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Lecture 05 - Design and Simulation of Heat Exchangers
WEBVTT Kind: captions Language: en
00:00:00.000 [Music] 00:00:14.25000:00:14.260 this talk is about the design and 00:00:17.50000:00:17.510 simulation of heat exchangers in 00:00:20.05000:00:20.060 particular we want to do the LM TD 00:00:24.92900:00:24.939 approach or the design of heat 00:00:28.45000:00:28.460 exchangers by LM TD method so before 00:00:32.95000:00:32.960 going into the details we want to look 00:00:35.26000:00:35.270 into the design and simulation what do 00:00:38.71000:00:38.720 we mean by design we want to see it with 00:00:42.36900:00:42.379 the example of a counter-current heat 00:00:45.31000:00:45.320 exchanger here this heat exchanger we 00:00:51.21900:00:51.229 know what are the fluid streams the 00:00:54.13000:00:54.140 process fluid streams or the process 00:00:56.29000:00:56.300 parameters the inlet temperature the 00:00:58.81000:00:58.820 exit temperature the exit and Inlet 00:01:01.53900:01:01.549 pressure the flow rates all these 00:01:03.88000:01:03.890 parameter process parameters are known 00:01:06.31000:01:06.320 to us and what we want to find out is 00:01:09.91000:01:09.920 the geometrical parameters like the 00:01:13.51000:01:13.520 length the inner and outer diameter of 00:01:17.26000:01:17.270 the tubes so in this counter current 00:01:21.91000:01:21.920 heat exchanger what we are basically 00:01:24.21900:01:24.229 trying to find out is the different 00:01:28.38000:01:28.390 dimensions or the geometric geometrical 00:01:31.20900:01:31.219 parameters or the dimensions of the heat 00:01:33.58000:01:33.590 exchanger and we often call it as sizing 00:01:37.69000:01:37.700 problem or the design problem because it 00:01:40.71900:01:40.729 is related to the size of the heat 00:01:42.27900:01:42.289 exchanger we call it sizing now in 00:01:45.66900:01:45.679 contrast to this design we are also 00:01:49.74900:01:49.759 often we know we try to solve the 00:01:52.14900:01:52.159 simulation problem what is simulation in 00:01:55.98900:01:55.999 simulation if we take the same heat 00:01:57.96900:01:57.979 exchanger example that counter-current 00:02:00.48900:02:00.499 heat exchanger in this case we know all 00:02:04.56900:02:04.57900:02:07.23900:02:07.249 length of the exchanger the diameter 00:02:09.37000:02:09.380 outer or inner all these parameters are 00:02:13.45000:02:13.460 known in addition to that we also know 00:02:18.19000:02:18.200 some of the process parameters like the 00:02:20.95000:02:20.960 inlet fluid temperature 00:02:23.77000:02:23.780 the flow rates are known that means we 00:02:27.46000:02:27.470 know this fluid flow temperature we know 00:02:31.69000:02:31.700 this Inlet temperature we know this 00:02:34.33000:02:34.340 Inlet temperature and what we intend to 00:02:38.05000:02:38.060 find out is the exit temperature of the 00:02:41.71000:02:41.720 hot and cold fluid stream so this is 00:02:44.82000:02:44.830 something related to the rating or the 00:02:54.24000:02:54.250 simulation of the heat exchanger we call 00:02:57.58000:02:57.590 it as simulation problem 00:03:08.25000:03:08.260 so now we want to try the similar design 00:03:18.53900:03:18.549 of a counter-current heat exchanger and 00:03:21.50000:03:21.510 here this is an example we want to try 00:03:26.16000:03:26.170 it with a counter-current heat exchanger 00:03:29.00900:03:29.019 where we will assume a simple situation 00:03:33.80900:03:33.819 by assuming C R or the C mean by C max 00:03:38.88000:03:38.890 equals to 1 when we assume that she 00:03:42.75000:03:42.760 recalls to 1 it will have a temperature 