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Transcript of HEN Design.pdf
7/27/2019 HEN Design.pdf
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DESIGN OF MAXIMUMENERGY RECOVERY
NETWORKS
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MER NETWORKS
• Networks featuring minimum utility usageare called MAXIMUM ENERGY
RECOVERY (MER) Networks.
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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 belowthe pinch a HEAT SOURCE region.
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Pinch
Minimum
heating utility
Minimum
cooling utilityHeat Source
Heat is released to
cooling utility
Heat Sink
Heat is obtained from
the heating utility
7.5
12.0
14.0
0.0
6.0
1.0
4.0
4.0
3.0
9.0
2.0
2.0
14.0
1.5
10.0
T
H
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CONCLUSION
• One can analyze the two systems separately,
that is,
• Heat exchangers will not contain heattransfer across the pinch.
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PINCH MATCHES• Consider two streams above the pinch
Q/FCpH > Q/FCpC
T p ∆TminTC,outQ
TH,inT p
TC,out= T p ∆Tmin + Q/FCpC
TH,in= T p+Q/FCpH
But TH,in> TC,out+ ∆Tmin.
Thus replacing one obtains
FCpH <FCpCGolden rule for pinch
matches above the pinch.
∆Tmin Violation when FCpH >FCpC∆Tmin
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EXERCISE
• A similar rule is obtained below the pinch. Derive it
and illustrate it.
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ANSWER TH,out
T p ∆Tmin
TC,inQ
T p
TC,in= T p ∆Tmin  Q/FCpC
TH,out= T pQ/FCpH
But TH,out> TC,in + ∆Tmin.
Thus replacing one obtains
FCpC <FCpHGolden rule for pinchmatches below the pinch.
∆Tmin∆Tmin
Violation when FCpC >FCpH
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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
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QUESTION
• Once a match has been selected how much heat should be exchanged?
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ANSWER • As much as possible!
• This means that one of the streams has its dutysatisfied!!
THIS IS CALLLED THE
TICKOFF RULE
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HANDS ON EXERCISE
Stream Type Supply T Target T ∆H F*Cp
(oC) (oC) (MW) (MW oC1)
Reactor 1 feed Cold 20 180 32.0 0.2
Reactor 1 product Hot 250 40 31.5 0.15
Reactor 2 feed Cold 140 230 27.0 0.3
Reactor 1 product Hot 200 80 30.0 0.25
T=140 0C
T=20 0C
∆H=27 MW
REACTOR 2
∆H=30 MW
∆H=32 MW
REACTOR 1
∆H=31.5 MW
T=230 0C
T=180 0CT=250 0C T=40 0C
T=200 0CT=80 0C
∆Tmin=10 oC PINCH=150 oC
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HANDS ON EXERCISE
Stream Type Supply T Target T ∆H F*Cp
(oC) (oC) (MW) (MW oC1)
Reactor 1 feed Cold 20 180 32.0 0.2
Reactor 1 product Hot 250 40 31.5 0.15
Reactor 2 feed Cold 140 230 27.0 0.3
Reactor 1 product Hot 200 80 30.0 0.25
∆Tmin=10 oC PINCH=150 oC
200 0C
20 0C
40 0C
180 0C
250 0CH1
H2
C2
C1
230 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3
140 0C
80 0C
150 0C
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ABOVE THE PINCH
• Which matches are possible?
200 0C
180 0C
250 0CH1
H2
C2
C1
230 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3
140 0C
150 0C
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ANSWER (above the pinch)
2000
C
180 0C
250 0CH1
H2
C2
C1
230 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3140 0C
150 0C
• The rule is that FCpH <FCpC . We therefore can onlymake the match H1C1 and H2C2.
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ANSWER (above the pinch)
• The tickoff rule says that a maximum of 8 MW is exchangedin the match H1C1 and as a result stream C1 reaches its targettemperature.
• Similarly 12.5 MW are exchanged in the other match and thestream H2 reaches the pinch temperature.
200 0C
180 0C
203.3 0CH1
H2
C2
C1
181.7 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3
140 0C
150 0C
8
12.5
230 0C
250 0C
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BELOW THE PINCH
• Which matches are possible?
20 0C
40 0CH
1
H2
C1
FCp=0.15
FCp=0.25
FCp=0.2 140 0C
80 0C
150 0C
The rule is that FCpC <FCp H . Thus, we can
only make the match H2C1
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ANSWER (below the pinch)
• The tickoff rule says that a maximum of 17.5 MW isexchanged in the match H2C1 and as a result stream H2reaches its target temperature.
52.5 0C
40 0CH1
H2
C1
FCp=0.15
FCp=0.25
FCp=0.2 140 0C
80 0C
150 0C
17.5
20 0C
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COMPLETE NETWORK AFTER PINCH MATCHES
• Streams with unfulfilled targets are colored.
200 0C
180 0C
203.3 0CH1
H2
C2
181.7 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3
140 0C
8
12.5
52.5 0C C1
140 0C
80 0C
150 0C
17.5
250 0C
230 0C
20 0C
40 0C
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WHAT TO DO NEXT?
• 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!!
2000
C
180 0C
203.3 0CH1
H2
C2
181.7 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3
140 0C
8
12.5
52.5 0C C1
140 0C
800
C
150 0C
17.5
250 0C
230 0C
20 0C
40 0C
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ANSWER
• We first note that we will use heating above the pinch. Thusall hot streams need to reach their inlet temperature. We are
then forced to look for a match for H1. Please locate it.
