13 0 4 MB
CHE 165 – Plant Design; D. Wagner
Heat Exchange Networks (HENs) Lecture 6 – Friday, September 27, 2019
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Heat Integration Schematics
Simple Heat Exchange Network •
Look at a simple system: T1 t1
Cold Stream
t2
Hot Stream Steam t3
T2 T3 CW
•
How can we determine the optimal values for T2 and t2?
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Simple Heat Exchange Network •
We can plot temperature vs. duty: Qhot t3
Temperature
T1 ∆Tmin
t2
T2 T3 t1 Qcold
Qrec
Duty © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Simple Heat Exchange Network •
The maximum possible heat recovery is when the two curves “pinch” and ΔTmin = 0
TEMPERATURE
Qhot min ∆Tmin= 0
T1
t3 t2
T2 T3 t1 Qcold min
Qrec max
DUTY © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Simple Heat Exchange Network •
What happens as ∆Tmin approaches 0? – Hot utility (steam) consumption is the lowest. – Cold utility (cooling water) consumption is the lowest. – We still need three heat exchangers • • •
1 process-process exchanger. 1 process-hot utility exchanger. 1 process-cold utility exchanger.
•
What is ∆Tlm for the process-process exchanger?
•
How big is the process-process heat exchanger?
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Simple Heat Exchange Network •
We can see that changing ∆Tmin affects – Utility requirements. – Heat exchanger areas
•
How can we find an optimum ∆Tmin? – Design and cost the system for a range of ∆Tmin . • •
Determine capital costs. Determine operating costs.
– Combine capital and operating costs to determine an annualized cost – Plot annualized cost vs. ∆Tmin – Select the minimum © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Simple Heat Exchange Network ∆Tmin OPTIMIZATION 140
6
Cost (10 $/y)
120 100
∆Tmin opt
80 60 40 20 0 0
Utility Costs Total Cost
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
20
∆Tmin
40
60
Annualized Capital Cost
Chemical Engineering Design
Multi-stream Problems: Composite Curves •
How do we handle multiple streams that have temperature overlap?
•
Stream data must be combined in such a way as to represent the total energy sources and total energy demands in each temperature range.
•
The pinch method creates what is called a composite curve.
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves •
Consider a two stage reactor with reheat:
Feed
550°
Rctr #1
510°
A
550°
B
560°
Rctr #2
520° To Next Reactor
Streams A and B have overlapping duties between 520° and 550°.
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves •
Multistage reactor example - stream data Range 1 2
3
T in 510 520 520 520 550
T out 520 550 550 550 560
Streams A A B A+B B
M*Cp 1 1 1 2 1
Q 10 30 30 60 10
Plot T vs. Q for each temperature range.
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves •
Multistage reactor example - composite curve
TEMPERATURE (T)
570 560 550 540 530 520 510 500 0
20
40
60
80
100
Duty (Q)
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves •
There is an easy way to plot the composite curves: just add up the Q values over each range of T
TEMPERATURE (T)
570 560 550 540 530 520 510 500 0
20
40
60
80
100
Duty (Q)
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves for UOP Platforming Process 1000
Temperature (F)
800 600 400 200 0
0
50
100
150
200
250
Duty (MMBtu/h)
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves QH
1000
Temperature (F)
800
• We can set targets for hot and cold utilities using the composites, while paying attention to the process pinch
Pinch
600 400 200 QC 0
0
50
100
150
200
250
Duty (MMBtu/h)
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves QH
1000
Temperature (F)
800
• Since the duty scale is a difference in enthalpy, we can slide the composite curves horizontally, increasing or decreasing ∆Tmin
Pinch
600 400 200 QC 0
0
50
100
150
200
250
Duty (MMBtu/h)
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Composite Curves QH
1000
Temperature (F)
800
• If we decrease ΔTmin then our utility targets are reduced • What is the effect on capital cost though?
