Pinch Technology and optimization of the use
of utilities – part two
Maurizio Fermeglia
maurizio.fermeglia@di3.units.it
www.mose.units.it
Introduction to HEN Synthesis – Summary of part 1
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, 13 April 2015 - slide 2
Introduction to HEN: Part two
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, 13 April 2015 - slide 3
Part Two: Objectives
This Unit on HEN synthesis serves to expand on what was
covered to more advanced topics.
Instructional Objectives - You should be able to:



Identify and eliminate “heat loops” in an MER design (lower the n. of
HEx)
Use stream splits to design for Umin and MER (minimize n. of HEx and
MER target)
Design a HEN for “Threshold Problems”
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 4
UNIT 4: Loops and Splits
The minimum number of units (Umin) in a network:
UMin = NStream + NUtil  1 (Hohman, 1971)
A HEN containing UHEX units (UHEX  Umin) has (UHEX  Umin)
independent “heat loops”.
C
H1
H
C1
H
C2
The HEN above has 2 “heat loops”
Normally, when heat loops are “broken”, heat flows across the pinch the number of heat exchangers is reduced, but the utility loads are
increased.
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 5
Class Exercise 5 (Linnhoff and Flower, 1978)
Example:
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
150
60
30
40
40
180
130
H
(kW)
280
440
360
260
Tmin = 10 oC.
CP
(kW/oC)
2.0
4.0
3.0
2.6
Step 1: Temperature Intervals
(subtract Tmin from hot temperatures)
Temperature intervals:
180oC  170oC 140oC 130oC 60oC 30oC
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 6
Class Exercise 5 (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
180
170
140
130
60
30
10
30
10
70
30
3.0
1.0
3.0
0.4
3.4
30
30
30
28
102
Trieste, 13 April 2015 - slide 7
Class Exercise 5 (Cont’d)
Step 3: Form enthalpy
cascade.
QH
Assume
QH = 0
Eliminate infeasible
(negative) heat transfer
QH = 60
o
T1 = 180 C
H = -30
Q1
-30
30
-60
0
-30
30
-2
58
100
160
o
T2 = 170 C
H = -30
Q2
o
T3 = 140 C
H = 30
Q3
o
T4 = 130 C
H = 28
This defines:
Cold pinch temp. = 140 oC
QHmin = 60 kW
QCmin = 160 kW
Progettazione di processo e di prodotto
Q4
T5 = 60oC
H = 102
QC
T6 = 30oC
Trieste, 13 April 2015 - slide 8
Class Exercise 5 (Cont’d)
MER Design above the pinch:
H1
CP
150o
180o
180o
H
160o
60
140o
UMin,MER = NStream + NUtil - 1
=2+1–1
=2
2.0
C1
3.0
60
MER Design below the pinch:
H1
H2
150o
CP
110o
150o
140o
140o
126.67o
110o
100o
80
40
130o
2.0
40o
4.0
C
160
60o
C1
3.0
C2
2.6
120
83.85
120
Progettazione di processo e di prodotto
80o
40o
o
30o
UMin,MER = 4 + 1 – 1
=4
MER design below
pinch has 6
exchangers!
i.e. There are two
loops below pinch.
140
Trieste, 13 April 2015 - slide 9
Class Exercise 5 (Cont’d)
Complete MER Design
H1
180o
150o
H2
180
o
H
160
60
CP
o
150o
140
60
o
110o
140o
110o
o
126.67
40
130o
80o
100
80
83.85o
120
C
40o
2.0
40o
4.0
160
60o
o
C1
3.0
C2
2.6
120
30o
140
= NStream + NUtil  1
= 4
+ 2 1
= 5
The MER network has 8 units. This implies 3 independent “heat load
loops”. We shall now identify and eliminate these
However, UMin
loops in order to design for UMin
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 10
Class Exercise 5 (Cont’d)
Identification and elimination of 1st loop:
CP
Tmin violation
H1
180o
150o
110
H2
180
o
H
H
60
160
o
150o
o
140
113.33
60
140
110o
140o
110o
o
126.67
40
40
130o
80o
120
120
2.0
40o
4.0
160
160
60o
o o
100
100
80
83.85o
C
40o
C1
C1
3.0
C2
2.6
120
30o
140
To reduce the number of units, the 80 kW exchanger is
merged with the 60 kW exchanger. This breaks the heat loop,
but also creates a Tmin violation in the network:
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 11
Class Exercise 5 (Cont’d)
Identification and elimination of 1st loop (Cont’d):
To restore Tmin, the loads of the exchangers must be adjusted
