Chapter 13.3, Problem 45P

(a)

To determine

The volume flow rate of the mixture using the ideal gas mixture.

The mass flow rate of the mixture using the ideal gas mixture.

Answer to Problem 45P

The volume flow rate of the mixture using the ideal gas mixture is $\underset{\_}{0.05454{\text{ft}}^{\text{3}}/\text{s}}$.

The mass flow rate of the mixture using the ideal gas mixture is $\underset{\_}{0.408\text{lbm/s}}$.

Explanation of Solution

Refer to Table A-1E, Obtain the molar masses of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$ as below:

$\begin{array}{l}{M}_{{\text{O}}_{\text{2}}}=32.0\text{kg/kmol}\\ M_{{\text{N}}_{\text{2}}}=28.0\text{kg/kmol}\\ M_{{\text{CO}}_{\text{2}}}=44.0\text{kg/kmol}\\ M_{{\text{CH}}_4}=16.0\text{kg/kmol}\end{array}$

Consider 100 lbmol of the mixture. Since the volume fractions are equal to the mole fractions, calculate the mass of each component.

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

$m_{{\text{N}}_{\text{2}}}=N_{{\text{N}}_2}{M}_{{\text{N}}_2}$ (II)

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

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

Here, the mole numbers of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$ are $N_{{\text{O}}_{\text{2}}},N_{{\text{N}}_2},N_{{\text{CO}}_{\text{2}}},\text{and}{N}_{{\text{CH}}_4}$ respectively.

Write the equation of total mass of the mixture.

$m_m=m_{{\text{CO}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}+m_{{\text{N}}_{\text{2}}}+m_{{\text{CH}}_4}$ (V)

Here, the mass of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$ are $m_{{\text{O}}_{\text{2}}},m_{{\text{N}}_{\text{2}}},m_{{\text{CO}}_{\text{2}}},\text{and}{m}_{{\text{CH}}_4}$ respectively.

Write the equation to calculate the apparent molecular weight of the mixture.

$\begin{array}{c}{M}_m=\frac{m_m}{N_m}\\ =\frac{m_m}{N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_2}+N_{{\text{CO}}_{\text{2}}}+N_{{\text{CH}}_4}}\end{array}$ (VI)

Write the equation to calculate the apparent gas constant of the mixture.

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

Here, the universal gas constant is $R_u$.

Write the equation of specific volume of the mixture.

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

Here, temperature of the mixture is T and atmospheric pressure is P.

Calculate the volume flow rate of the mixture.

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

Here, cross-sectional area of the pipe is A.

Calculate the mass flow rate of the mixture.

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

Conclusion:

Apply spreadsheet and substitute the given values of mole numbers and molar masses of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$ in Equations (I) to (IV) as in Table (I).

S.No	masses	Mole number (N), lbmol	Molar masses (M), lbm/lbmol	$M\times N\left(\text{lbm}\right)$
1	$m_{{\text{O}}_{\text{2}}}$	30	32	960
2	$m_{{\text{N}}_{\text{2}}}$	40	28	1120
3	$m_{{\text{CO}}_{\text{2}}}$	10	44	440
4	$m_{{\text{CH}}_{\text{4}}}$	20	16	320

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

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

Substitute 2840 lbm for $m_m$, 30 lbmol for $N_{{\text{O}}_{\text{2}}}$, 40 lbmol for $N_{{\text{N}}_2}$, 10 lbmol for $N_{{\text{CO}}_{\text{2}}}$, and 20 lbmol for $N_{{\text{CH}}_4}$ in Equation (VI).

$\begin{array}{c}{M}_m=\frac{2840\text{lbm}}{30\text{lbmol}+40\text{lbmol}+10\text{lbmol}+20\text{lbmol}}\\ =28.40\text{lbm/lbmol}\end{array}$

Substitute $10.73\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbmol}\cdot \text{R}$ for $R_u$ and $28.40\text{lbm/lbmol}$ for $M_m$ in Equation (VII).

$\begin{array}{c}R=\frac{10.73\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbmol}\cdot \text{R}}{28.40\text{lbm/lbmol}}\\ =0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R, $75°\text{F}$  for T, and 1500 psia for P in Equation (VIII).

