Chapter 13, Problem 13.49P

(a)

Interpretation Introduction

Interpretation:

The bubble point temperature for any one of the given binary systems in table $13.10$ is to be calculated using Wilson equation.

Concept Introduction:

Antoine equation is used to determine the vapor pressure of any substance at the given temperature by the equation:

  $\ln P^{\text{sat}}\left(\text{kPa}\right)=A-\frac{B}{\left(T^{\circ }\text{C}\right)+C}$ .......(1)

Here, $A,B\text{ and }C$ are the constants specific for a substance given in table B.2 of appendix B.

Equation $13.19$

to be used for Modified Raoult’s law is:

  $y_{i}P=x_i{\gamma }_i{P}_i{}^{\text{sat}}$ .......(2)

The Bubble point pressure for a binary system in vapor/liquid equilibrium is defined as the pressure where first bubble of vapor appears which is in equilibrium with the liquid present in the system. The equation which defines this pressure at this point is:

  $P_{\text{BUBL}}=x_{\text{1}}{\gamma }_1{P}_{\text{1}}^{\text{sat}}+x_{\text{2}}{\gamma }_2{P}_{\text{2}}^{\text{sat}}$ .......(3)

Wilson equations to be used are:

  $\begin{array}{c}\ln {\gamma }_1=-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\\ \ln {\gamma }_2=-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\end{array}$ .......(4)

Here, the parameters ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$

are calculated by the formula:

  $\begin{array}{c}{\Lambda }_{12}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{RT}\\ {\Lambda }_{21}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{RT}\end{array}$ .......(5)

Where, $V_1,V_2,a_{12}\text{ and }{a}_{21}$ are the Wilson equation parameters given in table $13.10$ for binary systems.

Answer to Problem 13.49P

The bubble point temperature for $1-\text{Propanol(1)/Water(2)}$

using Wilson equation is:

  $T_{\text{BUBL}}={87.95}^{\circ }\text{C}$

Explanation of Solution

Given information:

The pressure at which the bubble point temperature is to be calculated is $101.33\text{kPa}$ . The composition of the liquid phase at this point is given as $x_1=0.3$ .

Wilson equation parameters are given in Table 13.10 as shown below:

The binary system for which the bubble point temperature will be calculated is $1-\text{Propanol(1)/Water(2)}$ at $P=101.33\text{kPa}$

. The liquid phase composition is:

  $\begin{array}{c}{x}_1=0.3\\ x_2=0.7\end{array}$

From table B.2 of appendix B, the Antoine equation constants for $1-\text{Propanol(1) and Water(2)}$

are:

  $\begin{array}{c}{A}_1=16.1154\\ B_1=3483.67\\ C_1=205.807\\ A_2=16.3872\\ B_2=3885.70\\ C_2=230.170\end{array}$

Now, use equation (1) to calculate the vapor pressure of $1-\text{Propanol(1) and Water(2)}$ for bubble point temperature $T^{\circ }\text{C}$

as:

  $\begin{array}{c}\ln P_1{}^{\text{sat}}=16.1154-\frac{3483.67}{\left(T\right)+205.807}\\ P_1{}^{\text{sat}}=e^{\left(16.1154-\frac{3483.67}{\left(T\right)+205.807}\right)}\\ \ln P_2{}^{\text{sat}}=16.3872-\frac{3885.70}{\left(T\right)+230.170}\\ P_2{}^{\text{sat}}=e^{\left(16.3872-\frac{3885.70}{\left(T\right)+230.170}\right)}\end{array}$

From table $13.10$, the parameters needed for $1-\text{Propanol(1) and Water(2)}$

to be used in Wilson equation are:

  $\begin{array}{c}{V}_1=75.14{\text{ cm}}^3\text{/mol}\\ V_2=18.07{\text{ cm}}^3\text{/mol}\\ a_{12}=775.48\text{ cal/mol}\\ a_{21}=1351.90\text{ cal/mol}\end{array}$

The value of universal gas constant to be used is,

  $R=1.9859\text{ cal/}\left(\text{mol}\cdot \text{K}\right)$

Now, use equation (5) to calculate the values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$

as:

