Lecture 14 - Tubular Heat Exchanger - Shell - and - Tube Design

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Language: en

00:00:00.000
[Music]
00:00:13.740 00:00:13.750 welcome to this lecture in this lecture
00:00:18.280 00:00:18.290 we are trying to explain you the shell
00:00:21.070 00:00:21.080 and tube design approach in details you
00:00:25.269 00:00:25.279 may remember in the last class in the
00:00:28.060 00:00:28.070 last lecture we have talked about the
00:00:31.390 00:00:31.400 shell and tube heat exchanger design
00:00:33.729 00:00:33.739 approach and here with an example we
00:00:37.180 00:00:37.190 will try to elaborate the same now just
00:00:43.750 00:00:43.760 to recapitulate what we have talked in
00:00:46.480 00:00:46.490 the last class first of all we need to
00:00:51.520 00:00:51.530 define the problem like what is the
00:00:56.080 00:00:56.090 problem definition what are the
00:00:57.610 00:00:57.620 constraints what are the desired outputs
00:01:00.280 00:01:00.290 and based on that if we have that all
00:01:04.479 00:01:04.489 these informations with us we have to
00:01:08.230 00:01:08.240 first select the type of exchanger there
00:01:11.260 00:01:11.270 to here planning to design so we will
00:01:18.160 00:01:18.170 select the type of exchanger we are
00:01:21.969 00:01:21.979 going to design now for this case it's
00:01:26.559 00:01:26.569 already fixed and we have decided this
00:01:30.819 00:01:30.829 is the shell and cube type now once we
00:01:35.169 00:01:35.179 have decided the heat exchanger type
00:01:37.620 00:01:37.630 when now need to establish certain
00:01:42.910 00:01:42.920 design parameters like what would be
00:01:47.889 00:01:47.899 that expected heat transfer coefficient
00:01:51.609 00:01:51.619 for a typical heat exchangers like shell
00:01:55.179 00:01:55.189 and tube for a different kind of fluids
00:01:57.969 00:01:57.979 so once we have that knowledge with us
00:02:02.260 00:02:02.270 we will be able to find out using the
00:02:06.099 00:02:06.109 epsilon in T or the other techniques
00:02:09.669 00:02:09.679 that we have learned in the earlier
00:02:10.900 00:02:10.910 classes so basically we will be able to
00:02:14.290 00:02:14.300 find out some of the parameters or
00:02:16.210 00:02:16.220 design parameters so design parameters
00:02:20.320 00:02:20.330 by saying that we mean to say that
00:02:23.979 00:02:23.989 we come to know about the length maybe
00:02:26.470 00:02:26.480 we come to know about the diameter of
00:02:28.809 00:02:28.819 the tube we will be able to tell about
00:02:33.670 00:02:33.680 the shell diameter etc so once we know
00:02:40.240 00:02:40.250 all those parameters then what we do is
00:02:43.240 00:02:43.250 that now we go for a rating problem till
00:02:49.539 00:02:49.549 now we were trying to solve up to this
00:02:52.899 00:02:52.909 part we were trying to solve a sizing
00:02:56.199 00:02:56.209 problem that means where all the inlet
00:02:59.649 00:02:59.659 and flow conditions and the exit flow
00:03:03.550 00:03:03.560 conditions temperature flow rate
00:03:06.940 00:03:06.950 etcetera are known and we try to
00:03:09.699 00:03:09.709 estimate the heat exchanger length or
00:03:12.459 00:03:12.469 the heat exchanger size etcetera
00:03:14.649 00:03:14.659 etcetera so this was what was the sizing
00:03:18.159 00:03:18.169 problem as we have learned in the
00:03:22.300 00:03:22.310 earlier class so this was basically we
00:03:26.009 00:03:26.019 are trying to solve a sizing problem in
00:03:29.020 00:03:29.030 this one but up to this part we will try
00:03:33.909 00:03:33.919 to get an rough estimate of the sizing
00:03:38.649 00:03:38.659 problem now
00:03:41.219 00:03:41.229 once we know the rough estimate of the
00:03:45.640 00:03:45.650 design then we go for the rating problem
00:03:49.390 00:03:49.400 what is rating problem now as if we have
00:03:52.509 00:03:52.519 the geometry of the exchanger known the