00:03:45.93000:03:45.940 profile shown in the figure what we will 00:03:49.94900:03:49.959 find that the delta T the temperature 00:03:54.15000:03:54.160 difference between the hot and cold 00:03:56.30900:03:56.319 fluid it will remain same throughout the 00:03:59.22000:03:59.230 length of the heat exchanger so since we 00:04:03.56900:04:03.579 are trying to solve a counter-current 00:04:06.00000:04:06.010 heat exchanger design problem here the 00:04:09.56900:04:09.579 fluid temperature the Inlet and outlet 00:04:12.74000:04:12.750 all for the hot and cold fluids are 00:04:16.19900:04:16.209 known here we have assumed C R equals to 00:04:20.78900:04:20.799 1 now we want to find out what will be 00:04:24.02900:04:24.039 the length or diameter of this heat 00:04:26.12900:04:26.139 exchanger so what is C R it is basically 00:04:36.18000:04:36.190 a ratio between the heat capacity of the 00:04:39.33000:04:39.340 hot fluid and the cold fluid this CH or 00:04:44.01000:04:44.020 the hot capacity ratio heat capacity is 00:04:49.62000:04:49.630 basically a product of the hot fluid 00:04:53.33900:04:53.349 specific heat multiplied by the mass 00:04:56.76000:04:56.770 flow rate of the hot fluid and the cold 00:05:00.06000:05:00.070 sea C or the cold capacity is equals to 00:05:03.99000:05:04.000 a product of the specific heat of the 00:05:09.39000:05:09.400 cold fluid multiplied by the mass flow 00:05:14.85000:05:14.860 rate of the cold fluid so this 2 ratio 00:05:18.42000:05:18.430 is basically 00:05:21.30000:05:21.310 ratio between the hot and cold fluid is 00:05:23.92000:05:23.930 equals to 1 we have assumed so in that 00:05:27.49000:05:27.500 situation only we can expect this kind 00:05:30.13000:05:30.140 of temperature profile now in this case 00:05:33.79000:05:33.800 this is the hot in hot out and TC in and 00:05:38.47000:05:38.480 TC out now if we take a small elemental 00:05:43.03000:05:43.040 length DL across the length of the heat 00:05:46.36000:05:46.370 exchanger we find that this is th and 00:05:51.58000:05:51.590 this is TC and we also assume that the 00:05:55.72000:05:55.730 internal temperature I mean the wall 00:05:57.82000:05:57.830 temperature is T W so this TW is varying 00:06:02.56000:06:02.570 from this end to this end now in 00:06:07.33000:06:07.340 addition to this assumption that CR 00:06:11.14000:06:11.150 equals to 1 that is the simplest case we 00:06:14.26000:06:14.270 are making some more additional 00:06:16.03000:06:16.040 assumptions like there is no external 00:06:19.09000:06:19.100 heat in leak that means this fluid hot 00:06:22.87000:06:22.880 fluid is not connected with the 00:06:24.55000:06:24.560 environment neither this cold fluid is 00:06:27.31000:06:27.320 connected with the environment the heat 00:06:30.25000:06:30.260 is being supplied only from the hot 00:06:33.55000:06:33.560 fluid to the cold fluid via this 00:06:36.52000:06:36.530 separating wall the separating wall 00:06:41.68000:06:41.690 thickness we are assuming it to be 00:06:44.41000:06:44.420 negligible by saying that the wall 00:06:48.70000:06:48.710 thickness is negligible we mean that the 00:06:52.96000:06:52.970 heat flow from the hot fluid to the cold 00:06:55.30000:06:55.310 fluid is not affected by the resistance 00:07:00.28000:07:00.290 of the wall in addition to that we also 00:07:05.59000:07:05.600 assume that there is no axial conduction 00:07:08.35000:07:08.360 what is the actual conduction that it is 00:07:12.24000:07:12.250 that this wall is having a kind of 00:07:15.40000:07:15.410 temperature gradient but we are not 00:07:19.21000:07:19.220 assuming any heat flowing from this end 00:07:22.93000:07:22.940 of the wall to the