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Chen CL 1
Heat Exchanger Network Synthesis
Given a minimum temperature approach, the exactamount for minimum utility consumption can be
predicted prior to developing the network structure
Based on the pinch temperature for minimum utilityconsumption, the synthesis of the network can be
decomposed into subnetworks
It is possible to develop good a priori estimates of theminimum total area of heat exchange in a network
Q: explicit procedure for deriving configuration of
a heat exchanger network ?
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Chen CL 2
An optimal or near optimal network exhibits the
following characteristics:
Rule 1: minimum utility cost Rule 2: minimum number of units
Rule 3: minimum investment cost
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Chen CL 4
Heat Cascade Diagram
C C
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Chen CL 5
Heat Balances around each temp interval
R1+ 30 = Qs
R2+ 90 = R1+ 60
R3+ 357 = R2+ 480
Qw+ 78 = R3+ 180 (4 eq.s, 5 var.s)
LP Transshipment Problem:
min Z=Qs+ Qw
s.t. R1 Qs= 30
R2 R1= 30
R3 R2= 123
Qw R3= 102
Qs, Qw, R1, R2, R3 0
Qs= 60 MW, Qw= 225 MW
R1= 30, R2= 0, R3= 123
a pinch at temp interval 340o
320o
C
Ch CL 6
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Chen CL 6
Note:
min Z=Qs+ Qw
s.t. R1 Qs= 30
R2 R1= 30
R3 R2= 123
Qw R3= 102
min Z=Qs+ Qw = 2Qs+ 165 (?)
s.t. R1= Qs 30 0
R2= R1 30 =Qs 60 0
R3= R2+ 123 =Qs+ 63 0
Qw=R3+ 102 =Qs+ 165 0
Qs= 60Qw= 225
R1= 30
R2= 0
R3= 123
Ch CL 7
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Chen CL 7
General Transshipment Model forPredicting Min Utility Cost
Sets
Hk = {i| hot stream i supplies heat to interval k (k= 1, , K)}
Ck = {j| cold stream j demands heat from interval k}Sk = {m| hot utility m supplies heat to interval k}
Wk = {n| cold utility n extracts heat from interval k}
Par.s
QHik, QCjk heat content of hot/cold streams i/j in interval k
cm, cn unit cost of hot utility m and cold utility n
Var.s
QSm, QWn heat load of hot utilitym and cold utility n
Rk heat residual exiting intervalk
Ch CL 8
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Chen CL 8
Heat Flows in Interval k
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Chen CL 10
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Chen CL 10
Transshipment Model: Example
F Cp (MW/K) Tin (K) Tout (K)
H1 2.5 400 320
H2 3.8 370 320
C1 2.0 300 420
C2 2.0 300 370
HP Steam: 500K, $80/kW yr, LP Steam: 380K, $50/kW yrCW: 300K, $20/kW yr, Min recovery app temp = 10K
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Chen CL 12
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Chen CL 12
Expanded Transshipment Model:Min Utility Cost with Constrained Matches
Constrained Matches:
too far apart streams, for operational considerations,
Expanded Transshipment Model: Single overall heat residual Rk exiting at each temp interval k
individual heat residuals Rik, Rmk for each hot stream i and
each hot utility mthat are present at or above that
temp interval k Define variable Qijk to denote heat exchange between hot
stream i and cold stream j (and also QSjk, QiWk)
Chen CL 13
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Chen CL 13
Expanded Transshipment Model for Previous Example
Chen CL 14
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Chen CL 14
Interval k for Expanded Transshipment Model
Chen CL 15
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Chen CL 15
Chen CL 16
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Chen CL 16
The Expanded LP Transshipment Model:
Hk = {i| hot stream i is present at interval k or at a higher interval}
S
k
= {m| hot utility m is present at interval k or at a higher interval}
Qijk exahange of heat of hot streami and cold stream j at interval k
Qmjk exahange of heat of hot utility m and cold stream j at interval k
Qink exahange of heat of hot streami and cold utility m at interval k
Rik heat residual of hot stream i exiting interval k
Rmk heat residual of hot utility m exiting interval k
Chen CL 17
