Chapter 5.5, Problem 118P

(a)

To determine

The final pressure in the tank.

Answer to Problem 118P

The final pressure in the tank is $67.028\text{psia}$.

Explanation of Solution

At the final observation, the valve is closed and the tank composed with one-half water and vapor at the temperature of $300°\text{F}$. The temperature of the tank is kept constant.

Hence, the pressure $\left(P_2\right)$ of mixture in the tank at final state is equal to the saturation pressure $\left(P_{\text{sat}}\right)$ of that mixture.

$P_2=P_{\text{sat}\text{@}\text{300}\text{°F}}$

Refer Table A-4E, “Saturated water-Temperature table”.

The saturation pressure corresponding to the temperature of $300°\text{F}$ is $67.028\text{psia}$.

Conclusion:

Hence, the pressure of the mixture in the tank at the final state is $67.028\text{psia}$.

(b)

To determine

The amount of steam entered in the tank.

Answer to Problem 118P

The amount of steam entered in the tank is $85.7279\text{lbm}$.

Explanation of Solution

Write the equation of mass balance.

$m_{\text{in}}-m_{\text{e}}=\Delta m_{\text{system}}$ (I)

Here, the inlet mass is $m_{\text{in}}$, the exit mass is $m_{\text{e}}$ and the change in mass of the system is $\Delta m_{\text{system}}$.

The change in mass of the system for the control volume is expressed as,

$\Delta m_{\text{system}}={\left(m_2-m_1\right)}_{\text{cv}}$

Here, the suffixes 1 and 2 indicates the initial and final states of the system.

Consider the given rigid tank as the control volume.

At final state, the valve is close and the steam is not allowed to exit, i.e. $m_{\text{e}}=0$.

Rewrite the Equation (I) as follows.

$\begin{array}{l}{m}_{\text{in}}-0={\left(m_2-m_1\right)}_{\text{cv}}\\ m_{\text{in}}=m_2-m_1\end{array}$ (II)

At initial state, the tank consist of saturated water vapor $\left(g\right)$.

Write the formula for mass of steam $\left(m_1\right)$ at initial.

$m_1=\frac{{\nu }_1}{v_1}$ (III)

Here, the volume is $\nu$ and the specific volume is $v$; the suffix 1 indicates the initial state.

At final state, the tank consist of mixture of vapor $\left(g\right)$ and liquid $\left(f\right)$.

Write the formula for mass of steam $\left(m_2\right)$ at final.

$\begin{array}{c}{m}_2=m_{f,2}+m_{g,2}\\ =\frac{{\nu }_{f,2}}{v_{f,2}}+\frac{{\nu }_{g,2}}{v_{g,2}}\end{array}$ (IV)

Here, the suffixes $g,f$ indicates the gaseous (vapor) and fluid (liquid) condition and the suffix 1 and 2 indicates the initial and final states.

It is given that the tank consist of one-half of the volume of the tank is occupied by liquid water.

$\begin{array}{c}{\nu }_{f,2}=1.5{\text{ft}}^{\text{3}}\\ {\nu }_{g,2}=1.5{\text{ft}}^{\text{3}}\end{array}$

At the initial state (1):

The tank consist of saturated water vapor $\left(g\right)$.

Refer Table A-4E, “Saturated water-Temperature table”.

Obtain the initial specific volume $\left(v_1=v_g\right)$ corresponding to the temperature of $300°\text{F}$.

$v_g=v_1=6.4663{\text{ft}}^3\text{/lbm}$

At the final state (2):

The tank consist of mixture of vapor $\left(g\right)$ and liquid $\left(f\right)$.

Refer Table A-4E, “Saturated water-Temperature table”.

Obtain the final fluid and gaseous specific volume corresponding to the temperature of $300°\text{F}$.

$v_{f,2}=0.01745{\text{ft}}^3\text{/lbm}$

$v_{g,2}=6.4663{\text{ft}}^3\text{/lbm}$

Conclusion:

Substitute $3{\text{ft}}^3$ for ${\nu }_1$ and $6.4663{\text{ft}}^3\text{/lbm}$ for $v_1$ in Equation (III).

$\begin{array}{c}{m}_1=\frac{3{\text{ft}}^3}{6.4663{\text{ft}}^3\text{/lbm}}\\ =0.4639\text{lbm}\end{array}$

Substitute $1.5{\text{ft}}^3$ for ${\nu }_{f,2}$, $0.01745{\text{ft}}^3\text{/lbm}$ for $v_{f,2}$, $1.5{\text{ft}}^3$ for ${\nu }_{g,2}$ and $6.4663{\text{ft}}^3\text{/lbm}$ for $v_{g,2}$ in Equation (IV).

