explanation about techniues

1. CH EN 5253
Process Design II
Lecture 20
Heat and Power Integration
February 26, 2020

2. Books
Product and Process Design Principles: Synthesis, Analysis and Evaluation
by Warren D. Seider , J. D Seader, Daniel R. Lewin, and Soemantri Widagdo
• Chapter 11
Chemical Engineering Design: Principles, Practice and Economics of Plant and
Process Design
by Gavin Towler, and Ray Sinnott
• Section 3.5 ( 2nd Edition)
2

3. Outlines
•Minimum Utility Target/ Maximum Energy Recovery
 Temperature-Interval (TI) method
 Composite Curve Method ( graphical)
 Linear Programming Method
•Networks for Maximum Energy Recovery
•Minimum Numbers of Heat Exchangers
 Breaking Heat Loops
 Stream Splitting
•Heat Integrated Distillation Trains
3

4. Considerations in Heat Integration

5. Heat Integration
• Make a list of HX
• Instead of using utilities can you use another
stream to heat/cool any streams?
• How much of this can you do without causing
operational problems?
• Can you use air to cool?
– Air is a low cost coolant.
• Less utilities = smaller cost of operations

6. 2011
HPC REAC-2000
Q-2002
HPC VSSL-2000
2037
13
23
2024
XFS1
HPC SPLT-2001
30
2026
HPC PUMP-2000
2031
Q-2008
2012
HPC RCYL-2001
2032
6
HPC MIX-2000
2009
3
1
2004
HPC XCHG-2002
HPC MIX-2001
2010
HPC CMPR-2000
2002
Q-2001
7
2008
HPC XCHG- 2000
HPC XCHG-2001
2038
2039
2040
2041
HPC XCHG-2003
2014
2025
Q-2006
2050
2051
HPC XCHG-1008
HPC FAXR-2000
Q-2004
HPC FAXR-2002
Q-2005
HPC XCHG-2004
2042 2043
28
2052 2053
HPC XCHG-2007
2048 2049
HPC FAXR-2001
HPC SPLT-2000
2016
2015
HPC MIX-2002
2018
2017
Q-2003
HPC XCHG-2005
2044
2045
2019
HPC XCHG-2006
2020
2021
2046 2047
HPC VSSL-2001
HPC VSSL-2002
2022
HPC VSSL-2003
HPC XCHG-2009
Q-2007
2027
HPC RCYL-2002
2029
2101
HPC CMPR-2010
HPC MIX-2010
2102
HPC XCHG-2010
HPC XCHG-2011
2103
2104
2107
2108
HPC MIX-2011
2109 2110
HPC XCHG-2012
2114
2112
HPC VSSL-2010
HPC XCHG-2013
113
2105
2106
HPC SPLT-2010
HPC FAXR-2010
HPC FAXR-2011
HPC FAXR-2012
HPC MIX-2012
HPC VSSL-2011
HPC VSSL-2012
HPC VSSL-2013
HPC XCHG-2016
2115
2116
2117
2118
2119
2120
2121
2122
2146
2147
2123
125
2124
2129
2128
2127
HPC RCYL-2012
HPC SPLT-2011
2130
2126
HPC PUMP-2010
2131
Q-2108
2137
HPC RCYL-2011
2132
2005
2111
HPC MIX-2003
HPC CMPR-2001
2033
2034
Q-2009
HPC XCHG-2014
HPC XCHG-2015
Q-2104
Q-2103
Q-2105
HPC XCHG-2017
2142 2143
144 145
2148 2149
2138 2139
2140
2141
Q-2101
HPC SPLT-2003
2135
2035
Q-2106
HPC XCHG-2018
2150 2151
Q-107
HPC XCHG-2019
2152 2153
Q-2102
HPC REAC-2010
HPC RCYL-2000
HPC RCYL-2010
2036
2136
XFS2
MIX-100
2

7. Simple Heat Exchange Network
1 Hot Stream
1 Cold Stream

8. Simple Heat Exchange Network (HEN)
steam
cooling water
steam steam
water water
As ∆Tmin becomes less, utility demands become less. However, there is a
reasonable minimum ∆Tmin to provide a driving force for heat exchange (e.g. 10°C).

