00:00:00.180 [Music] 00:00:13.20900:00:13.219 hello students today we will learn that 00:00:15.25900:00:15.269 what is an heat exchanger and what are 00:00:19.18900:00:19.199 the basic attributes in regards of heat 00:00:21.38000:00:21.390 exchangers what other types of heat 00:00:22.67000:00:22.680 exchangers and we'll also learn that 00:00:24.65000:00:24.660 what is the concept of LMT D that is 00:00:27.20000:00:27.210 logarithmic mean temperature difference 00:00:29.83000:00:29.840 okay first of all remember this thing 00:00:33.29000:00:33.300 that heat exchanger is a device in which 00:00:36.22900:00:36.239 heat is transferred between two flowing 00:00:39.38000:00:39.390 fluids without mixing of each other 00:00:42.04900:00:42.059 means there are two fruits one is hot 00:00:45.02000:00:45.030 and other is cold okay and heat is 00:00:48.08000:00:48.090 transferred from the hot fruit to cold 00:00:50.66000:00:50.670 flu however they are not mixed with each 00:00:53.09000:00:53.100 other and heat exchangers are the 00:00:56.29900:00:56.309 devices which are widely used in 00:00:58.27900:00:58.289 industries for corresponding purpose the 00:01:01.18900:01:01.199 purpose is the main purpose in 00:01:03.59000:01:03.600 industries in the regards of this heat 00:01:05.71900:01:05.729 exchanger is that when we want one cold 00:01:08.96000:01:08.970 fluid to receive heat from some hot food 00:01:12.28900:01:12.299 for example suppose there is a dairy 00:01:14.00000:01:14.010 industry in which pasteurization process 00:01:15.62000:01:15.630 is happening so what happens generally 00:01:17.74900:01:17.759 the common practice of pasteurization of 00:01:19.67000:01:19.680 milk in dairy industry is there is a 00:01:21.49900:01:21.509 steam hot steam okay so by the help of 00:01:24.49900:01:24.509 hot steam the heat is transferred to the 00:01:27.20000:01:27.210 cold milk okay so heat of hot steam is 00:01:30.46900:01:30.479 transferring to the cold mnek the 00:01:32.63000:01:32.640 temperature of milk is rising and we 00:01:34.39900:01:34.409 noticing that pasteurization process is 00:01:35.89900:01:35.909 what that we have to increase the 00:01:38.06000:01:38.070 temperature of milk to 80 degree 00:01:40.01000:01:40.020 centigrade and by this way the bacterias 00:01:42.85900:01:42.869 and the other harmful germs contained in 00:01:45.59000:01:45.600 the milk can be killed 00:01:46.67000:01:46.680 so the common practice I told you that 00:01:49.49000:01:49.500 by the help of steam high temperature 00:01:51.35000:01:51.360 steam by the help of an heat exchanger 00:01:53.89900:01:53.909 the heat is supplied to the cold milk so 00:01:56.38900:01:56.399 the cold milk the temperature of Coleman 00:01:58.19000:01:58.200 Rises and by this way the pasteurization 00:02:00.46900:02:00.479 process completes and for an example 00:02:03.88900:02:03.899 real-life example making tea over the 00:02:07.16000:02:07.170 stove is also a sort of heat exchanging 00:02:10.16000:02:10.170 process over here I have surely 00:02:12.19900:02:12.209 the small diagram this is the utensil in 00:02:15.05000:02:15.060 which this is the liquid the T is liquid 00:02:17.83900:02:17.849 we know this thing okay so below it 00:02:20.66000:02:20.670 there is a stove so what is happening 00:02:22.30900:02:22.319 the flame consists of flue gases flue 00:02:24.86000:02:24.870 gases are fluids okay so these high 00:02:26.95900:02:26.969 temperature gases when touches this 00:02:29.27000:02:29.280 bottom of potential transfers its heat 00:02:32.44900:02:32.459 to this utensil outer surface ultimately 00:02:35.89900:02:35.909 heat conducts toward this T means T is 00:02:39.11000:02:39.120 acting as a cold fluid and hot flue 00:02:41.78000:02:41.790 gases as a hot food so heat exchanger is 00:02:44.50900:02:44.519 