LECTURE 6: INTRODUCTION TO
HEAT EXCHANGER NETWORK
SYNTHESIS
1
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Schedule – Introduction to HEN Synthesis
• Unit 1. Introduction: Capital vs. Energy
– What is an optimal HEN design
– A Simple Example (Class Exercise 1)
– Setting Energy Targets
• Unit 2. The Pinch and MER Design
– The Heat Recovery Pinch
– HEN Representation
– Class Exercise 2
• Unit 3. The Problem Table
– Class Exercises 3 and 4
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Schedule – Advanced HEN Synthesis
• Unit 4. Loops and Splits
–
–
–
–
Minimum Number of Units by Loop Breaking
Class Exercise 5
Stream Split Designs
Class Exercise 6
• Unit 5. Threshold Problems
– Class Exercise 7
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Schedule – Heat and Power Integration
• Unit 6. Data Extraction
– Class Exercise 8
• Unit 7. Heat Integration in Design
–
–
–
–
4
Grand Composite Curve
Heat-integrated Distillation
Heat Engines
Heat Pumps
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Part One: Objectives
• The first part of this three-part Unit on HEN synthesis
serves as an introduction to the subject, and covers:
– The “pinch”
– The design of HEN to meet Maximum Energy Recovery (MER)
targets
– The use of the Problem Table to systematically compute MER
targets
•
Instructional Objectives:
Given data on hot and cold streams, you should be able to:
– Compute the pinch temperatures
– Compute MER targets
– Design a simple HEN to meet the MER targets
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
UNIT 1: Introduction - Capital vs. Energy
• The design of Heat Exchanger Networks deals with the
following problem:
• Given:
– NH hot streams, with given heat capacity flowrate, each having
to be cooled from supply temperature THS to targets THT.
– NC cold streams, with given heat capacity flowrate, each having
to be heated from supply temperature TCS to targets TCT.
• Design:
An optimum network of heat exchangers, connecting between
the hot and cold streams and between the streams and
cold/hot utilities (furnace, hot-oil, steam, cooling water or
refrigerant, depending on the required duty temperature).
• What is optimal?
Implies a trade-off between CAPITAL COSTS (Cost of
equipment) and ENERGY COSTS (Cost of utilities).
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Example
Tout
Tout
Tout
H
H
H
in
T
Tin
Tin
Tin
Tin
Steam
Cooling
Water
out
C T
C
Tout
C
Tout
Tin
Tout
Network for minimal
equipment cost ?
Tout
Tout
Cooling
Water
out
T
Tin
Network for minimal
energy cost ?
Tin
Tout
Tin
Tout
Tin
7
Steam
Tin
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Tin
Intro HEN Synthesis - 6
Numerical Example
150o
150o
100
150o
CP = 1.0
300o
300o
0
50
CP = 1.0
50
0
CP = 1.0
Cooling
Water (90-110oF)
100
100
o
CP = 1.0 300
CP = 1.0
Steam (400oF)
50
100
200o
100
200o
100
200o
Design A:
(AREA) = 20.4
[ A = Q/UTlm ]
150o
0
CP = 1.0
CP = 1.0
200o
100
300o
200o
100
300o
100
500
500
CP = 1.0
8
150o
CP = 1.0
o
CP = 1.0 300
Design B:
(AREA) = 13.3
150o
CP = 1.0
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
200o
500
CP = 1.0
Intro HEN Synthesis - 6
Some Definitions
T
TT
TS
TT
H
CP
T
TS
H
9
H
= Stream supply temperature (oC)
= Stream target temperature (oC)
= Stream enthalpy (MW)
= m
  Cp (MW/ oC)