9. Example
2 Hot Streams
2 Cold Streams

10. Example: Minimize Utilities For 4 Streams
Total: 470
Total: 480
Note that total cooling demand is 10×104 Btu/hr more than total heating demand.
120
235
180
240
120
235
180
240
C1
C2
H1
H2

11. A possible Heat Exchange Network (HEN)
to Achieve Required Heating and Cooling
• Note that QCW – Qsteam = 10, which is required difference
Required hot utilities
(steam): 57.5
Required cold utilities
(cooling water): 67.5

12. Is this the minimum utility?
Required hot utilities (steam): 57.5 X 104 BTU/hr
Required cold utilities (cooling water): 67.5 X 104 BTU/hr

13. Minimum Utility Target/
Maximum Energy Recovery

14. Pinch Analysis
1.Temperature Interval Method
2.Composite Curve Method (Graphical)
3.Linear Programming

15. Temperature Interval Method

16. Pinch Analysis: Temperature Interval Method
Step 1: Adjust hot stream temperatures by subtracting ΔTmin
(This puts the streams in a common reference frame.)

17. Pinch Analysis: Temperature Interval Method
Step 2: Order the temperatures hot to cold:
250 : T0
240 : T1
235 : T2
180 : T3
150 : T4
120 : T5

18. Pinch Analysis: Temperature Interval Method
18
Step 3: For each temperature range (hot to next hottest),
determine which streams are in that interval and add up
the heat capacity rates of those streams
250
240
150
120 1.5

19. Pinch Analysis: Temperature Interval Method
19
Step 3: For each temperature range (hot to next
hottest), determine which streams are in that interval
and add up the heat capacity rates of those streams
250 – 240: H1
240 – 235: H1, H2, C2
235 – 180: H1, H2, C1, C2
180 – 150: H1, H2, C1
150 – 120: H2, C1

20. Pinch Analysis: Temperature Interval Method
20
Step 4: Multiply the combined heat capacity rate by the
temperature difference to get the energy needed to cool
or heat over that temperature range
250 : T0
240 : T1
235 : T2
180 : T3
150 : T4
120 : T5

21. Pinch Analysis: Temperature Interval Method
Step 5: Create a cascade of temperature intervals within which
energy balances are carried out
Pinch point
Note!
Energy can not flow from cold to hot.
Thus, we need to increase all of the levels
by 50 to make the pinch point = zero.

22. Pinch Analysis: Temperature Interval Method

23. Pinch Analysis
Actual Endpoint Temperatures!
ΔTapp
Hot side
One HEN
Cold side
One HEN

24. Graphical Method

25. Pinch Analysis: Graphical Method
Step 1: First, draw lines representing hot
streams being cooled, and cool streams being
heated, showing correct temperatures and with
a slope corresponding to the heat capacity rate
– Does not need to be lined up on x-axis yet
– Will be straight lines if C is not a function of
temperature; otherwise, they will bend
25

26. Pinch Analysis – Graphical Approach
• Initially, just get the temperatures and slope correct. Position on x-axis is
not yet important. Can be drawn so they don’t overlap.
26
260
100
120
140
160
180
200
220
240
0 2 4 6 8 10
Q (MMBtu/hr)
T
(°F)
H1
H2
C1
C2

27. Pinch Analysis – Graphical Approach
27
260
100
120
140
160
180
200
220
240
0 2 4 6 8 10
Q (MMBtu/hr)
T
(°F)
H1
H2
Common Temperature Region (160-250)
mCp= mCp (H1) + mCp (H2) = 3+1.5 =4.5
Step 2: Then, combine all the hot stream curves into one
composite curve with the lowest temperature crossing at Q = 0
H1+H2

