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VLE VAPOR LIQUID EQUILIBRIUM - Introduction

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VLE VAPOR LIQUID EQUILIBRIUM - Introduction with examples

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VLE VAPOR LIQUID EQUILIBRIUM - Introduction

  1. 1. VAPOR/LIQUID EQUILIBRIUM- Introduction ERT 206: Thermodynamics Miss Anis Atikah Ahmad Email: anis atikah@unimap.edu.my
  2. 2. OUTLINE 1. The Nature of Equilibrium 2. Duhem’s Theorem 3. Simple Models for VLE 4. VLE by Modified Raoult’s Law 5. VLE from K-value Correlations
  3. 3. 1. The Nature of Equilibrium • Equilibrium is a static condition in which no changes occur in the macroscopic properties of a system with time. – Eg: An isolated system consisting of liquid & vapor phase reaches a final state wherein no tendency exists for change to occur within the system. The temperature, pressure and phase compositions reach final values which thereafter remain fixed.
  4. 4. • At microscopic level, conditions are not static. – Molecules with high velocities near the interface overcome surface forces and pass into the other phase. – But the average rate of passage of molecules is the same in both directions & no net interphase transfer of material occurs.
  5. 5. Measures of Composition 1. Mass fraction: the ratio of the mass of a particular chemical species in a mixture or solution to the total mass of mixture or solution. 2. Mole fraction: the ratio of the number of moles of a particular chemical species in a mixture or solution to the number of moles of mixture or solution. m m m m x ii i    n n n n x ii i   
  6. 6. Measures of Composition 3. Molar concentration: the ratio of the mole fraction of a particular chemical species in a mixture or solution to the molar volume of mixture or solution. 4. Molar mass of mixture/solution: mole-fraction- weighted sum of the molar masses of all species present. q n V x C ii i   i i i MxM  in q Molar flow rate Volumetric flow rate
  7. 7. 2. Duhem’s Theorem • Duhem’s Theorem: for any closed system formed initially from given masses of prescribed chemical species, the equilibrium state is completely determined when any two independent variables are fixed. – Applies to closed systems at equilibrium – The extensive state and intensive state of system are fixed 22  NNF  Similar to phase rule, but it considers extensive state. No of equations No of variables
  8. 8. 3. SIMPLE MODELS FOR VAPOR/LIQUID EQUILIBRIUM • Vapor/liquid equilibrium (VLE): the state of coexistence of liquid and vapor phase. • VLE Model: to calculate temperatures, pressures and compositions of phases in equilibrium. • The two simplest models are: – Raoult’s law – Henry’s law
  9. 9. 3. SIMPLE MODELS FOR VAPOR/LIQUID EQUILIBRIUM 3.1 Raoult’s Law • Assumptions: – The vapor phase is an ideal gas (low to moderate pressure) – The liquid phase is an ideal solution (the system are chemically similar) *Chemically similar: the molecular species are not too different in size and are of the same chemical nature. eg: n-hexane/n-heptane, ethanol/propanol, benzene/toluene  NiPxPy sat iii ...,2,1 ix iy Liquid phase mole fraction Vapor phase mole fraction sat iP Vapor pressure of pure species i at system temperature
  10. 10. Pxy Diagram
