/ News & Press / Video / Lecture 35 (2013). 11.3 Analysis of Heat Exchangers. 11.4 Log Mean Temperature Difference Method
Lecture 35 (2013). 11.3 Analysis of Heat Exchangers. 11.4 Log Mean Temperature Difference Method
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00:00:01.280 ladies and gentlemen let's continue with 00:00:03.53000:00:03.540 chapter 11 chapter 11 is the chapter on 00:00:06.32000:00:06.330 heat exchanges that is the chapter where 00:00:09.20000:00:09.210 so many of the work that we've done 00:00:10.99000:00:11.000 comes together we've already looked at 00:00:14.41900:00:14.429 the different types of heat exchangers 00:00:16.26900:00:16.279 we've looked at the overall heat 00:00:19.25000:00:19.260 transfer coefficient the U and the 00:00:22.22000:00:22.230 combination of U multiplied by the area 00:00:24.74000:00:24.750 and today we're going to continue with 00:00:27.52900:00:27.539 paragraph eleven point three in the 00:00:30.91900:00:30.929 textbook of single and gauge or on the 00:00:33.88900:00:33.899 analysis of heat exchanges and then also 00:00:36.29000:00:36.300 paragraph eleven point four the LM TD 00:00:39.79900:00:39.809 method now in terms of when we look at 00:00:44.45000:00:44.460 it exchanges you'll see that in general 00:00:46.54900:00:46.559 there are at least two methods that can 00:00:49.70000:00:49.710 be used to analyze heat exchanges and 00:00:52.81900:00:52.829 these two methods are the most popular 00:00:55.31000:00:55.320 methods and are the ones which are been 00:00:58.70000:00:58.710 used the most most the first one is 00:01:02.72000:01:02.730 known as the LM TD method the lock mean 00:01:07.70000:01:07.710 temperature difference method and the 00:01:10.34000:01:10.350 second one is known as the X alone into 00:01:14.24000:01:14.250 you method or the effectiveness into you 00:01:18.20000:01:18.210 method we are not going to address that 00:01:21.08000:01:21.090 now we are going to address that with 00:01:22.91000:01:22.920 the next lecture but in general for both 00:01:27.32000:01:27.330 of these methods let's just consolidate 00:01:30.20000:01:30.210 some of the most important theories okay 00:01:33.35000:01:33.360 and the theory is that if we look ahead 00:01:37.28000:01:37.290 at the heat exchanger they are going to 00:01:39.77000:01:39.780 be two streams and the temperatures are 00:01:44.27000:01:44.280 going to change maybe something like 00:01:47.56900:01:47.579 that let's call that the temperature 00:01:51.23000:01:51.240 scale here and there that is X so that 00:01:56.17900:01:56.189 is how the temperature on the hot side 00:01:58.42900:01:58.439 is going to change and on the cold side 00:02:00.98000:02:00.990 it is maybe going to do 00:02:02.81000:02:02.820 something like that and although I've 00:02:07.40000:02:07.410 drawn the directions like that it 00:02:10.10000:02:10.110 doesn't have to be like that okay that 00:02:12.89000:02:12.900 is just a general schematic 00:02:15.13000:02:15.140 representation of an heat exchanger 00:02:17.63000:02:17.640 we've got a hot side and a cold side 00:02:19.94000:02:19.950 normally heat exchanges are well 00:02:22.34000:02:22.350 insulated we want to prevent the heat 00:02:25.07000:02:25.080 losses and it is equipment that or 00:02:30.53000:02:30.540 normally that normally runs for very 00:02:32.51000:02:32.520 long hours so normally we've got 00:02:34.22000:02:34.230 steady-state conditions also 00:02:36.29000:02:36.300 so therefore in general we can say that 00:02:39.40000:02:39.410 the heat transfer rate