$\begin{array}{c}{m}_2=\frac{1.5{\text{ft}}^3}{0.01745{\text{ft}}^3\text{/lbm}}+\frac{1.5{\text{ft}}^3}{6.4663{\text{ft}}^3\text{/lbm}}\\ =85.9599\text{lbm}+0.2319\text{lbm}\\ =86.1919\text{lbm}\end{array}$

Substitute $86.1919\text{lbm}$ for $m_2$ and $0.4639\text{lbm}$ for $m_1$ in Equation (II).

$\begin{array}{c}{m}_{\text{in}}=86.1919\text{lbm}-0.4639\text{lbm}\\ =85.7279\text{lbm}\end{array}$

Thus, the amount of steam entered in the tank is $85.7279\text{lbm}$.

(c)

To determine

The amount of the heat transfer.

Answer to Problem 118P

The amount of the heat transfer is $80896.0151\text{Btu}$.

Explanation of Solution

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{e}}+W_{\text{e}}+m_{\text{e}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{e}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (V)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

Since the tank is not insulated, the heat transfer occurs through the tank wall. In control volume, there is no work transfer, i.e. $\left(W_{\text{in}}=W_{\text{e}}=0\right)$. Neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$. There is no exit for the tank, the exit mass is neglected i.e. $\left(m_{\text{e}}=0\right)$.

The Equation (V) reduced as follows.

$\begin{array}{l}{Q}_{\text{in}}+m_{\text{in}}{h}_{\text{in}}=m_2{u}_2-m_1{u}_1\\ Q_{\text{in}}=m_2{u}_2-m_1{u}_1-m_{\text{in}}{h}_{\text{in}}\end{array}$ (VI)

At the final state, the tank is composed of vapor and liquid. Hence, the final state energy is expressed as follows.

$\begin{array}{c}{m}_2{u}_2=m_{f,2}{u}_{f,2}+m_{g,2}{u}_{g,2}\\ =\frac{{\nu }_{f,2}}{v_{f,2}}{u}_{f,2}+\frac{{\nu }_{g,2}}{v_{g,2}}{u}_{g,2}\\ =85.9599u_f+0.2319u_g\end{array}$ (VII)

At the line (while entering the tank):

The supply line consist of superheated water vapor.

Refer Table A-6E, “Superheated water”.

Obtain the line enthalpy $\left(h_{\text{in}}\right)$ corresponding to the pressure of $200\text{psia}$ and the temperature of $400°\text{F}$.

$h_{\text{in}}=1210.9\text{Btu/lbm}$

At the initial state (1):

The tank consist of saturated water vapor $\left(g\right)$.

Refer Table A-4E, “Saturated water-Temperature table”.

Obtain the initial internal energy $\left(u_1=u_g\right)$ corresponding to the temperature of $300°\text{F}$.

$u_g=u_1=1099.8\text{Btu/lbm}$

At the final state (2):

The tank consist of mixture of vapor $\left(g\right)$ and liquid $\left(f\right)$.

Refer Table A-4E, “Saturated water-Temperature table”.

Obtain the final fluid and gaseous internal energies corresponding to the temperature of $300°\text{F}$.

$u_{f,2}=269.51\text{Btu/lbm}$

$u_{g,2}=1099.8\text{Btu/lbm}$

Conclusion:

Substitute $269.51\text{Btu/lbm}$ for $u_{f,2}$ and $1099.8\text{Btu/lbm}$ for $u_{g,2}$ in Equation (VII).

$\begin{array}{c}{m}_2{u}_2=85.9599\left(269.51\text{Btu/lbm}\right)+0.2319\left(1099.8\text{Btu/lbm}\right)\\ =23167.0527\text{Btu}+255.0436\text{Btu}\\ =23422.0963\text{Btu}\end{array}$

Substitute $23422.0963\text{Btu}$ for $m_2{u}_2$, $0.4639\text{lbm}$ for $m_1$, $1099.8\text{Btu/lbm}$ for $u_1$,$85.7279\text{lbm}$ for $m_{\text{in}}$, and $1210.9\text{Btu/lbm}$ for $h_{\text{in}}$ in Equation (VI).

$\begin{array}{c}{Q}_{\text{in}}=\left[\begin{array}{l}23422.0963\text{Btu/lbm}-\left(0.4639\text{lbm}\right)\left(1099.8\text{Btu/lbm}\right)\\ -\left(85.7279\text{lbm}\right)\left(1210.9\text{Btu/lbm}\right)\end{array}\right]\\ =\left(23422.0963-510.1972-103807.9141\right)\text{Btu}\\ =-80896.0151\text{Btu}\end{array}$

Here, the negative sign indicates that the heat transfer occurs from the tank to the surrounding.

$Q_{\text{out}}=80896.0151\text{Btu}$

Thus, the amount of the heat transfer is $80896.0151\text{Btu}$.