00:03:56.710 00:03:56.720 inlet and of the two streams are known
00:04:01.749 00:04:01.759 and when we know the inlet condition of
00:04:05.199 00:04:05.209 both the fluid streams and the heat
00:04:06.939 00:04:06.949 exchanger configuration is known we can
00:04:09.670 00:04:09.680 find out what's happening to the exit
00:04:13.059 00:04:13.069 fluid streams or rather we try to find
00:04:16.659 00:04:16.669 out the performance of the heat
00:04:18.339 00:04:18.349 exchanger if we find that the heat
00:04:21.039 00:04:21.049 exchanger performance is the acceptable
00:04:23.830 00:04:23.840 like what is the pressure what is the
00:04:26.560 00:04:26.570 heat load etcetera that we have defined
00:04:29.409 00:04:29.419 earlier in the first problem definition
00:04:32.640 00:04:32.650 it is comparable with or it is except
00:04:37.360 00:04:37.370 to our problem definition or no or then
00:04:41.469 00:04:41.479 accordingly we will go for the
00:04:44.050 00:04:44.060 mechanical design if we find that it is
00:04:46.780 00:04:46.790 not in accordance to our desired output
00:04:50.350 00:04:50.360 or what we intend to do then we have to
00:04:53.710 00:04:53.720 make a change in the estimation so
00:04:57.040 00:04:57.050 basically you can understand that this
00:04:59.350 00:04:59.360 is a kind of trial and error so we are
00:05:02.620 00:05:02.630 starting with an initial or rough design
00:05:06.340 00:05:06.350 and we are trying to rectify that one
00:05:09.610 00:05:09.620 according to our need so let us try to
00:05:13.120 00:05:13.130 solve this one with a small example here
00:05:17.250 00:05:17.260 in this problem definition this is as
00:05:21.400 00:05:21.410 you can understand this is single cell
00:05:25.379 00:05:25.389 single tube heat exchanger here these
00:05:30.820 00:05:30.830 are the baffles and we have these are
00:05:34.480 00:05:34.490 the baffles we have these are the
00:05:37.210 00:05:37.220 baffles and one threat is entering the
00:05:41.110 00:05:41.120 condensed water is entering on this side
00:05:44.020 00:05:44.030 and it will be flowing like I'm sorry
00:05:47.740 00:05:47.750 this is supposed to this is not a good
00:05:50.890 00:05:50.900 design so it is supposed to come from
00:05:53.350 00:05:53.360 this end and it should come like this
00:05:55.629 00:05:55.639 and it is it is supposed to come another
00:05:58.690 00:05:58.700 baffle should be there so it should come
00:06:03.010 00:06:03.020 out from here so now though on the other
00:06:05.980 00:06:05.990 side the raw water is coming and it is
00:06:09.370 00:06:09.380 flowing like this and finally it will
00:06:12.460 00:06:12.470 flow like this from this outlet so here
00:06:19.560 00:06:19.570 what are the conditions or the problem
00:06:24.879 00:06:24.889 definition that is given the raw water
00:06:27.909 00:06:27.919 this is entering at 17 degree centigrade
00:06:32.830 00:06:32.840 and its CP is also given it is 4 1 8 4
00:06:40.110 00:06:40.120 Joule per kg Kelvin
00:06:45.280 00:06:45.290 and it's flow rate is also known this is
00:06:49.510 00:06:49.520 30 thousand kg per hour so we know the
00:06:57.550 00:06:57.560 flow rate of this water and we know the
00:07:04.750 00:07:04.760 inlet temperature of this raw water
00:07:09.240 00:07:09.250 what's about the condensed water it is
00:07:12.580 00:07:12.590 entering at 67 degree centigrade 67
00:07:22.450 00:07:22.460 degree centigrade and at Point 2 but now
00:07:29.950 00:07:29.960 it's flow rate has been specified as 50
00:07:34.710 00:07:34.720 thousand kg per hour now we know the
00:07:46.500 00:07:46.510 inlet temperature of the ro ro of water
00:07:49.960 00:07:49.970 we know the inlet temperature of the
00:07:54.430 00:07:54.440 condensed water what is expected that we
00:07:59.320 00:07:59.330 have to design a shell and tube heat
00:08:02.050 00:08:02.060 exchanger where the exit temperature
