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7/26/2019 relaxation technique.pdf http://slidepdf.com/reader/full/relaxation-techniquepdf 1/26 ENERGY RELAXATION NETWORKS By Siti Shawalliah Idris CPE 633 PROCESS ENGINEERING II

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Page 1: relaxation technique.pdf

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ENERGY RELAXATIONNETWORKS

By Siti Shawalliah Idris

CPE 633 PROCESS ENGINEERING II

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Energy relaxation is a name coined for theprocedure of allowing the energy usage toincrease in exchange for at least one of thefollowing effects :

 a reduction in area

A reduction in the heat exchangers

A reduction in complexity (typically less splitting)

Energy Relaxation 

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Energy Relaxation

This procedure has as goal to increase heating utility and inexchange reduce the number of units.

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Energy Relaxation

LOOP: A loop is a circuit through the network that

starts at one exchanger and ends in the same

exchanger

PATH: A path is a circuit through the network that starts at a heater and ends at a cooler

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175 0C CP=0.010 

280 

CP=0.040  65 0C 

155 0C  20 0C CP=0.020 

1 H 

1840 

360 

500  0 40 C 

C2 CP=0.015 1120C 

520 

560 

(*) Heat exchanger loads arein kW 

H2 

125 0C H1 

Energy Relaxation

Illustration of a loop

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125 0C 175 0C FCp=0.010  C 

280 

FCp=0.040 

65 0C 

0 20 C 155 0C FCp=0.020 H 

36

1840 500 FCp=0.015 

0 40 C 

112 0C 560  520 

(*) Heat exchanger loads arein kW 

C2 

H2 

H1 

Energy Relaxation

Illustration of a path

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125 0C 175 0C 

280 

65 0C 

155 0C 20 0C 

1840 

500 360 

0 40

-X  520 +X 560 

If one wants to eliminate one exchanger: X=500. Note that Xcould have beennegative, but we chose the smallest possible inabsolute value.

C2 

+X 

C1 

H2 

-X 

112 0C 

H1 

Energy Relaxation

Procedure: Find a loop and move around the heat from heatexchanger to exchanger until one exchanger is eliminated

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0 45 C 175 0C  C 

65 0C  280 0 65 C 

65 0C 

155 0C  20 0C 1 H 

2340 360 0 112 C  0 40 C 

60  1020 

108 0C 

This exchangeris in 

violation of the

minimum

a roach

C2 

H2 

H1 

Energy Relaxation

Result: Notice that there is an infeasible match!

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0 175 C C 

65 0C  280 

0 65 C 

65 0C 

0 1550C 

20 C H 

2340 360  0 C  0 40 C 

60  1020 

1080C 

Thevalue

of X needed to restore feasibilityis X=795

-X +X 

C2 

-X 

112 

C1 

+X 

H2 

+X 

H1 

ENERGY RELAXATION

Use a path to move heat around to restore feasibility45 C 

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• 

0  0 175 C  45 C C 

65 0C  1075 

65 C 0 

65 0C 

0 155 0C  20 C H 

1545 

112 C 1155 

0  0 40 C 

855  225 

55 0C 

C2 

C1 

H2 

H1 

Final Network

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Thank you for your attention

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Energy Relaxation Technique

Working Example

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Energy relaxation technique

Remove the smallest heat recovery unit from the network by exploiting the degree offreedom in a loop.

 Recalculate the network temperatures and identify any violationsof the Tmin = 10oCconstraint.

 Restore the original Tmin = 10oC throughout the network by exploiting a utility path.

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Maximum Energy Recovery Design

Identify the degrees of freedom for network optimisation

A

A

1

2

4

6

42oC

90o

C

160 oC

B

B

C

C

C

100o

C

140 oC

150 oCH

D

D

HP

H

3

100 oC

80 oC

80 oC

525o

C79 oC

25 oC

75 oC 40 oC

C

450

0

2100

160

00705

0

192

5

166

70

433

0

242

5

Tmin =o

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A

A

1

2

4

6

42oC

90 oC160 oC

B

B

C

C

C

100 oC

140 oC

150o

CH

D

D

H

3

100 oC

80 oC

80o

C

525 oC79 oC

25 oC

75 oC 40 oC

C

450

0

210

0

160

00705

0

192

5

16670

433

0

242

5

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Solution

A

A

1

2

4

6

42oC

90o

C

160 oC

B

B

C

C

C

100o

C

140 oC

150 oCH

D

D

HP

H

3

100 oC

80 oC

80 oC

525o

C

79 oC

25 oC

75 oC 40 oC

C

450

0

2100

160

00705

0

192

5

166

70

433

0

242

5

Loop

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1 A

A

1

2

4

6

42oC

90o

C

160 oC

B

B

C

C

C

100o

C

140 oC

150 oCH

D

D

H

3

100 oC

80 oC

80 oC

525o

C

79 oC

25 oC

75 oC 40 oC

C

450

0

210

0

160

00705

0

192

5

166

70

433

0

242

5

Loop

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A

A

1

2

4

6

42oC

90o

C

160 oC

B

B

C

C

C

100o

C

140 oC

150 oCH

D

D

H

3

100 oC

80 oC

80 oC

525o

C

79 oC

25 oC

75 oC 40 oC

C

450

0

210

0

160

00705

0

192

5

166

70

433

0

242

5

Utility path 1

Utility path 2

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A

A

1

2

4

6

42oC

90o

C

160 oC

B

B

C

C

C

100o

C

140 oC

150 oCH

D

D

HP

H

3

100 oC

80 oC

80 oC525

o

C

79 oC

25 oC

75 oC 40 oC

C

4500

+X 

2100

+X 

160

007050

-X 

1925 -

166

70

433

0

242

5

Loop

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