along a “heat path” by an unknown amount x. A “heat path” is a
path through the network that connects heaters with coolers.
H1
o
110
110o
180o
H2
180
o
H
160
6060
+x
o
150o
113.33
140
x
140
140o
o
40
40o
o
80
80o
110o
o
100
40
130o
C
C
o
40
40o
160+ x
160
o
60
60o
o
83.85oo
2.0
4.0
C1
C1
3.0
C2
2.6
120
o
30
30o
83.85
120
120 x
Progettazione di processo e di prodotto
CP
Tmin
violation
This
violates
Tmin
140
+x
140
Trieste, 13 April 2015 - slide 12
Class Exercise 5 (Cont’d)
Identification and elimination of 1st loop (Cont’d):
Performing a heat balance on H1 in the exchanger which exhibits
the Tmin violation:
140 - x = 2(180 - 113.33 - Tmin)  x = 26.66
H1
110oo
123.33
180o
H2
180
o
H
H
o
151.1
160o
60
+x
86.66
150o
140o
o
116.66
110o
113.33oo
100
113.33
140
x
113.33
CP
Tminviolates
violation
This
Tcorrected
min
40
130o
40oo
40
o
86.66
80o
160
186.66
+x
o
60
60o
o
o o
C
4.0
C1
3.0
C2
2.6
120120
o
30o
94.1
83.85
93.33
120
x
40oo
40
2.0
166.66
140 + x
This is called “energy relaxation”
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 13
Class Exercise 5 (Cont’d)
Identification and elimination of 2nd loop:
CP
H1
180o
H2
180
151.1o
o
H
H
86.66
o
40oo
40
123.33o
150o
113.33
140o
126.66o
116.66
o
113.33
100
40
130o
94.1oo
86.66o
C
186.66
60o
o
160
120
o
30
30
94.1
93.33
93.33
93.33
o
40
40o
2.0
4.0
C1
3.0
C2
2.6
2.6
166.66
166.66
166.66
Since there is no Tmin violation, no adjustment of the loads of
the exchangers is needed - we reduce the number of units by one
with no energy penalty.
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 14
Class Exercise 5 (Cont’d)
Identification and elimination of 3rd loop:
CP
H1
180o
H2
180o
151.1o
H
86.66
40oo
40
123.33o
113.33
126.66o
150o
o
86.66
C
40o
186.66
60oo
o
113.33
oo
o
160
o
30
30o
94.1
94.1
130
93.33
93.33
2.0
4.0
C1
3.0
C2
2.6
166.66
166.66
Shifting the load of the smallest exchanger (93.33 kW) around
the loop, the network is reduced to…
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 15
Class Exercise 5 (Cont’d)
Identification and elimination of 3rd loop:
CP
H1
180o
40o
170o
150o
H2
180
Tmin violation
151.1o
o
86.66o
20
130
o
C
186.66+x
60o
144.44o
H
86.66+x
40o
253  x
30o
2.0
4.0
C1
3.0
C2
2.6
260
We use the heat path to restore Tmin:
253.33 - x = 3(150 - Tmin- 60)  x = 13.33
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 16
Class Exercise 5 (Cont’d)
Therefore Umin Network is:
CP
H1
180o
H2
146.67o
o
180
20
90o
150o
40o
C
140
H
100
40o
170o
200
60o
o
240
30o
o
130
2.0
4.0
C1
3.0
C2
2.6
260
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 17
Loop Breaking - Summary
Step 1:
Perform MER Design for UHEX units. Try and ensure that design
meets UMin,MER separately above and below the pinch.
Step 2:
Compute the minimum number of units:
UMin = NStream + NUtil  1
This identifies UHEX  Umin independent “heat loops”, which can be
eliminated to reduce U.
Step 3:
For each loop, eliminate a unit.
If this causes a Tmin violation,
identify the “heat path” and perform “energy relaxation” by increasing
the duties of the cooler and heater on the heat path.
Loops improve energy recovery and heat load flexibility at the cost of
added units (>Umin)
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 18
Stream Split Designs
CP
Example.
500
o
300
o
3
H1
480o
180o
460o
1
C2
1
160o
QHmin = 0
Option 1.
C1
QCmin = 0
UMin = 2
CP
500
o
400
o
327
300
C
80
H1
480o
300
460o
H
80
Progettazione di processo e di prodotto
o
380o
o
3
180o
C1
1
C2
1
160o
220
Trieste, 13 April 2015 - slide 19
Stream Split Designs (Cont’d)
Option 2. Loops
CP
500
o
480
o
440
o
360
o
300
o
3
H1
480o
180o
420o
60
460o
340o
240
Option 3. Stream Splitting
C2
1
160o
CP = 1.5
500
o
CP
500o
480o
1
180
120
H1
C1
300
o
3
500o
180o
300
460o
C1
1
C2
1
160o
300
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 20
Loops vs. Stream Splits
Loops:

Improved energy recovery (normally)

Heat load flexibility (normally)

U > UMin (by definition)
Stream Splitting:

Maximum Energy recovery (always)

Branch flowrate flexibility (normally)

U = UMin (always)
Stream splitting is a powerful technique for better energy
recovery
BUT - Don’t split unless necessary
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 21
Stream Splitting Rules
1. Above the pinch (at the pinch):


Cold utilities cannot be used (for MER). So, if NH > NC, MUST
split COLD streams, since for feasibility NH  NC
Feasible matches must ensure CPH  CPC. If this is not possible
for every match, split HOT streams as needed. If Hot steams are
split, recheck 
2. Below the pinch (at the pinch):
 Hot utilities cannot be used (for MER). So, if NC > NH, MUST split
HOT streams, since for feasibility NC  NH
 Feasible matches must ensure CPC  CPH. If this is not possible for
every match, split COLD streams as needed. If Cold steams are
split, recheck 
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 22
Class Exercise 6
Design a hot-side HEN, given the stream data below:

200o
H1
150o
H2
H2
190o
CP
100
5

100o
4
T1
T2
200
o
x
90o
C1
10
500
10  x
Solution:
Since NH > NC, we must split C1. The split ratio is dictated by the rule:
CPH  CPC (necessary condition) and by a desire to minimize the
number of units (“tick off “streams)
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 23
Class Exercise 6 (Cont’d)
CP
200o
100
o
5
H1
150o
100o
H2
190o
4
T1
x
90o
C1
10
500
T2
200
10  x
x is determined by the following energy balances:
x(T1  90) = 500
(10  x)(T2  90) = 200
200 T1  Tmin = 10
150 T2  Tmin = 10
Best to make T1 = T2 . Here, this is not possible. Why?
We shall make T2 = 140 (why?)
subject to:
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 24
Class Exercise 6 (Cont’d)
A possible solution is therefore:
(10  x) (140  90) = 200  x = 6
T1 = 90 + 500/x = 173.33 (satisfies constraint)
Complete solution is:
CP
200o
H1
150o
H2