on the cold side 00:02:42.52000:02:42.530 on the cold side the cold side we just 00:02:46.58000:02:46.590 mean the stream with the lowest 00:02:48.44000:02:48.450 temperature so the it roles are right 00:02:51.83000:02:51.840 would obviously be from the stream which 00:02:55.88000:02:55.890 is at the higher temperature to the 00:02:57.41000:02:57.420 stream which is at the lower temperature 00:02:59.57000:02:59.580 so the heat roles for right cue from the 00:03:06.08000:03:06.090 cold side or to the cold side is equal 00:03:08.57000:03:08.580 to the mass flow rate of the cold side 00:03:10.69900:03:10.709 the CP of the cold side multiplied by 00:03:15.13000:03:15.140 the outlet temperature that increase of 00:03:18.89000:03:18.900 a minus TC in so this temperature is TC 00:03:25.82000:03:25.830 out and that temperature is equal to TC 00:03:28.97000:03:28.980 in the cold in a temperature and the 00:03:33.11000:03:33.120 cold outlet temperature 00:03:36.52000:03:36.530 and that's it should be the same on the 00:03:38.98000:03:38.990 hot side q on the hot side is equal to 00:03:41.80000:03:41.810 the mass flow rate on the hot side CP on 00:03:44.86000:03:44.870 the hot side x now th in minus th out so 00:03:56.71000:03:56.720 this temperature here is equal to gh in 00:04:00.67000:04:00.680 and thatis th out 00:04:06.19000:04:06.200 does that make sense okay so it's very 00:04:10.21000:04:10.220 simple okay now these terms have a 00:04:16.06000:04:16.070 specific meaning and the meaning is it 00:04:20.34900:04:20.359 is known as the heat capacity ratio the 00:04:26.53000:04:26.540 heat capacity and I shall and the heat 00:04:30.96900:04:30.979 capacity ratio is written as C for the 00:04:36.34000:04:36.350 hot side which is then equal to the mass 00:04:39.21900:04:39.229 flow rate of the hot side multiplied by 00:04:41.56000:04:41.570 CP of the hot side and on the cold side 00:04:48.31000:04:48.320 it is equal to the mass flow rate on the 00:04:50.74000:04:50.750 cold side x CP on the cold side 00:04:56.93000:04:56.940 and have and that product represents the 00:05:00.56000:05:00.570 amount of heat needed to increase the 00:05:03.62000:05:03.630 temperature with one degree Celsius so 00:05:06.59000:05:06.600 that is the heat capacity ratio okay if 00:05:09.83000:05:09.840 the ratio is very very large then it 00:05:12.77000:05:12.780 means I need a lot of heat to increase 00:05:15.80000:05:15.810 the temperature one one degree Celsius 00:05:17.71000:05:17.720 if it is a low value it means I do not 00:05:21.65000:05:21.660 leave need that much energy to increase 00:05:23.93000:05:23.940 the temperature with one degree Celsius 00:05:31.12000:05:31.130 now there are two special types of heat 00:05:34.25000:05:34.260 exchangers or two special categories two 00:05:42.05000:05:42.060 special categories of heat exchanges and 00:05:44.12000:05:44.130 those heat exchanges are known as and 00:05:47.96000:05:47.970 we've done it already 00:05:48.95000:05:48.960 as condenser or boilers 00:05:57.46000:05:57.470 condenser or a boiler so on a TS diagram 00:06:02.45000:06:02.460 that is a saturation line then with a 00:06:10.43000:06:10.440 condenser we will change a gas to a 00:06:16.85000:06:16.860 fluid okay 00:06:20.18000:06:20.190 and during that process they will be 00:06:23.96000:06:23.970 heat transferred to the environment or 00:06:28.67000:06:28.680 to the other stream in any case with a 00:06:32.75000:06:32.760 boiler it is the opposite and now it 00:06:36.98000:06:36.990 will be from fluid to a gas and to 