00:08:06.900 00:08:06.910 would be 40 degree so this raw water is
00:08:12.310 00:08:12.320 supposed to come out at 40 degree
00:08:17.080 00:08:17.090 centigrade it should not be less than 40
00:08:21.220 00:08:21.230 degree so this raw water is going to get
00:08:24.760 00:08:24.770 heated up and it will be heated up by
00:08:27.970 00:08:27.980 the condensed water and it's flow rate
00:08:31.510 00:08:31.520 and it's Inlet temperature is given and
00:08:34.380 00:08:34.390 what we intend to do is that the raw
00:08:38.560 00:08:38.570 water should always be it should not be
00:08:43.420 00:08:43.430 any time less than 40 degree centigrade
00:08:46.620 00:08:46.630 now what are the other condition that
00:08:49.060 00:08:49.070 has been already specified let us look
00:08:52.180 00:08:52.190 into that
00:08:55.670 00:08:55.680 it has been said that it should be a
00:08:57.920 00:08:57.930 shell and tube and it should be a single
00:09:01.700 00:09:01.710 cell and single tube pass not
00:09:06.260 00:09:06.270 necessarily also I mean but preferred is
00:09:09.920 00:09:09.930 single tube pass and coolant velocity is
00:09:14.829 00:09:14.839 maximum 1.5 meter per second that is the
00:09:22.100 00:09:22.110 maximum allowable velocity there is some
00:09:25.970 00:09:25.980 kind of form resistance or that means
00:09:29.900 00:09:29.910 there is a possibility that over the
00:09:32.570 00:09:32.580 time it will accumulate some kind of
00:09:35.630 00:09:35.640 scale and that resistance the found
00:09:39.199 00:09:39.209 resistance will be that fouled
00:09:45.949 00:09:45.959 resistance is equals to point triple 0
00:09:51.280 00:09:51.290 then 1 7 6 its unit would be meter
00:09:57.199 00:09:57.209 square Kelvin but what just inverse of
00:10:03.530 00:10:03.540 what we get for the heat transfer
00:10:06.079 00:10:06.089 coefficient now this is the foul
00:10:11.079 00:10:11.089 coefficient and fall resistance and the
00:10:15.710 00:10:15.720 tube length of this heat exchanger that
00:10:19.550 00:10:19.560 has already been specified because of
00:10:21.769 00:10:21.779 the length constant or space constant it
00:10:25.490 00:10:25.500 cannot be more than 5 meter so we will
00:10:29.680 00:10:29.690 work with 5 meter maximum length what it
00:10:34.970 00:10:34.980 has been said is the cube material this
00:10:40.850 00:10:40.860 tube materials are made of carbon steel
00:10:44.890 00:10:44.900 so that we know the thermal conductivity
00:10:51.730 00:10:51.740 so it becomes K equals to 60 watt per
00:10:57.069 00:10:57.079 meter Kelvin now
00:11:03.460 00:11:03.470 the tube diameter also has been
00:11:06.580 00:11:06.590 specified so the tube diameter is also
00:11:11.500 00:11:11.510 specified and it is basically OD is 19
00:11:17.050 00:11:17.060 mm and the ID that is equals to 16 mm
00:11:30.450 00:11:30.460 the tube arrangement that means how are
00:11:36.070 00:11:36.080 we going to arrange the tubes this is
00:11:39.280 00:11:39.290 the tube sheet in the tip sheet
00:11:41.770 00:11:41.780 there will be a density in the square
00:11:44.920 00:11:44.930 pitch so that means it will be inline
00:11:48.580 00:11:48.590 flow and this is how they're going to be
00:11:54.820 00:11:54.830 arranged
00:11:55.690 00:11:55.700 this is the four tubes if we look at the
00:11:59.380 00:11:59.390 flow is taking place like this and this
00:12:02.260 00:12:02.270 is the minimum area we have already
00:12:06.400 00:12:06.410 learned about it so this tube
00:12:09.210 00:12:09.220 arrangement has already been said it is
00:12:12.640 00:12:12.650 square pitch and the pitch ratio that is
00:12:18.330 00:12:18.340 pitch ratio PR that is
00:12:22.560 00:12:22.570 PT by D 0 where D 0 is the outside
00:12:28.900 00:12:28.910 diameter and this is PT so that P T by D
00:12:38.110 00:12:38.120 0 is equals to 1.25 so this has already
00:12:41.890 00:12:41.900 been specified and there is also the
