heat and power integration chen 4460 – process synthesis, simulation and optimization dr. mario...
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Heat and Power Integration
CHEN 4460 – Process Synthesis, Simulation and Optimization
Dr. Mario Richard EdenDepartment of Chemical Engineering
Auburn University
Lecture No. 8 – Heat and Power Integration: Targeting
October 23, 2006
Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel
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Lecture 8 – Objectives
Compute the pinch temperatures
Compute the Maximum Energy Recovery (MER) targets using graphical and/or algebraic methods
Given data on the hot and cold streams of a process, you should be able to:
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Motivating Example
• What is wrong with this process from an energy viewpoint?
C1C2
H1
H2
300 K 550 K
520 K
330 K
380 K
300 K
320 K
380 K
AdiabaticReactor
Washing
Purification
Separation
ToRecovery
ToStorage
ToFinishing
Impurities
No integration of energy!!!!
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Short Bibliography
• Early pioneers – Rudd @ Wisconsin (1968) – Hohmann @ USC (1971)
• Central figure – Linnhoff @ ICI/UMIST (1978)– Currently: President, Linnhoff-March
• Recommended text– Seider, Seader and Lewin (2004): Product and Process
Design Principles, 2 ed. Wiley and Sons, NY– Linnhoff et al. (1982): A User Guide on Process
Integration for the Efficient Use of Energy, I. Chem. E., London
• Most comprehensive review:– Gundersen, T. and Naess, L. (1988): The Synthesis of
Cost Optimal Heat Exchanger Networks: An Industrial Review of the State of the Art, Comp. Chem. Eng., 12(6), 503-530
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Capital vs. Energy 1:3
• The design of Heat Exchanger Networks (HENs) deals with the following problem:
Given:
NH hot streams, with given heat capacity flowrate, each having to be cooled from supply temperature TH
S to targets TH
T
NC cold streams, with given heat capacity flowrate, each having to be heated from supply temperature TC
S to targets TC
T
Design:
An optimum network of heat exchangers, connecting between the hot and cold streams and between the streams and cold/hot utilities (furnace, hot-oil, steam, cooling water or refrigerant, depending on the required duty temperature)
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Capital vs. Energy 2:3
• Optimality – Implies a trade-off between CAPITAL COSTS (cost of
equipment) and ENERGY COSTS (cost of utilities).
Network for minimal energy cost ?
Network for minimal equipment cost ?
H H H
C
C
C
CoolingWater
Steam
Tin
Tin
Tin
Tin Tin Tin
ToutTout Tout
Tout
Tout
Tout
CoolingWater
Steam
Tin
Tin
Tin
Tin Tin Tin
ToutTout Tout
Tout
Tout
Tout
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Capital vs. Energy 3:3
• Numerical Example
Design A: (AREA) = 13.3
[ A = Q/UTlm ]
Design B: (AREA) = 20.4 [ A = Q/UTlm ]
CoolingWater (90-110oF)CoolingWater (90-110oF)
Steam (400oF)
300o300o
500500
150o
200o200o
150o 150o
200o200o
200o200o
100100
100100
100100
300o300o
300o300o
500500 500500
CP = 1.0CP = 1.0
CP = 1.0CP = 1.0
CP = 1.0CP = 1.0
CP = 1.0 CP = 1.0 CP = 1.0
100 100 100
300o300o
500500
150o
200o200o
150o 150o
200o200o
200o200o
300o300o
300o300o
500500 500500
CP = 1.0CP = 1.0
CP = 1.0CP = 1.0
CP = 1.0CP = 1.0
CP = 1.0 CP = 1.0 CP = 1.0
100
100
100
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Some Definitions 1:3
T
HH
TS
TT
TTS = Supply temperature (oC)
TT = Target temperature (oC)
H = Stream enthalpy (MW) CP = Heat capacity flowrate (MW/
oC)= Flowrate x specific heat
capacity= m x Cp (MW/ oC)
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Some Definitions 2:3
• Minimum Allowable Temperature Driving Force Tmin
• Which of the two counter-current heat exchangers illustrated below violates T 20°F (i.e. Tmin = 20°F) ?
