part 6 synthesis of heat exchanger networks. 6.1 sequential synthesis minimum utility cost

Part 6

Synthesis of Heat Exchanger Networks

6.1 Sequential Synthesis

Minimum Utility Cost

Example 1

Fcp

(MW/C)

Tin

(C)

Tout

(C)

H1 1 400 120

H2 2 340 120

C1 1.5 160 400

C2 1.3 100 250

Steam: 500 CCooling water: 20 – 30 CMinimum recovery approach temperature (HRAT): 20 C

Heat Balances around Temperature Intervals

1

2 1

3 2

3

30

90 60

357 480

78 180

s

w

R Q

R R

R R

Q R

Transshipment Model

1

2 1

3 2

3

1 2 3

1 2 3

min

. .

30

30

123

102

, , , , 0

60 , 225

30 , 0 , 123

Pinch Point: 340 - 320

s w

s

w

s w

s w

Z Q Q

s t

R Q

R R

R R

Q R

Q Q R R R

Q MW Q MW

R MW R MW R MW

C

Remarks

• LP for minimum utility consumption leads to the same results as the Problem Table in Pinch method.

• The transshipment model can be generalized to consider multiple utilities to minimize total utility cost.

• This model can be expanded so as to handle constraints on matches.

• This model can also be expanded so as to predict the matches for minimizing the number of units.

• We can embed the equations of the transshipment model within an optimization model for synthesizing a process system where the flows of the process streams are unknown.

Index Sets

| hot stream supplies heat to interval

| cold stream demands heat from interval

| hot utility supplies heat to interval

| cold utility extracts heat from interval

k

k

k

k

H i i k

C j j k

S m m k

W n n k

Condensed Transshipment Model

1

0

min

. .

1,2, ,

0

0, 1,2, , 1

, 0

k k k k

S Wm m n n

m S n W

S W H Ck k m n ik jk

m S n W i H j C

K

k

S Wm n

Z c Q c Q

s t

R R Q Q Q Q

k K

R R

R k K

Q Q

Example 2

FCp

(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: 500 K, $80/kW-yrLP Steam: 380 K, $50/kW-yrCooling Water: 300 K, $20/kW-yrHRAT: 10K

1

2 1

3 2

3

1 2 3

1 2 3

min 80000 50000 20000

. .

60

10

15

75

, , , , , 0

6,550,000$ / yr

60 , 5 , 75

0, 10 , 0

HP LP CW

HP

LP

CW

HP LP CW

HP LP CW

Z Q Q Q

s t

R Q

R R

R R Q

R Q

R R R Q Q Q

Z

Q MW Q MW Q MW

R R MW R

Sequential Synthesis

Minimum Utility Cost with Constrained Matches

Basic Ideas

(1) Instead of assigning a single overall heat residual

exiting at each interval , we will assign individual

heat residuals and for each hot stream and each

hot utility t

k

ik mk

R

k

R R i

m hat are present at or above that interval .

(2) Within interval we will define the variable to

denote the heat exchange between hot stream and

cold stream .

ijk

k

k Q

i

j

Heat Exchange Options

• Hot stream i and cold stream j are present in interval k (see figure in the previous page).

• Cold stream j is present in interval k, but hot stream i is only present at higher temperature interval (see figure in the next page).

Index Sets

| hot stream is present at interval or at a higher interval

| hot utility is present at interval or at a higher interval

k

k

H i i k

S m m k

Expanded Transshipment Model

, 1

, 1

0 0

min

. .

A:

B: 0

C:

D: 0

1, 2, ,

k k

k

k k

k

S Wm m n

m S n W

Hik i k ijk ink ik k

j C n W

Smk m k mjk m k

j C

Cijk mjk jk k

i H m S

Wink n k

i H

i m

Z c Q cQ

s t

R R Q Q Q i H

R R Q Q m S

Q Q Q j C

Q Q n W

k K

R R

0

, , , , , , 0

iK mK

S Wik mk ijk mjk ink m n

R R

R R Q Q Q Q Q

Match Constraints

1

(1) Forbid match match between hot and cold

0 1,2, ,

(2) Impose a given match

ijk

KL Uij ijk ij

k

i j

Q k K

Q Q Q

Example 1

Fcp

(MW/C)

Tin

(C)

Tout

(C)

H1 1 400 120

H2 2 340 120

C1 1.5 160 400

C2 1.3 100 250

Steam: 500 C, $80/kW-yrCooling water: 20 – 30 C, $20/kW-yrMinimum recovery approach temperature (HRAT): 20 CThe match between H1 and C1 is forbidden.

