Transcript HEN Synthesis (Part 2)

Heat Exchanger Network
Synthesis, Part II
Ref: Seider, Seader and Lewin (2004), Chapter 10
1
Advanced HEN Synthesis -7
Part Two: Objectives
• This Unit on HEN synthesis serves to expand on
what was covered last week to more advanced
topics.
• Instructional Objectives - You should be able to:
– Identify and eliminate “heat loops” in an MER
design
– Use stream splits to design for Umin and MER
– Design a HEN for “Threshold Problems”
2
Advanced HEN Synthesis -7
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.
3
Advanced HEN Synthesis -7
Class Exercise 4 (Linnhoff and Flower, 1978)
Example:
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
150
60
30
40
40
180
130
CP
H
o
(kW) (kW/ C)
280
2.0
440
4.0
360
3.0
260
2.6
Tmin = 10 oC.
Step 1: Temperature Intervals
(subtract Tmin from hot temperatures)
Temperature intervals:
180oC  170oC 140oC 130oC 60oC 30oC
4
Advanced HEN Synthesis -7
Class Exercise 4 (Cont’d)
Step 2: Interval heat balances
For each interval, compute:
Hi = (Ti  Ti+1)(CPHot CPCold )
5
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
Advanced HEN Synthesis -7
Class Exercise 4 (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
6
Q4
T5 = 60oC
H = 102
QC
T6 = 30oC
Advanced HEN Synthesis -7
Class Exercise 4 (Cont’d)
MER Design above the pinch:
H1
150o
180o
180o
H
160o
60
140o
CP
UMin,MER = NStream + NUtil - 1
=2+1–1
=2
2.0
C1
3.0
60
MER Design below the pinch:
H1
H2
150o
110o
150o
140o
140o
126.67o
110o
100o
80
40
130o
40o
2.0
40o
4.0
C
160
60o
C1
3.0
C2
2.6
120
83.85
120
7
80o
CP
o
30o
UMin,MER = 4 + 1 – 1
=4
MER design below
pinch has 6
exchangers!
i.e. There are two
loops below pinch.
140
Advanced HEN Synthesis -7
Class Exercise 4 (Cont’d)
Complete MER Design
H1
180o
150o
H2
180
o
H
60
160
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
However, UMin = 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
loops in order to design for UMin
8
Advanced HEN Synthesis -7
Class Exercise 4 (Cont’d)
Identification and elimination of 1st loop:
CP
Tmin violation
H1
180o
150o
110
H2
180
o
H
60
160
o
150o
o
140
113.33
60
140
110o
140o
110o
o
126.67
40
40
130o
80o
120
2.0
40o
4.0
160
160
60o
o o
100
100
80
83.85o
C
40o
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 volation in the
network:
9
Advanced HEN Synthesis -7
Class Exercise 4 (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
o
40
40o
160+ x
160
o
60
60o
o
83.85oo
2.0
4.0
C1
3.0
C2
2.6
120
o
30
30o
83.85
120
120 x
10
CP
Tminviolates
violation
This
Tmin
140
140+ x
Advanced HEN Synthesis -7
Class Exercise 4 (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
30o
94.1
83.85
93.33
120
x
40oo
40
2.0
166.66
140 + x
This is called “energy relaxation”
11
Advanced HEN Synthesis -7
Class Exercise 4 (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
113.33
150o
113.33
140o
126.66o
116.66
o
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.
12
Advanced HEN Synthesis -7
Class Exercise 4 (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…
13
Advanced HEN Synthesis -7
Class Exercise 4 (Cont’d)
Identification and elimination of 3rd loop:
CP
H1
180o
40o
170o
150o
H2
180
Tmin violation
151.1o
o
40o
86.66o
C
186.66+x
60o
144.44o
H
86.66+x
20
130
o
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
14
Advanced HEN Synthesis -7
Class Exercise 4 (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
15
Advanced HEN Synthesis -7
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)
16
Advanced HEN Synthesis -7
Stream Split Designs
Example.
CP
500
o
300
o
3
H1
480o
180o
460o
Option 1.
1
C2
1
160o
QHmin = 0
QCmin = 0
UMin = 2
CP
500
o
400
o
327
o
300
C
80
H1
480o
300
460o
H
80
17
C1
380o
o
3
180o
C1
1
C2
1
160o
220
Advanced HEN Synthesis -7
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
C1
1
C2
1
160o
180
120
Option 3. Stream Splitting
CP = 1.5
500o
500
H1
480o
CP
o
300
o
3
500o
180o
300
460o
C1
1
C2
1
160o
300
18
Advanced HEN Synthesis -7
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
19
Advanced HEN Synthesis -7
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 
20
Advanced HEN Synthesis -7
Class Exercise 5
Design a hot-side HEN, given the stream data below:
200
H1
150o
H2
H2
190o
CP

o
100
o
5

100o
4
T1
x
90o
C1
10
500
T2
200
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)
21
Advanced HEN Synthesis -7
Class Exercise 5 (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
subject to:
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?)
22
Advanced HEN Synthesis -7
Class Exercise 5 (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
130o
140
10
4
5
200
This is an MER design which also satisfies UMin (UMin = 3).
23
Advanced HEN Synthesis -7
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
300o
200
o
1.5
H1
300o
H2
200o
250o
2.0
30o
C1
24
1.2
Advanced HEN Synthesis -7
Threshold Problems (Cont’d)
Assuming a value of
oC:
Tmin= 105
10 oC,
the
Problem 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
oo
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
25
Cold pinch temperature
195 oC
Advanced HEN Synthesis -7
Threshold Problems (Cont’d)
Threshold problems do not have a pinch, and have non-zero
utility duties only at one end.
26
Advanced HEN Synthesis -7
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.
27
Advanced HEN Synthesis -7
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
Pinched - treat as normal
pinched problem
28
Tmin,Experience
Threshold - must satisfy target
temperatures at the “no utility end”
Advanced HEN Synthesis -7
Class Exercise 6
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
29
o
Tmin, oF
Advanced HEN Synthesis -7
Class Exercise 6 (Cont’d)
Design a network for Umin and MER for the process.
Hint: Identify two essential matches by satisfying target
CP
temperatures oat the “no utility end”
o
590
400
2.376
H1
471o
200o
533o
150o
1.577
H2
1.32
H3
400o
200o
430o
1.6
C2
1.6
C3
4.128
C4
2.624
100o
400o
300o
150o
280o
QHmin = 217.5
30
C1
QCmin = 0
Advanced HEN Synthesis -7
Class Exercise 6 - Solution
CP
590
o
573.7

H1
471o
o
2.376
200o
416.3
416.3o o

H2
400
1.577
533o

1.32
200o
H
430o  195.1 38.7
H
400o
150o

H3
400o
o
22.4
416o
86.3
412.8
280o
QHmin = 217.5
341.1
1.6
C2
1.6
C3
4.128
C4
2.624
100o
505.6 300o

C1
150o
QCmin = 0
Note: UMin = NStreams + NUtilities  1 = 7
31
Advanced HEN Synthesis -7
Advanced HEN Synthesis - Summary
• Loops and Splits
– Minimum Number of Units by Loop Breaking Umin
– Stream Split Designs - Umin and MER
• Threshold Problems
– Problems with only hot or cold utilities (no
pinch!)
32
Advanced HEN Synthesis -7