$\begin{array}{c}v=\frac{\left(0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)70°\text{F}}{1500\text{psia}}\\ =\frac{\left(0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(70+460.67\right)\text{R}}{1500\text{psia}}\\ =0.13365{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Substitute $10\text{ft/s}$ for V and 1 in for D in Equation (IX).

$\begin{array}{c}\dot{V}=\frac{\pi }{4}{\left(\text{1}\text{in}\right)}^{2}10\text{ft/s}\\ =\frac{\pi }{4}{\left(\text{1}\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)}^{2}10\text{ft/s}\\ =0.05454{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Thus, the volume flow rate of the mixture using the ideal gas mixture is $\underset{\_}{0.05454{\text{ft}}^{\text{3}}/\text{s}}$.

Substitute $0.05454{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{V}$ and $0.13365{\text{ft}}^{\text{3}}/\text{lbm}$ for v in Equation (X).

$\begin{array}{c}\dot{m}=\frac{0.05454{\text{ft}}^{\text{3}}/\text{s}}{0.13365{\text{ft}}^{\text{3}}/\text{lbm}}\\ =0.408\text{lbm/s}\end{array}$

Thus, the mass flow rate of the mixture using the ideal gas mixture is $\underset{\_}{0.408\text{lbm/s}}$.

(b)

To determine

The volume flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes.

The mass flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes.

Answer to Problem 45P

The volume flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes is $\underset{\_}{0.05454{\text{ft}}^{\text{3}}/\text{s}}$.

The mass flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes is $\underset{\_}{0.4702\text{lbm/s}}$.

Explanation of Solution

Write the equation of reduced temperatures and pressures of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$.

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

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

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

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

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

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

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

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

Here, the critical temperature of ${\text{O}}_{\text{2}}$ is $T_{\text{cr},{\text{O}}_{\text{2}}}$, mixture temperature is $T_m$, mixture pressure is $P_m$, critical temperature of ${\text{N}}_2$ is $T_{\text{cr},{\text{N}}_2}$, critical temperature of ${\text{CO}}_{\text{2}}$ is $T_{\text{cr},{\text{CO}}_{\text{2}}}$, critical temperature of ${\text{CH}}_4$ is $T_{\text{cr},{\text{CH}}_4}$, critical pressure of ${\text{N}}_2$ is $P_{\text{cr},{\text{N}}_2}$, critical pressure of ${\text{CO}}_{\text{2}}$ is $P_{\text{cr},{\text{CO}}_{\text{2}}}$, critical pressure of ${\text{CH}}_4$ is $P_{\text{cr},{\text{CH}}_4}$,

Write the equation of compressibility factor of the mixture.

$\begin{array}{c}{Z}_m={\displaystyle \sum y_i{Z}_i}\\ =y_{{\text{O}}_{\text{2}}}{Z}_{{\text{O}}_{\text{2}}}+y_{{\text{N}}_2}{Z}_{{\text{N}}_2}+y_{{\text{CO}}_{\text{2}}}{Z}_{{\text{CO}}_{\text{2}}}+y_{{\text{CH}}_4}{Z}_{{\text{CH}}_4}\end{array}$ (XIX)

Here, the mole fraction of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$ are $y_{{\text{O}}_{\text{2}}},y_{{\text{N}}_{\text{2}}},y_{{\text{CO}}_{\text{2}}},\text{and}{y}_{{\text{CH}}_4}$ respectively.

Calculate the specific volume of the mixture.

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

Calculate the mass flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes.

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

Here, the volume flow rate of mixture using the compressibility factor based on Amagat’s law of additive volumes is $\dot{V}$.

Conclusion:

Refer to Table A-1E, obtain the critical temperatures and pressures of ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$.

$\begin{array}{l}{T}_{\text{cr},{\text{O}}_{\text{2}}}=278.6\text{R}\\ P_{\text{cr},{\text{O}}_{\text{2}}}=736\text{psia}\\ T_{\text{cr},{\text{N}}_2}=227.1\text{R}\\ P_{\text{cr},{\text{N}}_2}=492\text{psia}\end{array}$

$\begin{array}{l}{T}_{\text{cr},{\text{CO}}_{\text{2}}}=547.5\text{R}\\ P_{\text{cr},{\text{CO}}_{\text{2}}}=1071\text{psia}\\ T_{\text{cr},{\text{CH}}_4}=343.9\text{R}\\ P_{\text{cr},{\text{CH}}_4}=673\text{psia}\end{array}$

Substitute 530 R for $T_m$, 1500 psia for $P_m$ and obtained values from Table A-1E in Equations (XI) to (XVIII).

