Chapter 12.6, Problem 96RP

(a)

To determine

The mass of argon in the tank.

Answer to Problem 96RP

The mass of argon in the tank is $35.1\text{kg}$.

Explanation of Solution

Write formula for specific volume $\left(v\right)$  in terms of compressibility factor $\left(Z\right)$.

$v=\frac{ZRT}{P}$ (I)

Here, the gas constant of argon is $R$, the temperature is $T$, and the pressure is $P$.

Write formula for mass of the argon present in the tank.

$m=\frac{\nu }{v}$ (II)

Here, the volume of argon in the tank is $\nu$.

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

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

$\begin{array}{l}{T}_{\text{cr}}=151\text{K}\\ P_{\text{cr}}=4.86\text{MPa}\end{array}$

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

The gas constant $\left(R\right)$ of argon is $0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$.

The reduced pressure $\left(P_{R1}\right)$ and temperature $\left(T_{R1}\right)$ at initial state is expressed as follows.

$\begin{array}{c}{T}_{R1}=\frac{T_1}{T_{\text{cr}}}\\ =\frac{\left(-100+273\right)\text{K}}{151\text{K}}\\ =1.146\end{array}$

$\begin{array}{c}{P}_{R1}=\frac{P_1}{P_{\text{cr}}}\\ =\frac{1\text{MPa}}{4.86\text{MPa}}\\ =0.206\end{array}$

At initial:

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

The enthalpy departure factor $\left(Z_{h1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.95$.

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

The compressibility factor $\left(Z_1\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.18$.

Conclusion:

Substitute $0.95$ for $Z$, $0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $-100°\text{C}$ for $T$, and $1\text{MPa}$ for $P$ in Equation (I).

$\begin{array}{c}v=\frac{\left(0.95\right)\left(0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(-100°\text{C}\right)}{1\text{MPa}}\\ =\frac{\left(0.95\right)\left(0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(-100+273\right)\text{K}}{1\text{MPa}\times \frac{{10}^3\text{kPa}}{1\text{MPa}}}\\ =\frac{34.2505\text{kPa}\cdot {\text{m}}^3\text{/kg}}{1000\text{kPa}}\\ =0.0342{\text{m}}^3\text{/kg}\end{array}$

Substitute $1.2{\text{m}}^3$ for $\nu$ and $0.0342{\text{m}}^3\text{/kg}$ for $v$ in Equation (II).

$\begin{array}{c}m= \frac{1.2{\text{m}}^3}{0.0342{\text{m}}^3\text{/kg}}\\ =35.0877\text{kg}\\ \simeq 35.1\text{kg}\end{array}$

Thus, the mass of argon in the tank is $35.1\text{kg}$.

(b)

To determine

The final pressure.

Answer to Problem 96RP

The final pressure is $1531\text{kPa}$.

Explanation of Solution

The reduced pressure $\left(P_{R2}\right)$ and temperature $\left(T_{R2}\right)$ at final state is expressed as follows.

$\begin{array}{c}{T}_{R2}=\frac{T_2}{T_{\text{cr}}}\\ =\frac{\left(0+273\right)\text{K}}{151\text{K}}\\ =1.808\end{array}$

$\begin{array}{c}{P}_{R2}=\frac{P_2}{P_{\text{cr}}}\\ =\frac{P_2}{4.86\text{MPa}\times \frac{1000\text{kPa}}{1\text{MPa}}}\\ =\frac{P_2}{4860\text{kPa}}\end{array}$ (III)

Write the formula for reduced specific volume.

$v_{R2}=\frac{v_2}{R{T}_{\text{cr}}/P_{\text{cr}}}$ (IV)

Here, the subscript 2 indicates the final state.

Conclusion:

Here, the specific volume at initial and final state is constant.

$v_2=v_1=0.0342{\text{m}}^3\text{/kg}$.

Substitute $0.0342{\text{m}}^3\text{/kg}$ for $v_2$, $0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $151\text{}\text{K}$ for $P_{\text{cr}}$, and $4.86\text{MPa}$ for $P_{\text{cr}}$ in Equation (IV).

$\begin{array}{c}{v}_{R2}=\frac{0.0342{\text{m}}^3\text{/kg}}{\left(0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\frac{151\text{}\text{K}}{4.86\text{MPa}}}\\ =\frac{0.0342{\text{m}}^3\text{/kg}}{\left(0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\frac{151\text{}\text{K}}{4860\text{kPa}}}\\ =\frac{0.0342{\text{m}}^3\text{/kg}}{0.00646{\text{m}}^3\text{/kg}}\\ =5.29\end{array}$

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 $0.99$.

The reduced pressure $\left(P_R{}_2\right)$ corresponding the reduced volume $\left(v_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.315$.

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

The enthalpy departure factor $\left(Z_{h2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.00$.

Substitute $0.315$ for $P_{R2}$ in Equation (III).

$\begin{array}{l}0.315=\frac{P_2}{4860\text{kPa}}\\ P_2=\left(0.315\right)4860\text{kPa}\\ P_2=1530.9\text{kPa}\\ P_2=1531\text{kPa}\end{array}$

Thus, the final pressure is $1531\text{kPa}$.

