L20-Heat and Pdfffffffffffffffower Integration(1).pptx
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
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
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).
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
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.
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
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.
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.
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
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
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
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
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
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?
63. Improved design: Minimization of Lost work
63
Location of Heater moved for isothermal mixing
Mixing split streams isothermally minimizes lost work
Isothermal mixing
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
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.
