Chapter 6, Problem 117P

To determine

The final temperature and pressure in the tank.

Explanation of Solution

Given:

The rigid volume of tank $\left(v\right)$ is $0.15{\text{ m}}^3$.

The pressure of the tank $\left(P_1\right)$ is $3\text{ MPa}\text{.}$

The initial temperature of the tank $\left(T_1\right)$ is $130°\text{C}$.

Calculation:

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 tank as the control volume. Initially the tank is filled with helium and the valve is closed. No mass is allowed to inlet the tank i.e. $m_{\text{in}}=0$.

Rewrite the Equation (I) as follows.

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

It is given that the valve is opened until the initial mass of the helium reached into half (final mass).

  $m_2=\frac{1}{2}{m}_1$        (III)

Write the formula for initial and final masses.

  $m_{\text{1}}=\frac{P_1{\nu }_1}{R{T}_1}$        (IV)

  $m_{\text{2}}=\frac{P_2{\nu }_2}{R{T}_2}$        (V)

Here, the pressure is $P$, the volume is $\nu$, the gas constant of is $R$, the temperature is $T$ and the subscripts 1 and 2 indicates the initial and final states.

Write the energy balance equation.

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

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.

When the valve is opened the mass (helium) is allowed to escape and no work is done i.e. $W_{\text{in}}=W_{\text{e}}=0$. There is no heat transfer .i.e. $Q_{\text{in}}=Q_{\text{e}}=0$. Neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$.

The Equation (VI) reduced as follows.

  $\begin{array}{l}-m_{\text{e}}{h}_{\text{e}}=m_2{u}_2-m_1{u}_1\\ m_2{u}_2-m_1{u}_1+m_{\text{e}}{h}_{\text{e}}=0\end{array}$        (VII)

Write the general expressions for enthalpy and internal energy.

  $\begin{array}{l}h=c_{p}T\\ u=c_{v}T\end{array}$

Here, the specific heat at constant pressure is $c_p$, the specific heat at constant volume is $c_v$ and the temperature is $T$.

Here, the enthalpy of the helium changes continuously while leaving the tank. For simplicity, consider the properties of exiting helium as constant corresponding to the average temperature of initial and final states.

  $h_{\text{e}}=c_p\left(\frac{T_1+T_2}{2}\right)$

Rewrite the equation (VII) as follows with reference to the exit enthalpy and general expression of internal energy.

  $m_2{c}_v{T}_2-m_1{c}_v{T}_1+m_{\text{e}}{c}_p\left(\frac{T_1+T_2}{2}\right)=0$        (VIII)

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ is $5.1926\text{kJ/kg}\cdot \text{K}$ and the specific heat at constant volume $\left(c_v\right)$ is $3.1156\text{kJ/kg}\cdot \text{K}$. The specific heat ratio $(k)$ is $1.667$.

Substitute $\frac{1}{2}{m}_1$ for $m_2,m_e$ in Equation (VIII).

  $\begin{array}{l}\left(\frac{1}{2}{m}_1\right)c_v{T}_2-m_1{c}_v{T}_1+\left(\frac{1}{2}{m}_1\right)c_p\left(\frac{T_1+T_2}{2}\right)=0\\ \frac{1}{2}{m}_1{c}_v{T}_2-m_1{c}_v{T}_1+\frac{1}{4}{m}_1{c}_p\left(T_1+T_2\right)=0\end{array}$        (IX)

Divide the Equation (IX) by $\frac{m_1}{2}$ on both sides.

  $\begin{array}{l}\frac{\frac{1}{2}{m}_1{c}_v{T}_2-m_1{c}_v{T}_1+\frac{1}{4}{m}_1{c}_p\left(T_1+T_2\right)}{\frac{m_1}{2}}=\frac{0}{\frac{m_1}{2}}\\ c_v{T}_2-2c_v{T}_1+\frac{1}{2}{c}_p\left(T_1+T_2\right)=0\end{array}$        (X)

Divide the Equation (X) by $c_v$ on both sides.

  $\begin{array}{l}\frac{c_v{T}_2-2c_v{T}_1+\frac{1}{2}{c}_p\left(T_1+T_2\right)}{c_v}=\frac{0}{c_v}\\ T_2-2T_1+\frac{1}{2}\frac{c_p}{c_v}\left(T_1+T_2\right)=0\end{array}$

Substitute $k=\frac{c_p}{c_v}$ in the above equation.

  $\begin{array}{l}{T}_2-2T_1+\frac{1}{2}k\left(T_1+T_2\right)=0\\ 2T_2-4T_1+k\left(T_1+T_2\right)=0\\ 2T_2-4T_1+k{T}_1+k{T}_2=0\\ T_2\left(2+k\right)-T_1\left(4-k\right)=0\end{array}$

  $T_2=\frac{T_1\left(4-k\right)}{2+k}$        (XI)

Substitute $T_1=130°\text{C}$ and $k=1.667$ in Equation (XI).

  $\begin{array}{c}{T}_2=\frac{\left(130°\text{C}\right)\left(4-1.667\right)}{2+1.667}\\ =\frac{\left(130+273\right)\text{K}\left(4-1.667\right)}{2+1.667}\\ =\frac{403\text{K}\left(2.333\right)}{3.667}\\ =256.3946\text{K}\end{array}$

  $\simeq \text{257}\text{K}$

Thus, the final temperature in the tank is $\text{257}\text{K}$.

Divide the Equation (V) by Equation (IV).

  $\begin{array}{l}\frac{m_{\text{2}}}{m_{\text{1}}}=\frac{\frac{P_2{\nu }_2}{R{T}_2}}{\frac{P_1{\nu }_1}{R{T}_1}}\\ \frac{m_{\text{2}}}{m_{\text{1}}}=\frac{P_2{\nu }_2}{R{T}_2}\times \frac{R{T}_1}{P_1{\nu }_1}\\ \frac{m_{\text{2}}}{m_{\text{1}}}=\frac{P_2{\nu }_2{T}_2}{P_1{\nu }_1{T}_2}\\ P_2=\frac{m_2{P}_1{\nu }_1{T}_2}{m_1{\nu }_2{T}_1}\end{array}$        (XII)

Here, the volume of the tank cannot change. Hence, the initial and final volumes are equal.

  $v_1=v_2$

The Equation (XII) becomes as follows.

  $P_2=\frac{m_2{P}_1{T}_2}{m_1{T}_1}$        (XIII)

Substitute $m_2=\frac{1}{2}{m}_1$, $P_1=3\text{MPa}$, $T_1=130°\text{C}$ and $T_2=\text{257}\text{K}$  in Equation (XIII).

  $\begin{array}{c}{P}_2=\frac{\left(\frac{1}{2}{m}_1\right)\left(3\text{MPa}\right)\left(\text{257}\text{K}\right)}{m_1\left(130°\text{C}\right)}\\ =\frac{\left(3\text{MPa}\times \frac{1000\text{kPa}}{1\text{MPa}}\right)\left(\text{257}\text{K}\right)}{2\left(130+273\right)\text{K}}\\ =\frac{7710000\text{kPa}\cdot \text{K}}{806\text{K}}\\ =956.5757\text{kPa}\end{array}$

  $\simeq \text{956}\text{kPa}$

Thus, the final pressure in the tank is $\text{956}\text{kPa}$.