00:00:00.000 [Music] 00:00:13.74000:00:13.750 welcome to this lecture in this lecture 00:00:18.28000:00:18.290 we are trying to explain you the shell 00:00:21.07000:00:21.080 and tube design approach in details you 00:00:25.26900:00:25.279 may remember in the last class in the 00:00:28.06000:00:28.070 last lecture we have talked about the 00:00:31.39000:00:31.400 shell and tube heat exchanger design 00:00:33.72900:00:33.739 approach and here with an example we 00:00:37.18000:00:37.190 will try to elaborate the same now just 00:00:43.75000:00:43.760 to recapitulate what we have talked in 00:00:46.48000:00:46.490 the last class first of all we need to 00:00:51.52000:00:51.530 define the problem like what is the 00:00:56.08000:00:56.090 problem definition what are the 00:00:57.61000:00:57.620 constraints what are the desired outputs 00:01:00.28000:01:00.290 and based on that if we have that all 00:01:04.47900:01:04.489 these informations with us we have to 00:01:08.23000:01:08.240 first select the type of exchanger there 00:01:11.26000:01:11.270 to here planning to design so we will 00:01:18.16000:01:18.170 select the type of exchanger we are 00:01:21.96900:01:21.979 going to design now for this case it's 00:01:26.55900:01:26.569 already fixed and we have decided this 00:01:30.81900:01:30.829 is the shell and cube type now once we 00:01:35.16900:01:35.179 have decided the heat exchanger type 00:01:37.62000:01:37.630 when now need to establish certain 00:01:42.91000:01:42.920 design parameters like what would be 00:01:47.88900:01:47.899 that expected heat transfer coefficient 00:01:51.60900:01:51.619 for a typical heat exchangers like shell 00:01:55.17900:01:55.189 and tube for a different kind of fluids 00:01:57.96900:01:57.979 so once we have that knowledge with us 00:02:02.26000:02:02.270 we will be able to find out using the 00:02:06.09900:02:06.109 epsilon in T or the other techniques 00:02:09.66900:02:09.679 that we have learned in the earlier 00:02:10.90000:02:10.910 classes so basically we will be able to 00:02:14.29000:02:14.300 find out some of the parameters or 00:02:16.21000:02:16.220 design parameters so design parameters 00:02:20.32000:02:20.330 by saying that we mean to say that 00:02:23.97900:02:23.989 we come to know about the length maybe 00:02:26.47000:02:26.480 we come to know about the diameter of 00:02:28.80900:02:28.819 the tube we will be able to tell about 00:02:33.67000:02:33.680 the shell diameter etc so once we know 00:02:40.24000:02:40.250 all those parameters then what we do is 00:02:43.24000:02:43.250 that now we go for a rating problem till 00:02:49.53900:02:49.549 now we were trying to solve up to this 00:02:52.89900:02:52.909 part we were trying to solve a sizing 00:02:56.19900:02:56.209 problem that means where all the inlet 00:02:59.64900:02:59.659 and flow conditions and the exit flow 00:03:03.55000:03:03.560 conditions temperature flow rate 00:03:06.94000:03:06.950 etcetera are known and we try to 00:03:09.69900:03:09.709 estimate the heat exchanger length or 00:03:12.45900:03:12.469 the heat exchanger size etcetera 00:03:14.64900:03:14.659 etcetera so this was what was the sizing 00:03:18.15900:03:18.169 problem as we have learned in the 00:03:22.30000:03:22.310 earlier class so this was basically we 00:03:26.00900:03:26.019 are trying to solve a sizing problem in 00:03:29.02000:03:29.030 this one but up to this part we will try 00:03:33.90900:03:33.919 to get an rough estimate of the sizing 00:03:38.64900:03:38.659 problem now 00:03:41.21900:03:41.229 once we know the rough estimate of the 00:03:45.64000:03:45.650 design then we go for the rating problem 00:03:49.39000:03:49.400 what is rating problem now as if we have 00:03:52.50900:03:52.519 the geometry of the exchanger known the 00:03:56.71000:03:56.720 inlet and of the two streams are known 00:04:01.74900:04:01.759 