a device in which heat is transferred 00:02:45.89000:02:45.900 from a high temperature fuel to a load 00:02:48.44000:02:48.450 of the food without mixing with each 00:02:50.86900:02:50.879 other okay now let us talk that how many 00:02:54.37900:02:54.389 types of heat engines are there see 00:02:56.89900:02:56.909 these are some commonly used heat 00:03:00.34900:03:00.359 exchanger in Hinda cities first is 00:03:03.25900:03:03.269 double pipe or tube in tube heat 00:03:05.36000:03:05.370 exchanger if this type of heat exchange 00:03:07.61000:03:07.620 what happens there is a thin tube and it 00:03:10.84900:03:10.859 is enveloped inside a big diameter tube 00:03:13.78000:03:13.790 okay both are Co centric okay what 00:03:17.47900:03:17.489 happens from this inner tube one fuel 00:03:20.36000:03:20.370 flows freely hot food okay so from this 00:03:23.56900:03:23.579 inner 00:03:24.05000:03:24.060 fluid I am showing by this red colored 00:03:26.78000:03:26.790 food it is hot food and from this outer 00:03:29.21000:03:29.220 fuel cold fluid is flowing so what 00:03:31.81900:03:31.829 happens when these two fruits are 00:03:33.40900:03:33.419 flowing then the heat is transferred 00:03:35.68900:03:35.699 from hot fluid to the cold fruit through 00:03:38.21000:03:38.220 the walls of the tube means heat is 00:03:41.05900:03:41.069 first of all transferred to inner 00:03:43.28000:03:43.290 surface of the wall then it conducts 00:03:45.14000:03:45.150 through the thickness of wall and 00:03:46.28000:03:46.290 ultimately goes to the cold fluid it is 00:03:48.89000:03:48.900 like this it is sealed with the either 00:03:50.96000:03:50.970 end so that flute cannot skip out or 00:03:53.92900:03:53.939 leak out from the system so this type of 00:03:57.02000:03:57.030 heat exchanger is called as tube in tube 00:03:58.84900:03:58.859 or double pipe it is now in this type of 00:04:01.67000:04:01.680 utensil also there are two types one is 00:04:05.03000:04:05.040 we are using tube in tube heat exchanger 00:04:07.24900:04:07.259 in parallel flow mode other is we are 00:04:09.80000:04:09.810 using in counter flow mode now what is 00:04:11.50900:04:11.519 parallel flow see what happens in case 00:04:14.36000:04:14.370 both the fluids hot and cold are 00:04:16.89900:04:16.909 entering to this cube into petitioners 00:04:19.31000:04:19.320 from the same side okay in that case the 00:04:21.97900:04:21.989 direction of flow remains the same means 00:04:24.05000:04:24.060 hot fluid is moving 00:04:25.55000:04:25.560 for this it is in case I'm download then 00:04:27.62000:04:27.630 hot food is moving from left to right 00:04:29.12000:04:29.130 and cold is also moving from left to 00:04:30.89000:04:30.900 right so this type of arrangement this 00:04:33.29000:04:33.300 type of flow is called as parallel flow 00:04:35.60000:04:35.610 heat a senior the make is double five or 00:04:38.30000:04:38.310 even two but this arrangement in which 00:04:40.25000:04:40.260 the hot and cold foods are entering from 00:04:42.89000:04:42.900 the same side and ultimately comes out 00:04:45.80000:04:45.810 from the other opposite side then such 00:04:47.87000:04:47.880 type of flow is called external flow and 00:04:49.90900:04:49.919 in case one fluid is entering from one 00:04:52.64000:04:52.650 side suppose in digital I am swinging by 00:04:54.95000:04:54.960 arrow that cold fluid is entering to 00:04:57.44000:04:57.450 this outer peripheral tube from one side 00:04:59.65900:04:59.669 while hot fluid is entering from the 00:05:01.57000:05:01.580 opposite end to this inner central do so 00:05:04.64000:05:04.650 in this case what is happening both the 00:05:05.93000:05:05.940 fluids are moving opposite to each other 00:05:07.79000:05:07.800 so such type of flow is called as 00:05:09.56000:05:09.570 