Mod-05 Lec-01 Concepts and Basic Principles of Energy (or Heat) Integration -- Part 1

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Language: en

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Welcome, we will start today the fifth module
of our course that is heat integration of
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the process. So far we have studied the protocol
of design of a process. How the design evolves
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through several steps, the hierarchy of decisions
is first that we have to decide whether we
00:00:41.590 --> 00:00:46.140
have to go for batch process or continuous
process. Then we have to decide the input
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output structure of the flow sheet, then the
recycle structure of the flow sheet, then
00:00:50.980 --> 00:00:55.790
we have to design the separation processes
and finally is the heat integration of the
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process.
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Selection of the process and decision decisions
on major processing steps reactor separators
00:01:03.770 --> 00:01:11.950
and recycle steam gives us the complete energy
and material balance of the flow sheet. That
00:01:11.950 --> 00:01:18.490
is if we talk about the type of design, the
inner most is the reactor design, then comes
00:01:18.490 --> 00:01:23.240
the separation systems then comes the heat
integration. So, the design is evolving from
00:01:23.240 --> 00:01:30.820
center to outside of the design and then we
are now at the outermost layer that is heat
00:01:30.820 --> 00:01:36.030
integration of the process, but heat integration
of the process requires all the data.
00:01:36.030 --> 00:01:42.690
Now, we have completed all the previous steps
we have that data, so that is what we note
00:01:42.690 --> 00:01:46.710
that. Final stage in the process design is
the heat integration for which heating and
00:01:46.710 --> 00:01:53.080
cooling duties on all process steams should
be known. Then completing the design of heat
00:01:53.080 --> 00:01:59.560
integration network is not necessary in order
to assess the competent flow sheets. We are
00:01:59.560 --> 00:02:05.160
still at the conceptual design means we are
assessing the profitability of different flow
00:02:05.160 --> 00:02:11.090
sheets on pen and paper.
Therefore, while short listing some of the
00:02:11.090 --> 00:02:17.260
best flow sheets we do not have to design
the entargeted exchanger network, we can identify
00:02:17.260 --> 00:02:23.810
the targets. So, the targets can be set for
heat exchanger network to assess the performance
00:02:23.810 --> 00:02:28.900
of the complete design without actually carrying
out the network design. These, targets allow
00:02:28.900 --> 00:02:35.030
us to evaluate both energy and capital cost
of heat exchanger moreover the targets also
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allow the designer to suggest process changes
for reactor separation and recycle steam.
00:02:41.690 --> 00:02:48.799
Obviously, the goal is to minimize the energy
requirement of the process as possible because
00:02:48.799 --> 00:02:54.780
energy is costly whether we have to heat or
whether we have to cool we have to spend energy.
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So, the heat integration of the process means
essentially utilizing the excess heat surplus
00:03:00.360 --> 00:03:05.299
heat at one part of the process for meeting
the demands at the second part of the process
00:03:05.299 --> 00:03:07.630
where the heat is in deficit.
00:03:07.630 --> 00:03:13.340
The use of targets for heat exchanger network
rather than the complete design allow many
00:03:13.340 --> 00:03:20.049
designs for overall process to be screened
quickly and conveniently. We shall see as
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how to identify the targets for heat exchanger
network after integrating or after recovering
00:03:28.070 --> 00:03:35.770
as much heat as possible within the process
some heat still needs to be supplied externally
00:03:35.770 --> 00:03:44.870
in the form of utilities, so just steam liquid.
So, that is the heat requirement secondly
00:03:44.870 --> 00:03:52.769
when all of the heat is all of the excess
heat is absorbed in the process itself the
00:03:52.769 --> 00:03:58.389
process terms which need to be cooled to room
temperature or even below that those cause
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load on the cooling water requirement.
So, that also can be taken as a heating load
00:04:03.290 --> 00:04:08.079
because we have to circulate cooling water
then we have to operate cooling towers to
00:04:08.079 --> 00:04:13.479
restore the temperature of cooling water so
and so forth. However, as I said that we do
00:04:13.479 --> 00:04:18.239
not have to we are at a stage of screening
of process alternatives. We do not have to
00:04:18.239 --> 00:04:23.060
go for a complete heat exchanger design, we
have to only identify the targets that means
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we have to identify the minimum hot utility
requirement and the minimum cold utility requirement.
00:04:29.130 --> 00:04:34.710
Now, let us see how we can do it, I am giving
here an example, first of all we have to go
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for composite curves. The analysis of heat
exchanger network requires first the identification
00:04:39.900 --> 00:04:45.990
of sources of heat hot streams and strings
of it cold streams from the material and energy
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balance.
Now, consider the example that is appearing
00:04:49.210 --> 00:04:54.979
on the screen, now we have one hot stream
and one cold stream. The supply temperature
00:04:54.979 --> 00:05:01.710
of the cold stream is 40 degrees and the target
temperature is 100 and 10 degrees and heat
00:05:01.710 --> 00:05:06.081
that are available with this or the heat that
needs to be supplied to this steam, this steam
00:05:06.081 --> 00:05:14.180
is 14 mega Watts. Then we have another hot
screen which is available at 160 degrees and
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the target temperature it needs to be cooled
in 40 degrees. It has minus 12 mega volt that
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is minus indicate surplus or plus indicate
deficit, so minus 12 mega Watts of heat available.
00:05:26.729 --> 00:05:37.889
Now, we have to absorb as much heat from the
hot stream into the cold stream as possible.
00:05:37.889 --> 00:05:47.520
How we can achieve that, for that purpose
we plot the two steams on an H T diagram which
00:05:47.520 --> 00:05:55.169
means on the x axis, we plot enthalpy and
y axis we plot temperature. So, on the screen
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you see, now the two steams the hot stream
is available at 160 degrees it is getting
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cooled to 40 degrees. Then the cold stream
is available at 20 degrees and it is getting
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cool heated to 100 and 10 degrees. Let us
see, now how you can do it, now I am plotting
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the two curves and that you see on the screen,
now below here you will see the same steams
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plotted in the form of flow sheet.
