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
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:
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!!!!
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
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)
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
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
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)
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
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
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
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??
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
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
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
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!
!
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!
!
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
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!
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
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
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
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
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
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
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
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
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
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
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
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:
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)