  $\begin{array}{c}{\Lambda }_{12}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{RT}\\ =\frac{18.07\frac{{\text{cm}}^3}{\text{mol}}}{75.14\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-775.48\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(T+273.15\text{ K}\right)}\\ =\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}\\ {\Lambda }_{21}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{RT}\\ =\frac{75.14\frac{{\text{cm}}^3}{\text{mol}}}{18.07\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-1351.90\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(T+273.15\text{ K}\right)}\\ =\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}\end{array}$

Now, use these values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$ along with the values of $x_1\text{ and }{x}_2$ and calculate the values of ${\gamma }_1\text{ and }{\gamma }_2$

using equations set (4) as:

${\gamma }_1=\exp \left[-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]$

$=\exp \left[\begin{array}{l}-\ln \left(0.3+0.7\left(\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}\right)\right)+\\ 0.7\left(\frac{\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}}{0.3+0.7\left(\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}\right)}-\frac{\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}}{0.7+0.3\left(\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}\right)}\right)\end{array}\right]$

${\gamma }_2=\exp \left[-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]$

$=\exp \left[\begin{array}{l}-\ln \left(0.7+0.3\left(\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}\right)\right)+\\ 0.3\left(\frac{\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}}{0.3+0.7\left(\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}\right)}-\frac{\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}}{0.7+0.3\left(\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}\right)}\right)\end{array}\right]$

Now, use the Modified Raoult’s law equation to calculate the pressure at the given value of $x_1$

using the below mentioned formula as:

  $P=x_1{\gamma }_1{P}_1{}^{\text{sat}}+x_2{\gamma }_2{P}_2{}^{\text{sat}}$

At $x_1=0,P=P_2{}^{\text{sat}}$ and at $x_1=1,P=P_1{}^{\text{sat}}$ .

Make an initial guess for $T$ as $T_{\text{guess}}=333.15\text{ K}$ and using iterative method and the condition to stop the iteration as $P=101.33\text{ kPa}$, find $T_{\text{BUBL}}$ for the given value of $x_1$

as:

  $T_{\text{BUBL}}={87.95}^{\circ }\text{C}$

(b)

Interpretation Introduction

Interpretation:

The dew point temperature for any one of the given binary systems in table $13.10$ is to be calculated using Wilson equation.

Concept Introduction:

Equation $13.19$

to be used for Modified Raoult’s law is:

  $y_{i}P=x_i{\gamma }_i{P}_i{}^{\text{sat}}$ .......(2)

Wilson equations to be used are:

  $\begin{array}{c}\ln {\gamma }_1=-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\\ \ln {\gamma }_2=-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\end{array}$ .......(4)

Here, the parameters ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$

are calculated by the formula:

  $\begin{array}{c}{\Lambda }_{12}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{RT}\\ {\Lambda }_{21}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{RT}\end{array}$ .......(5)

Where, $V_1,V_2,a_{12}\text{ and }{a}_{21}$ are the Wilson equation parameters given in table $13.10$ for binary systems.

The Dew point pressure for a binary system in vapor/liquid equilibrium is defined as the pressure where first drop of liquid appears which is in equilibrium with the vapor present in the system at a particular temperature. The equation that defines this pressure at this point is:

  $P_{\text{DEW}}=\frac{1}{\frac{y_{\text{1}}}{{\gamma }_1{P}_{\text{1}}^{\text{sat}}}+\frac{y_{\text{2}}}{{\gamma }_2{P}_{\text{2}}^{\text{sat}}}}$ .......(6)

Answer to Problem 13.49P

The dew point temperature for $1-\text{Propanol(1)/Water(2)}$

using Wilson equation is:

  $T_{\text{DEW}}={91.13}^{\circ }\text{C}$

Explanation of Solution

Given information:

The pressure at which the bubble point temperature is to be calculated is $101.33\text{kPa}$ . The composition of the vapor phase at this point is given as $y_1=0.3$ .

Wilson equation parameters are given in Table 13.10 as shown below:

Use the values of $P_1{}^{\text{sat}},\text{ and }{P}_2{}^{\text{sat}}$ at $T^{\circ }\text{C}$

as calculated in part (a) as:

  $\begin{array}{c}\ln P_1{}^{\text{sat}}=16.1154-\frac{3483.67}{\left(T\right)+205.807}\\ P_1{}^{\text{sat}}=e^{\left(16.1154-\frac{3483.67}{\left(T\right)+205.807}\right)}\\ \ln P_2{}^{\text{sat}}=16.3872-\frac{3885.70}{\left(T\right)+230.170}\\ P_2{}^{\text{sat}}=e^{\left(16.3872-\frac{3885.70}{\left(T\right)+230.170}\right)}\end{array}$

