Chapter 13.3, Problem 42P

a)

To determine

The volumetric flow rate and mass flow rate of the mixture treating it as an ideal gas mixture.

Answer to Problem 42P

The volume flow rate is $0.001005{\text{m}}^{\text{3}}/\text{s}$.

The mass flow rate is $0.09535\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to obtain the mass of ${\text{O}}_{\text{2}}$ $\left(m_{{\text{O}}_{\text{2}}}\right)$.

$m_{{\text{O}}_{\text{2}}}=N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}$ (I)

Here, number of moles of ${\text{O}}_{\text{2}}$ is $N_{{\text{O}}_{\text{2}}}$ and molar mass of ${\text{O}}_{\text{2}}$ is $M_{{\text{O}}_{\text{2}}}$.

Write the expression to obtain the mass of ${\text{N}}_2$ $\left(m_{{\text{N}}_2}\right)$.

$m_{{\text{N}}_2}=N_{{\text{N}}_2}{M}_{{\text{N}}_2}$ (II)

Here, number of moles of ${\text{N}}_2$ is $N_{{\text{N}}_2}$ and molar mass of ${\text{N}}_2$ is $M_{{\text{N}}_2}$.

Write the expression to obtain the mass of ${\text{CO}}_{\text{2}}$ $\left(m_{{\text{CO}}_{\text{2}}}\right)$.

$m_{{\text{CO}}_{\text{2}}}=N_{{\text{CO}}_{\text{2}}}{M}_{{\text{CO}}_{\text{2}}}$ (III)

Here, number of moles of ${\text{CO}}_{\text{2}}$ is $N_{{\text{CO}}_{\text{2}}}$ and molar mass of ${\text{CO}}_{\text{2}}$ is $M_{{\text{CO}}_{\text{2}}}$.

Write the expression to obtain the mass of ${\text{CH}}_4$ $\left(m_{{\text{CH}}_4}\right)$.

$m_{{\text{CH}}_4}=N_{{\text{CH}}_4}{M}_{{\text{CH}}_4}$ (IV)

Here, number of moles of ${\text{CH}}_4$ is $N_{{\text{CH}}_4}$ and molar mass of ${\text{CH}}_4$ is $M_{{\text{CH}}_4}$.

Write the expression to obtain the total mass of each component $\left(m_m\right)$.

$m_m=m_{{\text{O}}_{\text{2}}}+m_{{\text{N}}_{\text{2}}}+m_{{\text{CO}}_{\text{2}}}+m_{{\text{CH}}_{\text{4}}}$ (V)

Write the expression to obtain the total number of moles $\left(N_m\right)$.

$N_m=N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{{\text{CO}}_{\text{2}}}+N_{{\text{CH}}_{\text{4}}}$ (VI)

Write the expression to obtain the molar mass of the mixture $\left(M_m\right)$.

$M_m=\frac{m_m}{N_m}$ (VII)

Write the expression to obtain the gas constant of the mixture $\left(R\right)$.

$R=\frac{R_u}{M_m}$ (VIII)

Here, universal gas constant is $R_u$.

Write the expression to obtain the specific volume of the mixture $\left(v\right)$.

$v=\frac{RT}{P}$ (IX)

Here, temperature and pressure is $T\text{and}P$.

Write the expression to obtain the volume flow rate $\left(\dot{V}\right)$.

$\begin{array}{c}\dot{V}=AV\\ =\frac{\pi D^2}{4}V\end{array}$ (X)

Here, velocity is V, area is A, and diameter is D.

Write the expression to obtain the mass flow rate $\left(\dot{m}\right)$.

$\dot{m}=\frac{\dot{V}}{v}$ (XI)

Conclusion:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of ${\text{O}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, ${\text{CO}}_{\text{2}}$, ${\text{CH}}_{\text{4}}$ as $32.0\text{kg}/\text{kmol}$, $28.0\text{kg}/\text{kmol}$, $44.0\text{kg}/\text{kmol}$, and $16.0\text{kg}/\text{kmol}$ respectively.

Substitute 30 kmol for $N_{{\text{O}}_{\text{2}}}$ and $32.0\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (I).

$\begin{array}{c}{m}_{{\text{O}}_{\text{2}}}=\left(30\text{kmol}\right)\left(32.0\text{kg}/\text{kmol}\right)\\ =960\text{kg}\end{array}$

Substitute 40 kmol for $N_{{\text{N}}_2}$ and $28.0\text{kg}/\text{kmol}$ for $M_{{\text{N}}_2}$ in Equation (II).

