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
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
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