Chapter 4.5, Problem 142RP

Two 10-ft³ adiabatic tanks are connected by a valve. Initially, one tank contains water at 450 psia with 10 percent quality, while the second contains water at 15 psia with 75 percent quality. The valve is now opened, allowing the water vapor from the high-pressure tank to move to the low-pressure tank until the pressure in the two becomes equal. Determine the final pressure and the final mass in each tank.

FIGURE P4–142E

To determine

The final pressure of each tank.

The final mass of each tank.

Answer to Problem 142RP

The final pressure of each tank is $\underset{\_}{313\text{psia}}$.

The final mass of each tank is $\underset{\_}{\text{41}\text{.61}\text{lbm}}$.

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 (V) and write energy balance relation of two adiabatic tanks.

$Q_{\text{in}}+W_{\text{b,out}}=\Delta U$ (II)

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

Substitute 0 for $Q_{\text{in}}$ and 0 for $W_{\text{b,out}}$ in Equation (II)

$\begin{array}{l}0+0=\Delta U\\ 0=U_2-U_1\\ U_1=U_2\\ m_{1,A}{u}_{1,A}+m_{1,B}{u}_{1,B}=m_{2,A}{u}_{2,A}+m_{2,B}{u}_{2,B}\end{array}$ (III)

Here, the initial mass of tank A is $m_{1,A}$, the initial specific internal energy of tank A is $u_{1,A}$, the initial mass of tank B is $m_{1,B}$, the initial specific internal energy of tank B is $u_{1,B}$, the final mass of tank A is $m_{2,A}$, the final specific internal energy of tank A is $u_{2,A}$, the final mass of tank B is $m_{2,B}$, and the final specific internal energy of tank B is $u_{2,B}$.

Write the expression for initial mass of tank A.

$m_{1,A}=\frac{V_A}{v_{1,A}}$ (IV)

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

Write the expression for initial mass of tank B.

$m_{1,B}=\frac{V_B}{v_{1,B}}$ (V)

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

Write the expression for total mass of tank.

$m=m_{1,A}+m_{1,B}$ (VI)

Write the expression for initial total internal energy contained in both tanks.

$U_1=m_{1,A}{u}_{1,A}+m_{1,B}{u}_{1,B}$ (VII)

Write the expression for initial is equal to final specific internal energy of tank.

$u_1=u_2=\frac{U_1}{m}$ (VIII)

Determine the total volume of both the tanks.

$V=V_A+V_B$ (IX)

Write the expression for initial is equal to final specific volume of tank.

$v_1=v_2=\frac{V}{m}$ (X)

Write the expression for final mass contained in both tanks.

$m_{2,A}=m_{2,B}=\frac{m}{2}$ (XI)

Conclusion:

At initial pressure and quality of initial state for tank A as 450 psia and 0.10, find the value of initial specific volume and specific internal energy of the tank.

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

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

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

Here, the specific internal energy of saturated liquid for tank A is $u_f$, the specific internal energy change upon vaporization for tank A is $u_{fg}$, and the quality of final state for tank A is $x$.

Substitute $0.01955$ for $v_f$, $0.10$ for $x$, and $1.0324{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_g$ in Equation (XII)

$\begin{array}{c}{v}_{1,A}=\left(0.01955\right)+\left(0.10\right)\left(1.0324{\text{ft}}^{\text{3}}/\text{lbm}-0.01955\right)\\ =\left(0.01955\right)+\left(0.10\right)\left(1.01285{\text{ft}}^{\text{3}}/\text{lbm}\right)\\ =0.120835{\text{ft}}^{\text{3}}/\text{lbm}\\ \cong 0.12084{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $435.67$ for $u_f$ $0.10$ for $x$, and $683.52\text{Btu}/\text{lbm}$ in Equation (XIII).

$\begin{array}{c}{u}_{1,A}=\left(435.67\right)+\left(0.10\right)\left(683.52\text{Btu}/\text{lbm}\right)\\ =\left(435.67\right)+\left(68.352\text{Btu}/\text{lbm}\right)\\ =504.022\text{Btu}/\text{lbm}\end{array}$

At initial pressure and quality of initial state for tank B as 15 psia and 0.75, find the value of initial specific volume and specific internal energy of the tank.

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

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 A is $v_{fg}$, and the quality of final state for tank B is $x$.