$\begin{array}{l}{T}_{R,{\text{O}}_{\text{2}}}=1.902\\ P_{R,{\text{O}}_{\text{2}}}=2.038\\ T_{R,{\text{N}}_2}=2.334\\ P_{R,{\text{N}}_2}=3.049\end{array}$

$\begin{array}{l}{T}_{R,{\text{CO}}_{\text{2}}}=0.968\\ P_{R,{\text{CO}}_{\text{2}}}=1.401\\ T_{R,{\text{CH}}_4}=1.541\\ P_{R,{\text{CH}}_4}=2.229\end{array}$

Refer to Figure A-15, obtain the compressibility factor for ${\text{O}}_2{\text{,N}}_{\text{2}},{\text{CO}}_2\text{,}\text{and}{\text{CH}}_4$ by reading the above values of reduced temperatures and reduced pressure.

$\begin{array}{l}{Z}_{{\text{O}}_{\text{2}}}=0.94\\ Z_{{\text{N}}_2}=0.99\\ Z_{{\text{CO}}_{\text{2}}}=0.21\\ Z_{{\text{CH}}_4}=0.85\end{array}$

Substitute 0.30 for $y_{{\text{O}}_{\text{2}}}$, 0.94 for $Z_{{\text{O}}_{\text{2}}}$, 0.40 for $y_{{\text{N}}_2}$, 0.99 for $Z_{{\text{N}}_2}$, 0.10 for $y_{{\text{CO}}_{\text{2}}}$, 0.21 for $Z_{{\text{CO}}_{\text{2}}}$, 0.20 for $y_{{\text{CH}}_4}$, and 0.85 for $Z_{{\text{CH}}_4}$ in Equation (XIX).

$\begin{array}{c}{Z}_m=\left(0.30\right)\left(0.94\right)+\left(0.40\right)\left(0.99\right)+\left(0.10\right)\left(0.21\right)+\left(0.20\right)\left(0.85\right)\\ =0.869\end{array}$

Substitute 1500 psia for P, 0.869 for $Z_m$, $0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R, and 530 R for T in Equation (XX).

$\begin{array}{c}v=\frac{0.869\left(0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)530\text{R}}{1500\text{psia}}\\ =0.1160{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Refer to part (a), the value calculated for volume flow rate is $\underset{\_}{0.05454{\text{ft}}^{\text{3}}/\text{s}}$.

Substitute $0.05454{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{V}$  and $0.1160{\text{ft}}^{\text{3}}/\text{lbm}$ for v in Equation (XXI).

$\begin{array}{c}\dot{m}=\frac{0.05454{\text{ft}}^{\text{3}}/\text{s}}{0.1160{\text{ft}}^{\text{3}}/\text{lbm}}\\ =0.4702\text{lbm/s}\end{array}$

Thus, the mass flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes is $\underset{\_}{0.4702\text{lbm/s}}$.

(c)

To determine

The volume flow rate of the mixture using Kay’s pseudocritical pressure and temperature.

The mass flow rate of the mixture using Kay’s pseudocritical pressure and temperature.

Answer to Problem 45P

The volume flow rate of the mixture using Kay’s pseudocritical pressure and temperature is $\underset{\_}{0.05454{\text{ft}}^{\text{3}}/\text{s}}$.

The mass flow rate of the mixture using Kay’s pseudocritical pressure and temperature.

is $\underset{\_}{0.4467\text{lbm}/\text{s}}$.

Explanation of Solution

Write the critical temperature of a gas mixture.

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{T}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{T}_{\text{cr},{\text{O}}_{\text{2}}}+y_{{\text{N}}_2}{T}_{\text{cr},{\text{N}}_2}+y_{{\text{CO}}_{\text{2}}}{T}_{\text{cr},{\text{CO}}_{\text{2}}}+y_{{\text{CH}}_4}{T}_{\text{cr},{\text{CH}}_4}\end{array}$ (XXII)

Write the critical pressure of a gas mixture.