other end only we are 00:07:25.75000:07:25.760 assuming the heat flow from the hot 00:07:29.38000:07:29.390 fluid to the cold fluid that's what is 00:07:32.92000:07:32.930 the assumption actually 00:07:34.69000:07:34.700 conduction heat is negligible so now 00:07:38.20000:07:38.210 what will happen the hot fluid will give 00:07:42.88000:07:42.890 heat to the separating wall the 00:07:46.39000:07:46.400 separating wall in turn will give heat 00:07:49.18000:07:49.190 to the cold fluid so how do we estimate 00:07:53.11000:07:53.120 this heat transfer we will take the 00:07:56.89000:07:56.900 difference between the hot fluid and the 00:08:00.22000:08:00.230 wall so T H minus TW and The Associated 00:08:04.51000:08:04.520 area with this fluid and the wall or the 00:08:09.43000:08:09.440 hot side heat transfer surface area how 00:08:13.09000:08:13.100 much is it that is say if it is a H then 00:08:18.52000:08:18.530 we have considered only DL length of 00:08:22.84000:08:22.850 this heat exchanger so if we assume that 00:08:27.58000:08:27.590 the heat transfer surface area is 00:08:29.71000:08:29.720 uniformly distributed over the entire 00:08:31.63000:08:31.640 length we will find that the heat 00:08:38.83000:08:38.840 transfer equation will look like 00:08:48.59000:08:48.600 DQ is he calls to as I told you that it 00:08:53.91000:08:53.920 is th minus TW and the area associated 00:08:58.59000:08:58.600 with this elemental length DL that is H 00:09:04.13900:09:04.149 by L and DL is the small element and 00:09:08.15000:09:08.160 over an above HH the heat transfer 00:09:11.07000:09:11.080 coefficient of the hot fluid side 00:09:12.88900:09:12.899 similarly we can equate it to the heat 00:09:17.61000:09:17.620 received from the wall to the cold fluid 00:09:20.28000:09:20.290 and the related heat transfer surface 00:09:24.81000:09:24.820 area and the heat transfer coefficient 00:09:27.19900:09:27.209 corresponding to the cold fluid side now 00:09:35.51000:09:35.520 if we simplify this equation or rather 00:09:41.34000:09:41.350 if we integrate this equation over the 00:09:43.56000:09:43.570 entire length we will find that this DQ 00:09:47.88000:09:47.890 is becoming Q and this heat transfer 00:09:52.01900:09:52.029 surface area is becoming a H so the 00:09:56.64000:09:56.650 overall equation becomes H h H into T H 00:10:01.92000:10:01.930 minus TW 00:10:02.91000:10:02.920 and on this side TW minus TC and that 00:10:06.03000:10:06.040 cold HC and AC now if we simplify this 00:10:11.04000:10:11.050 equation to find out the wall 00:10:13.56000:10:13.570 temperature we find this relation now 00:10:17.79000:10:17.800 what we can do is that we can put this 00:10:21.21000:10:21.220 wall temperature and replace it in 00:10:24.44900:10:24.459 either of these equations when we 00:10:28.17000:10:28.180 replace this equation or replace the 00:10:32.49000:10:32.500 wall temperature to this equation the 00:10:37.11000:10:37.120 earlier equation we find that the heat 00:10:40.65000:10:40.660 transfer becomes this like this and if 00:10:44.85000:10:44.860 we simplify this equation you will find 00:10:47.25000:10:47.260 that this will look like HH h HC AC 00:10:54.42000:10:54.430 divided by h HH HH CAC and the 00:10:59.43000:10:59.440 difference between the wall temperature 00:11:00.92000:11:00.930 this 00:11:02.10000:11:02.110 nothing but delta T now when I take this 00:11:07.00900:11:07.019 denominator and divide this part we find 00:11:11.55000:11:11.560 this will be simplified like 1 by u a 00:11:14.97000:11:14.980 equals to 1 by H h a h plus 1 by h CAC 00:11:21.59000:11:21.600 that is nothing but the this part is the 00:11:27.32900:11:27.339 hot fluid