The vapor phase composition is:

  $\begin{array}{c}{y}_1=0.3\\ y_2=0.7\end{array}$

From table $13.10$, the parameters needed for $1-\text{Propanol(1) and Water(2)}$

to be used in Wilson equation are:

  $\begin{array}{c}{V}_1=75.14{\text{ cm}}^3\text{/mol}\\ V_2=18.07{\text{ cm}}^3\text{/mol}\\ a_{12}=775.48\text{ cal/mol}\\ a_{21}=1351.90\text{ cal/mol}\end{array}$

The value of universal gas constant to be used is,

  $R=1.9859\text{ cal/}\left(\text{mol}\cdot \text{K}\right)$

Now, use the values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$

as calculated in part (a) as:

  $\begin{array}{c}{\Lambda }_{12}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{RT}\\ =\frac{18.07\frac{{\text{cm}}^3}{\text{mol}}}{75.14\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-775.48\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(T+273.15\text{ K}\right)}\\ =\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}\\ {\Lambda }_{21}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{RT}\\ =\frac{75.14\frac{{\text{cm}}^3}{\text{mol}}}{18.07\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-1351.90\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(T+273.15\text{ K}\right)}\\ =\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}\end{array}$

Now, use these values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$ along with the values of $x_1\text{ and }{x}_2$ and calculate the values of ${\gamma }_1\text{ and }{\gamma }_2$

using equations set (4) as:

  $\begin{array}{c}{\gamma }_1=\exp \left[-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]\\ {\gamma }_2=\exp \left[-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]\end{array}$

Now, use the Modified Raoult’s law equation to calculate the pressure at the guessed value of $x_1$

using the below mentioned formula as:

  $P=\frac{1}{\frac{y_{\text{1}}}{{\gamma }_1{P}_{\text{1}}^{\text{sat}}}+\frac{y_{\text{2}}}{{\gamma }_2{P}_{\text{2}}^{\text{sat}}}}$

Make an initial guess for $T$ as $T_{\text{guess}}=333.15\text{ K}$ and $x_1$ as $x_{1\left(\text{guess}\right)}=0.1$ and using iterative method and the condition to stop the iteration as $P=101.33\text{ kPa}$, find $T_{\text{Dew}}$ for the given value of $y_1$

as:

  $T_{\text{DEW}}={91.13}^{\circ }\text{C}$

(c)

Interpretation Introduction

Interpretation:

  $P,T-$ flash is to be performed for any one of the given binary systems in table $13.10$ using Wilson equation.

Concept Introduction:

Equation $13.19$

to be used for Modified Raoult’s law is:

  $y_{i}P=x_i{\gamma }_i{P}_i{}^{\text{sat}}$ .......(2)

The Bubble point pressure for a binary system in vapor/liquid equilibrium is defined as the pressure where first bubble of vapor appears which is in equilibrium with the liquid present in the system. The equation which defines this pressure at this point is:

  $P_{\text{BUBL}}=x_{\text{1}}{\gamma }_1{P}_{\text{1}}^{\text{sat}}+x_{\text{2}}{\gamma }_2{P}_{\text{2}}^{\text{sat}}$ .......(3)

The Dew point pressure for a binary system in vapor/liquid equilibrium is defined as the pressure where first drop of liquid appears which is in equilibrium with the vapor present in the system at a particular temperature. The equation that defines this pressure at this point is:

  $P_{\text{DEW}}=\frac{1}{\frac{y_{\text{1}}}{{\gamma }_1{P}_{\text{1}}^{\text{sat}}}+\frac{y_{\text{2}}}{{\gamma }_2{P}_{\text{2}}^{\text{sat}}}}$ .......(6)

  $P,T-$ flash calculations are done to find the values of $\mathcal{V}\text{, }\mathcal{L},y_i,\text{ and }{x}_i$ for a system which flashes to produce a two-phase system of vapor and liquid in equilibrium.

The equation for equilibrium ratio, $K_i$

also known as K-value is:

  $K_i\equiv \frac{y_i}{x_i}$ .......(7)

Here, $y_i$ is the vapor phase mole fraction and $x_i$ is the liquid phase mole fraction of species $i$ in the binary system in VLE.

