Lecture 35 (2013). 11.3 Analysis of Heat Exchangers. 11.4 Log Mean Temperature Difference Method

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Language: en

00:00:01.280
ladies and gentlemen let's continue with
00:00:03.530 00:00:03.540 chapter 11 chapter 11 is the chapter on
00:00:06.320 00:00:06.330 heat exchanges that is the chapter where
00:00:09.200 00:00:09.210 so many of the work that we've done
00:00:10.990 00:00:11.000 comes together we've already looked at
00:00:14.419 00:00:14.429 the different types of heat exchangers
00:00:16.269 00:00:16.279 we've looked at the overall heat
00:00:19.250 00:00:19.260 transfer coefficient the U and the
00:00:22.220 00:00:22.230 combination of U multiplied by the area
00:00:24.740 00:00:24.750 and today we're going to continue with
00:00:27.529 00:00:27.539 paragraph eleven point three in the
00:00:30.919 00:00:30.929 textbook of single and gauge or on the
00:00:33.889 00:00:33.899 analysis of heat exchanges and then also
00:00:36.290 00:00:36.300 paragraph eleven point four the LM TD
00:00:39.799 00:00:39.809 method now in terms of when we look at
00:00:44.450 00:00:44.460 it exchanges you'll see that in general
00:00:46.549 00:00:46.559 there are at least two methods that can
00:00:49.700 00:00:49.710 be used to analyze heat exchanges and
00:00:52.819 00:00:52.829 these two methods are the most popular
00:00:55.310 00:00:55.320 methods and are the ones which are been
00:00:58.700 00:00:58.710 used the most most the first one is
00:01:02.720 00:01:02.730 known as the LM TD method the lock mean
00:01:07.700 00:01:07.710 temperature difference method and the
00:01:10.340 00:01:10.350 second one is known as the X alone into
00:01:14.240 00:01:14.250 you method or the effectiveness into you
00:01:18.200 00:01:18.210 method we are not going to address that
00:01:21.080 00:01:21.090 now we are going to address that with
00:01:22.910 00:01:22.920 the next lecture but in general for both
00:01:27.320 00:01:27.330 of these methods let's just consolidate
00:01:30.200 00:01:30.210 some of the most important theories okay
00:01:33.350 00:01:33.360 and the theory is that if we look ahead
00:01:37.280 00:01:37.290 at the heat exchanger they are going to
00:01:39.770 00:01:39.780 be two streams and the temperatures are
00:01:44.270 00:01:44.280 going to change maybe something like
00:01:47.569 00:01:47.579 that let's call that the temperature
00:01:51.230 00:01:51.240 scale here and there that is X so that
00:01:56.179 00:01:56.189 is how the temperature on the hot side
00:01:58.429 00:01:58.439 is going to change and on the cold side
00:02:00.980 00:02:00.990 it is maybe going to do
00:02:02.810 00:02:02.820 something like that and although I've
00:02:07.400 00:02:07.410 drawn the directions like that it
00:02:10.100 00:02:10.110 doesn't have to be like that okay that
00:02:12.890 00:02:12.900 is just a general schematic
00:02:15.130 00:02:15.140 representation of an heat exchanger
00:02:17.630 00:02:17.640 we've got a hot side and a cold side
00:02:19.940 00:02:19.950 normally heat exchanges are well
00:02:22.340 00:02:22.350 insulated we want to prevent the heat
00:02:25.070 00:02:25.080 losses and it is equipment that or
00:02:30.530 00:02:30.540 normally that normally runs for very
00:02:32.510 00:02:32.520 long hours so normally we've got
00:02:34.220 00:02:34.230 steady-state conditions also
00:02:36.290 00:02:36.300 so therefore in general we can say that
00:02:39.400 00:02:39.410 the heat transfer rate on the cold side
00:02:42.520 00:02:42.530 on the cold side the cold side we just
00:02:46.580 00:02:46.590 mean the stream with the lowest
00:02:48.440 00:02:48.450 temperature so the it roles are right
00:02:51.830 00:02:51.840 would obviously be from the stream which
00:02:55.880 00:02:55.890 is at the higher temperature to the
00:02:57.410 00:02:57.420 stream which is at the lower temperature
00:02:59.570 00:02:59.580 so the heat roles for right cue from the
00:03:06.080 00:03:06.090 cold side or to the cold side is equal
00:03:08.570 00:03:08.580 to the mass flow rate of the cold side
00:03:10.699 00:03:10.709 the CP of the cold side multiplied by
00:03:15.130 00:03:15.140 the outlet temperature that increase of
00:03:18.890 00:03:18.900 a minus TC in so this temperature is TC
00:03:25.820 00:03:25.830 out and that temperature is equal to TC
00:03:28.970 00:03:28.980 in the cold in a temperature and the
00:03:33.110 00:03:33.120 cold outlet temperature
00:03:36.520 00:03:36.530 and that's it should be the same on the
00:03:38.980 00:03:38.990 hot side q on the hot side is equal to
00:03:41.800 00:03:41.810 the mass flow rate on the hot side CP on
00:03:44.860 00:03:44.870 the hot side x now th in minus th out so
00:03:56.710 00:03:56.720 this temperature here is equal to gh in
00:04:00.670 00:04:00.680 and thatis th out
00:04:06.190 00:04:06.200 does that make sense okay so it's very
00:04:10.210 00:04:10.220 simple okay now these terms have a
00:04:16.060 00:04:16.070 specific meaning and the meaning is it
