Chapter 10, Problem 10.27P

Interpretation Introduction

Interpretation:

Estimation of the fugacity for methane/ethane/propane in a mixture. Also estimate fugaciy coefficient for gases in the mixture.

Concept introduction:

The calculation of fugacity and fugacity coefficient of gases in the gas mixture is possible by using different correlation given as follows:

The fugacity coefficient for gas can be calculated by the equation given in the book as follows:

  $\widehat{{\varphi }_k}=\exp \left[\frac{P}{RT}\left[B_{k,k}+\frac{1}{2}{\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j\left(2{\delta }_{i,k}-{\delta }_{i,j}\right)}}\right]\right]$

Where,

  $\begin{array}{l}{\delta }_{i,j}=2B_{i,j}-B_{i,i}-B_{j,j}\\ B_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]\\ B_{i,j}^0=0.083-\frac{0.422}{{\left(T_{r_{i,j}}\right)}^{1.6}}\\ B_{i,j}^1=0.139-\frac{0.172}{{\left(T_{r_{i,j}}\right)}^{4.2}}\\ {\omega }_{i,j}=\frac{{\omega }_i+{\omega }_j}{2}\\ T_{c_{i,j}}=\sqrt{T_{c_i}.T_{c_j}}\\ Z_{c_{i,j}}=\frac{Z_{c_i}+Z_{c_j}}{2}\\ V_{C_{i,j}}={\left[\frac{{\left(V_{c_i}\right)}^{1/3}+{\left(V_{c_j}\right)}^{1/3}}{2}\right]}^3\\ P_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}\end{array}$

$\widehat{{\varphi }_i}$ = Fugacity coefficient of gases

$P_{c_{i,j}}$ = Critical pressure of gases

$V_{c_{i,j}}$ = Volume of gases

T = temperature

$B^0$ and $B^1$ are the correlation coefficient given in the book.

  $P_r$ = Reduced pressure

  $P_r=\frac{P_i^{sat}}{P_c}$

Where,

$P_c$ = Critical pressure

Similarly, $T_r$ = reduced temperature = $T_{r_{i,j}}=\frac{T}{T_{C_{i,j}}}$

The fugacity of gases can be calculated by the definition of fugacity coefficient that is fugacity coefficient is the ratio of fugacity and pressure.

  $\begin{array}{l}\widehat{{\varphi }_i}=\frac{\widehat{f_i}}{y_i\times P_i}\\ \widehat{f_i}=\widehat{{\varphi }_i}\times y_i\times P_i\end{array}$

Answer to Problem 10.27P

Fugacity and fugacity coefficient of methane are 7.4896 bar and 1.019, respectively.

Fugacity and fugacity coefficient of ethane are 13.259 bar and 0.881, respectively.

Fugacity and fugacity coefficient of propane are 9.765 bar and 0.775, respectively.

Explanation of Solution

Refer APPENDIX-B and Table-B.1 to determine critical properties and acentric factor of ethylene(1)/Propylene (2) as:

Component	Pc (bar)	Tc (K)	Vc (cm3/mol)	$\omega$	Zc
Methane (1)	45.99	190.6	98.6	0.012	0.286
Ethane (2)	48.72	305.3	145.5	0.100	0.279
Propane (3)	42.48	369.8	200	0.152	0.276

Where,

P_c = critical pressure

T_c = critical temperature

V_c = critical volume

Z_c = critical compressibility factor

  $\omega$ = eccentric factor

From the properties derived above, we can calculate the reduced pressure and temperature as follows:

Pressure P = 35 bar (given in question) and Temperature (T) = 100⁰C or 373.15K (given)

Y₁= 0.21 ; y₂ = 0.43 (given)

So,

$\begin{array}{l}{y}_3=1-y_1-y_2\\ y_3=1-0.21-0.43\\ y_3=0.36\end{array}$

Reduced temperature of gases can be calculated as:

  $\begin{array}{l}{T}_{c_{i,j}}=\sqrt{T_{c_i}.T_{c_j}}\\ T_{c_{1,1}}=\sqrt{T_{c_1}.T_{c_1}}\\ T_{c_{1,1}}=\sqrt{190.6\times 190.6}\\ T_{c_{1,1}}=190.6K\\ T_{c_{i,j}}=\left(\begin{array}{ccc}190.6& 241.23& 265.49\\ 241.226& 305.3& 336\\ 265.488& 336& 369.8\end{array}\right)\end{array}$

Similar to the above calculated matrix we will calculate the reduced temperature as follows:

  $\begin{array}{l}{T}_{r_{i,j}}=\frac{T}{T_{C_{i,j}}}\\ T_{r_{1,1}}=\frac{373.15}{190.6}=1.958\end{array}$

  $T_{r_{i,j}}=\left(\begin{array}{ccc}1.958& 1.547& 1.406\\ 1.547& 1.222& 1.11\\ 1.406& 1.11& 1.009\end{array}\right)$

Critical volume of both the gases in the mixture can be calculated as given below:

  $\begin{array}{l}{V}_{C_{i,j}}={\left[\frac{{\left(V_{c_i}\right)}^{1/3}+{\left(V_{c_j}\right)}^{1/3}}{2}\right]}^3\\ V_{C_{1,1}}=\left[\frac{{\left(V_{c_1}\right)}^{1/3}+{\left(V_{c_1}\right)}^{1/3}}{2}\right]\\ V_{C_{1,1}}=\left[\frac{{\left(98.6\right)}^{1/3}+{\left(98.6\right)}^{1/3}}{2}\right]=98.6\\ V_{C_{i,j}}=\left(\begin{array}{ccc}98.6& 120.53& 143.378\\ 120.53& 145.5& 171.308\\ 143.378& 171.3& 200\end{array}\right)\end{array}$

With the help of critical volume and critical temperature, critical pressure of gases in the mixture can be calculated as follows:

  $P_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}$

The reduced compressibility factor can be calculated as:

  $\begin{array}{l}{Z}_{c_{i,j}}=\frac{Z_{c_i}+Z_{c_j}}{2}\\ Z_{c_{1,1}}=\frac{Z_{c_1}+Z_{c_1}}{2}\\ Z_{c_{1,1}}=\frac{0.286+0.286}{2}=0.286\\ Z_{c_{i,j}}=\left(\begin{array}{ccc}0.286& 0.282& 0.281\\ 0.282& 0.279& 0.278\\ 0.281& 0.278& 0.276\end{array}\right)\end{array}$

After putting all values in the critical pressure formula, we will get:

  $\begin{array}{l}{P}_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}\\ P_{c_{1,1}}=\frac{Z_{c_{1,1}}R{T}_{c_{1,1}}}{V_{C_{1,1}}}\\ P_{c_{1,1}}=\frac{0.286\times 83.14\times 190.6}{98.6}=45.964bar\\ P_{c_{i,j}}=\left(\begin{array}{ccc}45.96& 47& 43.26\\ 47& 48.67& 45.25\\ 43.26& 45.25& 42.43\end{array}\right)\end{array}$

At this calculated value of reduced temperature, we can calculate correlation constant as:

  $\begin{array}{l}{B}_{i,j}^0=0.083-\frac{0.422}{{\left(T_{r_{i,j}}\right)}^{1.6}}\\ B_{1,1}^0=0.083-\frac{0.422}{{\left(T_{1,1}=1.958\right)}^{1.6}}=-0.061\\ B_{i,j}^0=\left(\begin{array}{ccc}-0.061& -0.127& -0.162\\ -0.127& -0.223& -0.274\\ -0.162& -0.274& -0.333\end{array}\right)\\ B_{i,j}^1=0.139-\frac{0.172}{{\left(T_{r_{i,j}}\right)}^{4.2}}\\ B_{1,1}^1=0.139-\frac{0.172}{{\left(T_{r_{1,1}}=1.958\right)}^{4.2}}=0.1287\\ B_{i,j}^1=\left(\begin{array}{ccc}0.1287& 0.1115& 0.0979\\ 0.1115& 0.0649& 0.028\\ 0.0979& 0.028& -0.0266\end{array}\right)\end{array}$

