2/21/99 HENS 1,Targeting 1
HEAT EXCHANGER NETWORK
SYNTHESIS 1,TARGETING
OBJECTIVES:
1) Determine the minimum energy inputs required
for a process
2) Determine the minimum number of heat exchangers
required to transfer this energy
3) Develop a methodology for determining where these
heat exchangers should be placed in the process
2/21/99 HENS 1,Targeting 2
SINGLE HEAT EXCHANGER - REVIEW
W HOT
T HOTin
T COLDout
W COLD
T COLDin
T HOTout
Stream W(kg/h) Cp (kJ/kg-K) Tin(K) Tout(K) Q(gJ/h)
Hot 20,000 4.2 180 112.5 -5.67
Cold 15,000 4.2 60 150 +5.67
2/21/99 HENS 1,Targeting 3
Heat Exchanger Q-T Diagram
020
4060
80100
120140
160180
200
0 1 2 3 4 5 6
Q (gJ/h)
Temperature (K)
Thot Tcold
2/21/99 HENS 1,Targeting 4
IMPORTANT POINTS
1) The First Law says that the heat transferred from the hot
stream must equal that transferred to the cold stream.
Therefore,of the six process variables (two flow rates
and four temperatures),only five can be specified
independently.
2) The Second Law says that heat can only be transferred
from a hotter fluid to a colder one,Therefore,the
temperature of the cold fluid must be less than that of the
hot fluid at all points along the length of the exchanger.
2/21/99 HENS 1,Targeting 5
IMPORTANT POINTS (Cont’d.)
3) As a practical matter,for heat to be transferred at an
acceptable rate,the temperature of the cold fluid must be
less than that of the hot fluid at each point in the
exchanger by a reasonable amount,We refer to the
minimum acceptable temperature difference as the
MINIMUM APPROACH TEMPERATURE or DTmin.
This is also known in the literature as the PINCH
TEMPERATURE,The methodology for synthesizing
heat exchanger networks is sometimes called
PINCH TECHNOLOGY
2/21/99 HENS 1,Targeting 6
THE HEAT EXCHANGER NETWORK
PROBLEM
A chemical plant generally has more than one hot
steam to be cooled and more than one cold stream
to be heated,To better understand this problem
let us consider the following simple example.
2/21/99 HENS 1,Targeting 7
1000 kg/h,30 C
Heat to 200 C
Cool to 30 C
10,000 kg/h,
250 C
9,000 kg/h,30 C
1,500 kg/h,100 C
16,000 kg/h
15,000 kg/h
~1 kg/h
500 kg/h
Cool to 30 C
Heat to 100 CREACTOR
ABSORBER STRIPPER
2/21/99 HENS 1,Targeting 8
THE PROBLEM
1) There are two hot streams that need to be cooled,
namely,
Reactor Effluent and Stripper Bottoms
2) There are two cold streams that need to be heated,
namely,Reactor Feed and Stripper Feed
What is the best way to do this?
2/21/99 HENS 1,Targeting 9
FIRST LAW ANALYSIS
PROCESS DATA:
Stream W(kg/h) Cp (kJ/kg-K) Tin(K) Tout(K) Q(gJ/h)
Hot 1 10,000 1.3 250 30 -2.860
Hot 2 15,000 4.2 100 30 -4.410
Cold 1 10,000 1.1 30 200 +1.870
Cold 2 16,000 4.2 30 100 +4.704
Total Heating Available = 2.860 + (4.410) = 7.270 gJ/h
Total Heating Required = -1.870 + (-4.704 ) = - 6.574 gJ/h
Net Heat Available = 0.696 gJ/h
(Subject to the Second Law)
2/21/99 HENS 1,Targeting 10
THE SIMPLE (BUT WASTEFUL) SOLUTION
The simplest solution is:
1) to cool each hot stream to its specified temperature
using a cold utility such as cooling tower water and
2) to heat each cold stream to its specified temperature
using a hot utility such as steam.
Typical utility costs (see Turton et al) are:
MP Steam (10 barg,184C)….$3.66/gJ
CW (15 C temperature rise),,$0.24/gJ
Cooling cost,(7.270)(8000)(0.24) = $14,000/yr
Heating cost,(6.574)(8000)(3.66) = $192,000/yr
.
2/21/99 HENS 1,Targeting 11
SECOND LAW CONSIDERATIONS
QUESTION,Can one devise a heat exchanger network
that will transfer 6.574 gJ/h from the hot streams to the
cold streams? If not,what is the best one can expect to do?