00:06:43.34000:06:43.350 change the fluid from from from a fluid 00:06:46.40000:06:46.410 or the fluid from a liquid to a gas we 00:06:50.78000:06:50.790 need to put in heat it has to be added 00:06:56.95000:06:56.960 this guys it is going to be released now 00:07:02.12000:07:02.130 for these special cases to special 00:07:05.54000:07:05.550 categories the heat transfer rate is not 00:07:09.71000:07:09.720 equal to any mass flow CP x delta T okay 00:07:15.58000:07:15.590 why because as you can see the 00:07:18.98000:07:18.990 temperature remains constant during the 00:07:21.50000:07:21.510 process the temperature remains constant 00:07:25.39000:07:25.400 it is equal to the mass flow rate 00:07:29.72000:07:29.730 multiplied by H FG 00:07:32.95000:07:32.960 the change in enthalpy between those two 00:07:37.64000:07:37.650 points 00:07:40.94000:07:40.950 okay so for this special category that 00:07:43.98000:07:43.990 we have the C which normally is equal to 00:07:49.56000:07:49.570 the mass flow rate multiplied by CP is 00:07:56.43000:07:56.440 infinite very very important they are 00:08:03.15000:08:03.160 going to be problems in the test on the 00:08:05.37000:08:05.380 exam heat exchanger is going to be given 00:08:08.07000:08:08.080 to you and then in the description it's 00:08:12.15000:08:12.160 going to say evaporation occurs or 00:08:15.21000:08:15.220 condensation occurs or something like 00:08:17.85000:08:17.860 that and one or both of these values are 00:08:22.95000:08:22.960 not going to be given and if you do not 00:08:24.87000:08:24.880 use C equal to infinite then you'll not 00:08:27.36000:08:27.370 be able to do the problem so it's a very 00:08:29.01000:08:29.020 very important concept that you do 00:08:31.29000:08:31.300 understand that CP normally as a gas you 00:08:37.92000:08:37.930 can use that type of calculation in that 00:08:40.74000:08:40.750 region you can use it in this region but 00:08:43.50000:08:43.510 not here 00:08:44.07000:08:44.080 CP doesn't exist here because it is 00:08:48.36000:08:48.370 infinite okay so the mass flow rate 00:08:50.46000:08:50.470 multiplied by the CP in the two-phase 00:08:53.19000:08:53.200 region is infinite okay now in general 00:09:01.96000:09:01.970 the heat transfer right now without 00:09:05.69000:09:05.700 referring to a hot site or a cold site 00:09:08.35000:09:08.360 we would also say is equal to u 00:09:11.18000:09:11.190 multiplied by the surface area 00:09:13.87000:09:13.880 multiplied by delta T and this delta T 00:09:18.05000:09:18.060 is now very important because it should 00:09:22.07000:09:22.080 be a representative a very 00:09:25.85000:09:25.860 representative temperature difference 00:09:31.24000:09:31.250 now if we look at this then you might 00:09:35.87000:09:35.880 say well let's use this okay and in many 00:09:38.78000:09:38.790 cases that might be a good assumption 00:09:40.32900:09:40.339 but they are going to be cases where it 00:09:43.49000:09:43.500 is not going to be a good assumption 00:09:45.59000:09:45.600 for example maybe the temperatures does 00:09:49.19000:09:49.200 that we've got condensation and we've 00:09:53.72000:09:53.730 got the other stream doing something 00:09:56.81000:09:56.820 like that okay 00:09:58.01000:09:58.020 so then that in creature difference is 00:10:00.05000:10:00.060 not a good representation and we've 00:10:02.57000:10:02.580 already looked at it we've said that it 00:10:04.34000:10:04.350 is better in those cases to use the LM 00:10:07.10000:10:07.110 TG temperature difference and here we 00:10:10.61000:10:10.620 