and when we know the inlet condition of 00:04:05.19900:04:05.209 both the fluid streams and the heat 00:04:06.93900:04:06.949 exchanger configuration is known we can 00:04:09.67000:04:09.680 find out what's happening to the exit 00:04:13.05900:04:13.069 fluid streams or rather we try to find 00:04:16.65900:04:16.669 out the performance of the heat 00:04:18.33900:04:18.349 exchanger if we find that the heat 00:04:21.03900:04:21.049 exchanger performance is the acceptable 00:04:23.83000:04:23.840 like what is the pressure what is the 00:04:26.56000:04:26.570 heat load etcetera that we have defined 00:04:29.40900:04:29.419 earlier in the first problem definition 00:04:32.64000:04:32.650 it is comparable with or it is except 00:04:37.36000:04:37.370 to our problem definition or no or then 00:04:41.46900:04:41.479 accordingly we will go for the 00:04:44.05000:04:44.060 mechanical design if we find that it is 00:04:46.78000:04:46.790 not in accordance to our desired output 00:04:50.35000:04:50.360 or what we intend to do then we have to 00:04:53.71000:04:53.720 make a change in the estimation so 00:04:57.04000:04:57.050 basically you can understand that this 00:04:59.35000:04:59.360 is a kind of trial and error so we are 00:05:02.62000:05:02.630 starting with an initial or rough design 00:05:06.34000:05:06.350 and we are trying to rectify that one 00:05:09.61000:05:09.620 according to our need so let us try to 00:05:13.12000:05:13.130 solve this one with a small example here 00:05:17.25000:05:17.260 in this problem definition this is as 00:05:21.40000:05:21.410 you can understand this is single cell 00:05:25.37900:05:25.389 single tube heat exchanger here these 00:05:30.82000:05:30.830 are the baffles and we have these are 00:05:34.48000:05:34.490 the baffles we have these are the 00:05:37.21000:05:37.220 baffles and one threat is entering the 00:05:41.11000:05:41.120 condensed water is entering on this side 00:05:44.02000:05:44.030 and it will be flowing like I'm sorry 00:05:47.74000:05:47.750 this is supposed to this is not a good 00:05:50.89000:05:50.900 design so it is supposed to come from 00:05:53.35000:05:53.360 this end and it should come like this 00:05:55.62900:05:55.639 and it is it is supposed to come another 00:05:58.69000:05:58.700 baffle should be there so it should come 00:06:03.01000:06:03.020 out from here so now though on the other 00:06:05.98000:06:05.990 side the raw water is coming and it is 00:06:09.37000:06:09.380 flowing like this and finally it will 00:06:12.46000:06:12.470 flow like this from this outlet so here 00:06:19.56000:06:19.570 what are the conditions or the problem 00:06:24.87900:06:24.889 definition that is given the raw water 00:06:27.90900:06:27.919 this is entering at 17 degree centigrade 00:06:32.83000:06:32.840 and its CP is also given it is 4 1 8 4 00:06:40.11000:06:40.120 Joule per kg Kelvin 00:06:45.28000:06:45.290 and it's flow rate is also known this is 00:06:49.51000:06:49.520 30 thousand kg per hour so we know the 00:06:57.55000:06:57.560 flow rate of this water and we know the 00:07:04.75000:07:04.760 inlet temperature of this raw water 00:07:09.24000:07:09.250 what's about the condensed water it is 00:07:12.58000:07:12.590 entering at 67 degree centigrade 67 00:07:22.45000:07:22.460 degree centigrade and at Point 2 but now 00:07:29.95000:07:29.960 it's flow rate has been specified as 50 00:07:34.71000:07:34.720 thousand kg per hour now we know the 00:07:46.50000:07:46.510 inlet temperature of the ro ro of water 00:07:49.96000:07:49.970 we know the inlet temperature of the 00:07:54.43000:07:54.440 condensed water what is expected that we 00:07:59.32000:07:59.330 have to design a shell and tube heat 00:08:02.05000:08:02.060 exchanger where the exit temperature 00:08:06.90000:08:06.910 would be 40 degree so this raw water is 00:08:12.31000:08:12.320 