Biegler/Grossmann/Westerberg: Systematic Methods of Chemical Process Design
Synthesis of Heat Exchanger Networks
Cheng-Liang Chen
PSE
LABORATORY
Department of Chemical Engineering National TAIWAN University
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Heat Exchanger Network Synthesis
Give Given n a mini minimu mum m temp temper erat atu ure app approa roach, ch, the the exa exact amou mount for for minimu can be minimum m utility utility consump onsumptio tion n ca predicted prior to developing the network structure
Based on the pinch temperature for minimum utility consumption , the synthesis of the network can be decomposed into subnetworks
It is possible to develop good a priori estimates of the minimum total area of heat exchange in a network
Q: explicit procedure for deriving configuration of a heat exchanger network ?
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Heat Exchanger Network Synthesis
Give Given n a mini minimu mum m temp temper erat atu ure app approa roach, ch, the the exa exact amou mount for for minimu can be minimum m utility utility consump onsumptio tion n ca predicted prior to developing the network structure
Based on the pinch temperature for minimum utility consumption , the synthesis of the network can be decomposed into subnetworks
It is possible to develop good a priori estimates of the minimum total area of heat exchange in a network
Q: explicit procedure for deriving configuration of a heat exchanger network ?
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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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Sequen Sequentia tiall Synthe Synthesis sis:: Min Utilit Utilityy Cost Cost H1 H2 C1 C2
F C p (MW/C)
T in in (C)
T out out (C)
1 .0 2 .0 1 .5 1 .3
400 340 160 100
120 120 400 250
Steam: 500oC , CW: 20 − 30oC , Mi Min n recov recover eryy app tem temp p = 20oC
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Heat Cascade Diagram
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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 = Q s + 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 − 320oC
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Note: min Z = Q s + Qw s.t. R1 − Qs = −30 R2 − R1 = −30 R3 − R2 = 123 Qw − R3 = 102
⇒
⇒
min Z = Q s + Qw s.t. R1 = Q s − 30 R2 = R 1 − 30 R3 = R 2 + 123
= 2Qs + 165 (?) ≥0 ≥0 = Q s − 60 = Q s + 63 ≥0 Qw = R 3 + 102 = Q s + 165 ≥ 0
Qs = 60 Qw = 225 R1 = 30 R2 = 0 R3 = 123
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General Transshipment Model for Predicting Min Utility Cost
Sets H k C k S k W k Par.s C QH , Q ik jk cm, cn Var.s W QS m , Qn Rk
= {i| hot stream i supplies heat to interval k (k = 1, · · · , K )} = { j | cold stream j demands heat from interval k } = {m| hot utility m supplies heat to interval k } = {n| cold utility n extracts heat from interval k } heat content of hot/cold streams i/ j in interval k unit cost of hot utility m and cold utility n heat load of hot utility m and cold utility n heat residual exiting interval k
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Heat Flows in Interval k
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LP Transshipment Model for Predicting Min Utility Cost min Z = s.t. Rk − Rk−1 −
QS m +
m∈S k
n∈W k
m∈S QW n =
cmQS m +
cnQW n
n∈W
QH ik −
i∈H k
QC jk
j ∈C k
k = 1, · · · , K
W QS m, Qn , Rk ≥ 0, k = 1, · · · , K − 1
R0 = RK = 0
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Transshipment Model: Example F Cp (MW/K)
T in (K)
T out (K)
H1 H2
2.5 3.8
400 370
320 320
C1 C2
2.0 2.0
300 300
420 370
HP Steam: 500K , $80/kW yr, CW: 300K , $20/kW yr,
LP Steam: 380K , $50/kW yr Min recovery app temp = 10K
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min Z = 80000QHP + 50000QLP + 20000QCW s.t. R1 − QHP = −60 R2 − R1 = 10 R3 − R2 − QLP = −15
− R3 + QCW = 75 R1, R2, R3, QHP , QLP , QCW ≥ 0
⇒ Z = $6, 550, 000/yr QHP = 60, QLP = 5, QCW = 75 MW R1 = 0, R2 = 10, R3 = 0 (two pinches) sub-network 1: above 400 − 390 K sub-network 2: betwn 400 − 390 K and 370 − 360 K sub-network 3: below 370 − 360 K
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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 m that 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 , QiW k )
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Expanded Transshipment Model for Previous Example