00:12:48.250 00:12:48.260 battle spacing the battle spacing is 0.6
00:12:52.300 00:12:52.310 of the cell diameter
00:13:00.340 00:13:00.350 and it is baffle cart is 25%
00:13:07.290 00:13:07.300 there is battle cart so if we look into
00:13:15.310 00:13:15.320 this total battle 25% this part is 25%
00:13:21.070 00:13:21.080 of the total one and as we have said
00:13:26.020 00:13:26.030 that there is the possibility that there
00:13:29.200 00:13:29.210 would be formation of scale so what we
00:13:32.500 00:13:32.510 need to anticipate that over the years
00:13:38.080 00:13:38.090 it will accumulate some kind of foul I
00:13:42.310 00:13:42.320 mean the darts on it and scale on it so
00:13:46.450 00:13:46.460 accordingly we have to slightly over
00:13:48.730 00:13:48.740 design the heat exchanger and it should
00:13:52.180 00:13:52.190 never be more than 30% so that we have
00:13:56.830 00:13:56.840 to taken into account that considering
00:14:00.280 00:14:00.290 the fall resistance our design should
00:14:02.500 00:14:02.510 never be more than 35% of more
00:14:06.300 00:14:06.310 over-designed so these are the
00:14:09.130 00:14:09.140 specification which is given now we have
00:14:11.140 00:14:11.150 to look for the actual sizing of the
00:14:16.390 00:14:16.400 shell and tube so how do we do so what
00:14:20.710 00:14:20.720 we need to keep it in mind that we have
00:14:23.380 00:14:23.390 to take an approach as we have discussed
00:14:26.860 00:14:26.870 in the in the previous schematic this is
00:14:31.930 00:14:31.940 the flow chart we have shown so we will
00:14:35.140 00:14:35.150 start with this is the problem
00:14:37.630 00:14:37.640 definition has already been told now we
00:14:40.000 00:14:40.010 have to already it is also selected now
00:14:42.760 00:14:42.770 you will try to make an rough estimation
00:14:44.860 00:14:44.870 of the design parameters like we will
00:14:47.320 00:14:47.330 try to find out what would be the number
00:14:48.940 00:14:48.950 of tubes on the shared side what will be
00:14:52.450 00:14:52.460 the diameter of the shell side and once
00:14:56.230 00:14:56.240 we know that we will come back to this
00:14:58.120 00:14:58.130 rating problem so let us try to look
00:15:01.480 00:15:01.490 into that and we have this kind of heat
00:15:07.150 00:15:07.160 exchanger so now we will start the
00:15:10.780 00:15:10.790 calculation so we know that
00:15:14.370 00:15:14.380 the heat transfer the flow rate of the
00:15:18.610 00:15:18.620 cold fluid CPC the specific heat of the
00:15:24.430 00:15:24.440 cold fluid multiplied by the difference
00:15:27.280 00:15:27.290 in temperature so if we look at all this
00:15:31.090 00:15:31.100 temperature all these parameters are
00:15:32.860 00:15:32.870 known on the right hand side so that we
00:15:35.410 00:15:35.420 would be able to find out the amount of
00:15:37.840 00:15:37.850 heat transferred so this is just nothing
00:15:42.310 00:15:42.320 but if you try to calculate this is
00:15:46.180 00:15:46.190 30,000 divided by three six zero zero
00:15:50.340 00:15:50.350 multiplied by four 179 is the CP and
00:15:57.930 00:15:57.940 then you have the difference in
00:16:00.070 00:16:00.080 temperature 14 minus 40 minus 1740 is
00:16:04.150 00:16:04.160 the desired temperature and this is 17
00:16:07.510 00:16:07.520 is the inlet temperature of the cold
00:16:09.610 00:16:09.620 fluid so it is coming to be eight zero
00:16:12.460 00:16:12.470 one kilowatt amount of heat that is
00:16:16.630 00:16:16.640 going to be transferred and this is
00:16:19.150 00:16:19.160 equal to obviously m dot H CPH thi minus
00:16:25.960 00:16:25.970 T H zero so here this equation will be
00:16:28.900 00:16:28.910 used to find out the exit temperature of
00:16:31.330 00:16:31.340 the hot fluid and if we look at this
00:16:35.170 00:16:35.180 will come out to be because already we
00:16:39.010 00:16:39.020 have calculated Q equals to H zero one
