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MULTISTAGE SEPARATION
PROCESSES
CHE 452
ENG. AMAL MAGDY
SECTION (4)
Multistage Distillation
Calculations
• Binary System
• Sorel plate to plate calculations
• Ponchon Sevarit method
• Lewis plate to plate calculations
• McCabe Thiele method
General Laws for sorel
plate to plate calculations
• Overall Material Balance (OMB)
F = D + W
• Component Material Balance (CMB)
𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘
• Overall Heat Balance (OHB)
𝑭. 𝒉𝒇 + 𝑸𝒓 = 𝑫. 𝒉𝑫 + 𝑸𝒄 + 𝑾. 𝒉𝒘
• Equilibrium Relation
𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Or 𝒚𝑨 =
𝜶.𝒙𝑨
𝟏+(𝜶 −𝟏) 𝒙𝑨
Lewis plate to plate
calculations
• Calculation steps:
1. Overall calculations
2. Top Section
3. Bottom Section
Lewis plate to plate
calculations
• Overall calculations:
1. OMB  F = D + W
2. CMB  𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘
3. Reflux ratio  R =
𝑳𝒐
𝑫
4. 𝑳′𝒎 = 𝑭 + 𝑳𝒏
5. 𝑽′𝒎 = 𝑽𝒏
Lewis plate to plate
calculations
• Top Section (Rectifying section):
OMB  𝑽𝒏+𝟏= 𝑳𝒏 + D
CMB  𝑽𝒏+𝟏. 𝒚𝒏+𝟏 = 𝑳𝒏 . 𝒙𝒏 + 𝑫. 𝒙𝑫  Op. line eq.
For first plate  n = 1
Equilibrium Relation:
𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Or 𝒚𝒊 =
𝜶.𝒙𝒊
𝟏+(𝜶 −𝟏) 𝒙𝒊
• These calculations should be repeated until
𝒙𝒏 ≤ 𝒙𝒇
Lewis plate to plate
calculations
• Bottom Section (Stripping Section):
OMB  𝑳′𝒎+𝟏 = 𝑽′𝒎 + W
CMB 𝑳′𝒎+𝟏 . 𝒙′𝒎+𝟏 = 𝑽′𝒎. 𝒚′𝒎 + 𝑾. 𝒙𝒘  Op. line eq.
For first plate (Reboiler) m = 0
Equilibrium Relation:
𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Or 𝒚𝒊 =
𝜶.𝒙𝒊
𝟏+(𝜶 −𝟏) 𝒙𝒊
• These calculations should be repeated until
𝒚𝒎 ≥ 𝒙𝒇
Lewis plate to plate
calculations
• General Tips for Calculations:
• 𝒚𝟏 = 𝒙𝒐 = 𝒙𝑫
• 𝒚𝒎 is in eq. with 𝒙𝒘
• If you need to get a relation between the
composition of liquid and vapor on the same
plate use the equilibrium relation
• If you need to get a relation between the
composition of liquid and vapor on different
plates use the operating line equation
• Operating line equation = CMB equation
Eq.
Eq.
SHEET (4)
Example (1)
100 kmole/h mixture of benzene and toluene containing 40% benzene is to be
separated to give a product containing 90% benzene at the top, and a bottom product
containing not more than 10% benzene. The feed is liquid at its boiling point with
relative volatility of 2.5. The separation process is done using a distillation column with
total condenser and the reflux ratio is 3.
Find:
• The number of theoretical plates needed.
• The position of entry for the feed from the top.