supposed to come out at 40 degree 00:08:17.08000:08:17.090 centigrade it should not be less than 40 00:08:21.22000:08:21.230 degree so this raw water is going to get 00:08:24.76000:08:24.770 heated up and it will be heated up by 00:08:27.97000:08:27.980 the condensed water and it's flow rate 00:08:31.51000:08:31.520 and it's Inlet temperature is given and 00:08:34.38000:08:34.390 what we intend to do is that the raw 00:08:38.56000:08:38.570 water should always be it should not be 00:08:43.42000:08:43.430 any time less than 40 degree centigrade 00:08:46.62000:08:46.630 now what are the other condition that 00:08:49.06000:08:49.070 has been already specified let us look 00:08:52.18000:08:52.190 into that 00:08:55.67000:08:55.680 it has been said that it should be a 00:08:57.92000:08:57.930 shell and tube and it should be a single 00:09:01.70000:09:01.710 cell and single tube pass not 00:09:06.26000:09:06.270 necessarily also I mean but preferred is 00:09:09.92000:09:09.930 single tube pass and coolant velocity is 00:09:14.82900:09:14.839 maximum 1.5 meter per second that is the 00:09:22.10000:09:22.110 maximum allowable velocity there is some 00:09:25.97000:09:25.980 kind of form resistance or that means 00:09:29.90000:09:29.910 there is a possibility that over the 00:09:32.57000:09:32.580 time it will accumulate some kind of 00:09:35.63000:09:35.640 scale and that resistance the found 00:09:39.19900:09:39.209 resistance will be that fouled 00:09:45.94900:09:45.959 resistance is equals to point triple 0 00:09:51.28000:09:51.290 then 1 7 6 its unit would be meter 00:09:57.19900:09:57.209 square Kelvin but what just inverse of 00:10:03.53000:10:03.540 what we get for the heat transfer 00:10:06.07900:10:06.089 coefficient now this is the foul 00:10:11.07900:10:11.089 coefficient and fall resistance and the 00:10:15.71000:10:15.720 tube length of this heat exchanger that 00:10:19.55000:10:19.560 has already been specified because of 00:10:21.76900:10:21.779 the length constant or space constant it 00:10:25.49000:10:25.500 cannot be more than 5 meter so we will 00:10:29.68000:10:29.690 work with 5 meter maximum length what it 00:10:34.97000:10:34.980 has been said is the cube material this 00:10:40.85000:10:40.860 tube materials are made of carbon steel 00:10:44.89000:10:44.900 so that we know the thermal conductivity 00:10:51.73000:10:51.740 so it becomes K equals to 60 watt per 00:10:57.06900:10:57.079 meter Kelvin now 00:11:03.46000:11:03.470 the tube diameter also has been 00:11:06.58000:11:06.590 specified so the tube diameter is also 00:11:11.50000:11:11.510 specified and it is basically OD is 19 00:11:17.05000:11:17.060 mm and the ID that is equals to 16 mm 00:11:30.45000:11:30.460 the tube arrangement that means how are 00:11:36.07000:11:36.080 we going to arrange the tubes this is 00:11:39.28000:11:39.290 the tube sheet in the tip sheet 00:11:41.77000:11:41.780 there will be a density in the square 00:11:44.92000:11:44.930 pitch so that means it will be inline 00:11:48.58000:11:48.590 flow and this is how they're going to be 00:11:54.82000:11:54.830 arranged 00:11:55.69000:11:55.700 this is the four tubes if we look at the 00:11:59.38000:11:59.390 flow is taking place like this and this 00:12:02.26000:12:02.270 is the minimum area we have already 00:12:06.40000:12:06.410 learned about it so this tube 00:12:09.21000:12:09.220 arrangement has already been said it is 00:12:12.64000:12:12.650 square pitch and the pitch ratio that is 00:12:18.33000:12:18.340 pitch ratio PR that is 00:12:22.56000:12:22.570 PT by D 0 where D 0 is the outside 00:12:28.90000:12:28.910 diameter and this is PT so that P T by D 00:12:38.11000:12:38.120 0 is equals to 1.25 so this has already 00:12:41.89000:12:41.900 been specified and there is also the 00:12:48.25000:12:48.260 battle spacing the battle spacing is 0.6 