= Heat capacity flowrate (MW/ oC)
= Stream flowrate  specific heat capacity
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Tmin
- Example
Tmin = Lowest permissible temperature difference
Which of the two counter-current heat exchangers
illustrated below violates T  20 oF (i.e. Tmin = 20 oF) ?
20o
100
30o
80o
60o
o
50o
A
10o
100
70o
60o
o
40o
20o
B
Clearly, exchanger A violates the Tmin constraint.
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Definitions (Cont’d)
Exchanger Duty.
Data: Hot stream CP = 0.3 MW/ oC
Cold stream CP = 0.4 MW/ oC
T1 = 70o
OK
60o
100o
Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC 
40o
OK
Q = 0.4(70 - 40) = 0.3(100 - 60) = 12 MW
Heat Transfer Area (A): A = Q/(UTlm)
Data: Overall heat transfer coefficient, U=1.7 kW/m2 oC
(Alternative formulation in terms of film coefficients)
Tlm = (30 - 20)/loge(30/20) = 24.66
So, A = Q/(UTlm) = 12000/(1.724.66) = 286.2 m2
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Class Exercise 1
Tmin = 10 oC
H=100
Cond
180o
o
60
80o
R2
C1
H=160
Reb
100o
130
40o
TT
(oC)
H1
H2
C1
C2
180
130
60
30
80
40
100
120
CP
H
o
(kW) (kW/ C)
100
1.0
180
2.0
160
4.0
162
1.8
Utilities. Steam@150 oC, CW@25oC
o
H=180
R1
Stream
TS
(oC)
120o
H=162
Design a network of steam heaters,
water coolers and exchangers for the
o
30 process streams. Where possible, use
exchangers in preference to utilities.
.
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Setting Energy Targets
Cond
80
60
180o
o
Summary of proposed design:
100
o
R2
C1
60
H
130
Reb
o
CW
Units
60 kW
18 kW
4
R1
100o
162
C 18
Steam
Are 60 kW of Steam
Necessary?
120o
30o
40o
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
The Temperature-Enthalpy Diagram
T
T
130oC
200oC
40oC
100oC
H=180
One hot stream
14
H=300
H
H
H=100
Two hot streams
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
The Temperature-Enthalpy Diagram
C
CW
Tmin = 20
10
Steam
Steam
T
H
120oC
110oC
100oC
CW
QCmin = 30
50
H
50
QHmin = 70
Correlation between Tmin, QHmin and QCmin
More in, More out! QHmin + x  QCmin + x
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
The Composite Curve
Hot Composite Curve
H interval
o
180 C
H=100
Cond
o
180
60o
80o
R2
C1
CP=1.0
50
130oC
150
o
80 C
CP=2.0
o
40 C
80
H=160
Reb
100o
120o 130 C
H=180
16
180oC
o
130o
40o
R1
H=162
CP=1.0
50
CP=3.0
150
o
80 C
30o
o
40 C
CP=2.0
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
80
Intro HEN Synthesis - 6
The Composite Curve (Cont’d)
Cold Composite Curve
H interval
o
120 C
H=100
Cond
180o
o
60
CP=1.8
100oC
232
80o
60oC
R2
C1
CP=4.0
30oC
H=160
Reb
100o
130o
40
17
o
120oC
R1
CP=1.8
o
100 C
120o
H=180
36
H=162
CP=5.8
60oC
30
o
30oC
CP=1.8
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
54
36
232
54
Intro HEN Synthesis - 6
The Composite Curve (Cont’d)
T
QHmin = 54
48
130oC
Result:
QCmin and QHmin for
desired Tmin
MER Target
100oC
o
80 C
o
T
=
20
min
Tmin = 10oC
C
60oC
QQ
Cmin
Cmin==6 12
H
Here,
hot pinch is at 70 oC,
cold pinch is at 60 oC
QHmin = 48 kW and
QCmin = 6 kW
Method: manipulate hot and cold composite curves until
required Tmin is satisfied.
This defines hot and cold pinch temperatures.
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
UNIT 2: The Pinch
QHmin
T
Heat
Source
QCmin
QHmin
+x
Tmin
“PINCH”
x
Heat
Sink
+x
H
H
The “pinch” separates the HEN problem into two parts:
– Heat sink - above the pinch, where at least QHmin utility must be
used
– Heat source - below the pinch, where at least QCmin utility must be
used.
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Significance of the Pinch
Do not transfer heat across pinch
Do not use cold utilities above the pinch
Do not use hot utilities below the pinch
Cond
100
Summary of modified design:
R2
C1
Steam
CW
Units
~49 kW
~7 kW
5
R1
Reb
H 49
111
62
C 7
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
HEN Representation
100
Cond
60oC
180oC
80oC
H1
R2
C1
H2
130oC
100oC
o
100 C
Reb
o
130 C
R1
120oC
H 49
111
120oC
40 C
21
C 7
30oC
40oC
C
7
60oC
H
49
111
100
62
o
80oC
180oC
30oC
C1
C2
62
Where is the pinch ?
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
HEN Representation with the Pinch
Thot
H1
H2
H
Thot
Thot
Tcold
Tcold
Tcold
Tcold
C
C1
C2
The pinch divides the HEN into two parts:
 the left hand side (above the pinch)
 the right hand side (below the pinch)
At the pinch, ALL hot streams are hotter than ALL cold
streams by Tmin.
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Class Exercise 2
•
•
•
For this network, draw the
grid representation
Given pinch temperatures at
480 oC /460 oC, and MER
targets: QHmin= 40, QCmin=
106, redraw the network
separating the sections above
and below the pinch.
Why is QH > QHmin ?
H1
320o
C1 240
o
140
o
C2
100
170
S
50
500o
320o
CP = 1.0
CP = 1.5
116
CW
290o
200o
CP = 1.8 CP = 2.0
H1
480oC
500oC
320oC
23
480o
210
320oC
H2
H2
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
CP
200oC
1.8
290oC
2.0
240oC
140oC
C1
1.0
C2
1.5
Intro HEN Synthesis - 6
Class Exercise 2 - Solution
H2
H1
320o
C1 240
o
140
o
C2
480o
210
100
S
50
170
500o
320o
CP = 1.0
CP = 1.5
116
CW
290o
200o
CP = 1.8 CP = 2.0
CP
320oC
H1
H2
500oC
H
40
480oC
290oC
460oC
240oC
H
10
210
320oC
24
o
C 200 C 1.8
116
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
140oC
170
2.0
C1
1.0
C2
1.5
100
Intro HEN Synthesis - 6
Class Exercise 2 - Solution (Cont’d)
This can be fixed by reducing the cooling duty by 10 units, and
eliminate the excess 10 units of heating below the pinch.
CP
H1
H2
320oC
C
116
106
480oC
500oC
320oC
H
H
40
10
1.8
290oC
2.0
240oC
460o
450
210
220
140oC
170
160
25
200oC
C1
1.0
C2
1.5
100
110
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for Maximum Energy Recovery(MER)
Example
CP
170oC
60oC
3.0
150oC
30oC
1.5
H1
H2
135oC
140oC
20oC
80oC
C1
2.0
C2
4.0
Step 1: MER Targeting.
Pinch at 90o (Hot) and 80o (Cold)
Energy Targets:
Total Hot Utilities:
20 kW
Total Cold Utilities:
60 kW
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Step 2: Divide the problem at the pinch
170oC
90oC
90oC
60oC
3.0
150oC
90oC
90oC
30oC
1.5
80oC
80oC
20oC
H1
H2
135oC
140oC
27
CP
80oC
C2
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
C1
2.0
4.0
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Step 3: Design hot-end, starting at the pinch:
Pair up exchangers according to CP-constraints.
Immediately above the pinch, pair up streams
such that: CPHOT  CPCOLD
(This ensures that TH TC  Tmin)
CP
H1
3.0
H2
1.5
C1
2.0
C2
4.0
Tmin
Meets Tmin
Violates
Tminconstraint
constraint
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Step 3 (Cont’d): Complete hot-end design, by ticking-off
streams.
CP
H1
QHmin = 20 kW 
H2