The equations for flash calculations to be used are:

  $\mathcal{L}+\mathcal{V}=1$ .......(8)

Here, $\mathcal{V}$ is the total moles of vapor and $\mathcal{L}$ is the total moles of liquid.

In terms of $K_i$, the equation relating $y_i\text{ and }{z}_i$

is:

  $y_i=\frac{z_i{K}_i}{1+\mathcal{V}\left(K_i-1\right)}$ .......(9)

Here, $z_i$ is the total mole fraction of component $i$ in the system.

Answer to Problem 13.49P

The result of the $P,T-$

flash calculations is:

  $\begin{array}{c}\mathcal{V}=0.807\\ \mathcal{L}=0.193\\ y_1=0.35\\ y_2=0.65\\ x_1=0.09\\ x_2=0.91\end{array}$

Explanation of Solution

Given information:

The flash pressure at which the $P,T-$ flash is to be calculated is $P=101.33\text{ kPa}$ . The total composition of component 1 at this point is given as $z_1=0.3$ .

The condition for the flash temperature for this system is,

  $T=\frac{1}{2}\left(T_{\text{BUBL}}+T_{\text{DEW}}\right)$

Wilson equation parameters are given in Table 13.10 as shown below:

Use the values of $P_1{}^{\text{sat}},\text{ and }{P}_2{}^{\text{sat}}$ at $T^{\circ }\text{C}$

as calculated in part (a) as:

  $\begin{array}{c}\ln P_1{}^{\text{sat}}=16.1154-\frac{3483.67}{\left(T\right)+205.807}\\ P_1{}^{\text{sat}}=e^{\left(16.1154-\frac{3483.67}{\left(T\right)+205.807}\right)}\\ \ln P_2{}^{\text{sat}}=16.3872-\frac{3885.70}{\left(T\right)+230.170}\\ P_2{}^{\text{sat}}=e^{\left(16.3872-\frac{3885.70}{\left(T\right)+230.170}\right)}\end{array}$

To perform $P,T-$ flash calculations, first find the bubble point and dew point temperature for the binary system $1-\text{Propanol(1) and Water(2)}$ at pressure $P=101.33\text{ kPa}$ .

To calculate bubble point temperature, let

  $\begin{array}{c}{z}_1=x_1=0.3\\ z_2=x_2=0.7\end{array}$

Since, the given conditions are same as in part (a), the calculated value of $T_{\text{BUBL}}$

as in part (a) is:

  $T_{\text{BUBL}}={87.95}^{\circ }\text{C}$

To calculate dew point temperature, let

  $\begin{array}{c}{z}_1=y_1=0.3\\ z_2=y_2=0.7\end{array}$

Since, the given conditions are same as in part (a), the calculated value of $T_{\text{DEW}}$

as in part (a) is:

  $T_{\text{DEW}}={91.13}^{\circ }\text{C}$

From the given condition of the flash temperature, it is calculated as:

  $\begin{array}{c}T=\frac{1}{2}\left(T_{\text{BUBL}}+T_{\text{DEW}}\right)\\ =\frac{1}{2}\left(87.95+91.13\right)\\ ={89.54}^{\circ }\text{C}\end{array}$

Use this temperature to get the values of $P_1{}^{\text{sat}},\text{ and }{P}_2{}^{\text{sat}}$

as:

  $\begin{array}{c}{P}_1{}^{\text{sat}}=e^{\left(16.1154-\frac{3483.67}{\left(T\right)+205.807}\right)}=e^{\left(16.1154-\frac{3483.67}{\left(89.54\right)+205.807}\right)}=75.21\text{ kPa}\\ P_2{}^{\text{sat}}=e^{\left(16.3872-\frac{3885.70}{\left(T\right)+230.170}\right)}=e^{\left(16.3872-\frac{3885.70}{\left(89.54\right)+230.170}\right)}=68.95\text{ kPa}\end{array}$

Now, use the values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$

as calculated in part (a) as:

  $\begin{array}{c}{\Lambda }_{12}=\left(0.2405\right)\exp \frac{-390.493}{\left(T+273.15\right)}=\left(0.2405\right)\exp \frac{-390.493}{\left(89.54+273.15\right)}=0.08195\\ {\Lambda }_{21}=\left(4.1583\right)\exp \frac{-680.749}{\left(T+273.15\right)}=\left(4.1583\right)\exp \frac{-680.749}{\left(89.54+273.15\right)}=0.6365\end{array}$

