Chapter 5, Problem 112RQ

To determine

The final pressure and The amount of heat transfer to the two rigid tanks.

Explanation of Solution

Given:

The volume of the Tank A $\left({\nu }_{\text{A}}\right)$ is 0.2 m³ .

The pressure of water at tank A${\left(P_1\right)}_A$  is 400 kPa.

The quality of the tank A $\left(x\right)$ is 80 percent.

The volume of the tank B$\left({\nu }_{\text{B}}\right)$  is 0.5 m^3.

The pressure of tank B ${\left(P_1\right)}_B$ is 200 kPa

The temperature of the tank B${\left(T_1\right)}_B$  is 250°C.

The temperature of the surroundings $\left(T\right)$ is 25°C.

Calculation:

For Tank A:

From the Table A-5, “Saturated water-Pressure”, at 400 kPa of pressure and 0.80 of dryness fraction of water in tank A as:

The value of the specific volume of liquid $\left(v_f\right)$ is $0.001084{\text{m}}^{\text{3}}/\text{kg}$.

The specific volume of vapour $\left(v_g\right)$ is $0.46242{\text{m}}^{\text{3}}/\text{kg}$.

The specific internal energy of liquid $\left(u_f\right)$ is $604.22\text{kJ}/\text{kg}$.

The specific internal energy change upon vaporization $\left(u_{fg}\right)$ is $1948.9\text{kJ}/\text{kg}$.

Calculate the initial specific volume of the tank A.

  $v_{1,\text{A}}=v_f+x_1\left(v_g-v_f\right)$

  $\begin{array}{c}{v}_{1,\text{A}}=\left(0.001084{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.8\right)\left(0.46242{\text{m}}^{\text{3}}/\text{kg}-0.001084{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(0.001084{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.8\right)\left(0.461336{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(0.001084{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.369069{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.370153{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Calculate the initial internal energy of the tank A.

  $u_{1,\text{A}}=u_f+x_1{u}_{fg}$

  $\begin{array}{c}{u}_{1,\text{A}}=\left(604.22\text{kJ}/\text{kg}\right)+\left(0.8\right)\left(1948.9\text{kJ}/\text{kg}\right)\\ =\left(604.22\text{kJ}/\text{kg}\right)+\left(1559.12\text{kJ}/\text{kg}\right)\\ =2163.34\text{kJ}/\text{kg}\end{array}$

For tank B:

The unit conversion of pressure from kPa to MPa.

  $\begin{array}{c}{P}_1=200\text{kPa}\times \left(\frac{{10}^{-3}\text{MPa}}{1\text{kPa}}\right)\\ =0.2\text{MPa}\end{array}$

From the Table A-5, “Superheated water-Pressure”, and at 0.2 MPa of pressure and $250°\text{C}$ of temperature of water in tank B as:

  The initial specific volume of liquid $\left(v_{\text{1,B}}\right)$ is $1.1989{\text{m}}^{\text{3}}/\text{kg}$

  The initial specific internal energy of liquid $\left(u_{\text{1,B}}\right)$ is $2731.4\text{kJ}/\text{kg}$.

Calculate the total mass of the two rigid tanks.

  $\begin{array}{c}{m}_t=m_{1,\text{A}}+m_{1,\text{B}}\\ =\left(\frac{{\nu }_{\text{A}}}{v_{\text{1,A}}}\right)+\left(\frac{{\nu }_{\text{B}}}{v_{\text{1,B}}}\right)\end{array}$

  $\begin{array}{c}{m}_t=\left(\frac{0.2{\text{m}}^{\text{3}}}{0.37015{\text{m}}^{\text{3}}/\text{kg}}\right)+\left(\frac{0.5{\text{m}}^{\text{3}}}{1.1989{\text{m}}^{\text{3}}/\text{kg}}\right)\\ =0.540321\text{kg}+0.417049\text{kg}\\ =\text{0}\text{.95737}\text{kg}\end{array}$

Calculate the final specific volume of the two rigid tanks.

  $v_2=\frac{{\nu }_t}{m_t}$

  $\begin{array}{c}{v}_2=\frac{0.7{\text{m}}^{\text{3}}}{0.95737\text{kg}}\\ =0.731169{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Refer the Table A-4, “Saturated water-Temperature”, obtain the value of following value at $25°\text{C}$ of temperature and $0.731169{\text{m}}^{\text{3}}/\text{kg}$ of final specific volume of water in tank B as:

The value of the specific volume of liquid $\left(v_f\right)$ of mixture is $0.001003{\text{m}}^{\text{3}}/\text{kg}$.

