1. CH EN 5253 – Process Design II
Heat Integration, Part 2
February 16, 2018

2. Books
Product and Process Design Principles: Synthesis, Analysis and Evaluation
by J. D Seader, Warren D. Seider and Daniel R. Lewin
Chapter 9
Chemical Engineering Design: Principles, Practice and Economics of Plant and Process Design
by Gavin Towler, and Ray Sinnott
Section 3.5

3. Minimum Utility Target
Temperature Interval (TI) Method
Graphical Method: Composite heating and cooling curves
Linear Programming (LP) method

4. Linear Programming (LP) method
Qsteam - R = ….. (LP.1)
R1 - R = ….. (LP.2)
R2 - R = ….. (LP.3)
R3-R4+75= ….. (LP.4)
R4-Qcw-15= ….. (LP.5)
Qsteam ,Qcw , R1 , R2 , R3 ,R4  …..(LP.6)
Qsteam=0, R3=-50

5. Linear Form All values are positive
When Qsteam is minimum, Qcw is also minimum
Solution 1
Qsteam =0
R1 =30
R2 =32.5
R3 =-50
R4 =25
Qcw =10
Solution 2
Qsteam =50
R1 =80
R2 =82.5
R3 =0
R4 =75
Qcw =60
Solution 3
Qsteam =100
R1 =130
R2 =132.5
R3 =50
R4 =125
Qcw =110

6. Comparison ( Minimum Utilities)
Simple HEN
HEN with Min. Utilities
HE = 6 ( 3 Interior + 3 Auxiliary)
HE = 7 ( 4 Interior + 3 Auxiliary)
Saves
CW e4 BTU/hr
Steam 7.5e4 BTU/hr

7. Solution to reduce number of HE
NHx,min= Ns + NU – NNW
Break Heat Exchanger Loops
Stream Splitting
4+2-2=4

8. Solution 1: Break Heat Exchanger Loops
Simplest Change
Eliminate HE 1
Transfer Heat duty to HE 2
Heat duty
C1 stream + H @Heater
C2 stream - H @ Heater

9. Solution 1: Break Heat Exchanger Loops
Attack small Heat Exchangers First
Cp=K an where 1>n  0.6
Case 1:
small HE = 20 m2, large HE = 80 m2
Cp = K K = 19.9 K
Case 2:
Single HE= 100 m2 ( combined area)
Cp= K =15.8 K

10. Solution 2: Stream Splitting
Two streams created from one
In principle, splitting of heat capacity mass flowrate, mCp
Example: Design minimum number of HE
ΔTmin =10 OC
MER target for hot utility : 300 KW

11. No Splitting HEN NHx,min= Ns + NU – NNW =3+1-1=3
Simple Network without splitting
Where is the problem?

12. Violation of 2nd law of thermodynamics
Cold Stream, C1
Hot Stream, H2

13. Stream Splitting X (T1-90)=500 (10-x) (T2-90)=200 and
200-T1  ΔTmin i.e., T1  190
150-T2  ΔTmin i.e., T2  140
T2= 140
T1= 173.3
x= 4
T2= 135
T1= 180
x= 5.56
T2= 130
T1= 190
x= 5
Many Solutions

14. Hot Stream, H2
Cold Stream, C1

15. Stream Splitting: SolutionOC OC
Non-isothermal mixing == maximize lost work

16. Improved design: Minimization of Lost work
Isothermal mixing
Location of Heater moved for isothermal mixing
Mixing split streams isothermally minimizes lost work

17. Optimization of HEN
CP=K(Area)0.6

18. Effect of ΔTmin on total cost
True pinch is approached
Area of heat transfer  ∞
Utility  minimum
ΔTmin  ∞
Area of heat transfer  0
Utility  maximum
Tradeoff between Capital cost and Utility cost

19. ΔTmin > ΔTthres Threshold ΔTmin
If ΔTmin is such that no pinch exists
Either hot or cold utility to be used, not both
ΔTthres = minimum ΔTmin below no pinch exists
A chemical company wants to design a HEN with a pinch so that both hot/cold can be used
ΔTmin > ΔTthres

20. Threshold ΔTmin ΔTmin=10C ΔTmin=105C Case 1 Case 2 S T(C) T(C) C Q(kW)
Qsteam, min = 0 Kw
Qcw, min = 46 Kw
Case 2
Qsteam, min = 6 Kw
Qcw, min = 52 Kw

21. Sensitivity of ΔTmin CP=K(Area)0.6 Area=Q/(UF ΔTmin)
ΔT =UA/Q
Utilities increase due to Lost work since it increases as ΔT increases
Tradeoff between Capital cost and Utility cost

22. Optimization based on total cost
Minimum
Capital cost goes down when A is less. This is caused by delta T being larger for Q to remain the same.

23. Distillation Columns F HF+ Qreb= D HD+B HB+Qcond Q  Qreb  Qcond
F HF-D HD-B HB 0
If Q is reduced
Utility cost reduced
number of trays / height of packing increased
Tradeoff between operating cost and capital cost

24. Heuristic “Position a Distillation Column Between Composite Heating and Cooling Curves”
When utility cost is high
Adjust pressure level to position T-Q between hot and cold composite curves
Hot streams to reboiler
Condenser to cold streams
Difficult to position
When utility cost is high and separating close boiling points
Hot utility to reboiler
Condenser to cold streams
Difficult to position

25. Multi-effect Distillation
When position a Distillation Column Between Composite Heating and Cooling Curves not possible
Feed split
Two towers operating at different pressures

26. Adjust Pressure in C2 for ΔTmin
Good when utility cost high
Down side
Purchase cost for two towers
pump cost
process complexity

27. Variation on two-effect distillation
Feed Splitting (FS)
Light Splitting/
forward heat-integration (LSF)
Light Splitting/
reverse heat-integration (LSR)