are going on to the same pause okay so 00:10:14.06000:10:14.070 what immature difference should be used 00:10:16.67000:10:16.680 there you're going to see is going to be 00:10:18.80000:10:18.810 the LM TD okay and that is what 00:10:22.61000:10:22.620 paragraph eleven point four is all about 00:10:25.39000:10:25.40000:10:29.06000:10:29.070 the LNT d the Lachlan integer difference 00:10:35.97000:10:35.980 okay and it starts by considering a 00:10:39.31000:10:39.320 parallel heat exchanger a parallel heat 00:10:46.03000:10:46.040 exchanger if exchanger looks like that 00:10:54.31900:10:54.329 and because it is parallel all the flow 00:10:58.25900:10:58.269 directions are in the same direction 00:11:07.75900:11:07.769 parallel flow heat exchanger so the flow 00:11:11.10000:11:11.110 is going in there it's going out there 00:11:13.82900:11:13.839 that flow is going in there and it goes 00:11:16.76900:11:16.779 out there and this graph gives the 00:11:22.37900:11:22.389 temperature as a function of X 00:11:36.67000:11:36.680 now if we look at the parallel heat 00:11:38.93000:11:38.940 exchanger typically the temperature 00:11:41.21000:11:41.220 profile does something like that that 00:11:44.60000:11:44.610 temperature we call th in on the hot 00:11:49.19000:11:49.200 side so that is th and that would be th 00:11:52.85000:11:52.860 out and the direction is from left to 00:11:58.31000:11:58.320 right on the cold side typically the 00:12:08.00000:12:08.010 temperature distribution is going to 00:12:09.62000:12:09.630 look like that so that is now going to 00:12:12.80000:12:12.810 be TC in and that is going to be TC out 00:12:24.62000:12:24.630 and this temperature difference is going 00:12:28.16000:12:28.170 to we can call it delta T 1 is equal to 00:12:31.85000:12:31.860 TIG minus TC in so this temperature 00:12:38.60000:12:38.610 difference th in minus TC in this 00:12:43.55000:12:43.560 temperature difference is equal to delta 00:12:47.03000:12:47.040 T 2 and that is equal to d HLT minus TC 00:12:55.46000:12:55.470 out 00:13:06.04000:13:06.050 okay now mathematically very correctly 00:13:08.73900:13:08.749 what is there being done in the textbook 00:13:11.01900:13:11.029 is a control volume is considered it is 00:13:16.41900:13:16.429 the control volume and you know how 00:13:18.99900:13:19.009 things goes with a control volume 00:13:20.85900:13:20.869 approach that is normally being followed 00:13:23.28900:13:23.299 and without showing all the detail what 00:13:27.51900:13:27.529 is being considered is the heat transfer 00:13:29.49900:13:29.509 right to the control volume which is 00:13:33.63900:13:33.649 equal to minus the mass flow rate h 00:13:36.93900:13:36.949 multiplied by CP h multiplied by dt h 00:13:41.37900:13:41.389 and we use the minus sign in terms of 00:13:45.69900:13:45.709 the directions now and then on the cold 00:13:50.97900:13:50.989 side is equal to minus the mass flow 00:13:53.43900:13:53.449 rate on the cold side CP on the cold 00:13:55.74900:13:55.759 side multiplied by dtc 00:14:04.03000:14:04.040 we have done the derivation previously 00:14:06.92000:14:06.930 of this equation so we are not going to 00:14:10.31000:14:10.320 do it in detail again 00:14:11.66000:14:11.670 just go and look on in the chapter on 00:14:14.60000:14:14.610 eternal force convection but the result 00:14:18.20000:14:18.210 of that is as you know that it can be 00:14:21.50000:14:21.510 shown that the heat transfer rate is 00:14:23.93000:14:23.940 equal to u multiplied by the area x ln 00:14:29.42000:14:29.430 