chen 4460 – process synthesis, simulation and optimization
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Heat and Power Integration. CHEN 4460 – Process Synthesis, Simulation and Optimization Dr. Mario Richard Eden Department of Chemical Engineering Auburn University Lab Lecture No. 6 – Heat and Power Integration November 6, 2012. Example 1. - PowerPoint PPT PresentationTRANSCRIPT
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