Chapter 12.6, Problem 101RP

An adiabatic 0.2-m³ storage tank that is initially evacuated is connected to a supply line that carries nitrogen at 225 K and 10 MPa. A valve is opened, and nitrogen flows into the tank from the supply line. The valve is closed when the pressure in the tank reaches 10 MPa. Determine the final temperature in the tank (a) treating nitrogen as an ideal gas, and (b) using generalized charts. Compare your results to the actual value of 293 K.

FIGURE P12–101

To determine

The final temperature in the tank by treating nitrogen as an ideal gas and compare the result to the actual value of $293\text{}\text{K}$.

Answer to Problem 101RP

The final temperature in the tank by treating nitrogen as an ideal gas is $314.8\text{K}$.

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 insulated tank as the control volume.

The valve is closed when the pressure in tank reaches to $10\text{MPa}$ and the nitrogen is not allowed to exit i.e. $m_{\text{e}}=0$. And also the initially the storage tank is evacuated i.e. $m_1=0$.

Rewrite the Equation (I) as follows.

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

Write the energy balance equation.

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

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 initial and final states of the system.

Since the tank is adiabatic, there is no heat transfer i.e. $\left(Q_{\text{in}}=Q_{\text{e}}=0\right)$, 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. $\left(m_{\text{e}}=0\right)$. At initial state, the tank is evacuated i.e. $\left(m_1=0\right)$.

The Equation (III) reduced as follows.

$m_{\text{in}}{h}_{\text{in}}=m_2{u}_2$ (IV)

Substitute $m_2$ for $m_{\text{in}}$ in Equation (IV).

$\begin{array}{l}{m}_2{h}_{\text{in}}=m_2{u}_2\\ h_{\text{in}}=u_2\end{array}$ (V)

Express the Equation (V) in molar basis.

$\begin{array}{l}M\left(h_{\text{in}}\right)=M\left(u_2\right)\\ \overline{h_{\text{in}}}=\overline{u_2}\end{array}$ (VI)

Here, the molar mass of nitrogen is $M$, the molar enthalpy is $\overline{h_{\text{in}}}$, and molar internal energy is $\overline{u_2}$.

Conclusion:

The inlet condition of the nitrogen is $225\text{K}$ and $10\text{MPa}$.

While considering the nitrogen as the ideal gas, its enthalpy is solely depends on temperature.

Refer Table A-18, “Ideal-gas properties of nitrogen, ${\text{N}}_{\text{2}}$ ”.

The molar enthalpy of nitrogen corresponding to the temperature of $225\text{K}$ is $6537\text{kJ/kmol}$.

Refer Equation (VI).

$\overline{h_{\text{in}}}=\overline{u_2}=6537\text{kJ/kmol}$

The final temperature of the nitrogen is expressed as follows.

$T_2=T_{@\overline{u_2}=6537\text{kJ/kmol}}$

Refer Table A-18, “Ideal-gas properties of nitrogen, ${\text{N}}_{\text{2}}$ ”.

The final temperature $\left(T_2\right)$ of nitrogen corresponding to the internal energy of $6537\text{kJ/kmol}$ is $314.8\text{K}$.

Thus, the final temperature in the tank by treating nitrogen as an ideal gas is $314.8\text{K}$.

The percentage error with the actual temperature value of $293\text{}\text{K}$ is expressed as follows.

$\begin{array}{c}\%\text{Errror}=\frac{314.8\text{K}-293\text{}\text{K}}{293\text{}\text{K}}\times 100\\ =0.0744\times 100\\ =7.44\%\end{array}$

The error associated is $7.44\%$.

To determine

The final temperature in the tank by using generalized departure charts.

Answer to Problem 101RP

The final temperature in the tank by using generalized departure charts is $294.7\text{K}$.

Explanation of Solution

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and pressure of nitrogen gas is as follows.

$\begin{array}{l}{T}_{\text{cr}}=126.2\text{K}\\ P_{\text{cr}}=3.39\text{MPa}\end{array}$

The reduced pressure $\left(P_{R,\text{i}}\right)$ and temperature $\left(T_{R,\text{i}}\right)$ at inlet state is expressed as follows.

$\begin{array}{c}{T}_{R,\text{i}}=\frac{T_{\text{i}}}{T_{\text{cr}}}\\ =\frac{225\text{K}}{126.2\text{K}}\\ =1.78\end{array}$

$\begin{array}{c}{P}_{R,\text{i}}=\frac{P_{\text{i}}}{P_{\text{cr}}}\\ =\frac{10\text{MPa}}{3.39\text{MPa}}\\ =2.95\end{array}$

At inlet:

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h,\text{i}}\right)$ corresponding the reduced pressure $\left(P_{R,\text{i}}\right)$ and reduced temperature $\left(T_{R,\text{i}}\right)$ is $0.9$.

