Transcript Heat Integration

An Introduction to
Heat Exchanger Network (HEN)
Design
By:
Anwaruddin Hisyam
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In this lecture we will learn how to set energy
recovery targets for a process.
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Pinch identification
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Base case
178 C
1620
880
160 C
180 C
130 C
60 C
2640
210 C
Reactor
270 C
160 C
210 C
149 C
1220
50 C
1980
Pinch technology series
220 C
4
Data Extraction
178 C
1620
880
160 C
180 C
130 C
60 C
2640
210 C
Reactor
270 C
160 C
210 C
149 C
1220
50 C
1980
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220 C
5
…..from data extraction
160 C
130 C
210 C
Reactor
270 C
160 C
210 C
50 C
220 C
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60 C
6
Stream Data (Problem Table)
T source,
C
T target,
C
1
220
60
3520
22
Hot
2
270
160
1980
18
Hot
3
50
210
3200
20
Cold
4
160
210
2500
50
Cold
No
Heat duty,
kW
CP
Type
CP = Heat duty/ABS(T source – T target)
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Composite Curve
Composite Curve
Temp interval (K)
300
DT min
250
200
150
100
50
0
0
1000
2000
3000
4000
5000
6000
7000
Heat duty (kW)
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Set DTmin = 20 C
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Shifted Stream Data
Hot - ½DTmin; Cold + ½DTmin
T source,
C
T target,
C
1
210
50
3520
22
Hot
2
260
150
1980
18
Hot
3
60
220
3200
20
Cold
4
170
220
2500
50
Cold
No
Heat duty,
kW
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CP
Type
10
Shifted Composite Curve
Temp interval (K)
Shifted Composite Curve
300
250
200
150
100
50
0
0
1000
2000
3000
4000
5000
6000
7000
Heat Duty (kW)
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Cascade Diagram
Stream population
Cold
Hot
ColdHot
260
0
720
-720
SURPLUS
700
180
520
DEFICIT
2800 1600
1200
DEFICIT
400
-400
SURPLUS
1800 1980
-180
SURPLUS
0
-220
SURPLUS
220
50
210
18
170
20
800
150
22
60
220
50
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Heat balance in the interval
Hot utility
0
260
-720
Heat
flow
720
220
520
200
210
1200
170
-400
150
- 1000
- 600
-180
60
-220
50
- 420
- 200
Cold utility
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The heat flow must NOT be
negative
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Normalization
Hot utility
0
260
-720
720
220
520
200
210
1200
170
-400
150
- 1000
Need
additional
heat
- 600
-180
60
-220
50
- 420
- 200
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Cold utility
15
Original Grand Composite Curve
300
Unfeasible
region
250
200
150
100
Feasible
region
50
0
-1200 -1000 -800
-600
-400
-200
0
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200
400
600
800
1000
16
…finding pinch
Hot utility
1000
260
-720
1720
220
520
1200
210
1200
170
-400
150
0
No heat flow
at this point
400
-180
60
-220
50
580
800
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Cold utility
17
Grand Composite Curve
Temp interval (K)
300
Qh min
250
200
150
100
50
Qc min
0
0
500
1000
1500
2000
Heat duty (kW)
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finally….
The PINCH POINT = 170 C
which means that
Hot stream PINCH = 170+½DTmin = 180
Cold stream PINCH = 170-½DTmin = 160
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Heat source and sink

Composite Curve
Temp interval (K)
300
PINCH
250

200
150
100
50
Heat Source
0
0
1000
2000
Heat source 
this part releases
heat
Heat sink  this
part requires heat
Heat Sink
3000
4000
5000
6000
7000
Heat duty (kW)
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and,
how can we design HEN
based on the pinch?
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Base case….the existing network
2640
60
880
220
C
1980
160
270
50
210
H H
1220
160
1620
H
210
Heat recovery = 1980 + 880 = 2860 kW
Cold utility = 2640 kW
Hot utility = 1220 + 1620 = 2840 kW
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Let’s start from
the pinch
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PINCH
AT 170
BELOW
CP
ABOVE
180
2640
22
60
220
880
180
360
18
160
270
160
2200
20
50
1640
210
1000
210
2500
160
0
50
160
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Rules….

CP in ≤ CP out


N stream IN ≤ N stream out




Start finding partners for streams OUT (with streams IN,
away from pinch, or utility)
If Ns IN > Ns OUT, split stream(s) OUT
If CP in > CP out (no match), try to split stream(s) IN
Set maximum heat recovery
The remaining heat duty is covered by heater or
cooler
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Step 1: Below the PINCH
Connect S1(22) and S3(20)
CP in < CP out
180
60
220
1
180
160
2200
50
√
1
2200
270
160
210
160
160
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210
26
Step 2: Above the PINCH
Connect S2(18) and S4(50)
CP in < CP out
180
60
220
1
180
160
2200
50
√
1
2200
1620
2
160
√
270
1620
210
160
160
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2
210
27
Step 3: Above the PINCH
Connect S1(22) and S4(50)
CP in < CP out
880
180
60
1
3
1620
180
160
2200
50
√
1
2200
√
√
2
160
880
270
1620
210
160
160
220
3
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2
√
210
2500
28
Step 4: Above the PINCH
Install Heater at S3(20)
880
180
60
1
1620
180
160
2200
50
√
1
2200
√
3
2
880
270
1620
1000
160
H
160
160
√
√
√
220
3
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2
210
1000
210
2500
29
Step 5: Below the PINCH
Install Cooler at S1(22) and S2(18)
440
2640
60
C
1
√
2200
160
50
√
1
2200
√
3
360
360
880
180
C
1620
180
√
2
880
270
1620
1000
160
H
160
160
√
√
√
220
3
2
210
1000
210
2500
All heat requirements have been met !!!
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…finally…
Heat Exchanger Network (HEN)
880
60
C
220
1620
440
160
C
270
360
1000
H
50
210
2200
160
210
Maximum Energy Recovery (MER) = 2200 + 880 + 1620 = 4700 kW
Minimum cooling heat duty (Qc min) = 440 + 360 = 800 kW
Minimum heating heat duty (Qh min) = 1000 kW
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Then draw the flowsheet…
177.6 C
880
160 C
1620
180 C
130 C
360
210 C
Reactor
270 C
160 C
210 C
160 C
220 C
180 C
1000
50 C
440
2200
60 C
80 C
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Possible modifications
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Grand Composite Curve
Temp interval (K)
300
MP steam
250
200
150
100
Heat
generation
50
Cooling water
0
0
500
1000
1500
2000
Heat duty (kW)
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Working Session
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How will the HEN be….?
Feed 2
H= 27 MW
230 C
H=-30 MW
140 C
200 C
Product 2
80 C
Reactor 2
Feed 1
20 C
180 C
250 C
40 C
H=32 MW
40 C
Reactor 1
H= -31.5 MW
40 C
Product 1
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