Solution (1)
Givens:
𝑥𝑓 = 0.4 , 𝑥𝐷 = 0.9 & 𝑥𝑤 = 0.1
𝛼 = 2.5 , 𝑅 = 3 & 𝐹 = 100 𝑘𝑚𝑜𝑙𝑒/ℎ
Answer:
1. Overall calculations:
F = D + W
D + W = 100  (1)
𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘
100 ∗ 0.4 = 0.9 𝐷 + 0.1 𝑊  (2)
From (1) & (2)
D = 37.5 kmole/h
W = 62.5 kmole/h
Solution (1)
R =
𝑳𝒐
𝑫
= 3
𝑳𝒐 = 3 D = 112.5 kmole/h = 𝑳𝒏
𝑽𝟏= 𝑳𝒐 + D = 150 kmole/h
𝑳′𝒎 = 𝑭 + 𝑳𝒏 = 100 + 112.5 = 212.5 kmole/h
𝑽′𝒎 = 𝑽𝒏 = 150 kmole/h
𝒚𝟏 = 𝒙𝒐 = 𝒙𝑫 = 0.9
Eq. relation  𝒚𝟏 =
𝜶.𝒙𝟏
𝟏+(𝜶 −𝟏) 𝒙𝟏
𝒙𝟏 = 0.78
Solution (1)
2. Top section:
For the first plate  n=1
𝑽𝟐. 𝒚𝟐 = 𝑳𝟏 . 𝒙𝟏 + 𝑫. 𝒙𝑫
150 𝒚𝟐 = 112.5 𝒙𝟏 + 37.5 * 0.9
Op. Line eq.  𝒚𝟐 = 0.75 𝒙𝟏 + 0.225 , 𝒙𝟏 = 0.78
𝒚𝟐 = 0.81
For the second plate  n=2
Eq. relation  𝒚𝟐 =
𝜶.𝒙𝟐
𝟏+(𝜶 −𝟏) 𝒙𝟐
𝒙𝟐 = 0.63
From Op. line equation  𝒚𝟑 = 0.7
Solution (1)
2. Top section:
For the third plate  n=3
From eq. relation  𝒙𝟑 = 0.48
From Op. line equation  𝒚𝟒 = 0.585
For the forth plate  n=4
From eq. relation  𝒙𝟒 = 0.36 < 𝒙𝒇
∴ The top section contains four stages
Solution (1)
2. Bottom section:
For the first plate (Reboiler)  m = 0
Eq. relation  𝒚′𝒐 =
𝜶.𝒙𝒘
𝟏+(𝜶 −𝟏) 𝒙𝒘
𝒚′𝒐 = 0.217
𝑳′𝟏 . 𝒙′𝟏 = 𝑽′𝒐. 𝒚′𝒐 + 𝑾. 𝒙𝒘
𝟐𝟏𝟐. 𝟓 𝒙′𝟏 = 𝟏𝟓𝟎 ∗ 𝒚′𝒐 + 𝟔𝟐. 𝟓 ∗ 𝟎. 𝟏
Op. Line eq.  𝒙′𝟏 = 0.706 𝒚′𝒐 + 0.029
𝒙′𝟏 = 0.182
Solution (1)
2. Bottom section:
For the second plate  m = 1
Eq. relation  𝒚′𝟏 =
𝜶.𝒙′𝟏
𝟏+(𝜶 −𝟏) 𝒙′𝟏
𝒚′𝟏 = 0.357
From Op. line equation  𝒙′𝟐 = 0.281
For the third plate  m = 2
From eq. relation  𝒚′𝟐 = 0.49
∴ The Bottom section has two stages + the reboiler stage
Solution (1)
Section Stage Vapor phase composition (y) Liquid phase composition (x)
Top Section
1st 0.9 0.78
2nd 0.81 0.63
3rd 0.7 0.48
4th 0.585 0.36
Bottom Section
2nd 0.49 0.281
1st 0.357 0.182
Reboiler stage 0.217 0.1
Section 4   multistage separation processes

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Section 4 multistage separation processes

  • 2. Multistage Distillation Calculations • Binary System • Sorel plate to plate calculations • Ponchon Sevarit method • Lewis plate to plate calculations • McCabe Thiele method
  • 3. General Laws for sorel plate to plate calculations • Overall Material Balance (OMB) F = D + W • Component Material Balance (CMB) 𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘 • Overall Heat Balance (OHB) 𝑭. 𝒉𝒇 + 𝑸𝒓 = 𝑫. 𝒉𝑫 + 𝑸𝒄 + 𝑾. 𝒉𝒘 • Equilibrium Relation 𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Or 𝒚𝑨 = 𝜶.𝒙𝑨 𝟏+(𝜶 −𝟏) 𝒙𝑨
  • 4. Lewis plate to plate calculations • Calculation steps: 1. Overall calculations 2. Top Section 3. Bottom Section
  • 5. Lewis plate to plate calculations • Overall calculations: 1. OMB  F = D + W 2. CMB  𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘 3. Reflux ratio  R = 𝑳𝒐 𝑫 4. 𝑳′𝒎 = 𝑭 + 𝑳𝒏 5. 𝑽′𝒎 = 𝑽𝒏
  • 6. Lewis plate to plate calculations • Top Section (Rectifying section): OMB  𝑽𝒏+𝟏= 𝑳𝒏 + D CMB  𝑽𝒏+𝟏. 𝒚𝒏+𝟏 = 𝑳𝒏 . 𝒙𝒏 + 𝑫. 𝒙𝑫  Op. line eq. For first plate  n = 1 Equilibrium Relation: 𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Or 𝒚𝒊 = 𝜶.𝒙𝒊 𝟏+(𝜶 −𝟏) 𝒙𝒊 • These calculations should be repeated until 𝒙𝒏 ≤ 𝒙𝒇
  • 7. Lewis plate to plate calculations • Bottom Section (Stripping Section): OMB  𝑳′𝒎+𝟏 = 𝑽′𝒎 + W CMB 𝑳′𝒎+𝟏 . 𝒙′𝒎+𝟏 = 𝑽′𝒎. 𝒚′𝒎 + 𝑾. 𝒙𝒘  Op. line eq. For first plate (Reboiler) m = 0 Equilibrium Relation: 𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Or 𝒚𝒊 = 𝜶.𝒙𝒊 𝟏+(𝜶 −𝟏) 𝒙𝒊 • These calculations should be repeated until 𝒚𝒎 ≥ 𝒙𝒇
  • 8. Lewis plate to plate calculations • General Tips for Calculations: • 𝒚𝟏 = 𝒙𝒐 = 𝒙𝑫 • 𝒚𝒎 is in eq. with 𝒙𝒘 • If you need to get a relation between the composition of liquid and vapor on the same plate use the equilibrium relation • If you need to get a relation between the composition of liquid and vapor on different plates use the operating line equation • Operating line equation = CMB equation Eq. Eq.
