PART 4
PINCH DESIGN METHOD
MAXIMUM ENERGY
RECOVERY NETWORKS
MER NETWORKS
• Networks featuring minimum utility usage are
called MAXIMUM ENERGY RECOVERY
(MER) Networks.
DIVISION AT THE PINCH
RECALL THAT
• No heat is transferred through the pinch.
• This makes the region above the pinch a
HEAT SINK region and the region below
the pinch a HEAT SOURCE region.
Heat Sink
Heat is obtained from
the heating utility
7.5
Minimum
heating utility
1.5
9.0
-6.0
T
3.0
1.0
4.0
-4.0
0.0
Pinch
14.0
14.0
-2.0
12.0
H
-2.0
Heat Source
Heat is released to
cooling utility
10.0
Minimum
cooling utility
CONCLUSION
• One can analyze the two systems separately,
that is,
• Heat exchangers will not match streams
above the pinch with streams below the
pinch
PINCH MATCHES
• Consider two streams above the pinch
Tp
TH,in
TC,out= Tp- ΔTmin + Q/FCpC
TH,in= Tp+Q/FCpH
But TH,in> TC,out+ ΔTmin.
Thus replacing one obtains
Q
TC,out
ΔTmin
Tp- ΔTmin
FCpH < FCpC
Q/FCpH > Q/FCpC
Golden rule for pinch
matches above the pinch.
ΔTmin
Violation when FCpH > FCpC
Below the Pinch
TC,in= Tp- ΔTmin - Q/FCpC
Tp
TH,out
TH,out= Tp-Q/FCpH
But TH,out> TC,in + ΔTmin.
Thus replacing one obtains
Tp- ΔTmin
Q
TC,in
ΔTmin
FCpC < FCpH
Golden rule for pinch
matches below the pinch.
ΔTmin
Violation when FCpC > FCpH
CONCLUSION
• Since matches at the pinch need to satisfy these rules,
one should start locating these matches first. Thus, our
first design rule:
START BY MAKING PINCH
MATCHES
TICK-OFF RULE
Once a match has been selected how much heat
should be exchanged?
• As much as possible! (We want to minimize units, so
once we have one, we make the biggest use of it
possible)
• This means that one of the streams has its duty
satisfied!!
HANDS ON EXERCISE
ΔH=27 MW
ΔH=-30 MW
REACTOR 2
T=230 0C
T=140 0C
T=200
ΔH=32 MW
T=80 0C
ΔH=-31.5 MW
REACTOR 1
T=20 0C
T=180
Stream
0C
Type
Reactor 1 feed
Cold
Reactor 1 product
T=250 0C
0C
T=40 0C
Supply T
Target T
ΔH
F*Cp
(oC)
(oC)
(MW)
(MW oC-1)
20
180
32.0
0.2
Hot
250
40
-31.5
0.15
Reactor 2 feed
Cold
140
230
27.0
0.3
Reactor 2 product
Hot
200
80
-30.0
0.25
ΔTmin=10 oC
PINCH=150 oC
HANDS ON EXERCISE
FCp=0.15
H1
250 0C
40 0C
150 0C
200 0C
FCp=0.25
80 0C
H2
140 0C
180 0C
20 0C
FCp=0.2
FCp=0.3
Stream
230 0C
Type
Reactor 1 feed
Cold
Reactor 1 product
C1
C2
Supply T
Target T
ΔH
F*Cp
(oC)
(oC)
(MW)
(MW oC-1)
20
180
32.0
0.2
Hot
250
40
-31.5
0.15
Reactor 2 feed
Cold
140
230
27.0
0.3
Reactor 2 product
Hot
200
80
-30.0
0.25
ΔTmin=10 oC
PINCH=150 oC
ABOVE THE PINCH
FCp=0.15
H1
FCp=0.25
H2
250 0C
150 0C
200 0C
180 0C
C1
FCp=0.2
FCp=0.3
230 0C
140 0C
• Which matches are possible?
C2
Pinch matches above the Pinch
FCp=0.15
H1
FCp=0.25
H2
FCp=0.2
FCp=0.3
250 0C
150 0C
200 0C
180 0C
C1
230 0C
140 0C
C2
• The rule is that FCpH < FCpC .