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Interval k for Expanded Transshipment Model
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The Expanded LP Transshipment Model: H k = {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 stream i 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 stream i and cold utility m at interval k Rik Rmk
heat residual of hot stream i exiting interval k heat residual of hot utility m exiting interval k
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min Z =
cmQS m +
m∈S
s.t. Rik − Ri,k−1 +
Qijk +
j ∈C k
Qink =
cnQW n
n∈W QH ik i
∈ H k
n∈W k
Rmk − Rm,k−1 +
Qmjk = Q S m
m ∈ S k
Qmjk = QC jk
j ∈ C k
j ∈C k
Qijk +
i∈H k
Qink =
m∈S k QW n n
∈ W k ,
k = 1, · · · , K
i∈H k W Rik , Rmk , Qijk , Qmjk , Qink , QS n , Qn ≥ 0
Ri0 = RiK = 0 K
QL ij ≤
k=1
Qijk ≤ QU ij
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Expanded LP Transshipment Model with Restricted Match: Example F Cp (MW/C)
T in (C)
T out (C)
1.0 2.0 1.5 1.3
400 340 160 100
120 120 400 250
H1 H2 C1 C2
Steam: 500oC , CW: 20 − 30oC , Min recovery app temp = 20oC Note: match for H1 and C1 is forbidden (Q112 = Q 113 = 0)
Minimum utility cost Heating utility load
Z = $9, 300, 000/yr → $15, 300, 000/yr QS =
60 MW →
120 MW
Cooling utility load QW =
225 MW →
285 MW
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Prediction of Matches for Minimizing Number of Units
q subnetworks, K 1 temperature intervals in subnetwork q q yij
=
1 hot stream i, cold stream j exchange heat 0 hot stream i, cold stream j do not exchange heat ⇒
min
q yij
i∈H j ∈C
Rik − Ri,k−1 +
Qijk = Q H ik
i ∈ H k , k = 1, · · · , K q
Qijk = Q C jk
j ∈ C k
j ∈C k
K q
k=1
i∈H k q ≤0 Qijk − U ij yij
Rik , Qijk ≥ 0
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Matches for Minimizing Number of Units: Example
F Cp (MW/C)
T in (C)
T out (C)
H1 H2
1.0 2.0
400 340
120 120
C1 C2
1.5 1.3
160 100
400 250
Steam: 500oC , CW: 20 − 30oC , Min recovery app temp = 20oC ⇒ QS = 60, QW = 225 MW
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Results: Above Pinch: A Match Steam-C1 60 MW yS 1 = 1, QS 11 = 30, QS 12 = 30 A Match H1-C1 60 MW y11 = 1, Q112 = 60 Below Pinch: B Match H1-C1 25 MW y11 = 1, Q113 = 25 B 195 MW y12 = 1, Q123 = 117, Q124 = 78 Match H1-C2 B Match H2-C1 215 MW y21 = 1, Q213 = 215 B Match H2-CW 225 MW y2W = 1, Q2W 4 = 225
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B Alternative Structure: if let y11 = 0 (forbit H1-C1 below
pinch)
Above Pinch: A Match Steam-C1 60 MW yS 1 = 1, QS 11 = 30, QS 12 = 30 A Match H1-C1 60 MW y11 = 1, Q112 = 60 Below Pinch: B Match H1-C2 195 MW y12 = 1, Q123 = 117, Q124 = 78 B Match H2-C1 240 MW y21 = 1, Q213 = 240 B Match H1-CW 25 MW y1W = 1, Q1W 4 = 25 B Match H2-CW 225 MW y2W = 1, Q2W 4 = 200
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Alternative Structure: if without partitioning into subnetworks match H1-C1 is denoted by y11, Q112 + Q113 − 220y11 ≤ 0 Match Steam-C1 Match H1-C1
60 MW 85 MW
Match H1-C2 Match H2-C1
195 MW 215 MW
Match H2-CW
225 MW
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Automatic Derivation of Network Structures Superstructure of A 1H2C Example
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Alternatives Embedded in Previous Superstructure
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Variables for Superstructure with Two Matches
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Objective and Constraints
min C = C 1Aβ11 + C 2Aβ12 = C 1 + C 2
Q11 1/3
U 11 [θ11θ21 (θ11 + θ21) /2] Q12
1/3
U 12 [θ12θ22 (θ12 + θ22) /2]
Mass balance for initial splitter F 1 + F 2 = F
β
β
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Mass and heat balances for mixers at inlet of two units F 1 + F 8 − F 3 = 0 F 1T in + F 8T 78 − F 3T 3 = 0 F 2 + F 6 − F 4 = 0 F 2T in + F 6T 56 − F 4T 4 = 0
Mass balances for splitters at outlet of exchangers F 3 − F 6 − F 5 = 0
F 4 − F 7 − F 8 = 0
Heat balances in exchangers Q11 − F 3(T 3 − T 56) = 0
Q12 − F 4(T 4 − T 78) = 0
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Temperature differences θ11 = T 3 − tout 1
θ21 = T 56 − tin 1
θ12 = T 4 − tout 2
θ22 = T 78 − tin 2
Feasibility constraints for temperatures θ11, θ21, θ12, θ22 ≥ ∆T min
Nonnegativity conditions on heat capacity flow rates F j ≥ 0
j = 1, · · · , 8
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Automatic Derivation of Network Structures: Example F Cp (kW/K)
T in (K)
T out (K)
h (kW/m2K)
Cost ($/kW-yr)
H1 C1
22. 20.