00:12:52.30000:12:52.310 of the cell diameter 00:13:00.34000:13:00.350 and it is baffle cart is 25% 00:13:07.29000:13:07.300 there is battle cart so if we look into 00:13:15.31000:13:15.320 this total battle 25% this part is 25% 00:13:21.07000:13:21.080 of the total one and as we have said 00:13:26.02000:13:26.030 that there is the possibility that there 00:13:29.20000:13:29.210 would be formation of scale so what we 00:13:32.50000:13:32.510 need to anticipate that over the years 00:13:38.08000:13:38.090 it will accumulate some kind of foul I 00:13:42.31000:13:42.320 mean the darts on it and scale on it so 00:13:46.45000:13:46.460 accordingly we have to slightly over 00:13:48.73000:13:48.740 design the heat exchanger and it should 00:13:52.18000:13:52.190 never be more than 30% so that we have 00:13:56.83000:13:56.840 to taken into account that considering 00:14:00.28000:14:00.290 the fall resistance our design should 00:14:02.50000:14:02.510 never be more than 35% of more 00:14:06.30000:14:06.310 over-designed so these are the 00:14:09.13000:14:09.140 specification which is given now we have 00:14:11.14000:14:11.150 to look for the actual sizing of the 00:14:16.39000:14:16.400 shell and tube so how do we do so what 00:14:20.71000:14:20.720 we need to keep it in mind that we have 00:14:23.38000:14:23.390 to take an approach as we have discussed 00:14:26.86000:14:26.870 in the in the previous schematic this is 00:14:31.93000:14:31.940 the flow chart we have shown so we will 00:14:35.14000:14:35.150 start with this is the problem 00:14:37.63000:14:37.640 definition has already been told now we 00:14:40.00000:14:40.010 have to already it is also selected now 00:14:42.76000:14:42.770 you will try to make an rough estimation 00:14:44.86000:14:44.870 of the design parameters like we will 00:14:47.32000:14:47.330 try to find out what would be the number 00:14:48.94000:14:48.950 of tubes on the shared side what will be 00:14:52.45000:14:52.460 the diameter of the shell side and once 00:14:56.23000:14:56.240 we know that we will come back to this 00:14:58.12000:14:58.130 rating problem so let us try to look 00:15:01.48000:15:01.490 into that and we have this kind of heat 00:15:07.15000:15:07.160 exchanger so now we will start the 00:15:10.78000:15:10.790 calculation so we know that 00:15:14.37000:15:14.380 the heat transfer the flow rate of the 00:15:18.61000:15:18.620 cold fluid CPC the specific heat of the 00:15:24.43000:15:24.440 cold fluid multiplied by the difference 00:15:27.28000:15:27.290 in temperature so if we look at all this 00:15:31.09000:15:31.100 temperature all these parameters are 00:15:32.86000:15:32.870 known on the right hand side so that we 00:15:35.41000:15:35.420 would be able to find out the amount of 00:15:37.84000:15:37.850 heat transferred so this is just nothing 00:15:42.31000:15:42.320 but if you try to calculate this is 00:15:46.18000:15:46.190 30,000 divided by three six zero zero 00:15:50.34000:15:50.350 multiplied by four 179 is the CP and 00:15:57.93000:15:57.940 then you have the difference in 00:16:00.07000:16:00.080 temperature 14 minus 40 minus 1740 is 00:16:04.15000:16:04.160 the desired temperature and this is 17 00:16:07.51000:16:07.520 is the inlet temperature of the cold 00:16:09.61000:16:09.620 fluid so it is coming to be eight zero 00:16:12.46000:16:12.470 one kilowatt amount of heat that is 00:16:16.63000:16:16.640 going to be transferred and this is 00:16:19.15000:16:19.160 equal to obviously m dot H CPH thi minus 00:16:25.96000:16:25.970 T H zero so here this equation will be 00:16:28.90000:16:28.910 used to find out the exit temperature of 00:16:31.33000:16:31.340 the hot fluid and if we look at this 00:16:35.17000:16:35.180 will come out to be because already we 00:16:39.01000:16:39.020 have calculated Q equals to H zero one 00:16:41.38000:16:41.390 and if we put that