Now, we have to decide the minimum if these
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two teams are contacted in heat exchanger
what should be the delta T min at any end
00:07:02.129 --> 00:07:08.409
of the heat exchanger. We know from basic
heat exchanger theory that the area of the
00:07:08.409 --> 00:07:14.940
heat exchanger is proportional to, sorry inversely
proportional to the overall heat transfer
00:07:14.940 --> 00:07:21.620
coefficient and the L M T D the delta T min.
So, we have the relation Q is equal to u a
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delta T min for a given heat duty Q the area
of heat exchanger is Q divided by u into delta
00:07:29.080 --> 00:07:37.719
T min. Now, the delta T min we set two as
a minimum, so we assume that the minimum delta
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T min at any end of the heat exchanger should
be about 10 degrees. Now, this 10 degree has
00:07:42.470 --> 00:07:46.139
come out of experience it is an empirical
value.
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So, we have the hot utility steam available
at 180 degree and cooling water at 120 degrees
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then heating all cold streams with steam and
cooling with all hot streams with water would
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be waste of energy. So, we have to see as
much coupling between the hot and cold stream
00:08:08.090 --> 00:08:14.710
that we are given as possible, so that the
load on steam and cooling water reduces. Now,
00:08:14.710 --> 00:08:19.080
the scope of heat recovery is identified by
plotting both the steams on the diagram that
00:08:19.080 --> 00:08:26.249
I just said and then we set delta T minimum
at any end of the heat exchanger were as 10
00:08:26.249 --> 00:08:29.400
degrees.
Then, you can see that the reason of overlap
00:08:29.400 --> 00:08:36.409
between the two curves gives us the amount
of heat that could be recovered and for delta
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T minimum equal to 10 degree centigrade. We
have a heat recovery of 11 mega Watts, we
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have a heat recovery of 11 mega Watts, but
despite this even after recovering you see
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how much heat was available with hot stream.
It was 12 mega Watts, how much heat was to
00:09:00.389 --> 00:09:07.000
be was required for heating the cold stream
of 14 mega Watts out of which 11 mega Watts
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is coming from the coupling between the two
steams.
00:09:11.860 --> 00:09:19.839
This leaves 3 mega Watts to be supplied to
the cold stream to meet the 14 mega Watt demand
00:09:19.839 --> 00:09:26.220
and the hot stream is still left with 1 mega
Watt of energy. Now, these two could be these
00:09:26.220 --> 00:09:31.170
two energy requirements could be met with
steam and cooling water as I am showing here
00:09:31.170 --> 00:09:36.910
the hot stream. The cold stream after absorbing
11 mega Watts from hot stream is still left
00:09:36.910 --> 00:09:41.320
with 3 mega Watt.
So, that is supplied by steam the cold stream
00:09:41.320 --> 00:09:48.760
after taking the hot stream the hot stream
after giving 11 mega Watts to the cold stream
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is still left with 1 mega Watt and that could
be taken out using cooling water. Now, this
00:09:54.209 --> 00:10:01.589
delta T minimum was an empirical value as
I said now how we can see the effect of this
00:10:01.589 --> 00:10:03.750
particular empirical parameter.
00:10:03.750 --> 00:10:11.440
Suppose, we change the delta T minimum to
20 degrees, now how we can do that this we
00:10:11.440 --> 00:10:19.100
can do by shifting the curves horizontally
we can change the enthalpy, we can change
00:10:19.100 --> 00:10:25.610
the relative enthalpy by shifting the curves
horizontally. Then in the second graph which
00:10:25.610 --> 00:10:32.810
is now shown the right hand side of the screen
we have delta T minimum is 20 degrees. Now,
00:10:32.810 --> 00:10:41.959
after you shift the cold curve on to right
hand side that decreases the overlap between
00:10:41.959 --> 00:10:47.620
the two lines.
The two streams here the Q recovery or the
00:10:47.620 --> 00:10:53.980
heat that is recovered through the coupling
of the two streams is reduced from 11 mega
00:10:53.980 --> 00:11:00.860
Watt to 10 mega Watt, when we have delta T
minimum equal to 20 degrees. Then this is
00:11:00.860 --> 00:11:06.820
this particular feature leaves 4 mega Watt
deficit with the cold stream which has to
00:11:06.820 --> 00:11:12.980
be met with steam and 2 mega Watts of excess
with hot stream which has to be taken out
00:11:12.980 --> 00:11:18.920
using cooling water.
So, if you compare these values that Q c min
00:11:18.920 --> 00:11:24.899
in case of delta T min minimum equal to 10
degrees and Q H min minimum hot utility for
00:11:24.899 --> 00:11:32.250
delta T min equal to 20 degrees and you see
that as delta T minimum value increases the
00:11:32.250 --> 00:11:38.740
load on the utilities also increases. So,
that point I have noted here the features
00:11:38.740 --> 00:11:45.730
e temperature or enthalpy change for stream
and hence their slopes that cannot be changed,
00:11:45.730 --> 00:11:49.830
but relative position of the streams can be
changed by horizontal movement.
00:11:49.830 --> 00:11:55.870
This is possible since the reference enthalpy
for hot streams can be changed independently
00:11:55.870 --> 00:12:00.769
from the reference enthalpy of the cold streams
when delta T minimum equal to 20 degrees as
00:12:00.769 --> 00:12:04.820
the cold stream is moved horizontally away
from the hot stream the overlap between the
00:12:04.820 --> 00:12:10.270
streams. Hence, the heat recovery decreases
the new values for delta minimum 20 degrees
00:12:10.270 --> 00:12:15.370
are Q c, Q recovery equal to 10 mega Watt
Q H min equal to 4 mega Watt. Q c min equal
00:12:15.370 --> 00:12:20.050
to 2 mega Watt, the minimum cold utility Q
c min is equal to minimum cold utility q H
00:12:20.050 --> 00:12:25.800
min is minimum hot utility and the values
are 2 and 4 mega Watt respectively.