side heat transfer resistance 00:11:31.07900:11:31.089 and this is the resistance offered by 00:11:33.96000:11:33.970 the cold fluid side convective heat 00:11:36.69000:11:36.700 transfer resistance and this one can be 00:11:39.54000:11:39.550 represented by 1 by u way that is the 00:11:42.84000:11:42.850 overall resistance offered by this fluid 00:11:47.51900:11:47.529 so now this can be finally expressed in 00:11:52.47000:11:52.480 the form of u way into delta T where 00:11:56.40000:11:56.410 this delta T is the difference between 00:11:59.34000:11:59.350 the two fluids and u way is the overall 00:12:03.38900:12:03.399 heat transfer coefficient given by this 00:12:06.38900:12:06.399 expression so if we now go to the next 00:12:17.55000:12:17.560 slide 00:12:24.94000:12:24.950 so in summary we can say that the Q is 00:12:30.43000:12:30.440 given by you a delta T and so this is 00:12:37.56900:12:37.579 the final outcome we have seen where you 00:12:41.82900:12:41.839 a is given by 1 by a is related to the 00:12:45.85000:12:45.860 individual heat transfer I mean 00:12:48.28000:12:48.290 resistances offered by the cold and the 00:12:51.10000:12:51.110 hot fluid and over an above we have this 00:12:56.06900:12:56.079 individual mass balance for the hot and 00:12:59.37900:12:59.389 the cold fluid the hot fluid mass 00:13:02.98000:13:02.990 balance is given by CH and th in minus 00:13:07.75000:13:07.760 th out similarly for the cold fluid we 00:13:11.23000:13:11.240 can have this equation now as we are 00:13:15.79000:13:15.800 trying to solve a design problem we know 00:13:19.11000:13:19.120 the hot inlet we know the hot outlet we 00:13:23.68000:13:23.690 know the cold inlet and the cold outlet 00:13:27.06900:13:27.079 also so we know essentially the amount 00:13:31.44900:13:31.459 of heat getting transferred so in this 00:13:34.18000:13:34.190 equation we know Q we know the delta T 00:13:37.72000:13:37.730 we have assumed a simple situation where 00:13:40.99000:13:41.000 C R equals to 1 delta T is throughout 00:13:44.11000:13:44.120 the length uniform so from this relation 00:13:48.10000:13:48.110 we can calculate you way so this away 00:13:51.97000:13:51.980 will now come to this equation where it 00:13:56.47000:13:56.480 is the combination of 1 by H h a h and h 00:14:00.85000:14:00.860 c AC now the hot side heat transfer 00:14:04.75000:14:04.760 coefficient can also be obtained because 00:14:07.21000:14:07.220 we know the mass flow rate h dot we can 00:14:13.24000:14:13.250 also find out HC because the 00:14:16.72000:14:16.730 corresponding mass flow rate is also 00:14:20.29000:14:20.300 known MC dot is also known we can also 00:14:24.28000:14:24.290 try to find out this HC and then we can 00:14:30.04000:14:30.050 try to find out the a H and AC but this 00:14:35.05000:14:35.060 is only one equation where the number of 00:14:37.99000:14:38.000 unknowns 00:14:38.98000:14:38.990 many like the length internal diameter 00:14:44.91000:14:44.920 outside diameter so in order to find out 00:14:48.54000:14:48.550 those numbers we have to get other 00:14:53.14000:14:53.150 equations and by putting some other 00:14:58.48000:14:58.490 constants like the pressure drop then 00:15:01.36000:15:01.370 the sock over all sides then we get 00:15:04.30000:15:04.310 other equations and then we can try to 00:15:07.30000:15:07.310 find out all this aah and AC so this is 00:15:12.79000:15:12.800 the simplest situation when we have 00:15:15.75000:15:15.760 assumed C R equals to 1 and we have 00:15:19.12000:15:19.130 tried to design a simple tube in tube 00:15:23.23000:15:23.240 heat exchanger now we may not expect 00:15:28.18000:15:28.190 always that