Chapter 4.5, Problem 44P

Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2 kg of steam at 1 MPa and 300°C while Tank B contains 3 kg of saturated liquid–vapor mixture at 150°C with a vapor mass fraction of 50 percent. The partition is now removed and the two sides are allowed to mix until mechanical and thermal equilibrium are established. If the pressure at the final state is 300 kPa, determine (a) the temperature and quality of the steam (if mixture) at the final state and (b) the amount of heat lost from the tanks.

FIGURE P4–44

To determine

The final temperature of the steam for a tank.

The quality of the steam at the final state for a tank.

Answer to Problem 44P

The final temperature of the steam is $\underset{\_}{133.5°\text{C}}$.

The quality of the steam at the final state is $\underset{\_}{0.3641}$.

Explanation of Solution

Write the expression for the energy balance equation.

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

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (I) and write energy balance relation of two tanks.

$W_{\text{in}}-Q_{\text{out}}=\Delta U$ (II)

Here, the boundary work to be done into the system is $W_{\text{in}}$, the heat to be transfer by the system is $Q_{\text{out}}$, and the change in the internal energy is $\Delta U$.

Substitute 0 for $W_{\text{in}}$ in Equation (II)

$\begin{array}{l}0-Q_{\text{out}}=\Delta U_A+\Delta U_B\\ -Q_{\text{out}}={\left[m\left(u_2-u_1\right)\right]}_A+{\left[m\left(u_2-u_1\right)\right]}_B\end{array}$ (III)

Here, the initial mass of tank A is $m_A$, the initial specific internal energy of tank A is $u_A$, the initial mass of tank B is $m_B$, the initial specific internal energy of tank B is $u_B$, the final mass of tank A is $m_A$, the final specific internal energy of tank A is $u_A$, the final mass of tank B is $m_B$, and the final specific internal energy of tank B is $u_B$.

Write the expression for volume of tank A.

${\nu }_A=m_A\cdot v_{1,A}$ (IV)

Here, the volume of the tank A is ${\nu }_A$ and the initial specific volume of tank A is $v_{1,A}$.

Write the expression for volume of tank B.

${\nu }_B=m_B\cdot v_{1,B}$ (V)

Here, the volume of the tank B is ${\nu }_B$ and the initial specific volume of tank B is $v_{1,B}$.

Write the expression for total mass of tank.

$m=m_A+m_B$ (VI)

Write the expression for total volume contained in both tanks.

$\nu ={\nu }_A+{\nu }_B$ (VII)

Write the expression for final specific volume of tank.

$v_2=\frac{\nu }{m}$ (VIII)

Conclusion:

Convert the unit of pressure from kPa to MPa.

$\begin{array}{c}P=1000\text{kPa}\\ =1000\text{kPa}\times {10}^{-3}\frac{\text{MPa}}{\text{kPa}}\\ =1.0\text{MPa}\end{array}$

From the Table A-6, “Superheated water”, obtain the value of initial specific volume $\left(v_{1,A}\right)$ and specific internal energy $\left(u_{1,A}\right)$ for tank A at the initial pressure of tank A of 1.0 MPa and temperature of tank A of 300 C.

$v_{1,A}=0.25799{\text{m}}^{\text{3}}/\text{kg}$.

$u_{1,A}=2793.7\text{kJ}/\text{kg}$.

From the Table A-4, “Superheated water”, obtain the value of initial specific volume and specific internal energy for tank B.

At the initial temperature and quality of the steam at the final state of tank B as 150 C and 0.50.

$v_f=0.001091{\text{m}}^{\text{3}}/\text{kg}$.

$v_g=0.39248{\text{m}}^{\text{3}}/\text{kg}$.

$u_f=631.66\text{kJ}/\text{kg}$.

$u_{fg}=1927.4\text{kJ}/\text{kg}$.

Determine the specific volume of a two-phase system for tank B.

$\begin{array}{c}{v}_{1,A}=v_f+x{v}_{fg}\\ =v_f+x\left(v_g-v_f\right)\end{array}$ (IX)

Here, the specific volume of saturated liquid for tank B is $v_f$, the specific volume of saturated vapour for tank B  is $v_g$, the specific volume change upon vaporization for tank B is $v_{fg}$, and the quality of final state for tank B is $x$.

$u_{1,A}=u_f+x{u}_{fg}$ (X)

Here, the specific internal energy of saturated liquid for tank B is $u_f$ and the specific internal energy change upon vaporization for tank B is $u_{fg}$.

Substitute $0.001091{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.50$ for $x$, and $0.39248{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (IX)

$\begin{array}{c}{v}_{1,B}=\left(0.001091{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.50\right)\left(0.39248{\text{m}}^{\text{3}}/\text{kg}-0.001091{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(0.001091{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.50\right)\left(0.391389{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.196786{\text{m}}^{\text{3}}/\text{kg}\\ \cong 0.19679{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $631.66\text{kJ}/\text{kg}$ for $u_f$ $0.50$ for $x$, and $1927.4\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (X).

$\begin{array}{c}{u}_{1,B}=\left(631.66\text{kJ}/\text{kg}\right)+\left(0.50\right)\left(1927.4\text{kJ}/\text{kg}\right)\\ =\left(631.66\text{kJ}/\text{kg}\right)+\left(963.7\text{kJ}/\text{kg}\right)\\ =1595.36\text{kJ}/\text{kg}\end{array}$

Substitute $2\text{kg}$ for $m_A$ and $0.25799{\text{m}}^{\text{3}}/\text{kg}$ for $v_{1,A}$ in Equation (IV).