00:04:20.349 00:04:20.359 is known as the heat capacity ratio the
00:04:26.530 00:04:26.540 heat capacity and I shall and the heat
00:04:30.969 00:04:30.979 capacity ratio is written as C for the
00:04:36.340 00:04:36.350 hot side which is then equal to the mass
00:04:39.219 00:04:39.229 flow rate of the hot side multiplied by
00:04:41.560 00:04:41.570 CP of the hot side and on the cold side
00:04:48.310 00:04:48.320 it is equal to the mass flow rate on the
00:04:50.740 00:04:50.750 cold side x CP on the cold side
00:04:56.930 00:04:56.940 and have and that product represents the
00:05:00.560 00:05:00.570 amount of heat needed to increase the
00:05:03.620 00:05:03.630 temperature with one degree Celsius so
00:05:06.590 00:05:06.600 that is the heat capacity ratio okay if
00:05:09.830 00:05:09.840 the ratio is very very large then it
00:05:12.770 00:05:12.780 means I need a lot of heat to increase
00:05:15.800 00:05:15.810 the temperature one one degree Celsius
00:05:17.710 00:05:17.720 if it is a low value it means I do not
00:05:21.650 00:05:21.660 leave need that much energy to increase
00:05:23.930 00:05:23.940 the temperature with one degree Celsius
00:05:31.120 00:05:31.130 now there are two special types of heat
00:05:34.250 00:05:34.260 exchangers or two special categories two
00:05:42.050 00:05:42.060 special categories of heat exchanges and
00:05:44.120 00:05:44.130 those heat exchanges are known as and
00:05:47.960 00:05:47.970 we've done it already
00:05:48.950 00:05:48.960 as condenser or boilers
00:05:57.460 00:05:57.470 condenser or a boiler so on a TS diagram
00:06:02.450 00:06:02.460 that is a saturation line then with a
00:06:10.430 00:06:10.440 condenser we will change a gas to a
00:06:16.850 00:06:16.860 fluid okay
00:06:20.180 00:06:20.190 and during that process they will be
00:06:23.960 00:06:23.970 heat transferred to the environment or
00:06:28.670 00:06:28.680 to the other stream in any case with a
00:06:32.750 00:06:32.760 boiler it is the opposite and now it
00:06:36.980 00:06:36.990 will be from fluid to a gas and to
00:06:43.340 00:06:43.350 change the fluid from from from a fluid
00:06:46.400 00:06:46.410 or the fluid from a liquid to a gas we
00:06:50.780 00:06:50.790 need to put in heat it has to be added
00:06:56.950 00:06:56.960 this guys it is going to be released now
00:07:02.120 00:07:02.130 for these special cases to special
00:07:05.540 00:07:05.550 categories the heat transfer rate is not
00:07:09.710 00:07:09.720 equal to any mass flow CP x delta T okay
00:07:15.580 00:07:15.590 why because as you can see the
00:07:18.980 00:07:18.990 temperature remains constant during the
00:07:21.500 00:07:21.510 process the temperature remains constant
00:07:25.390 00:07:25.400 it is equal to the mass flow rate
00:07:29.720 00:07:29.730 multiplied by H FG
00:07:32.950 00:07:32.960 the change in enthalpy between those two
00:07:37.640 00:07:37.650 points
00:07:40.940 00:07:40.950 okay so for this special category that
00:07:43.980 00:07:43.990 we have the C which normally is equal to
00:07:49.560 00:07:49.570 the mass flow rate multiplied by CP is
00:07:56.430 00:07:56.440 infinite very very important they are
00:08:03.150 00:08:03.160 going to be problems in the test on the
00:08:05.370 00:08:05.380 exam heat exchanger is going to be given
00:08:08.070 00:08:08.080 to you and then in the description it's
00:08:12.150 00:08:12.160 going to say evaporation occurs or
00:08:15.210 00:08:15.220 condensation occurs or something like
00:08:17.850 00:08:17.860 that and one or both of these values are
00:08:22.950 00:08:22.960 not going to be given and if you do not
00:08:24.870 00:08:24.880 use C equal to infinite then you'll not
00:08:27.360 00:08:27.370 be able to do the problem so it's a very
00:08:29.010 00:08:29.020 very important concept that you do
00:08:31.290 00:08:31.300 understand that CP normally as a gas you
00:08:37.920 00:08:37.930 can use that type of calculation in that
00:08:40.740 00:08:40.750 region you can use it in this region but
00:08:43.500 00:08:43.510 not here
00:08:44.070 00:08:44.080 CP doesn't exist here because it is
00:08:48.360 00:08:48.370 infinite okay so the mass flow rate
00:08:50.460 00:08:50.470 multiplied by the CP in the two-phase
00:08:53.190 00:08:53.200 region is infinite okay now in general
00:09:01.960 00:09:01.970 the heat transfer right now without
00:09:05.690 00:09:05.700 referring to a hot site or a cold site
00:09:08.350 00:09:08.360 we would also say is equal to u
00:09:11.180 00:09:11.190 multiplied by the surface area
00:09:13.870 00:09:13.880 multiplied by delta T and this delta T
00:09:18.050 00:09:18.060 is now very important because it should
00:09:22.070 00:09:22.080 be a representative a very
00:09:25.850 00:09:25.860 representative temperature difference
00:09:31.240 00:09:31.250 now if we look at this then you might
00:09:35.870 00:09:35.880 say well let's use this okay and in many
00:09:38.780 00:09:38.790 cases that might be a good assumption