00:12:25.800 --> 00:12:31.209
This approach can determine the hot and cold
utility for a given value of delta T min importance
00:12:31.209 --> 00:12:37.940
of the delta T mini delta T minimum is that
it sets the relative locations of the hot
00:12:37.940 --> 00:12:43.760
and cold streams. Therefore, the amount of
heat recovery, Let us extend the same theme
00:12:43.760 --> 00:12:46.589
for multiple streams.
00:12:46.589 --> 00:12:52.170
Now, we shall have two hot streams and two
cold streams and then we shall see how we
00:12:52.170 --> 00:12:57.970
can couple the streams to have as maximum
heat recovery as possible for a given delta
00:12:57.970 --> 00:13:04.199
T minimum. Now, what you see on the screen
is a simple process, we have two reactors
00:13:04.199 --> 00:13:14.650
with feeds feed 1 is at enters at 20 degrees
and it is heated to 180 degrees. Then, the
00:13:14.650 --> 00:13:21.350
feed 1 enters at 20 degrees and is heated
to 180 degrees to reactor 1 so that requires
00:13:21.350 --> 00:13:29.190
32 mega Watt of energy feed the feed for reactor
2 is at 140 degrees. It needs to be heated
00:13:29.190 --> 00:13:34.490
to 230 degrees for entering the reactor, so
that delta H is the heat requirement is 87
00:13:34.490 --> 00:13:41.110
mega Watt the output of reactor 1 also enters
reactor 2 part is sent to reactor 2 and part
00:13:41.110 --> 00:13:49.810
is sent for separation and the product of
reactor 2 is comes out at 200 degrees. It
00:13:49.810 --> 00:13:57.290
has to be cooled to 80 degrees, so it gives
out about 30 mega Watts of heat. The product
00:13:57.290 --> 00:14:02.779
of reactor 1 is splitting in two parts as
I just said one part goes to reactor 2 another
00:14:02.779 --> 00:14:08.519
part is cooled and that gives out 31.5 mega
Watt of heat.
00:14:08.519 --> 00:14:16.589
So, we have essentially two hot streams and
two cold streams reactor feed 1 cold stream
00:14:16.589 --> 00:14:22.300
supply temperature 20 degrees target temperature
180 degrees heat capacity flow rate. Now,
00:14:22.300 --> 00:14:31.480
heat capacity flow rate essentially mass rate
into heat capacity mass or molar flow rate
00:14:31.480 --> 00:14:37.730
that depends and the corresponding heat capacity.
So, this is because we have the basic relation
00:14:37.730 --> 00:14:48.399
Q is equal to m c p delta T and here we are
coupling the m into c p, so as to get the
00:14:48.399 --> 00:14:54.509
heat capacity flow rate. So, m into c p is
the flow rate capacity and the units of that
00:14:54.509 --> 00:14:59.940
as mega Watt per degree centigrade.
Now, m value you can define either in molls
00:14:59.940 --> 00:15:07.910
or mass, so c p will also correspondingly
in either mol per degrees like the joules
00:15:07.910 --> 00:15:13.070
per molls degree Kelvin or Joules per kg per
degree Kelvin depending on what you need for
00:15:13.070 --> 00:15:17.319
them. But for heat capacity flow rate will
always have units of mega Watt per degree
00:15:17.319 --> 00:15:22.699
centigrade or mega Watt per Kelvin depending
on the units that you need. So, we have the
00:15:22.699 --> 00:15:27.731
stream data is given reactor product one is
a hot stream supply temperature 250 target
00:15:27.731 --> 00:15:34.920
temperature 40 heat capacity flow rate 0.15.
So, it gives out minus 31.5 mega Watt of heat
00:15:34.920 --> 00:15:41.089
reactor feed 2 is a cold stream and reactor
product two is a hot stream. So, that has
00:15:41.089 --> 00:15:47.790
energy requirement of 27 mega Watt and energy
surplus of 30 mega Watts each. The two sources
00:15:47.790 --> 00:15:54.209
are the two streams are sources of heat and
two are sinks if the heat capacities of the
00:15:54.209 --> 00:15:59.279
streams are constant the heat content in the
hot and cold stream can be determined using
00:15:59.279 --> 00:16:04.500
the heat capacity flow rate which is a product
of mass molar flow rate as I just said.
00:16:04.500 --> 00:16:13.519
Now, what wee plot are hot streams the left
hand side plot this plot use the H T diagram
00:16:13.519 --> 00:16:19.019
for the two hot streams and these hot streams
are plotted separately one stream goes from
00:16:19.019 --> 00:16:25.170
250 to 40. Second stream goes from 200 to
80, 31.5 mega Watt of surplus with first stream
00:16:25.170 --> 00:16:30.560
30 mega Watt with second.
Now, these two streams can be combined to
00:16:30.560 --> 00:16:39.720
form a composite hot stream here what will
happen is that temperatures will remain the
00:16:39.720 --> 00:16:44.500
same. But c p values will get added like for
example, in the temperature interval of 80
00:16:44.500 --> 00:16:50.509
to 200 will have two streams because the stream
one is going from 40 to 250. So, it is available
00:16:50.509 --> 00:16:55.939
in the temperature interval of 80 to 200.
So, in this particular temperature interval
00:16:55.939 --> 00:17:01.540
the two streams are available, so they add
together and the total c p for that is 0.4
00:17:01.540 --> 00:17:06.060
between 40 to 80 interval, temperature interval
40 to 80 degree centigrade.
00:17:06.060 --> 00:17:12.980
We have the c p only one stream the c p 0.5
and above 200 degree again we have only the
00:17:12.980 --> 00:17:20.430
first stream. So, there c p is on front and
then we can have a composite H T diagram which
00:17:20.430 --> 00:17:26.180
gives us the available heat in a particular
temperature interval. For example, between
00:17:26.180 --> 00:17:31.910
40 to 80, we have total 6 mega Watts of heat
available between 80 to 200 we have 48 mega
00:17:31.910 --> 00:17:37.550
Watts of heat available and from 200 to 250
we have 7.5 mega Watt of heat available so
00:17:37.550 --> 00:17:43.070
that adds up totally 261.5 mega Watt.