170o
150o
90o
3.0

90o

135o
140o
H
20
80o
90

80o
1.5
C1
2.0
C2
4.0
240
Add heating utilities as needed (MER target)
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Step 4: Design cold-end, starting at the pinch:
Pair up exchangers according to CP-constraints.
Immediately above the pinch, pair up streams
such that: CPHOT  CPCOLD
(This ensures that TH TC  Tmin)
CP
H1
3.0
H2
1.5
C1
Tmin
2.0
Violates
Meets
TT
constraint
constraint
minmin
30
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Step 4 (Cont’d): Complete cold-end design, by ticking-off
streams.
CP
H1
H2

90o
60o
90o
C
60
80o

35o
90

30o
20o
C1
3.0
1.5
QCmin = 60 kW 
2.0
30
Add cooling utilities as needed (MER target)
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Completed Design:
CP
H1
H2
170o
90o
150o
135o
H
140o
60o
90o
20
80o
125o
90
30o
70o
C
60
20o
35o
90
30
80o
3.0
1.5
C1
2.0
C2
4.0
240
Note that this design meets the MER targets:
QHmin = 20 kW and QCmin = 60 kW
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Design for MER (Cont’d)
Design for MER - Summary:
 MER Targeting. Define pinch temperatures, Qhmin and QCmin
 Divide problem at the pinch
 Design hot-end, starting at the pinch: Pair up exchangers
according to CP-constraints. Immediately above the pinch, pair up
streams such that: CPHOT  CPCOLD. “Tick off” streams in order to
minimize costs. Add heating utilities as needed (up to QHmin). Do
not use cold utilities above the pinch.
 Design cold-end, starting at the pinch: Pair up exchangers
according to CP-constraints. Immediately below the pinch, pair up
streams such that: CPHOT  CPCOLD. “Tick off” streams in order to
minimize costs. Add heating utilities as needed (up to QCmin). Do
not use hot utilities below the pinch.
 Done!
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Class Exercise 3
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
130
60
30
80
40
100
120
CP
H
o
(kW/
C)
(kW)
100
1.0
180
2.0
160
4.0
162
1.8
Tmin = 10 oC.
Utilities:
Steam@150 oC, CW@25oC
Design a network of steam heaters,
water coolers and exchangers for
the process streams. Where
possible, use exchangers in
preference to utilities.
CP
H1
QHmin=48

H2
o

180oC
130oC

100 C
120oC 40

34
80oC

80oC
70oC
43oC
60 C
H
H
8
120 60oC
100
40 C
C
o
6
o
1.0

54
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
2.0
o
60 C
30oC
C1
4.0
C2
1.8
QCmin=6
Intro HEN Synthesis - 6
UNIT 3: The Problem Table
Example:
H
Stream
TS
(oF)
TT
(oF)
(kBtu/h)
(kBtu/h oF)
H1
H2
C1
C2
260
250
120
180
160
130
235
240
3000
1800
2300
2400
30
15
20
40
CP
Tmin = 10 oF.
Step 1: Temperature Intervals
(subtract Tmin from hot temperatures)
Temperature intervals:
250F  240F  235F  180F  150F  120F
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
UNIT 3: The Problem Table (Cont’d)
Step 2: Interval heat balances
For each interval, compute:
Hi = (Ti  Ti+1)(CPHot CPCold )
36
Interval
Ti
Ti  Ti+1
CPHot
CPCold
Hi
1
2
3
4
5
6
250
240
235
180
150
120
10
5
55
30
30
30
5
15
25
5
300
25
825
750
150
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
UNIT 3: The Problem Table (Cont’d)
Step 3: Form enthalpy
cascade.
QH
Assume
QH = 0
Eliminate infeasible
(negative) heat transfer
QH = 500
T1 = 250oF
H = 300
Q1
300
800
325
825
-500
0
250
750
100
600
o
T2 = 240 F
H = 25
Q2
o
T3 = 235 F
H = -825
Q3
o
T4 = 180 F
H = 750
Q4
This defines:
Cold pinch temp. = 180 oF
QHmin = 500 kBtu/h
QCmin = 600 kBtu/h
37
o
T5 = 150 F
H = -150
QC
T6 = 120oF
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Class Exercise 4 - Now try again!
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
130
60
30
80
40
100
120
CP
H
o
(kW/
C)
(kW)
100
1.0
180
2.0
160
4.0
162
1.8
Tmin = 10 oC.
Calculate the Problem Table.
Predict QHmin and QCmin.
Draw the Enthalpy Cascade.
Step 1: Temperature Intervals
(subtract Tmin from hot temperatures)
Temperature intervals:
38
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Class Exercise 4 (Cont’d)
Step 2: Interval heat balances
For each interval, compute:
Hi = (Ti  Ti+1)(CPHot CPCold )
Interval
Ti
Ti  Ti+1
CPHot
CPCold
Hi
1
2
3
4
5
6
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Class Exercise 4 (Cont’d)
Step 3: Form enthalpy
T
cascade.
QH
Assume
QH = 0
Eliminate infeasible
(negative) heat transfer
QH =
1=
H =
Q1
T2 =
H =
Q2
T3 =
H =
Q3
T4 =
H =
This defines:
Cold pinch temp. =
QHmin =
kW
QCmin =
kW
40
Q4
T5 =
oC
H =
QC
QC
T6 =
DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6
Introduction to HEN Synthesis - Summary
• Unit 1. Introduction: Capital vs. Energy
– What is an optimal HEN design
– Setting Energy Targets
• Unit 2. The Pinch and MER Design
– The Heat Recovery Pinch
– HEN Representation
– MER Design: (a) MER Target; (b) Hot- and cold-side designs
• Unit 3. The Problem Table
– for MER Targeting
Next week: Advanced HEN Synthesis
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DESIGN AND ANALYSIS II - (c) Daniel R. Lewin
Intro HEN Synthesis - 6