100o 60o
50o
80o
100o 60o
40o
70o
A B
20o 10o
20o 30o
Clearly, exchanger A violates the Tmin
constraint
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Some Definitions 3:3
100o 60o
40o
T1 = 70oOK
OK
Exchanger Duty (Q):
Data: Hot stream CP = 0.3 MW/ oC Cold stream CP = 0.4 MW/ oC
Heat Transfer Area (A):Data: Overall heat transfer coefficient, U=1.7 kW/m2 oC
(Alternative formulation in terms of film coefficients)
Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC Q = 0.4(70 - 40) = 0.3(100 - 60) = 12 MW
Tlm = (30 - 20)/loge(30/20) = 24.66
So, A = Q/(UTlm) = 12000/(1.724.66) = 286.2 m2
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Simple Example
Stream TS
(oC) TT
(oC) H
(kW) CP
(kW/oC)
H1 180 80 100 1.0 H2 130 40 180 2.0 C1 60 100 160 4.0 C2 30 120 162 1.8
Design a network of steam heaters, water coolers and exchangers for the process streams. Where possible, use exchangers in preference to utilities.
Utilities:
Steam @ 150 oC, CW @ 25oC
30° 120°
180° 80°
130° 40°
60° 100°
ΔH=162
ΔH=160
ΔH=100
ΔH=180
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Simple Example - Targets
30° 120°
180° 80°
40°
60°
ΔH=162
ΔH=60ΔH=1813
0°
ΔH=100
100°
Units: 4Steam: 60 kWCooling water: 18 kW
Are these numbers optimal??
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Temperature-Enthalpy Diagram
Correlation between Tmin, QHmin and QCmin
More in, More out! QHmin + x QCmin + x
Tmin = 10
Steam
CWC
H
110oC
100oC
T
HQCmin = 30 QHmin = 50
Steam
CW
Tmin = 20
Steam
CWC
H
120oC
100oC
T
HQCmin = 50 QHmin = 70
Steam
CW
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The Composite Curve 1:2Temperature
Enthalpy
T1
T2
T3
T4
T5
CP
= A
CP = B
C P =
C
H Interval
(T1 - T2)*B
(T2 - T3)*(A+B+C)
(T3 - T4)*(A+C)
(T4 - T5)*A
Three (3) hot streams
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The Composite Curve 2:2
Three (3) hot streams
Temperature
Enthalpy
T1
T2
T3
T4
T5
H Interval
H1
H2
H3
H4
CP = B
CP = A + B + C
CP = A + B
CP = A
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H=150
H
180
130
CP = 3.0
80
40
H=50
H=80
CP
= 1
.0
C P = 2
.0
T
Simple Ex. – Hot Composite
30° 120°
180° 80°
130° 40°
60° 100°
ΔH=162
ΔH=160
ΔH=100
ΔH=180
H=150
T
H
180
130
CP
= 1
.0
C P = 2
.0
80
40
H=50
H=80Not to scale!
!
Not to scale!
!
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H=232
T
H
120
100
CP = 5.8
60
30
H=36
H=54
CP
= 1
.8
CP
= 1
.8
Simple Ex. – Cold Composite
30° 120°
180° 80°
130° 40°
60° 100°
ΔH=162
ΔH=160
ΔH=100
ΔH=180
H=232
T
H
120
100
CP
= 1
.8
CP = 4.0
60
30
H=36
H=54Not to scale!
!
Not to scale!
!
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Thermal Pinch Diagram
T
H
QC,min
QH,min
Tmin
Tpinch,hot
Tpinch,cold
Move cold composite
horizontally until the two curves are
exactly ΔTmin apart
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Simple Ex. - Pinch Diagram
0
20
40
60
80
100
120
140
160
180
200
0 50 100 150 200 250 300 350
Enthalpy
Te
mp
era
ture
QCmin = 6 kW QHmin = 48 kW
TCpinch = 60
THpinch = 70
Maximum Energy Recovery (MER) Targets!
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The Pinch
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.