00:17:43.070 --> 00:17:50.560
Similarly, we can do for cold streams also
what you see on the left hand side is the
00:17:50.560 --> 00:17:56.390
individual plot the two cold streams plotted
separately, the first stream going from 20
00:17:56.390 --> 00:18:02.130
to 180, second stream going from 140 to 230.
Now, in a similar way we find that in the
00:18:02.130 --> 00:18:07.590
temperature interval 140 to 180 both streams
are present. So, when we make a composite
00:18:07.590 --> 00:18:14.630
diagram of the cold streams, we can see that
between 20 to 140 only first stream is available.
00:18:14.630 --> 00:18:20.570
So, c p 0.2 between 140 to 180 two streams
are available, so their heat capacity flow
00:18:20.570 --> 00:18:26.930
rates get added and then we have c p equal
to 0.5. Then after 180 degrees again we have
00:18:26.930 --> 00:18:35.850
first stream that is c p equal to 0.3. So,
this gives us again distribution of heat requirement
00:18:35.850 --> 00:18:41.600
against temperature interval between temperature
intervals 20 to 140. We need 24 mega Watt
00:18:41.600 --> 00:18:52.690
between 140 to 180 we need 20 mega Watt and
180 to 230 we need 15 mega Watts.
00:18:52.690 --> 00:19:00.491
So, the composite diagrams gives us an overview
of the process hot streams are the overall
00:19:00.491 --> 00:19:05.030
behavior of the hot streams can be quantified
by combining them together in the given temperature
00:19:05.030 --> 00:19:10.750
range. All this points which I just said I
have noted in this slide the temperature ranges
00:19:10.750 --> 00:19:19.110
in question are defined where an alteration
occurs in overall rate of change of enthalpy
00:19:19.110 --> 00:19:24.170
which temperature for constant heat capacities
alterations occur only when stream start or
00:19:24.170 --> 00:19:28.710
finish. Thus, the temperature axis is divided
into ranges defined by supply and target temperature
00:19:28.710 --> 00:19:34.160
of streams within each temperature range the
streams are combined to produce the composite
00:19:34.160 --> 00:19:35.550
hot streams.
00:19:35.550 --> 00:19:41.780
The c p of composite hot streams is a sum
of individual streams in any temperature range
00:19:41.780 --> 00:19:47.540
the enthalpy change of the two of the composite
streams is the sum of individual streams.
00:19:47.540 --> 00:20:02.670
Now, what we see here is the two composite
streams plotted together and the same diagram.
00:20:02.670 --> 00:20:18.690
Now, this is the hot composite stream and
the lower one is cold composite stream and
00:20:18.690 --> 00:20:26.340
these are plotted on the same H T diagram.
Now, we will see that for delta T min is equal
00:20:26.340 --> 00:20:33.750
to 10 degrees the region of overlap between
the two streams which is essentially the heat
00:20:33.750 --> 00:20:43.080
recovery the one which I am marking.
Now, is 51.5 mega Watt Q recovery is 51.5
00:20:43.080 --> 00:20:48.200
Watt, now we shall calculate these values
later for time being I am just giving you
00:20:48.200 --> 00:20:53.250
the direct answer, but we are going to treat
the same problem later and then we shall actually
00:20:53.250 --> 00:21:25.540
calculate these values. The after meeting
51.5 mega Watts of heat recovery, we
00:21:25.540 --> 00:21:49.670
have the cold streams left with energy requirement
of 7.5 mega Watt which I am marking. Now,
00:21:49.670 --> 00:21:56.100
the hot composite curve extends beyond, sorry
the cold composite curve extend beyond the
00:21:56.100 --> 00:22:03.380
hot composite curves to this much extend which
I am marking and this is the minimum hot utility
00:22:03.380 --> 00:22:09.990
requirement Q H min this is this 7.5 mega
Watt.
00:22:09.990 --> 00:22:16.890
Now, the part of hot composite curves that
extends beyond the cold composite curve which
00:22:16.890 --> 00:22:23.630
is already marked here, I have written here
Q c min that is the minimum cold utility requirement
00:22:23.630 --> 00:22:34.860
and this turns out to be 10 mega Watt. So,
we have both complete heat exchanger network
00:22:34.860 --> 00:22:58.000
target available here. We have Q recovery,
Q recovery ranging in this ray the one the
00:22:58.000 --> 00:23:08.790
marked the arrows in red that is the region
of recovery that is 51.5 mega Watt after this
00:23:08.790 --> 00:23:15.640
heat recovery we are left with 7.5 recovery
of heat requirement for cold streams that
00:23:15.640 --> 00:23:19.980
is met with hot utility.
So, that is Q H min and after this heat recovery
00:23:19.980 --> 00:23:25.120
we are left with ten mega Watt of excess heat
with hot composite curves which is taken off
00:23:25.120 --> 00:23:32.080
from cold streams, so that is what the overall
energy target is. Now, as we did in the previous
00:23:32.080 --> 00:23:39.800
case we can change the delta T min delta T
min here was assumed to be 10 degrees, we
00:23:39.800 --> 00:23:46.920
can change delta T min to 20 degrees by shifting
this curve the cold composite curve horizontally
00:23:46.920 --> 00:23:54.520
to right. Now, what will happen, obviously
the region of heat recovery will go down the
00:23:54.520 --> 00:23:58.880
heat recovery in this now is only in this
range the one that is marked red. So, this
00:23:58.880 --> 00:24:06.300
is the region of heat recovery, Q recovery
which is which has now reduced this region
00:24:06.300 --> 00:24:12.420
has released.
