Question

Q3 A binary vapor mixture contains 65.00% ethanol (1) in water (2) at 50 oC. Determine the pressure in which this vapor develops the first drop of liquid b) a) the liquid composition Use NRTL to calculate the activity coefficient a12 3437 cal/mol, a2 1022.9466 cal/mol and a12-0.2935 Note: To solve this exercise, you will need to assign a number to x and calculate for psat XY Psat y1.cal =-iriri sat Then, goalseek 이.cal-y,actual) to zero by changing xỉ.

Answer 1

We will be using iterative procedure for calculating X_i , NRTL eqns are given as;

$\ln \gamma _{1}=x_{2}^{2}\left [ \tau _{21}\left (\frac{G_{21}}{x_{1}+x_{2}G_{21}} \right )^{2}+\frac{G_{12}\tau _{12}}{(x_{2}+x_{1}G_{12})^{2}}\right ]$ ......... activity coefficient of component 1

$\ln \gamma _{2}=x_{1}^{2}\left [ \tau _{12}\left (\frac{G_{12}}{x_{2}+x_{1}G_{12}} \right )^{2}+\frac{G_{21}\tau _{21}}{(x_{1}+x_{2}G_{21})^{2}}\right ]$

$G_{12}=\exp (-\alpha \tau _{12}) ............. G_{21}=\exp (-\alpha \tau _{21})$

$\tau _{21}=\frac{a_{21}}{RT}..........\tau _{12}=\frac{a_{12}}{RT}$

Using all these formulas and given information    $y_{1}=0.65..............y_{2}=0.35$

Pressure at which first drop appears is dew point pressure;

first calculate vapou pressure of components at 50^o C

$\ln P^{Sat}(KPa)=A-\frac{B}{T+C}........... a,b,c ..parameters$

for ethanol A = 16.8958 B = 3795.17 C = -42.232

for Water A = 16.3872 B = 3885.7 C = -42.98

P^sat (ethanol)= 29.545 KPa P^sat (water)= 12.405 KPa  

Intially Assume x₁ = x₂ = 0.5

$\tau _{21}=1.594.........\tau _{12}=0.0176$

$\gamma _{1}=1.2665....\gamma _{2}=1.4646$

$P^{Tot}=\sum \gamma _{i}x_{i}P_{i}^{Sat}\Rightarrow 27.79 KP a$

$x_{i}=\frac{y_{i}P^{Tot}}{\gamma _{i}P_{i}^{Sat}}\Rightarrow x_{1}=0.497..and..x_{2}=0.503$

Using these values again repeat the procedure;

$\gamma _{1}=1.6368....\gamma _{2}=1.46$

$P^{Tot}=\sum \gamma _{i}x_{i}P_{i}^{Sat}\Rightarrow 33.144 KP a$

$x_{i}=\frac{y_{i}P^{Tot}}{\gamma _{i}P_{i}^{Sat}}\Rightarrow x_{1}=0.493..and..x_{2}=0.507$ \

again iterating;

$\gamma _{1}=1.2827....\gamma _{2}=1.4527$

$P^{Tot}=\sum \gamma _{i}x_{i}P_{i}^{Sat}\Rightarrow 27.81 KP a$

$x_{i}=\frac{y_{i}P^{Tot}}{\gamma _{i}P_{i}^{Sat}}\Rightarrow x_{1}=0.493..and..x_{2}=0.507$

Therefore ; the dew point pressure is

Liquid composition