CHEN 4460 – Process Synthesis, Simulation and Optimization

Dr. Mario Richard EdenDepartment of Chemical Engineering

Auburn University

Lab Lecture No. 6 – Heat and Power Integration

November 6, 2012

Heat and Power Integration

Example 1

• Given the stream information below, design the MER network

Stream Ts Tt Cp [kW/oC]

1 180 40 2

2 150 40 4

3 60 180 3

4 30 105 2.6

Tmin = 10 oC

Example 1 – Solution

Tmin = 10 oCStream Ts Tt Cp [kW/oC]

1 180 40 2

2 150 40 4

3 60 180 3

4 30 105 2.6

Temperature Driving Force

TH-ΔT

Example 1 – Solution

Tmin = 10 oCStream Ts Tt Cp [kW/oC]

1 170 30 2

2 140 30 4

3 60 180 3

4 30 105 2.6

Temperature Driving Force

TH-ΔT

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 180

2 170

3 140

4 105

5 60

6 30

Example 1 – Solution

Tmin = 10 oCStream Ts Tt Cp [kW/oC]

1 170 30 2

2 140 30 4

3 60 180 3

4 30 105 2.6

Temperature Driving Force

TH-ΔT

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 180 10

2 170 30

3 140 35

4 105 45

5 60 30

6 30

Example 1 – Solution

Tmin = 10 oCStream Ts Tt Cp [kW/oC]

1 170 30 2

2 140 30 4

3 60 180 3

4 30 105 2.6

Temperature Driving Force

TH-ΔT

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 180 10 -3

2 170 30 2-3=-1

3 140 35 2+4-3=3

4 105 45 2+4-3-2.6=0.4

5 60 30 2+4-2.6=3.4

6 30

Example 1 – Solution

Tmin = 10 oCStream Ts Tt Cp [kW/oC]

1 170 30 2

2 140 30 4

3 60 180 3

4 30 105 2.6

Temperature Driving Force

TH-ΔT

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 180 10 -3 -30

2 170 30 2-3=-1 -30

3 140 35 2+4-3=3 105

4 105 45 2+4-3-2.6=0.4 18

5 60 30 2+4-2.6=3.4 102

6 30

Example 1 – Solution

• Enthalpy Cascade Diagram

H=-30

H=-30

H=105

H=18

T1 =180

T2 =170

T3 =140

T4 =105

T5 =60

QH=0

QC=165

H=102

T6 =30

Q1=-30

Q2=-60

Q3=45

Q4=63

QH=60

QC=225

Q1=30

Q2=0

Q3=105

Q4=123

T Cold pinch : 140QH min : 60QC min : 225

Example 1 – Solution

• HEN Grid Diagram

1

2

3

4

180oC

150oC

180oC

105oC 30oC

60oC

40oC

40oC 2

4

3

2.6

Cp

140oC

140oC

150oC

150oC

140oC

140oC

150oC

150oC

T Cold pinch : 140QH min : 60QC min : 225

Example 1 – Solution

• HEN Grid Diagram

T Cold pinch : 140QH min : 60QC min : 225

1

2

3

4

180oC

150oC

180oC

105oC 30oC

60oC

40oC

40oC 2

4

3

2.6

Cp

140oC

140oC

150oC

150oC

140oC

105oC

150oC

150oC

Example 1 – Solution

• HEN Grid Diagram

T Cold pinch : 140QH min : 60QC min : 225

1

2

3

4

180oC

150oC

180oC

105oC 30oC

60oC

40oC

40oC 2

4

3

2.6

Cp

140oC

140oC

150oC

150oC

140oC

105oC

150oC

150oC

60 kW

160oCH

60 kW

Example 1 – Solution

• HEN Grid Diagram

T Cold pinch : 140QH min : 60QC min : 225

1

2

3

4

180oC

150oC

180oC

105oC 30oC

60oC

40oC

40oC 2

4

3

2.6

Cp

140oC

140oC

150oC

150oC

140oC

105oC

150oC

150oC

60 kW

160oC60 kW

240 kW

195 kW

C200 kW

H

90oC

52.5oCC

25 kW

Example 2

• Given the stream information below, design the MER network

Stream Ts Tt Cp [kW/oC]

1 180 60 3

2 150 30 1

3 30 135 2

4 80 140 5

Tmin = 10 oC

Example 2 – Solution

Tmin = 10 oC

Temperature Driving Force

TH-ΔT

Stream Ts Tt Cp [kW/oC]