Overall correlation constant value in gas mixture can be calculated by equation:

  $B_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]$

  $\begin{array}{l}{\omega }_{i,j}=\frac{{\omega }_i+{\omega }_j}{2}\\ {\omega }_{1,1}=\frac{{\omega }_1+{\omega }_2}{2}=\frac{0.012+0.012}{2}0.012\\ {\omega }_{i,j}=\left(\begin{array}{ccc}0.012& 0.056& 0.082\\ 0.056& 0.1& 0.126\\ 0.082& 0.126& 0.152\end{array}\right)\end{array}$

  $B_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]$

  $\begin{array}{l}{\delta }_{i,j}=2B_{i,j}-B_{i,i}-B_{j,j}\\ {\delta }_{i,j}=\left(\begin{array}{ccc}0& 30.442& 107.809\\ 30.442& 0& 23.482\\ 107.809& 23.482& 0\end{array}\right)\end{array}$

Using above calculated values fugacity coefficient of gases calculated as given below:

  $\begin{array}{l}\widehat{{\varphi }_k}=\exp \left[\frac{P}{RT}\left[B_{k,k}+\frac{1}{2}{\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j\left(2{\delta }_{i,k}-{\delta }_{i,j}\right)}}\right]\right]\\ \widehat{{\varphi }_1}=\exp \left[\frac{P}{RT}\left[B_{1,1}+\frac{1}{2}\left[\begin{array}{l}{y}_1{y}_1\left(2{\delta }_{1,1}-{\delta }_{1,1}\right)+y_1{y}_2\left(2{\delta }_{1,1}-{\delta }_{1,2}\right)\\ +y_2{y}_1\left(2{\delta }_{2,1}-{\delta }_{2,1}\right)+y_2{y}_2\left(2{\delta }_{2,1}-{\delta }_{2,2}\right)+\\ y_1{y}_3\left(2{\delta }_{1,1}-{\delta }_{1,3}\right)+y_1{y}_3\left(2{\delta }_{1,1}-{\delta }_{1,3}\right)\\ +y_3{y}_1\left(2{\delta }_{3,1}-{\delta }_{3,1}\right)+y_3{y}_1\left(2{\delta }_{3,1}-{\delta }_{3,1}\right)\end{array}\right]\right]\right]\\ \widehat{{\varphi }_1}=1.019\end{array}$

Similarly, we can calculate fugacity coefficient for second species ethane and for third one propane as:

  $\begin{array}{l}\widehat{{\varphi }_k}=\exp \left[\frac{P}{RT}\left[B_{k,k}+\frac{1}{2}{\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j\left(2{\delta }_{i,k}-{\delta }_{i,j}\right)}}\right]\right]\\ \widehat{{\varphi }_2}=0.881\\ \widehat{{\varphi }_3}=0.775\end{array}$

Now finally we can estimate the fugacity of all gases by just multiplying P with derived fugacity coefficient values:

  $\begin{array}{l}\widehat{{\varphi }_i}=\frac{\widehat{f_i}}{y_i\times P_i}\\ \widehat{f_1}=\widehat{{\varphi }_1}\times y_1\times P\\ \widehat{f_1}=1.019\times 0.21\times 35\\ \widehat{f_1}=7.4896bar\\ \widehat{f_2}=\widehat{{\varphi }_2}\times y_2\times P\\ \widehat{f_2}=0.881\times 0.43\times 35\\ \widehat{f_2}=13.259bar\\ \widehat{f_3}=\widehat{{\varphi }_3}\times y_3\times P\\ \widehat{f_3}=0.775\times 0.36\times 35\\ \widehat{f_3}=9.765bar\end{array}$

Therefore fugacity of methane is 7.4896 bars, fugacity of ethane is 13.259 bar and fugacity of propane is 9.765 bar in the gas mixture.