counter flow heat Arsenal means double 00:05:11.65900:05:11.669 pipe counter flow Here I am showing the 00:05:14.42000:05:14.430 magnified view of this to be witticism 00:05:17.15000:05:17.160 you can see this is the center tube 00:05:19.04000:05:19.050 through which hot fluid is flowing and 00:05:20.78000:05:20.790 this is the outer peripheral tube 00:05:22.40000:05:22.410 through which code through this flow and 00:05:23.84000:05:23.850 heat from this central tube from this 00:05:26.33000:05:26.340 fluid which is flowing through the 00:05:27.50000:05:27.510 central tube is transferred through the 00:05:29.45000:05:29.460 walls of this inner tube to the cold 00:05:32.12000:05:32.130 fluid you can see this the direction of 00:05:33.40900:05:33.419 flow of heat it is like this other type 00:05:35.71900:05:35.729 of make is plate type it is here by the 00:05:38.39000:05:38.400 help of arrow I am showing it plate type 00:05:39.98000:05:39.990 it is here it is consisting of one plate 00:05:43.07000:05:43.080 or another okay this is a magnified view 00:05:45.50000:05:45.510 okay suppose it is consisting of five 00:05:48.40900:05:48.419 states 1 2 3 4 5 ok so you can see 00:05:52.99000:05:53.000 between first 2 plates cold fluid is 00:05:55.96900:05:55.979 flowing and in the immediate next 2 00:05:58.67000:05:58.680 plate hot fluid is flowing then again 00:06:00.62000:06:00.630 cold food and again hot food so 00:06:02.60000:06:02.610 arrangement is of such kind that between 00:06:04.64000:06:04.650 two plates some cold fluid is flowing 00:06:07.43000:06:07.440 and immediate next the two plates hot 00:06:10.04000:06:10.050 fluid is flowing so the arrangement is 00:06:11.90000:06:11.910 of such time that a hot fluid layer is 00:06:15.05000:06:15.060 sandwiched between two cold food layer 00:06:17.15000:06:17.160 and one cold food layer is sandwich 00:06:19.12900:06:19.139 between two what fruit layers by the 00:06:21.05000:06:21.060 help of arrangement of plates so such 00:06:23.24000:06:23.250 type of it as singer is called as plate 00:06:25.21900:06:25.229 type eternal it is widely used type of 00:06:27.46900:06:27.479 attention in industries okay so this 00:06:29.71900:06:29.729 blue colored food is cold food and this 00:06:33.32000:06:33.330 red colored food is hot food blue 00:06:36.64000:06:36.650 again code hard so between two hot fuel 00:06:41.14000:06:41.150 strings there is one cold fluid stream 00:06:42.87900:06:42.889 and between two co2 stream there is one 00:06:44.92000:06:44.930 hot photostream 00:06:45.61000:06:45.620 again this type of heater cell can also 00:06:47.86000:06:47.870 be parallel so in case fuels are 00:06:49.71900:06:49.729 entering from same end and coming out 00:06:51.90900:06:51.919 from the same opposite end then well 00:06:53.89000:06:53.900 also in case the fluid flow is of such 00:06:56.40900:06:56.419 kind that a hot fluid is entering from 00:06:59.37900:06:59.389 one end while cold food is entering from 00:07:00.79000:07:00.800 other opposite end then it is plate type 00:07:02.95000:07:02.960 counter flow in case one fluid is 00:07:05.40900:07:05.419 flowing suppose I am taking this heaters 00:07:07.42000:07:07.430 inner as shown in the diagram from left 00:07:09.27900:07:09.289 to right other fluid is flowing inside 00:07:11.32000:07:11.330 the screen okay in such case the two 00:07:14.02000:07:14.030 fluids are moving making an angle 90 00:07:16.12000:07:16.130 degree to each other first type of 00:07:17.74000:07:17.750 arrangement is called as cross flow type 00:07:19.68900:07:19.699 of heater snare now this is the shell 00:07:24.15900:07:24.169 and you with a fringe inshalla tube heat 00:07:26.14000:07:26.150 exchanger what happens that there is a 00:07:28.27000:07:28.280 big shell okay through which one fluid 