Transcript PPT - Auburn University

Heat and Power Integration
CHEN 4460 – Process Synthesis,
Simulation and Optimization
Dr. Mario Richard Eden
Department of Chemical Engineering
Auburn University
Lecture No. 9 – Heat and Power Integration: Targeting
October 23, 2012
Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel
Lecture 8 – Objectives
Given data on the hot and cold streams of a
process, you should be able to:
 Compute the pinch temperatures
 Compute the Maximum Energy Recovery
(MER) targets using graphical and/or algebraic
methods
Motivating Example
•
What is wrong with this process from an energy
viewpoint?
H1
520 K
C1
300 K
330 K To
Recovery
H2
300 K To
Storage
C2
380 K
550 K
Adiabatic
Reactor
320 K
Separation
380 K
To
Finishing
Washing
No integration
of energy!!!!
Purification
Impurities
Short Bibliography
•
Early pioneers
–
–
•
Central figure
–
–
•
Linnhoff @ ICI/UMIST (1978)
Currently: President, Linnhoff-March
Recommended text
–
–
•
Rudd @ Wisconsin (1968)
Hohmann @ USC (1971)
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 THS to targets THT

NC cold streams, with given heat capacity flowrate, each having
to be heated from supply temperature TCS to targets TCT
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).
Tout
Tout
Tout
H
H
H
in
T
Tin
Tin
Tin
in
T
Steam
Cooling
Water
out
C T
C
Tout
C
Tout
Network for minimal
equipment cost ?
Tout
Tin
Tout
Tout
Cooling
Water
Tout
Tin
Network for minimal
energy cost ?
Steam
Tin
Tout
Tin
Tout
Tin
Tin
Tin
Capital vs. Energy 3:3
•
Numerical Example
150o
100
150o
100
150o
CP = 1.0
300o
300o
500
CP = 1.0
500
CP = 1.0
Cooling
Water (90-110oF)
100
o
CP = 1.0 300
CP = 1.0
Steam (400oF)
100
200o
100
200o
100
200o
Design A:
(AREA) = 13.3
[ A = Q/UTlm ]
500
150o
CP = 1.0
Design B:
(AREA) = 20.4
[ A = Q/UTlm ]
o
CP = 1.0 300
CP = 1.0
CP = 1.0
150o
150o
200o
100
300o
200o
100
300o
100
500
CP = 1.0
500
CP = 1.0
500
CP = 1.0
200o
Some Definitions 1:3
T
TT
T
TS = Supply temperature (oC)
TT = Target temperature (oC)
TS
H
H = Stream enthalpy (MW)
H
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) ?
20o
100
30o
80o
60o
o
50o
A
10o
100
70o
60o
o
40o
20o
B
Clearly, exchanger A violates the Tmin constraint
Some Definitions 3:3
Exchanger Duty (Q):
Data:
OK
MW/ oC
Hot stream CP = 0.3
Cold stream CP = 0.4 MW/ oC
T1 = 70o
60o
100o
Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC 
40o
Q = 0.4(70 - 40) = 0.3(100 - 60) = 12 MW
Heat Transfer Area (A):
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
OK
Simple Example
30°
60°
120°
100°
ΔH=160
ΔH=162
180°
80°
ΔH=100
Utilities:
Steam @ 150 oC, CW @ 25oC
Stream
TS
(oC)
TT
(oC)
H1
H2
C1
C2
180
130
60
30
80
40
100
120
CP
H
o
(kW) (kW/ C)
100
1.0
180
2.0
160
4.0
162
1.8
130°
40°
ΔH=180
Design a network of steam heaters,
water coolers and exchangers for the
process streams. Where possible, use
exchangers in preference to utilities.
Simple Example - Targets
40°
100°
ΔH=18
ΔH=60
30°
60°
120°
130°
ΔH=162
180°
80°
ΔH=100
Units:
Steam:
Cooling water:
4
60 kW
18 kW
Are these numbers optimal??
Temperature-Enthalpy Diagram
Steam
T
C
CW
120oC
110oC
Tmin = 20
10
Steam
H
100oC
CW
30
QCmin = 50
H
50
QHmin = 70
Correlation between Tmin, QHmin and QCmin
More in, More out! QHmin + x  QCmin + x
The Composite Curve 1:2
Temperature
H Interval
T1
C
T2
=B
P
(T1 - T2)*B
(T2 - T3)*(A+B+C)
=
P
C
CP =
A
C
T3
(T3 - T4)*(A+C)
T4
(T4 - T5)*A
T5
Enthalpy
Three (3) hot streams
The Composite Curve 2:2
Temperature
H Interval
T1
CP = B
H1
T2
CP = A + B + C
H2
T3
CP = A + B
H3
T4
CP = A
H4
T5
Enthalpy
Three (3) hot streams
Simple Ex. – Hot Composite
30°
60°
120°
100°
ΔH=160
ΔH=162
180°
80°
ΔH=100
130°
40°
ΔH=180
T
T
1.0
1.0
180
180
CP =
CP =
H=50
130
H=150
C
C
P
=
2.
0
130
=3
P
.0
H=150
P
=
H=80
C
Not to
scale!!