value we will be able 00:16:44.35000:16:44.360 to estimate it to be fifty three point 00:16:48.25000:16:48.260 two degree centigrade so now you see 00:16:51.10000:16:51.110 already we know the all inlet and exit 00:16:55.35000:16:55.360 temperature of the heat exchanger so now 00:17:00.87000:17:00.880 at this point we have to make some kind 00:17:05.74000:17:05.750 of assumption regarding the heat 00:17:08.53000:17:08.540 transfer coefficient and so that we can 00:17:11.80000:17:11.810 try to estimate the overall heat 00:17:14.23000:17:14.240 transfer coefficient on the shell side 00:17:16.36000:17:16.370 so how do we do that 00:17:20.29000:17:20.300 in order to do that we have to depend on 00:17:23.59000:17:23.600 certain realistic values of the 00:17:27.15900:17:27.169 internal and external heat transfer 00:17:30.51900:17:30.529 coefficient so this is the overall heat 00:17:32.83000:17:32.840 transfer coefficient where we find this 00:17:36.48900:17:36.499 is the wall resistance this is the 00:17:39.29900:17:39.309 following distance inside and at the 00:17:42.27900:17:42.289 outside then we have the overall 00:17:45.70000:17:45.710 resistance of the external one and this 00:17:49.35900:17:49.369 is the overall heat transfer coefficient 00:17:51.46000:17:51.470 on the inverse of the overall heat 00:17:55.57000:17:55.580 transfer coefficient this is one by one 00:17:58.81000:17:58.820 by u 0 0 so this is what we get in 00:18:02.88900:18:02.899 general and for the specific case where 00:18:06.31000:18:06.320 we have R 0 and R I as the radius of the 00:18:13.28900:18:13.299 two tubes we have 1 by H 0 then we have 00:18:19.01900:18:19.029 our G ro by RI this is defined in terms 00:18:24.22000:18:24.230 of the external heat transfer area so 00:18:27.70000:18:27.710 this is the what is called the following 00:18:33.03900:18:33.049 resistance this is already been given 00:18:35.59000:18:35.600 this is this takes account of the heat 00:18:40.89900:18:40.909 transfer resistance offered by the wall 00:18:43.98900:18:43.999 of the cube and here this is we have 00:18:50.79900:18:50.809 taken a rough estimate of this sell side 00:18:55.38900:18:55.399 heat transfer coefficient to be 500 watt 00:18:59.04900:18:59.059 per meter square Kelvin and we have 00:19:02.97900:19:02.989 taken it as 4,000 watt per meter square 00:19:06.72900:19:06.739 Kelvin this has been given from taken 00:19:10.50900:19:10.519 from the heat exchanger selection by s 00:19:13.77900:19:13.789 cogito so they have given a chart of 00:19:17.51900:19:17.529 these values which typically which can 00:19:23.59000:19:23.600 be considered for different kind of 00:19:25.35900:19:25.369 fluids and different kind of heat 00:19:26.83000:19:26.840 exchanger configuration so based on this 00:19:30.58000:19:30.590 value if we now estimate we will find 00:19:33.87900:19:33.889 that already we know h0 if we can 00:19:37.59900:19:37.609 substitute r0 already we have talked 00:19:40.89900:19:40.909 about r0 is already known to us this is 00:19:44.22900:19:44.239 and this is r0 is say if we define it in 00:19:50.34900:19:50.359 terms of this can also be 0 by D I so d0 00:19:55.21000:19:55.220 is basically 00:19:56.58000:19:56.590 d0 is equals to 19 mm and di is 16 mm 00:20:06.57000:20:06.580 inch I is already given are fi is 00:20:11.52900:20:11.539 already known that is equals to point 00:20:14.91900:20:14.929 triple 0 1 7 6 meter squared Kelvin per 00:20:20.85900:20:20.869 watt and we have d0 or sorry are 0 and 00:20:27.36900:20:27.379 you have to put the value of K and R 0 00:20:30.84900:20:30.859 by RI and Ln so if we put all these 00:20:33.82000:20:33.830 values we will find you if this yep 00:20:39.15900:20:39.169 where we consider the fouling resistance 00:20:43.34900:20:43.359 this comes out to be 14 28.4 watt per 00:20:49.41900:20:49.429 meter square