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Chen CL 17
min Z=
mScmQ
Sm+
nWcnQ
Wn
s.t. Rik Ri,k1+jCk
Qijk +nWk
Qink=QHik i Hk
Rmk Rm,k1+jCk
Qmjk =QSm m S
k
iHk
Qijk +mSk
Qmjk =QCjk j Ck
iHk
Qink=QWn n Wk, k= 1, , K
Rik, Rmk, Qijk, Qmjk, Qink, Q
S
n, Q
W
n 0Ri0=RiK= 0
QLij K
k=1Qijk Q
Uij
Chen CL 18
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Chen CL 18
Expanded LP Transshipment Model withRestricted Match: Example
F Cp (MW/C) Tin (C) Tout (C)
H1 1.0 400 120
H2 2.0 340 120
C1 1.5 160 400
C2 1.3 100 250
Steam: 500oC, CW: 20 30oC, Min recovery app temp = 20oCNote: match for H1 and C1 is forbidden (Q112=Q113= 0)
Minimum utility cost Z = $9, 300, 000/yr $15, 300, 000/yr
Heating utility load QS = 60 MW 120MW
Cooling utility load QW = 225 MW 285MW
Chen CL 19
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Chen CL 19
Chen CL 20
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Chen CL 20
Prediction of Matches for MinimizingNumber of Units
qsubnetworks, K1 temperature intervals in subnetwork q
yqij =
1 hot streami, cold stream j exchange heat
0 hot streami, cold stream j do not exchange heat
miniH
jC
yqij
Rik Ri,k1+
jCkQijk =Q
Hik i H
k, k= 1, , Kq
iHk
Qijk =QCjk j Ck
Kqk=1
Qijk Uijyqij 0 Rik, Qijk 0
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Chen CL 22
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Chen CL 23
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Chen CL 24
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Results:
Above Pinch:
Match Steam-C1 60 MW yAS1= 1, QS11= 30, QS12= 30
Match H1-C1 60 MW yA11= 1, Q112= 60Below Pinch:
Match H1-C1 25 MW yB11= 1, Q113= 25
Match H1-C2 195 MW yB12= 1, Q123= 117, Q124= 78
Match H2-C1 215 MW yB21= 1, Q213= 215
Match H2-CW 225 MW yB2W = 1, Q2W4= 225
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Chen CL 26
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Chen CL 27
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Alternative Structure:if without partitioning into subnetworksmatch H1-C1 is denoted by y11, Q112+ Q113 220y11 0
Match Steam-C1 60 MW
Match H1-C1 85 MW
Match H1-C2 195 MW
Match H2-C1 215 MW
Match H2-CW 225 MW
Chen CL 28
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Automatic Derivation of NetworkStructures
Superstructure of A 1H2C Example
Chen CL 29
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Alternatives Embedded in PreviousSuperstructure
Chen CL 30
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Variables for Superstructure with TwoMatches
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Chen CL 32
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Mass and heat balances for mixers at inlet of two units
F1+ F8 F3= 0
F1Tin + F8T78 F3T3= 0F2+ F6 F4= 0
F2Tin + F6T56 F4T4= 0
Mass balances for splitters at outlet of exchangers
F3 F6 F5= 0 F4 F7 F8= 0
Heat balances in exchangers
Q11 F3(T3 T56) = 0 Q12 F4(T4 T78) = 0
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Chen CL 34
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Automatic Derivation of NetworkStructures: Example
F Cp (kW/K) Tin (K) Tout (K) h (kW/m2K) Cost ($/kW-yr)
H1 22. 440 350 2.0 -
C1 20. 349 430 2.0 -
C2 7.5 320 368 .67 -
S1 500 500 1.0 120W1 300 320 1.0 20
Min recovery app temp = 1 KExchanger Cost = 6, 600 + 670(area)0.83
Chen CL 35
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Chen CL 36
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Simultaneous Optimization Model forHeat Exchanger Network Synthesis
One Problem with 2-Hot-2-Cold Streams
Stream Tin Tout F Cp (kW/K) h (KW/m2K) Cost ($/KW-yr)
H1 650 370 10.0 1.0 -
H2 590 370 20.0 1.0 -
C1 410 650 15.0 1.0 -
C2 353 500 13.0 1.0 -
S1 680 680 5.0 80
W1 300 320 1.0 15
AssumeTmin= 10K, Exchanger cost = $5500 + 150A (area, m2)
Chen CL 37
S S O
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HENS: Simultaneous OptimizationA Typical Stage-wise Superstructure
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Chen CL 39
HENS Si l O i i i
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HENS: Simultaneous OptimizationParameters
