Pinch Technology and optimization of the use
of utilities – part one
Maurizio Fermeglia
[email protected]
www.mose.units.it
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

Class Exercises 3 and 4


Unit 3. The Problem Table
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 2
Heat and Power Integration
Unit 6. Data Extraction

Class Exercise 8
Unit 7. Heat Integration in Design




Grand Composite Curve
Heat-integrated Distillation
Heat Engines
Heat Pumps
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 3
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 4
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).
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 6
Impact of process integration
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 7
Example
Tout
Tout
Tout
H
H
H
Tin
Tin
Tin
Tin
Tin
Steam
Cooling
Water
out
C T
C
Tout
C
Tout
Network for minimal
equipment cost ?
Tout
Tin
Tout
Tout
Cooling
Water
out
T
Tin
Network for minimal
energy cost ?
Tin
Tout
Tin
Tout
Tin
Progettazione di processo e di prodotto
Steam
Tin
Tin
Trieste lunedì 5 marzo 2012 - slide 8
Numerical Example
150o
100
150o
150o
CP = 1.0
300o
300o
500
CP = 1.0
500
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 ]
500
150o
CP = 1.0
Design B:
(AREA) = 13.3
o
CP = 1.0 300
CP = 1.0
CP = 1.0
150o
200o
100
300o
200o
100
300o
100
500
CP = 1.0
Progettazione di processo e di prodotto
150o
500
CP = 1.0
200o
500
CP = 1.0
Trieste lunedì 5 marzo 2012 - slide 9
Which option requires more capital?
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 10
Energy efficient process can also be more capital
efficient process (saves energy AND capital)
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 11
Energy efficient design reduces investment in the
utility infrastructure
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 12
Some Definitions
T
TT
TS
TT
H
CP
T
TS
H
Progettazione di processo e di prodotto
H
= Stream supply temperature (oC)
= Stream target temperature (oC)
= Stream enthalpy (MW)
  Cp (MW/ oC)
= m
= Heat capacity flowrate (MW/ oC)
= Stream flowrate specific heat capacity
Trieste lunedì 5 marzo 2012 - slide 13
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 10o
A
70o
100
60o
o
40o 20o
B
Clearly, exchanger A violates the Tmin constraint.
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 14
Definitions (Cont’d)
Exchanger Duty.
Data:
T1 = 70o
OK
Hot stream CP = 0.3 MW/ oC
Cold stream CP = 0.4 MW/ oC
Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC 
100
60o
o
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 15
Pinch technology basics
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 16
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 17
How can we identify appropriate
process design changes?
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 19
Sustainable Development by Process Integration
Reduce
Environmental
Impact
Optimise
Production
Plan
Improve
Utility System
Performance
Minimise
Operating
Cost
All Aspects of
Debottlenecking
Capital Cost
Avoidance
Energy, Water, Hydrogen
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 23
Introduction to Pinch technology
Whenever the design is considered limits exits that
constraints the design

Mechanical constraints
 Length and diameter of towers
 Diameters of heat exchangers

Thermodynamic constrains
 First principles and second principles
 Close approach in heat exchanger  large surface area
 Reflux ratio close to the minimum  number of stages grows
When driving force becomes small  area becomes large


We say that the design has a PINCH
Applies to heat and mass
In a network (mass or heat) there is a point in which the
driving force is minimum  PINCH POINT
A succesful design involves defining where the pinch is
Using the information at the PINCH POINT is named
PINCH TECHNOLOGY
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 24
Introduction to Pinch technology
Pinch technology applications



Both heat and mass transfer
New processes
Existing processes for retrofitting
Pinch technology = optimization
For new and existing processes an algorithm is used





Design of heat and mass exchanger network
… that consumes the minimum amount of utilities HEN (MEN)
… that requires the minimum number of equipments (exchangers)
MUMNE
The solution may not be optimal in the economic sense
… it is a starting point close enough to the economic minimum.
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 25
Pinch technology and heat integration
Growing importance of heat integration is driving force
Formalization of heat integration theory  pinch technology