190o
173.3o

100

H
o
5
100o
4
6
5
90o
C1
300
H
300
500
140
130o
10
4
5
200
This is an MER design which also satisfies UMin (UMin = 3).
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 25
UNIT 5: Threshold Problems
Networks with excess heat supply or heat demand may have MER
targets with only one utility (i.e., either QHmin = 0 or QCmin = 0). Such
designs are not separated at the pinch, and are called “Threshold
Problems”
Example - Consider the problem
CP
300
o
200
o
1.5
H1
300o
H2
200o
250o
2.0
30o
C1
Progettazione di processo e di prodotto
1.2
Trieste, 13 April 2015 - slide 28
Threshold Problems (Cont’d)
Assuming a value of
oC:the Problem
Tmin= 105
10 oC,
Table gives the following
result.
QHH
Assume
QHH = 0
Eliminate infeasible
(negative) heat transfer
QH = 6
290ooC
T11 = 200
H = -6
175
Q11
175
-6
0
235
109
115
238
124
130
46
52
oo
T22 = 195
240 C
60
H = 115
Q22
oo
T33 = 145
200 C
3
H = 15
Q33
o o
190CC
T44 = 95
-192
H = -78
QCC
oo
T55 = 30 C
= 6 kW
QHMin Q
= HMin
0 kW
= 52 kW
QCMin Q
= CMin
46 kW
Progettazione di processo e di prodotto
Cold pinch temperature
195 oC
Trieste, 13 April 2015 - slide 29
Threshold Problems (Cont’d)
Threshold problems do not have a pinch, and have non-zero utility
duties only at one end.
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 30
Threshold Problems (Cont’d)
Steam
T
Utility
Heat Loads
o
TT
= 10
=o20o
T = 14o
Cooling Water
Steam
CW
CW
CW
H
Tmin
14o
However, increasing driving forces beyond the Threshold Value leads to
additional utility requirements.
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 31
Threshold Design Guidelines
1. Establish the threshold Tmin
2. Note the common practice values for Tmin:
Application
Refrigeration
Process
Boiler
Tmin, Experience
1-2 oC
10 oC
20-30 oC
3. Compare the threshold Tmin to Tmin,Experience
Classify as one of the following:
Utilities
Utilities
Tmin
Tmin
Tmin,Experience
Tmin,Experience
Pinched - treat as normal pinched Threshold - must satisfy target
problem
temperatures at the “no utility end”
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 32
Class Exercise 7
The graph shows the effect of Tmin on the required levels of QHmin
and QCmin for a process consisting of 3 hot and 4 cold streams.
Q (104 Btu/h)
Hot Utility
QHmin
217.5
Cold Utility
QCmin
50
Progettazione di processo e di prodotto
o
Tmin, oF
Trieste, 13 April 2015 - slide 33
Introduction to HEN software
Ref. Turton et al.
Analysis, Synthesis and Design of Chemical Processes
The MUMNE algorithm
The Minimum Utility steps:






Choose a minimum approach temperature (parametric optimization)
Construct a temperature interval diagram
Construct a cascade diagram and determine the minimum utility
requirements and the pinch temperatures
Calculate the minimum number of heat exchangers above pinch
Calculate the minimum number of heat exchangers below pinch
Construct the heat exchanger network
The object is to obtain an heat exchanger network


…That exchange the minimum energy between the streams and the
utilities
…That uses the minimum number of equipment
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 38
Algorithm: initial condition and step 1
Minimum Temperature Approach = Smallest T of two streams leaving
or entering an heat exchanger = 10°C
Hot Stream Data
Mass Flow Cp
Temp In
kg/s
kJ/kg/°C
°C
10.00
2.500
3.000
.8000
.8000
1.000
300.0
150.0
200.0
Temp Out
°C
Stream Enthalpy
kW
Film Heat Transf. Coef
W/m2/°C
150.0
50.00
50.00
1200.
200.0
450.0
400.0
270.0
530.0
Cumulative Hot Stream Energy Available =
Cold Stream Data
Mass Flow Cp
Temp In
kg/s
kJ/kg/°C
°C
6.250
10.00
4.000
.8000
.8000
1.000
190.0
90.00
40.00
Temp Out
°C
290.0
190.0
190.0
1850.0
Stream Enthalpy
kW
Film Heat Transf. Coef
W/m2/°C
-500.0
-800.0
-600.0
100.0
250.0
80.00
Cumulative Cold Stream Energy Available = -1900.0
Progettazione di processo e di prodotto
kW
kW
Trieste, 13 April 2015 - slide 39
Algorithm: construct Temperature interval
diagram (step 2)
Process streams represented by vertical lines
Axes are shifted by the minimum T approach
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 40
Algorithm: construct a cascade diagram (1)
Shows the net amount of energy in each interval diagram
Cascade  if there is an excess energy in a T interval we may
“cascade” it down
Energy cannot be transferred up (II law)
Line is the point at which no more energy can cascade down
We need to resort to utilities
NOTE: not all problems have a pinch condition: the algorithm is
still valid
Pinch temperature
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 41
Algorithm: construct a cascade diagram (2)
Additional heat is transferred to the C interval (yellow line)
Energy is cascaded down through the pinch and rejected to the cold
utility
If heat is transferred across the pinch, the net result will be that more
heat will have to be added from the hot utility and rejected to the
cold utility
To minimize the hot and cold utility requirements, energy
should NOT be transferred across the pinch
Pinch temperature
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 42
Algorithm: minimum n. of exchangers
Above the pinch
Draw boxes representing energy in the hot and cold process
streams and utilities
Transfer energy is indicated by lines (with the amount)
For each line an heat exchanger is required
The problem is split into two sub problems
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 43
Algorithm: minimum n. of exchangers
Below the pinch
Draw boxes representing energy in the hot and cold process
streams and utilities
Transfer energy is indicated by lines (with the amount )
For each line an heat exchanger is required
The problem is not split into sub problems
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 44
Algorithm: minimum n. of exchangers
General relationship



