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Design Method for Heat ExchangerNetworks
2006
DISEÑO DE LA RED DEINTERCAMBIO
©2000 AspenTech. All Rights Reserved.
2006
ROAD MAP
PROJECTS
SYNTHESIS
ANALYSIS
Project Selectionfor Implementation DETAILDESIGNENVIRONMENT
PROCE
SS SIMULATION ENVIRONMENTData Extraction
Network Design
ProcessImprovement Road Map
GenerateEvaluateSelect Options
THREE DAYCOURSE
I know what is thepotential for energyrecovery in my process.But, how should I designa process that achievesthe energy target ?
Energy Rec
Process Mod
¿Cómo consigo losobjetivos de mínimo
consumo de energía?
2©2000 AspenTech. All Rights Reserved.
ROAD MAP
PROJECTS
SYNTHESIS
ANALYSIS
Project Selectionfor Implementation DETAILDESIGNENVIRONMENT
PROCE
SS SIMULATION ENVIRONMENTData Extraction
Network Design
ProcessImprovement Road Map
GenerateEvaluateSelect Options
THREE DAYCOURSE
I know what is thepotential for energyrecovery in my process.But, how should I designa process that achievesthe energy target ?
Energy Rec
Process Mod
Our Example Problem
REACTOR
FEEDSTEAM
STEAMRECYCLE
PRODUCT
DistillationColumn
DISTVAP
CW
REAC.OUT
880
19801220
1620
2640
160°
160°
130°
220°
220°
210°
210° 270°
149°
178°
50°
60°160°
180°
= EXCHANGER DUTY, kW
TEMPERATURE, Cº
Tmin = 20°Q = 1000, Q = 800H Cmin min
Para nuestro ejemplo:
3©2000 AspenTech. All Rights Reserved.
REACTOR
FEEDSTEAM
STEAMRECYCLE
PRODUCT
DistillationColumn
DISTVAP
CW
REAC.OUT
880
19801220
1620
2640
160°
160°
130°
220°
220°
210°
210° 270°
149°
178°
50°
60°160°
180°
= EXCHANGER DUTY, kW
TEMPERATURE, Cº
Tmin = 20°Q = 1000, Q = 800H Cmin min
Let’s design the network
210° 160° 160° 50°
220° 180° 180° 60°
210° 160°
270° 180° 180° 160°
Product
Reac. Out
Feed
Recycle
Where should we start?
Diseñemos la red
4©2000 AspenTech. All Rights Reserved.
210° 160° 160° 50°
220° 180° 180° 60°
210° 160°
270° 180° 180° 160°
Product
Reac. Out
Feed
Recycle
Where should we start?¿Por dónde comenzar?
Divide the problemat the Pinch
Start at the Pinchand move away
This will ensure no Cross Pinch heat transfer
T T
H H
Pinch Pinch
PINCH BelowAbove
Identifique elpunto de corte
Comience en el punto de cortey luego muévase hacia afuera
5©2000 AspenTech. All Rights Reserved.
Divide the problemat the Pinch
Start at the Pinchand move away
This will ensure no Cross Pinch heat transfer
T T
H H
Pinch Pinch
PINCH BelowAbove
How should we match streams at the Pinch? T T
HH
Pinch Pinch
Hot Stream
ColdStre
amCold Stream
Hot Stream
BelowC CPPcold hot
AboveC CPPhot cold
C CP
(The CP Rule)
PIN OUT
¿Cómo definir los intercambios entrecorrientes en el punto de corte?
6©2000 AspenTech. All Rights Reserved.
T T
HH
Pinch Pinch
Hot Stream
ColdStre
amCold Stream
Hot Stream
BelowC CPPcold hot
AboveC CPPhot cold
C CP
(The CP Rule)
PIN OUT
Abajo Arriba
Which Streams should we match first?
Below Above
} Tmin Tmin {
Cold Utility
Hot Utility
1. Match with a stream INPossible if
CP CPIN OUT
2. Match with a streamaway from the Pinch
Possible
3. Match with utility
Possible
Finding partners for streams OUT ofthe Pinch is easy
Let’s examine streams OUT¿Qué corrientes se deben intercambiar primero?
Examinemos las corrientes de salidaAbajo Arriba
1. Match con una corriente de entradaPosible si
2. Match con una corrientelejos del punto de corte
7©2000 AspenTech. All Rights Reserved.
Below Above
} Tmin Tmin {
Cold Utility
Hot Utility
1. Match with a stream INPossible if
CP CPIN OUT
2. Match with a streamaway from the Pinch
Possible
3. Match with utility
Possible
Finding partners for streams OUT ofthe Pinch is easy
Let’s examine streams OUT
2. Match con una corrientelejos del punto de corte
Posible
3. Match con servicios auxiliares
Posible
Encontrar los intercambios para las corrientesque salen del punto de corte es sencillo
Below Above
} Tmin Tmin {
Cold Utility
Hot Utility
1. Match with a stream OUTPossible if
CP CPIN OUT
2. Match with a Streamaway from the Pinch
Not Possible( T violation)
3. Match with utilityNot Possible(Cross-Pinch)
Let’s examine streams IN
Finding partners for streams goingIN the Pinch is more difficult
Veamos las corrientes deentrada
1. Match con una corriente de salidaPosible si
2. Match con una corrientelejos del punto de corte
8©2000 AspenTech. All Rights Reserved.
