Chapter 4.5, Problem 131RP

To determine

The final pressure of the two rigid tanks.

The amount of heat transfer to the two rigid tanks.

Answer to Problem 131RP

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

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

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 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 $u_1$.

Determine the initial specific volume of the tank A.

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

Here, the specific volume of the saturated liquid phase is $v_f$, the initial dryness fraction is $x_1$, and the specific volume of the saturated vapour phase is $v_g$.

Determine the initial internal energy of the tank A.

$u_{1,\text{A}}=u_f+x_1{u}_{fg}$ (V)

Here, the specific internal energy of the saturated liquid phase is $u_f$, the initial dryness fraction is $x_1$, and the specific internal energy change upon vaporization is $v_{fg}$.

Determine the total mass of the two rigid tanks.

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

Determine the final specific volume of the two rigid tanks.

$v_2=\frac{{\nu }_t}{m_t}$ (VII)

Here, the total volume of the two tanks is ${\nu }_t$.

Determine the final dryness fraction of the two rigid tanks.

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

Here, the specific volume change upon vaporization is $v_{fg}$.

Determine the final internal energy of the tanks.

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

Conclusion:

For Tank A:

From the Table A-5, “Saturated water-Pressure”, obtain the value of the specific volume of liquid, the specific volume of vapour, the specific internal energy of liquid, and the specific  internal energy change upon vaporization at 400 kPa of pressure and 0.80 of dryness fraction of water in tank A as:

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

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

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

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

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

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”, obtain the value of the initial specific volume of liquid and the initial specific internal energy of liquid at 0.2 MPa of pressure and $250°\text{C}$ of temperature of water in tank B as:

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

Substitute $0.2{\text{m}}^{\text{3}}$ for ${\nu }_{\text{A}}$, $0.37015{\text{m}}^{\text{3}}/\text{kg}$ for $v_{\text{1,A}}$, $0.5{\text{m}}^{\text{3}}$ for ${\nu }_{\text{B}}$, and $1.1989{\text{m}}^{\text{3}}/\text{kg}$ for $v_{\text{1,B}}$ in Equation (VI).

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

Substitute $0.7{\text{m}}^{\text{3}}$ for ${\nu }_t$ and $0.95737\text{kg}$ for $m_t$ in Equation (VII).

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

From the Table A-4, “Saturated water-Temperature”, obtain the value of the specific volume of liquid, the specific volume of vapour, the specific internal energy of liquid, the specific  internal energy change upon vaporization, and the final pressure of the saturated mixture of liquid-vapour 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:

$\begin{array}{l}{v}_f=0.001003{\text{m}}^{\text{3}}/\text{kg}\\ v_g=43.340{\text{m}}^{\text{3}}/\text{kg}\\ u_f=104.83\text{kJ}/\text{kg}\\ u_{fg}=2304.3\text{kJ}/\text{kg}\\ P_2=3.1698\text{kPa}\end{array}$

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

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).

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

Substitute $104.83\text{kJ}/\text{kg}$ for $u_f$, $0.01645$ for $x_2$, and $2304.3\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (IX).

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

Substitute $0.95737\text{kg}$ for $m_t$, $142.7357\text{kJ}/\text{kg}$ for $u_2$, $0.540321\text{kg}$ for $m_{\text{1,A}}$, $2163.34\text{kJ}/\text{kg}$ for $u_{\text{1,A}}$, $0.417049\text{kg}$ for $m_{\text{1,B}}$, and $2731.4\text{kJ}/\text{kg}$ for $u_{\text{1,B}}$ 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 $\underset{\_}{2171\text{kJ}}$.