28. Pinch Analysis: Graphical Method
28
260
100
120
140
160
180
200
220
240
0 2 4 6 8 10
Q (MMBtu/hr)
T
(°F)
H2
H1+H2
H1
C1
C2

29. Pinch Analysis: Graphical Method
29
260
100
120
140
160
180
200
220
240
0 2 4 6 8 10
Q (MMBtu/hr)
T
(°F)
Step 3: Combine the cold stream curves into a composite.
Horizontal position will be set in the next step.
H2
H1+H2
H1
C1
C2
C1+C2

30. Pinch Analysis: Graphical Method
30
260
100
120
140
160
180
200
220
240
0 2 4 6 8 10
Q (MMBtu/hr)
T
(°F)
10°
Step 4: Position the cold stream curve so that the pinch is ∆Tmin
(vertically) from the hot stream.

31. Pinch Analysis: Graphical Method
31
260
100
120
140
160
180
200
220
240
2 4 6 8 10
Q (MMBtu/hr)
T
(°F)
0.5
0.6
Step 5: Measure the hot and cold utilities requirements.

32. Linear Programming

33. 33
Linear Programming (LP) method
Qsteam - R1 + 30 =0 ….. (LP.1)
R1 - R2+ 2.5 = 0 ….. (LP.2)
R2 - R3 - 82.5 = 0 ….. (LP.3)
R3-R4+75=0 ….. (LP.4)
R4-Qcw-15=0 ….. (LP.5)
Qsteam ,Qcw , R1 ,R2 , R3 ,R4 0 Constraints

34. Linear Form
• All values are positive
34

35. Solution Region
35
Feasible Space
Qsteam ,Qcw , R1 , R2 , R3 ,R4 0

36. Solution
36
Minimum Utility
When Qsteam is minimum, Qcw is also minimum
Solution
Qsteam =50
R1 =80
R2 =82.5
R3 =0
R4 =75
Qcw =60

37. Solution
Is this the minimum utility?
Required hot utilities (steam): 57.5 X 104 BTU/hr
Required cold utilities (cooling water): 67.5 X 104 BTU/hr
Required hot utilities (steam): 50 X 104 BTU/hr
Required cold utilities (cooling water): 60 X 104 BTU/hr
No
Minimum utility
Find the HEN for this minimum utility or maximum energy recovery.

38. Networks for Maximum
Energy Recovery

39. HEN For Maximum Energy Recovery
• Goal is to minimize the cost (number) of heat exchangers
while keeping duty required by external utilities to a
minimum
– Should be able to achieve theoretical minimum utility duty
• Sequence heat exchangers to achieve maximum overall
efficiency
• Use relation Q = C∆T to help simplify analysis
39

40. HEN For Maximum Energy Recovery
• Start with HEN for hot side of pinch
• Seek a matching stream to provide
cooling for hot stream
– Important that C of cold stream is
larger than or equal to C of hot stream!
• Calculate resulting exit temperature of
exchanger on cold stream
• Continue with hot streams, finally
making up any additional required duty
for heating the cold streams with
utilities (e.g. steam)
• Repeat for cold side, making up
required cooling with e.g. cooling water
40

41. Resulting HEN with Pinch
41
Hot side Cold side

42. Resulting HEN Layout
42
Hot side
Cold side

43. 43
Comparison ( Minimum Utilities)
• Simple HEN • HEN with Min. Utilities
Saves
CW 7.5e4 BTU/hr
Steam 7.5e4 BTU/hr
HE = 7 ( 4 Interior + 3 Auxiliary)
HE = 6 ( 3 Interior + 3 Auxiliary)