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

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

Substitute $0.01672$ for $v_f$, $0.75$ for $x$, and $26.297{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_g$ in Equation (XIV)

$\begin{array}{c}{v}_{1,B}=\left(0.01672\right)+\left(0.75\right)\left(26.297{\text{ft}}^{\text{3}}/\text{lbm}-0.01672\right)\\ =\left(0.01672\right)+\left(0.75\right)\left(26.28028{\text{ft}}^{\text{3}}/\text{lbm}\right)\\ =19.72693{\text{ft}}^{\text{3}}/\text{lbm}\\ \cong 19.727{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $181.16$ for $u_f$ $0.75$ for $x$, and $896.52\text{Btu}/\text{lbm}$ in Equation (XV).

$\begin{array}{c}{u}_{1,A}=\left(181.16\right)+\left(0.75\right)\left(896.52\text{Btu}/\text{lbm}\right)\\ =\left(181.16\right)+\left(672.36\text{Btu}/\text{lbm}\right)\\ =853.55\text{Btu}/\text{lbm}\end{array}$

Substitute $10{\text{ft}}^{\text{3}}$ for $V_A$ and $0.12084{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_{1,A}$ in Equation (IV).

$\begin{array}{c}{m}_{1,A}=\frac{10{\text{ft}}^{\text{3}}}{0.12084{\text{ft}}^{\text{3}}/\text{lbm}}\\ =82.75\text{lbm}\end{array}$

Substitute $10{\text{ft}}^{\text{3}}$ for $V_B$ and $19.727{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_{1,B}$ in Equation (V).

$\begin{array}{c}{m}_{1,B}=\frac{10{\text{ft}}^{\text{3}}}{19.727{\text{ft}}^{\text{3}}/\text{lbm}}\\ =0.506919\text{lbm}\\ \cong 0.50692\text{lbm}\end{array}$

Substitute $82.75\text{lbm}$ for $m_{1,A}$ and $504.02\text{lbm}$ for $m_{1,B}$ in Equation (VI).

$\begin{array}{c}m=82.75\text{lbm}+0.5069\text{lbm}\\ =83.256\text{lbm}\\ \cong 83.26\text{lbm}\end{array}$

Substitute $82.75\text{lbm}$ for $m_{1,A}$, $504.02\text{Btu}/\text{lbm}$ for $u_{1,A}$, $0.50692\text{lbm}$ for $m_{1,B}$, and $853.55\text{Btu}/\text{lbm}$ for $u_{1,B}$ in Equation (VII).

$\begin{array}{c}{U}_1=\left(82.75\text{lbm}\right)\left(504.02\text{Btu}/\text{lbm}\right)+\left(0.50692\text{lbm}\right)\left(853.55\text{Btu}/\text{lbm}\right)\\ =\left(41707.66\text{Btu}\right)+\left(432.6816\text{Btu}\right)\\ =42140.34\text{Btu}\end{array}$

Substitute $42140.34\text{Btu}$ for $U_1$ and $83.26\text{lbm}$ for $m$ in Equation (VIII).

$\begin{array}{c}{u}_{1,2}=\frac{42140.34\text{Btu}}{83.26\text{lbm}}\\ =506.1294\text{Btu}/\text{lbm}\\ \cong 506.13\text{Btu}/\text{lbm}\end{array}$

Substitute $10{\text{ft}}^{\text{3}}$ for $V_A$ and $10{\text{ft}}^{\text{3}}$ for $V_A$ in Equation (IX).

$\begin{array}{c}V=10{\text{ft}}^{\text{3}}+10{\text{ft}}^{\text{3}}\\ =20{\text{ft}}^{\text{3}}\end{array}$

Substitute $20{\text{ft}}^{\text{3}}$ for $V$ and $83.26\text{lbm}$ for $m$ in Equation (X).

$\begin{array}{c}{v}_{1,2}=\frac{20{\text{ft}}^{\text{3}}}{83.26\text{lbm}}\\ =0.2402{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

By above calculation from Table A-5E “saturated water” the final pressure of both tanks as $313\text{psia}$

Thus, the final pressure of each tank is $\underset{\_}{313\text{psia}}$.

Substitute $83.26\text{lbm}$ for $m$ in Equation (XI).

$\begin{array}{c}{m}_{2,A,B}=\frac{83.26\text{lbm}}{2}\\ =41.61\text{lbm}\end{array}$

Thus, the final mass of each tank is $\underset{\_}{\text{41}\text{.61}\text{lbm}}$.