TD where this LM TD is equal to LM TD is 00:14:42.86000:14:42.870 equal to this temperature difference 00:14:45.29000:14:45.300 which is delta T one minus this 00:14:53.15000:14:53.160 temperature difference which is delta T 00:14:55.49000:14:55.500 2 divided by the limb of delta T 1 00:15:00.59000:15:00.600 divided by delta T 2 00:15:12.02000:15:12.030 okay so this is the correct way of doing 00:15:16.46000:15:16.470 it of using the alum TD if you would say 00:15:25.22000:15:25.230 let's consider the average temperature 00:15:28.79000:15:28.800 as 1/2 times delta T 1 plus delta T 2 00:15:36.37000:15:36.380 then you can actually go in show 00:15:38.69000:15:38.700 mathematically that Delta TM will always 00:15:46.31000:15:46.320 be smaller than delta T of average 00:15:55.40000:15:55.410 which means that if you use delta T 00:15:58.06900:15:58.079 average you're going to over estimate 00:16:00.57900:16:00.589 the transfer right isn't it okay that is 00:16:06.85900:16:06.869 why it is so dangerous to use this 00:16:08.83900:16:08.849 approach so do not do that okay that is 00:16:15.19900:16:15.209 not so accurate some cases the error is 00:16:18.46900:16:18.479 not so large that in general be careful 00:16:22.78900:16:22.799 for that now this derivation take note 00:16:25.78900:16:25.799 has been done for parallel flow heat 00:16:27.67900:16:27.689 exchanger you could do the same for 00:16:30.52900:16:30.539 counter flow and go and do the 00:16:37.90900:16:37.919 derivation for counter flow for counter 00:16:41.53900:16:41.549 flow as you know means that the flow on 00:16:45.01900:16:45.029 the inside would be in that direction 00:16:46.51900:16:46.529 and in the others it would be in the 00:16:49.48900:16:49.499 opposite direction that is the 00:16:51.85900:16:51.869 definition of a counter flow heat 00:16:54.19900:16:54.209 exchanger okay and it can be proven that 00:17:01.21000:17:01.220 the delta T pelham TD of take note this 00:17:09.67900:17:09.689 means counter flow see if counter flow 00:17:12.74000:17:12.750 heat exchanger is always larger than the 00:17:16.15900:17:16.169 Delta TLM TD of the parallel flow heat 00:17:20.26900:17:20.279 exchanger 00:17:49.97000:17:49.980 okay now why would this be important 00:17:52.63000:17:52.640 Aling TD of a counter flow heat 00:17:54.80000:17:54.810 exchanger is always larger that the LM 00:17:56.99000:17:57.000 TD of the parallel flow heat exchanger 00:17:59.03000:17:59.040 it would mean that just by changing the 00:18:03.71000:18:03.720 configuration of the streams the counter 00:18:07.73000:18:07.740 flow heat exchanger is always going to 00:18:09.74000:18:09.750 give a higher heat transfer rate than a 00:18:11.75000:18:11.760 parallel flow heat exchanger okay that 00:18:15.08000:18:15.090 is what it means okay so in industry 00:18:19.13000:18:19.140 I've never seen a parallel flow heat 00:18:20.39000:18:20.400 exchanger and if that happens then it is 00:18:23.96000:18:23.970 wrongly connected most probably okay as 00:18:28.61000:18:28.620 you can see we're parallel flow heat 00:18:30.02000:18:30.030 exchanger the delta T just becomes 00:18:32.27000:18:32.280 smaller and smaller and smaller so this 00:18:36.44000:18:36.450 is a very important conclusion that can 00:18:41.24000:18:41.250 be made from this derivation and then 00:18:44.06000:18:44.070 the other category of problems is let's 00:18:46.49000:18:46.500 suppose we now consider a multi pause 00:18:49.93000:18:49.940 and cross flow heat exchangers 00:18:57.51000:18:57.520 Crossland exchanges