Tini , Tinj : inlet temperature of hot (cold) streami (j)
Touti , Toutj : outlet temperature of hot (cold) streami (j)
Fi, Fj : heat capacity flow rate for hot (cold) stream i (j)
TinHU
: inlet temperature of hot utility
ToutHU : outlet temperature of hot utility
TinCU : inlet temperature of cold utility
ToutCU : outlet temperature of cold utility
: upper bound for heat exchange
: upper bound for temperature differenceC, B : area cost coefficient and exponent
CF : fixed charge for exchangers
Chen CL 40
HENS Si l O i i i
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HENS: Simultaneous OptimizationVariables
dtijk : temperature approach (TA) for match (i, j) at stage k
dtcui : TA for match of hot streami and cold utility
dthuj : TA for match of cold streamj and hot utility
qcui
: heat exchanger between hot streami and cold utility
qhuj : HE between cold streamj and hot utility
qijk : HE between hot streami, cold stream j, stage k
tik : temperature of hot stream i at hot end of stage k
tjk : temperature of cold streamj at hot end of stage k
zijk : {0, 1} denoting existence of match (i, j) in stage kzcui : {0, 1} denoting cold utility exchanges heat with stream i
zhuj : {0, 1} denoting hot utility exchanges heat with stream j
Chen CL 41
HENS Si l O i i i
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HENS: Simultaneous OptimizationConstraints
Overall heat balance for each stream:
(Tini Touti )Fi = kST jCP
qijk + qcui i HP
(Toutj Tinj )Fj =
kST
iHP
qijk+ qhuj j CP
Heat balance at each stage:
(tik ti,k+1)Fi =jCP
qijk k S T , i HP
(tjk tj,k+1)Fj = iHP
qijk k S T , j CP
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Chen CL 43
HENS Si lt O ti i ti
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HENS: Simultaneous OptimizationConstraints
Hot and cold utility load:
(ti,Ns+1 Touti )Fi =qcui i HP
(Toutj tj1)Fj =qhuj j CP
Logical constraints:
qijk zijk 0 i H P , j C P , k ST
qcui zcui 0 i HP
qhuj zhuj 0 j CP
zijk,zcui,zhuj {0, 1}
Chen CL 44
HENS Si lt O ti i ti
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HENS: Simultaneous OptimizationConstraints
Calculation of approach temperatures:
dtijk tik tjk + (1 zijk) i H P , j C P , k STdtij,k+1 ti,k+1 tj,k+1+ (1 zijk) i H P , j C P , k ST
dtcui ti,Ns+1 ToutCU+ (1 zcui) i HP
dthuj ToutHU tj1+ (1 zhuj) j CP
dtijk Tmin i H P , j C P , k ST
dtcui Tmin i HP
dthuj Tmin j CP
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Chen CL 47
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=
x
(Tini Touti )Fi =
kST
jCP
qijk + qcui
(Toutj Tinj )Fj =
kST
iHP
qijk + qhuj
(tik ti,k+1)Fi =
jCP
qijk
(tjk tj,k+1)Fj =
iHP
qijk
Tini =ti,1
Tinj =tj,1
tik ti,k+1
tjk t
j,k+1Touti ti,Ns+1
Toutj tj1
(ti,Ns+1 Touti )Fi =qcui
(Toutj tj1)Fj =qhuj
qijk zijk 0
qcui zcui 0
qhuj zhuj 0
dtijk tik tjk + (1 zijk)
dtij,k+1 ti,k+1 tj,k+1+ (1 zijk)
dtcui ti,Ns+1 ToutCU+ (1 zcui)
dthuj ToutHU tj1+ (1 zhuj)
dtijk, dtcui, dthuj Tmin
Chen CL 48
HENS Si lta eo s O ti i atio
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HENS: Simultaneous OptimizationObjective: Min (Utility + Fixed + Area) Cost
MINLP Problem:
minx
iHP
CCUqcui+jCP
CHUqhuj
+
iHP jCP kSTCFijzijk +
iHPCFi,CUzcui+
jCPCFj,HUzhuj
+iHP
jCP
kST
Cij(Aijk)Bij +
iHP
Ci,CU(Ai,CU)Bi,CU
+ jCP
Cj,HU(Aj,HU)Bj,HU
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Chen CL 50
HENS: Simultaneous Optimization
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HENS: Simultaneous OptimizationOptimal Network Structure
155, 000/yr total cost
(71, 400 for utility cost and 83, 600 for capital cost)
Chen CL 51
Simultaneous MINLP Model: Example
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Simultaneous MINLP Model: Example
F Cp (kW/K) Tin (K) Tout (K) h (kW/m2K) Cost ($/kW-yr)
H1 22. 440 350 2.0 -
C1 20. 349 430 2.0 -
C2 7.5 320 368 .67 -
S1 500 500 1.0 120
W1 300 320 1.0 20
Min recovery app temp = 1 K
Exchanger Cost = 6, 600 + 670(area)0.83
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Chen CL 53
Simultaneous MINLP Model: Same
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Simultaneous MINLP Model: SameExample with No Stream Splitting