Condensed Transshipment Model

1

2 1

3 2

3

1 2 3

1 2 3

min

. .

30

30

123

102

, , , , 0

60 , 225

30 , 0 , 123

Pinch Point: 340 - 320

s w

s

w

s w

s w

Z Q Q

s t

R Q

R R

R R

Q R

Q Q R R R

Q MW Q MW

R MW R MW R MW

C

The annual utility cost: $9,300,000.

Expanded Transshipment Model

,1 ,1,1

,1,1

1,2 1,1,2

,2 ,1 ,1,2

,1,2 1,1,2

1,3

Utility Cost: min 80000 20000

Interval 1: 0

30

Interval 2: 60

0

90

Interval 3:

S W

S S S

S

S S S

S

Z Q Q

R Q Q

Q

R Q

R R Q

Q Q

R

1,2 1,1,3 1,2,3

2,3 2,1,3 2,2,3

,3 ,2 ,1,3 ,2,3

1,1,3 2,1,3 ,2,3

1,2,3 2,2,3 ,2,3

160

320

0

240

117

Interval 4: -

S S S S

S

S

R Q Q

R Q Q

R R Q Q

Q Q Q

Q Q Q

1,3 1,2,4 1, ,4

2,3 2,2,4 2, ,4

,3 ,2,4

1,2,4 2,2,4 ,2,4

1, ,4 2, ,4

1,1,2 1,1,3

60

120

- 0

78

0

Forbidden Match:

W

W

S S

S

W W W

R Q Q

R Q Q

R Q

Q Q Q

Q Q Q

Q Q

0

Annual Utility Cost: $15,300,000Heating Utility Load: 120 MWCooling Utility Load: 285 MW

Sequential Synthesis

Prediction of matches for minimizing the unit number

Objective FunctionLet us partition the HEN into several subnetworks

according to the pinch points. Each subnetwork

has an associated set of temperature intervals.

The objective for the subnetwork can be written as

q

q

K

min

where

1 hot stream exchange heat with cold stream

0 otherwise

qij

i H j C

qij

y

i jy

Heat Balances

The constraints in the expanded transshipment model can be modified for the present model:

1. The heat contents of the utility streams are given.

2. The common index i can be used for hot process and utility streams; The common index j can be used for cold process and utility streams.

Heat Balances

, , 1 , , ,

, , ,

, , ,

, 0

k

k

Hi k i k i j k i k k

j C

Ci j k j k k

i H

i k i j k

R R Q Q i H

Q Q j C

R Q

Logical Constraints

, , , ,1

,

where the upper bound is the smaller of the

heat contents of stream and stream .

qKq

i j k i j i jk

i j

Q U y

U

i j

Solution

This MILP problem can be solved independently for each subnetwork

or simultaneously over all subnetworks.

The following results can be obtained:

(1) Matches that take place: 1

(2) Heat exchanged at

qijy

1

each match: qK

ijkk

Q

Example 1

Fcp

(MW/C)

Tin

(C)

Tout

(C)

H1 1 400 120

H2 2 340 120

C1 1.5 160 400

C2 1.3 100 250

Steam: 500 CCooling water: 20 – 30 CMinimum recovery approach temperature (HRAT): 20 C

Condensed Transshipment Model

1

2 1

3 2

3

1 2 3

1 2 3

min

. .

30

30

123

102

, , , , 0

60 , 225

30 , 0 , 123

Pinch Point: 340 - 320

s w

s

w

s w

s w

Z Q Q

s t

R Q

R R

R R

Q R

Q Q R R R

Q MW Q MW

R MW R MW R MW

C

MILP (i),1 1,1 1,1 1,2 1,

2,1 2,2 2,

,1 ,1,1

,1,1

1,2 1,1,2

Number of units: min

Interval 1: 60

30

Interval 2: 60

A A B B BS W

B B BW

S S

S

Z y y y y y

y y y

R Q

Q

R Q

,2 ,1 ,1,2

,1,2 1,1,2

1,3 1,2 1,1,3 1,2,3

2,3 2,1,3 2,2,3

1,1,3 2,1,3 ,2,3

0

90

Interval 3: 160

320

240

S S S

S

S

R R Q

Q Q

R R Q Q

R Q Q

Q Q Q

1,2,3 2,2,3 ,2,3

1,3 1,2,4 1, ,4

2,3 2,2,4 2, ,4

1,2,4 2,2,4 ,2,4

1, ,4 2, ,4

117

Interval 4: - 60

120

78

225

S

W

W

S

W W

Q Q Q

R Q Q

R Q Q

Q Q Q

Q Q

MILP (ii)