ANSWER,In general,not all the heat available to be
transferred can be,How much can depends upon two
factors:
The problem structure,i.e.,the flow rates and temperature
levels of the process streams of interest,and
The Minimum Approach Temperature one elects to use.
The first is set by the process; the second by the designer.
2/21/99 HENS 1,Targeting 12
SECOND LAW ANALYSIS
Let us chose a value for DTmin,10 C is a reasonable
value based on experience,Later on we can chose other
values to see how much difference it makes.
We then construct a temperature interval diagram as
shown next,Keep in mind that TCOLD must be 10 C less
than THOT at every point,Thus,there are two
temperature scales side by side,one 10 C lower than the
other.
2/21/99 HENS 1,Targeting 13
250
200
150
100
50
0
240
190
140
90
40
-10
H1
H2
C1 C2
TEMPERATURE INTERVAL DIAGRAM
2/21/99 HENS 1,Targeting 14
TEMPERATURE INTERVALS
A temperature interval is created every time a stream
enters or leaves the Temperature Interval Diagram:
For the problem at hand the temperature intervals are,
starting at the top of the diagram:
Interval Thigh Tlow DT
1 250 210 40
2 210 110 100
3 110 100 10
4 100 40 60
5 40 30 10
2/21/99 HENS 1,Targeting 15
THE ENERGY CASCADE CONCEPT
In each temperature interval there is at least one stream
can either absorb heat (cold stream) or provide heat (hot
stream),The amount a given stream can absorb or provide
per degreeis merely its WCp,The amount that can be
provided is the sum of the WCp’s for the hot streams active
in the interval,Similarly,the amount that can be absorbed
is the sum of the WCp’s for the cold streams active in the
interval.
Using this concept we can construct an Energy Cascade as
shown on the next slide.
2/21/99 HENS 1,Targeting 16
ENERGY CASCADE
(K) (gJ/h-K) (gJ/h-K) (gJ/h-K) (gJ/h)
Interval DT SWCp(hot) SWCp(cold) DSWCp Q
1 40 0.013 - 0 = 0.013 0.52
2 100 0.013 - 0.011 = 0.002 0.20
3 10 0.013 - 0.0782 = -0.0652 -0.652
4 60 0.076 - 0.0782 = -0.0022 -0.132
5 10 0.076 - 0 = 0.076 0.760
Note,The sum of Q over the five temperature intervals is
0.595 gJ/h which is what the First Analysis gave.
The next step is to show this information in an Energy
Cascade Diagram per the next slide.
2/21/99 HENS 1,Targeting 17
0.20
-0.652
-0.132
0.52
0.760
0.52
0.72
0.068
-0.0654
0.696
Energy Cascade Diagram
To Cold Utility
What’s wrong
with this
picture?
2/21/99 HENS 1,Targeting 18
0.20
-0.652
-0.132
0.52
0.760
0.52
0.72
0.068
0
0.760
Modified
Energy Cascade Diagram
0.064
Hot Utility
To Cold Utility
2/21/99 HENS 1,Targeting 19
WHAT THE DIAGRAM SAYS
There are several things to be learned from the previous
diagram:
1) Where the heat transferred down from one interval
to the next equals 0 is the pinch point,i.e.,the
temperature at which the minimum approach
temperature occurs,For this cascade it is 40 C (hot
side) or 30 C (cold side)
2) The pinch divides the process into two separate
networks,one above the pinch and a second below it.
2/21/99 HENS 1,Targeting 20
THERMODYNAMIC INTERPRETATION
The Modified Cascade Diagram allows us to determine
the minimum hot and cold utility requirements for the
chosen value of the Minimum Approach Temperature.
We cannot do any better than this without violating the
Second Law,
For the example problem,Qhot = 0.064 gJ/h and
Qcold = 0.760 gJ/h
The annual cost = [(0.064)(3.66) + (0.760)(0.24)](8000)
= $3300/yr
(Which is considerably less than the $204,000/yr that the
simple solution costs!)
2/21/99 HENS 1,Targeting 21
WHAT IT ALL MEANS
What this analysis does is tell us what the minimum
energy requirements are for our process without ever
sizing a heat exchanger,laying out a network,or even
knowing how many heat exchangers are required,
What it gives us is a target to shoot for,We know at the
outset what the best that we can do is and can make
design compromises accordingly,For this reason this
part of the Heat Exchanger Network Synthesis procedure
is knows as TARGETING.