitute $273\text{K}$ for $T_{\text{0}}$ and $1.234\text{kJ}/\text{K}$ for $S_{\text{gen}}$ in Equation (XI).

$\begin{array}{c}{X}_{\text{dest}}=273\text{K}\times 1.234\text{kJ}/\text{K}\\ =337\text{kJ}\end{array}$

Thus, the amount of work potential wasted during the process is $337\text{kJ}$.