Chapter 10, Problem 10.25P

(a)

Interpretation Introduction

Interpretation:

Estimation of the fugacity for ethylene and propylene in a mixture. Also estimate fugaciy coefficient for both gases in the mixture through the application of Eq. 10.63

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.

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

Answer to Problem 10.25P

Fugacity and fugacity coefficient of ethylene are 10.0485 bar and 0.957, respectively.

Fugacity and fugacity coefficient of propylene are 17.0625 bar and 0.875, respectively.

Explanation of Solution

Refer APPENDIX-B and Table-B.1 to determine critical properties and acentric factor of ethylene(1)/Propylene2) as:

Component	Pc (bar)	Tc (K)	Vc (cm3/mol)	$\omega$	Zc
Ethylene (1)	50.40	282.3	131	0.087	0.281
Propylene (2)	46.65	365.6	188.4	0.140	0.289

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 = 30 bar (given in question) and Temperature (T) = 150⁰C or 423.15K (given)

Y₁= 0.35 (given)

So,

$\begin{array}{l}{y}_2=1-y_1\\ y_2=1-0.35\\ y_2=0.65\end{array}$

Reduced temperature for both gases can be calculated as:

  $\begin{array}{l}{T}_{c_{i,j}}=\sqrt{T_{c_i}.T_{c_j}}\\ T_{c_{1,1}}=\sqrt{T_{c_1}.T_{c_1}}\\ T_{c_{1,1}}=\sqrt{282.3\times 282.3}\\ T_{c_{1,1}}=282.3K\\ T_{c_{1,2}}=\sqrt{T_{c_1}.T_{c_2}}\\ T_{c_{1,2}}=\sqrt{282.3\times 365.6}\\ T_{c_{1,2}}=321.26K\\ T_{c_{2,1}}=\sqrt{T_{c_1}.T_{c_2}}\\ T_{c_{2,1}}=\sqrt{282.3\times 365.6}\\ T_{c_{2,1}}=321.26K\\ T_{c_{2,2}}=\sqrt{T_{c_2}.T_{c_2}}\\ T_{c_{2,2}}=\sqrt{365.6\times 365.6}\\ T_{c_{2,2}}=365.6K\\ T_{c_{i,j}}=\left(\begin{array}{cc}282.3& 321.26\\ 321.26& 365.6\end{array}\right)\end{array}$

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

  $\begin{array}{l}{T}_{r_{i,j}}=\left(\begin{array}{cc}\frac{423.15}{282.3}& \frac{423.15}{321.26}\\ \frac{423.15}{321.26}& \frac{423.15}{365.6}\end{array}\right)\\ T_{r_{i,j}}=\left(\begin{array}{cc}1.499& 1.317\\ 1.317& 1.157\end{array}\right)\end{array}$

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

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:

  $\begin{array}{l}{Z}_{c_{i,j}}=\frac{Z_{c_i}+Z_{c_j}}{2}\\ Z_{c_{i,j}}=\left(\begin{array}{cc}\frac{Z_{c_1}+Z_{c_1}}{2}& \frac{Z_{c_1}+Z_{c_2}}{2}\\ \frac{Z_{c_2}+Z_{c_1}}{2}& \frac{Z_{c_2}+Z_{c_2}}{2}\end{array}\right)\\ Z_{c_{i,j}}=\left(\begin{array}{cc}\frac{0.281+0.281}{2}& \frac{0.281+0.289}{2}\\ \frac{0.289+0.281}{2}& \frac{0.289+0.289}{2}\end{array}\right)\\ Z_{c_{i,j}}=\left(\begin{array}{cc}0.281& 0.285\\ 0.285& 0.289\end{array}\right)\end{array}$

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

  $\begin{array}{l}{P}_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}\\ P_{c_{i,j}}=\left(\begin{array}{cc}50.345& 48.19\\ 48.19& 46.63\end{array}\right)\end{array}$

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

  $\begin{array}{l}{B}_{i,j}^0=0.083-\frac{0.422}{{\left(T_{r_{i,j}}\right)}^{1.6}}\\ B_{i,j}^0=\left(\begin{array}{cc}-0.138& -0.189\\ -0.189& -0.251\end{array}\right)\\ B_{i,j}^1=0.139-\frac{0.172}{{\left(T_{r_{i,j}}\right)}^{4.2}}\\ B_{i,j}^1=\left(\begin{array}{cc}0.108& 0.085\\ 0.085& 0.046\end{array}\right)\end{array}$