00:07:30.85000:07:30.860 is passing for example by the help of 00:07:32.83000:07:32.840 arrow I am showing a type that is single 00:07:36.55000:07:36.560 shell 2 to power shell and tube type of 00:07:38.80000:07:38.810 addiction here suppose cold fluid is 00:07:40.83900:07:40.849 entering from this end and ultimately 00:07:42.55000:07:42.560 coming out of the other end okay and 00:07:44.62000:07:44.630 there is a tube pass inside the shell 00:07:47.44000:07:47.450 you can see through this tube 00:07:49.06000:07:49.070 hot fluid is flowing okay so actually 00:07:52.06000:07:52.070 what is happening that cold fluid is 00:07:53.92000:07:53.930 flowing through a big shell while hot 00:07:55.75000:07:55.760 fluid is flowing through a tube 00:07:57.31000:07:57.320 so this is the reason it is called a 00:07:58.60000:07:58.610 shell and tube heat exchanger okay so it 00:08:01.06000:08:01.070 also consists of its various types for 00:08:04.08900:08:04.099 example this is the case of single shell 00:08:06.73000:08:06.740 and to to pass the reason it the reason 00:08:09.12900:08:09.139 is that this is the shell and cube is 00:08:12.33900:08:12.349 going in and then coming out so it is to 00:08:14.89000:08:14.900 tube pass okay once going in and then 00:08:17.17000:08:17.180 coming out one pass to pass slowly this 00:08:20.50000:08:20.510 type is to shell for to pass because it 00:08:23.14000:08:23.150 consists of two shell cold food is 00:08:25.51000:08:25.520 flowing through this shell and hot food 00:08:28.96000:08:28.970 is passing ones going in and then coming 00:08:32.17000:08:32.180 out and then from this other shell going 00:08:35.35000:08:35.360 in and then coming out so it is two 00:08:37.08900:08:37.099 shell because shells are two 00:08:38.44000:08:38.450 however the hot fluid is passing four 00:08:41.68000:08:41.690 times that is in out and then in out so 00:08:44.19900:08:44.209 two shell for to pass and this third one 00:08:47.65000:08:47.660 is single shell for to pass because it 00:08:49.69000:08:49.700 consists of 00:08:50.44000:08:50.450 one shell only old fluid is passing 00:08:52.81000:08:52.820 through this shell and then what fruit 00:08:55.30000:08:55.310 is going and then turning and then again 00:08:58.00000:08:58.010 turning and then again Turing and then 00:08:59.29000:08:59.300 going out so it is called as single 00:09:01.51000:09:01.520 shell for tube pass this is also shell 00:09:04.48000:09:04.490 and I would beautiful in which old food 00:09:07.51000:09:07.520 is flowing through the shell suppose 00:09:09.76000:09:09.770 from left hand side to right hand side 00:09:11.23000:09:11.240 while the hot fluid is passing in a 00:09:14.14000:09:14.150 cross flow arrangement to this flow of 00:09:16.66000:09:16.670 cold food because see these are the hot 00:09:19.30000:09:19.310 flue tubes so these tubes are running 00:09:21.76000:09:21.770 inside the screen of this computer okay 00:09:24.70000:09:24.710 so cold fluid is flowing from left to 00:09:27.52000:09:27.530 right while hot fluid is flowing at an 00:09:29.65000:09:29.660 angle 90 degree to the cold fluid such 00:09:32.20000:09:32.210 type of eternal is called a shell and 00:09:34.18000:09:34.190 tube cross flow type of heat exchanger 00:09:36.61000:09:36.620 however these diagrams are just 00:09:38.14000:09:38.150 indicating just to make you understand 00:09:40.66000:09:40.670 the concept of these different types of 00:09:43.57000:09:43.580 heat exchangers now next case you will 00:09:45.82000:09:45.830 talk about the heat transfer rate you 00:09:47.38000:09:47.390 know what is heat afraid the amount of 00:09:48.85000:09:48.860 heat transferred per unit time is called 00:09:50.59000:09:50.600 as heat transfer rate okay remember in 00:09:53.17000:09:53.180 case there are two bodies