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MULTISTAGE SEPARATION
PROCESSES
CHE 452
ENG. AMAL MAGDY
SECTION (1)
Binary system phase diagrams
• Temperature – composition diagram
𝑇𝐴 : Boiling point of the more volatile component
𝑇𝐵 : Boiling point of the less volatile component
T
x,y
Liquid
L+V
100% A
0%B
0% A
100%B
Vapour
𝑻𝑩
𝑻𝑨
Mole fraction
𝑥𝐴 : mole fraction of component (A) in liquid phase
𝑦𝐴 : mole fraction of component (A) in vapor phase
Dalton’s law:
𝑷𝑨 = 𝒚𝑨 ∗ 𝑷𝑻
Raoult’s law:
𝑷𝑨 = 𝒙𝑨 ∗ 𝑷𝑨
𝒐
From Dalton’s and Raoult’s law:
𝒙𝑨 =
𝑷𝑻 − 𝑷𝑩
𝒐
𝑷𝑨
𝒐
− 𝑷𝑩
𝒐
Total pressure:
𝑷𝑻 = 𝑷𝑨 + 𝑷𝑩
SHEET (1)
Example (1)
The vapor pressures of n-heptane and toluene at 373 K are 106 and 73.7 kPa
respectively, while the total pressure is 101.3 kPa. Determine the mole fractions of n-
heptane in the vapor and in the liquid phase at 373 K.
Solution (1)
Find : 𝑥𝐴 & 𝑦𝐴
𝒙𝑨 = 0.85
Substitute in Raoult’s equation
𝑃𝐴 = 90.1 kPa
From Dalton’s law
𝒚𝑨 = 0.889
A :
T :
𝐏𝐀
𝐨
:
n-heptane
373 K
106 kPa
B :
𝐏𝐓 :
𝐏𝐁
𝐨
:
Toluene
101.3 kPa
73.7 kPa
𝒙𝑨 =
𝑷𝑻 − 𝑷𝑩
𝒐
𝑷𝑨
𝒐
− 𝑷𝑩
𝒐
𝑷𝑨 = 𝒙𝑨 ∗𝑷𝑨
𝒐
𝑷𝑨 = 𝒚𝑨 ∗𝑷𝑻
Vapor pressure
Antoine Equation:
𝒍𝒐𝒈𝑷𝒐
= 𝑨 −
𝑩
𝑪 + 𝑻
𝑷𝒐 = 𝟏𝟎𝑨 −
𝑩
𝑪+𝑻
Where T in °C and 𝑃𝑜 in mmHg
Example (2)
Determine the vapor phase composition of a mixture in equilibrium with a liquid
mixture of 0.5 mole fraction benzene and 0.5 mole fraction toluene at 338 K. Do you
think the liquid vaporize at pressure of 101.3 kPa?
The Antoine equation constants are given below:
For benzene: A = 6.906 B = 1211.033 C = 220.79
For toluene: A = 6.953 B = 1343.943 C = 219.377
Solution:
𝑦𝐴 , 𝑦𝐵 ???
𝒚𝑨 = 0.734
𝒚𝑩 = 0.266
The liquid will vaporize if the total pressure is higher than the given pressure
𝑷𝒐
= 𝟏𝟎 𝑨 −
𝑩
𝑪+𝑻
𝑷𝑨 = 𝒚𝑨 ∗𝑷𝑻
Bubble point & Dew point
• For a mixture:
The boiling range starts at the bubble point and
ends at the dew point
• To get the values of bubble and dew points:
For Bubble point (Tb)  𝒚𝒊 = 𝟏
For Dew point (Td)  𝒙𝒊 = 𝟏
T
x,y
Liquid
L+V
100% A
0%B
0% A
100%B
Vapour
𝑻𝑩
𝑻𝑨
𝑻𝑩𝒖𝒃𝒃𝒍𝒆
𝑻𝑫𝒆𝒘
Example (3)
What is the bubble point of an equimolar mixture of benzene and toluene at 101.3
kPa?