00:16:41.380 00:16:41.390 and if we put that value we will be able
00:16:44.350 00:16:44.360 to estimate it to be fifty three point
00:16:48.250 00:16:48.260 two degree centigrade so now you see
00:16:51.100 00:16:51.110 already we know the all inlet and exit
00:16:55.350 00:16:55.360 temperature of the heat exchanger so now
00:17:00.870 00:17:00.880 at this point we have to make some kind
00:17:05.740 00:17:05.750 of assumption regarding the heat
00:17:08.530 00:17:08.540 transfer coefficient and so that we can
00:17:11.800 00:17:11.810 try to estimate the overall heat
00:17:14.230 00:17:14.240 transfer coefficient on the shell side
00:17:16.360 00:17:16.370 so how do we do that
00:17:20.290 00:17:20.300 in order to do that we have to depend on
00:17:23.590 00:17:23.600 certain realistic values of the
00:17:27.159 00:17:27.169 internal and external heat transfer
00:17:30.519 00:17:30.529 coefficient so this is the overall heat
00:17:32.830 00:17:32.840 transfer coefficient where we find this
00:17:36.489 00:17:36.499 is the wall resistance this is the
00:17:39.299 00:17:39.309 following distance inside and at the
00:17:42.279 00:17:42.289 outside then we have the overall
00:17:45.700 00:17:45.710 resistance of the external one and this
00:17:49.359 00:17:49.369 is the overall heat transfer coefficient
00:17:51.460 00:17:51.470 on the inverse of the overall heat
00:17:55.570 00:17:55.580 transfer coefficient this is one by one
00:17:58.810 00:17:58.820 by u 0 0 so this is what we get in
00:18:02.889 00:18:02.899 general and for the specific case where
00:18:06.310 00:18:06.320 we have R 0 and R I as the radius of the
00:18:13.289 00:18:13.299 two tubes we have 1 by H 0 then we have
00:18:19.019 00:18:19.029 our G ro by RI this is defined in terms
00:18:24.220 00:18:24.230 of the external heat transfer area so
00:18:27.700 00:18:27.710 this is the what is called the following
00:18:33.039 00:18:33.049 resistance this is already been given
00:18:35.590 00:18:35.600 this is this takes account of the heat
00:18:40.899 00:18:40.909 transfer resistance offered by the wall
00:18:43.989 00:18:43.999 of the cube and here this is we have
00:18:50.799 00:18:50.809 taken a rough estimate of this sell side
00:18:55.389 00:18:55.399 heat transfer coefficient to be 500 watt
00:18:59.049 00:18:59.059 per meter square Kelvin and we have
00:19:02.979 00:19:02.989 taken it as 4,000 watt per meter square
00:19:06.729 00:19:06.739 Kelvin this has been given from taken
00:19:10.509 00:19:10.519 from the heat exchanger selection by s
00:19:13.779 00:19:13.789 cogito so they have given a chart of
00:19:17.519 00:19:17.529 these values which typically which can
00:19:23.590 00:19:23.600 be considered for different kind of
00:19:25.359 00:19:25.369 fluids and different kind of heat
00:19:26.830 00:19:26.840 exchanger configuration so based on this
00:19:30.580 00:19:30.590 value if we now estimate we will find
00:19:33.879 00:19:33.889 that already we know h0 if we can
00:19:37.599 00:19:37.609 substitute r0 already we have talked
00:19:40.899 00:19:40.909 about r0 is already known to us this is
00:19:44.229 00:19:44.239 and this is r0 is say if we define it in
00:19:50.349 00:19:50.359 terms of this can also be 0 by D I so d0
00:19:55.210 00:19:55.220 is basically
00:19:56.580 00:19:56.590 d0 is equals to 19 mm and di is 16 mm
00:20:06.570 00:20:06.580 inch I is already given are fi is
00:20:11.529 00:20:11.539 already known that is equals to point
00:20:14.919 00:20:14.929 triple 0 1 7 6 meter squared Kelvin per
00:20:20.859 00:20:20.869 watt and we have d0 or sorry are 0 and
00:20:27.369 00:20:27.379 you have to put the value of K and R 0
00:20:30.849 00:20:30.859 by RI and Ln so if we put all these
00:20:33.820 00:20:33.830 values we will find you if this yep
00:20:39.159 00:20:39.169 where we consider the fouling resistance
00:20:43.349 00:20:43.359 this comes out to be 14 28.4 watt per