For any sub problem
With or without a pinch
Above or below the pinch
Min. No. of exchangers = No. of hot streams + No. of cold streams +
No. of utilities – 1
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 45
Algorithm: Design the network above the pinch
Start from the design at the
pinch
To make sure that T min is
not violated, match streams
such that
 c p, IN  m
 c p,OUT
m
 c p,hot  m
 c p,cold
m
Note that we consider ONLY
streams present at the pinch
Each exchanger is
represented by two circles
connected with a line, each
circle represent a side
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 46
Algorithm: Design the network above the pinch
Move away from the pinch
Look at the remaining
streams
Criterion used at pinch not
necessarily holds away from
the pinch
The following constraints are
not violated:



The minimum approach T is used
throughout the design
The number of exchangers must
be that calculated in step 3
Heat is added form the coolest
possible source
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 47
Algorithm: Design the network below the pinch
Similar to previous one: start from
the design at the pinch
To make sure that T min is not
violated, match streams such that
 c p,hot  m
 c p,cold
m
 c p, IN  m
 c p,OUT
m
 c p,hot  m
 c p,cold
m
Note that we consider ONLY
streams present at the pinch
Each exchanger is represented by
two circles connected with a line,
each circle represent a side
What happens if the T min is
violated (see figure)
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 48
Algorithm: Design the network below the pinch
Split stream into
substreams to meet
the T min criterion
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 49
Algorithm: Design the network below the pinch
Move away from the pinch
Look at the remaining streams
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 50
Final result
The final network of heat
exchangers is the
following



It has the minimum n. of
exchangers (8)
Minimum utility
requirement (Qh = 100 kW
and Qc = 50 kW)
Using a minimum approach
T = 10°
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 51
Heat exchangers design: area and costs
Up to now, emphasis on the topology of the network
… to complete the design, it is necessary to


Estimate the heat transfer area (A= Q / (U Tln F)
And the cost estimate
If heat transfer coefficients are known (including fouling)…

Transfer coefficients form literature (Seider – Tate, Donahue, …)
… exchanger area can be calculated (for streams exchangers)
Exchanger DT ln
U
Q
F factor
Area.
°C
W/m2/°C
kW
m2
1
24.1
129.8
100
0.8
40.0
2
20.0
69.5
300
0.8
270.3
3
47.5
153.8
700
0.8
119.7
4
24.1
80.0
500
0.8
324.6
6
10.0
61.7
100
0.8
202.5
7
17.0
69.5
100
0.8
195.8
TOTAL
1063.0
Exchanger 5 requires an hot utility (steam): T = 76.8 °C, U= 76.9 W/m2/°C, Area = 16.9 m2
Exchanger 8 requires cooling water: T = 23.2 °C, U= 346 W/m2/°C, Area = 7.8 m2
TOTAL Area: 1087.7 m2
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 52
Effect of the minimum approach temperature
Calculations must be repeated for different approach T (step 1)
Problem: step 5 (matching streams and exchanging energy) cannot
be programmed easily
An approximate approach is necessary for investigating the effect
of the approach temperature on the total cost
Based on the Composite temperature enthalpy diagram