440 349
350 430
2.0 2.0
-
C2 S1
7.5 − −
320 500
368 500
.67 1.0
120
300
320
1.0
20
W1
Min recovery app temp = 1 K Exchanger Cost = 6, 600 + 670(area)0.83
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Simultaneous Optimization Model for Heat Exchanger Network Synthesis One Problem with 2-Hot-2-Cold Streams Stream
T in
T out F Cp (kW/K) h (KW/m2K) Cost ($/KW-yr)
H 1
650
370
10.0
1.0
-
H 2
590
370
20.0
1.0
-
C 1
410
650
15.0
1.0
-
C 2
353
500
13.0
1.0
-
S 1
680
680
−
5.0
80
W 1
300
320
−
1.0
15
Assume ∆T min = 10K ,
Exchanger cost = $5500 + 150A (area, m2)
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HENS: Simultaneous Optimization A Typical Stage-wise Superstructure
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HENS: Simultaneous Optimization Indices and Sets
i = index for hot process stream j = index for cold process stream k = index for superstructure stage HP = {i|i is a hot process stream} CP = { j | j is a cold process stream} HU = {h|h is a hot utility} CU = {c|c is a cold utility} ST = {k|k is a stage in the superstructure, [ST ] = N s or N OK }
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HENS: Simultaneous Optimization Parameters T iin, T jin T iout, T jout F i, F j in T HU out T HU in T CU out T CU Ω Γ C, B CF
: : : : : : : : : : :
inlet temperature of hot (cold) stream i ( j) outlet temperature of hot (cold) stream i ( j) heat capacity flow rate for hot (cold) stream i ( j) inlet temperature of hot utility outlet temperature of hot utility inlet temperature of cold utility outlet temperature of cold utility upper bound for heat exchange upper bound for temperature difference area cost coefficient and exponent fixed charge for exchangers
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HENS: Simultaneous Optimization Variables dtijk dtcui dthuj qcu i qhu j q ijk tik tjk zijk zcu i zhu j
: : : : : : : : : : :
temperature approach (TA) for match (i, j) at stage k TA for match of hot stream i and cold utility TA for match of cold stream j and hot utility heat exchanger between hot stream i and cold utility HE between cold stream j and hot utility HE between hot stream i, cold stream j, stage k temperature of hot stream i at hot end of stage k temperature of cold stream j at hot end of stage k ∈ {0, 1} denoting existence of match (i, j) in stage k ∈ {0, 1} denoting cold utility exchanges heat with stream i ∈ {0, 1} denoting hot utility exchanges heat with stream j
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HENS: Simultaneous Optimization Constraints
Overall heat balance for each stream: (T iin − T iout)F i =
q ijk + qcui
i ∈ HP
k ∈ST j ∈CP
(T jout − T jin)F j =
q ijk + qhuj j ∈ CP
k ∈ST i∈HP
Heat balance at each stage: (tik − ti,k+1)F i =
q ijk k ∈ S T , i ∈ HP
j ∈CP
(tjk − tj,k+1)F j =
i∈HP
q ijk k ∈ S T , j ∈ CP
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HENS: Simultaneous Optimization Constraints
Assignment of superstructure inlet temperature: T iin = ti1 T jin = tj,N s+1