Linhoff and Flower
Hohman
Umeda et al.
Douglas
Turton et al (text book)
Different operative configurations of the same process may
result in




Same conditions (composition, temperature, pressure, flow rate)
Different Fixed capital investment
Different cost of utilities
Different Net present value
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 26
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
H
(kW)
100
180
160
162
CP
(kW/oC)
1.0
2.0
4.0
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 process
o
30 streams. Where possible, use exchangers in
preference to utilities.
.
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 28
Setting Energy Targets
Cond
60
180o
o
Summary of proposed design:
100
80o
R2
C1
60
H
130
Reb
o
Steam
CW
Units
60 kW
18 kW
4
R1
100o
Are 60 kW of Steam Necessary?
162
C 18
120
o
30o
40o
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 29
The Temperature-Enthalpy Diagram
T
T
130oC
200oC
40oC
100oC
H=180
One hot stream
Progettazione di processo e di prodotto
H=300
H
H
H=100
Two hot streams
Trieste lunedì 5 marzo 2012 - slide 30
The Temperature-Enthalpy Diagram
C
CW
Steam
Steam
T
Tmin = 20
10
H
120oC
110oC
100oC
CW
30
QCmin = 50
H
50
QHmin = 70
Correlation between Tmin, QHmin and QCmin
More in, More out! QHmin + x  QCmin + x
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 31
The Composite Curve
Hot Composite Curve
H interval
o
180 C
H=100
Cond
o
180
60o
80o
R2
C1
50
CP=1.0
130oC
150
o
80 C
CP=2.0
o
40 C
80
H=160
Reb
100o
40o
Progettazione di processo e di prodotto
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
80
Trieste lunedì 5 marzo 2012 - slide 32
The Composite Curve (Cont’d)
Cold Composite Curve
H interval
o
120 C
H=100
Cond
100oC
180o
o
60
36
CP=1.8
232
80o
60oC
R2
C1
CP=4.0
30oC
H=160
Reb
100o
130o
H=180
o
40
Progettazione di processo e di prodotto
120oC
R1
o
CP=1.8
100 C
120o
H=162
CP=5.8
60oC
o
30
30oC
CP=1.8
54
36
232
54
Trieste lunedì 5 marzo 2012 - slide 33
The Composite Curve (Cont’d)
T
QHmin = 54
48
130oC
100oC
o
80 C
o
T
=
20
min
Tmin = 10oC
C
60oC
QQ
==6 12
Cmin
Cmin
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.
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 34
UNIT 2: The Pinch
QHmin
T
Heat
Source
QCmin
+x
QHmin
+x
Tmin
Heat
Sink
“PINCH”
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.
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 35
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 36
HEN Representation
100
Cond
60oC
180oC
80oC
H1
R2
C1
H2
130oC
100oC
100 C
Reb
o
130 C
120oC
H 49
111
R1
o
120 C
C 7
Progettazione di processo e di prodotto
60oC
H
49
111
100
30oC
40oC
C
7
30oC
o
62
40oC
80oC
180oC
C1
C2
62
Where is the pinch ?
Trieste lunedì 5 marzo 2012 - slide 37
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.
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 38
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
C2
H2
500oC
320oC
Progettazione di processo e di prodotto
170
50
500o
320o
CP = 1.0
CP = 1.5
CW
290o
200o
CP = 1.8 CP = 2.0
320 C
480oC
100
S
116
o
H1
480o
210
o
140o
H2
CP
200oC
1.8
290oC
2.0
240oC
140oC
C1
1.0
C2
1.5
Trieste lunedì 5 marzo 2012 - slide 39
Class Exercise 2 - Solution
H1
320o
C1 240
C2
480o
210
o
140o
H2
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
200oC 1.8
480oC
290oC
460oC
240oC
H
10
210
320oC
Progettazione di processo e di prodotto
C
116
140oC
170
2.0
C1
1.0
C2
1.5
100
Trieste lunedì 5 marzo 2012 - slide 40
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
Progettazione di processo e di prodotto
200oC
C1
1.0
C2
1.5
100
110
Trieste lunedì 5 marzo 2012 - slide 41
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 42
Design for MER (Cont’d)
Step 2: Divide the problem at the pinch
CP
170oC
90oC
90oC
60oC
3.0
150oC
90oC
90oC
30oC
1.5
80oC
80oC
20oC
H1
H2
135oC
140oC
Progettazione di processo e di prodotto
80oC
C2
C1
2.0
4.0
Trieste lunedì 5 marzo 2012 - slide 43
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 44
Design for MER (Cont’d)
Step 3 (Cont’d): Complete hot-end design, by ticking-off
streams.
CP
H1
QHmin = 20 kW 
H2
170
150o