00:09:40.329 00:09:40.339 but they are going to be cases where it
00:09:43.490 00:09:43.500 is not going to be a good assumption
00:09:45.590 00:09:45.600 for example maybe the temperatures does
00:09:49.190 00:09:49.200 that we've got condensation and we've
00:09:53.720 00:09:53.730 got the other stream doing something
00:09:56.810 00:09:56.820 like that okay
00:09:58.010 00:09:58.020 so then that in creature difference is
00:10:00.050 00:10:00.060 not a good representation and we've
00:10:02.570 00:10:02.580 already looked at it we've said that it
00:10:04.340 00:10:04.350 is better in those cases to use the LM
00:10:07.100 00:10:07.110 TG temperature difference and here we
00:10:10.610 00:10:10.620 are going on to the same pause okay so
00:10:14.060 00:10:14.070 what immature difference should be used
00:10:16.670 00:10:16.680 there you're going to see is going to be
00:10:18.800 00:10:18.810 the LM TD okay and that is what
00:10:22.610 00:10:22.620 paragraph eleven point four is all about
00:10:25.390 00:10:25.400 00:10:29.060 00:10:29.070 the LNT d the Lachlan integer difference
00:10:35.970 00:10:35.980 okay and it starts by considering a
00:10:39.310 00:10:39.320 parallel heat exchanger a parallel heat
00:10:46.030 00:10:46.040 exchanger if exchanger looks like that
00:10:54.319 00:10:54.329 and because it is parallel all the flow
00:10:58.259 00:10:58.269 directions are in the same direction
00:11:07.759 00:11:07.769 parallel flow heat exchanger so the flow
00:11:11.100 00:11:11.110 is going in there it's going out there
00:11:13.829 00:11:13.839 that flow is going in there and it goes
00:11:16.769 00:11:16.779 out there and this graph gives the
00:11:22.379 00:11:22.389 temperature as a function of X
00:11:36.670 00:11:36.680 now if we look at the parallel heat
00:11:38.930 00:11:38.940 exchanger typically the temperature
00:11:41.210 00:11:41.220 profile does something like that that
00:11:44.600 00:11:44.610 temperature we call th in on the hot
00:11:49.190 00:11:49.200 side so that is th and that would be th
00:11:52.850 00:11:52.860 out and the direction is from left to
00:11:58.310 00:11:58.320 right on the cold side typically the
00:12:08.000 00:12:08.010 temperature distribution is going to
00:12:09.620 00:12:09.630 look like that so that is now going to
00:12:12.800 00:12:12.810 be TC in and that is going to be TC out
00:12:24.620 00:12:24.630 and this temperature difference is going
00:12:28.160 00:12:28.170 to we can call it delta T 1 is equal to
00:12:31.850 00:12:31.860 TIG minus TC in so this temperature
00:12:38.600 00:12:38.610 difference th in minus TC in this
00:12:43.550 00:12:43.560 temperature difference is equal to delta
00:12:47.030 00:12:47.040 T 2 and that is equal to d HLT minus TC
00:12:55.460 00:12:55.470 out
00:13:06.040 00:13:06.050 okay now mathematically very correctly
00:13:08.739 00:13:08.749 what is there being done in the textbook
00:13:11.019 00:13:11.029 is a control volume is considered it is
00:13:16.419 00:13:16.429 the control volume and you know how
00:13:18.999 00:13:19.009 things goes with a control volume
00:13:20.859 00:13:20.869 approach that is normally being followed
00:13:23.289 00:13:23.299 and without showing all the detail what
00:13:27.519 00:13:27.529 is being considered is the heat transfer
00:13:29.499 00:13:29.509 right to the control volume which is
00:13:33.639 00:13:33.649 equal to minus the mass flow rate h
00:13:36.939 00:13:36.949 multiplied by CP h multiplied by dt h
00:13:41.379 00:13:41.389 and we use the minus sign in terms of
00:13:45.699 00:13:45.709 the directions now and then on the cold
00:13:50.979 00:13:50.989 side is equal to minus the mass flow
00:13:53.439 00:13:53.449 rate on the cold side CP on the cold
00:13:55.749 00:13:55.759 side multiplied by dtc
00:14:04.030 00:14:04.040 we have done the derivation previously
00:14:06.920 00:14:06.930 of this equation so we are not going to
00:14:10.310 00:14:10.320 do it in detail again
00:14:11.660 00:14:11.670 just go and look on in the chapter on
00:14:14.600 00:14:14.610 eternal force convection but the result
00:14:18.200 00:14:18.210 of that is as you know that it can be
00:14:21.500 00:14:21.510 shown that the heat transfer rate is
00:14:23.930 00:14:23.940 equal to u multiplied by the area x ln
00:14:29.420 00:14:29.430 TD where this LM TD is equal to LM TD is
00:14:42.860 00:14:42.870 equal to this temperature difference
00:14:45.290 00:14:45.300 which is delta T one minus this
00:14:53.150 00:14:53.160 temperature difference which is delta T
00:14:55.490 00:14:55.500 2 divided by the limb of delta T 1
00:15:00.590 00:15:00.600 divided by delta T 2
00:15:12.020 00:15:12.030 okay so this is the correct way of doing
00:15:16.460 00:15:16.470 it of using the alum TD if you would say
00:15:25.220 00:15:25.230 let's consider the average temperature
00:15:28.790 00:15:28.800 as 1/2 times delta T 1 plus delta T 2
00:15:36.370 00:15:36.380 then you can actually go in show