(c)

To determine

The heat transfer.

Answer to Problem 96RP

The heat transfer is $1251\text{kJ}$.

Explanation of Solution

Write formula for enthalpy departure factor $\left(Z_h\right)$.

$Z_h=\frac{{\left(h_{\text{ideal}}-h\right)}_{T,P}}{R{T}_{\text{cr}}}$ (V)

Here, the enthalpy at ideal gas state is $h_{\text{ideal}}$, the enthalpy and normal state is $h$, the gas constant of propane is $R$, 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 $h$.

$h=h_{\text{ideal}}-Z_{h}R{T}_{\text{cr}}$ (VI)

Refer Equation (II) express as two states of enthalpy difference (final – initial).

$h_2-h_1={\left(h_2-h_1\right)}_{\text{ideal}}-\left(Z_{h2}-Z_h{}_1\right)R{T}_{\text{cr}}$ (VII)

The enthalpy difference at ideal gas state is expressed as follows.

${\left(h_2-h_1\right)}_{\text{ideal}}=c_p\left(T_2-T_1\right)$ (VIII)

Here, the specific heat at constant pressure is $c_p$.

Write the energy balance equation for the system (piston-cylinder).

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ Q_{\text{in}}-0=\Delta u\\ Q_{\text{in}}=m\left(u_2-u_1\right)\end{array}$ (IX)

Here, the net energy in is $E_{\text{in}}$, the net energy out is $E_{\text{out}}$ and the change in net energy of the system is $\Delta E_{\text{system}}$.

The internal energy is expressed as follows.

$u=h-P\nu$

Here, the enthalpy is $h$, the pressure is $P$, and the volume is $\nu$.

The change in internal energy is expressed as follows.

$\begin{array}{c}{u}_2-u_1=h_2-h_1-\left(P_2{\nu }_2-P_1{\nu }_1\right)\\ =h_2-h_1-\left(Z_{2}R{T}_2-Z_{1}R{T}_1\right)\\ =h_2-h_1-R\left(Z_2{T}_2-Z_1{T}_1\right)\end{array}$ (X)

Substitute $h_2-h_1-R\left(Z_2{T}_2-Z_1{T}_1\right)$ for $u_2-u_1$ in Equation (X).

$Q_{\text{in}}=m\left[h_2-h_1-R\left(Z_2{T}_2-Z_1{T}_1\right)\right]$ (XI)

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

The specific heat at constant pressure $\left(c_p\right)$ is $0.5203\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $0.5203\text{kJ/kg}\cdot \text{K}$ for $c_p$, $0°\text{C}$ for $T_2$, and $-100°\text{C}$ in Equation (VIII).

$\begin{array}{c}{\left(h_2-h_1\right)}_{\text{ideal}}=0.5203\text{kJ/kg}\cdot \text{K}\left[0°\text{C}-\left(-100°\text{C}\right)\right]\\ =0.5203\text{kJ/kg}\cdot \text{K}\left[\left(0+273\right)\text{K}-\left(-100+273\right)\text{K}\right]\\ =52.03\text{kJ/kg}\end{array}$

Substitute $52.03\text{kJ/kg}$ for ${\left(h_2-h_1\right)}_{\text{ideal}}$, $0$ for $Z_{h2}$, $0.18$ for $Z_{h1}$, $0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $151\text{K}$ for $T_{\text{cr}}$ in Equation (VII).

$\begin{array}{c}{h}_2-h_1=\left\{\begin{array}{l}52.03\text{kJ/kg}\\ -\left[\left(0-0.18\right)\left(0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(151\text{K}\right)\right]\end{array}\right\}\\ =52.03\text{kJ/kg}+5.6561\text{kJ/kg}\\ =57.6861\text{kJ/kg}\\ \simeq 57.69\text{kJ/kg}\end{array}$

Substitute $57.69\text{kJ/kg}$ for $h_2-h_1$, $0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $0.99$ for $Z_2$, $273\text{K}$ for $T_2$, $0.95$ for $Z_1$, and $173\text{K}$ for $T_1$ in Equation (X).

$\begin{array}{c}{u}_2-u_1=\left\{\begin{array}{l}57.69\text{kJ/kg}\\ -\left(0.2081\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left[\left(0.99\times 273\text{K}\right)-\left(0.95\times 173\text{K}\right)\right]\end{array}\right\}\\ =57.69\text{kJ/kg}-22.0419\text{kJ/kg}\\ =35.648\text{kJ/kg}\end{array}$

Substitute $35.1\text{kg}$ for $m$ and $35.648\text{kJ/kg}$ for $u_2-u_1$ in Equation (XI).

$\begin{array}{c}{Q}_{\text{in}}=35.1\text{kg}\left(35.648\text{kJ/kg}\right)\\ =1251.2465\text{kJ}\\ \simeq 1251\text{kJ}\end{array}$

Thus, the heat transfer is $1251\text{kJ}$.