Pinch
600 400 200 QC 0
0
50
100
150
200
250
Duty (MMBtu/h)
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Optimization of ΔTmin
•
What happens as ∆Tmin is increased?
– More heat exchangers are required (extra cost) – Log mean temperature differences are greater • •
Each heat exchanger is smaller The cost for each heat exchanger decreases (cost savings)
– More utilities are consumed • • • •
•
Cooling water demand increases Steam demand increases Utility costs increase Note: hot utility increase = cold utility increase
How do we decide on the appropriate ∆Tmin? – Same as the two-stream problem – Plot Total Annualized Cost vs. ∆Tmin for the process © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Optimization of ΔTmin ∆Tmin OPTIMIZATION 140
6
Cost (10 $/y)
120 100
∆Tmin opt
80 60 40 20 0 0
Utility Costs Total Cost
20
∆Tmin
40
60
Annualized Capital Cost
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Energy Costs •
Energy prices are often assumed to be well known •
•
In practice, energy prices are affected by: • • • •
•
See Ch6 & lecture on operating costs
Commodity nature of fuels Fuel mix Flaring of waste products (“fuel value” vs. disposal cost) Capital cost implications of fuel substitution
So the actual energy price varies with time and is seldom properly captured
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
THE CAPITAL – ENERGY TRADEOFF
Heat Exchanger Networks (HENs) design addresses the following problem:
Given:
Design:
NH hot streams, each with given heat-capacity flow rate, Ch supply temperature, Ths, and target temperature Tht NC cold streams, each with given heat-capacity flow rate, Cc supply temperature, Tcs, and target temperature Tct An optimal network of heat exchangers, connecting between the hot and cold streams and between the streams and cold/hot utilities.
What is optimal?
Implies a trade-off between CAPITAL COSTS (Cost of equipment) and ENERGY COSTS (Cost of utilities).
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STREAM REPRESENTATION IN Q-T DIAGRAMS T
Hot stream
T
∆H
∆H
TT
TS ∆H ∆T C = ∆T TT
Cold stream
H
∆T TS
TS = Stream supply temperature (oC) TT = Stream target temperature (oC) H = Stream enthalpy (MW) ⋅ Cp (MW/ oC) C = m = Heat-capacity flow rate (MW/ oC) = Stream flow rate × specific heat capacity
H
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HEX REPRESENTATION IN Q-T DIAGRAMS
60o
∆Tmin = 10o 50o
∆H
10o
Enthalpy
100
60o
∆Tmin = 20o 40o
60
(a)
70o 100o
80
40 0
Temperature
100o
Temperature
100
80o
(b)
80 60 40 0
∆H
Enthalpy
20o
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Heuristics: Heat exchangers and furnaces 24
CHE 165 – Plant Design; D. Wagner
25.
26.
Unless required as part of the design of the separator or reactor, provide necessary heat exchange for heating or cooling process fluid streams, with or without utilities, in an external shell-and-tube heat exchanger using countercurrent flow. However, if a process stream requires heating above 750∘F, use a furnace unless the process fluid is subject to chemical decomposition. Near-optimal minimum temperature approaches in heat exchangers depend on the temperature level as follows: 10∘F or less for temperatures below ambient. 20∘F for temperatures at or above ambient up to 300∘F. 50∘F for high temperatures. 250 to 350∘F in a furnace for flue gas temperature above inlet process fluid temperature.