$\begin{array}{c}{m}_{{\text{N}}_{\text{2}}}=\left(40\text{kmol}\right)\left(28.0\text{kg}/\text{kmol}\right)\\ =1,120\text{kg}\end{array}$

Substitute 10 kmol for $N_{{\text{CO}}_{\text{2}}}$ and $44.0\text{kg}/\text{kmol}$ for $M_{{\text{CO}}_{\text{2}}}$ in Equation (III).

$\begin{array}{c}{m}_{{\text{CO}}_{\text{2}}}=\left(10\text{kmol}\right)\left(44.0\text{kg}/\text{kmol}\right)\\ =440\text{kg}\end{array}$

Substitute 20 kmol for $N_{{\text{CH}}_4}$ and $16.0\text{kg}/\text{kmol}$ for $M_{{\text{CH}}_4}$ in Equation (IV).

$\begin{array}{c}{m}_{{\text{CH}}_4}=\left(20\text{kmol}\right)\left(16.0\text{kg}/\text{kmol}\right)\\ =320\text{kg}\end{array}$

Substitute 960 kg for $m_{{\text{O}}_{\text{2}}}$, 1120 kg for $m_{{\text{N}}_{\text{2}}}$, 440 kg for $m_{{\text{CO}}_{\text{2}}}$, and 320 kg for $m_{{\text{CH}}_{\text{4}}}$ in Equation (V).

$\begin{array}{c}{m}_m=960\text{kg}+1120\text{kg}+\text{440}\text{kg}+\text{320}\text{kg}\\ =2840\text{kg}\end{array}$

Substitute $30\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $40\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $10\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$ and $20\text{kmol}$ for $N_{{\text{CH}}_{\text{4}}}$ in Equation (VI).

$\begin{array}{c}{N}_m=30\text{kmol}+40\text{kmol}+10\text{kmol}+20\text{kmol}\\ =100\text{kmol}\end{array}$

Substitute 2,840 kg for $m_m$ and 100 kmol for $N_m$ in Equation (VII).

$\begin{array}{c}{M}_m=\frac{2,840\text{kg}}{100\text{kmol}}\\ =28.40\text{kg}/\text{kmol}\end{array}$

Substitute $8.314\text{kJ}/\text{kmol}\cdot \text{K}$ for $R_u$ and $28.40\text{kg}/\text{kmol}$ for $M_m$ in Equation (VIII).

$\begin{array}{c}{R}_m=\frac{8.314\text{kJ}/\text{kmol}\cdot \text{K}}{28.40\text{kg}/\text{kmol}}\\ =0.2927\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.2927\text{kJ}/\text{kg}\cdot \text{K}$ for R, $15°\text{C}$ for T, and 8 MPa for P in Equation (IX).

$\begin{array}{c}v=\frac{\left(0.2927\text{kJ}/\text{kg}\cdot \text{K}\right)\left(15°\text{C}+273\right)\text{K}}{8\text{MPa}}\\ =\frac{\left(0.2927\text{kJ}/\text{kg}\cdot \text{K}\right)288\text{K}}{8\text{MPa}\times \text{1,000}\left(\frac{\text{kPa}}{1\text{MPa}}\right)}\\ =0.01054{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $0.016{\text{m}}^{\text{2}}$ for D, and $5\text{m}/\text{s}$ for V in Equation (X).

$\begin{array}{c}\dot{V}=\frac{\pi {\left(0.016{\text{m}}^{\text{2}}\right)}^2}{4}\left(5\text{m}/\text{s}\right)\\ =0.001005{\text{m}}^{\text{3}}/\text{s}\end{array}$

Thus, the volume flow rate is $0.001005{\text{m}}^{\text{3}}/\text{s}$.

Substitute $0.001005{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ and $0.01054{\text{m}}^{\text{3}}/\text{kg}$ for v in Equation (XI).

$\begin{array}{c}\dot{m}=\frac{0.001005{\text{m}}^{\text{3}}/\text{s}}{0.01054{\text{m}}^{\text{3}}/\text{kg}}\\ =0.09535\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate is $0.09535\text{kg}/\text{s}$.

b)

To determine

The volumetric flow rate and mass flow rate using a compressibility factor based on Amagad’s law of additive volume.