Now, use these values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$ along with the guessed values of $x_1\text{ and }{x}_2$ and calculate the values of ${\gamma }_1\text{ and }{\gamma }_2$

using equations set (4) as:

  $\begin{array}{c}{\gamma }_1=\exp \left[-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]\\ =\exp \left[-\ln \left(x_1+x_2\left(0.08195\right)\right)+x_2\left(\frac{\left(0.08195\right)}{x_1+x_2\left(0.08195\right)}-\frac{\left(0.6365\right)}{x_2+x_1\left(0.6365\right)}\right)\right]\\ {\gamma }_2=\exp \left[-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]\\ =\exp \left[-\ln \left(x_2+x_1\left(0.6365\right)\right)+x_1\left(\frac{\left(0.08195\right)}{x_1+x_2\left(0.08195\right)}-\frac{\left(0.6365\right)}{x_2+x_1\left(0.6365\right)}\right)\right]\end{array}$

Now, using the modified Raoult’s law, calculate the values of equilibrium ratio of component 1 and 2 using equations (2) and (7) as:

  $\begin{array}{c}{K}_i=\frac{y_i}{x_i}\\ =\frac{{\gamma }_i{P}_i{}^{\text{sat}}}{P}\\ K_1=\frac{{\gamma }_1{P}_1{}^{\text{sat}}}{P}\\ K_2=\frac{{\gamma }_2{P}_2{}^{\text{sat}}}{P}\end{array}$

Now, use equation (9) and write it for both the component, 1 and 2 as shown below:

  $\begin{array}{c}{y}_1=\frac{0.3\left(K_1\right)}{1+\mathcal{V}\left(K_1-1\right)}\\ y_2=\frac{0.7\left(K_2\right)}{1+\mathcal{V}\left(K_2-1\right)}\end{array}$ .......(10)

Since, $y_1+y_2=1$, the above equations are solved to calculate $\mathcal{V}$

as:

  $\frac{0.3\left(K_1\right)}{1+\mathcal{V}\left(K_1-1\right)}+\frac{0.7\left(K_2\right)}{1+\mathcal{V}\left(K_2-1\right)}=1$

Now, use equation (8) to calculate the value of $\mathcal{L}$

as:

  $\begin{array}{c}\mathcal{L}+\mathcal{V}=1\\ \mathcal{L}=1-\mathcal{V}\end{array}$

Also, use the calculated value of $\mathcal{V}$ in equation set (10) and find the values of $y_1\text{ and }{y}_2$ .

Using these values and the calculated values of $K_1\text{ and }{K}_2$, find the values of $x_1\text{ and }{x}_2$

by equation (7) as:

  $\begin{array}{c}{K}_i=\frac{y_i}{x_i}\\ x_i=\frac{y_i}{K_i}\\ x_1=\frac{y_1}{K_1}\\ x_2=\frac{y_2}{K_2}\end{array}$

Again, substitute these calculated values of $x_1\text{ and }{x}_2$ as guessed values $x_{1\left(\text{guess}\right)}\text{ and }{x}_{2\left(\text{guess}\right)}$ in calculating ${\gamma }_1\text{ and }{\gamma }_2$ and follow the steps given until $x_{1\left(\text{guess}\right)}\approx x_1$ . After a considerable time of iterations, the result of the $P,T-$

flash calculations is:

  $\begin{array}{c}\mathcal{V}=0.807\\ \mathcal{L}=0.193\\ y_1=0.35\\ y_2=0.65\\ x_1=0.09\\ x_2=0.91\end{array}$

(d)

Interpretation Introduction

Interpretation:

The values of the azeotropic temperature and composition of the system is to be calculated if it exists for the given binary system.

Concept Introduction:

Antoine equation is used to determine the vapor pressure of any substance at the given temperature by the equation:

  $\ln P^{\text{sat}}\left(\text{kPa}\right)=A-\frac{B}{\left(T^{\circ }\text{C}\right)+C}$ .......(1)

Here, $A,B\text{ and }C$ are the constants specific for a substance given in table B.2 of appendix B.