C H will be equal to the CC 00:15:33.28000:15:33.290 or C R will be equals to 1 there may be 00:15:37.36000:15:37.370 situation when CH is greater than C C or 00:15:41.74000:15:41.750 it may also happen that CH is less than 00:15:46.09000:15:46.100 C C that means C are either it will be 00:15:48.82000:15:48.830 greater than 1 or it will be less than 1 00:15:52.08000:15:52.090 in case C H is greater than C C we will 00:15:56.74000:15:56.750 find that the temperature profile is 00:15:59.86000:15:59.870 similar to the one like shown here for 00:16:04.48000:16:04.490 the counter current heat exchanger when 00:16:08.02000:16:08.030 CH is greater than C C we will find that 00:16:11.83000:16:11.840 the cold fluid outlet temperature is 00:16:15.64000:16:15.650 almost approaching the hot Inlet 00:16:20.47000:16:20.480 temperature on the contrary if CH is 00:16:24.82000:16:24.830 less than C C we will find that this hot 00:16:30.73000:16:30.740 outlet is approaching the cold Inlet 00:16:36.22000:16:36.230 temperature so the temperature profile 00:16:40.03000:16:40.040 will look like this shown in this figure 00:16:43.60000:16:43.610 so now let us consider for time being 00:16:47.17000:16:47.180 that CH is greater than C C and let us 00:16:50.68000:16:50.690 try to see 00:16:52.48000:16:52.490 the or design problem for this situation 00:16:58.80000:16:58.810 if we consider again a small elemental 00:17:02.53000:17:02.540 length DL and then we can expect there 00:17:07.57000:17:07.580 would be a small amount of heat DQ 00:17:10.63000:17:10.640 getting transferred from the hot fluid 00:17:13.36000:17:13.370 to the cold fluid through the separating 00:17:16.75000:17:16.760 valve and all this assumptions which has 00:17:20.23000:17:20.240 been made earlier except C R equals to 00:17:23.05000:17:23.060 one all assumptions are still valid here 00:17:26.80000:17:26.810 in this case also and only exception is 00:17:30.73000:17:30.740 that we have assumed C H is greater than 00:17:33.22000:17:33.230 C C now in this case what we will find 00:17:37.87000:17:37.880 that the heat transfer coefficient or 00:17:40.39000:17:40.400 the overall heat transfer between the 00:17:43.39000:17:43.400 hot fluid to the cold fluid can be 00:17:46.03000:17:46.040 related to the overall heat transfer 00:17:48.49000:17:48.500 coefficient u and if we assume that the 00:17:53.08000:17:53.090 area is distributed uniformly over the 00:17:55.87000:17:55.880 entire length then a by L multiplied by 00:17:59.20000:17:59.210 DL and th minus TC th is the hot site 00:18:04.98000:18:04.990 temperature TC is the cold site 00:18:08.17000:18:08.180 temperature at that point on in that 00:18:11.35000:18:11.360 small elemental L length DL so now if we 00:18:15.64000:18:15.650 make an energy balance for the hot fluid 00:18:19.72000:18:19.730 we will find DK because 2 minus CH into 00:18:23.56000:18:23.570 D th where D th is the change in 00:18:26.68000:18:26.690 temperature within that small length DL 00:18:30.24000:18:30.250 DQ is in terms of the cold fluid we can 00:18:36.22000:18:36.230 write it that DQ is equal to minus CC d 00:18:40.27000:18:40.280 TC where this DC DT C is the change in 00:18:44.83000:18:44.840 temperature that is happening in the 00:18:48.25000:18:48.260 small elemental length DL in the cold 00:18:52.39000:18:52.400 fluid we can write this first equation 00:18:56.47000:18:56.480 in this form where D H can be written as 00:18:59.95000:18:59.960 DQ by CH with a negative sign and it is 00:19:03.91000:19:03.920 C can be written as 00:19:06.19000:19:06.200 - DQ by CC now let us try to write or 00:19:12.19000:19:12.200 express D of th - DTC this is nothing 