$\begin{array}{c}{\nu }_A=\left(2\text{kg}\right)\times \left(0.25799{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.51598{\text{m}}^3\end{array}$

Substitute $3\text{kg}$ for $m_B$ and $0.19679{\text{m}}^{\text{3}}/\text{kg}$ for $v_{1,B}$ in Equation (V).

$\begin{array}{c}{\nu }_B=\left(3\text{kg}\right)\times \left(0.19679{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.59037{\text{m}}^3\end{array}$

Substitute $2\text{kg}$ for $m_A$ and $3\text{kg}$ for $m_B$ in Equation (VI).

$\begin{array}{c}m=2\text{kg}+\text{5}\text{kg}\\ =\text{5}\text{kg}\end{array}$

Substitute $0.51598{\text{m}}^{\text{3}}$ for ${\nu }_A$ and $0.59037{\text{m}}^{\text{3}}$ for ${\nu }_B$ in Equation (VIII).

$\begin{array}{c}\nu =\left(0.51598{\text{m}}^{\text{3}}\right)+\left(0.59037{\text{m}}^{\text{3}}\right)\\ =1.106{\text{m}}^{\text{3}}\end{array}$

Substitute $1.106{\text{m}}^{\text{3}}$ for $\nu$ and $5\text{kg}$ for $m$ in Equation (X).

$\begin{array}{c}{v}_2=\frac{1.106{\text{m}}^{\text{3}}}{5\text{kg}}\\ =0.22127{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Refer Table A-5, “Saturated water-pressure”, obtain the final temperature of the steam.

At final pressure of the steam is $133.5°\text{C}$.

Thus, the final temperature of the steam is $\underset{\_}{133.5°\text{C}}$.

From the Table A-5, “Saturated water-pressure”, obtain the value of quality of the steam at the final state and final specific internal energy for tank B.

At the final pressure and specific volume at the final state as 300 kPa and $0.22127{\text{m}}^{\text{3}}/\text{kg}$.

$v_f=0.001073{\text{m}}^{\text{3}}/\text{kg}$.

$v_g=0.60582{\text{m}}^{\text{3}}/\text{kg}$.

$u_f=561.11\text{kJ}/\text{kg}$.

$u_{fg}=1982.1\text{kJ}/\text{kg}$.

Determine the specific volume of a two-phase system for steam.

$\begin{array}{l}{\nu }_2=v_f+x_2\left(v_g-v_f\right)\\ x_2=\frac{{\nu }_2-v_f}{v_g-v_f}\end{array}$ (XI)

Here, the specific volume of saturated liquid for steam is $v_f$, the specific volume of saturated vapour for steam is $v_g$, and the quality of final state for steam is $x_2$.

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

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

Substitute $0.001073{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.22127{\text{m}}^{\text{3}}/\text{kg}$ for ${\nu }_2$, and $0.60582{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (XI)

$\begin{array}{c}{x}_2=\frac{\left(0.22127{\text{m}}^{\text{3}}/\text{kg}-0.001073{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.60582{\text{m}}^{\text{3}}/\text{kg}-0.001073{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =\frac{0.220197{\text{m}}^{\text{3}}/\text{kg}}{0.604747{\text{m}}^{\text{3}}/\text{kg}}\\ =0.3641\end{array}$

Thus, the quality of the steam at the final state is $\underset{\_}{0.3641}$.

Substitute $561.11\text{kJ}/\text{kg}$ for $u_f$ $0.3641$ for $x_2$, and $1982.1\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (XII).

$\begin{array}{c}{u}_2=\left(561.11\text{kJ}/\text{kg}\right)+\left(0.3641\right)\times \left(1982.1\text{kJ}/\text{kg}\right)\\ =\left(561.11\text{kJ}/\text{kg}\right)+\left(721.68\text{kJ}/\text{kg}\right)\\ =1282.79\text{kJ}/\text{kg}\\ \cong 1282.8\text{kJ}/\text{kg}\end{array}$

To determine

The amount of heat lost from the tanks.

Answer to Problem 44P

The amount of heat lost from the tanks is $\underset{\_}{3959\text{kJ}}$.

Explanation of Solution

Conclusion:

Substitute $2\text{kg}$ for $m_A$, $1282.8\text{kJ}/\text{kg}$ for $u_{2,A}$, $2793.7\text{kJ}/\text{kg}$ for $u_{1,A}$, $3\text{kg}$ for $m_B$, $1282.8\text{kJ}/\text{kg}$ for $u_{2,B}$, and $1595.4\text{kJ}/\text{kg}$ for $u_{1,B}$ in Equation (III).

$\begin{array}{c}-Q_{\text{out}}={\left[\left(2\text{kg}\right)\left(1282.8\text{kJ}/\text{kg}-2793.7\text{kJ}/\text{kg}\right)\right]}_A+{\left[\left(3\text{kg}\right)\left(1282.8\text{kJ}/\text{kg}-1595.4\text{kJ}/\text{kg}\right)\right]}_B\\ -Q_{\text{out}}={\left[\left(2\text{kg}\right)\left(-1510.9\text{kJ}/\text{kg}\right)\right]}_A+{\left[\left(3\text{kg}\right)\left(-312.6\text{kJ}/\text{kg}\right)\right]}_B\\ -Q_{\text{out}}=\left[\left(-3021\text{kJ}\right)+\left(-937.8\text{kJ}\right)\right]\\ -Q_{\text{out}}=-3959.6\text{kJ}\end{array}$           $\begin{array}{l}-Q_{\text{out}}=-3959.6\text{kJ}\\ Q_{\text{out}}=3959.6\text{kJ}\end{array}$

Thus, the amount of heat lost from the tanks is $\underset{\_}{3959\text{kJ}}$.