remember what this 00:19:00.46000:19:00.470 all is all about is is with this 00:19:03.28000:19:03.290 calculation what temperature 00:19:05.47000:19:05.480 distribution should we use so tourette 00:19:08.32000:19:08.330 eclis you can show for the parallel heat 00:19:10.33000:19:10.340 exchanger it should be the LM TD that 00:19:12.97000:19:12.980 that is the one that gives us the exact 00:19:15.13000:19:15.140 answer for the parallel flow the same 00:19:18.12000:19:18.130 but it can be shown that the LM TD for a 00:19:21.91000:19:21.920 counter flow is higher than that of the 00:19:25.18000:19:25.190 LM TD of a parallel flow now let's 00:19:28.21000:19:28.220 suppose we've got a multi pause or a 00:19:30.13000:19:30.140 cross flow type of heat exchanger what 00:19:32.80000:19:32.810 happens then 00:19:33.58000:19:33.590 well unfortunately it is not so simple 00:19:36.84000:19:36.850 but what happens is that we need to jump 00:19:39.49000:19:39.500 over things a little bit okay and we 00:19:42.31000:19:42.320 Gypo it with a correction factor and we 00:19:44.74000:19:44.750 say the LM TD is now equal to the 00:19:48.78000:19:48.790 correction factor multiplied by the LM 00:19:52.18000:19:52.190 TD take note very very importantly of 00:19:57.58000:19:57.590 the counter flow heat exchanger 00:20:03.08000:20:03.09000:20:05.69000:20:05.700 TD of counterflow okay now where do we 00:20:14.74900:20:14.759 get the friction factor of this 00:20:16.15900:20:16.169 correction factor from if the correction 00:20:18.86000:20:18.870 factor well the correction factor is 00:20:21.11000:20:21.120 given in graphs I'm going to show you 00:20:23.29900:20:23.309 one just now and usually the this 00:20:26.88900:20:26.899 correction factor is a function of P 00:20:30.58000:20:30.590 which might be on that scale as a 00:20:33.86000:20:33.870 function of R now what is P and or equal 00:20:38.41900:20:38.429 to P and are typically just as an 00:20:44.98900:20:44.999 example P would maybe be equal to small 00:20:48.20000:20:48.210 T 2 minus small T 1 divided by T 1 minus 00:20:53.18000:20:53.190 T 1 something like that and then to show 00:20:56.14900:20:56.159 it to you just now an R is might be 00:21:00.01900:21:00.029 equal to t1 minus t2 divided by t2 minus 00:21:05.41900:21:05.429 t1 00:21:09.60000:21:09.610 now those temperatures the capital T's 00:21:13.42000:21:13.430 and the small T's they will be a sketch 00:21:16.12000:21:16.130 given to you 00:21:17.17000:21:17.180 maybe the sketch should look something 00:21:18.94000:21:18.950 like that and on the sketch it's going 00:21:24.61000:21:24.620 to show you t1 and t2 for example okay 00:21:30.70000:21:30.710 and t1 like that and t2 like that so you 00:21:37.09000:21:37.100 have to use that you have to align that 00:21:40.96000:21:40.970 with the practical situation in terms of 00:21:43.39000:21:43.400 the heat exchanger given to you then you 00:21:45.61000:21:45.620 calculate the P and the order value and 00:21:47.47000:21:47.480 from that you can get the fit the 00:21:49.33000:21:49.340 correction factor 00:21:57.03000:21:57.040 and I just as an example Lloyd if you 00:21:59.95000:21:59.960 can go up to the ogre it project today 00:22:03.03000:22:03.040 in figure 11 point 18 in your textbook 00:22:06.40000:22:06.410 there are a few examples are given four 00:22:08.71000:22:08.720 of them I think because it is too small 00:22:11.32000:22:11.330 I'm going to try to zoom in a little bit 00:22:18.63000:22:18.640 okay 00:22:20.92000:22:20.930 now look at that one the bee Grove it 