00:15:38.690 00:15:38.700 mathematically that Delta TM will always
00:15:46.310 00:15:46.320 be smaller than delta T of average
00:15:55.400 00:15:55.410 which means that if you use delta T
00:15:58.069 00:15:58.079 average you're going to over estimate
00:16:00.579 00:16:00.589 the transfer right isn't it okay that is
00:16:06.859 00:16:06.869 why it is so dangerous to use this
00:16:08.839 00:16:08.849 approach so do not do that okay that is
00:16:15.199 00:16:15.209 not so accurate some cases the error is
00:16:18.469 00:16:18.479 not so large that in general be careful
00:16:22.789 00:16:22.799 for that now this derivation take note
00:16:25.789 00:16:25.799 has been done for parallel flow heat
00:16:27.679 00:16:27.689 exchanger you could do the same for
00:16:30.529 00:16:30.539 counter flow and go and do the
00:16:37.909 00:16:37.919 derivation for counter flow for counter
00:16:41.539 00:16:41.549 flow as you know means that the flow on
00:16:45.019 00:16:45.029 the inside would be in that direction
00:16:46.519 00:16:46.529 and in the others it would be in the
00:16:49.489 00:16:49.499 opposite direction that is the
00:16:51.859 00:16:51.869 definition of a counter flow heat
00:16:54.199 00:16:54.209 exchanger okay and it can be proven that
00:17:01.210 00:17:01.220 the delta T pelham TD of take note this
00:17:09.679 00:17:09.689 means counter flow see if counter flow
00:17:12.740 00:17:12.750 heat exchanger is always larger than the
00:17:16.159 00:17:16.169 Delta TLM TD of the parallel flow heat
00:17:20.269 00:17:20.279 exchanger
00:17:49.970 00:17:49.980 okay now why would this be important
00:17:52.630 00:17:52.640 Aling TD of a counter flow heat
00:17:54.800 00:17:54.810 exchanger is always larger that the LM
00:17:56.990 00:17:57.000 TD of the parallel flow heat exchanger
00:17:59.030 00:17:59.040 it would mean that just by changing the
00:18:03.710 00:18:03.720 configuration of the streams the counter
00:18:07.730 00:18:07.740 flow heat exchanger is always going to
00:18:09.740 00:18:09.750 give a higher heat transfer rate than a
00:18:11.750 00:18:11.760 parallel flow heat exchanger okay that
00:18:15.080 00:18:15.090 is what it means okay so in industry
00:18:19.130 00:18:19.140 I've never seen a parallel flow heat
00:18:20.390 00:18:20.400 exchanger and if that happens then it is
00:18:23.960 00:18:23.970 wrongly connected most probably okay as
00:18:28.610 00:18:28.620 you can see we're parallel flow heat
00:18:30.020 00:18:30.030 exchanger the delta T just becomes
00:18:32.270 00:18:32.280 smaller and smaller and smaller so this
00:18:36.440 00:18:36.450 is a very important conclusion that can
00:18:41.240 00:18:41.250 be made from this derivation and then
00:18:44.060 00:18:44.070 the other category of problems is let's
00:18:46.490 00:18:46.500 suppose we now consider a multi pause
00:18:49.930 00:18:49.940 and cross flow heat exchangers
00:18:57.510 00:18:57.520 Crossland exchanges remember what this
00:19:00.460 00:19:00.470 all is all about is is with this
00:19:03.280 00:19:03.290 calculation what temperature
00:19:05.470 00:19:05.480 distribution should we use so tourette
00:19:08.320 00:19:08.330 eclis you can show for the parallel heat
00:19:10.330 00:19:10.340 exchanger it should be the LM TD that
00:19:12.970 00:19:12.980 that is the one that gives us the exact
00:19:15.130 00:19:15.140 answer for the parallel flow the same
00:19:18.120 00:19:18.130 but it can be shown that the LM TD for a
00:19:21.910 00:19:21.920 counter flow is higher than that of the
00:19:25.180 00:19:25.190 LM TD of a parallel flow now let's
00:19:28.210 00:19:28.220 suppose we've got a multi pause or a
00:19:30.130 00:19:30.140 cross flow type of heat exchanger what
00:19:32.800 00:19:32.810 happens then
00:19:33.580 00:19:33.590 well unfortunately it is not so simple
00:19:36.840 00:19:36.850 but what happens is that we need to jump
00:19:39.490 00:19:39.500 over things a little bit okay and we
00:19:42.310 00:19:42.320 Gypo it with a correction factor and we
00:19:44.740 00:19:44.750 say the LM TD is now equal to the
00:19:48.780 00:19:48.790 correction factor multiplied by the LM
00:19:52.180 00:19:52.190 TD take note very very importantly of
00:19:57.580 00:19:57.590 the counter flow heat exchanger
00:20:03.080 00:20:03.090 00:20:05.690 00:20:05.700 TD of counterflow okay now where do we
00:20:14.749 00:20:14.759 get the friction factor of this
00:20:16.159 00:20:16.169 correction factor from if the correction
00:20:18.860 00:20:18.870 factor well the correction factor is
00:20:21.110 00:20:21.120 given in graphs I'm going to show you
00:20:23.299 00:20:23.309 one just now and usually the this
00:20:26.889 00:20:26.899 correction factor is a function of P
00:20:30.580 00:20:30.590 which might be on that scale as a
00:20:33.860 00:20:33.870 function of R now what is P and or equal
00:20:38.419 00:20:38.429 to P and are typically just as an
00:20:44.989 00:20:44.999 example P would maybe be equal to small
