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1. Separation Process-II
(ChE-306)
1

2. Ponchon-Savarit Method
• For non-ideal systems
• Molar latent heat is no more constant
• Heat of mixing will be non-zero
• Enthalpy balance
• Using an Enthalpy-Composition diagram

3. • Phase
A quantity of mixture in any physical
state.
• Phase C between the boiling point
and dew point curve will be:
𝑚𝐴
𝑚𝐵
=
𝐶𝐵
𝐶𝐴

4. • Combination of phases
• Separation of phases
• Addition of heat
• Adding the phases A and B with mass
m, composition x and H as enthalpy per
unit mass to give phase C will result:
𝑚𝐴 + 𝑚𝐵 = 𝑚𝐶
𝑚𝐴𝑥𝐴 + 𝑚𝐵𝑥𝐵 = 𝑚𝐶𝑥𝐶
𝑚𝐴𝐻𝐴 + 𝑚𝐵𝐻𝐵 = 𝑚𝐶𝐻𝐶
• If heat Q is added to A then increase in
enthalpy from A to C will be:
𝐻𝐴 +
𝑄
𝑚𝐴𝐵
= 𝐻𝐶

5. Enthalpy balance for a continuous Distillation
Column
• Feed, F (xF)
• Distillate (xD)
• Bottom Product (xW)
• HV and HL represent enthalpy per unit mass
of vapor and liquid
• QC = heat removed in condenser
• QB = heat added in reboiler

6. Material and Heat Balance
• Overall Material balance across section I
𝑉
𝑛 = 𝐿𝑛+1 + 𝐷
• Material balance with respect to MVC
𝑉
𝑛𝑦𝑛 = 𝐿𝑛+1𝑥𝑛+1 + 𝐷𝑥𝑑
• Substitute the value of Vn and simplify
𝐿𝑛+1 + 𝐷 𝑦𝑛 = 𝐿𝑛+1𝑥𝑛+1 + 𝐷𝑥𝑑
𝐿𝑛+1
𝐷
=
𝑥𝑑 − 𝑦𝑛
𝑦𝑛 − 𝑥𝑛+1

7. • Heat Balance
𝑉
𝑛𝐻𝑛
𝑉 = 𝐿𝑛+1𝐻𝑛+1
𝐿
+ 𝐷𝐻𝑑
𝐿
+ 𝑄𝑐
• Putting 𝐻𝑑
′
= 𝐻𝑑
𝐿
+
𝑄𝐶
𝐷
the above equation will simplify to:
𝑉
𝑛𝐻𝑛
𝑉 = 𝐿𝑛+1𝐻𝑛+1
𝐿
+ 𝐷𝐻𝑑
′
• Substitute the value of Vn from first equation and simplify
𝐿𝑛+1
𝐷
=
𝐻𝑑
′
− 𝐻𝑛
𝑉
𝐻𝑛
𝑉
− 𝐻𝑛+1
𝐿

8. 𝐿𝑛+1
𝐷
=
𝑥𝑑 − 𝑦𝑛
𝑦𝑛 − 𝑥𝑛+1
𝐿𝑛+1
𝐷
=
𝐻𝑑
′
− 𝐻𝑛
𝑉
𝐻𝑛
𝑉
− 𝐻𝑛+1
𝐿
As left hand sides are same so equating right hand sides
𝐻𝑑
′
− 𝐻𝑛
𝑉
𝐻𝑛
𝑉
− 𝐻𝑛+1
𝐿 =
𝑥𝑑 − 𝑦𝑛
𝑦𝑛 − 𝑥𝑛+1
Simplification gives:
𝑦𝑛 =
𝐻𝑑
′
− 𝐻𝑛
𝑉
𝐻𝑛
𝑉
− 𝐻𝑛+1
𝐿 𝑥𝑛+1 +
𝐻𝑛
𝑉 − 𝐻𝑛+1
𝐿
𝐻𝑑
′
− 𝐻𝑛+1
𝐿 𝑥𝑑
• Equation of top operating line
• Relation between the compositions of vapor and liquid streams
between any two plates
• The line passes through a common pole N of coordinates (xd , Hd’)

9. • Overall Material balance across section II
𝐿𝑚 = 𝑉
𝑚 + 𝑊
• Material balance with respect to MVC
−𝑉
𝑚𝑦𝑚 + 𝐿𝑚𝑥𝑚+1 = 𝑊𝑥𝑤
• Substitute the value of Vm and simplify
𝐿𝑚+1
𝑊
=
−𝑥𝑊 + 𝑦𝑚
𝑦𝑚 − 𝑥𝑚+1
9

10. • Heat Balance
−𝑉
𝑚𝐻𝑚
𝑉 + 𝐿𝑚+1𝐻𝑚+1
𝐿
= 𝑊𝐻𝑊
𝐿
− 𝑄𝐵
• Putting 𝐻𝑊
′
= 𝐻𝑊
𝐿
−
𝑄𝐵
𝑊
the above equation will simplify to:
−𝑉
𝑚𝐻𝑚
𝑉
+ 𝐿𝑚+1𝐻𝑚+1
𝐿
= 𝑊𝐻𝑊
′
Substitute the value of Vm from first equation and simplify
𝐿𝑚+1
𝑊
=
𝐻𝑊
′
+ 𝐻𝑚
𝑉
𝐻𝑚
𝑉
− 𝐻𝑚+1
𝐿

11. 𝐿𝑚+1
𝑊
=
−𝑥𝑊+𝑦𝑚
𝑦𝑚−𝑥𝑚+1
𝐿𝑚+1
𝑊
=
𝐻𝑊
′
+𝐻𝑚
𝑉
𝐻𝑚
𝑉 −𝐻𝑚+1
𝐿
−𝑥𝑊 + 𝑦𝑚
𝑦𝑚 − 𝑥𝑚+1
=
𝐻𝑊
′
+ 𝐻𝑚
𝑉
𝐻𝑚
𝑉
− 𝐻𝑚+1
𝐿
• Equation of operating line below the feed plate
• Line passes through pole M of coordinates (xW ,
HW’)
• Overall sum of phases
𝐹 = 𝑀 + 𝑁
𝐹𝑥𝐹 = 𝑀𝑥𝑊 + 𝑁𝑥𝑑
• Phases F, M and N lie on a straight line.

12. Determination of number of plates on H-x
diagram
• Locate xd, xf and xw
• Locate pole N
• Join N to F
• Extend backwards till xw, this will give pole M
• Starting from pole N, draw the lines to get the vapor and liquid
compositions on each plate
• Switch to pole M, when feed composition xf is crossed
• Draw the tie lines to count steps
• Stop when composition reached to xw
• Feed plate will be the one where the pole has been switched from N to M

13. L1
L2
L3
V1
V2
V3

15. Self Study
• Example 11.10