00:19:18.13000:19:18.140 but the delta T over the length D DL so 00:19:23.38000:19:23.390 d of delta T can now be expressed as 00:19:26.23000:19:26.240 minus DQ if we take this one and this 00:19:29.28900:19:29.299 expression I mean here in this equation 00:19:32.83000:19:32.840 if we put we find that D th - DT C this 00:19:38.16900:19:38.179 is we have taken from here this we have 00:19:40.77900:19:40.789 taken from here and we are able to 00:19:44.16900:19:44.179 express D of delta T is equals to minus 00:19:48.70000:19:48.710 DQ into 1 by C H minus 1 by CC now we 00:19:58.45000:19:58.460 can substitute this heat transfer from 00:20:04.29900:20:04.309 here to this expression if we do that we 00:20:10.45000:20:10.460 will find the equation is taking this 00:20:15.63900:20:15.649 form what is that form d of d th d TC is 00:20:23.13900:20:23.149 equals to u into a by LD l th minus TC 00:20:27.82000:20:27.830 and 1 by CH - CC now what we are trying 00:20:31.96000:20:31.970 we can now integrate it from this l 00:20:37.09000:20:37.100 equals to 0 to l equals to l and when l 00:20:43.96000:20:43.970 equals to 0 the corresponding difference 00:20:47.52900:20:47.539 in temperature is delta T that is the 00:20:51.73000:20:51.740 cost of delta T smaller one and this is 00:20:55.69000:20:55.700 delta T larger one this is because we 00:21:00.66900:21:00.679 have assumed t c h greater than CC so we 00:21:06.10000:21:06.110 have the delta T is smaller on this side 00:21:09.25000:21:09.260 and delta T large sorry this is delta T 00:21:12.97000:21:12.980 L large on this side so if I integrate 00:21:18.76000:21:18.770 it now over this 00:21:19.84000:21:19.850 length what we will find is that log of 00:21:24.96000:21:24.970 natural log of delta T large divided by 00:21:28.96000:21:28.970 delta T small is equals to minus e way 00:21:31.77900:21:31.789 into 1 by CH and minus 1 by CC now we 00:21:37.60000:21:37.610 have this equation still remaining with 00:21:41.91900:21:41.929 us cuba cost of CH into th n minus th 00:21:46.57000:21:46.580 out that is the mass balance of the cold 00:21:49.72000:21:49.730 fluid and this is the mass balance of 00:21:53.04900:21:53.059 the study hot fluid and the cold fluid 00:21:55.63000:21:55.640 this is for the hot fluid and this is 00:21:57.97000:21:57.980 for the cold fluid this is known so if 00:22:00.90900:22:00.919 we know this one we can take CH you know 00:22:05.89000:22:05.900 this expression and put it here 00:22:08.95000:22:08.960 we can take out CC and put this 00:22:13.33000:22:13.340 expression in this one so if we do that 00:22:16.62000:22:16.630 what you will find is an expression like 00:22:22.48000:22:22.490 this we have replaced that one by CH 00:22:26.59000:22:26.600 with this expression and 1 by CC with 00:22:30.58000:22:30.590 this expression then with little algebra 00:22:36.01000:22:36.020 if we do I mean we have rearranged with 00:22:38.83000:22:38.840 this term we have taken th same as this 00:22:42.94000:22:42.950 one we have taken you know this TC out 00:22:46.60000:22:46.610 from there to here and this one is 00:22:50.52900:22:50.539 somewhere here and this is like this 00:22:53.62000:22:53.630 this is just to you know express this 00:22:57.15900:22:57.169 temperature in terms of Delta TS and 00:23:00.58000:23:00.590 Delta TL so the overall equation now 00:23:05.08000:23:05.090 becomes Q equals to UA Delta TL minus 00:23:10.57000:23:10.580 delta T s divided by natural log Delta 00:23:15.13000:23:15.140 TL by delta T s so if we put this 00:23:20.32000:23:20.330 expression as the natural log or sorry 00:23:25.75000:23:25.760 the log mean temperature difference so 00:23:29.64900:23:29.659 then it