Answer to Problem 42P

The volume flow rate is $0.001005{\text{m}}^{\text{3}}/\text{s}$.

The mass flow rate is $0.1095\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to obtain the reduced temperature of ${\text{O}}_{\text{2}}$$\left(T_{R{\text{,O}}_{\text{2}}}\right)$.

$T_{R{\text{,O}}_{\text{2}}}=\frac{T_m}{T_{\text{cr},{\text{O}}_{\text{2}}}}$ (XII)

Here, critical temperature of ${\text{O}}_{\text{2}}$ is $T_{\text{cr},{\text{O}}_{\text{2}}}$.

Write the expression to obtain the reduced pressure of ${\text{O}}_{\text{2}}$$\left(P_{R{\text{,O}}_{\text{2}}}\right)$.

$P_{R{\text{,O}}_{\text{2}}}=\frac{P_m}{P_{\text{cr},{\text{O}}_{\text{2}}}}$ (XIII)

Here, critical temperature of ${\text{O}}_{\text{2}}$ is $P_{\text{cr},{\text{O}}_{\text{2}}}$.

Write the expression to obtain the reduced temperature of ${\text{N}}_{\text{2}}$$\left(T_{R{\text{,N}}_2}\right)$.

$T_{R{\text{,N}}_2}=\frac{T_m}{T_{\text{cr},{\text{N}}_2}}$ (XIV)

Here, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$.

Write the expression to obtain the reduced pressure of ${\text{N}}_2$$\left(P_{R{\text{,N}}_2}\right)$.

$P_{R{\text{,N}}_2}=\frac{P_m}{P_{\text{cr},{\text{N}}_2}}$ (XV)

Here, critical temperature of ${\text{N}}_2$ is $P_{\text{cr},{\text{N}}_2}$.

Write the expression to obtain the reduced temperature of ${\text{CO}}_2$$\left(T_{R{\text{,CO}}_2}\right)$.

$T_{R{\text{,CO}}_2}=\frac{T_m}{T_{\text{cr},{\text{CO}}_2}}$ (XVI)

Here, critical temperature of ${\text{CO}}_2$ is $T_{\text{cr},{\text{CO}}_2}$.

Write the expression to obtain the reduced pressure of ${\text{CO}}_2$$\left(P_{R{\text{,CO}}_2}\right)$.

$P_{R{\text{,CO}}_2}=\frac{P_m}{P_{\text{cr},{\text{CO}}_2}}$ (XVII)

Here, critical temperature of ${\text{CO}}_2$ is $P_{\text{cr},{\text{CO}}_2}$.

Write the expression to obtain the reduced temperature of ${\text{CH}}_4$$\left(T_{R{\text{,CH}}_4}\right)$.

$T_{R{\text{,CH}}_4}=\frac{T_m}{T_{\text{cr},{\text{CH}}_4}}$ (XVIII)

Here, critical temperature of ${\text{CH}}_4$ is $T_{\text{cr},{\text{CH}}_4}$.

Write the expression to obtain the reduced pressure of ${\text{CH}}_4$$\left(P_{R{\text{,CH}}_4}\right)$.

$P_{R{\text{,CH}}_4}=\frac{P_m}{P_{\text{cr},{\text{CH}}_4}}$ (XIX)

Here, critical temperature of ${\text{CH}}_4$ is $P_{\text{cr},{\text{CH}}_4}$.

Write the expression to obtain the compressibility factor of a mixture $\left(Z_m\right)$.

$\begin{array}{c}{Z}_m={\displaystyle \sum y_i{Z}_i}\\ =y_{{\text{O}}_{\text{2}}}{Z}_{{\text{O}}_{\text{2}}}+y_{{\text{N}}_{\text{2}}}{Z}_{{\text{N}}_{\text{2}}}+y_{{\text{CO}}_{\text{2}}}{Z}_{{\text{CO}}_{\text{2}}}+y_{{\text{CH}}_{\text{4}}}{Z}_{{\text{CH}}_{\text{4}}}\end{array}$ (XX)

Here, compressibility factor of ${\text{O}}_{\text{2}}$ is $Z_{{\text{O}}_{\text{2}}}$, compressibility factor of ${\text{N}}_2$ is $Z_{{\text{N}}_{\text{2}}}$, compressibility factor of ${\text{CO}}_{\text{2}}$ is $Z_{{\text{CO}}_{\text{2}}}$, and compressibility factor of ${\text{CH}}_4$ is $Z_{{\text{CH}}_4}$.