00:22:23.77000:22:23.780 gives the correction factor for what 00:22:26.41000:22:26.420 type of heat exchanger the correction 00:22:30.37000:22:30.380 factor for a two shell pause so there 00:22:32.86000:22:32.870 are two shells and for a 12 etc any 00:22:37.39000:22:37.400 multiple of four two pulses you see so 00:22:41.53000:22:41.540 schematically if it has been given to 00:22:44.71000:22:44.720 you in the exam it is a two shell pause 00:22:47.44000:22:47.450 and an 82 pause heat exchanger then this 00:22:51.70000:22:51.710 graph would be valid you see and you 00:22:54.82000:22:54.830 would use this graph then in terms of 00:22:56.62000:22:56.630 what flows through the shell and there 00:22:59.11000:22:59.120 you can see this the shell there is T 1 00:23:01.96000:23:01.970 capital T 1 is then the imitating 00:23:04.18000:23:04.190 picture of the shell and T 2 is the 00:23:06.85000:23:06.860 outlet temperature of the shell you see 00:23:09.01000:23:09.020 and the tubes in a temperature small T 1 00:23:13.15000:23:13.160 and the outlet temperature is small T 2 00:23:16.26000:23:16.270 based on that you can calculate P which 00:23:20.44000:23:20.450 is then given on this axis and the 00:23:22.75000:23:22.760 arrows are which is those lines there so 00:23:25.56000:23:25.570 maybe if P is equal to 0.5 and R is 00:23:30.97000:23:30.980 equal to 1.5 then we can get the 00:23:33.73000:23:33.740 correction factor as 0.85 you see that 00:23:37.32000:23:37.330 example okay now 00:23:45.12000:23:45.130 there is however one very important 00:23:48.45000:23:48.460 thing and that is let's suppose in one 00:23:53.15900:23:53.169 of those lines we've got a fluid which 00:23:56.19000:23:56.200 is being condensed or which boils okay 00:24:00.84000:24:00.850 what happens then okay so if you look at 00:24:05.31000:24:05.320 that if one of them contains has or 00:24:08.39900:24:08.409 boils then the temperatures is going to 00:24:11.66900:24:11.679 be the same t1 small t1 is going to be 00:24:14.87900:24:14.889 small t2 a capital T 1 is going to be 00:24:17.90900:24:17.919 capital T 2 and then you might have the 00:24:20.85000:24:20.860 problem that some of those Peas or or or 00:24:23.85000:24:23.860 equal to zero or infinite and then 00:24:27.69000:24:27.700 students look at it and they say oh I 00:24:29.39900:24:29.409 don't know what to do now okay 00:24:31.04900:24:31.059 what is important to know is that if we 00:24:35.66900:24:35.679 have boiling or condensation 00:24:43.00000:24:43.010 for boiling or conversation if is equal 00:24:46.21000:24:46.220 to one so if is always equal to one if 00:24:51.52000:24:51.530 we have boiling or condensation on one 00:24:55.12000:24:55.130 of the sides okay ladies and gentlemen 00:24:58.87000:24:58.880 that's the theory I ready to do a 00:25:02.11000:25:02.120 problem okay let's do example 11.5 in 00:25:12.66900:25:12.679 your textbook 00:25:23.74000:25:23.750 this 00:25:26.85000:25:26.860 and in this example what is given to you 00:25:30.53900:25:30.549 is a to shout pause and a full tube 00:25:33.60000:25:33.610 pause heat exchanger okay just 00:25:38.03900:25:38.049 schematically they are the two shells 00:25:46.88000:25:46.890 there's the init of the first shell like 00:25:51.12000:25:51.130 there and this is the outlet of the 00:25:53.00900:25:53.019 second show 00:25:57.84000:25:57.850 and we've got 32 pauses going through so 00:26:02.01000:26:02.020 it is one two three and four and this 00:26:12.06000:26:12.070 one inside the tube we've got water at 00:26:16.73000:26:16.740 80 