Transcript HEN Synthesis (Part 1) - ????????

Heat Exchanger Network Synthesis
Part I: Introduction
Ref: Seider, Seader and Lewin (2004), Chapter 10
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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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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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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
4
Steam
Tin
Tin
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Numerical Example
150o
150o
100
150o
CP = 1.0
300o
300o
0
50
CP = 1.0
0
50
CP = 1.0
Cooling
Water (90-110oF)
100
100
o
CP = 1.0 300
CP = 1.0
Steam (400oF)
100
200o
100
200o
100
200o
Design A:
(AREA) = 20.4
[ A = Q/UTlm ]
150o
0
CP = 1.0
CP = 1.0
CP = 1.0
200o
100
300o
200o
100
300o
100
500
CP = 1.0
5
150o
50
o
CP = 1.0 300
Design B:
(AREA) = 13.3
150o
500
CP = 1.0
200o
500
CP = 1.0
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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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Class Exercise 1
Tmin = 10 oC
H=100
Cond
60
180o
o
80o
R2
C1
H=160
Reb
130
100o
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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Setting Energy Targets
Cond
180o
o
60
Summary of proposed design:
100
o
80
R2
C1
60
H
Reb
o
130
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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The Temperature-Enthalpy Diagram
T
T
130oC
200oC
40oC
100oC
H=180
One hot stream
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H=300
H
H
H=100
Two hot streams
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The Composite Curve
Hot Composite Curve
H interval
o
180 C
H=100
Cond
180
60o
80o
o
R2
C1
CP=1.0
50
130oC
150
o
80 C
CP=2.0
o
40 C
80
H=160
Reb
100o
40o
10
180oC
o
130o
H=180
R1
120o 130 C
H=162
CP=1.0
50
CP=3.0
150
o
80 C
30o
o
40 C
CP=2.0
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The Composite Curve (Cont’d)
Cold Composite Curve
H interval
o
120 C
H=100
Cond
60
100oC
180o
o
CP=1.8
232
80o
60oC
R2
C1
CP=4.0
30oC
H=160
Reb
100o
130o
H=180
40
11
o
36
120oC
R1
o
CP=1.8
100 C
120o
H=162
CP=5.8
60oC
30
o
30oC
CP=1.8
54
36
232
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The Composite Curve (Cont’d)
T
QHmin = 54
48
130oC
100oC
o
80 C
o
T
=
20
min
Tmin = 10oC
C
60oC
QQ
Cmin
Cmin==6 12
H
Result:
QCmin and QHmin for
desired Tmin
MER Target
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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The Pinch
T
T
QHmin
Heat
Source
QCmin
Cmin
Q
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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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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Class Exercise 2
H1
320o
C1 240
C2
480o
210
o
140o
H2
100
170
S
50
500o
320o
CP = 1.0
CP = 1.5
116
CW
290o
200o
CP = 1.8 CP = 2.0
•
•
•
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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.
How many energy can be recovered?
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Class Exercise 2 - Solution
H1
320o
C1 240
o
140
o
C2
H2
480o
210
100
S
50
170
500o
320o
•
CP = 1.0
•
CP = 1.5
116
pinch temperatures;
480 oC /460 oC
MER targets:
QHmin= 40, QCmin= 106
CW
290o
200o
CP = 1.8 CP = 2.0
CP
320oC
H1
H2
500oC
H
40
480oC
290oC
460oC
240oC
H
10
210
320oC
16
o
C 200 C 1.8
116
140oC
170
2.0
C1
1.0
C2
1.5
100
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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
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200oC
C1
1.0
C2
1.5
100
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Design for Maximum Energy Recovery
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 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
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CP
80oC
C2
C1
2.0
4.0
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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 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 for MER (Cont’d)
Step 4: 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
(This ensures that TH TC  Tmin)
CP
H1
3.0
H2
1.5
C1
Tmin
2.0
Violates
Meets
TT
constraint
constraint
minmin
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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 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 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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Class Exercise 3
Stream
TS
(oC)
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
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

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80oC

80oC
70oC
43oC
60 C
H
H
8
120 60oC
100

54
40 C
C
o
6
o
1.0
2.0
o
60 C
30oC
C1
4.0
C2
1.8
QCmin=6
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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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The Problem Table (Cont’d)
Step 2: Interval heat balances
For each interval, compute:
Hi = (Ti  Ti+1)(CPHot CPCold )
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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
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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
29
o
T5 = 150 F
H = -150
QC
T6 = 120oF
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Introduction to HEN Synthesis - Summary
1. Introduction: Capital vs. Energy
What is an optimal HEN design
Setting Energy Targets
2. The Pinch and MER Design
– The Heat Recovery Pinch
– HEN Representation
– MER Design: (a) MER Target; (b) Hot- and cold-side designs
3. The Problem Table
– for MER Targeting
Next Lecture: Advanced HEN Synthesis
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