Write the expression to obtain the specific volume of the mixture $\left(v\right)$.

$v=\frac{Z_{m}RT}{P}$ (XXI)

Write the expression to obtain the volume flow rate $\left(\dot{V}\right)$.

$\begin{array}{c}\dot{V}=AV\\ =\frac{\pi D^2}{4}V\end{array}$ (XXII)

Here, velocity is V, area is A, and diameter is D.

Write the expression to obtain the mass flow rate $\left(\dot{m}\right)$.

$\dot{m}=\frac{\dot{V}}{v}$ (XXIII)

Conclusion:

Substitute 288 K for $T_m$ and 154.8 K for $T_{\text{cr},{\text{O}}_{\text{2}}}$ in Equation (XII).

$\begin{array}{c}{T}_{R{\text{,O}}_{\text{2}}}=\frac{288\text{}\text{K}}{154.8\text{K}}\\ =1.860\end{array}$

Substitute 8 MPa for $P_m$ and 5.08 MPa for $P_{\text{cr},{\text{O}}_{\text{2}}}$ in Equation (XIII).

$\begin{array}{c}{P}_{R{\text{,O}}_{\text{2}}}=\frac{8\text{MPa}}{5.08\text{MPa}}\\ =1.575\end{array}$

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for ${\text{O}}_{\text{2}}$ as 0.95.

Substitute 288 K for $T_m$ and 126.2 K for $T_{\text{cr},{\text{N}}_2}$ in Equation (XIV).

$\begin{array}{c}{T}_{R{\text{,N}}_2}=\frac{288\text{}\text{K}}{126.2\text{K}}\\ =2.282\end{array}$

Substitute 8 MPa for $P_m$ and 3.39 MPa for $P_{\text{cr},{\text{N}}_2}$ in Equation (XV).

$\begin{array}{c}{P}_{R{\text{,N}}_2}=\frac{8\text{MPa}}{3.39\text{MPa}}\\ =2.360\end{array}$

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for ${\text{N}}_2$ as 0.99.

Substitute 288 K for $T_m$ and 304.2 K for $T_{\text{cr},{\text{CO}}_2}$ in Equation (XVI).

$\begin{array}{c}{T}_{R{\text{,CO}}_2}=\frac{288\text{}\text{K}}{304.2\text{K}}\\ =0.947\end{array}$

Substitute 8 MPa for $P_m$ and 7.39 MPa for $P_{\text{cr},{\text{CO}}_2}$ in Equation (XVII).

$\begin{array}{c}{P}_{R{\text{,CO}}_2}=\frac{8\text{MPa}}{7.39\text{MPa}}\\ =1.083\end{array}$

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for ${\text{CO}}_2$ as 0.199.

Substitute 288 K for $T_m$ and 191.1 K for $T_{\text{cr},{\text{CH}}_4}$ in Equation (XVIII).

$\begin{array}{c}{T}_{R{\text{,CH}}_4}=\frac{288\text{}\text{K}}{191.1\text{K}}\\ =1.507\end{array}$

Substitute 8 MPa for $P_m$ and 4.64 MPa for $P_{\text{cr},{\text{CH}}_4}$ in Equation (XIX).

$\begin{array}{c}{P}_{R{\text{,CH}}_4}=\frac{8\text{MPa}}{4.64\text{MPa}}\\ =1.724\end{array}$

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for ${\text{CH}}_4$ as 0.85.

Substitute 0.30 for $y_{{\text{O}}_{\text{2}}}$, 0.95 for $Z_{{\text{O}}_{\text{2}}}$, 0.40 for $y_{{\text{N}}_{\text{2}}}$, 0.99 for $Z_{{\text{N}}_{\text{2}}}$, 0.10 for $y_{{\text{CO}}_{\text{2}}}$, 0.199 for $Z_{{\text{CO}}_{\text{2}}}$, 0.20 for $y_{{\text{CH}}_{\text{4}}}$, and 0.85 for $Z_{{\text{CH}}_4}$ in Equation (XX).