00:20:48.200 00:20:48.210 T 2 minus small T 1 divided by T 1 minus
00:20:53.180 00:20:53.190 T 1 something like that and then to show
00:20:56.149 00:20:56.159 it to you just now an R is might be
00:21:00.019 00:21:00.029 equal to t1 minus t2 divided by t2 minus
00:21:05.419 00:21:05.429 t1
00:21:09.600 00:21:09.610 now those temperatures the capital T's
00:21:13.420 00:21:13.430 and the small T's they will be a sketch
00:21:16.120 00:21:16.130 given to you
00:21:17.170 00:21:17.180 maybe the sketch should look something
00:21:18.940 00:21:18.950 like that and on the sketch it's going
00:21:24.610 00:21:24.620 to show you t1 and t2 for example okay
00:21:30.700 00:21:30.710 and t1 like that and t2 like that so you
00:21:37.090 00:21:37.100 have to use that you have to align that
00:21:40.960 00:21:40.970 with the practical situation in terms of
00:21:43.390 00:21:43.400 the heat exchanger given to you then you
00:21:45.610 00:21:45.620 calculate the P and the order value and
00:21:47.470 00:21:47.480 from that you can get the fit the
00:21:49.330 00:21:49.340 correction factor
00:21:57.030 00:21:57.040 and I just as an example Lloyd if you
00:21:59.950 00:21:59.960 can go up to the ogre it project today
00:22:03.030 00:22:03.040 in figure 11 point 18 in your textbook
00:22:06.400 00:22:06.410 there are a few examples are given four
00:22:08.710 00:22:08.720 of them I think because it is too small
00:22:11.320 00:22:11.330 I'm going to try to zoom in a little bit
00:22:18.630 00:22:18.640 okay
00:22:20.920 00:22:20.930 now look at that one the bee Grove it
00:22:23.770 00:22:23.780 gives the correction factor for what
00:22:26.410 00:22:26.420 type of heat exchanger the correction
00:22:30.370 00:22:30.380 factor for a two shell pause so there
00:22:32.860 00:22:32.870 are two shells and for a 12 etc any
00:22:37.390 00:22:37.400 multiple of four two pulses you see so
00:22:41.530 00:22:41.540 schematically if it has been given to
00:22:44.710 00:22:44.720 you in the exam it is a two shell pause
00:22:47.440 00:22:47.450 and an 82 pause heat exchanger then this
00:22:51.700 00:22:51.710 graph would be valid you see and you
00:22:54.820 00:22:54.830 would use this graph then in terms of
00:22:56.620 00:22:56.630 what flows through the shell and there
00:22:59.110 00:22:59.120 you can see this the shell there is T 1
00:23:01.960 00:23:01.970 capital T 1 is then the imitating
00:23:04.180 00:23:04.190 picture of the shell and T 2 is the
00:23:06.850 00:23:06.860 outlet temperature of the shell you see
00:23:09.010 00:23:09.020 and the tubes in a temperature small T 1
00:23:13.150 00:23:13.160 and the outlet temperature is small T 2
00:23:16.260 00:23:16.270 based on that you can calculate P which
00:23:20.440 00:23:20.450 is then given on this axis and the
00:23:22.750 00:23:22.760 arrows are which is those lines there so
00:23:25.560 00:23:25.570 maybe if P is equal to 0.5 and R is
00:23:30.970 00:23:30.980 equal to 1.5 then we can get the
00:23:33.730 00:23:33.740 correction factor as 0.85 you see that
00:23:37.320 00:23:37.330 example okay now
00:23:45.120 00:23:45.130 there is however one very important
00:23:48.450 00:23:48.460 thing and that is let's suppose in one
00:23:53.159 00:23:53.169 of those lines we've got a fluid which
00:23:56.190 00:23:56.200 is being condensed or which boils okay
00:24:00.840 00:24:00.850 what happens then okay so if you look at
00:24:05.310 00:24:05.320 that if one of them contains has or
00:24:08.399 00:24:08.409 boils then the temperatures is going to
00:24:11.669 00:24:11.679 be the same t1 small t1 is going to be
00:24:14.879 00:24:14.889 small t2 a capital T 1 is going to be
00:24:17.909 00:24:17.919 capital T 2 and then you might have the
00:24:20.850 00:24:20.860 problem that some of those Peas or or or
00:24:23.850 00:24:23.860 equal to zero or infinite and then
00:24:27.690 00:24:27.700 students look at it and they say oh I
00:24:29.399 00:24:29.409 don't know what to do now okay
00:24:31.049 00:24:31.059 what is important to know is that if we
00:24:35.669 00:24:35.679 have boiling or condensation
00:24:43.000 00:24:43.010 for boiling or conversation if is equal
00:24:46.210 00:24:46.220 to one so if is always equal to one if
00:24:51.520 00:24:51.530 we have boiling or condensation on one
00:24:55.120 00:24:55.130 of the sides okay ladies and gentlemen
00:24:58.870 00:24:58.880 that's the theory I ready to do a
00:25:02.110 00:25:02.120 problem okay let's do example 11.5 in
00:25:12.669 00:25:12.679 your textbook
00:25:23.740 00:25:23.750 this
00:25:26.850 00:25:26.860 and in this example what is given to you
00:25:30.539 00:25:30.549 is a to shout pause and a full tube
00:25:33.600 00:25:33.610 pause heat exchanger okay just
00:25:38.039 00:25:38.049 schematically they are the two shells
00:25:46.880 00:25:46.890 there's the init of the first shell like
00:25:51.120 00:25:51.130 there and this is the outlet of the
00:25:53.009 00:25:53.019 second show
00:25:57.840 00:25:57.850 and we've got 32 pauses going through so