  • 10. Example (1) 100 kmole/h mixture of benzene and toluene containing 40% benzene is to be separated to give a product containing 90% benzene at the top, and a bottom product containing not more than 10% benzene. The feed is liquid at its boiling point with relative volatility of 2.5. The separation process is done using a distillation column with total condenser and the reflux ratio is 3. Find: • The number of theoretical plates needed. • The position of entry for the feed from the top.
  • 11. Solution (1) Givens: 𝑥𝑓 = 0.4 , 𝑥𝐷 = 0.9 & 𝑥𝑤 = 0.1 𝛼 = 2.5 , 𝑅 = 3 & 𝐹 = 100 𝑘𝑚𝑜𝑙𝑒/ℎ Answer: 1. Overall calculations: F = D + W D + W = 100  (1) 𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘 100 ∗ 0.4 = 0.9 𝐷 + 0.1 𝑊  (2) From (1) & (2) D = 37.5 kmole/h W = 62.5 kmole/h
  • 12. Solution (1) R = 𝑳𝒐 𝑫 = 3 𝑳𝒐 = 3 D = 112.5 kmole/h = 𝑳𝒏 𝑽𝟏= 𝑳𝒐 + D = 150 kmole/h 𝑳′𝒎 = 𝑭 + 𝑳𝒏 = 100 + 112.5 = 212.5 kmole/h 𝑽′𝒎 = 𝑽𝒏 = 150 kmole/h 𝒚𝟏 = 𝒙𝒐 = 𝒙𝑫 = 0.9 Eq. relation  𝒚𝟏 = 𝜶.𝒙𝟏 𝟏+(𝜶 −𝟏) 𝒙𝟏 𝒙𝟏 = 0.78
  • 13. Solution (1) 2. Top section: For the first plate  n=1 𝑽𝟐. 𝒚𝟐 = 𝑳𝟏 . 𝒙𝟏 + 𝑫. 𝒙𝑫 150 𝒚𝟐 = 112.5 𝒙𝟏 + 37.5 * 0.9 Op. Line eq.  𝒚𝟐 = 0.75 𝒙𝟏 + 0.225 , 𝒙𝟏 = 0.78 𝒚𝟐 = 0.81 For the second plate  n=2 Eq. relation  𝒚𝟐 = 𝜶.𝒙𝟐 𝟏+(𝜶 −𝟏) 𝒙𝟐 𝒙𝟐 = 0.63 From Op. line equation  𝒚𝟑 = 0.7
  • 14. Solution (1) 2. Top section: For the third plate  n=3 From eq. relation  𝒙𝟑 = 0.48 From Op. line equation  𝒚𝟒 = 0.585 For the forth plate  n=4 From eq. relation  𝒙𝟒 = 0.36 < 𝒙𝒇 ∴ The top section contains four stages
  • 15. Solution (1) 2. Bottom section: For the first plate (Reboiler)  m = 0 Eq. relation  𝒚′𝒐 = 𝜶.𝒙𝒘 𝟏+(𝜶 −𝟏) 𝒙𝒘 𝒚′𝒐 = 0.217 𝑳′𝟏 . 𝒙′𝟏 = 𝑽′𝒐. 𝒚′𝒐 + 𝑾. 𝒙𝒘 𝟐𝟏𝟐. 𝟓 𝒙′𝟏 = 𝟏𝟓𝟎 ∗ 𝒚′𝒐 + 𝟔𝟐. 𝟓 ∗ 𝟎. 𝟏 Op. Line eq.  𝒙′𝟏 = 0.706 𝒚′𝒐 + 0.029 𝒙′𝟏 = 0.182
  • 16. Solution (1) 2. Bottom section: For the second plate  m = 1 Eq. relation  𝒚′𝟏 = 𝜶.𝒙′𝟏 𝟏+(𝜶 −𝟏) 𝒙′𝟏 𝒚′𝟏 = 0.357 From Op. line equation  𝒙′𝟐 = 0.281 For the third plate  m = 2 From eq. relation  𝒚′𝟐 = 0.49 ∴ The Bottom section has two stages + the reboiler stage
  • 17. Solution (1) Section Stage Vapor phase composition (y) Liquid phase composition (x) Top Section 1st 0.9 0.78 2nd 0.81 0.63 3rd 0.7 0.48 4th 0.585 0.36 Bottom Section 2nd 0.49 0.281 1st 0.357 0.182 Reboiler stage 0.217 0.1