,1,1 ,1,2 ,1

1,1,2 1,1

1,1,3 1,1

1,2,3 1,2,4 1,2

1, ,4 1,

2,1,3 2,1

2,2,3 2,2,4 2,2

2, ,4 2,

Matches above the pinch:

60

60

Matches below the pinch:

220

195

220

240

60

225

AS S S

A

B

B

BW W

B

B

BW W

Q Q y

Q y

Q y

Q Q y

Q y

Q y

Q Q y

Q y

Solution

,1 ,1,1 ,1,2

1,1 1,1,2

1,1 1,1,3

1,2 1,2,3 1,2,4

2,1 2,1,3

2, 2, ,4

1, 30, 30

1, 60

1, 25

1, 117, 78

1, 215

1, 225

AS S S

A

B

B

B

BW W

y Q Q

y Q

y Q

y Q Q

y Q

y Q

Alternative Solution

1,1set 0By

Solve MILP without Partition

,1,1 ,1,2 ,1

1,1,2 1,1,3 1,1

1,2,3 1,2,4 1,2

1, ,4 1,

2,1,3 2,1

2,2,3 2,2,4 2,2

2, ,4 2,

60

220

195

220

240

60

225

S S S

W W

W W

Q Q y

Q Q y

Q Q y

Q y

Q y

Q Q y

Q y

Only 5 units! One less than the previous two!

Sequential Synthesis

Automatic Generation of Network Structures

Basic Ideas

• Each exchanger in the superstructure corresponds to a match predicted by the MILP model (with or without pinch partition). Each exchanger will also have as heat load the one predicted by MILP.

• The superstructure will contain those stream interconnections among the units that can potentially define all configurations. The stream interconnections will be treated as unknowns that must be determined.

Superstructure for one hot stream and two cold streams

Embedded Alternative Configurations

• H1-C1 and H1-C2 in series

• H1-C2 and H1-C1 in series

• H1-C1 and H1-C2 in parallel

• H1-C1 and H1-C2 in parallel with bypass to H1-C2

• H1-C1 and H1-C2 in parallel with bypass to H1-C1

Parameters and Unknowns

1 1 1

2 2 2

11 12

1 8

3 4 56 78

Parameters---

For stream H1: , ,

For stream C1: , ,

For stream C2: , ,

Heat loads: ,

Unknowns---

, , ,

in out

in out

in out

F T T

f t t

f t t

Q Q

F F

T T T T

Objective Function

1 11 2 12

11 121 2

11 1 12 2

1/32 1

2 1 2 12

1

11 121 21/3

1 1 1 1 2 2 2 211 2 1 2 1 12 2 1 2 1

min

Since

/ 2ln

min/ 2 /

i ii i i i

i i

i

C c A c A

Q Qc cU LMTD U LMTD

LMTD

Q QC c c

U U

1/3

2

Equality Constraints

1 2

1 8 3

1 8 78 3 3

2 6 4

2 6 56 4 4

1. Mass balance for initial splitter

2. Mass and heat balances for inlet mixers

0

0

- 0

0

3. Mass balances for outlet splitters

in

in

F F F

F F F

FT FT FT

F F F

F T F T F T

3 6 5

4 7 8

11 3 3 56

12 4 4 78

0

0

4. Heat balances of exchangers

0

0

F F F

F F F

Q F T T

Q F T T

Inequality Constraints1 1 2 21 2 1 2 min

1 8

11 3 1

12 56 1

21 4 2

22 78 2

, , ,

, , 0

whereout

in

out

in

T

F F

T t

T t

T t

T t

Example 3

Stream Tin

(K)

Tout

(K)

Fcp

(kW/K)

h

(kW/m^2K)

Cost

($/kW-yr)

H1 440 350 22 2.0 -

C1 349 430 20 2.0 -

C2 320 368 7.5 0.67 -

S1 500 500 - 1.0 120

W1 300 320 - 1.0 20

Minimum temperature approach = 1 KExchanger cost = 6600+670(Area)^0.83

Solution