Equation $13.19$

to be used for Modified Raoult’s law is:

  $y_{i}P=x_i{\gamma }_i{P}_i{}^{\text{sat}}$ .......(2)

Wilson equations to be used are:

  $\begin{array}{c}\ln {\gamma }_1=-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\\ \ln {\gamma }_2=-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\end{array}$ .......(4)

Here, the parameters ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$

are calculated by the formula:

  $\begin{array}{c}{\Lambda }_{12}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{RT}\\ {\Lambda }_{21}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{RT}\end{array}$ .......(5)

Where, $V_1,V_2,a_{12}\text{ and }{a}_{21}$ are the Wilson equation parameters given in table $13.10$ for binary systems.

Relative volatility is defined by,

  ${\alpha }_{12}\equiv \frac{y_1/x_1}{y_2/x_2}$ .......(11)

When ${\left({\alpha }_{12}\right)}_{x_1=0}>1\text{ and }{\left({\alpha }_{12}\right)}_{x_1=1}<1$, there exists an azeotrope in the given system.

At the azeotropic point, ${\alpha }_{12}=1\text{ and }{x}_1=y_1$ .

Answer to Problem 13.49P

The azeotropic values of temperature and composition for the binary system is calculated as:

  $\begin{array}{c}{T}^{\text{az}}={87.73}^{\circ }\text{C}\\ x_1=y_1=0.4546\end{array}$

Explanation of Solution

Given information:

The pressure at which the azeotrope of the system may exists is $101.33\text{ kPa}$ .

Wilson equation parameters are given in Table 13.10 as shown below:

Use the given value of $P$ using equation (1) to calculate $T_{\text{b1}}\text{ and }{T}_{\text{b2}}$

as:

  $\begin{array}{c}\ln P=16.1154-\frac{3483.67}{\left(T_{\text{b1}}\right)+205.807}\\ \ln \left(101.33\right)=16.1154-\frac{3483.67}{\left(T_{\text{b1}}\right)+205.807}\\ T_{\text{b1}}={97.2}^{\circ }\text{C}\\ T_{\text{b1}}=370.35\text{ K}\\ \ln P=16.3872-\frac{3885.70}{\left(T_{\text{b2}}\right)+230.170}\\ \ln \left(101.33\right)=16.3872-\frac{3885.70}{\left(T_{\text{b2}}\right)+230.170}\\ T_{\text{b2}}={99.99}^{\circ }\text{C}\\ T_{\text{b2}}=373.14\text{ K}\end{array}$

From table $13.10$, the parameters needed for $1-\text{Propanol(1) and Water(2)}$

to be used in Wilson equation are:

  $\begin{array}{c}{V}_1=18.07{\text{ cm}}^3\text{/mol}\\ V_2=75.14{\text{ cm}}^3\text{/mol}\\ a_{12}=775.48\text{ cal/mol}\\ a_{21}=1351.90\text{ cal/mol}\end{array}$

The value of universal gas constant to be used is,

  $R=1.9859\text{ cal/}\left(\text{mol}\cdot \text{K}\right)$

Now, use equation (5) to calculate the values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$ for $T_{\text{b1}}\text{ and }{T}_{\text{b2}}$

as:

${\left({\Lambda }_{12}\right)}_{\text{b1}}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{R{T}_{\text{b1}}}$

$=\frac{18.07\frac{{\text{cm}}^3}{\text{mol}}}{75.14\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-775.48\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(370.35\text{ K}\right)}$

$=0.08379$

${\left({\Lambda }_{12}\right)}_{\text{b2}}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{R{T}_{\text{b2}}}$

$=\frac{18.07\frac{{\text{cm}}^3}{\text{mol}}}{75.14\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-775.48\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(373.14\text{ K}\right)}$

$=0.08445$

${\left({\Lambda }_{21}\right)}_{\text{b1}}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{R{T}_{\text{b1}}}$

$=\frac{75.14\frac{{\text{cm}}^3}{\text{mol}}}{18.07\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-1351.90\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(370.35\text{ K}\right)}$

$=0.6616$

${\left({\Lambda }_{21}\right)}_{\text{b2}}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{R{T}_{\text{b2}}}$

$=\frac{75.14\frac{{\text{cm}}^3}{\text{mol}}}{18.07\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-1351.90\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(373.14\text{ K}\right)}$

$=0.6708$

Now, use these values of ${\Lambda }_{12}\text{ and }{\Lambda }_{21}$ along with the values of $x_1=0\text{ and }{x}_2=1$ and calculate the values of ${\gamma }_1\text{ and }{\gamma }_2\text{ for }{T}_{\text{b1}}$

using equations set (4) as:

  $\begin{array}{c}{\gamma }_1=\exp \left[-\ln \left(x_1+x_2{\left({\Lambda }_{12}\right)}_{\text{b1}}\right)+x_2\left(\frac{{\left({\Lambda }_{12}\right)}_{\text{b1}}}{x_1+x_2{\left({\Lambda }_{12}\right)}_{\text{b1}}}-\frac{{\left({\Lambda }_{21}\right)}_{\text{b1}}}{x_2+x_1{\left({\Lambda }_{21}\right)}_{\text{b1}}}\right)\right]\\ =\exp \left[-\ln \left(0+1\left(0.08379\right)\right)+1\left(\frac{0.08379}{0+1\left(0.08379\right)}-\frac{0.6616}{1+0\left(0.6616\right)}\right)\right]\\ =16.741\\ {\gamma }_2=\exp \left[-\ln \left(x_2+x_1{\left({\Lambda }_{21}\right)}_{\text{b1}}\right)+x_1\left(\frac{{\left({\Lambda }_{12}\right)}_{\text{b1}}}{x_1+x_2{\left({\Lambda }_{12}\right)}_{\text{b1}}}-\frac{{\left({\Lambda }_{21}\right)}_{\text{b1}}}{x_2+x_1{\left({\Lambda }_{21}\right)}_{\text{b1}}}\right)\right]\\ =\exp \left[-\ln \left(1+0\left(0.6616\right)\right)+0\left(\frac{0.08379}{0+1\left(0.08379\right)}-\frac{0.6616}{1+0\left(0.6616\right)}\right)\right]\\ =1.00\end{array}$

Calculate $P_1^{\text{sat}}\text{ at }{T}_{\text{b2}}\text{ and }{P}_2^{\text{sat}}\text{ at }{T}_{\text{b1}}$

using equation (1) as:

  $\begin{array}{c}\ln P_1{}^{\text{sat}}=16.1154-\frac{3483.67}{\left(T_{\text{b2}}\right)+205.807}\\ P_1{}^{\text{sat}}=e^{\left(16.1154-\frac{3483.67}{\left(99.99\right)+205.807}\right)}\\ P_1{}^{\text{sat}}=112.54\text{ kPa}\\ \ln P_2{}^{\text{sat}}=16.3872-\frac{3885.70}{\left(T_{\text{b1}}\right)+230.170}\\ P_2{}^{\text{sat}}=e^{\left(16.3872-\frac{3885.70}{\left(97.2\right)+230.170}\right)}\\ P_2{}^{\text{sat}}=91.63\text{ kPa}\end{array}$

Using equation (11) along with the modified Raoult’s law, calculate the value of relative volatility at $x_1=0$

as:

  $\begin{array}{c}{\left({\alpha }_{12}\right)}_{x_1=0}\equiv \frac{y_1/x_1}{y_2/x_2}\\ =\frac{{\gamma }_1{P}_1{}^{\text{sat}}/P}{{\gamma }_2{P}_2{}^{\text{sat}}/P}\\ =\frac{{\gamma }_1{P}_1{}^{\text{sat}}}{{\gamma }_2{P}_2{}^{\text{sat}}}\\ =\frac{\left(16.741\right)\left(112.54\right)}{\left(1.00\right)\left(91.63\right)}\\ =20.56\end{array}$

For $x_1=1\text{ and }{x}_2=0$ and calculate the values of ${\gamma }_1\text{ and }{\gamma }_2\text{ for }{T}_{\text{b2}}$

using equations set (4) as:

  $\begin{array}{c}{\gamma }_1=\exp \left[-\ln \left(x_1+x_2{\left({\Lambda }_{12}\right)}_{\text{b2}}\right)+x_2\left(\frac{{\left({\Lambda }_{12}\right)}_{\text{b2}}}{x_1+x_2{\left({\Lambda }_{12}\right)}_{\text{b2}}}-\frac{{\left({\Lambda }_{21}\right)}_{\text{b2}}}{x_2+x_1{\left({\Lambda }_{21}\right)}_{\text{b2}}}\right)\right]\\ =\exp \left[-\ln \left(1+0\left(0.08445\right)\right)+0\left(\frac{0.08445}{1+0\left(0.08445\right)}-\frac{0.6708}{0+1\left(0.6708\right)}\right)\right]\\ =1.00\\ {\gamma }_2=\exp \left[-\ln \left(x_2+x_1{\left({\Lambda }_{21}\right)}_{\text{b2}}\right)+x_1\left(\frac{{\left({\Lambda }_{12}\right)}_{\text{b2}}}{x_1+x_2{\left({\Lambda }_{12}\right)}_{\text{b2}}}-\frac{{\left({\Lambda }_{21}\right)}_{\text{b2}}}{x_2+x_1{\left({\Lambda }_{21}\right)}_{\text{b2}}}\right)\right]\\ =\exp \left[-\ln \left(0+1\left(0.6708\right)\right)+1\left(\frac{0.08445}{0+1\left(0.08445\right)}-\frac{0.6708}{1+0\left(0.6708\right)}\right)\right]\\ =2.072\end{array}$

Using equation (11) along with the modified Raoult’s law, calculate the value of relative volatility at $x_1=1$

as:

  $\begin{array}{c}{\left({\alpha }_{12}\right)}_{x_1=1}\equiv \frac{y_1/x_1}{y_2/x_2}\\ =\frac{{\gamma }_1{P}_1{}^{\text{sat}}/P}{{\gamma }_2{P}_2{}^{\text{sat}}/P}\\ =\frac{{\gamma }_1{P}_1{}^{\text{sat}}}{{\gamma }_2{P}_2{}^{\text{sat}}}\\ =\frac{\left(1.00\right)\left(112.54\right)}{\left(2.072\right)\left(91.63\right)}\\ =0.593\end{array}$

Since ${\left({\alpha }_{12}\right)}_{x_1=0}>1\text{ and }{\left({\alpha }_{12}\right)}_{x_1=1}<1$, there exists an azeotrope for this binary system.

To calculate the azeotropic pressure, consider the condition ${\alpha }_{12}=1\text{ and }{x}_1=y_1$ .

Consider the following set of equations in the given order as:

$\begin{array}{c}\text{step 1:}\\ {\Lambda }_{12}=\frac{V_2}{V_1}\exp \frac{-a_{12}}{R{T}^{\text{az}}}\end{array}$

$=\frac{18.07\frac{{\text{cm}}^3}{\text{mol}}}{75.14\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-775.48\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(T^{\text{az}}+273.15\text{ K}\right)}$

$=\left(0.2405\right)\exp \frac{-390.493}{\left(T^{\text{az}}+273.15\right)}$

${\Lambda }_{21}=\frac{V_1}{V_2}\exp \frac{-a_{21}}{R{T}^{\text{az}}}$

$=\frac{75.14\frac{{\text{cm}}^3}{\text{mol}}}{18.07\frac{{\text{cm}}^3}{\text{mol}}}\exp \frac{-1351.90\frac{\text{cal}}{\text{mol}}}{\left(1.9859\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(T^{\text{az}}+273.15\text{ K}\right)}$

$=\left(4.1583\right)\exp \frac{-680.749}{\left(T^{\text{az}}+273.15\right)}$

$\text{Step 2:}$

${\gamma }_1=\exp \left[-\ln \left(x_1+x_2{\Lambda }_{12}\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]$

${\gamma }_2=\exp \left[-\ln \left(x_2+x_1{\Lambda }_{21}\right)+x_1\left(\frac{{\Lambda }_{12}}{x_1+x_2{\Lambda }_{12}}-\frac{{\Lambda }_{21}}{x_2+x_1{\Lambda }_{21}}\right)\right]$

$\text{Step 3:}$

$P^{\text{az}}=x_1{\gamma }_1{P}_1{}^{\text{sat}}+x_2{\gamma }_2{P}_2{}^{\text{sat}}$

Now, use the values of $x_1\text{ and }{x}_2$ guessed as $0.4\text{ and 0}\text{.8}$, and the values of $y_1\text{ and }{y}_2$ guessed as $0.4\text{ and 0}\text{.8}$, respectively. Also, make an initial guess for $T^{\text{az}}$ as $T_{\text{guess}}=333.15\text{ K}$ and using iterative method and the condition to stop the iteration as $P^{\text{az}}=101.33\text{ kPa}$, find $T^{\text{az}}$ for the given value of $x_1$

as:

  $\begin{array}{c}{T}^{\text{az}}={87.73}^{\circ }\text{C}\\ x_1=y_1=0.4546\end{array}$