1 180 60 3

2 150 30 1

3 30 135 2

4 80 140 5

Example 2 – Solution

Stream Ts Tt Cp [kW/oC]

1 170 50 3

2 140 20 1

3 30 135 2

4 80 140 5

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 170

2 140

3 135

4 80

5 50

6 30

7 20

Tmin = 10 oC

Temperature Driving Force

TH-ΔT

Example 2 – Solution

Stream Ts Tt Cp [kW/oC]

1 170 50 3

2 140 20 1

3 30 135 2

4 80 140 5

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 170 30

2 140 5

3 135 55

4 80 30

5 50 20

6 30 10

7 20

Tmin = 10 oC

Temperature Driving Force

TH-ΔT

Example 2 – Solution

Stream Ts Tt Cp [kW/oC]

1 170 50 3

2 140 20 1

3 30 135 2

4 80 140 5

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 170 30 3

2 140 5 3+1-5=-1

3 135 55 3+1-2-5=-3

4 80 30 3+1-2=2

5 50 20 1-2=-1

6 30 10 1

7 20

Tmin = 10 oC

Temperature Driving Force

TH-ΔT

Example 2 – Solution

Stream Ts Tt Cp [kW/oC]

1 170 50 3

2 140 20 1

3 30 135 2

4 80 140 5

Interval Ti Ti-Ti+1 CpH- CpC ΔHi

1 170 30 3 90

2 140 5 3+1-5=-1 -5

3 135 55 3+1-2-5=-3 -165

4 80 30 3+1-2=2 60

5 50 20 1-2=-1 -20

6 30 10 1 10

7 20

Tmin = 10 oC

Temperature Driving Force

TH-ΔT

Example 2 – Solution

• Temperature Interval Diagram

T t

H1 CP = 3

H2 CP = 1

C1 CP = 2

C2 CP = 5

180 170

150 140

145 135

90 80

60 50

40 30

30 20

Interval CP H

1: 30o

2: 5o

3: 55o

4: 30o

5: 20o

6: 10o

3 90

1+3-5=-1 -5

1+3-2-5=-3 -165

1+3-2=2 60

1-2=-1 -20

1 10

Example 2 – Solution

• Enthalpy Cascade Diagram

T Cold pinch : 80QH min : 80QC min : 50

H=-5

H=-165

H=60

T1 =170

T2 =140

T3 =135

T4 =80

T5 =50

QH=0

Q5=-40

H=-20

T6 =30

T4 =80

Q1=90

Q2=85

Q3=-80

Q4=-20

H=90

H=10

T5 =50 Q4=-20

QC=-30T7 =20

Q3=-80

QH=80

Q5=40

Q1=170

Q2=165

Q3=0

Q4=60Q4=

QC=50

Example 2 – Solution

• HEN Grid Diagram

T Cold pinch : 80QH min : 80QC min : 50

1

2

3

4

180oC

150oC

135oC

140oC 80oC

30oC

30oC

60oC 3

1

2

5

Cp

80oC

80oC

90oC

90oC

80oC

80oC

90oC

90oC

Example 2 – Solution

• HEN Grid Diagram

T Cold pinch : 80QH min : 80QC min : 50

1

2

3

4

80oC

80oC

90oC

90oC

80oC

80oC

90oC

90oC

80oC

30oC

30oC

60oC 3

1

2

5

Cp

180oC

150oC

135oC

140oC

Example 2 – Solution

• HEN Grid Diagram

T Cold pinch : 80QH min : 80QC min : 50

1

2

3

4

270 kW

134oCH

30 kW

80oC

80oC

90oC

90oC

80oC

80oC

90oC

90oC

80oC

30oC

30oC

60oC 3

1

2

5

Cp

60 kW

110oCH

50 kW

180oC

150oC

135oC

140oC

Example 2 – Solution

• HEN Grid Diagram

T Cold pinch : 80QH min : 80QC min : 50

1

2

3

4

270 kW

134oCH

30 kW

80oC

80oC

90oC

90oC

80oC

80oC

90oC

90oC

80oC

30oC

30oC

60oC 3

1

2

5

Cp

60 kW

110oCH

50 kW

90 kW

35oC

10 kW80oC C

50 kW

180oC

150oC

135oC

140oC