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

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

  $\begin{array}{l}{\omega }_{i,j}=\frac{{\omega }_i+{\omega }_j}{2}\\ {\omega }_{i,j}=\left(\begin{array}{cc}0.087& 0.114\\ 0.114& 0.14\end{array}\right)\end{array}$

  $\begin{array}{l}{B}_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]\\ B_{i,j}=\left(\begin{array}{cc}-59.892& -99.181\\ -99.181& -159.43\end{array}\right)\end{array}$

  $\begin{array}{l}{\delta }_{i,j}=2B_{i,j}-B_{i,i}-B_{j,j}\\ {\delta }_{i,j}=\left(\begin{array}{cc}0& 20.96\\ 20.96& 0\end{array}\right)\end{array}$

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

  $\begin{array}{l}\widehat{{\varphi }_k}=\exp \left[\frac{P}{RT}\left[B_{k,k}+\frac{1}{2}{\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j\left(2{\delta }_{i,k}-{\delta }_{i,j}\right)}}\right]\right]\\ \widehat{{\varphi }_1}=\exp \left[\frac{P}{RT}\left[B_{1,1}+\frac{1}{2}\left[\begin{array}{l}{y}_1{y}_1\left(2{\delta }_{1,1}-{\delta }_{1,1}\right)+y_1{y}_2\left(2{\delta }_{1,1}-{\delta }_{1,2}\right)\\ +y_2{y}_1\left(2{\delta }_{2,1}-{\delta }_{2,1}\right)+y_2{y}_2\left(2{\delta }_{2,1}-{\delta }_{2,2}\right)\end{array}\right]\right]\right]\\ \widehat{{\varphi }_1}=\exp \left[\frac{30}{83.14\times 423.15}\left[-59.892+\frac{1}{2}\left[\begin{array}{l}\left(0.35\times 0.35\right)\left(\left(2\times 0\right)-0\right)\\ +\left(0.35\times 0.65\right)\left(\left(2\times 0\right)-20.96\right)\\ +\left(0.65\times 0.35\right)\left(\left(2\times 20.96\right)-20.96\right)\\ +\left(0.65\times 0.65\right)\left(\left(2\times 20.96\right)-0\right)\end{array}\right]\right]\right]\\ \widehat{{\varphi }_1}=0.957\end{array}$

Similarly, we can calculate fugacity coefficient for second species propylene:

  $\begin{array}{l}\widehat{{\varphi }_k}=\exp \left[\frac{P}{RT}\left[B_{k,k}+\frac{1}{2}{\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j\left(2{\delta }_{i,k}-{\delta }_{i,j}\right)}}\right]\right]\\ \widehat{{\varphi }_2}=\exp \left[\frac{P}{RT}\left[B_{1,1}+\frac{1}{2}\left[\begin{array}{l}{y}_1{y}_1\left(2{\delta }_{1,2}-{\delta }_{1,1}\right)+y_1{y}_2\left(2{\delta }_{1,2}-{\delta }_{1,2}\right)\\ +y_2{y}_1\left(2{\delta }_{2,2}-{\delta }_{2,1}\right)+y_2{y}_2\left(2{\delta }_{2,2}-{\delta }_{2,2}\right)\end{array}\right]\right]\right]\\ \widehat{{\varphi }_2}=0.875\end{array}$

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

  $\begin{array}{l}\widehat{{\varphi }_i}=\frac{\widehat{f_i}}{y_i\times P_i}\\ \widehat{f_1}=\widehat{{\varphi }_1}\times y_1\times P_1\\ \widehat{f_1}=0.957\times 0.35\times 30\\ \widehat{f_1}=10.0485bar\\ \widehat{f_2}=\widehat{{\varphi }_2}\times y_2\times P_2\\ \widehat{f_2}=0.875\times \left(1-0.35\right)\times 30\\ \widehat{f_2}=17.0625bar\end{array}$

Therefore fugacity of ethylene is 10.0485 bar and fugacity of propylene is 17.0625 bar in the gas mixture.

(b)

Interpretation Introduction

Interpretation:

Estimation of the fugacity for ethylene and propylene in a ideal solution. Also estimate fugaciy coefficient for both gases in the solution by assuming that mixture as an ideal solution.