Constructed by plotting enthalpy of all streams as a function of T
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 53
Construction of Composite T-H diagram
Temperature
Interval
D
T °C
Enthalpy of HOT streams in
Temperature interval (kW)
50 0
Cumulative Enthalpy
of HOT streams (kW)
0
C
100 (2+3)(100-50)
= 250
250
B
150 (2+3)(150-100)
= 250
500
A
200 (8+3)(200-150)
= 50
1050
300 (8)(300-200)
= 800
1850
Temperature
Interval
T °C
Enthalpy of COLD streams in
Temperature interval (kW)
D
40 0
C
90 (4)(90-40)
Cumulative Enthalpy
of COLD streams (kW)
0
= 200
200
B
140 (8+4)(140-90)
= 600
800
A
190 (8+4)(190-140)
= 600
1400
290 (5)(290-190)
= 500
Progettazione di processo e di prodotto
1900
Trieste, 13 April 2015 - slide 54
Construction of Composite T-H diagram
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 55
Using the T-H diagram to estimate heat
exchanger area
The working equation is (A= Q / (U Tln F)
Consider a portion of the diagram (figure)
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 57
Results of the heat transfer area calculations
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 58
Finalize the design
Consider the F factor


By calculating the number of shells in a 1-2 geometry exchanger
The effect of the ‘economy of scale’: the cost of two 1-2 S&T in series is
greater than the equivalent 1-2 S&T with the same total area
Calculate the cost of equipments



Knowing the number of shells for each exchanger
Using cost correlations
Applying an economic criteria (such as Equivalent Annual Operating Cost
– EAOC)
Approximations





All heat exchangers have the same area  over estimation of the costs
No effect of material of construction (corrections available)
No effect of operating pressure (corrections available)
No multiple utilities (alternative methods)
No streams with phase change (correction available)
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 59
Calculation of the costs for the network
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 60
Typical relationship for heat transfer area,
utilities and EAOC for a HEN
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 61
Mass – exchange networks
Exchangers to use energy more
efficiently
Temp Interval Diagram
Cascade diagram – pinch
Min n. of eq. above and below
Composite T exchange diag.
Final T exchange network
Utility: cold and hot source of
energy
Hot and cold
Temperature
Progettazione di processo e di prodotto
Separators to use mass more
efficiently
Composition Interval Diagram
Cascade diagram – pinch
Min n. of eq. above and below
Composite mass exchange diag.
Final mass exchange network
Utility: separation - addition of
solute (from source or to sink)
Rich and lean
Concentration
Trieste, 13 April 2015 - slide 62
Esercizio
Le CORRENTI
• Correnti calde:
NOME
T in
(°C)
T out
(°C)
MCp
(kW/°C)
H1
400
150
8
H2
200
50
5
H3
250
100
6
NOME
T in
(°C)
T out
(°C)
MCp
(kW/°C)
C4
190
390
10
C5
40
140
8
C6
90
240
5
• Correnti fredde:
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 64
Gli INTERVALLI di TEMPERATURE
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 65
La CURVA COMPOSITA
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 66
Il CALORE CUMULATO
Ti
Ti-Ti+1
COLD PINCH
TEMPERATURE
390
150
MCpHOT MCpCOLD
-2
240
50
190
190
+
T-1
min
50
14
140
50
90
50
Qi
Q
Q
cumulato
cumulato
(QH=350
(QH=0)
)
HOT PINCH
TEMPERATURE
-300
-300
-50
50
-350
0
700
350
700
-3
-100
250
600
-2
-150
100
450
200
40
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 67
RISOLUZIONE MANUALE
Sopra il Pinch
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 68
RISOLUZIONE MANUALE
Sotto il Pinch
Progettazione di processo e di prodotto
Trieste, 13 April 2015 - slide 69