  11. 11. 3.2 Dewpoint & Bubblepoint Calculations with Raoult’s Law 4 Calculations • BUBL P : Calculate {yi} and P, given {xi} and T • DEW P : Calculate {xi} and P, given {yi} and T • BUBL T : Calculate {yi} and T, given {xi} and P • DEW T : Calculate {xi} and T, given {yi} and P If the vapor-phase composition is unknown, may be assumed; thus  i sat ii PxP  i iy 1 sat iii PxPy  For bubble point calculation
  12. 12. 3.2 Dewpoint & Bubblepoint Calculations with Raoult’s Law If the liquid-phase composition is unknown, may be assumed; thus   i sat ii Py P / 1  i ix 1 sat iii PxPy  For dew point calculation
  13. 13. 3.2.1 BUBL P CALCULATION (Calculate {yi} and P, given {xi} and T) Find P1 sat & P2 sat using Antoine equation Find P Calculate yi  i sat ii PxP satsat PxPxP 2211    satsat PxPxP 2111 1   1212 xPPPP satsatsat  sat iii PxPy  sat PxPy 111  P Px y sat 11 1 
  14. 14. Example 1 Binary system acetronitrile (1)/ nitromethane (2) conforms closely to Raoult’s law. Vapor pressure for the pure species are given by the following Antoine equations: Prepare a graph showing P vs. X1 and P vs. Y1 for a temperature of 75°C. 00.224/ 47.2945 2724.14/ln 1   Ct kPaPsat 00.209/ 64.2972 2043.14/ln 2   Ct kPaPsat
  15. 15. 3.2.1 BUBL P CALCULATION (Calculate {yi} and P, given {xi} and T) At 75°C, by Antoine Equations, 00.22475 47.2945 2724.14/ln 1   C kPaPsat 00.20975 64.2972 2043.14/ln 2   C kPaPsat kPaPsat 21.831  kPaPsat 98.412  Find P1 sat & P2 sat using Antoine equation Find P Calculate yi
  16. 16. 3.2.1 BUBL P CALCULATION (Calculate {yi} and P, given {xi} and T)   1212 xPPPP satsatsat    198.4121.8398.41 xP  Taking at any value of x1, say x1=0.6,   6.098.4121.8398.41 P kPa72.66 Find P1 sat & P2 sat using Antoine equation Find P Calculate yi
  17. 17. 3.2.1 BUBL P CALCULATION (Calculate {yi} and P, given {xi} and T) Find P1 sat & P2 sat using Antoine equation Find P Calculate yi P Px y sat 11 1    7483.0 72.66 21.836.0  At 75°C, a liquid mixture of 60 mol-% acetonitrile and 40 mol-% nitromethane is in equilibrium with a vapor containing 74.83 mol-% acetonitrile at a pressure of 66.72 kPa
  18. 18. To draw P-x-y graph, repeat the calculation with different values of x; x1 y1 P/kPa 0.0 0.0000 41.98 0.2 0.3313 50.23 0.4 0.5692 58.47 0.6 0.7483 66.72 0.8 0.8880 74.96 1.0 1.0000 83.21 P-x-y Diagram
  19. 19. P x y diagram for acetonitrile/nitromethane at 75°C as given by Raoult’s law 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 P/kPa x1, y1 P2 sat = 41.98 P1 sat = 83.21T= 75°C Subcooled liquid Superheated vapor
  20. 20. P x y diagram for acetonitrile/nitromethane at 75°C as given by Raoult’s law 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 P/kPa x1, y1 P2 sat = 41.98 P1 sat = 83.21T= 75°C Subcooled liquid Superheated vapor a b b' c d c’ Point a is a subcooled liquid mixture of 60 mol- % acetonitrile and 40 mol-% of nitromethane at 75°C. Point b is saturated liquid. Points lying between b and c are in two phase region, where saturated liquid and saturated vapor coexist in equilibrium. Saturated liquid and saturated vapor of the pure species coexist at vapor pressure P1 sat and P2 sat
  21. 21. P x y diagram for acetonitrile/nitromethane at 75°C as given by Raoult’s law 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 P/kPa x1, y1 P2 sat = 41.98 P1 sat = 83.21T= 75°C Subcooled liquid Superheated vapor a b b' c d c’ Point b: bubblepoint P-x1 is the locus of bubblepoints As point c is approached, the liquid phase has almost disappeared, with only droplets (dew) remaining. Point c: dewpoint P-y1 is the locus of dewpoints.