Below Above
} Tmin Tmin {
Cold Utility
Hot Utility
1. Match with a stream OUTPossible if
CP CPIN OUT
2. Match with a Streamaway from the Pinch
Not Possible( T violation)
3. Match with utilityNot Possible(Cross-Pinch)
Let’s examine streams IN
Finding partners for streams goingIN the Pinch is more difficult
2. Match con una corrientelejos del punto de corte
No Posible
3. Match con servicios auxiliaresNo Posible
Encontrar los intercambios para las corrientesque entran al punto de corte es máscomplicado
How big should we make each match?
HH = 20
H = 100H
100
120
Try to maximize load to minimize number of matches(The “Tick Off” heuristic)
¿Qué cantidad de calor se debe intercambiaren cada match?
9©2000 AspenTech. All Rights Reserved.
HH = 20
H = 100H
100
120
Try to maximize load to minimize number of matches(The “Tick Off” heuristic)
Lo máximo posible!!!!!
H CP
210° 160°
220° 180°
210° 160°177.6°
270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
PRODUCT is the biggest stream INApply the CP Rule and Tick Off heuristic
880
Reac. Out
Above the Pinch
Apliquemos el procedimiento de diseño ennuestro ejemplo
Arriba PC
10©2000 AspenTech. All Rights Reserved.
H CP
210° 160°
220° 180°
210° 160°177.6°
270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
PRODUCT is the biggest stream INApply the CP Rule and Tick Off heuristic
880
Reac. Out
Above the Pinch
PRODUCTO es la mayor corrienteentrante
H CP
210° 160°
220° 180°
210° 160°177.6°
235.6°270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
REAC. OUT is the next biggest stream INAgain apply the CP Rule and Tick Off heuristic
880
1000Reac. Out
11©2000 AspenTech. All Rights Reserved.
H CP
210° 160°
220° 180°
210° 160°177.6°
235.6°270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
REAC. OUT is the next biggest stream INAgain apply the CP Rule and Tick Off heuristic
880
1000Reac. Out
REACTOR OUT es la siguientecorriente entrante
H CP
210° 160°
220° 180°
210° 160°177.6°
235.6°270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
Now place the matches away from the PinchFollow Tick Off heuristic
880
1000
190°
620Reac. Out
12©2000 AspenTech. All Rights Reserved.
H CP
210° 160°
220° 180°
210° 160°177.6°
235.6°270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
Now place the matches away from the PinchFollow Tick Off heuristic
880
1000
190°
620Reac. Out
Ahora ubique los matches lejosdel punto de corte
H CP
210° 160°
220° 180°
210° 160°177.6°
235.6°270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
Last place the heaters
880
1000
1000
190°
620
H
Reac. Out
13©2000 AspenTech. All Rights Reserved.
H CP
210° 160°
220° 180°
210° 160°177.6°
235.6°270° 180°
Product
Feed
Recycle
880 22
1620 18
1000 20
2500 50
Last place the heaters
880
1000
1000
190°
620
H
Reac. Out
Al último ubique los servicios auxiliares
160° 50°
180° 60°
180° 160°
Product
Feed
H CP
2640 22
360 18
2200 20
FEED is the only stream INApply the CP rule and Tick Off heuristic
Next, below the Pinch
220080°
Reac. Out
Ahora, debajo del punto de corte
14©2000 AspenTech. All Rights Reserved.
160° 50°
180° 60°
180° 160°
Product
Feed
H CP
2640 22
360 18
2200 20
FEED is the only stream INApply the CP rule and Tick Off heuristic
Next, below the Pinch
220080°
Reac. Out
FEED es la única corrienteentrante
160° 50°
180° 60°
180° 160°
Product
Feed
H CP
2640 22
360 18
2200 20
There are no more streams INNow we can place the coolers
440
360C
C
80°2200
Reac. Out
15©2000 AspenTech. All Rights Reserved.
160° 50°
180° 60°
180° 160°
Product
Feed
H CP
2640 22
360 18
2200 20
There are no more streams INNow we can place the coolers
440
360C
C
80°2200
Reac. Out
No hay más corrientes de entrada,ahora podemos ubicar los refrigeradores
Linking Above and Below together
210° 160° 50°
220° 180° 60°80°
210° 190° 177.6°
235.6°
160°
270° 180° 160°
Product
Feed
RecycleH
C
C
880
440
360
22001000
6201000
Q = 1000Hmin Q = 800Cmin
Design is on Target
Reac. Out
Uniendo las dos partes...