$\begin{array}{c}{Z}_m=0.30\left(0.95\right)+0.40\left(0.99\right)+0.10\left(0.199\right)+0.20\left(0.85\right)\\ =0.8709\end{array}$

Substitute 0.8709 for $Z_m$, $0.2927\text{kJ}/\text{kg}\cdot \text{K}$ for R, $15°\text{C}$ for T, and 8 MPa for P in Equation (XXI).

$\begin{array}{c}v=\frac{0.8709\left(0.2927\text{kJ}/\text{kg}\cdot \text{K}\right)\left(15°\text{C}+273\right)\text{K}}{8\text{MPa}\left(\frac{{\text{10}}^3\text{kPa}}{1\text{MPa}}\right)}\\ =0.009178{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $0.016{\text{m}}^{\text{2}}$ for D, and $5\text{m}/\text{s}$ for V in Equation (XXII).

$\begin{array}{c}\dot{V}=\frac{\pi {\left(0.016{\text{m}}^{\text{2}}\right)}^2}{4}\left(5\text{m}/\text{s}\right)\\ =0.001005{\text{m}}^{\text{3}}/\text{s}\end{array}$

Thus, the volume flow rate is $0.001005{\text{m}}^{\text{3}}/\text{s}$.

Substitute $0.001005{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ and $0.009178{\text{m}}^{\text{3}}/\text{kg}$ for v in Equation (XXIII).

$\begin{array}{c}\dot{m}=\frac{0.001005{\text{m}}^{\text{3}}/\text{s}}{0.009178{\text{m}}^{\text{3}}/\text{kg}}\\ =0.1095\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate is $0.1095\text{kg}/\text{s}$.

c)

To determine

The volumetric flow rate and mass flow rate using Key’s pseudocritical pressure and temperature.

Answer to Problem 42P

The volume flow rate is $0.001005{\text{m}}^{\text{3}}/\text{s}$.

The mass flow rate is $0.1037\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to obtain the pseudo-critical temperature of the mixture $\left({T^{\prime }}_{\text{cr},m}\right)$.

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{T}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{T}_{{\text{cr,O}}_{\text{2}}}+y_{{\text{N}}_{\text{2}}}{T}_{{\text{cr,N}}_{\text{2}}}+y_{{\text{CO}}_{\text{2}}}{T}_{{\text{cr,CO}}_{\text{2}}}+y_{{\text{CH}}_{\text{4}}}{T}_{{\text{cr,CH}}_{\text{4}}}\end{array}$ (XXIV)

Here, critical temperature of ${\text{O}}_{\text{2}},{\text{N}}_{\text{2}},{\text{CO}}_{\text{2}},\text{and}{\text{CH}}_{\text{4}}$ are $T_{{\text{cr,O}}_{\text{2}}},T_{{\text{cr,N}}_{\text{2}}},T_{{\text{cr,CO}}_{\text{2}}},and{T}_{{\text{cr,CH}}_{\text{4}}}$.

Write the expression to obtain the pseudo-critical pressure of the mixture $\left({P^{\prime }}_{\text{cr},m}\right)$.

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{P}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{P}_{{\text{cr,O}}_{\text{2}}}+y_{{\text{N}}_{\text{2}}}{P}_{{\text{cr,N}}_{\text{2}}}+y_{{\text{CO}}_{\text{2}}}{P}_{{\text{cr,CO}}_{\text{2}}}+y_{{\text{CH}}_{\text{4}}}{P}_{{\text{cr,CH}}_{\text{4}}}\end{array}$ (XXV)

Here, critical pressure of ${\text{O}}_{\text{2}},{\text{N}}_{\text{2}},{\text{CO}}_{\text{2}},\text{and}{\text{CH}}_{\text{4}}$ are $P_{{\text{cr,O}}_{\text{2}}},P_{{\text{cr,N}}_{\text{2}}},P_{{\text{cr,CO}}_{\text{2}}},and{P}_{{\text{cr,CH}}_{\text{4}}}$.

Write the expression to obtain the reduced temperature $\left(T_R\right)$.

$T_R=\frac{T_m}{{T^{\prime }}_{\text{cr},m}}$ (XXVI)

Write the expression to obtain the reduced pressure $\left(P_R\right)$.

$P_R=\frac{P_m}{{P^{\prime }}_{\text{cr},m}}$ (XXVII)

Write the expression to obtain the specific volume of the mixture $\left(v\right)$.