  22. 22. P x y diagram for acetonitrile/nitromethane at 75°C as given by Raoult’s law 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 P/kPa x1, y1 P2 sat = 41.98 P1 sat = 83.21T= 75°C Subcooled liquid Superheated vapor a b b' c d c’ Once the dew has evaporated, only saturated vapor at point c remains. Further pressure reduction leads to superheated vapor at point d
  23. 23. 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 P/kPa x1, y1 P2 sat = 41.98 P1 sat = 83.21T= 75°C Subcooled liquid Superheated vapor a b b' c d c’ 3.2.2 DEW P CALCULATION (DEW P : Calculate {xi} and P, given {yi} and T) What is x1 & P at point c’? Step 1: Calculate P Step 2: Calculate x1 satsat PyPy P 2211 // 1   kPa74.59 98.41/4.021.83/6.0 1   sat P Py x 1 1 1    21.83 74.596.0  4308.0
  24. 24. 3.2.2 DEW P CALCULATION (DEW P : Calculate {xi} and P, given {yi} and T) Find P from Raoult’s Law assuming Calculate xi sat iii PxPy  sat PxPy 111  sat P Py x 1 1 1   i ix 1   i sat ii Py P / 1 satsat PyPy P 2211 // 1  
  25. 25. T-x-y Diagram Find T1 sat & T2 sat using Antoine equation Find P1 sat & P2 sat using T btween T1 sat & T2 sat Calculate xi Calculate yi i i isat i C PA B T    ln sat iii PxPy  sat PxPy 111  P Px y sat 11 1   i sat ii PxP satsat PxPxP 2211    satsat PxPxP 2111 1   1212 xPPPP satsatsat   satsat sat PP PP x 21 2 1   
  26. 26. Example 2 Binary system acetronitrile (1)/ nitromethane (2) conforms closely to Raoult’s law. Vapor pressure for the pure species are given by the following Antoine equations: Prepare a graph showing T vs. X1 and T vs. Y1 for a pressure of of 70kPa. 00.224/ 47.2945 2724.14/ln 1   Ct kPaPsat 00.209/ 64.2972 2043.14/ln 2   Ct kPaPsat
  27. 27. i i isat i C PA B T    ln CT sat    84.69224 70ln2724.14 47.2945 1 CT sat    58.89209 70ln2043.14 64.2972 2 Find T1 sat & T2 sat using Antoine equation Find P1 sat & P2 sat using T btween T1 sat & T2 sat Calculate xi Calculate yi T-x-y Diagram
  28. 28. Find T1 sat & T2 sat using Antoine equation Find P1 sat & P2 sat using T btween T1 sat & T2 sat Calculate xi Calculate yi T1 sat = 69.84°C, T2 sat = 89.58°C Let T=78°C, kPaP C kPaP sat sat 76.91 00.22478 47.2945 2724.14/ln 1 1    kPaP C kPaP sat sat 84.46 00.20978 64.2972 2043.14/ln 2 2    T-x-y Diagram
  29. 29. Find T1 sat & T2 sat using Antoine equation Find P1 sat & P2 sat using T btween T1 sat & T2 sat Calculate xi Calculate yi P1 sat = 91.76kPa, P2 sat = 46.84kPa  satsat sat PP PP x 21 2 1    5156.0 84.4676.91 84.4670     T-x-y Diagram
  30. 30. Find T1 sat & T2 sat using Antoine equation Find P1 sat & P2 sat using T btween T1 sat & T2 sat Calculate xi Calculate yi P1 sat = 91.76kPa, x = 0.5156 P Px y sat 11 1    6759.0 70 76.915156.0  T-x-y Diagram
  31. 31. To draw T-x-y graph, repeat the calculation with different values of T; x1 y1 T/°C 0.0000 0.0000 89.58 (T2 sat) 0.1424 0.2401 86 0.3184 0.4742 82 0.5156 0.6759 78 0.7378 0.8484 74 1.0000 1.0000 69.84 (T1 sat) T-x-y Diagram
  32. 32. 65 70 75 80 85 90 0 0.2 0.4 0.6 0.8 1 T/°C x1, y1 Subcooled liquid Superheated vapor T2 sat = 89.58°C T1 sat = 69.84°c T x y diagram for acetonitrile/nitromethane at 70 kPa as given by Raoult’s law
  33. 33. What is y1 and T at point b’ (with x1=0.6 and P= 70 kPa)? 65 70 75 80 85 90 0 0.2 0.4 0.6 0.8 1 T/°C x1, y1 Subcooled liquid Superheated vapor T2 sat = 89.58°C T1 sat = 69.84°c 3.2.3 BUBL T CALCULATION (Calculate {yi} and T, given {xi} and P) c’ c b b'
  34. 34. 3.2.3 BUBL T CALCULATION (Calculate {yi} and T, given {xi} and P) 21 2 xx P Psat    sat sat P P 2 1  C PA B T sat    2ln 00.209 64.2972 00.224 47.2945 0681.0ln     tt  The substraction of ln P1 sat & P2 sat from Antoine Equation 00.209 ln2043.14 64.2972 2    sat P Start with α=1, find P2 sat Find T using Antoine eq & substitute P2 sat obtained in step 1 Find new α by substituting T Repeat step 1 by using new α until similar value of α is obtained Find P1 sat & find y1 using Raoult’s law satsat PxPxP 2211  2 2 11 2 x P Px P P sat sat sat 