Therefore, a lot of heat requirement is left
00:24:12.420 --> 00:24:20.410
out the amount of heat that needs to be supplied
to the cold streams through hot utility is
00:24:20.410 --> 00:24:25.390
now 11.5 mega Watt. Now, this again values
we are going to calculate, so the Q H min
00:24:25.390 --> 00:24:32.400
increases from 7.5 mega Watt to 11.5 mega
Watt. Then the Q c min lot of heat remains
00:24:32.400 --> 00:24:37.520
in the hot composite stream after absorption
into the cold stream and then the Q c min
00:24:37.520 --> 00:24:47.140
is now 40 mega Watt. So, that is how the picture
changes with increasing delta T min with the
00:24:47.140 --> 00:24:53.490
load on hot utility and cold utility changes,
so all those points I have noted here.
00:24:53.490 --> 00:24:57.530
The left diagram gives plot of composite curves
for delta T min equal to 10 degrees while
00:24:57.530 --> 00:25:01.380
the right diagram gives plot of composite
curves for delta T min equal to 20 degrees
00:25:01.380 --> 00:25:06.640
where the curves overlap. The heat can be
rejected vertically from hot streams into
00:25:06.640 --> 00:25:12.570
the cold streams the way in which composite
curves are constructed monotonically increasing
00:25:12.570 --> 00:25:17.370
hot composite curve. Monotonically decreasing
cold composite curve allows maximum overlap
00:25:17.370 --> 00:25:22.730
between the curves. Hence, the maximum heat
recovery for delta T min equal to 10 degrees
00:25:22.730 --> 00:25:26.520
Q c, Q recovery is 51.5 mega Watt.
00:25:26.520 --> 00:25:32.000
When the composite curve extends beyond the
start of hot composite curve and where the
00:25:32.000 --> 00:25:39.100
hot composite curve extends beyond the start
of cold composite curve. The heat recovery
00:25:39.100 --> 00:25:44.010
is not possible and utilities must be used
and then for delta T min equal to 10 degrees
00:25:44.010 --> 00:25:50.860
the minimum hot utility is 7.5 mega Watt and
minimum cold utility is 10 mega Watt.
00:25:50.860 --> 00:25:59.100
Now, there are three variables here Q c min
Q H min delta T min specify any of the three
00:25:59.100 --> 00:26:04.680
variables fixes the relative positions of
the composite curves, but usually delta T
00:26:04.680 --> 00:26:10.330
min is used as a variable rather than Q c
min and Q H min. But depending on situation
00:26:10.330 --> 00:26:16.360
either you can also specify Q c min or Q H
min and then that fixes both any fixing any
00:26:16.360 --> 00:26:23.330
of these three variable fixes the other two
variables. So, as per as two stream cases
00:26:23.330 --> 00:26:30.400
concern the relative position of the curves
also degree of freedom at our disposal.
00:26:30.400 --> 00:26:34.600
The relative positions of the two curves can
be changed by moving them horizontally for
00:26:34.600 --> 00:26:39.470
feasible heat transfer from hot streams into
cold streams, hot composite curves must always
00:26:39.470 --> 00:26:45.490
be above the cold composite curve. If the
delta T min is increased to 20 degrees by
00:26:45.490 --> 00:26:51.350
shifting the cold curve to right the hot and
cold utility targets increase to 11.5 mega
00:26:51.350 --> 00:26:57.270
Watt and 14 mega Watt respectively.
Now, correct setting of delta T min is fixed
00:26:57.270 --> 00:27:02.581
by the economic tradeoff between operating
and capital cost obviously if delta T min
00:27:02.581 --> 00:27:14.750
increases the Q c min and Q H min increase,
but the area of individual heat exchanger
00:27:14.750 --> 00:27:22.340
reduces the area of heat exchangers in which
the hot and cold streams are coupled. So,
00:27:22.340 --> 00:27:29.260
Q c min and Q H min indicate basically the
operating cost and delta T min indicates the
00:27:29.260 --> 00:27:34.970
fixed or capital cost delta T min increases
area decreases capital cost decreases. But
00:27:34.970 --> 00:27:39.480
at the same time Q c min and Q H min increase
which means the operating cost decreases.
00:27:39.480 --> 00:27:46.390
So, there has to be an economic trade off
for deciding the optimum level of delta T
00:27:46.390 --> 00:27:47.390
min.
00:27:47.390 --> 00:27:53.120
So, that is what is plotted here increasing
delta T reduces the heat exchanger area. Hence,
00:27:53.120 --> 00:27:59.860
the capital cost of the exchanger at the same
time increase in delta T min increases both
00:27:59.860 --> 00:28:05.150
Q H min and Q c min. Hence, the load on cold
and hot utilities that increases the operating
00:28:05.150 --> 00:28:11.900
cost when the hot composite and the cold composite
curves just touch then the driving force becomes
00:28:11.900 --> 00:28:15.830
0 and the heat exchanger area requirement
goes to infinity.
00:28:15.830 --> 00:28:21.540
Thus, there is always a tradeoff between energy
and capital cost, so you can see here that
00:28:21.540 --> 00:28:27.690
the total goes through a minimum at delta
T min at with respect to delta T min. There
00:28:27.690 --> 00:28:36.710
is always a delta T optimum your experience
tells that delta T minimum is 10 degrees for
00:28:36.710 --> 00:28:42.540
70 heat exchangers, but for other exchangers
other type of the here could be the this,
00:28:42.540 --> 00:28:47.490
this could vary, so that thing you see on
the screen.
00:28:47.490 --> 00:28:54.590
Now, that is also an economic degree of heat
recovery we cannot go there are some constants
00:28:54.590 --> 00:29:01.480
where which restrict the amount of heat that
could be recovered between hot and cold steps.
00:29:01.480 --> 00:29:06.710
So, the practical constraints on delta T min
are to achieve small delta T min in design
00:29:06.710 --> 00:29:11.210
heat exchanger should exhibit pure counter
current flow. That is a basic aspect that
00:29:11.210 --> 00:29:16.330
we have learnt in heat transfer process with
shell and tube exchanger time is not purely
00:29:16.330 --> 00:29:22.340
counter current even with shell and pass on
shell and tube side. So, the delta T is reduced
00:29:22.340 --> 00:29:27.550
delta T min less than ten degrees is not advised
unless special circumstances prevail.