Concept introduction:

The calculation of fugacity and fugacity coefficient of gases in the ideal solution mixture is similar to a pure species mixture or an ideal gas mixture only difference is the replacement of y_i with x_i and calculation possible by using different correlation given as follows:

Fugacity coefficient for gases in ideal solution can be calculated by the equation given in the book as follows:

  ${\widehat{\varphi }}_k^{id}=\exp \left[\frac{P_{r_k}}{T_{r_{k,k}}}\left[B_{{}_{k,k}}^0+{\omega }_{kk}{B}_{{}_{k,k}}^1\right]\right]$

Where,

  $\begin{array}{l}{B}_{k,k}^0=0.083-\frac{0.422}{{\left(T_{r_{k,k}}\right)}^{1.6}}\\ B_{k,k}^1=0.139-\frac{0.172}{{\left(T_{r_{k,k}}\right)}^{4.2}}\end{array}$

$\widehat{{\varphi }_{{}_k}^{id}}$ = Fugacity coefficient of gases in ideal solution

$T_{r_{kk}}$ = reduced temperature = $T_{r_{k,k}}=\frac{T}{T_{C_{k,k}}}$

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

  $\begin{array}{l}\widehat{{\varphi }_{{}_i}^{id}}=\frac{\widehat{f_{{}_i}^{id}}}{x_{i}P}\\ \widehat{f_{{}_i}^{id}}=\widehat{{\varphi }_{{}_i}^{id}}\times x_i\times P\end{array}$

Answer to Problem 10.25P

Fugacity and fugacity coefficient of ethylene are 9.975 bar and 0.95, respectively.

Fugacity and fugacity coefficient of propylene are 17.0235 bar and 0.873, respectively.

Explanation of Solution

Refer APPENDIX-B and Table-B.1 to determine critical properties and acentric factor of ethylene(1)/Propylene2) as:

Component	Pc (bar)	Tc (K)	Vc (cm3/mol)	$\omega$	Zc
Ethylene (1)	50.40	282.3	131	0.087	0.281
Propylene (2)	46.65	365.6	188.4	0.140	0.289

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 = 30 bar (given in question) and Temperature (T) = 150⁰C or 423.15K (given)

x₁= 0.35 (given)

So,

$\begin{array}{l}{x}_2=1-x_1\\ x_2=1-0.35\\ x_2=0.65\end{array}$

Reduced temperature of gases in the mixture can be calculated as:

  $\begin{array}{l}{T}_{c_{k,k}}=\sqrt{T_{c_k}.T_{c_k}}\\ T_{c_{1,1}}=\sqrt{T_{c_1}.T_{c_1}}\\ T_{c_{1,1}}=\sqrt{282.3\times 282.3}\\ T_{c_{1,1}}=282.3K\\ T_{c_{2,2}}=\sqrt{T_{c_2}.T_{c_2}}\\ T_{c_{2,2}}=\sqrt{365.6\times 365.6}\\ T_{c_{2,2}}=365.6K\\ T_{c_{k,k}}=\left(\begin{array}{l}282.3\\ 365.6\end{array}\right)\end{array}$

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

  $\begin{array}{l}{T}_{r_{k,k}}=\left(\begin{array}{l}\frac{423.15}{282.3}\\ \frac{423.15}{365.6}\end{array}\right)\\ T_{r_{k,k}}=\left(\begin{array}{l}1.4989\\ 1.1574\end{array}\right)\end{array}$

The critical volume of both the gases in the ideal mixture can be calculated as given below:

  $\begin{array}{l}{V}_{C_{k,k}}={\left[\frac{{\left(V_{c_k}\right)}^{1/3}+{\left(V_{c_k}\right)}^{1/3}}{2}\right]}^3\\ V_{C_{k,k}}=\left(\begin{array}{l}{\left[\frac{{\left(V_{c_1}\right)}^{1/3}+{\left(V_{c_1}\right)}^{1/3}}{2}\right]}^3\\ {\left[\frac{{\left(V_{c_2}\right)}^{1/3}+{\left(V_{c_2}\right)}^{1/3}}{2}\right]}^3\end{array}\right)\\ V_{C_{k,k}}=\left(\begin{array}{l}131\\ 188.4\end{array}\right)\end{array}$

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

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

The reduced compressibility factor can be calculated as:

  $\begin{array}{l}{Z}_{c_{k,k}}=\frac{Z_{c_k}+Z_{c_k}}{2}\\ Z_{c_{k,k}}=\left(\begin{array}{l}\frac{Z_{c_1}+Z_{c_1}}{2}\\ \frac{Z_{c_2}+Z_{c_2}}{2}\end{array}\right)\\ Z_{c_{k,k}}=\left(\begin{array}{l}\frac{0.281+0.281}{2}\\ \frac{0.289+0.289}{2}\end{array}\right)\\ Z_{c_{k,k}}=\left(\begin{array}{l}0.281\\ 0.289\end{array}\right)\end{array}$