Adapted from SSLW
HEX REPRESENTATION IN Q-T DIAGRAMS
60o
∆Tmin = 10o 50o
∆H
10o
Enthalpy
100
60o
∆Tmin = 20o 40o
60
(a)
70o 100o
80
40 0
Temperature
100o
Temperature
100
80o
(b)
80 60 40 0
∆H
Enthalpy
20o
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HEX REPRESENTATION IN Q-T DIAGRAMS
60o
100o
∆Tmin = 10o
Temperature
100
80o
∆H
10o
Enthalpy
60o
∆Tmin = 10o
Temperature
100
80o
50o
60 40 0
50o
100o
80
80 60 40 0
∆H
Enthalpy
10o
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HEX REPRESENTATION IN Q-T DIAGRAMS
60o
100o
∆Tmin = 10o
Temperature
100
80o
80 60
∆H
40 0
50o
10o
Enthalpy
80o H
67.5o 100o 50o
70o
C
60o
∆Tmin = 20o
Temperature
100 80
20o
60 QH
40 0
∆H - QH
Enthalpy
QC
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DEFINITIONS
70o
Exchanger Duty. Data: Hot stream CH = 0.3 MW/ oC Cold stream CC = 0.4 MW/ oC
60o
100o 40o
∆Tmin = 20o
Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC Q = 0.4(70 - 40) = 0.3(100 - 60) = 12 MW Heat Transfer Area (A): A = Q/(U⋅∆Tlm) Data: Overall heat transfer coefficient, U=1.7 kW/m2 oC (Alternative formulation in terms of film coefficients) ∆Tlm = (30 - 20)/loge(30/20) = 24.66 So, A = Q/(U⋅∆Tlm) = 12000/(1.7×24.66) = 286.2 m2
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EXAMPLE PROBLEM ∆ H = 160
o
60 C ∆ H = 100 o
80 C 180 oC C-1 R-1 100 oC o
130 C
120 oC
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC. Utilities. Steam@150 oC, CW@25oC Design a network of steam heaters, water coolers and exchangers for the process streams. Where possible, use exchangers in preference to utilities.
∆ H = 162 ∆ H = 180
40 oC
TS Stream (oC) H1 180 H2 130 C1 60 C2 30
30 oC
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POSSIBLE SOLUTION TO EXAMPLE PROBLEM 60 oC
Summary of proposed design:
100
80 oC
180 oC
H
C-1
60
100 oC 130 oC
R-1
120 oC
Steam
CW
Units
60 kW
18 kW
4
Are 60 kW of Steam Necessary?
162
C 40 oC
18 30 oC
30
MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC. (oC)
H2
∆T
H1
Tend (oC)
C2
Tstart (oC)
C1
Interval
CP
(kW/oC)
∆H
(kW)
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MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC.
Cold-1
120
100
20
H2
(oC)
H1
Tend (oC)
C2
Tstart (oC)
C1
∆T
Interval
CP
∆H
(kW/oC)
(kW)
1.8
36
32
MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC.
Cold-1
120
100
Cold-2
100
60
H2
(oC)
H1
Tend (oC)
C2
Tstart (oC)
C1
∆T
Interval
CP
∆H
(kW/oC)
(kW)
20
1.8
36
40
5.8
232
33
MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC.
Cold-1
120
100
Cold-2
100
Cold-3
60
H2
(oC)
H1
Tend (oC)
C2
Tstart (oC)
C1
∆T
Interval
CP
∆H
(kW/oC)
(kW)
20
1.8
36
60
40
5.8
232
30
30
1.8
54
34
MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC.
Cold-1
120
100
Cold-2
100
Cold-3 Hot-1
H2
(oC)
H1
Tend (oC)
C2
Tstart (oC)
C1
∆T
Interval
CP
∆H
(kW/oC)
(kW)
20
1.8
36
60
40
5.8
232
60
30
30
1.8
54
180
130
50
1.0
50 35
MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC.
Cold-1
120
100
Cold-2
100
Cold-3
H2
(oC)
H1
Tend (oC)
C2
Tstart (oC)
C1
∆T
Interval
CP
∆H
(kW/oC)
(kW)
20
1.8
36
60
40
5.8
232
60
30
30
1.8
54
Hot-1
180
130
50
1.0
50
Hot-2
130
80
50
3.0
150
36
MER TARGETING WITH COMPOSITE CURVE Example:
TS Stream (oC) H1 180 130 H2 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10 oC.
