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Methanol-Benzene Phase Equilibrium

The document summarizes the calculation of various properties for a binary mixture of methanol and benzene at 60°C using the Wilson equation. It provides: 1) The bubble point pressure of 1.281×105 Pa and vapor phase compositions of 0.467 methanol and 0.533 benzene for a liquid composition of 0.3 methanol. 2) The dew point pressure of 1.547×105 Pa and liquid phase compositions of 0.11 methanol and 0.89 benzene for a vapor composition of 0.3 methanol. 3) The flash calculation at 60°C and 1/2(bubble point pressure + dew point pressure) is needed to find the composition if an azeot

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262 views3 pages

Methanol-Benzene Phase Equilibrium

The document summarizes the calculation of various properties for a binary mixture of methanol and benzene at 60°C using the Wilson equation. It provides: 1) The bubble point pressure of 1.281×105 Pa and vapor phase compositions of 0.467 methanol and 0.533 benzene for a liquid composition of 0.3 methanol. 2) The dew point pressure of 1.547×105 Pa and liquid phase compositions of 0.11 methanol and 0.89 benzene for a vapor composition of 0.3 methanol. 3) The flash calculation at 60°C and 1/2(bubble point pressure + dew point pressure) is needed to find the composition if an azeot

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happywhewmi
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© © All Rights Reserved
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For Methanol-Benzene system, based on Wilson Equation make the following

calculations:
a) BUBLP:T=60 C , x1=0.3:
b) DEW P: T=60 C, y1=0.3;
c) P,T flash: t=60 C, P=1/2(Pbubble +Pdew ) and z1=0.3
az and x1az = y1az
d) If and azeotrope exists at T=60 C find P

The system is a binary mixture of Methanol(1) and Benzene(2):

3 cal
kPa 10 Pa R := 1.987 M mol
molK

The Antoine equation constants for methanol and benzene are:

A := 16.5938 B := 3644.30 C := 239.76 methanol


a := 13.8594 b := 2773.78 c := 220.07 benzene


Psat1 ( T) := exp A
BK kPa

( T 273.15 K) + C K
We can write the expressions for
saturated pressure as a function of

Psat2 ( T) := exp a
b K kPa
temperature.
( T 273.15 K) + c K

For methanol (1) and benzene (2), the Wilson equation parameters are:

3 3
cm cm
V1 := 40.73 V2 := 89.41
mol mol
cal cal
a12 := 1734.42 a21 := 183.04
mol mol

12 ( T) :=
V2 a12
exp
V1 RT

21 ( T) :=
V1 a21
exp
V2 RT
Calculation of Activity Coefficients from Wilson Equation:

12 ( T) 21 ( T)
expx2
1 ( x1 , x2 , T) :=
x1 + x212 ( T) x2 + x121 ( T)
( x1 + x212 ( T) )
12 ( T) 21 ( T)
expx1
2 ( x1 , x2 , T) :=
x1 + x212 ( T) x2 + x121 ( T)
( x1 + x221 ( T) )

For this problem, the temperature is constant

T := ( 60 + 273.15) K

a) BUBL P Calculation:
The liquid phase mole
x1 := 0.3 x2 := 1 x1 fraction is given in the
problem statement.

Guess values:
We can guess what the bubblepoint
P := 101.33 kPa y1 := 0.4 y2 := 1 y1 pressure and vapor phase mole fractions
will be.

Given
y1P x11 ( x1 , x2 , T) Psat1 ( T) The first two equations for the vapor phase mole
fractions are written directly from the modified
y2P x22 ( x1 , x2 , T) Psat2 ( T) form of Raoult's Law.

y1 + y2 1
Pbubl

y1 := Find( P , y1 , y2)
y2

5
Pbubl = 1.281 10 Pa y1 = 0.467 y2 = 0.533
b) DEW P Calculation:
The vapor phase mole
y1 := 0.3 y2 := 1 y1 fraction is given in the
problem statement.

Guess Values:

P := 101.33 kPax1 := 0.1 x2 := 1 x1

Given
y1P x11 ( x1 , x2 , T) Psat1 ( T)
We follow the exact same procedure
y2P x22 ( x1 , x2 , T) Psat2 ( T) as in calcualting the bubblepoint pressure.
Using Raoult's Law, we get the desired result.
x1 + x2 1

Pdew

x1 := Find( P , x1 , x2)
x2

5
Pdew = 1.547 10 Pa x1 = 0.11 x2 = 0.89

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