16©2000 AspenTech. All Rights Reserved.
210° 160° 50°
220° 180° 60°80°
210° 190° 177.6°
235.6°
160°
270° 180° 160°
Product
Feed
RecycleH
C
C
880
440
360
22001000
6201000
Q = 1000Hmin Q = 800Cmin
Design is on Target
Reac. Out
Se consiguen la metas dediseño!!!!
Sometimes, there are problems …..
HOT2
HOT1
COLD1
COLD2
COLD3
CP8
4
5
1
3
?
Missing a stream OUT
A veces, se presentan problemas...
17©2000 AspenTech. All Rights Reserved.
HOT2
HOT1
COLD1
COLD2
COLD3
CP8
4
5
1
3
?
Missing a stream OUTFalta una corriente de
salida!!!!
Let’s look at the problem more generally
HOT2
HOT3
HOT1
COLD1
COLD2
COLD3
?
?
For feasibilityN NIN OUT
T>90°
T >100°
100°
100°
100°
100°
100°
90°
90° 90°
90°
90°
CurrentlyN > NIN OUT
CurrentlyN > NIN OUT
AboveNo Cold Utility
BelowNo Hot Utility
Miremos un problema más general
18©2000 AspenTech. All Rights Reserved.
HOT2
HOT3
HOT1
COLD1
COLD2
COLD3
?
?
For feasibilityN NIN OUT
T>90°
T >100°
100°
100°
100°
100°
100°
90°
90° 90°
90°
90°
CurrentlyN > NIN OUT
CurrentlyN > NIN OUT
AboveNo Cold Utility
BelowNo Hot Utility
HOT2
HOT1
COLD1
COLD2
COLD3
CP8
4
5
1
3
Split a stream OUT of the Pinch
If N > NIN OUT
CP = 2
CP = 6
1
2
3
SI
19©2000 AspenTech. All Rights Reserved.
HOT2
HOT1
COLD1
COLD2
COLD3
CP8
4
5
1
3
Split a stream OUT of the Pinch
If N > NIN OUT
CP = 2
CP = 6
1
2
3
Divida una corriente de salidadel punto de corte
Another Example
HOT2
HOT1
COLD1
COLD2
COLD3
CP4
9
3
8
5
There are enough streams OUT butCP CP not possible for every Pinch MatchIN OUT
?
Otro ejemplo...
20©2000 AspenTech. All Rights Reserved.
HOT2
HOT1
COLD1
COLD2
COLD3
CP4
9
3
8
5
There are enough streams OUT butCP CP not possible for every Pinch MatchIN OUT
?
Hay suficientes corrientes de salida pero...no es posible para cada match
HOT2
HOT1
COLD1
COLD2
COLD3
CP4
9
3
8
5
We split a stream IN
What do we do ?
CP = 2
CP = 7
1
2
3
¿Qué debemos hacer?
21©2000 AspenTech. All Rights Reserved.
HOT2
HOT1
COLD1
COLD2
COLD3
CP4
9
3
8
5
We split a stream IN
What do we do ?
CP = 2
CP = 7
1
2
3
Dividimos una corriente entrante
However do not follow the rules blindly …..
HOT2
HOT2
HOT1
HOT1
COLD1
COLD1
COLD2
COLD2CP
CP
4
4
10
10
3
3
17
17
CP = 12
CP = 5
1
2
HOT2
HOT1
COLD1
COLD2
CP4
10
3
17
CP = 2
CP = 2
CP = 13
CP = 4
1
2
3
Splitting a stream INcauses the need tosplit a stream OUT
Splitting a streamOUT results in asimpler solution
Always try to minimize stream splits to reduce complexity
Another Example
Sin embargo, no debemos seguir las reglasciegamente...
22©2000 AspenTech. All Rights Reserved.
HOT2
HOT2
HOT1
HOT1
COLD1
COLD1
COLD2
COLD2CP
CP
4
4
10
10
3
3
17
17
CP = 12
CP = 5
1
2
HOT2
HOT1
COLD1
COLD2
CP4
10
3
17
CP = 2
CP = 2
CP = 13
CP = 4
1
2
3
Splitting a stream INcauses the need tosplit a stream OUT
Splitting a streamOUT results in asimpler solution
Always try to minimize stream splits to reduce complexity
Another Example
Siempre intentemos minimizar las bifurcaciones para reducirla complejidad
Resumen
Ubique el punto de corte y divida elproblema en dos partes
Realice losintercambiosen el punto decorte
Nin≤Nout
CPin≤CPout
Divida unacorriente de
salida
Divida unacorriente de
entrada
No
No
23©2000 AspenTech. All Rights Reserved.
Realice los intercambios remanentes
Ubique los serviciosauxiliares
Realice losintercambiosen el punto decorte
CPin≤CPout
Ubique la mayorcorriente de
entrada
CPin≤CPout
Maximice elintercambio
Divida unacorriente de
entrada
Paracada
match