00:26:02.010 00:26:02.020 it is one two three and four and this
00:26:12.060 00:26:12.070 one inside the tube we've got water at
00:26:16.730 00:26:16.740 80 degrees Celsius and the exit
00:26:20.850 00:26:20.860 temperature of the water V is equal to
00:26:23.040 00:26:23.050 40 degrees Celsius the side we've got
00:26:27.480 00:26:27.490 the serene and the temperature Inlet
00:26:34.500 00:26:34.510 temperature is 20 degrees Celsius and
00:26:37.520 00:26:37.530 here we've got the outlet temperature of
00:26:40.320 00:26:40.330 the glycerine which is 50 degrees
00:26:42.270 00:26:42.280 Celsius it is also given that the tube
00:26:50.340 00:26:50.350 diameters are 20 millimeters and that
00:26:53.910 00:26:53.920 the tubes often let some tubes
00:27:00.560 00:27:00.570 and the heat transfer coefficient in the
00:27:06.259 00:27:06.269 shell is equal to 25 watts per square
00:27:10.159 00:27:10.169 meter degree Celsius and the heat
00:27:13.129 00:27:13.139 transfer coefficient inside the tubes
00:27:14.839 00:27:14.849 his 160 watts per square meter degree
00:27:18.649 00:27:18.659 Celsius and they asked us to determine
00:27:26.079 00:27:26.089 the heat transfer rate firstly without
00:27:31.310 00:27:31.320 any fouling and then secondly the heat
00:27:35.209 00:27:35.219 transfer rate if the fouling on the
00:27:39.619 00:27:39.629 outside of the tubes is equal to 0.2
00:27:46.759 00:27:46.769 Palsy row six square meters screaming
00:27:52.369 00:27:52.379 this Kelvin for what
00:28:27.570 00:28:27.580 okay we are going to need the surface
00:28:29.340 00:28:29.350 area of the off the tubes and it's just
00:28:32.460 00:28:32.470 calculated to make things easier it is
00:28:35.460 00:28:35.470 PI multiplied by the diameter multiplied
00:28:38.160 00:28:38.170 by the links and sorry I forgot to give
00:28:41.190 00:28:41.200 you the length the length is 60 meters
00:28:44.870 00:28:44.880 okay the length of the tubes 60 meters
00:28:51.620 00:28:51.630 okay so the surface area is equal to PI
00:28:54.800 00:28:54.810 multiplied by the diameter which is 20
00:28:57.690 00:28:57.700 moles multiplied by 60 and that gives us
00:29:02.370 00:29:02.380 a surface area of three point seven
00:29:05.850 00:29:05.860 seven square meters
00:29:17.970 00:29:17.980 now so from the theory we have seen what
00:29:21.450 00:29:21.460 it has been shown to us that we can
00:29:23.789 00:29:23.799 calculate the heat roles for right as
00:29:25.799 00:29:25.809 you multiplied by the area multiplied by
00:29:29.370 00:29:29.380 L MTD or sorry you multiplied by the
00:29:33.450 00:29:33.460 area multiplied by the correction factor
00:29:35.970 00:29:35.980 multiplied by L MTD of a counter flow
00:29:40.139 00:29:40.149 heat exchanger
00:29:48.090 00:29:48.100 now what we also know is that the heat
00:29:51.159 00:29:51.169 rolls for right on the cold side is
00:29:53.230 00:29:53.240 equal to the mass flow rate multiplied
00:29:55.239 00:29:55.249 by CP multiplied by the delta T and on
00:29:59.889 00:29:59.899 the hot side it is equal to the mass
00:30:03.159 00:30:03.169 flow rate multiplied by CP delta T and
00:30:10.259 00:30:10.269 we could have used one of these
00:30:12.430 00:30:12.440 equations also to get the heat transfer
00:30:14.649 00:30:14.659 right but the mass flow rates are not
00:30:17.499 00:30:17.509 given to us so that is the reason why we
00:30:20.499 00:30:20.509 can't use these two equations otherwise
00:30:23.470 00:30:23.480 we could have calculated it directly
00:30:25.869 00:30:25.879 from there
00:30:31.510 00:30:31.520 okay now take a note we have to get the
00:30:33.880 00:30:33.890 Delta TLM TD of a counter flow heat
00:30:37.630 00:30:37.640 exchanger now I want you to try for me
00:30:41.080 00:30:41.090 and see please calculate that quickly
00:30:43.270 00:30:43.280 for me don't have to do this final
00:30:45.310 00:30:45.320 calculation but write down all the terms
00:30:47.950 00:30:47.960 for me in terms of how you would
00:30:49.270 00:30:49.280 calculate the ln TD of that heat
00:30:51.610 00:30:51.620 exchanger counter flow
00:31:19.170 00:31:19.180 okay we do it as follows you don't need
00:31:21.880 00:31:21.890 to do it physically temperature as a
00:31:26.950 00:31:26.960 function of X there must be a hot stream
00:31:30.940 00:31:30.950 in a cold stream so the hot stream
00:31:33.130 00:31:33.140 obviously must be on top
00:31:34.900 00:31:34.910 okay the hot stream always does
00:31:37.990 00:31:38.000 something like that isn't it okay now I
00:31:40.990 00:31:41.000 choose for no specific reason
00:31:44.520 00:31:44.530 normally that as the inlet temperature
00:31:47.040 00:31:47.050 the outside which is 80 come outside
00:31:52.840 00:31:52.850 then it is 80 and then the outlet
00:31:55.660 00:31:55.670 temperature there is equal to 40 you
00:31:58.000 00:31:58.010 agree okay and then there must be a cold
00:32:01.630 00:32:01.640 site okay now if that is the inlet
00:32:05.920 00:32:05.930 temperature 80 and 40 the counter flow