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

  $\begin{array}{l}{P}_{c_{k,k}}=\frac{Z_{c_{k,k}}R{T}_{c_{k,k}}}{V_{C_{k,k}}}\\ P_{c_{k,k}}=\left(\begin{array}{l}50.345\\ 46.63\end{array}\right)\end{array}$

Reduced pressure will be:

  $\begin{array}{l}{P}_{r_{k,k}}=\frac{P}{P_{c_{k,k}}}\\ P_{r_{k,k}}=\left(\begin{array}{l}0.595\\ 0.643\end{array}\right)\end{array}$

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

  $\begin{array}{l}{B}_{k,k}^0=0.083-\frac{0.422}{{\left(T_{r_{k,k}}\right)}^{1.6}}\\ B_{k,k}^0=\left(\begin{array}{l}-0.138\\ -0.251\end{array}\right)\\ B_{k,k}^1=0.139-\frac{0.172}{{\left(T_{r_{k,k}}\right)}^{4.2}}\\ B_{k,k}^1=\left(\begin{array}{l}0.108\\ 0.046\end{array}\right)\end{array}$

Accentric factor for both the species in ideal gas mixture calculated by the equation given below:

  $\begin{array}{l}{\omega }_{i,j}=\frac{{\omega }_i+{\omega }_j}{2}\\ {\omega }_{i,j}=\left(\begin{array}{l}0.087\\ 0.14\end{array}\right)\end{array}$

Using above calculated values fugacity coefficient of gases in ideal solution can be calculated as given below:

  $\begin{array}{l}{\widehat{\varphi }}_k^{id}=\exp \left[\frac{P_{r_k}}{T_{r_{k,k}}}\left[B_{{}_{k,k}}^0+{\omega }_{kk}{B}_{{}_{k,k}}^1\right]\right]\\ {\widehat{\varphi }}_1^{id}=\exp \left[\frac{P_{r_1}}{T_{r_{1,1}}}\left[B_{{}_{1,1}}^0+{\omega }_{1,1}{B}_{{}_{1,1}}^1\right]\right]\\ {\widehat{\varphi }}_1^{id}=\exp \left[\frac{0.595}{1.4989}\left[-0.138+\left(0.087\times 0.108\right)\right]\right]\\ {\widehat{\varphi }}_1^{id}=0.95\end{array}$

Similarly, we can calculate fugacity coefficient for second species propylene:

  $\begin{array}{l}{\widehat{\varphi }}_k^{id}=\exp \left[\frac{P_{r_k}}{T_{r_{k,k}}}\left[B_{{}_{k,k}}^0+{\omega }_{kk}{B}_{{}_{k,k}}^1\right]\right]\\ {\widehat{\varphi }}_2^{id}=\exp \left[\frac{P_{r_2}}{T_{r_{2,2}}}\left[B_{{}_{2,2}}^0+{\omega }_{2,2}{B}_{{}_{2,2}}^1\right]\right]\\ {\widehat{\varphi }}_2^{id}=\exp \left[\frac{0.643}{1.157}\left[-0.251+\left(0.14\times 0.046\right)\right]\right]\\ {\widehat{\varphi }}_2^{id}=0.873\end{array}$

Now finally we can estimate the fugacity of both gases in ideal solution using equation (10.52) as follows:

  $\begin{array}{l}\widehat{{\varphi }_{{}_i}^{id}}=\frac{\widehat{f_{{}_i}^{id}}}{x_{i}P}\\ \widehat{f_{{}_i}^{id}}=\widehat{{\varphi }_{{}_i}^{id}}\times x_i\times P\\ \widehat{f_{{}_i}^{id}}=0.35\times 0.95\times 30\\ \widehat{f_{{}_1}^{id}}=9.975bar\\ \widehat{f_{{}_2}^{id}}=\widehat{{\varphi }_{{}_2}^{id}}\times x_2\times P\\ \widehat{f_{{}_2}^{id}}=\left(1-0.35\right)\times 0.873\times 30\\ \widehat{f_{{}_2}^{id}}=17.0235bar\end{array}$

Therefore fugacity of ethylene is 9.975 bar and fugacity of propylene is 17.0235 bar in the ideal solution.