Kelvin 00:20:53.15900:20:53.169 now we also try to find out if it is a 00:20:59.46900:20:59.479 clean surface that means if there is no 00:21:05.48900:21:05.499 resistance given or offered by the 00:21:10.06000:21:10.070 fouling so in that case what will happen 00:21:12.87900:21:12.889 we will find that this term is not there 00:21:15.39900:21:15.409 in that case we define it let us look 00:21:19.45000:21:19.460 into that condition here in this case as 00:21:26.46000:21:26.470 you can see that this part is not there 00:21:30.46000:21:30.470 so when this part is not that 00:21:33.02900:21:33.039 corresponding R value is not appearing 00:21:36.82000:21:36.830 here and in this case all the other 00:21:39.58000:21:39.590 parameters being known we can try to 00:21:41.88900:21:41.899 find out you see and this will come out 00:21:44.68000:21:44.690 to be 1 9 0 8 point 0 9 watt per meter 00:21:51.78900:21:51.799 square Kelvin 00:21:54.67000:21:54.680 so you can understand that this is in 00:21:56.86000:21:56.870 contrary to or in contrast to our 00:22:00.33000:22:00.340 earlier value where we were considering 00:22:03.61000:22:03.620 the falling resistance and that was 00:22:07.12000:22:07.130 appearing fourteen twenty eight point 00:22:10.51000:22:10.520 four watt per meter square Kelvin so 00:22:14.38000:22:14.390 this resistance I'm sorry not this this 00:22:18.97000:22:18.980 is the following resistance this falling 00:22:21.01000:22:21.020 resistance contribute I mean we are 00:22:24.90000:22:24.910 considering when we are considering the 00:22:27.37000:22:27.380 falling resistance we are finding that 00:22:30.00000:22:30.010 the heat transfer overall heat transfer 00:22:33.34000:22:33.350 coefficient is substantially small as 00:22:36.55000:22:36.560 compared to the clean heat exchanger so 00:22:42.79000:22:42.800 now we will try to find out based on 00:22:46.69000:22:46.700 this information we have already have 00:22:49.78000:22:49.790 the knowledge about the overall heat 00:22:52.48000:22:52.490 transfer coefficient then we have the 00:22:55.15000:22:55.160 knowledge about the heat transfer so 00:23:01.98000:23:01.990 next is we have the knowledge of the 00:23:06.33000:23:06.340 overall temperature profile or rather 00:23:10.69000:23:10.700 the exit and Inlet temperatures so we 00:23:13.51000:23:13.520 will be able to find out what is the log 00:23:17.05000:23:17.060 mean temperature difference if the 00:23:18.85000:23:18.860 fluids are an density in a 00:23:20.68000:23:20.690 counter-current exchanger configuration 00:23:24.25000:23:24.260 so if we look into it we will find that 00:23:29.74000:23:29.750 the hot fluid is coming from sixty seven 00:23:34.15000:23:34.160 degree centigrade and it is coming 00:23:37.44000:23:37.450 degree centigrade it is coming to fifty 00:23:41.28000:23:41.290 three point two whereas the cold fluid 00:23:46.42000:23:46.430 is entering from 17 degree centigrade 00:23:51.37000:23:51.380 and it is coming out at 40 degree 00:23:56.14000:23:56.150 centigrade so now we can find out the 00:24:00.40000:24:00.410 delta T 1 and Delta T two and we will 00:24:04.30000:24:04.310 find that this difference 00:24:07.84000:24:07.850 and this difference will be appearing in 00:24:10.96000:24:10.970 this expression to give us a delta T LM 00:24:16.45000:24:16.460 off we'll be able to find out this one 00:24:22.00000:24:22.010 this will come out if you put all these 00:24:24.52000:24:24.530 values you will find that this is coming 00:24:26.79900:24:26.809 as thirty-one point four degree 00:24:29.35000:24:29.360 centigrade as the delta T log bin so in 00:24:34.48000:24:34.490 reality this is this is true only for 00:24:37.69000:24:37.700 the purely counter-current exchanger but 00:24:42.13000:24:42.140 in reality it is not so so this is 00:24:44.74000:24:44.750 thirty-one point four degree