00:32:11.050 00:32:11.060 direction the counter flow direction
00:32:13.420 00:32:13.430 then on the cold side what is the inlet
00:32:15.910 00:32:15.920 temperature the inlet temperature on the
00:32:18.400 00:32:18.410 cold side is equal to 20 and the outlet
00:32:21.400 00:32:21.410 temperature on the cold side is equal to
00:32:23.320 00:32:23.330 50 okay now you've got a counter flow
00:32:25.750 00:32:25.760 heat exchanger you forced it almighty to
00:32:28.540 00:32:28.550 be a counter flow heat exchanger
00:32:33.210 00:32:33.220 does that make sense you happy lab okay
00:32:38.100 00:32:38.110 so therefore we can L say LM TD is equal
00:32:45.760 00:32:45.770 to this temperature difference which is
00:32:48.610 00:32:48.620 equal to 30 conserve a temperature
00:32:53.830 00:32:53.840 difference is 30 and that temperature
00:32:56.020 00:32:56.030 difference is 20 okay so it is 30 minus
00:33:00.250 00:33:00.260 20 divided by the limb of the T divided
00:33:04.659 00:33:04.669 by 20 which is then equal to twenty four
00:33:10.840 00:33:10.850 point seven degree Celsius
00:33:20.890 00:33:20.900 00:33:22.460 00:33:22.470 now we need to get the correction factor
00:33:24.680 00:33:24.690 if the correction factor is there is the
00:33:28.220 00:33:28.230 two shall pause and the four to process
00:33:31.879 00:33:31.889 as we have okay so based on that draw
00:33:35.720 00:33:35.730 off which is the figure 11 point 18 B P
00:33:45.350 00:33:45.360 is equal to T 2 minus T 1 divided by T 1
00:33:50.240 00:33:50.250 minus T 1 okay so if we look at that
00:33:57.259 00:33:57.269 then the shell is T 1 okay the shell in
00:34:04.279 00:34:04.289 a temperature is equal to t 1 so that is
00:34:07.460 00:34:07.470 equal to t 1 and that is equal to t 2
00:34:12.069 00:34:12.079 degrees
00:34:14.620 00:34:14.630 and in terms of the tubes what is going
00:34:18.220 00:34:18.230 in is that temperature t-1 there and
00:34:21.550 00:34:21.560 that is equal to t2 there you happy that
00:34:28.140 00:34:28.150 okay so P is equal to 40 minus 80
00:34:38.460 00:34:38.470 divided by 20
00:34:47.540 00:34:47.550 41:20 - 80 and that is equal to 0.67 and
00:34:58.410 00:34:58.420 R is equal to t1 minus t2 divided by
00:35:04.950 00:35:04.960 small t2 minus small t1 which is then
00:35:09.780 00:35:09.790 equal to 20 minus 50 divided by 40 minus
00:35:14.849 00:35:14.859 80 which is equal to 0.75
00:35:25.100 00:35:25.110 so based on that the friction factor is
00:35:27.770 00:35:27.780 of the correction factor is equal to 0.9
00:35:30.950 00:35:30.960 one
00:35:40.270 00:35:40.280 okay now we've got the LM TD we've got
00:35:43.190 00:35:43.200 the correction factor now we need to get
00:35:45.890 00:35:45.900 you multiplied by the area we actually
00:35:48.140 00:35:48.150 already have the surface area so we
00:35:50.870 00:35:50.880 actually only need to get the overall
00:35:52.520 00:35:52.530 heat transfer coefficient we can
00:35:54.260 00:35:54.270 calculate it separately or we can
00:35:57.470 00:35:57.480 calculate it as you multiplied by the
00:35:59.690 00:35:59.700 area whichever we want to do remember
00:36:03.880 00:36:03.890 one divided by UI is equal to one
00:36:10.910 00:36:10.920 divided by the heat transfer coefficient
00:36:12.710 00:36:12.720 on the inside multiplied by the area on
00:36:15.140 00:36:15.150 the inside plus the fouling on the
00:36:18.470 00:36:18.480 inside multiplied by the area plus the
00:36:21.650 00:36:21.660 resistance of the tube to PI KL plus the
00:36:29.120 00:36:29.130 fouling on the outside plus the flow
00:36:34.760 00:36:34.770 resistance on the outside that is
00:36:38.630 00:36:38.640 overall heat transfer coefficient
00:36:39.859 00:36:39.869 correlation we've done that with a
00:36:42.080 00:36:42.090 previous lecture
00:36:52.030 00:36:52.040 okay the fowling at this takes for part
00:36:56.710 00:36:56.720 a it is without any fouling the second
00:37:00.430 00:37:00.440 part of the problem is with the fouling
00:37:02.850 00:37:02.860 so firstly we can delete those two terms
00:37:08.040 00:37:08.050 it has been given that it is a
00:37:11.050 00:37:11.060 20-millimeter - but it is very very thin
00:37:13.660 00:37:13.670 okay if it is thin then it means that
00:37:17.140 00:37:17.150 that the diameter and that diameter is
00:37:19.330 00:37:19.340 approximately the same the limb of one
00:37:22.150 00:37:22.160 is equal to zero okay three four that
00:37:25.930 00:37:25.940 term is also negligible okay and because
00:37:33.370 00:37:33.380 it is thin that area and that area and
00:37:36.040 00:37:36.050 that area is the same okay so in this
00:37:39.730 00:37:39.740 specific case we can say 1 divided by u
00:37:42.400 00:37:42.410 is equal to 1 divided by the heat
00:37:45.430 00:37:45.440 transfer coefficient on the outside
00:37:47.160 00:37:47.170 divided by 1 divided by the heat
00:37:49.660 00:37:49.670 transfer coefficient on the inside plus
