/ News & Press / Video / Mod-05 Lec-01 Concepts and Basic Principles of Energy (or Heat) Integration -- Part 1
Mod-05 Lec-01 Concepts and Basic Principles of Energy (or Heat) Integration -- Part 1
WEBVTT Kind: captions Language: en
00:00:20.680 --> 00:00:26.340 Welcome, we will start today the fifth module of our course that is heat integration of 00:00:26.340 --> 00:00:35.840 the process. So far we have studied the protocol of design of a process. How the design evolves 00:00:35.840 --> 00:00:41.590 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 00:00:46.140 --> 00:00:50.980 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 00:00:55.790 --> 00:00:56.790 process. 00:00:56.790 --> 00:01:03.770 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 00:02:35.030 --> 00:02:41.690 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. 00:02:54.780 --> 00:03:00.360 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 00:03:20.049 --> 00:03:28.070 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 00:03:58.389 --> 00:04:03.290 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 00:04:23.060 --> 00:04:29.130 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 00:04:34.710 --> 00:04:39.900 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 00:04:45.990 --> 00:04:49.210 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 00:05:14.180 --> 00:05:20.370 the target temperature it needs to be cooled in 40 degrees. It has minus 12 mega volt that 00:05:20.370 --> 00:05:26.729 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 00:05:55.169 --> 00:06:03.180 you see, now the two steams the hot stream is available at 160 degrees it is getting 00:06:03.180 --> 00:06:10.259 cooled to 40 degrees. Then the cold stream is available at 20 degrees and it is getting 00:06:10.259 --> 00:06:42.340 cool heated to 100 and 10 degrees. Let us see, now how you can do it, now I am plotting 00:06:42.340 --> 00:06:49.460 the two curves and that you see on the screen, now below here you will see the same steams 00:06:49.460 --> 00:06:55.259 plotted in the form of flow sheet. Now, we have to decide the minimum if these 00:06:55.259 --> 00:07:02.129 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 00:07:21.620 --> 00:07:29.080 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 00:07:37.719 --> 00:07:42.470 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. 00:07:46.139 --> 00:07:54.860 So, we have the hot utility steam available at 180 degree and cooling water at 120 degrees 00:07:54.860 --> 00:08:02.629 then heating all cold streams with steam and cooling with all hot streams with water would 00:08:02.629 --> 00:08:08.090 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 00:08:36.409 --> 00:08:45.321 T minimum equal to 10 degree centigrade. We have a heat recovery of 11 mega Watts, we 00:08:45.321 --> 00:08:52.970 have a heat recovery of 11 mega Watts, but despite this even after recovering you see 00:08:52.970 --> 00:09:00.389 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 00:09:07.000 --> 00:09:11.860 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 00:09:48.760 --> 00:09:54.209 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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