which are at 00:09:55.57000:09:55.580 temperatures t1 and t2 and t1 is greater 00:09:58.90000:09:58.910 than t2 okay then what happens heat 00:10:01.57000:10:01.580 transfers always from high temperature 00:10:03.61000:10:03.620 body to low down visibility 00:10:05.02000:10:05.030 okay this temperature this body is at 00:10:07.81000:10:07.820 higher temperature and this body is at 00:10:09.73000:10:09.740 lower temperature t2 t1 is higher then 00:10:11.83000:10:11.840 t2 ok so heat will is transfer from 00:10:14.44000:10:14.450 items it will double now the heat 00:10:16.63000:10:16.640 transfer rate formula the general 00:10:18.31000:10:18.320 formula for heat transfer rate in case 00:10:20.11000:10:20.120 heat is transferred between two bodies 00:10:21.37000:10:21.380 is given by Q Q stands for heat transfer 00:10:23.74000:10:23.750 it is equals to you a t1 minus t2 here U 00:10:28.33000:10:28.340 is called as overall heat transfer 00:10:30.01000:10:30.020 coefficient which is a constant in 00:10:31.69000:10:31.700 regards of those heat transferring 00:10:33.93000:10:33.940 bodies a is the area through which heat 00:10:38.05000:10:38.060 is transferring okay a is the area 00:10:41.02000:10:41.030 through which heat is transferring and 00:10:42.55000:10:42.560 t1 minus c2 is the temperature 00:10:44.23000:10:44.240 difference okay so U is overall heat 00:10:47.62000:10:47.630 transfer coefficient a is the area of 00:10:49.18000:10:49.190 exposure through which heat is 00:10:50.11000:10:50.120 transferring and TN minus between the 00:10:51.43000:10:51.440 temperature difference this is what the 00:10:52.78000:10:52.790 heat transfer force formula but let us 00:10:55.51000:10:55.520 take an example of parallel flow heat 00:10:58.24000:10:58.250 exchanger and cube and cube type okay 00:11:00.85000:11:00.860 so hot flow is entering inside this 00:11:03.10000:11:03.110 center 00:11:03.67000:11:03.680 suppose from left hand side and cold 00:11:05.82900:11:05.839 fluid is entering from the same left 00:11:07.75000:11:07.760 hand side in this outer periphery tube 00:11:10.00000:11:10.010 and both are moving parallel to each 00:11:12.04000:11:12.050 other 00:11:12.31000:11:12.320 now when hot fluid is entering just 00:11:15.28000:11:15.290 entering this heat exchanger at that 00:11:17.62000:11:17.630 time the hot foods temperature is 00:11:20.05000:11:20.060 highest suppose it is thi the hot to 00:11:22.81000:11:22.820 temperatures inland similarly when cold 00:11:25.09000:11:25.100 dude is entering to this outer 00:11:26.35000:11:26.360 peripherals to do at that time it is 00:11:28.96000:11:28.970 having its lowest temperature because no 00:11:31.24000:11:31.250 heat it has gained it is just entering 00:11:33.34000:11:33.350 inside the system now when hot fluid 00:11:35.71000:11:35.720 will move inside this tube by and bias 00:11:38.47000:11:38.480 since it is giving its heat to cold fuel 00:11:40.78000:11:40.790 so hot fluids temperature will by an by 00:11:44.13900:11:44.149 degrees and ultimately the hot fluid 00:11:46.57000:11:46.580 will come out this temperature th oh 00:11:48.66000:11:48.670 okay similarly since this cold fluid is 00:11:51.88000:11:51.890 gaining heat receiving heat from this 00:11:54.31000:11:54.320 hot food so continuously by and by its 00:11:56.74000:11:56.750 temperature will increase and it will 00:11:58.54000:11:58.550 come out ultimately from this opposite 00:12:00.61000:12:00.620 end with temperature TC you okay so this 00:12:03.16000:12:03.170 is the graph the temperature gradient of 00:12:06.43000:12:06.440 the two fluids for the case of parallel 00:12:08.82900:12:08.839 flow heat exchanger okay now heat 00:12:11.41000:12:11.420 transfer rate formula Q equals to u 00:12:14.11000:12:14.120 eight t1 minus t2 where t1 - t2 