becomes Q because to UA delta T 00:23:32.95000:23:32.960 LM or log mean so this is the log mean 00:23:37.81000:23:37.820 temperature of this inter temperature 00:23:42.13000:23:42.140 difference and then we can write the 00:23:47.47000:23:47.480 heat transfer as this is the design 00:23:54.52000:23:54.530 summary now what is known to us the Q 00:24:01.00000:24:01.010 this is known because we know all the 00:24:03.67000:24:03.680 exit temperatures now because this 00:24:07.48000:24:07.490 exists and Paris is unknown we can also 00:24:10.12000:24:10.130 find out this log mean temperature 00:24:13.00000:24:13.010 difference this Q is known in this 00:24:16.48000:24:16.490 equation the log mean temperature is 00:24:18.28000:24:18.290 known so we can find out the overall 00:24:22.27000:24:22.280 heat transfer coefficient which is 00:24:24.76000:24:24.770 nothing but 1 by h HH plus 1 by h CAC so 00:24:32.20000:24:32.210 again as we have discussed earlier once 00:24:35.83000:24:35.840 we know this expression we can try to 00:24:38.86000:24:38.870 find out the hot side and the cold side 00:24:43.62000:24:43.630 parameters but this may be it may be 00:24:47.11000:24:47.120 necessary that there will be certain 00:24:49.33000:24:49.340 number of iterations now we have 00:24:55.20000:24:55.210 discussed about this particular 00:24:58.27000:24:58.280 situation 00:24:59.08000:24:59.090 CH greater than CC and we have seen that 00:25:02.95000:25:02.960 Q equals to you way into delta T of LM 00:25:11.56000:25:11.570 now is it valid for CH less than C see 00:25:17.52000:25:17.530 it can be shown that this is also valid 00:25:21.31000:25:21.320 for this situation also only thing is 00:25:24.70000:25:24.710 that this time that delta T large is 00:25:30.48000:25:30.490 occurring on this side and delta T small 00:25:34.87000:25:34.880 is occurring on this side in contrast to 00:25:37.96000:25:37.970 this situation other things will remain 00:25:41.17000:25:41.180 same the sign etc will also be taken 00:25:44.92000:25:44.930 care accordingly 00:25:46.39000:25:46.400 and the same expression also remains 00:25:49.27000:25:49.280 valid for this situation now we want to 00:25:53.89000:25:53.900 look for the counter current already 00:26:04.30000:26:04.310 situation we have already looked for we 00:26:07.39000:26:07.400 have already discussed now if we go for 00:26:10.15000:26:10.160 a parallel flow heat exchanger or 00:26:12.13000:26:12.140 co-current heat exchanger again by the 00:26:15.67000:26:15.680 same way if we take a elemental length 00:26:18.22000:26:18.230 over this heat exchangers length if we 00:26:22.72000:26:22.730 take and small Delta L if we consider 00:26:25.99000:26:26.000 and if we consider the temperature 00:26:27.67000:26:27.680 difference if we write the energy 00:26:29.50000:26:29.510 balance we will find that the heat 00:26:32.38000:26:32.390 transfer overall heat transfer is given 00:26:35.20000:26:35.210 by mu way into Delta TM this Delta TM in 00:26:39.34000:26:39.350 this case we will have this as delta T 00:26:44.22000:26:44.230 large and the difference on this side is 00:26:49.42000:26:49.430 delta T small on this side other things 00:26:54.13000:26:54.140 are remaining same so in a nutshell we 00:26:57.22000:26:57.230 can see that all the counter current and 00:27:03.18000:27:03.190 parallel flow or co-current heat 00:27:05.38000:27:05.390 exchanger can be expressed by Q equals 00:27:10.84000:27:10.850 to the heat transfer in this cases can 00:27:13.30000:27:13.310 be expressed as Q equals to UA Delta TLM 00:27:17.31000:27:17.320 in fact for all counter current parallel 00:27:23.17000:27:23.180 flow as well as for situations