Write formula for enthalpy departure factor $\left(Z_{h,\text{i}}\right)$ at inlet condition in molar basis.

$Z_{h,\text{i}}=\frac{{\left({\overline{h}}_{\text{i,ideal}}-\overline{h_{\text{i}}}\right)}_{T,P}}{R_u{T}_{\text{cr}}}$ (VII)

Here, the inlet molar enthalpy at ideal gas state is ${\overline{h}}_{\text{i,ideal}}$, the inlet molar enthalpy and normal state is $\overline{h_{\text{i}}}$, the universal gas constant is $R_u$, and the critical temperature is $T_{\text{cr}}$; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (I) to obtain $\overline{h_{\text{i}}}$.

$\overline{h_{\text{i}}}={\overline{h}}_{\text{i,ideal}}-Z_{h,\text{i}}{R}_u{T}_{\text{cr}}$ (VIII)

Write the formula for molar enthalpy at final state

$\overline{h_2}={\overline{h}}_{\text{2,ideal}}-Z_{h,2}{R}_u{T}_{\text{cr}}$ (IX)

Write the formula for molar internal energy at final state.

$\overline{u_2}=\overline{h_2}-Z{R}_u{T}_{\text{2}}$ (X)

Here, the compressibility factor is $Z$.

The universal gas constant $\left(R_u\right)$ is $8.314\text{kJ/kmol}\cdot \text{K}$.

Conclusion:

Refer part (a) answer for ${\overline{h}}_{\text{i,ideal}}$.

${\overline{h}}_{\text{i,ideal}}=6537\text{kJ/kmol}$

Substitute $6537\text{kJ/kmol}$ for ${\overline{h}}_{\text{i,ideal}}$, $0.9$ for $Z_{h,\text{i}}$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $126.2\text{K}$ for $T_{\text{cr}}$ in Equation (VIII).

$\begin{array}{c}\overline{h_{\text{i}}}=6537\text{kJ/kmol}-\left(0.9\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(126.2\text{K}\right)\\ =6537\text{kJ/kmol}-944.3041\text{kJ/kmol}\\ =5592.6958\text{kJ/kmol}\\ \simeq 5593\text{kJ/kmol}\end{array}$

Refer Equation (VI).

$\overline{h_{\text{in}}}=\overline{u_2}=5593\text{kJ/kmol}$

It is given that the actual final temperature of nitrogen is $293\text{K}$ and it is lies in the temperature band of $280\text{K}$ to $300\text{K}$.

Consider the exit temperature $\left(T_2\right)$ as $280\text{K}$ and calculate the final internal energy.

The reduced pressure $\left(P_{R,2}\right)$ and temperature $\left(T_{R,2}\right)$ at inlet state is expressed as follows.

$\begin{array}{c}{T}_{R,2}=\frac{T_2}{T_{\text{cr}}}\\ =\frac{280\text{K}}{126.2\text{K}}\\ =2.22\end{array}$

$\begin{array}{c}{P}_{R,2}=\frac{P_2}{P_{\text{cr}}}\\ =\frac{10\text{MPa}}{3.39\text{MPa}}\\ =2.95\end{array}$

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h,2}\right)$ corresponding the reduced pressure $\left(P_{R,2}\right)$ and reduced temperature $\left(T_{R,2}\right)$ is $0.55$.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor $\left(Z_2\right)$ corresponding the reduced volume $\left(v_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.98$.

Refer Table A-18, “Ideal-gas properties of nitrogen, ${\text{N}}_{\text{2}}$ ”.

The final molar enthalpy of nitrogen $\left({\overline{h}}_{2,\text{ideal}}\right)$ at ideal gas state corresponding to the temperature of $280\text{K}$ is $8141\text{kJ/kmol}$.

Substitute $8141\text{kJ/kmol}$ for ${\overline{h}}_{2,\text{ideal}}$, $0.55$ for $Z_{h,2}$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $126.2\text{K}$ for $T_{\text{cr}}$ in Equation (IX).

$\begin{array}{c}\overline{h_2}=8141\text{kJ/kmol}-\left(0.55\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(126.2\text{K}\right)\\ =8141\text{kJ/kmol}-577.0747\text{kJ/kmol}\\ =7563.9253\text{kJ/kmol}\\ \simeq 7564\text{kJ/kmol}\end{array}$

Substitute $7564\text{kJ/kmol}$ for $\overline{h_2}$, $0.98$ for $Z_2$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $280\text{K}$ for $T_{\text{2}}$ in Equation (X).