is the 00:12:16.30000:12:16.310 temperature difference okay but in this 00:12:18.73000:12:18.740 case what we see that from beginning to 00:12:21.67000:12:21.680 the end the temperature difference 00:12:24.25000:12:24.260 between the two fluids is continuously 00:12:26.07900:12:26.089 changing okay so what we can keep in 00:12:29.80000:12:29.810 this place in place of T 1 minus C 2 00:12:31.93000:12:31.940 what temperature difference we have to 00:12:33.94000:12:33.950 keep this is the question initially 00:12:35.71000:12:35.720 there is greatest temporal difference by 00:12:38.14000:12:38.150 and by the hot fluid is cooling down and 00:12:40.84000:12:40.850 cold food is heating up so the 00:12:42.69900:12:42.709 temperature difference is decreasing so 00:12:44.17000:12:44.180 in place of t1 minus t2 what we can keep 00:12:46.81000:12:46.820 the concept is that since the 00:12:50.74000:12:50.750 temperature difference is continuously 00:12:52.90000:12:52.910 decreasing so we have to take the 00:12:55.99000:12:56.000 average of all those temperature 00:12:58.54000:12:58.550 differences called as LMT 00:13:00.76000:13:00.770 the full form is logarithmic mean 00:13:03.16000:13:03.170 temperature difference the significance 00:13:04.87000:13:04.880 of logarithmic mean temperature 00:13:06.46000:13:06.470 difference is that however the 00:13:08.86000:13:08.870 temperature difference of the two fluids 00:13:10.48000:13:10.490 is continuously decreasing but we can 00:13:13.93000:13:13.940 assume that from beginning to end of 00:13:16.48000:13:16.490 heat exchanger 00:13:17.44000:13:17.450 the temperature difference is constant 00:13:19.72000:13:19.730 that is the logarithmic mean temperature 00:13:22.66000:13:22.670 difference which is the hypothetical 00:13:25.03000:13:25.040 temporal difference means it is mean of 00:13:26.68000:13:26.690 all those LC temper the difference is 00:13:29.02000:13:29.030 continually decreasing but the mean of 00:13:30.76000:13:30.770 all those separate events we can assume 00:13:33.07000:13:33.080 that both the fluids are transferring 00:13:35.29000:13:35.300 their heat with this constant 00:13:36.91000:13:36.920 temperature differences hypothetical 00:13:38.53000:13:38.540 assumption which is called as LM TD 00:13:40.90000:13:40.910 logarithmic means of the difference and 00:13:42.70000:13:42.710 the formula of logarithm in term of the 00:13:45.04000:13:45.050 difference is LM TD equals to theta 2 00:13:48.10000:13:48.110 minus theta 1 upon log theta 2 beta 1 00:13:51.43000:13:51.440 here what is Theta 2 theta 2 is the 00:13:54.19000:13:54.200 finite temperature difference of the two 00:13:56.47000:13:56.480 fruits you know that fluid is hot food 00:13:59.11000:13:59.120 is coming out with temperature th oh and 00:14:00.85000:14:00.860 cold food is coming out with a media TCO 00:14:02.77000:14:02.780 so this final temperature difference is 00:14:04.60000:14:04.610 thi minus TC oh it is what theta 2 is 00:14:07.00000:14:07.010 and theta 1 is the initial temperature 00:14:09.61000:14:09.620 difference thi minus TCI okay so in 00:14:13.06000:14:13.070 place of theta 2 and theta 1 we can keep 00:14:15.52000:14:15.530 these corresponding values to get the 00:14:17.59000:14:17.60000:14:19.06000:14:19.070 once again what is the significance of 00:14:20.98000:14:20.990 logarithm interval difference however 00:14:22.81000:14:22.820 the temperature difference of the two 00:14:24.70000:14:24.710 fluid is continuously changing 00:14:26.92000:14:26.930 but this is an hypothetical constant 00:14:30.52000:14:30.530 temperature difference that we can 00:14:32.50000:14:32.510 assume that heat is transferring some 00:14:35.23000:14:35.240 hot fruit to cold food with this 00:14:37.51000:14:37.520 constant temperature difference through 