00:29:27.550 --> 00:29:35.560
For plate heat exchangers delta T min as low
as 5 degrees could be possible or could be
00:29:35.560 --> 00:29:41.951
achieved and this value can go down to 1 or
2 degrees with plate and fin design. These
00:29:41.951 --> 00:29:47.770
constraints apply to only those exchangers
that are placed around the point of closest
00:29:47.770 --> 00:29:53.970
approach between composite curves. Remember,
that these constants apply this; we shall
00:29:53.970 --> 00:29:59.710
come back again when we study the pinch technology
additional constraints apply if vaporization
00:29:59.710 --> 00:30:07.560
condensation occurs at the point of closest
approach. Now, the heat recovery pinch correct
00:30:07.560 --> 00:30:12.340
setting of composite curves is determined
by economic tradeoff between the corresponding
00:30:12.340 --> 00:30:15.290
tradeoff corresponding to economic delta T
00:30:15.290 --> 00:30:19.790
Assume that a correct delta T min is assumed
and this fixes the relative positions of the
00:30:19.790 --> 00:30:25.830
curves. Hence, the energy targets the delta
T min for composite curves and its location
00:30:25.830 --> 00:30:31.510
is important if the energy target is to be
achieved in design of heat exchanger network
00:30:31.510 --> 00:30:37.470
delta T min is normally, but not always observed
at only one point between the hot and cold
00:30:37.470 --> 00:30:44.100
composite curves called as the heat recovery
pinch. The pinch point has the special significance
00:30:44.100 --> 00:30:49.580
in the design of heat exchanger network. We
shall come back to this point again as I said
00:30:49.580 --> 00:30:54.250
when we shall study the pinch technology of
heat exchanger network design.
00:30:54.250 --> 00:31:01.700
How we can design the coupling of the streams,
so has to have maximum heat recovery. Now,
00:31:01.700 --> 00:31:05.700
the trade off suggests that no individual
exchanger should have the delta T min less
00:31:05.700 --> 00:31:11.810
than 10 degrees. Now, this is the good initialization
in heat exchanger network design assume that
00:31:11.810 --> 00:31:17.480
delta T min less than 10 degrees never occurs
the whole process can be divided at pinch
00:31:17.480 --> 00:31:22.100
where the two curves are closest. So, this
point is the pinch point where the delta T
00:31:22.100 --> 00:31:27.870
min is 10 degrees, so the whole process can
be divided at pinch as follows one is the
00:31:27.870 --> 00:31:32.860
below pinch this portion, this envelop and
above pinch which is this envelop.
00:31:32.860 --> 00:31:40.000
Above pinch in terms of temperature the process
is in heat balance with Q H min that is minimum
00:31:40.000 --> 00:31:47.080
hot utility the heat is received from the
utility, but it is not rejected. Thus, a process
00:31:47.080 --> 00:31:55.680
is a heat sink above pinch which means in
this region above pinch is this region here.
00:31:55.680 --> 00:32:01.150
The process is heat sink because heat is been
absorbed from the hot utility that is Q H
00:32:01.150 --> 00:32:06.420
min, but not rejected below pinch process
in a heat balance with Q c min or minimum
00:32:06.420 --> 00:32:10.760
cold utility no heat is recovered, but rejected
to cold utility.
00:32:10.760 --> 00:32:18.720
Thus, the process is a heat source here we
are not absorbing any heat from outside, but
00:32:18.720 --> 00:32:25.530
only rejecting the heat to cold utility Q
c min. So, the process is heat source consider
00:32:25.530 --> 00:32:30.940
possibility of transferring heat between two
systems if it possible to transfer any heat
00:32:30.940 --> 00:32:37.380
from hot streams above pinch into colder streams
below pinch by contrast. The transfer of heat
00:32:37.380 --> 00:32:42.980
from hot streams below pinch into cold streams
above pinch is not possible without violating
00:32:42.980 --> 00:32:48.480
delta T min constraint.
So, what we see now is that suppose you want
00:32:48.480 --> 00:32:55.900
to this is the here the process is divided
at the pinch. This is below pinch, this is
00:32:55.900 --> 00:33:03.950
above pinch, and if you want to pass heat
from the streams above pinch to streams means
00:33:03.950 --> 00:33:10.580
hot streams above pinch to cold streams below
pinch it is possible, but above is not possible.
00:33:10.580 --> 00:33:18.620
The reverse is not possible, you cannot pass
this streams the heat from hot streams below
00:33:18.620 --> 00:33:25.060
pinch to the cold streams above pinch why
because as you go above pinch here. Let us
00:33:25.060 --> 00:33:31.900
say the pinch occurs at delta some T the temperature
here increases and the temperature here decreases.
00:33:31.900 --> 00:33:42.280
So, the delta T min constraint is highlighted
as you go in a heat exchanger from one end
00:33:42.280 --> 00:33:47.590
to the another.
00:33:47.590 --> 00:33:54.410
If a quantity of heat x p is Transferred from
system above pinch to system below pinch there
00:33:54.410 --> 00:34:01.490
is the heat deficit of x p above pinch the
only way to rectify this problem is to import
00:34:01.490 --> 00:34:11.160
an extra x p of heat from hot utility that
will increase Q H min. If you pass some heat
00:34:11.160 --> 00:34:18.560
from above pinch to below pinch streams then
there will be a heat deficit here.
00:34:18.560 --> 00:34:30.389
Then, that deficit can only be fulfilled by
importing more from hot utility Q H min likewise
00:34:30.389 --> 00:34:39.450
an excess of x p below the pinch leads to
export of an extra x p to cold utility analogous
00:34:39.450 --> 00:34:45.250
effects can be caused by improper use of utilities
example is given here. If some cooling water
00:34:45.250 --> 00:34:51.549
is used to cool hot streams above pinch to
satisfy this enthalpy importance above pinch
00:34:51.549 --> 00:35:00.060
Q H min plus x p. It needs to be imported
from the steam or hot utility or moreover
00:35:00.060 --> 00:35:06.799
Q c min plus x p cooling water has to be used,
so that is that is the imbalance between the
00:35:06.799 --> 00:35:12.759
processes.