$v=\frac{Z_{m}RT}{P}$ (XXVIII)

Write the expression to obtain the volume flow rate $\left(\dot{V}\right)$.

$\begin{array}{c}\dot{V}=AV\\ =\frac{\pi D^2}{4}V\end{array}$ (XXIX)

Write the expression to obtain the mass flow rate $\left(\dot{m}\right)$.

$\dot{m}=\frac{\dot{V}}{v}$ (XXX)

Conclusion:

Substitute 0.30 for $y_{{\text{O}}_{\text{2}}}$, 154.8 K for $T_{{\text{cr,O}}_{\text{2}}}$, 0.40 for $y_{{\text{N}}_{\text{2}}}$, 126.2 K for $T_{{\text{cr,N}}_{\text{2}}}$, 0.10 for $y_{{\text{CO}}_{\text{2}}}$, 304.2 K for $T_{{\text{cr,CO}}_{\text{2}}}$, 0.20 for $y_{{\text{CH}}_{\text{4}}}$, and 191.1 K for $T_{{\text{cr,CH}}_4}$ in Equation (XXIV).

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}=0.30\left(154.8\text{K}\right)+0.40\left(126.2\text{K}\right)+0.10\left(304.2\text{K}\right)+0.20\left(191.1\text{K}\right)\\ =165.6\text{K}\end{array}$

Substitute 0.30 for $y_{{\text{O}}_{\text{2}}}$, 5.08 MPa for $P_{{\text{cr,O}}_{\text{2}}}$, 0.40 for $y_{{\text{N}}_{\text{2}}}$, 3.39 MPa for $P_{{\text{cr,N}}_{\text{2}}}$, 0.10 for $y_{{\text{CO}}_{\text{2}}}$, 7.39 MPa for $P_{{\text{cr,CO}}_{\text{2}}}$, 0.20 for $y_{{\text{CH}}_{\text{4}}}$, and 4.64 MPa for $P_{{\text{cr,CH}}_4}$ in Equation (XXV).

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}=0.30\left(5.08\text{MPa}\right)+0.40\left(3.39\text{MPa}\right)+0.10\left(7.39\text{MPa}\right)+0.20\left(4.64\text{MPa}\right)\\ =4.547\text{MPa}\end{array}$

Substitute 288 K for $T_m$ and 165.6 K for ${T^{\prime }}_{\text{cr},m}$ in Equation (XXVI).

$\begin{array}{c}{T}_R=\frac{288\text{K}}{165.6\text{K}}\\ =1.739\end{array}$

Substitute 8 MPa for $P_m$ and 4.547 MPa for ${P^{\prime }}_{\text{cr},m}$ in Equation (XXVII).

$\begin{array}{c}{P}_R=\frac{8\text{MPa}}{4.547\text{MPa}}\\ =1.759\end{array}$

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, $Z_m$ as 0.92.

Substitute 0.92 for $Z_m$, $0.2927\text{kJ}/\text{kg}\cdot \text{K}$ for R, $15°\text{C}$ for T, and 8 MPa for P in Equation (XXVIII).

$\begin{array}{c}v=\frac{0.92\left(0.2927\text{kJ}/\text{kg}\cdot \text{K}\right)\left(15°\text{C}+273\right)\text{K}}{8\text{MPa}\left(\frac{{\text{10}}^3\text{kPa}}{1\text{MPa}}\right)}\\ =0.009694{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $0.016{\text{m}}^{\text{2}}$ for D, and $5\text{m}/\text{s}$ for V in Equation (XXIX).

$\begin{array}{c}\dot{V}=\frac{\pi {\left(0.016{\text{m}}^{\text{2}}\right)}^2}{4}\left(5\text{m}/\text{s}\right)\\ =0.001005{\text{m}}^{\text{3}}/\text{s}\end{array}$

Thus, the volume flow rate is $0.001005{\text{m}}^{\text{3}}/\text{s}$.

Substitute $0.001005{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ and $0.009694{\text{m}}^{\text{3}}/\text{kg}$ for v in Equation (XXX).

$\begin{array}{c}\dot{m}=\frac{0.001005{\text{m}}^{\text{3}}/\text{s}}{0.009694{\text{m}}^{\text{3}}/\text{kg}}\\ =0.1037\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate is $0.1037\text{kg}/\text{s}$.