The specific volume of vapour $\left(v_g\right)$ of mixture is $43.340{\text{m}}^{\text{3}}/\text{kg}$.

The specific internal energy of liquid $\left(u_f\right)$ of mixture is $104.83\text{kJ}/\text{kg}$.

The specific internal energy change upon vaporization $\left(u_{fg}\right)$ of mixture is $2304.3\text{kJ}/\text{kg}$.

The final pressure of the saturated mixture of liquid-vapour $\left(P_2\right)$ is $3.1698\text{kPa}$.

Thus, the final pressure of the two rigid tanks is $\underset{\_}{3.1698\text{kPa}}$.

Calculate the final dryness fraction of the two rigid tanks.

  $\begin{array}{c}{x}_2=\frac{v_2-v_f}{v_{fg}}\\ =\frac{v_2-v_f}{v_g-v_f}\end{array}$

Substitute $0.731169{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$, $0.001003{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $43.340{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (VIII).

  $\begin{array}{c}{x}_2=\frac{\left(0.731169{\text{m}}^{\text{3}}/\text{kg}\right)-\left(0.001003{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(43.340{\text{m}}^{\text{3}}/\text{kg}-0.001003{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =\frac{0.730166{\text{m}}^{\text{3}}/\text{kg}}{43.339{\text{m}}^{\text{3}}/\text{kg}}\\ =0.016848\\ \cong 0.01645\end{array}$

Calculate the final internal energy of the tanks.

  $u_2=u_f+x_2{u}_{fg}$

  $\begin{array}{c}{u}_2=\left(104.83\text{kJ}/\text{kg}\right)+\left(0.01645\right)\left(2304.3\text{kJ}/\text{kg}\right)\\ =\left(104.83\text{kJ}/\text{kg}\right)+\left(37.90574\text{kJ}/\text{kg}\right)\\ =142.7357\text{kJ}/\text{kg}\end{array}$

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 two rigid tanks.

  $W_{\text{in}}-Q_{\text{out}}=\Delta U_{\text{A}}+\Delta U_{\text{B}}$        (II)

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

Take the two rigid tanks as the system.

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

  $\begin{array}{c}\left(0\right)-Q_{\text{out}}=U_{\text{2,A+B}}-U_{\text{1,A}}+U_{\text{1,B}}\\ Q_{\text{out}}=-\left[U_{\text{2,A+B}}-U_{\text{1,A}}+U_{\text{1,B}}\right]\\ =-\left[m_{2\text{,total}}{u}_2-{\left(m_1{u}_1\right)}_{\text{A}}-{\left(m_1{u}_1\right)}_{\text{B}}\right]\end{array}$        (III)

Here, the total mass of the two rigid tank is $m_{2\text{,total}}$, the final specific internal energy of the two rigid tank is $u_2$, the initial mass of the tank A is $m_{\text{1,A}}$, the initial specific internal energy of the tank A is $u_1$, the initial mass of the tank B is $m_{\text{1,B}}$, and the initial specific internal energy of the tank B is $u_1$.

Substitute $m_t=0.95737\text{kg}$, $u_2=142.7357\text{kJ}/\text{kg}$, $m_{\text{1,A}}=0.540321\text{kg}$, $u_{\text{1,A}}=2163.34\text{kJ}/\text{kg}$, $m_{\text{1,B}}=0.417049\text{kg}$, and $u_{\text{1,B}}=2731.4\text{kJ}/\text{kg}$ in Equation (III).

$\begin{array}{c}{Q}_{\text{out}}=-\left[\begin{array}{l}\left(0.95737\text{kg}\right)\left(142.7357\text{kJ}/\text{kg}\right)\\ -\left(0.540321\text{kg}\right)\left(2163.34\text{kJ}/\text{kg}\right)\\ -\left(0.417049\text{kg}\right)\left(2731.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ =-\left[\left(136.6509\text{kJ}\right)-\left(1168.898\text{kJ}\right)-\left(1139.128\text{kJ}\right)\right]\\ =-\left[\left(-2171.37\text{kJ}\right)\right]\\ =2171\text{kJ}\end{array}$

Thus, the amount of heat transfer to the two rigid tanks is $\underset{\_}{2171\text{kJ}}$.