Knowing that:
For benzene: A = 6.906 B = 1211.033 C = 220.79
For toluene: A = 6.953 B = 1343.943 C = 219.377
Liquid phase mixture
Solution (3)
• To get the bubble point: 𝒚𝒊 = 𝟏
𝑦𝑖 =
𝑃𝑖
𝑃𝑇
=
𝑥𝑖 ∗ 𝑃𝑖
𝑜
𝑃𝑇
𝑇𝐵𝑢𝑏𝑏𝑙𝑒 = 92.1°C
𝑷𝒐
= 𝟏𝟎 𝑨 −
𝑩
𝑪+𝑻
Example (4)
What is the dew point of an equimolar mixture of benzene and toluene at 101.3 kPa?
Knowing that:
For benzene: A = 6.906 B = 1211.033 C = 220.79
For toluene: A = 6.953 B = 1343.943 C = 219.377
Vapor phase mixture
Solution (4)
To get the dew point: 𝒙𝒊 = 𝟏
𝑥𝑖 =
𝑃𝑖
𝑃𝑖
𝑜 =
𝑦𝑖 ∗ 𝑃𝑇
𝑃𝑖
𝑜
𝑇𝐷𝑒𝑤 = 98.78°C
The boiling range is 92.1 - 98.78°C
Relative volatility
• For a binary system:
A : More Volatile Component
B : Less Volatile Component
∴ 𝑃𝐴
𝑜
> 𝑃𝐵
𝑜
𝜶 =
𝑷𝑨
𝒐
𝑷𝑩
𝒐
𝛼 : Relative volatility
Equilibrium curve
• For binary mixtures there are three ways to get the
equilibrium curve:
1. Using the temperature – composition diagram
2. Using the vapor pressure values and the general
equilibrium relation
𝒙𝑨 =
𝑷𝑻 − 𝑷𝑩
𝒐
𝑷𝑨
𝒐
− 𝑷𝑩
𝒐 & 𝒚𝑨 =
𝑷𝑨
𝒐
𝑷𝑻
. 𝒙𝑨
3. Using the relative volatility (𝛼)
𝒚𝑨 =
𝜶. 𝒙𝑨
𝟏 + (𝜶 − 𝟏) 𝒙𝑨
T
y
x
x
• For multi-component systems:
1. The ideal systems
𝒚𝒊 =
𝑷𝒊
𝒐
𝑷𝑻
. 𝒙𝒊 = 𝒌𝒊 . 𝒙𝒊
2. The non-ideal systems
𝒚𝒊 =
𝑷𝒊
𝒐
𝑷𝑻
. 𝜸𝒊 . 𝒙𝒊 = 𝒌𝒊 . 𝜸𝒊. 𝒙𝒊
Equilibrium curve
SHEET (1)
Example (5)
The following vapor pressures were obtained for benzene and toluene:
Construct the following diagrams at 1 atmosphere:
• A temperature-composition diagram
• The equilibrium curve
• Relative volatility against mole fraction of toluene In liquid
Temperature, °C Vapor pressure of benzene, mmHg Vapor pressure of toluene, mmHg
80 760 300
92 1078 432
100 1344 554
110.4 1748 760
Solution (5)
• Temperature-Composition Diagram
𝑥𝐴 =
𝑃𝑇 − 𝑃𝐵
𝑜
𝑃𝐴
𝑜
− 𝑃𝐵
𝑜
𝑦𝐴 =
𝑃𝐴
𝑜
𝑃𝑇
. 𝑥𝐴
0
20
40
60
80
100
120
0 0.2 0.4 0.6 0.8 1
T,°C
xA , yA
Solution (5)
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
A
xA
• EQUILIBRIUM CURVE
Solution (5)
2.3
2.3
2.4
2.4
2.5
2.5
2.6
0 0.2 0.4 0.6 0.8 1
a
xB
• Relative Volatility vs XB diagram
Binary system phase diagrams
• Pressure – composition diagram
As pressure increases, separation becomes more
difficult.