where CR 00:27:27.40000:27:27.410 equals to zero that is because two 00:27:29.20000:27:29.210 condenser or RIBA I mean boiler 00:27:31.57000:27:31.580 situations also we will find that this 00:27:34.72000:27:34.730 relation remains valid now if we go to 00:27:39.94000:27:39.950 the next light I mean if we find that 00:27:44.56000:27:44.570 there is a cross flow heat exchanger do 00:27:48.16000:27:48.170 we think that still we can use this 00:27:51.43000:27:51.440 correlation we can use this correlation 00:27:54.25000:27:54.260 I mean relation Q equals to UA Delta TLM 00:27:57.94000:27:57.950 but with small 00:28:00.05000:28:00.060 section factor because in this situation 00:28:03.05000:28:03.060 this Delta TM or the mean temperature or 00:28:06.68000:28:06.690 the log mean temperature that is not the 00:28:09.53000:28:09.540 same in case of cross counter flow and 00:28:15.67000:28:15.680 parallel flow heat exchanger if was 00:28:18.71000:28:18.720 taken as 1 and if we put that relation 00:28:22.19000:28:22.200 here we will find that this relation is 00:28:25.73000:28:25.740 valid for the counter current and the 00:28:29.41000:28:29.420 co-current heat exchanger but as we are 00:28:32.69000:28:32.700 going to talk about a cross flow heat 00:28:34.94000:28:34.950 exchanger we see that this will become 00:28:37.85000:28:37.860 less than 1 so how to find out that 00:28:43.58000:28:43.590 relation we will see that for different 00:28:52.24000:28:52.250 heat exchanger configuration this has 00:28:56.00000:28:56.010 already been calculated for our is and 00:28:59.68000:28:59.690 this is related to the it is expressed 00:29:06.44000:29:06.450 in terms of r and p two parameters which 00:29:11.00000:29:11.010 are related to the two temperature 00:29:14.57000:29:14.580 differences here this is T Inlet minus T 00:29:18.17000:29:18.180 exit and this is the tube side that is T 00:29:23.06000:29:23.070 0 minus TI this is one ratio the other 00:29:26.75000:29:26.760 ratio is T 0 minus TI divided by the 00:29:31.22000:29:31.230 maximum temperature difference that is T 00:29:33.50000:29:33.510 I capital T I minus small T I this 00:29:36.92000:29:36.930 expression has been taken from this 00:29:39.23000:29:39.240 graph has been taken from AP entropy de 00:29:41.72000:29:41.730 and David heat and mass transfer book so 00:29:49.24000:29:49.250 similarly if the geometry is changing 00:29:55.54000:29:55.550 from sailin tube this is multiple to 00:30:00.86000:30:00.870 tube passes I mean it will be either two 00:30:05.39000:30:05.400 tube or two tube multiples and there is 00:30:09.17000:30:09.180 two shell side this is one cell side 00:30:12.17000:30:12.180 another cell side 00:30:13.94000:30:13.950 so this is the expression for now this 00:30:16.70000:30:16.710 is the correction factor corresponding 00:30:18.40900:30:18.419 to that configuration similarly the 00:30:27.62000:30:27.630 other configurations will be like when 00:30:32.89900:30:32.909 we have a cross flow heat exchanger with 00:30:36.16900:30:36.179 single paths where both fluids are 00:30:38.18000:30:38.190 unmixed or both fluids are mixed one 00:30:41.48000:30:41.490 through it makes another fluid unmixed 00:30:43.49000:30:43.500 this will correspond to a different 00:30:45.68000:30:45.690 connection factor so based on this 00:30:48.20000:30:48.210 values of the F factor we have to add 00:30:52.51900:30:52.529 some correction to that ln TD method to 00:30:55.82000:30:55.830 solve the Q equals to you way if Delta 00:30:59.99000:31:00.000 TLM and then we can design the cross 00:31:03.50000:31:03.510 flow heat exchanger thank you
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