Feasibility of temperatures: tik ≥ ti,k+1
k ∈ S T , i ∈ HP
tjk ≥ tj,k+1
k ∈ S T , j ∈ CP
T iout ≤ ti,N s+1 i ∈ HP T jout ≥ tj 1
j ∈ CP
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HENS: Simultaneous Optimization Constraints
Hot and cold utility load: (ti,N s+1 − T iout)F i = qcui (T jout − tj1)F j = qhuj
i ∈ HP j ∈ CP
Logical constraints: q ijk − Ωzijk ≤ 0
i ∈ H P , j ∈ C P , k ∈ ST
qcu i − Ωzcui ≤ 0
i ∈ HP
qhu j − Ωzhuj ≤ 0
j ∈ CP
zijk ,zcui,zhuj ∈ {0, 1}
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HENS: Simultaneous Optimization Constraints
Calculation of approach temperatures: dtijk
≤ tik − tjk + Γ(1 − zijk )
dtij,k+1 ≤ ti,k+1 − tj,k+1 + Γ(1 − zijk )
i ∈ H P , j ∈ C P , k ∈ ST i ∈ H P , j ∈ C P , k ∈ ST
dtcui
out + Γ(1 − zcui) i ∈ HP ≤ ti,N s+1 − T CU
dthuj
out ≤ T HU − tj 1 + Γ(1 − zhu j )
j ∈ CP
dtijk
≥ ∆T min
i ∈ H P , j ∈ C P , k ∈ ST
dtcui
≥ ∆T min
i ∈ HP
dthuj
≥ ∆T min
j ∈ CP
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Avoid stream splits (optional)
zijk ≤ 1 j ∈ CP , k ∈ ST
i∈HP j ∈CP
zijk ≤ 1 i ∈ HP , k ∈ ST
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HENS: Simultaneous Optimization Objective: Minimum Utility Cost
MILP Problem: min J = x∈Ω
CCUqcui +
i∈HP
x
≡
CHUqhu j
j ∈CP
zijk ,zcui,zhuj ,
tik , tjk , dtijk ,dtcui,dthuj , q ijk ,qcui,qhuj ; i ∈ HP,j ∈ CP,k ∈ ST
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Ω =
x
(T iin − T iout)F i =
qijk + qcui
k ∈ST j ∈CP
(T jout − T jin)F j
=
qijk + qhuj
k∈ST i∈HP
(tik − ti,k+1 )F i =
qijk
j ∈CP
(tjk − tj,k+1 )F j = T iin T jin
qijk
i∈HP
= t i,1 = t j,1
tik ≥ ti,k+1 tjk ≥ tj,k+1
T iout ≤ ti,N s +1 T jout ≥ tj1
(ti,N s +1 − T iout)F i = qcui (T jout − tj1 )F j = qhuj qijk − Ωzijk ≤ 0 qcui − Ωzcui ≤ 0 qhuj − Ωzhuj ≤ 0 dtijk ≤ tik − tjk + Γ(1 − zijk ) dtij,k+1 ≤ ti,k+1 − tj,k+1 + Γ(1 − zijk ) out + Γ(1 − zcui ) dtcui ≤ ti,N s +1 − T CU out dthuj ≤ T HU − tj1 + Γ(1 − zhu j ) dtijk , dtcui , dthuj ≥ ∆T min
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HENS: Simultaneous Optimization Objective: Min (Utility + Fixed + Area) Cost
MINLP Problem:
min x∈Ω
+
CCUqcui +
i∈HP
j ∈CP
CF ij zijk +
i∈HP j ∈CP k ∈ST
+
Bij
C ij (Aijk )
CF i,CU zcui +
+
i∈HP Bj,HU
C j,HU (Aj,HU )
j ∈CP
i∈HP
i∈HP j ∈CP k ∈ST
+
CHUqhuj
CF j,HU zhuj
j ∈CP Bi,CU
C i,CU (Ai,CU )
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where Aijk
q ijk
=
U ij (dtijk ) (dtij,k+1) 1 U ij
=
1 hi
+ h1j
dtijk +dtij,k+1 2
1/3
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HENS: Simultaneous Optimization Optimal Network Structure
155, 000/yr total cost (71, 400 for utility cost and 83, 600 for capital cost)
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Simultaneous MINLP Model: Example
H1 C1 C2 S1 W1
F Cp (kW/K)
T in (K)
T out (K)
h (kW/m2K)
Cost ($/kW-yr)
22. 20. 7.5 −
440 349 320 500 300
350 430 368 500 320
2.0 2.0
-
.67 1.0 1.0
120 20
−
Min recovery app temp = 1 K Exchanger Cost = 6, 600 + 670(area)0.83
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