  35. 35. 3.2.3 BUBL T CALCULATION (Calculate {yi} and T, given {xi} and P) 1 kPaPsat 702  CT  58.89 88.1 88.1 kPaPsat 81.452  CT  38.77 96.1 Iteration 1 Iteration 2 96.1 kPaPsat 41.442  CT  53.76 97.1 Iteration 3 97.1 CT  43.76 97.1 Iteration 4 kPaPsat 24.442  Start with α=1, find P2 sat Find T using Antoine eq & substitute P2 sat obtained in step 1 Find new α by substituting T Repeat step 1 by using new α until similar value of α is obtained Find P1 sat & find y1 using Raoult’s law satsat PP 21   24.4497.1 kPa17.87 P Px y sat 11 1    70 17.876.0  7472.0
  36. 36. What is x1 and T at point c’ (with y1=0.6 and P= 70 kPa)? 65 70 75 80 85 90 0 0.2 0.4 0.6 0.8 1 T/°C x1, y1 Subcooled liquid Superheated vapor T2 sat = 89.58°C T1 sat = 69.84°c 3.2.4 DEW T CALCULATION (Calculate {xi} and T, given {yi} and P) c’ c b b'
  37. 37. 3.2.4 DEW T CALCULATION (Calculate {xi} and T, given {yi} and P) Start with α=1, find P1 sat Find T using Antoine eq & substitute P1 sat obtained in step 1 Find new α by substituting T Repeat step 1 by using new α until similar value of α is obtained Find x1  211 yyPPsat  sat sat P P 2 1  C PA B T sat    1ln 00.209 64.2972 00.224 47.2945 0681.0ln     tt  00.224 ln2724.14 47.2945 1    sat P satsat PyPy P 2211 1     2211 1 yPPy P P satsat sat  
  38. 38. 3.3 Henry’s Law • Used for a species whose critical temperature is less than the temperature of application, in which Raoult’s Law could not be applied (since Raoult’s Law requires a value of Pi sat). iii xPy  Where Hi is Henry’s constant and obtained from experiment.
  39. 39. 4. VLE by Modified Raoult’s Law • Used when the liquid phase is not an ideal solution. sat iiii PxPy  Where ɣi is an activity coefficient (deviation from solution ideality in liquid phase).
  40. 40. 4. VLE by Modified Raoult’s Law • For bubblepoint calculation, (assuming ) • For dewpoint calculation, (assuming )  i sat iii PxP   i iy 1  i ix 1   i sat iii Py P  1
  41. 41. 5. VLE from K-value Correlations • Equilibrium ratio, Ki • When Ki > 1, species exhibits a higher concentration of vapor phase • When Ki < 1, species exhibits a higher concentration of liquid phase (is considered as heavy constituent.) i i i x y K 
  42. 42. 5. VLE from K-value Correlations • K value for Raoult’s Law • K value for modified Raoult’s Law P P x y K sat i i i i  sat iii PxPy since P P K sat ii i   sat iiii PxPy since
  43. 43. 5. VLE from K-value Correlations • For bubblepoint calculations, • For dewpoint calculations i i i x y K   i iy 1 1i ii xK  i ix 1 i i i x y K  1i i i K y
  44. 44. Example For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F, determine: (a) The dewpoint pressure (b)The bubblepoint pressure
  45. 45. Example (a) The dewpoint pressure When the system at its dewpoint, only an insignificant amount of liquid is present. Thus 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane are the values of yi. assuming, thus, 1i i i K y  i ix 1 For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F, determine: By trial, find the value of pressure that satisfy 1i i i K y
  46. 46. Species yi P=100psia P=150psia P=126psia Ki yi/Ki Ki yi/Ki Ki yi/Ki Methane 0.10 20.0 0.005 13.2 0.008 16.0 0.006 Ethane 0.20 3.25 0.062 2.25 0.089 2.65 0.075 Propane 0.70 0.92 0.761 0.65 1.077 0.762 0.919 828.0i ii Ky 174.1i ii Ky 000.1i ii Ky Thus, the dewpoint pressure is 126 psia. Example (a) The dewpoint pressure For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F, determine:
  47. 47. Example (b)The bubblepoint pressure assuming , thus 1i ii xK  i iy 1 For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F, determine: By trial, find the value of pressure that satisfy 1i ii xK
  48. 48. Species xi P=380psia P=400psia P=385psia Ki Kixi Ki Kixi Ki Kixi Methane 0.10 5.60 0.560 5.25 0.525 5.49 0.549 Ethane 0.20 1.11 0.222 1.07 0.214 1.10 0.220 Propane 0.70 0.335 0.235 0.32 0.224 0.33 0.231 017.1i ii xK 963.0i ii xK 000.1i ii xK Thus, the bubblepoint pressure is 385 psia. Example (b) The bubble point pressure For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F, determine:

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