If you if you give out some portion x p heat
00:35:12.759 --> 00:35:18.900
x p from above pinch to below pinch there
will be heat deficit here. So, here it will
00:35:18.900 --> 00:35:26.420
be Q H min plus x p this much of heat needs
to be imported from hot utility that it will
00:35:26.420 --> 00:35:32.339
just load on hot utility. Similarly, the heat
that is has to be rejected to cold utility
00:35:32.339 --> 00:35:39.180
also goes up by same amount, so Q H min and
Q c min both increase if you transfer heat
00:35:39.180 --> 00:35:46.571
across pinch. Therefore, when we are designing
the heat exchanger network we have to see
00:35:46.571 --> 00:35:52.529
the steams above pinch, we have to see the
streams below pinch. Then make a match between
00:35:52.529 --> 00:35:58.950
the two that that thing we shall come again
we shall come again when we shall see the
00:35:58.950 --> 00:36:00.529
enthalpy intervals.
00:36:00.529 --> 00:36:07.230
Other inappropriate use heating of cold streams
below pinch by steam say by amount x p, this
00:36:07.230 --> 00:36:12.480
is one another improper use in this case the
Q H min must still be supplied above pinch
00:36:12.480 --> 00:36:17.880
to satisfy enthalpy imbalance above pinch.
So, the total utility in this case will again
00:36:17.880 --> 00:36:24.009
be Q H min plus x p above pinch and Q c min
plus x p below pinch to achieve the energy
00:36:24.009 --> 00:36:29.210
targets set by composite curves. The designer
must not transfer heat across pinch by process
00:36:29.210 --> 00:36:33.140
to process heat transfer and inappropriate
use of utilities.
00:36:33.140 --> 00:36:44.849
So, that is how that gives you the essence
of the pinch the heat recovery pinch. So,
00:36:44.849 --> 00:36:49.519
the process is divided at this pinch that
should not be process to process transfer
00:36:49.519 --> 00:36:56.440
across pinch where also should not be inappropriate
use of utilities above and below pinch no
00:36:56.440 --> 00:37:13.559
hot utility below pinch no cold utility above
pinch.
00:37:13.559 --> 00:37:24.930
These rules are both necessary and sufficient
to ensure that the energy target is achieved,
00:37:24.930 --> 00:37:30.710
provided the initialization value is adhered
to that individual heat exchanger operates
00:37:30.710 --> 00:37:35.170
on driving force less than delta T min.
00:37:35.170 --> 00:37:47.529
So, that completes the basic theory of the
heat exchange, heat recovery principle. Now,
00:37:47.529 --> 00:37:53.369
let us see some special cases like threshold
problems not all of the process have a pinch
00:37:53.369 --> 00:38:03.630
dividing the process in parts consider the
curve a in the figure that is shown on left
00:38:03.630 --> 00:38:11.740
hand side of the screen which is constant
coupled to three curves one two and three.
00:38:11.740 --> 00:38:25.890
Now, a is a hot composite curve and we have
only one stream or let us say cold composite
00:38:25.890 --> 00:38:32.800
stream which in with constant slope and that
is shown in three parts. Now, as I said you
00:38:32.800 --> 00:38:40.349
cannot move the cold composite stream horizontally
that will change the delta T min. Now, initially
00:38:40.349 --> 00:38:49.700
position three now in position three you have
both you have Q C min, but not Q H min.
00:38:49.700 --> 00:38:59.440
If you couple a with one coupling A 1 coupling
A 1 both steam and cooling water are required
00:38:59.440 --> 00:39:10.559
as the low curve moves closer to curve a lower
curve means the cold composite curve. As the
00:39:10.559 --> 00:39:14.490
cold composite curve moves closer to curve
a both steam and cooling water requirement
00:39:14.490 --> 00:39:22.791
decrease until the coupling A 2 is achieved
like here A 2 coupling. Now, here you can
00:39:22.791 --> 00:39:32.410
see that a Q H min here becomes 0, because
the target temperature of, sorry let the target
00:39:32.410 --> 00:39:36.550
temperature different the enthalpies are exactly
matching when you have coupling A 2.
00:39:36.550 --> 00:39:43.569
There, is no hot utility required moving the
curves closer decreases the cold utility demand
00:39:43.569 --> 00:39:50.270
at the colder end, but opens up for cold utility
at hot end correspondent to decrease at the
00:39:50.270 --> 00:39:58.279
cold end. So, you require cold utility both
above and below pinch, you can see here that
00:39:58.279 --> 00:40:04.520
once the curve goes to for a three coupling
after a three coupling even after the what
00:40:04.520 --> 00:40:07.740
you say the cold composite curve is absorbed
all the heat.
00:40:07.740 --> 00:40:14.089
There is still heat left with the hot composite
curve that becomes a load on the Q c min moving
00:40:14.089 --> 00:40:18.400
the curves closer decreases cold utility demand
at colder end, but opens up demand for cold
00:40:18.400 --> 00:40:24.059
utility at hot end corresponding to decrease
at the cold end. In other words, as the curves
00:40:24.059 --> 00:40:31.880
move closer beyond A 2 the utility demand
remains constant, so that is how is shown
00:40:31.880 --> 00:40:38.279
in this figure.
This is for case one, case two, case three
00:40:38.279 --> 00:40:48.029
to the three positions as you go beyond three
positions there is no hot utility, but the
00:40:48.029 --> 00:40:53.369
cold utility demand goes up. So, here the
hot stream between this temperature and this
00:40:53.369 --> 00:41:00.920
temperature the hot streams have to be cooled
using cold utility. That is very strange that
00:41:00.920 --> 00:41:06.190
we are using cold utility above pinch because
as I just said that you cannot use cold utility
00:41:06.190 --> 00:41:12.190
above pinch, but in this particular case you
have to use because does constraint of the
00:41:12.190 --> 00:41:13.970
cold composite curve.