P1 < P5
T
x,y
P1
P2
P3
P4
P5
Abnormal mixtures
T
y
x
• Azeotropic mixtures
o Maximum azeotrope
Abnormal mixtures
T
y
x
• Azeotropic mixtures
o Maximum azeotrope
o Minimum azeotrope
Abnormal mixtures
T
y
x
• Azeotropic mixtures
o Maximum azeotrope
o Minimum azeotrope
• Partial liquid miscibility
Abnormal mixtures
• Azeotropic mixtures
o Maximum azeotrope
o Minimum azeotrope
• Partial liquid miscibility
• Complete immiscibility mixtures
T
x
Section 1   multistage separation processes

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Section 1 multistage separation processes

  • 2. Binary system phase diagrams • Temperature – composition diagram 𝑇𝐴 : Boiling point of the more volatile component 𝑇𝐵 : Boiling point of the less volatile component T x,y Liquid L+V 100% A 0%B 0% A 100%B Vapour 𝑻𝑩 𝑻𝑨
  • 3. Mole fraction 𝑥𝐴 : mole fraction of component (A) in liquid phase 𝑦𝐴 : mole fraction of component (A) in vapor phase Dalton’s law: 𝑷𝑨 = 𝒚𝑨 ∗ 𝑷𝑻 Raoult’s law: 𝑷𝑨 = 𝒙𝑨 ∗ 𝑷𝑨 𝒐 From Dalton’s and Raoult’s law: 𝒙𝑨 = 𝑷𝑻 − 𝑷𝑩 𝒐 𝑷𝑨 𝒐 − 𝑷𝑩 𝒐 Total pressure: 𝑷𝑻 = 𝑷𝑨 + 𝑷𝑩
  • 5. Example (1) The vapor pressures of n-heptane and toluene at 373 K are 106 and 73.7 kPa respectively, while the total pressure is 101.3 kPa. Determine the mole fractions of n- heptane in the vapor and in the liquid phase at 373 K.
  • 6. Solution (1) Find : 𝑥𝐴 & 𝑦𝐴 𝒙𝑨 = 0.85 Substitute in Raoult’s equation 𝑃𝐴 = 90.1 kPa From Dalton’s law 𝒚𝑨 = 0.889 A : T : 𝐏𝐀 𝐨 : n-heptane 373 K 106 kPa B : 𝐏𝐓 : 𝐏𝐁 𝐨 : Toluene 101.3 kPa 73.7 kPa 𝒙𝑨 = 𝑷𝑻 − 𝑷𝑩 𝒐 𝑷𝑨 𝒐 − 𝑷𝑩 𝒐 𝑷𝑨 = 𝒙𝑨 ∗𝑷𝑨 𝒐 𝑷𝑨 = 𝒚𝑨 ∗𝑷𝑻
  • 7. Vapor pressure Antoine Equation: 𝒍𝒐𝒈𝑷𝒐 = 𝑨 − 𝑩 𝑪 + 𝑻 𝑷𝒐 = 𝟏𝟎𝑨 − 𝑩 𝑪+𝑻 Where T in °C and 𝑃𝑜 in mmHg
  • 8. Example (2) Determine the vapor phase composition of a mixture in equilibrium with a liquid mixture of 0.5 mole fraction benzene and 0.5 mole fraction toluene at 338 K. Do you think the liquid vaporize at pressure of 101.3 kPa? The Antoine equation constants are given below: For benzene: A = 6.906 B = 1211.033 C = 220.79 For toluene: A = 6.953 B = 1343.943 C = 219.377 Solution: 𝑦𝐴 , 𝑦𝐵 ??? 𝒚𝑨 = 0.734 𝒚𝑩 = 0.266 The liquid will vaporize if the total pressure is higher than the given pressure 𝑷𝒐 = 𝟏𝟎 𝑨 − 𝑩 𝑪+𝑻 𝑷𝑨 = 𝒚𝑨 ∗𝑷𝑻
  • 9. Bubble point & Dew point • For a mixture: The boiling range starts at the bubble point and ends at the dew point • To get the values of bubble and dew points: For Bubble point (Tb)  𝒚𝒊 = 𝟏 For Dew point (Td)  𝒙𝒊 = 𝟏 T x,y Liquid L+V 100% A 0%B 0% A 100%B Vapour 𝑻𝑩 𝑻𝑨 𝑻𝑩𝒖𝒃𝒃𝒍𝒆 𝑻𝑫𝒆𝒘
  • 10. Example (3) What is the bubble point of an equimolar mixture of benzene and toluene at 101.3 kPa? Knowing that: For benzene: A = 6.906 B = 1211.033 C = 220.79 For toluene: A = 6.953 B = 1343.943 C = 219.377 Liquid phase mixture