centigrade 00:24:46.72000:24:46.730 and we have a assuming a correction 00:24:52.36000:24:52.370 factor of 0.9 so this we have already 00:24:57.10000:24:57.110 learned that for a cross-flow type or if 00:25:03.76000:25:03.770 I mean when we have a single pass single 00:25:07.02900:25:07.039 tube heat exchanger single cell what 00:25:10.18000:25:10.190 would be the corresponding correction 00:25:14.14000:25:14.150 factor but we are just assuming it at 00:25:17.38000:25:17.390 this moment this assumption has to be 00:25:20.77000:25:20.780 later on you know rectified so this is 00:25:25.60000:25:25.610 an assumption and based on that 00:25:27.07000:25:27.080 assumption we are now trying to find out 00:25:29.79900:25:29.809 the if we now look we have the knowledge 00:25:33.94000:25:33.950 of Q we have the knowledge of Delta TM 00:25:36.78900:25:36.799 we have the knowledge of UF and we have 00:25:40.69000:25:40.700 assumed correction factor so accordingly 00:25:44.02000:25:44.030 we would be able to calculate that if a 00:25:47.23000:25:47.240 if corresponding to the overall heat 00:25:51.01000:25:51.020 transfer coefficient where this is 00:25:55.11000:25:55.120 considering the falling and here this is 00:25:59.89000:25:59.900 where we have not considered the fouling 00:26:03.58000:26:03.590 and this is basically a clean heat 00:26:05.95000:26:05.960 exchanger so accordingly we will be able 00:26:09.88000:26:09.890 to find out another heat transfer 00:26:12.78900:26:12.799 surface area and obviously as you can 00:26:15.03900:26:15.049 understand that this will be higher than 00:26:18.70000:26:18.710 this 00:26:20.51000:26:20.520 so now we will try to estimate this one 00:26:26.66900:26:26.679 and you will find that this is coming to 00:26:30.96000:26:30.970 be twenty point zero five meter squared 00:26:37.25000:26:37.260 whereas the other one will come to be 00:26:43.97000:26:43.980 fifteen point if you put all these 00:26:48.57000:26:48.580 values meter squared so the queue 00:26:52.35000:26:52.360 remains same the delta T Ln F remains 00:26:56.34000:26:56.350 same and only thing is changing is UF 00:27:00.45000:27:00.460 and you see that is giving rise to 00:27:03.53900:27:03.549 different kind of overall area which is 00:27:09.32000:27:09.330 to be taken into consideration for the 00:27:13.16900:27:13.179 hit X near calculation know once we know 00:27:20.07000:27:20.080 this heat transfer surface area which is 00:27:23.94000:27:23.950 necessary and obviously we will be going 00:27:25.86000:27:25.870 by this surface area which corresponds 00:27:29.03900:27:29.049 to twenty point zero five a meter square 00:27:33.21000:27:33.220 so our calculation will be based upon 00:27:36.41900:27:36.429 this one because eventually we find that 00:27:39.63000:27:39.640 the fouling is going to occur and this 00:27:42.75000:27:42.760 falling is going to be there so that if 00:27:48.12000:27:48.130 we do not take that extra area the 00:27:51.62000:27:51.630 design performance of the exchanger is 00:27:54.53900:27:54.549 cannot be expected after certain time so 00:27:59.43000:27:59.440 now if we look into the area of this a F 00:28:03.99000:28:04.000 by a C so this is what is the clean 00:28:09.86000:28:09.870 surface area and this is anticipating 00:28:13.59000:28:13.600 some kind of falling what is this ratio 00:28:16.68000:28:16.690 this comes out to be one point three 00:28:19.32000:28:19.330 four so now if you look at that we have 00:28:23.37000:28:23.380 been told that the over design should be 00:28:26.43000:28:26.440 less than thirty five percent so as we 00:28:30.36000:28:30.370 can understand that this is less 00:28:32.97900:28:32.989 1.35 so this design is or this 00:28:37.99900:28:38.009 oversizing is within the limit 00:28:47.34900:28:47.359 thank you
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