00:14:39.46000:14:39.470 the entire run of this heat exchanger 00:14:41.71000:14:41.720 now we have a fixed temperature 00:14:44.35000:14:44.360 difference at a lambda T so here Q we go 00:14:47.41000:14:47.420 so you ATO not see too in case of T nos 00:14:49.54000:14:49.550 - I can keep Q equals to UA 00:14:51.58000:14:51.590 lmpd which is the average constant 00:14:54.01000:14:54.020 temperature difference of the two foods 00:14:55.99000:14:56.000 which are running through the heat 00:14:57.10000:14:57.110 exchanger okay now we will take the case 00:15:01.00000:15:01.010 of counterfeit exchanger suppose this is 00:15:03.94000:15:03.950 a tube in tube double pipe counterfeit 00:15:07.03000:15:07.040 Effinger so hot fluid is entering from 00:15:10.15000:15:10.160 left hand side to this inner central 00:15:12.76000:15:12.770 tube it is flowing toward right hand 00:15:15.16000:15:15.170 side and cold fluid is entering from 00:15:17.47000:15:17.480 right hand side to this outer 00:15:20.41000:15:20.420 peripheral tube and ultimately coming 00:15:23.08000:15:23.090 out from this left hand side okay so 00:15:26.05000:15:26.060 when hot fluid is entering inside this 00:15:27.94000:15:27.950 inner center tube it is super hot 00:15:30.50000:15:30.510 it is having with this highest 00:15:31.82000:15:31.830 temperature okay and by and by when it 00:15:37.28000:15:37.290 will move inside the tube it will lose 00:15:38.96000:15:38.970 its heat so it will come out from the 00:15:40.73000:15:40.740 opposite end with temperature Thu which 00:15:43.67000:15:43.680 is lower than thi okay 00:15:45.35000:15:45.360 similarly cold fluid is entering from 00:15:47.47000:15:47.480 right-hand side so it has temperature TC 00:15:50.57000:15:50.580 a lowest temperature since by-and-by it 00:15:52.91000:15:52.920 is gaining heat from this high 00:15:54.77000:15:54.780 temperature so ultimately it comes out 00:15:57.08000:15:57.090 from the opposite end is temperature TC 00:15:58.85000:15:58.860 OH 00:15:59.36000:15:59.370 okay so these are the temperature 00:16:01.94000:16:01.950 profiles with respect to the advance of 00:16:04.46000:16:04.470 food flowing inside this double pipe 00:16:06.68000:16:06.690 fitter fringes okay now in this case 00:16:09.41000:16:09.420 also the heat transfer rate formula 00:16:11.30000:16:11.310 remains the same that is equal to UA LM 00:16:13.52000:16:13.530 TD LM TD corresponds to the mean 00:16:15.77000:16:15.780 temperature difference logarithmic be 00:16:17.72000:16:17.730 interpretations because over here also 00:16:19.55000:16:19.560 the case is same that from one end to 00:16:23.69000:16:23.700 other okay the temperature difference is 00:16:26.54000:16:26.550 continuously changing however in this 00:16:28.04000:16:28.050 diagram it might be appearing the 00:16:30.59000:16:30.600 temporal difference is same but 00:16:32.21000:16:32.220 practically at every point of this heat 00:16:35.36000:16:35.370 exchanger the temperature difference can 00:16:38.15000:16:38.160 be different okay so again we have to 00:16:40.58000:16:40.590 take the concept of logarithmic mean 00:16:42.26000:16:42.270 temperature difference which is the 00:16:43.25000:16:43.260 hypothetical constant temperature 00:16:45.20000:16:45.210 difference which we can assume for the 00:16:46.91000:16:46.920 flowing fruits in the heat exchanger so 00:16:49.10000:16:49.110 the formula of LM Katti remains the same 00:16:52.22000:16:52.230 as like that of parallel flow 00:16:53.60000:16:53.610 temperature that is equals to theta 2 00:16:56.66000:16:56.670 minus theta 1 identity equal to Z 2 00:16:58.25000:16:58.260 minus theta 1 upon log 1002 Waseda 1 but 00:17:00.50000:17:00.510 the definition of theta 1 and theta 2 00:17:02.81000:17:02.820 changes over here what you have to take 00:17:04.55000:17:04.560 