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}={\displaystyle \sum y_i{P}_{\text{cr},i}}\\ =y_{{\text{O}}_{\text{2}}}{P}_{\text{cr},{\text{O}}_{\text{2}}}+y_{{\text{N}}_2}{P}_{\text{cr},{\text{N}}_2}+y_{{\text{CO}}_{\text{2}}}{P}_{\text{cr},{\text{CO}}_{\text{2}}}+y_{{\text{CH}}_4}{P}_{\text{cr},{\text{CH}}_4}\end{array}$ (XXIII)

Write the equation of reduced temperature and pressure.

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

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

Conclusion:

Substitute 0.30 for $y_{{\text{O}}_{\text{2}}}$, 278.6 R for $T_{\text{cr},}{}_{{\text{O}}_{\text{2}}}$, 0.40 for $y_{{\text{N}}_2}$, 227.1 R for $T_{\text{cr},}{}_{{\text{N}}_2}$, 0.10 for $y_{{\text{CO}}_{\text{2}}}$, 547.5 R for $T_{\text{cr},}{}_{{\text{CO}}_{\text{2}}}$, 0.20 for $y_{{\text{CH}}_4}$, and 343.9 R for $T_{\text{cr},}{}_{{\text{CH}}_4}$ in Equation (XXII).

$\begin{array}{c}{T^{\prime }}_{\text{cr},m}=\left(0.30\right)\left(278.6\text{R}\right)+\left(0.40\right)\left(227.1\text{R}\right)+\left(0.10\right)\left(547.5\text{R}\right)+\left(0.20\right)\left(343.9\text{R}\right)\\ =298.0\text{R}\end{array}$

Substitute 0.30 for $y_{{\text{O}}_{\text{2}}}$, 278.6 R for $P_{\text{cr},}{}_{{\text{O}}_{\text{2}}}$, 0.40 for $y_{{\text{N}}_2}$, 227.1 R for $P_{\text{cr},}{}_{{\text{N}}_2}$, 0.10 for $y_{{\text{CO}}_{\text{2}}}$, 547.5 R for $P_{\text{cr},}{}_{{\text{CO}}_{\text{2}}}$, 0.20 for $y_{{\text{CH}}_4}$, and 343.9 R for $P_{\text{cr},}{}_{{\text{CH}}_4}$ in Equation (XXIII).

$\begin{array}{c}{P^{\prime }}_{\text{cr},m}=\left(0.30\right)\left(736\text{psia}\right)+\left(0.40\right)\left(492\text{psia}\right)+\left(0.10\right)\left(1071\text{psia}\right)+\left(0.20\right)\left(673\text{psia}\right)\\ =659.3\text{psia}\end{array}$

Substitute 530 R for $T_m$, 1500 psia for $P_m$, 298.0 R for ${T^{\prime }}_{\text{cr},m}$, and 659.3 psia for ${P^{\prime }}_{\text{cr},m}$ in Equations (XXIV) and (XXV).

$\begin{array}{l}{T}_R=1.779\\ P_R=2.275\end{array}$

Refer to Figure A-15, obtain the compressibility factor for gas mixture by reading the values of $T_R\text{and}{P}_R$.

$Z_m=0.915$

Substitute 1500 psia for P, 0.915 for $Z_m$, $0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for R, and 530 R for T in Equation (XX).

$\begin{array}{c}v=\frac{0.915\left(0.3778\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)530\text{R}}{1500\text{psia}}\\ =0.1221{\text{ft}}^{\text{3}}/\text{lbm}\end{array}$

Refer to part (a), the value calculated for volume flow rate is $\underset{\_}{0.05454{\text{ft}}^{\text{3}}/\text{s}}$.

Substitute $0.05454{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{V}$  and $0.1221{\text{ft}}^{\text{3}}/\text{lbm}$ for v in Equation (XXI).

$\begin{array}{c}\dot{m}=\frac{0.05454{\text{ft}}^{\text{3}}/\text{s}}{0.1221{\text{ft}}^{\text{3}}/\text{lbm}}\\ =0.4467\text{lbm/s}\end{array}$

Thus, the mass flow rate of the mixture using the compressibility factor based on Amagat’s law of additive volumes is $\underset{\_}{0.4467\text{lbm/s}}$.