Cold-1
120
100
Cold-2
100
Cold-3
H2
(oC)
H1
Tend (oC)
C2
Tstart (oC)
C1
∆T
Interval
CP
∆H
(kW/oC)
(kW)
20
1.8
36
60
40
5.8
232
60
30
30
1.8
54
Hot-1
180
130
50
1.0
50
Hot-2
130
80
50
3.0
150
Hot-3
80
40
40
2.0
80
37
180 160
Temperature (oC)
140 120
Int
Tstart (oC)
Tend (oC)
∆T (oC)
CP (kW/oC)
∆H (kW)
Cold-1
120
100
20
1.8
36
Cold-2
100
60
40
5.8
232
Cold-3
60
30
30
1.8
54
Hot-1
180
130
50
1.0
50
Hot-2
130
80
50
3.0
150
Hot-3
80
40
40
2.0
80
100 80 60 40 20 0
0
50
100
150
200
Enthalpy (kW)
250
300
350
38
180 160
Temperature (oC)
140 120 100 80 60 40 20 0
0
50
100
150
200
Enthalpy (kW)
250
300
350
39
30 kW 180 160
Temperature (oC)
140 120 100
“Pinch”
80 60 40 20 0
0
50
100
150
200
Enthalpy (kW)
250
300
350
40
100 kW
180 160
Temperature (oC)
140
∆Tmin = 30o
120 100 80 60 40 20 0
0
50
100
150
200
Enthalpy (kW)
250
300
350
41
MER Targeting (∆Tmin = 20oC): Hot pinch temperature = 80oC Cold pinch temperature = 60oC QH,min = 54 kW QC,min = 12 kW
54 kW
180 160
Temperature (oC)
140 120
∆Tmin = 20o
100 80 60 40 20 0
12 kW 0
50
100
150
200
Enthalpy (kW)
250
300
350
42
180
MER Targeting (∆Tmin = 10oC): Hot pinch temperature = 70oC Cold pinch temperature = 60oC QH,min = 48 kW QC,min = 6 kW
48 kW
160
Temperature (oC)
140 120
∆Tmin = 10o
100 80 60 40 20 0
6 kW 0
50
100
150
200
Enthalpy (kW)
250
300
350
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MER TARGETING USING THE TI METHOD The temperature-interval method (Linnhoff and Flower, 1978), a systematic procedure for determining the minimum utility requirements over all possible HENs, consists of the following steps: Adjusting the hot and cold stream temperatures to bring them to the same reference. Determining the temperature intervals and carrying out enthalpy balances in each interval. Computing the enthalpy cascade, the residual enthalpy flows, determining the location of the "pinch," and computing MER targets. 44
MER TARGETING WITH TI : EXAMPLE PROBLEM ∆ H = 160
o
60 C ∆ H = 100 o
80 C 180 oC C-1 R-1 100 oC 130 oC
120 oC
TS Stream (oC) H1 180 H2 130 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10oC. Utilities. Steam@150oC, CW@25oC
∆ H = 162
Compute the MER targets for this process using the TI Method.
∆ H = 180
40 oC
30 oC
45
MER TARGETING WITH TI : STEP 1 Adjusting the hot and cold stream temperatures to bring them to the same reference. o Need to adjust temperatures so that both hot and cold streams are on the same terms of reference for a given desired ∆Tmin. Arbitrarily, this is accomplished by subtracting ∆Tmin from all hot stream temperatures.