00:20:49.419 00:20:49.429 meter square Kelvin
00:20:53.159 00:20:53.169 now we also try to find out if it is a
00:20:59.469 00:20:59.479 clean surface that means if there is no
00:21:05.489 00:21:05.499 resistance given or offered by the
00:21:10.060 00:21:10.070 fouling so in that case what will happen
00:21:12.879 00:21:12.889 we will find that this term is not there
00:21:15.399 00:21:15.409 in that case we define it let us look
00:21:19.450 00:21:19.460 into that condition here in this case as
00:21:26.460 00:21:26.470 you can see that this part is not there
00:21:30.460 00:21:30.470 so when this part is not that
00:21:33.029 00:21:33.039 corresponding R value is not appearing
00:21:36.820 00:21:36.830 here and in this case all the other
00:21:39.580 00:21:39.590 parameters being known we can try to
00:21:41.889 00:21:41.899 find out you see and this will come out
00:21:44.680 00:21:44.690 to be 1 9 0 8 point 0 9 watt per meter
00:21:51.789 00:21:51.799 square Kelvin
00:21:54.670 00:21:54.680 so you can understand that this is in
00:21:56.860 00:21:56.870 contrary to or in contrast to our
00:22:00.330 00:22:00.340 earlier value where we were considering
00:22:03.610 00:22:03.620 the falling resistance and that was
00:22:07.120 00:22:07.130 appearing fourteen twenty eight point
00:22:10.510 00:22:10.520 four watt per meter square Kelvin so
00:22:14.380 00:22:14.390 this resistance I'm sorry not this this
00:22:18.970 00:22:18.980 is the following resistance this falling
00:22:21.010 00:22:21.020 resistance contribute I mean we are
00:22:24.900 00:22:24.910 considering when we are considering the
00:22:27.370 00:22:27.380 falling resistance we are finding that
00:22:30.000 00:22:30.010 the heat transfer overall heat transfer
00:22:33.340 00:22:33.350 coefficient is substantially small as
00:22:36.550 00:22:36.560 compared to the clean heat exchanger so
00:22:42.790 00:22:42.800 now we will try to find out based on
00:22:46.690 00:22:46.700 this information we have already have
00:22:49.780 00:22:49.790 the knowledge about the overall heat
00:22:52.480 00:22:52.490 transfer coefficient then we have the
00:22:55.150 00:22:55.160 knowledge about the heat transfer so
00:23:01.980 00:23:01.990 next is we have the knowledge of the
00:23:06.330 00:23:06.340 overall temperature profile or rather
00:23:10.690 00:23:10.700 the exit and Inlet temperatures so we
00:23:13.510 00:23:13.520 will be able to find out what is the log
00:23:17.050 00:23:17.060 mean temperature difference if the
00:23:18.850 00:23:18.860 fluids are an density in a
00:23:20.680 00:23:20.690 counter-current exchanger configuration
00:23:24.250 00:23:24.260 so if we look into it we will find that
00:23:29.740 00:23:29.750 the hot fluid is coming from sixty seven
00:23:34.150 00:23:34.160 degree centigrade and it is coming
00:23:37.440 00:23:37.450 degree centigrade it is coming to fifty
00:23:41.280 00:23:41.290 three point two whereas the cold fluid
00:23:46.420 00:23:46.430 is entering from 17 degree centigrade
00:23:51.370 00:23:51.380 and it is coming out at 40 degree
00:23:56.140 00:23:56.150 centigrade so now we can find out the
00:24:00.400 00:24:00.410 delta T 1 and Delta T two and we will
00:24:04.300 00:24:04.310 find that this difference
00:24:07.840 00:24:07.850 and this difference will be appearing in
00:24:10.960 00:24:10.970 this expression to give us a delta T LM
00:24:16.450 00:24:16.460 off we'll be able to find out this one
00:24:22.000 00:24:22.010 this will come out if you put all these
00:24:24.520 00:24:24.530 values you will find that this is coming
00:24:26.799 00:24:26.809 as thirty-one point four degree
00:24:29.350 00:24:29.360 centigrade as the delta T log bin so in
00:24:34.480 00:24:34.490 reality this is this is true only for
00:24:37.690 00:24:37.700 the purely counter-current exchanger but
00:24:42.130 00:24:42.140 in reality it is not so so this is