135o
140o

o
H
20
90o
3.0

90o
80o
90

80o
1.5
C1
2.0
C2
4.0
240
Add heating utilities as needed (MER target)
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 45
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 46
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)
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 47
Design for MER (Cont’d)
Completed Design:
CP
H1
H2
170o
90o
150o
135o
H
140o
60o
90o
20
90
30o
70o
80o
125o
C
60
20o
35o
90
30
3.0
80o
1.5
C1
2.0
C2
4.0
240
Note that this design meets the MER targets:
QHmin = 20 kW and QCmin = 60 kW
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 48
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!
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 49
Class Exercise 3
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
130
60
30
80
40
100
120
H
(kW)
100
180
160
162
CP
(kW/oC)
1.0
2.0
4.0
1.8
Tmin = 10 oC.
- Pinch Temperature 70° - 60°
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

180oC
130oC

100oC

120oC 40
8
Progettazione di processo e di prodotto
80oC
80oC
70oC
43oC
o
60 C
H
120 60oC
H
100

1.0
 40 C
o
C
6
2.0
QCmin=6 
o
60 C
30oC
C1
4.0
C2
1.8
54
Trieste lunedì 5 marzo 2012 - slide 50
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 51
UNIT 3: The Problem Table (Cont’d)
Step 2: Interval heat balances
For each interval, compute:
Hi = (Ti  Ti+1)(CPHot CPCold )
Progettazione di processo e di prodotto
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
Trieste lunedì 5 marzo 2012 - slide 52
UNIT 3: The Problem Table (Cont’d)
Step 3: Form enthalpy
cascade.
QH
Assume
QH = 0
Eliminate infeasible
(negative) heat transfer
QH = 500
o
T1 = 250 F
H = 300
Q1
300
800
325
825
-500
0
250
750
100
600
T2 = 240oF
H = 25
Q2
o
T3 = 235 F
H = -825
Q3
T4 = 180oF
H = 750
This defines:
Cold pinch temp. = 180 oF
QHmin = 500 kBtu/h
QCmin = 600 kBtu/h
Q4
T5 = 150oF
H = -150
QC
o
T6 = 120 F
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 53
Class Exercise 4 - Now try again!
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
130
60
30
80
40
100
120
H
(kW)
100
180
160
162
CP
(kW/oC)
1.0
2.0
4.0
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:
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 54
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 55
Class Exercise 4 (Cont’d)
Step 3: Form enthalpy
cascade.
QH
Assume
QH = 0
Eliminate infeasible
(negative) heat transfer
QH =
T1 =
H =
Q1
T2 =
H =
Q2
T3 =
H =
Q3
T4 =
H =
This defines:
Cold pinch temp. =
QHmin =
kW
QCmin =
kW
Progettazione di processo e di prodotto
Q4
T5 =
oC
H =
QC
QC
T6 =
Trieste lunedì 5 marzo 2012 - slide 56
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
Progettazione di processo e di prodotto
Trieste lunedì 5 marzo 2012 - slide 57