  • 11. Solution (3) • To get the bubble point: 𝒚𝒊 = 𝟏 𝑦𝑖 = 𝑃𝑖 𝑃𝑇 = 𝑥𝑖 ∗ 𝑃𝑖 𝑜 𝑃𝑇 𝑇𝐵𝑢𝑏𝑏𝑙𝑒 = 92.1°C 𝑷𝒐 = 𝟏𝟎 𝑨 − 𝑩 𝑪+𝑻
  • 12. Example (4) What is the dew point of an equimolar mixture of benzene and toluene at 101.3 kPa? Knowing that: For benzene: A = 6.906 B = 1211.033 C = 220.79 For toluene: A = 6.953 B = 1343.943 C = 219.377 Vapor phase mixture
  • 13. Solution (4) To get the dew point: 𝒙𝒊 = 𝟏 𝑥𝑖 = 𝑃𝑖 𝑃𝑖 𝑜 = 𝑦𝑖 ∗ 𝑃𝑇 𝑃𝑖 𝑜 𝑇𝐷𝑒𝑤 = 98.78°C The boiling range is 92.1 - 98.78°C
  • 14. Relative volatility • For a binary system: A : More Volatile Component B : Less Volatile Component ∴ 𝑃𝐴 𝑜 > 𝑃𝐵 𝑜 𝜶 = 𝑷𝑨 𝒐 𝑷𝑩 𝒐 𝛼 : Relative volatility
  • 15. Equilibrium curve • For binary mixtures there are three ways to get the equilibrium curve: 1. Using the temperature – composition diagram 2. Using the vapor pressure values and the general equilibrium relation 𝒙𝑨 = 𝑷𝑻 − 𝑷𝑩 𝒐 𝑷𝑨 𝒐 − 𝑷𝑩 𝒐 & 𝒚𝑨 = 𝑷𝑨 𝒐 𝑷𝑻 . 𝒙𝑨 3. Using the relative volatility (𝛼) 𝒚𝑨 = 𝜶. 𝒙𝑨 𝟏 + (𝜶 − 𝟏) 𝒙𝑨 T y x x
  • 16. • For multi-component systems: 1. The ideal systems 𝒚𝒊 = 𝑷𝒊 𝒐 𝑷𝑻 . 𝒙𝒊 = 𝒌𝒊 . 𝒙𝒊 2. The non-ideal systems 𝒚𝒊 = 𝑷𝒊 𝒐 𝑷𝑻 . 𝜸𝒊 . 𝒙𝒊 = 𝒌𝒊 . 𝜸𝒊. 𝒙𝒊 Equilibrium curve
  • 18. Example (5) The following vapor pressures were obtained for benzene and toluene: Construct the following diagrams at 1 atmosphere: • A temperature-composition diagram • The equilibrium curve • Relative volatility against mole fraction of toluene In liquid Temperature, °C Vapor pressure of benzene, mmHg Vapor pressure of toluene, mmHg 80 760 300 92 1078 432 100 1344 554 110.4 1748 760
  • 19. Solution (5) • Temperature-Composition Diagram 𝑥𝐴 = 𝑃𝑇 − 𝑃𝐵 𝑜 𝑃𝐴 𝑜 − 𝑃𝐵 𝑜 𝑦𝐴 = 𝑃𝐴 𝑜 𝑃𝑇 . 𝑥𝐴 0 20 40 60 80 100 120 0 0.2 0.4 0.6 0.8 1 T,°C xA , yA
  • 20. Solution (5) 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y A xA • EQUILIBRIUM CURVE
  • 21. Solution (5) 2.3 2.3 2.4 2.4 2.5 2.5 2.6 0 0.2 0.4 0.6 0.8 1 a xB • Relative Volatility vs XB diagram
  • 22. Binary system phase diagrams • Pressure – composition diagram As pressure increases, separation becomes more difficult. P1 < P5 T x,y P1 P2 P3 P4 P5
  • 23. Abnormal mixtures T y x • Azeotropic mixtures o Maximum azeotrope
  • 24. Abnormal mixtures T y x • Azeotropic mixtures o Maximum azeotrope o Minimum azeotrope
  • 25. Abnormal mixtures T y x • Azeotropic mixtures o Maximum azeotrope o Minimum azeotrope • Partial liquid miscibility
  • 26. Abnormal mixtures • Azeotropic mixtures o Maximum azeotrope o Minimum azeotrope • Partial liquid miscibility • Complete immiscibility mixtures T x