200 0C
180 0C
203.3 0CH
1
H2
C2
181.7 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.3
140 0C
8
12.5
52.5 0C C1
140 0C
80 0C
150 0C
17.5
250 0C
230 0C
20 0C
40 0C
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ANSWER
140 0C
• The match is H1C1. We finally put a heater on the
cold stream
200 0C
180 0C
H1
H2
C2
230 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.38
12.5
52.5 0C C1
140 0C
80 0C
150 0C
17.5
77.5
H
250 0C
200
C
40 0C
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ANSWER • Below the pinch we try to have the cold streams start at
their inlet temperatures and we later locate coolers (one in
this case).
140 0C
200 0C
180 0C
H1
H2
C2
230 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.38
12.5
20 0C C1
140 0C
80 0C
150 0C
17.5
77.5
H6.5
C
10
40 0C250 0C
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DISCUSSION
• State what are your objections to the pinch design
method, so far!!
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THESE ARE MINE
• It seems to imply building expertise• It does not seem to guarantee that one will always
obtain a good result.
• What if there are unequal number of streams at the pinch?
• What if the inequalities are not satisfied at the
pinch?
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GENERAL FORMULA FOR UNIT TARGETING
Nmin= (SP)above pinch+ (SP) below pinch
If we do not consider two separate problems, above
and below the pinch we can get misleading results.
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EXAMPLE
Nmin= (SP)above pinch+ (SP) below pinch =
=(51) + (41) = 7
If we do not consider two separate problems
Nmin= (61)= 5, which is wrong
Note: A heat exchanger network with 5 exchangers exists, but it is impractical andcostly. This is beyond the scope of this course.
140 0C
200 0C
180 0C
H1
H2
C2
230 0C
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.38
12.5
20 0C C1
140 0C
80 0C
150 0C
17.5
77.5
H6.5
C
10
250 0C 40 0C
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UNEQUAL NUMBER OFSTREAMS AT THE PINCH
Above the pinch, we notice the following rule
SH≤SC
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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.
C2
C1
H1
H2
H3
Assume the matches H1C1 and
the matches H2C2 have beenselected. 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. Suchmatching would be infeasible.
150 OC 100 OCFCp=0.15
FCp=0.25
FCp=0.2
FCp=0.4
FCp=0.1
90 OC
7.5
140 OC
10
Target=170 OC127.5 OC
122.5 OCTarget=140 OC
130 OC
What is then, the solution?
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ANSWER
Split cold stream until the inequality is satisfied.
Notice that different combinations of flowrates in the split
satisfy the inequality.
C2
C1
H1
H2
H3
150O
C100 OC
FCp=0.15
FCp=0.25
FCp=0.2
FCp=0.25
FCp=0.1
90 OC
7.5
140 OC
10
Target=170 OC127.5 OC
130 OCTarget=140 OC
130 OC
FCp=0.15 110 OC
3
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UNEQUAL NUMBER OFSTREAMS AT THE PINCH
A similar rule can be discussed below the pinch,
that is,
SH≥SC
and similar splitting of hot streams may benecessary.
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INEQUALITY NOTSATISFIED
Consider the following case:
C2
C1
H1
150 OC 100 OCFCp=0.5
FCp=0.2
FCp=0.4
90 OCTarget=170 OC
Target=140 OC
Suggest a solution.
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ANSWER Split the hot stream
C2
C1
H1 150O
C 100O
CFCp=0.2
FCp=0.2
FCp=0.4
90 OC
15
Target=170 OC
Target=140 OC
FCp=0.3
10
140 OC
127.5 OC
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SOLVE THE FOLLOWING
PROBLEM
Below the Pinch :
C1
H2
H1
100 OCFCp=0.5
FCp=0.3
FCp=0.790 OC
Target=40 OC
Target=20 OC
30 OC
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ANSWER
Below the Pinch :
30
C1
H2
H1
100 OCFCp=0.5
FCp=0.3
FCp=0.590 OC
Target=40 OC
Target=20 OC
30 OC
FCp=0.2
12
40 OC
60 OC
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COMPLETE PROCEDUREABOVE THE PINCH
Start
SH≤SC
Split Cold
Stream
FCpH≤FCpC
at pinch?
Place
matchesSplit Hot
Stream
YesYes
NoNo
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COMPLETE PROCEDUREBELOW THE PINCH
SH≥SC
Split Hot
Stream
FCpH≥ FCpC
at pinch?
Split Cold
Stream
Start
Place
matches
YesYes
No No
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HANDS ON EXERCISEType Supply T Target T F*Cp
(oC) (oC) (MW oC1)
Hot 750 350 0.045
Hot 550 250 0.04
Cold 300 900 0.043
Cold 200 550 0.02
∆Tmin=50 oC
Minimum Heating Utility= 9.2 MW
Minimum Cooling Utility= 6.4 MW
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ANSWER
750 0C
900 0C
H1
H2
C2
FCp=0.045
FCp=0.04
FCp=0.043
FCp=0.02
500 0C
8
C1
250 0C
61
9.2
300 0C
8.6
0.4
550 0C
550 0C
6
3500
C
200 0C
FCp=0.04
686.05 0C 400 0C
358.89 0C
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ANSWER
We are minimizing the number of matches. Wesaw already that different answers can be
obtained if separate regions are not considered.