00:37:52.210 00:37:52.220 1 divided by the transfer coefficient on
00:37:54.070 00:37:54.080 the outside okay which is equal to 1
00:38:00.100 00:38:00.110 00:38:01.840 00:38:01.850 on the inside which is equal to hundred
00:38:06.270 00:38:06.280 160 plus 1 divided by the heat transfer
00:38:09.760 00:38:09.770 coefficient on the outside which is 25
00:38:13.080 00:38:13.090 so the overall heat transfer coefficient
00:38:16.300 00:38:16.310 is going to be very close to the
00:38:19.870 00:38:19.880 smallest deterrence or coefficient which
00:38:21.400 00:38:21.410 is 25 and if we calculated it is going
00:38:24.490 00:38:24.500 to be 20 1.6 watts per square meter
00:38:28.200 00:38:28.210 Kelvin
00:38:35.530 00:38:35.540 right so now we can calculate the heat
00:38:38.420 00:38:38.430 transfer rate we can say the heat
00:38:40.010 00:38:40.020 transfer rate is equal to u multiplied
00:38:42.349 00:38:42.359 by the area multiplied by the correction
00:38:45.380 00:38:45.390 factor multiplied by L MTD for a counter
00:38:50.690 00:38:50.700 flow heat exchanger
00:39:14.420 00:39:14.430 you is equal to 21.7 the surface area
00:39:20.220 00:39:20.230 we've calculated as 3.77
00:39:24.589 00:39:24.599 the correction factor is 0.9 1 and the
00:39:29.309 00:39:29.319 LM TD is equal to twenty four point
00:39:32.069 00:39:32.079 seven and that would give us a heat
00:39:36.630 00:39:36.640 transfer rate of one thousand eight
00:39:38.760 00:39:38.770 hundred and fifty watts for one point
00:39:42.780 00:39:42.790 eight three kilowatts
00:39:55.160 00:39:55.170 you happy with that okay let's do Part B
00:39:59.090 00:39:59.100 or B is now with fouling of 0.2 palsy
00:40:05.340 00:40:05.350 row so we need to recalculate the
00:40:10.170 00:40:10.180 overall heat transfer coefficient so one
00:40:12.510 00:40:12.520 divided by the overall heat transfer
00:40:14.040 00:40:14.050 coefficient is equal to one divided by
00:40:17.460 00:40:17.470 the heat transfer coefficient multiplied
00:40:19.290 00:40:19.300 by the area fouling on the inside plus
00:40:24.180 00:40:24.190 the learning of the diameter I chose
00:40:27.840 00:40:27.850 divided by 2 pi KL plus the resistance
00:40:32.640 00:40:32.650 on the outside plus 1 divided by the
00:40:37.140 00:40:37.150 heat transfer coefficient on the outside
00:40:38.910 00:40:38.920 multiplied by the area on outside
00:40:54.939 00:40:54.949 okay again the tube is very thin so we
00:40:59.299 00:40:59.309 can say that the resistance through the
00:41:01.039 00:41:01.049 tube is negligible they didn't give us
00:41:03.529 00:41:03.539 any fouling information on the inside
00:41:08.049 00:41:08.059 for this special case where the area on
00:41:12.049 00:41:12.059 the inside is equal to the area on the
00:41:14.239 00:41:14.249 outside we can cancel all the areas and
00:41:21.789 00:41:21.799 the result is one divided by the overall
00:41:24.919 00:41:24.929 heat transfer coefficient is equal to
00:41:27.049 00:41:27.059 one divided by the heat transfer
00:41:29.419 00:41:29.429 coefficient on the inside plus the
00:41:32.179 00:41:32.189 fouling resistance on the outside plus
00:41:35.650 00:41:35.660 00:41:38.239 00:41:38.249 coefficient on the outside
00:41:47.280 00:41:47.290 the transfer coefficient on the inside
00:41:51.210 00:41:51.220 is equal to 160 the troweling resistance
00:41:56.950 00:41:56.960 is equal to 0.2 'pl 0 6 plus 1 divided
00:42:03.820 00:42:03.830 by the heat transfer coefficient on the
00:42:05.440 00:42:05.450 outside which is equal to 25 and the
00:42:11.560 00:42:11.570 result is if we calculate the overall
00:42:14.020 00:42:14.030 heat transfer coefficient now is going
00:42:16.360 00:42:16.370 to be 21.3 watts per square meter kelvin
00:42:25.110 00:42:25.120 should it be lower than the other one
00:42:29.490 00:42:29.500 obviously yes because of the fouling
00:42:32.580 00:42:32.590 should decrease its efficiency we can
00:42:35.860 00:42:35.870 see it is but it's not much not much so
00:42:40.030 00:42:40.040 if you go and recalculate the overall
00:42:43.480 00:42:43.490 heat transfer rate which is equal to UA
00:42:46.320 00:42:46.330 x if x LM t-d-z which is now equal to 21
00:42:54.340 00:42:54.350 point 3 multiplied by the surface area
00:42:57.940 00:42:57.950 which is three point seven seven
00:43:00.480 00:43:00.490 multiplied by the correction factor 0.9
00:43:03.580 00:43:03.590 one multiply 5 and TD twenty four point
00:43:06.790 00:43:06.800 seven and the result is the heat
00:43:12.880 00:43:12.890 transfer rate of 1805 watts
00:43:18.990 00:43:19.000 so if we could compare as it rolls right
00:43:21.510 00:43:21.520 to that one we can see that they either
00:43:24.480 00:43:24.490 decrease but at this stage it's not a
00:43:29.550 00:43:29.560 lot after 10 years
00:43:34.190 00:43:34.200 might be very significant any questions
00:43:38.580 00:43:38.590 ladies and gentlemen if not thank you
00:43:41.190 00:43:41.200 very much
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