• The candidates are: H1-C1, H1-C2 and H2-C2.
• Because all hot streams at the pinch need to participate
in a pinch match, we therefore can only choose the
matches H1-C1 and H2-C2.
Pinch matches above the Pinch
FCp=0.15
FCp=0.25
H1
H2
250 0C
150 0C
203.3 0C
200 0C
180 0C
FCp=0.2
FCp=0.3
C1
230 0C
181.7 0C
8
12.5
140 0C
C2
• The tick-off rule says that a maximum of 8 MW is exchanged
in the match H1-C1 and as a result stream C1 reaches its target
temperature.
• Similarly 12.5 MW are exchanged in the other match and the
stream H2 reaches the pinch temperature.
Pinch matches below the Pinch
FCp=0.15
H1
FCp=0.25
H2
FCp=0.2
150 0C
40 0C
80 0C
140 0C
20 0C
C1
• The rule is that FCpC < FCpH .
• Only one match qualifies: H2-C1
• Below the pinch all cold streams need to
participate in pinch matches
ANSWER (below the pinch)
FCp=0.15
H1
FCp=0.25
H2
FCp=0.2
40 0C
150 0C
80 0C
52.5 0C
140 0C
20 0C
C1
17.5
• The tick-off rule says that a maximum of 17.5 MW is
exchanged in the match H2-C1 and as a result stream H2
reaches its target temperature.
COMPLETE NETWORK
AFTER PINCH MATCHES
FCp=0.15
H1
FCp=0.25
H2
FCp=0.2
250 0C
150 0C
203.3 0C
40 0C
80 0C
200 0C
140 0C
180 0C
52.5 0C
C1
20
FCp=0.3
17.5
8
0
230 0C 181.7 C
140
0C
C2
12.5
• Streams with unfulfilled targets are colored.
0C
NON-PINCH MATCHES
FCp=0.15
H1
FCp=0.25
H2
250 0C
40 0C
150 0C
203.3 0C
80 0C
200 0C
140 0C
180 0C
FCp=0.2
52.5 0C
C1
20 0C
FCp=0.3
230 0C
17.5
8
181.7 0C
140 0C
C2
12.5
• Away from the pinch, there is more flexibility to make
matches, so the inequalities do not have to hold.
• The pinch design method leaves you now on your own!!!!!
• Therefore, use your judgment as of what matches to select!!
NON-PINCH MATCHES
FCp=0.15
H1
FCp=0.25
H2
250 0C
40 0C
150 0C
203.3 0C
80 0C
200 0C
140 0C
180 0C
FCp=0.2
52.5 0C
C1
20
FCp=0.3
230 0C
17.5
8
181.7 0C
0C
140 0C
C2
12.5
• We first note that we will use heating above the pinch. Thus
all hot streams need to reach their inlet temperature. We are
then forced to look for a match for H1.
NON-PINCH MATCHES
• The match is H1-C1. We finally put a heater on the
cold stream
FCp=0.15
H1
FCp=0.25
H2
250 0C
150 0C
40 0C
80 0C
200 0C
140 0C
180 0C
FCp=0.2
52.5 0C
C1
20 0C
FCp=0.3
230
0C
17.5
8
H
140 0C
7.5
7
12.5
C2
NON-PINCH MATCHES
• Below the pinch we try to have the cold streams start at
their inlet temperatures and we later locate coolers (one in
this case).
FCp=0.15
H1
150 0C
250 0C
40 0C
C
10
FCp=0.25
H2
200
0C
80 0C
FCp=0.2
FCp=0.3
140 0C
180 0C
230 0C
8
140 0C
7.5
17.5
C2
H
7
12.5
20 0C
6.5
C1
UNEQUAL NUMBER OF
STREAMS AT THE PINCH
Indeed, if the number of hot streams is larger than the
number of cold streams, then no pinch matches are
possible. Consider this (new) example:
FCp=0.15
H1
FCp=0.25
H2
FCp=0.1
Target=170 OC
Target=140 OC
100 OC
150 OC
140 OC
130 OC
H3
FCp=0.2
FCp=0.4
90 OC
127.5 OC
C1
122.5
OC
7.5
C2
Assume the matches H1-C1 and
the matches H2-C2 have been
selected. Since H3 needs to go to
the pinch temperature, there is no
cold stream left to match, even if
there is portions of C1 or C2 that
are left for matching. Such
matching would be infeasible.
