1. Vapor – Liquid Equilibrium
Lecture 12
10/17/2014

2. Upcoming Schedule
Today: Vapor-Liquid Equilibrium – K-values Monday: Vapor-Liquid Equilibrium – Correlations Wednesday: Review Lecture Friday: In-Class Problem Session PLEASE me with suggestions, areas of concern so I can cover them Monday: Mid-term exam, 50 minutes, 2 note sheets

3. Vapor-Liquid Equilbrium
Gas
Liquid

4. Vapor-Liquid Equilbrium
Gas
Liquid

5. Vapor-Liquid Equilbrium
Gas
Gas
Liquid
Liquid

6. Vapor-Liquid Equilbrium
Gas
Gas
Gas
Liquid
Liquid
Liquid

7. Primary Reservoir Questions
Can we compute reservoir properties if we only know:
Total mole fraction of the components in liquid and gas
P and T in reservoir (and/or) separator
We want to know:
Composition of the liquid and gas produced at the surface
°API gravity and Rs of the produced stream
How can we maximize the value of the produced stream?

8. Definitions
n total moles of both gas and liquid present nL total moles of mixture in liquid phase nV total moles of mixture in vapor phase zj mole fraction of component j in total mixture xj mole fraction of component j in liquid phase yj mole fraction of component j in vapor phase pj partial pressure of component j pvj vapor pressure of component j

9. VLE Equations
Dalton’s Law: pj = yjp where yj is the mole fraction in the gas phase Raoult’s Law: pj = xjpvj where xj is the mole fraction in the liquid phase This indicates: The partial pressure exerted by a component in an ideal liquid solution is equal to the vapor pressure of that component times the liquid mole fraction of that component

10. Dalton and Raoult’s Law
Two-component, two-phase system p = 14.7 psia, pv2 = 10 psia, y1 = 0.4 What is the liquid mole fraction of component 2?

11. Dalton and Raoult’s Law
Two-component, two-phase system p = 14.7 psia, pv2 = 10 psia, y1 = 0.4 What is the liquid mole fraction of component 2? Use Dalton’s Law p2 = y2p = 0.6 × 14.7 psia = 8.8 psia

12. Dalton and Raoult’s Law
Two-component, two-phase system p = 14.7 psia, pv2 = 10 psia, y1 = 0.4 What is the liquid mole fraction of component 2? Use Dalton’s Law p2 = y2p = 0.6 × 14.7 psia = 8.8 psia Use Raoult’s Law p2 = x2pv2 X2 = p2/pv2 = 8.8/10 = 0.88

13. K-Values
If the liquid and gas are in equilibrium, then: pj = yjp = xjpvj yj/xj = pvj/p This ratio is known as the K-value of component j, or Kj Since: Kj = yj/xj = pvj/p Then: log Kj = log pvj – log p This will not work close or beyond the critical point of any of the individual components, as the vapor pressure does not exist there.

14. K-Values for a Mixture at Tc
Kj = 1 at critical point

15. Problem Setup

16. Class Exercise T = 200 °F P = 700 psia N = 1 mole
Component
Mole Fraction C1
0.339 C2
0.064 C3
0.099 C4
0.043 C5
0.032 C6
0.031
C7+
0.392
T = 200 °F
P = 700 psia
N = 1 mole
We want to know the fraction of the fluid that is gas.

17. Class Exercise Component z K C1 0.339 7 C2 0.064 2 C3 0.099 1 C4 0.043
0.5 C5
0.032
0.2 C6
0.031
0.12
C7+
0.392
0.01

18. Class Exercise
Solve 7 or 8:

19. Class Exercise
Solve 7 or 8: But we need nL, so estimate, calculate and iterate.

20. Class Exercise

21. Solution Method Solve (8.7) or (8.8) in an iterative fashion.
1. Find Kj for the given p, T and composition
2. Guess nL
3. Evaluate the left hand side of (8.7)
4. If this is not equal to 1, repeat steps 2-4 with new guess

22. Primary Issue with Approach
What does it mean if the iterative approach does not converge?

23. Primary Issue with Approach
What does it mean if the iterative approach does not converge? The fluid is not a two-phase system. P
liquid
2-phase
region
gas T

24. Primary Issue with Approach
What does it mean if the iterative approach does not converge? The fluid is not a two-phase system. P
liquid
2-phase
region
gas T

25. Primary Issue with Approach
What does it mean if the iterative approach does not converge? The fluid is not a two-phase system.
At bubble point:
nV = 0  zj = xj  yj = zj Kj
At dew point:
nL = 0  zj = yj  xj = zj /Kj P
liquid
2-phase
region
gas T

26. Determining Bubble and Dew Point Pressures
Step 1. Guess the (bubble/dew point) pressure Step 2. Determine the Kj’s Step 3. Evaluate If not equal to 1, repeat At bubble point, we know that Mixture will be all liquid if T < Tb, or p > pb This is case if At dew point, we know that Mixture will be all gas if T > Td, or p < pd

27. Single Phase Systems?

28. Single Phase Systems?

29. Single Phase Systems?

30. A Closer Look at K-values
Equation holds for ideal mixtures at low pressures and temps Can extrapolate to other conditions (but not very far) Better: use fugacity Fugacity defines the change in free energy (or chemical potential) required to pass from one phase to another Chemical potential can be seen as measure of escaping tendency of a component in a solution

31. Fugacity
A thermodynamic property of a real gas that, if substituted for the pressure or partial pressure in the equations for an ideal gas, gives equations applicable to the real gas

32. K-value Curves
Normally given on a log-log plot against pressure Slope at low pressures equal to -1 At Tc, K-values converge to 1 at pc K-values converge to a value near 1.0 for other temps at a pressure called the convergence pressure Convergence pressure measures effect of type and quantities of molecules present

33. Summary
Primary goal is to compute reservoir parameters for 2-phase systems Must compute the composition of the liquid and gas phase Use Dalton’s law and Raoult’s law to derive equilibrium equations K-values (ratio of gas mole fraction over liquid mole fraction) crucial in equilibrium computations K-values are determined experimentally and depend on temperature, pressure and the convergence pressure