Selfstudy Thermo 4a VLE calcs Solutions PDF

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Thermo & Unit Ops 3 – Self-study Themo4a. VLE diagrams.

A) For the system n-pentane (1) and methanol (2) the activity coefficients can be described using the Wilson activity coefficient model:   12 21 ln 1 ln 1 2 12 2  2 12 1 21 2  1   21 12 ln 2 ln 2 1 21 1  1 21 2 12 1  2 Where    2 1 21  exp 12  ; exp  12 21     1 2 and the pure component molar volumes are v1 = 116.11 and v2 =40.73, while the binary interaction parameters are a12 = 236.6 K and a 21 = 1413 K. The vapour pressure can be calculated from the Antoine equation (constants for npentane A=6.88098; B=1080.48; C=233.9 – constants for methanol A=8.08097; B=1582.271; C=239.726). For this system: a) Determine the appropriate form and units of the Antoine equation. b) Calculate the P vs x,y and y vs x diagrams at 350 K. c) Calculate the T vs x,y and y vs x diagrams at 1 atm (101325 Pa) HINT: use Excel (or Matlab/Mathcad) and generate 101 points, i.e. from 0 to 1 with a 0.01 interval in x. B) For the same system as in part A, calculate the azeotrope pressure and composition at 350 K. This is the solution of a single nonlinear equation, followed by the calculation of the total pressure. [4.58 bar; 0.703] C) For the same system as in part A, calculate the azeotrope temperature and composition at 1 atm. This is now a nested solution of single nonlinear equations. Hint: Use the solution to part B in a loop where you vary the temperature – easier to do in Matlab or Mathcad. Actually very straightforward in Mathcad. [303.98 K; 0.809] D) Derive the condition for the existence of an azeotrope for a binary system with positive deviations (use the convention that component 1 is the more volatile, as a result  >  ) HINT: You have to show that the condition for an azeotrope 2

1

1

2

has a solution. Remember that the RHS is now a constant and that the ratio of the activity coefficients is a continuous function of the mole fraction of component 1.

Solution to parts A-C Find Tb for methanol and n-Pentane from NIST to find the correct Antoine equation. See Mathcad spreadsheets for graphs and numerical solution of the azeotrope equation. NOTE that for some of the calculations the pressures are above 2 atm and it would be useful to check the validity of the approximation that the ratio of the fugacity coefficients is approximately equal to 1. For this we need an equation of state.

Solution to part D We need to consider the values of

2

at

1

2

1)

0 and

1

1.

1

For

0 , ie pure component 2 (

1

2

1

1

1

1

The final inequality is obtained from the fact that we have positive deviations: or  > 1 ln  > 0 Note also that 

 =



=

 

and at infinite dilution





 =   > 1 

This means that in the first region (from  = 0 to the azeotrope, where  = 1) and 1 −  < 1 −  or  <  ie  < 1  >  For

1 , ie pure component 1 (

1

1

1)

2 2 1

Therefore if 1 2 2

there must be at least one concentration where the condition for the azeotrope has a solution, ie 2

1

1

2

According to the convention that component 1 is the more volatile

1 2

1 2

1

2

1 1

If we take the natural log of this 

ln  > ln  > 0 > − ln   

The first two terms are the condition given in the lecture.

1 and therefore...