11 Synthesis of Heat Exchanger Networks.pdf

8/11/2019 11 Synthesis of Heat Exchanger Networks.pdf

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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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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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Heat Flows in Interval k


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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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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)

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Expanded Transshipment Model for Previous Example

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Interval k for Expanded Transshipment Model

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Chen CL 16


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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

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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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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 20


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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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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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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

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Automatic Derivation of NetworkStructures

Superstructure of A 1H2C Example

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Alternatives Embedded in PreviousSuperstructure

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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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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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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)

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S S O


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HENS: Simultaneous OptimizationA Typical Stage-wise Superstructure


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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

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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

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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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HENS Si lt O ti i ti


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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}

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HENS Si lt O ti i ti


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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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=

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

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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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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)

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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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Simultaneous MINLP Model: Same


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Simultaneous MINLP Model: SameExample with No Stream Splitting

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11 Synthesis of Heat Exchanger Networks.pdf