Original stream table TS Stream (oC) H1 180 H2 130 C1 60 C2 30
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
Adjusted stream table TS Stream (oC) H1 170 H2 120 C1 60 C2 30
TT (oC) 70 30 100 120 46
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
1 2 3 4 5
47
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
1
170 – 120 = 50
2
120 – 100 = 20
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
TT (oC) 70 30 100 120
C2
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
48
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
1
170 – 120 = 50
2
120 – 100 = 20
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
49
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
1
170 – 120 = 50
2
120 – 100 = 20
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
50
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
1
170 – 120 = 50
2
120 – 100 = 20
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
51
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
1
170 – 120 = 50
2
120 – 100 = 20
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
52
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
1
170 – 120 = 50
2
120 – 100 = 20
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC) 1
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW) 50
53
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
1
170 – 120 = 50
1
50
2
120 – 100 = 20
1 + 2 –1.8 = 1.2
24
3
100 – 70 = 30
4
70 – 60 = 10
5
60 – 30 = 30
54
MER TARGETING WITH TI : STEP 2 Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval
Ti-1 –Ti (oC)
H1 H2 C1
TS Stream (oC) H1 170 H2 120 C1 60 C2 30
C2
TT (oC) 70 30 100 120
∆H (kW) 100 180 160 162
ΣCh –ΣCc (kW/oC)
C (kW/oC) 1.0 2.0 4.0 1.8
∆H (kW)
1
170 – 120 = 50
1
50
2
120 – 100 = 20
1 + 2 –1.8 = 1.2
24
3
100 – 70 = 30
1 + 2 –1.8 –4 = –2.8
–84
4
70 – 60 = 10
2 –1.8 –4 = –3.8
–38
5
60 – 30 = 30
2 –1.8 = 0.2
6
55
MER TARGETING WITH TI : STEP 3 Energy Flows Between Computing the enthalpy cascade, the residual enthalpy flows, determining the location of the "pinch," and computing MER targets.
Intervals (kW)
First Pass T0 = 170 oC
QH
R0 = 0
50 R1 = 50
T1 = 120 oC 24
R2 =74
T2 = 100 oC -84
R3 =-10
T3 = 70 oC -38
R4 =-48
T4 = 60 oC 6
T5 = 30 oC
QC
56 R5 =-42
MER TARGETING WITH TI : STEP 3 Energy Flows Between Computing the enthalpy cascade, the residual enthalpy flows, determining the location of the "pinch," and computing MER targets.
Intervals (kW)
T0 = 170 oC
QH
First Pass
Final Pass
R0 = 0
48
R1 = 50
98
R2 =74
122
50 T1 = 120 oC 24 T2 = 100 oC -84 R3 =-10
T3 = 70 oC
MER Targeting: Cold pinch temp. = 60oC
QHmin
38
-38 R4 =-48
T4 = 60 oC
Hot pinch temp. = 70oC
No heat transfers across the pinch
0
6
57
QH,min = 48 kW QC,min = 6 kW
T5 = 30 oC
QC
R5 =-42
6
QCmin
PINCH
DECOMPOSITION
58
THE SIGNIFICANCE OF THE PINCH QHmin T
QHmin Heat sink Pinch
Pinch Heat source
QCmin
QCmin
The “pinch” separates the HEN problem into two parts:
Heat sink - above the pinch, where at least QHmin utility must be used. Heat source - below the pinch, where at least QCmin utility must be used. At MER no heat passes across the pinch.
H
59
HEN DESIGN RULES FOR MER MER Targeting. Define pinch temperatures, QHmin and QCmin. Divide problem at the pinch. Design hot-end, starting at the pinch: Immediately above the pinch, pair up streams such that: CH ≤ CC. “Tick off” streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch. Design cold-end, starting at the pinch: Immediately below the pinch, pair up streams such that: CH ≥ CC. “Tick off” streams to minimize costs. Add cooling utilities as needed (up to QCmin). Do not use hot utilities below the pinch. Done!
60
EXAMPLE MER SOLUTION: PROBLEM STATEMENT ∆ H = 160
o
60 C ∆ H = 100 o
80 C 180 oC C-1 R-1 100 oC o
130 C
120 oC
TT (oC) 80 40 100 120
∆H (kW) 100 180 160 162
C (kW/oC) 1.0 2.0 4.0 1.8
∆Tmin = 10oC. Utilities. Steam@150oC, CW@25oC Design a network of steam heaters, water coolers and exchangers for the process streams. Where possible, use exchangers in preference to utilities.