H
T
QCmin
QHmin
“PI NCH”
H
T
QCmin
QHmin
HeatSource Heat
Sink
Tmin
+x
x
+x
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Significance of the Pinch
• Do not transfer heat across pinch
• Do not use cold utilities above the pinch
• Do not use hot utilities below the pinch
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Algebraic Targeting Method
• Temperature scales– Hot stream temperatures (T)– Cold stream temperatures (t)
• Thermal equilibrium– Achieved when T = t
• Inclusion of temperature driving force ΔTmin
– T = t + ΔTmin
– Thus substracting ΔTmin from the hot temperatures will ensure thermal feasibility at all times
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Algebraic Targeting Method
• Exchangeable load of the u’th hot stream passing through the z’th temperature interval:
• Exchangeable capacity of the v’th cold stream passing through the z’th temperature interval:
, 1( )Hu z u z zQ C T T
, 1 1 min min
, 1
( ) (( ) ( ))
( )
Cv z v z z v z z
Cv z v z z
Q C t t C T T T T
Q C T T
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Algebraic Targeting Method
• Collective load of the hot streams passing through the z’th temperature interval is:
• Collective capacity of the cold streams streams passing through the z’th temperature interval is:
,H Hz u z
u
H Q
,C Cz v z
u
H Q
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Algebraic Targeting Method
• Heat balance around each temperature interval:
1H C
z z z zr H H r
zHeat added by hot streams
Heat removed by cold streams
Residual heat from preceding interval
Residual heat to subsequent interval
HzH
1zr
CzH
zr
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Algebraic Targeting Method
• The enthalpy cascade– r0 is zero (no hot streams exist above the first
interval)
– Feasibility is insured when all the rz's are nonnegative
– The most negative rz corresponds to the minimum heating utility requirement (QHmin) of the process
– By adding an amount (QHmin) to the top interval a revised enthalpy cascade is obtained
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Algebraic Targeting Method
• The revised enthalpy cascade– On the revised cascade the location of rz=0
corresponds to the heat-exchange pinch point
– Overall energy balance for the network must be realized, thus the residual load leaving the last temperature interval is the minimum cooling utility requirement (QCmin) of the process
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Algebraic Targeting Method
Stream TS
(oF)
TT
(oF)
H
(kBtu/h)
CP
(kBtu/h oF)
H1 260 160 3000 30
H2 250 130 1800 15
C1 120 235 2300 20
C2 180 240 2400 40
• Example– Two hot streams and two cold streams
– ΔTmin = 10°F
Step 1: Temperature intervals– Substract ΔTmin from hot temperatures
– 250°F 240°F 235°F 180°F 150°F 120°F
Stream TS
(oF)
TT
(oF)
H
(kBtu/h)
CP
(kBtu/h oF)
H1 250 150 3000 30
H2 240 120 1800 15
C1 120 235 2300 20
C2 180 240 2400 40
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Stream TS
(oF)
TT
(oF)
H
(kBtu/h)
CP
(kBtu/h oF)
H1 260 160 3000 30
H2 250 130 1800 15
C1 120 235 2300 20
C2 180 240 2400 40
Algebraic Targeting Method
Step 2: Interval heat balances– For each interval calculate the enthalpy load Hi = (Ti Ti+1)(CPHot CPCold )
I nterval T i T i T i+1
CPHot
CPCold Hi
1 250 10 30 300 2 240 5 5 25 3 235 55 15 825 4 180 30 25 750 5 150 30 5 150 6 120
Stream CP
(kBtu/h oF)
H1 30
H2 15
C1 20
C2 40
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Algebraic Targeting Method
Step 3: Enthalpy cascade
ΔH = 300
QH = 0T1 = 250°F
T2 = 240°F
ΔH = 25
T3 = 235°F
ΔH = -825
T4 = 180°F
ΔH = 750
T5 = 150°F
ΔH = -150
T6 = 120°F
Q1 = 300
Q2 = 325
Q3 = -500
Q4 = 250
QC = 100
ΔH = 300
QH = 500T1 = 250°F
T2 = 240°F
ΔH = 25
T3 = 235°F
ΔH = -825
T4 = 180°F
ΔH = 750
T5 = 150°F
ΔH = -150
T6 = 120°F
Q1 = 800
Q2 = 825
Q3 = 0
Q4 = 750
QC = 600
Most negative residual
TCpinch = 180°F
QHmin
QCmin
I nterval T i T i T i+1
CPHot
CPCold Hi
1 250 10 30 300 2 240 5 5 25 3 235 55 15 825 4 180 30 25 750 5 150 30 5 150 6 120
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Summary – Heat Integration
Compute the pinch temperatures
Compute the Maximum Energy Recovery (MER) targets using graphical and/or algebraic methods
On completion of this part, given data on the hot and cold streams of a process, you should be able to:
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Other Business
• Review of Midterm Exam– Tuesday October 24 during lab– Will meet in Ross Hall Auditorium– You will get your tests back to look at during solution
review – Test MUST be returned after presentation
• Next Lecture – October 30– Heat and Power Integration: Network Design (SSL pp. 316-
346)