that theta 1 is the temperature 00:17:08.48000:17:08.490 difference over the left hand side of 00:17:11.09000:17:11.100 this heat exchanger means over this and 00:17:14.41000:17:14.420 the leftmost end the temperature of hot 00:17:18.53000:17:18.540 food is thi while the cold food is TCO 00:17:20.84000:17:20.850 so theta 1 would be thi minus TCI OH 00:17:23.75000:17:23.760 similarly theta 2 would be what it is 00:17:26.66000:17:26.670 the temperature difference of the 00:17:28.93000:17:28.940 rightmost side of this heater when this 00:17:32.54000:17:32.550 code food is just entering and hot food 00:17:35.09000:17:35.100 is just a fitting so that is equal to 00:17:36.77000:17:36.780 thi minus TCI okay this theta 2 so the 00:17:40.07000:17:40.080 only change is you will use the same 00:17:42.11000:17:42.120 formula for heat transfer it gives us 2 00:17:44.28000:17:44.290 LM trading birds up in formula of 00:17:46.65000:17:46.660 elaborate e the definition of theta 2 00:17:48.36000:17:48.370 and theta 1 is this theta 1 equals to 00:17:50.01000:17:50.020 thi minus TC oh and theta 2 is th over 00:17:53.34000:17:53.350 Seca so you can keep remember is this 00:17:56.13000:17:56.140 thing by this way that we have to take 00:17:59.10000:17:59.110 the temperature difference of the either 00:18:00.96000:18:00.970 n for parallel flow heat exchanger theta 00:18:03.24000:18:03.250 1 equals to thi minus 3 say for counter 00:18:05.73000:18:05.740 flow it is equals to thi minus EC you 00:18:07.62000:18:07.630 the definition of theta 1 and for 00:18:09.66000:18:09.670 parallel flow theta 2 equals to 1 th o 00:18:12.00000:18:12.010 minus T Co for counter flow it is equals 00:18:14.34000:18:14.350 to thi minus TCI so we have to take the 00:18:16.08000:18:16.090 difference temperate differences of the 00:18:17.97000:18:17.980 either ends for the case of theta 1 and 00:18:19.83000:18:19.840 theta 2 so that will be the definition 00:18:21.63000:18:21.640 of LMT D so the value of those theta 1 00:18:24.54000:18:24.550 theta 2 can be kept in this for another 00:18:26.34000:18:26.350 lambda T to get variability and to where 00:18:28.05000:18:28.060 the heat transfer rate Q goes to you al 00:18:30.21000:18:30.220 n Creedy here you is what you is the 00:18:33.30000:18:33.310 overall heat transfer coefficient of 00:18:34.95000:18:34.960 this particular type of it official and 00:18:36.63000:18:36.640 a is the area of exposure through which 00:18:39.51000:18:39.520 it is transferring for example for this 00:18:42.06000:18:42.070 case for this double pipe heat exchanger 00:18:43.83000:18:43.840 case the area of exposure is the outer 00:18:46.91000:18:46.920 peripheral area of this inner tube outer 00:18:51.18000:18:51.190 peripheral area that would be what if L 00:18:53.43000:18:53.440 is the length of this tube then 2 pi RL 00:18:55.86000:18:55.870 is the area of exposure for this double 00:18:58.86000:18:58.870 pipe it is ginger in which counter flow 00:19:00.99000:19:01.000 in which counter flow motion is 00:19:04.17000:19:04.180 happening with the flute same for this 00:19:06.51000:19:06.520 case also the area of exposure is what 00:19:08.07000:19:08.080 for parallel for additional the area of 00:19:10.11000:19:10.120 exposure reward the outer peripheral 00:19:13.02000:19:13.030 area of this central tube through which 00:19:16.26000:19:16.270 ultimately heat is transferring from hot 00:19:17.97000:19:17.980 fluid food so hope you would have 00:19:19.56000:19:19.570 understood the concept of exchanges and 00:19:21.27000:19:21.280 the types of attacks injure and you 00:19:23.28000:19:23.290 would have understood the concept of the 00:19:25.14000:19:25.150 LM TT by this lecture thank you 00:19:35.50000:19:35.510 you
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