∆ H = 162 ∆ H = 180
40 oC
TS Stream (oC) H1 180 H2 130 C1 60 C2 30
30 oC
61
EXAMPLE MER SOLUTION – STEP 1 MER Targeting.
H1
H2
180oC
80oC
40oC
130oC
100oC
60oC
o
30oC
120 C
C (kW/ oC) 1.0
2.0
C1
4.0
C2
1.8
62
EXAMPLE MER SOLUTION – STEP 1 MER Targeting. For ∆Tmin = 10oC: Pinch temperatures 70oC and 60oC, QH,min = 48 kW and QC,min = 6 kW
H1
H2
180oC
80oC
40oC
130oC
100oC
60oC
o
30oC
120 C
QH,min = 48 kW
C (kW/ oC) 1.0
2.0
C1
4.0
C2
1.8
QC,min = 6 kW
63
EXAMPLE MER SOLUTION – STEP 2 Divide problem at the pinch.
H1
H2
180oC
130oC
100oC
o
120 C
QH,min = 48 kW
C (kW/ oC)
80oC
1.0
70oC
40oC
60oC
C1
4.0
C2
1.8
60oC
30oC
QC,min = 6 kW
2.0
64
On hot @ pinch: EXAMPLE MER SOLUTION – side STEP 3 Design hot-end, starting at the pinch: Immediately above Pairing rule is ≤ min CC. “Tick off” the pinch, pair CH ≤up CCstreams such Th,out –that: Tc,in C=H∆T streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.
pinch
H1
H2
80oC
180oC
130oC
1
100oC
o
120 C
QH,min = 48 kW
C (kW/ oC)
1
1.0
70oC
40oC
60oC
C1
4.0
C2
1.8
60oC
30oC
QC,min = 6 kW
2.0
65
On hot @ pinch: EXAMPLE MER SOLUTION – side STEP 3 Design hot-end, starting at the pinch: Immediately above Pairing rule is ≤ min CC. “Tick off” the pinch, pair CH ≤up CCstreams such Th,out –that: Tc,in C=H∆T streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.
pinch
H1
H2
80oC
180oC
130oC
100oC
C (kW/ oC)
1
1
1.0
70oC
40oC
60oC
C1
4.0
C2
1.8
2.0
120 o
120 C
QH,min = 48 kW
60oC
30oC
QC,min = 6 kW
66
On hot @ pinch: EXAMPLE MER SOLUTION – side STEP 3 Design hot-end, starting at the pinch: Immediately above Pairing rule is ≤ min CC. “Tick off” the pinch, pair CH ≤up CCstreams such Th,out –that: Tc,in C=H∆T streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.
pinch
H1
H2
180oC
80oC
2
130oC
C (kW/ oC)
100oC
1
1
1.0
70oC
40oC
60oC
C1
4.0
C2
1.8
2.0
120 o
120 C
QH,min = 48 kW
2
100
60oC
30oC
QC,min = 6 kW
67
EXAMPLE MER SOLUTION – STEP 3 Design hot-end, starting at the pinch: Immediately above the pinch, pair up streams such that: CH ≤ CC. “Tick off” streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch. H1
H2
180oC
100oC
o
120 C
80oC
2
130oC
1
H
1
40
120
H
2
8
100
QH,min = 48 kW
C (kW/ oC) 1.0
70oC
40oC
60oC
C1
4.0
C2
1.8
60oC
30oC
QC,min = 6 kW
2.0
68
EXAMPLE MER SOLUTION – Sside TEP@4pinch: On cold Design cold-end, starting at the pinch: Immediately below up streams such Pairing that: Crule the pinch,Tpair isC. “Tick off” H ≥ C h,in – Tc,out = ∆Tmin streams to minimize costs. Add cooling as needed CH ≥ utilities CC hot utilities below the pinch. (up to QCmin). Do not use pinch H1
H2
180oC
100oC
o
120 C
80oC
2
130oC
1
H
1
40
120
H
2
8
100
QH,min = 48 kW
C (kW/ oC) 1.0
70oC
40oC
3
60oC
60oC
3
54
30oC
2.0
C1
4.0
C2
1.8
QC,min = 6 kW
69
EXAMPLE MER SOLUTION – STEP 4 Design cold-end, starting at the pinch: Immediately below the pinch, pair up streams such that: CH ≥ CC. “Tick off” streams to minimize costs. Add cooling utilities as needed (up to QCmin). Do not use hot utilities below the pinch. H1
H2
180oC
80oC
2
130oC
C (kW/ oC)
1
1.0
70oC
3
C
40oC
2.0
6 100oC
o
120 C
H
1
40
120
H
2
8
100
QH,min = 48 kW
60oC
60oC
3
54
30oC
C1
4.0
C2
1.8
QC,min = 6 kW
70
PREVIOUS SOLUTION TO EXAMPLE PROBLEM 60 oC
Summary of previous design:
100
80 oC
180 oC
H
C-1
60
100 oC 130 oC
R-1
120 oC
Steam
CW
Units
60 kW
18 kW
4
Are 60 kW of Steam Necessary?