10
What is then, the solution?
UNEQUAL NUMBER OF
STREAMS AT THE PINCH
Split cold stream until the inequality is
satisfied.
FCp=0.15
H1
FCp=0.25
H2
FCp=0.1
Target=170
OC
Target=140 OC
FCp=0.2
H3
100 OC
150 OC
140 OC
130 OC
127.5 OC
90 OC
C1
FCp=0.25
130 OC
FCp=0.15
110 OC
7.5
C2
10
3
Notice that different combinations of flowrates in the split
satisfy the inequality.
UNEQUAL NUMBER OF
STREAMS AT THE PINCH
Above the pinch, we notice the following rule
SH ≤SC
If that is NOT the case, we split a cold stream until SH=SC
A similar rule can be discussed below the pinch, that is,
SH ≥SC
If that is NOT the case, we split a cold stream until SH=SC
INEQUALITIES NOT SATISFIED
Consider the following caseabove the pinch
We notice that FCpH >FCpC (needs to be FCpH ≤FCpC)
FCp=0.5
Target=170
OC
Target=140 OC
FCp=0.2
FCp=0.4
H1
150 OC
100 OC
90 OC
C1
C2
INEQUALITIES NOT SATISFIED
The hot stream needs to be split
FCp=0.2
H1
150 OC
100 OC
140 OC
90 OC
FCp=0.3
Target=170 OC
FCp=0.2
Target=140 OC
FCp=0.4
C1
10
127.5 OC
C2
15
INEQUALITIES NOT SATISFIED
Below the Pinch :
FCp=0.5
H1
FCp=0.3
H2
FCp=0.7
100 OC
Target=40 OC
Target=20 OC
90 OC
30 OC
C1
INEQUALITIES NOT SATISFIED
The cold stream needs to be split
FCp=0.5
H1
FCp=0.3
H2
FCp=0.5
FCp=0.2
100 OC
40 OC
Target=40 OC
Target=20 OC
60 OC
90 OC
30 OC
C1
30
12
COMPLETE PROCEDURE
ABOVE THE PINCH
Start
Yes
FCpH≤FCpC
Yes
SH≤SC
at pinch?
No
No
Split Cold
Stream
Place
matches
Split Hot
Stream
COMPLETE PROCEDURE
BELOW THE PINCH
Start
Yes
FCpH ≥ FCpC
at pinch?
No
Yes
SH≥SC
No
Split Hot
Stream
Place
matches
Split Cold
Stream
HANDS ON EXERCISE
Type
Supply T
(oC)
Target T
(oC)
F*Cp
(MW oC-1)
Hot
750
350
0.045
Hot
550
250
0.04
Cold
300
900
0.043
Cold
200
550
0.02
ΔTmin=50 oC
Minimum Heating Utility= 9.2 MW
Minimum Cooling Utility= 6.4 MW
ANSWER
FCp=0.04
550 0C
750 0C
358.89 0C
350 0C
FCp=0.045
H1
0.4
FCp=0.04
H2
686.05 0C
6
400 0C
300 0C
FCp=0.043
FCp=0.02
900
0C
550 0C
200 0C
1
C1
8.6
8
9.2
250 0C
500 0C
6
C2
NOTE ON UNIT TARGETING
Nmin= (S-P)above pinch+ (S-P)below pinch
If we do not consider two separate problems,
above and below the pinch we can get
misleading results.
EXAMPLE
FCp=0.15
H1
150 0C
250 0C
C
40 0C
10
FCp=0.25
H2
200 0C
80 0C
FCp=0.2
FCp=0.3
140 0C
180 0C
8
230 0C
17.5
C1
6.5
C2
H
140
7.5
20 0C
7
0C
12.5
Nmin= (S-P)above pinch+ (S-P)below pinch
=
=(5-1)
+ (4-1)
=
7
If we do not consider two separate problems
Nmin= (6-1)= 5, which is most of the time wrong
Note: A heat exchanger network with 5 exchangers exists, but it is impractical and
costly. This is beyond the scope of this course.