44. Economical HENs
• CP,Ii
= Cost of HEs in the interior networks ( cold-hot streams)
• CP,Aj
= Cost of HEs in the auxiliary networks ( steam/cooling
water – hot/cold streams)
• Fs = Annual flow rate of steam ( kilograms/year)
• Fcw = Annual flow rate of cooling water ( kilograms/year)
• s = unit cost of steam ($/kilogram)
• cw = unit cost of cooling water ($/kilogram)
• im=return on investment

45. Optimization based on total cost
Minimum
CP=K(Area)0.6
Area=Q/(UF ΔTmin)

46. Effect of ΔTmin on total cost
ΔTmin  0
• 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

47. 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

48. Minimum Numbers of Heat
Exchangers

49. Too Many Heat Exchangers
• Sometimes fewer Heat exchangers and increased
utilities leads to a lower annual cost
• Determine minimum number of HEs
NHx,min= Ns + NU – NNW
– s=No. streams
– U=No. discrete Utilities
• Hot utilities: Fuel, HPS, IPS, LPS,
• Cold utilities: BFW, CW, refrigeration
– NW=No. independent Networks (1 above the pinch, 1 below the pinch)

50. Too Many Heat Exchangers
• Minimum number of HEs
NHx,min = Ns + NU – NNW
= 4+2-2
= 4
Simple HFN: 2 excess HEs
HEN with Min. Utilities = 3 excess HEs

51. Minimize Number of HEs
Method to reduce number of HE
1. Break Heat Exchanger Loops
2. Stream Splitting

52. Break Heat Exchanger Loops

53. Break Heat Exchanger Loops
53

54. Break Heat Exchanger Loops
Simplest Change
1. Eliminate HE 1
2. Transfer Heat duty to HE 2
3. Heat duty
i. C1 stream + H @Heater
ii. C2 stream - H @ Heater

55. Break Heat Exchanger Loops
• Attack small Heat Exchangers First
Cp=Kan where n<1 ( 0.6)
Case 1:
• small HE = 20 m2, large HE = 80 m2
• Cp = K200.6+K800.6 = 19.9 K
Case 2:
• Single HE= 100 m2 ( combined area)
• Cp= K 1000.6 =15.8 K

56. Stream Splitting

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

58. No Splitting HEN
58
• NHx,min= Ns + NU – NNW
=3+1-1=3
Simple Network without splitting
Where is the problem?

59. Violation of 2nd law of thermodynamics
59

60. Stream Splitting
60
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= 6
Many Solutions
T2= 130
T1= 190
x= 5
T2= 135
T1= 180
x= 5.56

61. 61
T2

62. Stream Splitting: Solution
62
OC
OC
Non-isothermal mixing == maximize lost work

63. Improved design: Minimization of Lost work
63
Location of Heater moved for isothermal mixing
Mixing split streams isothermally minimizes lost work
Isothermal mixing

64. Heat Integrated Distillation Trains
64

65. Distillation Columns
F HF+ Qreb= D HD+B HB+Qcond
Energy balance

66. Distillation Columns
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

67. 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
• Cold stream to Condenser
• Close boiling point species

68. Heuristic “Position a Distillation Column Between
Composite Heating and Cooling Curves”
Difficult to use both hot and cold streams
• Hot utility to reboiler
• Cold streams to condenser

69. Multi-effect Distillation
• When position a Distillation Column Between
Composite Heating and Cooling Curves not
possible
• Feed split
• Two towers operating at different pressures
69

70. Heat Integration for Direct Distillation
Sequence

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

72. Variation on two-effect distillation
72
Feed Splitting (FS) Light Splitting/
forward heat-integration (LSF)
Light Splitting/
reverse heat-integration (LSR)

73. Heat Pumps in Distillation

74. Heat Pumps/Heat Engines Heurisitcs
• When positioning heat engines, to reduce the
cold utilities, place them entirely above or
below the pinch
• When positioning heat pumps, to reduce the
total utilities, place them across the pinch.

75. Heat Pump Location
75

76. Heat Engine Location
76
Tp