00:41:13.970 --> 00:41:19.109
Setting A 2 marks threshold and a problem
exhibiting this feature are known as threshold,
00:41:19.109 --> 00:41:24.880
problems in some problems hot utility requirement
disappears. In some cases the cold utility
00:41:24.880 --> 00:41:28.529
requirement disappears cold utility requirement
disappears. So, what I have plotted here is
00:41:28.529 --> 00:41:34.650
the cost the energy cost decreases as delta
T min decreases as the second curve moves
00:41:34.650 --> 00:41:44.430
closer to the curve A. Then after delta T
threshold it remains constant but the capital
00:41:44.430 --> 00:41:52.799
cost keeps on increasing as delta T min decreases
the capital cost keeps on increasing.
00:41:52.799 --> 00:41:58.930
So, from here you have an optimum whether
the delta T min optimum will be the delta
00:41:58.930 --> 00:42:04.991
T threshold that depends on particular case
like for example in this case in case a the
00:42:04.991 --> 00:42:11.530
T optimum and T threshold are same. But in
certain cases they may not be same like the
00:42:11.530 --> 00:42:18.299
total cost may show least at some T optimum,
but the threshold shows the energy requirement
00:42:18.299 --> 00:42:25.170
becomes constant and after certain delta T
min a threshold. So, in this case the optimum
00:42:25.170 --> 00:42:29.800
delta T optimum as per the cost is not same
as delta T threshold, but in some cases it
00:42:29.800 --> 00:42:33.599
is ok.
00:42:33.599 --> 00:42:42.670
Now, let us see the capital and energy trade
off for threshold problem the two possible
00:42:42.670 --> 00:42:48.789
outcomes case A and case B, which I just discussed
for capital energy trade off below threshold
00:42:48.789 --> 00:42:54.369
delta T min utility demand. Energy cost is
constant graph A shows this situation here
00:42:54.369 --> 00:42:59.880
the optimum occurs at threshold delta T min.
In the second case, the same optimum occurs
00:42:59.880 --> 00:43:05.490
above threshold delta T min the flat profile
of energy costs below threshold delta T min
00:43:05.490 --> 00:43:13.990
means that optimum can never occur below threshold
value it can occur only at or above the threshold
00:43:13.990 --> 00:43:15.410
00:43:15.410 --> 00:43:23.280
Now, in case A where that optimum delta T
min is at threshold, there is pinch. On the
00:43:23.280 --> 00:43:29.700
other hand, for case B with optimum value
above threshold value there is demand for
00:43:29.700 --> 00:43:36.869
both utilities and then and there is a pinch,
now threshold problems are quite common. In
00:43:36.869 --> 00:43:44.970
this case, we can introduce a pinch although
when we actually plot the process curves.
00:43:44.970 --> 00:43:51.789
We absorbed no pinch, but we can introduce
a pinch by addition of utilities like for
00:43:51.789 --> 00:43:59.559
example, in which case A which we have shown
the hot composite curve remains the same for
00:43:59.559 --> 00:44:05.570
the cold composite curve. We divided into
two parts, the first part and then here we
00:44:05.570 --> 00:44:08.750
shall have intermediate, we shall have steam
generation.
00:44:08.750 --> 00:44:19.579
Then, we have the cold utility, so that generates
a pinch, another this is so likely utility
00:44:19.579 --> 00:44:26.800
addition on cold composite curve in some cases
like here, sorry not steam generation means
00:44:26.800 --> 00:44:36.849
we are using steam here to keep the to heat
up. The cold stream in between in some cases
00:44:36.849 --> 00:44:48.259
you can use the steam generation, so this
is steam supply here we have steam generation
00:44:48.259 --> 00:44:56.589
hp steam generation here. What will happen
before coupling the hot stream to the cold
00:44:56.589 --> 00:45:02.220
stream, we shall first use it for high pressure
steam generation to reduce the enthalpy add
00:45:02.220 --> 00:45:06.119
constraint temperature and then in certain
portion we shall couple.
00:45:06.119 --> 00:45:14.749
Then, here we introduce the pinch and there
after again we have low pressure steam generation
00:45:14.749 --> 00:45:21.099
after taking up certain heat, then we shall
couple to the composite stream. So, these
00:45:21.099 --> 00:45:39.490
are the actual heat recovery areas, so these
are the Q recovery areas and here we have
00:45:39.490 --> 00:45:44.799
intermediate utility generation.
00:45:44.799 --> 00:45:53.349
So, these points I have noted which I just
discussed case A shows the composite curves,
00:45:53.349 --> 00:45:59.339
but with two levels of utility instead of
one the second cold utility is steam generation
00:45:59.339 --> 00:46:07.130
introduction of second utility causes a pinch
known as utility pinch. Then case B shows
00:46:07.130 --> 00:46:11.829
composite curves with two levels of steam
introduction of the second steam level causes
00:46:11.829 --> 00:46:17.809
utility pinch. Same rules must be obeyed around
utility pinch as around the process pinch
00:46:17.809 --> 00:46:22.549
the heat should not be transferred across
the pinch by process to process heat transfer
00:46:22.549 --> 00:46:27.819
and there should not be an inappropriate use
of utilities.
00:46:27.819 --> 00:46:35.299
This means, that in case a the only utility
is to be used above utility pinch is the steam
00:46:35.299 --> 00:46:43.569
generation and cooling water below in case
B, the only utility is to be used only utility
00:46:43.569 --> 00:46:51.089
is to be used above pinch is the high pressure
steam and only low pressure steam. So, that
00:46:51.089 --> 00:46:58.499
completes the basic discussion on principles
of heat integration. In the next lecture,
00:46:58.499 --> 00:47:05.460
we shall see has how we can determine the
minimum hot and cold utility requirement for
00:47:05.460 --> 00:47:13.200
a particular process using problem table algorithm.
Then we shall also see some practical constraints
00:47:13.200 --> 00:47:15.559
on the amount of heat that is recovered.
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