$\begin{array}{c}\overline{u_2}=7564\text{kJ/kmol}-\left(0.98\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(280\text{K}\right)\\ =7564\text{kJ/kmol}-2281.3616\text{kJ/kmol}\\ =5282.6384\text{kJ/kmol}\\ \simeq 5283\text{kJ/kmol}\end{array}$

Consider the exit temperature $\left(T_2\right)$ as $300\text{K}$ and calculate the final internal energy.

The reduced pressure $\left(P_{R,2}\right)$ and temperature $\left(T_{R,2}\right)$ at inlet state is expressed as follows.

$\begin{array}{c}{T}_{R,2}=\frac{T_2}{T_{\text{cr}}}\\ =\frac{300\text{K}}{126.2\text{K}}\\ =2.38\end{array}$

$\begin{array}{c}{P}_{R,2}=\frac{P_2}{P_{\text{cr}}}\\ =\frac{10\text{MPa}}{3.39\text{MPa}}\\ =2.95\end{array}$

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h,2}\right)$ corresponding the reduced pressure $\left(P_{R,2}\right)$ and reduced temperature $\left(T_{R,2}\right)$ is $0.50$.

Refer Figure A-15, “Nelson–Obert generalized compressibility chart”.

The compressibility factor $\left(Z_2\right)$ corresponding the reduced volume $\left(v_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $1.0$.

Refer Table A-18, “Ideal-gas properties of nitrogen, ${\text{N}}_{\text{2}}$ ”.

The final molar enthalpy of nitrogen $\left({\overline{h}}_{2,\text{ideal}}\right)$ at ideal gas state corresponding to the temperature of $300\text{K}$ is $8723\text{kJ/kmol}$.

Substitute $8723\text{kJ/kmol}$ for ${\overline{h}}_{2,\text{ideal}}$, $0.50$ for $Z_{h,2}$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $126.2\text{K}$ for $T_{\text{cr}}$ in Equation (IX).

$\begin{array}{c}\overline{h_2}=8723\text{kJ/kmol}-\left(0.50\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(126.2\text{K}\right)\\ =8723\text{kJ/kmol}-524.6134\text{kJ/kmol}\\ =8198.3866\text{kJ/kmol}\\ \simeq 8198\text{kJ/kmol}\end{array}$

Substitute $8198\text{kJ/kmol}$ for $\overline{h_2}$, $1.0$ for $Z_2$, $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$, and $300\text{K}$ for $T_{\text{2}}$ in Equation (X).

$\begin{array}{c}\overline{u_2}=8198\text{kJ/kmol}-\left(1.0\right)\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\left(300\text{K}\right)\\ =8198\text{kJ/kmol}-2494.2\text{kJ/kmol}\\ =5703.8\text{kJ/kmol}\\ \simeq 5704\text{kJ/kmol}\end{array}$

Express interpolation formula to determine the final temperature $\left(T_2\right)$ corresponding to the molar internal energy of $5593\text{kJ/kmol}$.

$\frac{5593\text{kJ/kmol}-{\overline{u_2}}_{@280\text{K}}}{{\overline{u_2}}_{@300\text{K}}-{\overline{u_2}}_{@280\text{K}}}=\frac{T_2-280\text{K}}{300-280\text{K}}$

Substitute $5283\text{kJ/kmol}$ for ${\overline{u_2}}_{@280\text{K}}$ and $5704\text{kJ/kmol}$ for ${\overline{u_2}}_{@300\text{K}}$.

$\begin{array}{l}\frac{5593\text{kJ/kmol}-5283\text{kJ/kmol}}{5704\text{kJ/kmol}-5283\text{kJ/kmol}}=\frac{T_2-280\text{K}}{300-280\text{K}}\\ \frac{310\text{kJ/kmol}}{421\text{kJ/kmol}}=\frac{T_2-280\text{K}}{20\text{K}}\\ T_2=20\text{K}\left(0.7363\right)+280\text{K}\\ T_2=294.7268\text{K}\end{array}$

$T_2\simeq 294.7\text{K}$

Thus, the final temperature in the tank by using generalized departure charts is $294.7\text{K}$.

The percentage error with the actual temperature value of $293\text{}\text{K}$ is expressed as follows.

$\begin{array}{c}\%\text{Errror}=\frac{294.7\text{K}-293\text{}\text{K}}{293\text{}\text{K}}\times 100\\ =0.00580\times 100\\ =0.58\%\end{array}$

The error associated is $0.58\%$.