00:24:44.740 00:24:44.750 thirty-one point four degree centigrade
00:24:46.720 00:24:46.730 and we have a assuming a correction
00:24:52.360 00:24:52.370 factor of 0.9 so this we have already
00:24:57.100 00:24:57.110 learned that for a cross-flow type or if
00:25:03.760 00:25:03.770 I mean when we have a single pass single
00:25:07.029 00:25:07.039 tube heat exchanger single cell what
00:25:10.180 00:25:10.190 would be the corresponding correction
00:25:14.140 00:25:14.150 factor but we are just assuming it at
00:25:17.380 00:25:17.390 this moment this assumption has to be
00:25:20.770 00:25:20.780 later on you know rectified so this is
00:25:25.600 00:25:25.610 an assumption and based on that
00:25:27.070 00:25:27.080 assumption we are now trying to find out
00:25:29.799 00:25:29.809 the if we now look we have the knowledge
00:25:33.940 00:25:33.950 of Q we have the knowledge of Delta TM
00:25:36.789 00:25:36.799 we have the knowledge of UF and we have
00:25:40.690 00:25:40.700 assumed correction factor so accordingly
00:25:44.020 00:25:44.030 we would be able to calculate that if a
00:25:47.230 00:25:47.240 if corresponding to the overall heat
00:25:51.010 00:25:51.020 transfer coefficient where this is
00:25:55.110 00:25:55.120 considering the falling and here this is
00:25:59.890 00:25:59.900 where we have not considered the fouling
00:26:03.580 00:26:03.590 and this is basically a clean heat
00:26:05.950 00:26:05.960 exchanger so accordingly we will be able
00:26:09.880 00:26:09.890 to find out another heat transfer
00:26:12.789 00:26:12.799 surface area and obviously as you can
00:26:15.039 00:26:15.049 understand that this will be higher than
00:26:18.700 00:26:18.710 this
00:26:20.510 00:26:20.520 so now we will try to estimate this one
00:26:26.669 00:26:26.679 and you will find that this is coming to
00:26:30.960 00:26:30.970 be twenty point zero five meter squared
00:26:37.250 00:26:37.260 whereas the other one will come to be
00:26:43.970 00:26:43.980 fifteen point if you put all these
00:26:48.570 00:26:48.580 values meter squared so the queue
00:26:52.350 00:26:52.360 remains same the delta T Ln F remains
00:26:56.340 00:26:56.350 same and only thing is changing is UF
00:27:00.450 00:27:00.460 and you see that is giving rise to
00:27:03.539 00:27:03.549 different kind of overall area which is
00:27:09.320 00:27:09.330 to be taken into consideration for the
00:27:13.169 00:27:13.179 hit X near calculation know once we know
00:27:20.070 00:27:20.080 this heat transfer surface area which is
00:27:23.940 00:27:23.950 necessary and obviously we will be going
00:27:25.860 00:27:25.870 by this surface area which corresponds
00:27:29.039 00:27:29.049 to twenty point zero five a meter square
00:27:33.210 00:27:33.220 so our calculation will be based upon
00:27:36.419 00:27:36.429 this one because eventually we find that
00:27:39.630 00:27:39.640 the fouling is going to occur and this
00:27:42.750 00:27:42.760 falling is going to be there so that if
00:27:48.120 00:27:48.130 we do not take that extra area the
00:27:51.620 00:27:51.630 design performance of the exchanger is
00:27:54.539 00:27:54.549 cannot be expected after certain time so
00:27:59.430 00:27:59.440 now if we look into the area of this a F
00:28:03.990 00:28:04.000 by a C so this is what is the clean
00:28:09.860 00:28:09.870 surface area and this is anticipating
00:28:13.590 00:28:13.600 some kind of falling what is this ratio
00:28:16.680 00:28:16.690 this comes out to be one point three
00:28:19.320 00:28:19.330 four so now if you look at that we have
00:28:23.370 00:28:23.380 been told that the over design should be
00:28:26.430 00:28:26.440 less than thirty five percent so as we
00:28:30.360 00:28:30.370 can understand that this is less
00:28:32.979 00:28:32.989 1.35 so this design is or this
00:28:37.999 00:28:38.009 oversizing is within the limit
00:28:47.349 00:28:47.359 thank you
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