2.
0
80
80
40
H=50
40
H
Not to
scale!!
H=80
H
Simple Ex. – Cold Composite
30°
60°
120°
100°
ΔH=160
ΔH=162
180°
80°
ΔH=100
130°
40°
ΔH=180
T
T
CP =
CP =
H=36
100
100
H=232
60
C
=5
P
.8
H=36
H=232
60
Not to
scale!!
1.8
=
P
4 .0
H=54
CP =
C
30
1. 8
120
1.8
120
30
H
Not to
scale!!
H=54
H
Thermal Pinch Diagram
T
QH,min
Move cold
composite
horizontally until
the two curves are
exactly ΔTmin apart
Tpinch,hot
Tmin
Tpinch,cold
QC,min
H
Temperature
Simple Ex. - Pinch Diagram
200
180
160
140
120
100
80
60
40
20
0
QHmin = 48 kW
QCmin = 6 kW
Maximum Energy Recovery
(MER) Targets!
THpinch = 70
TCpinch = 60
0
50
100
150
200
Enthalpy
250
300
350
The Pinch
QHmin
T
Heat
Source
QCmin
+x
QHmin
+x
Tmin
Heat
Sink
“PINCH”
x
H
H
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.
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:
QuH, z  C u (Tz 1  T z )
•
Exchangeable capacity of the v’th cold stream
passing through the z’th temperature interval:
QvC, z  Cv (t z 1  t z )  Cv ((Tz 1  Tmin )  (Tz  Tmin ))
QvC, z  Cv (Tz 1  Tz )
Algebraic Targeting Method
•
Collective load of the hot streams passing through
the z’th temperature interval is:
 H zH   QuH, z
u
•
Collective capacity of the cold streams streams
passing through the z’th temperature interval is:
 H zC   QvC, z
u
Algebraic Targeting Method
•
Heat balance around each temperature interval:
rz   H zH   H zC  rz 1
Residual heat from
preceding interval
rz  1
Heat added by
hot streams
HzH
HzC
z
rz
Residual heat to
subsequent interval
Heat removed
by cold streams
Algebraic Targeting Method
•
The enthalpy cascade
– r0 is zero (no hot streams exist above the first
interval)
– Feasibility is
nonnegative
insured
when
all
the
rz's
are
– 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
•
Example
– Two hot streams and two cold streams
– ΔTmin = 10°F
H
Stream
TS
(oF)
TT
(oF)
(kBtu/h)
H1
H2
C1
C2
260
250
120
180
160
130
235
240
3000
1800
2300
2400
H
(kBtu/h F)
Stream
TS
(oF)
TT
(oF)
(kBtu/h)
(kBtu/h oF)
30
15
20
40
H1
H2
C1
C2
250
240
120
180
150
120
235
240
3000
1800
2300
2400
30
15
20
40
CP
o
CP
Step 1: Temperature intervals
– Substract ΔTmin from hot temperatures
– 250°F  240°F  235°F  180°F  150°F  120°F
Algebraic Targeting Method
Step 2: Interval heat balances
– For each interval calculate the enthalpy load
– Hi = (Ti  Ti+1)(CPHot CPCold )
Stream
H1
H2
C1
C2
H
TSCP TT
(kBtu/h
(oF) oF)
(oF)
(kBtu/h)
(kBtu/h oF)
26030
25015
12020
18040
3000
1800
2300
2400
30
15
20
40
160
130
235
240
CP
Interval
Ti
1
2
3
4
5
6
250
240
235
180
150
120
Ti  Ti+1
CPHot
CPCold
Hi
10
5
55
30
30
30
5
15
25
5
300
25
825
750
150
Algebraic Targeting Method
Step 3: Enthalpy cascade
Interval
Ti
Ti  Ti+1
CPHot
CPCold
Hi
1
2
3
4
5
6
250
240
235
180
150
120
10
5
55
30
30
30
5
15
25
5
300
25
825
750
150
QH = 0
T1 = 250°F
QH = 500
T1 = 250°F
ΔH = 300
ΔH = 300
Q1 = 800
Q1 = 300
T2 = 240°F
T2 = 240°F
ΔH = 25
ΔH = 25
Q2 = 825
Q2 = 325
T3 = 235°F
T3 = 235°F
Most
negative
residual
ΔH = -825
ΔH = -825
Q3 = 0
Q3 = -500
T4 = 180°F
T4 = 180°F
TCpinch = 180°F
ΔH = 750
ΔH = 750
Q4 = 750
Q4 = 250
T5 = 150°F
T5 = 150°F
ΔH = -150
ΔH = -150
QC = 600
QC = 100
T6 = 120°F
QHmin
T6 = 120°F
QCmin
Summary – Heat Integration
On completion of this part, given data on the
hot and cold streams of a process, you should
be able to:
 Compute the pinch temperatures
 Compute the Maximum Energy Recovery
(MER) targets using graphical and/or algebraic
methods
Other Business
•
No lecture next week (October 30)
–
•
I will be in Pittsburgh for the AIChE Annual Meeting
Next Lecture – November 6
–
Heat and Power Integration: Network Design (SSLW p. 261-280)