162
C 40 oC
18 30 oC
71
MER SOLUTION TO EXAMPLE PROBLEM 60 oC
Summary of MER design:
100
80 oC
180 oC
H
C-1
60
100 oC 130 oC
R-1
Steam
CW
Units
48 kW
6 kW
6
120 oC
162
C 40 oC
18 30 oC
72
MER SOLUTION TO EXAMPLE PROBLEM 60 oC 100
80 oC
180 oC
120
C-1 R-1
H
Summary of MER design: Steam
CW
Units
48 kW
6 kW
6
40
100 oC
120 oC
o
130 C
H
8
54
C 6
40 oC
73
30 oC
Capital Targets •
How do we get the capital cost without designing the heat exchange network (HEN)?
Temperature
Qhot
Qcold
Qrec Duty
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Thermodynamic Significance of the Pinch “pinch” Temperature
Qhot min
Qrec max Qcold min Duty
•
When the process is pinched it is decomposed into two sub problems
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Pinch Decomposition “pinch” Temperature
Qhot min
Qrec max Qcold min
Above the pinch we only put in utility heat and the process acts as a heat sink
Below the pinch we only reject heat to cold utility and the process acts as a heat source
Duty
•
When the process is pinched it is decomposed into two sub problems
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Pinch Decomposition •
What if we put in extra heat above the pinch? Qextra “pinch” Temperature
Qhot min
Qextra
Heat sink is now out of energy balance and we have to reject Qextra to a lower temperature
Qrec max Qcold min Duty © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Pinch Decomposition •
What if we put in extra heat above the pinch? Qextra “pinch” Temperature
Qhot min
Qextra
Qextra Qrec max Qcold min
Now the heat source is also out of energy balance and we have to reject Qextra to cold utility
Duty © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Pinch Decomposition The overall effect is that both hot and cold utility are increased by the amount of heat transferred across the pinch = Qextra Qextra “pinch” Temperature
Qhot min
Qextra
Qextra Qcold min
Qrec max Duty
So a simple rule for achieving energy targets is don’t transfer heat across the pinch! © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Pinch Design Method •
Use composite curves to find the pinch temperature
•
Optimize ∆Tmin, making process changes
•
Decompose the network design problem using the pinch temperatures
•
Design the Heat Exchanger Network (HEN) for each subsystem separately: – Transfer no heat “across” the pinch – ∆T > ∆Tmin for all heat exchangers – Start at the pinch and work outward
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design
Pinch Analysis: Overall Conclusions • The pinch techniques can give useful insights into process energy efficiency • The main value lies in identifying good process modifications and choosing the right utilities • It is seldom (if ever) necessary to work through all of the details of the analysis – most benefit is gained with a high-level view • Different techniques are used for retrofit of existing networks and these can have more value, particularly in petrochemicals
© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy
Chemical Engineering Design