degrees Celsius and the exit 00:26:20.85000:26:20.860 temperature of the water V is equal to 00:26:23.04000:26:23.050 40 degrees Celsius the side we've got 00:26:27.48000:26:27.490 the serene and the temperature Inlet 00:26:34.50000:26:34.510 temperature is 20 degrees Celsius and 00:26:37.52000:26:37.530 here we've got the outlet temperature of 00:26:40.32000:26:40.330 the glycerine which is 50 degrees 00:26:42.27000:26:42.280 Celsius it is also given that the tube 00:26:50.34000:26:50.350 diameters are 20 millimeters and that 00:26:53.91000:26:53.920 the tubes often let some tubes 00:27:00.56000:27:00.570 and the heat transfer coefficient in the 00:27:06.25900:27:06.269 shell is equal to 25 watts per square 00:27:10.15900:27:10.169 meter degree Celsius and the heat 00:27:13.12900:27:13.139 transfer coefficient inside the tubes 00:27:14.83900:27:14.849 his 160 watts per square meter degree 00:27:18.64900:27:18.659 Celsius and they asked us to determine 00:27:26.07900:27:26.089 the heat transfer rate firstly without 00:27:31.31000:27:31.320 any fouling and then secondly the heat 00:27:35.20900:27:35.219 transfer rate if the fouling on the 00:27:39.61900:27:39.629 outside of the tubes is equal to 0.2 00:27:46.75900:27:46.769 Palsy row six square meters screaming 00:27:52.36900:27:52.379 this Kelvin for what 00:28:27.57000:28:27.580 okay we are going to need the surface 00:28:29.34000:28:29.350 area of the off the tubes and it's just 00:28:32.46000:28:32.470 calculated to make things easier it is 00:28:35.46000:28:35.470 PI multiplied by the diameter multiplied 00:28:38.16000:28:38.170 by the links and sorry I forgot to give 00:28:41.19000:28:41.200 you the length the length is 60 meters 00:28:44.87000:28:44.880 okay the length of the tubes 60 meters 00:28:51.62000:28:51.630 okay so the surface area is equal to PI 00:28:54.80000:28:54.810 multiplied by the diameter which is 20 00:28:57.69000:28:57.700 moles multiplied by 60 and that gives us 00:29:02.37000:29:02.380 a surface area of three point seven 00:29:05.85000:29:05.860 seven square meters 00:29:17.97000:29:17.980 now so from the theory we have seen what 00:29:21.45000:29:21.460 it has been shown to us that we can 00:29:23.78900:29:23.799 calculate the heat roles for right as 00:29:25.79900:29:25.809 you multiplied by the area multiplied by 00:29:29.37000:29:29.380 L MTD or sorry you multiplied by the 00:29:33.45000:29:33.460 area multiplied by the correction factor 00:29:35.97000:29:35.980 multiplied by L MTD of a counter flow 00:29:40.13900:29:40.149 heat exchanger 00:29:48.09000:29:48.100 now what we also know is that the heat 00:29:51.15900:29:51.169 rolls for right on the cold side is 00:29:53.23000:29:53.240 equal to the mass flow rate multiplied 00:29:55.23900:29:55.249 by CP multiplied by the delta T and on 00:29:59.88900:29:59.899 the hot side it is equal to the mass 00:30:03.15900:30:03.169 flow rate multiplied by CP delta T and 00:30:10.25900:30:10.269 we could have used one of these 00:30:12.43000:30:12.440 equations also to get the heat transfer 00:30:14.64900:30:14.659 right but the mass flow rates are not 00:30:17.49900:30:17.509 given to us so that is the reason why we 00:30:20.49900:30:20.509 can't use these two equations otherwise 00:30:23.47000:30:23.480 we could have calculated it directly 00:30:25.86900:30:25.879 from there 00:30:31.51000:30:31.520 okay now take a note we have to get the 00:30:33.88000:30:33.890 Delta TLM TD of a counter flow heat 00:30:37.63000:30:37.640 exchanger now I want you to try for me 00:30:41.08000:30:41.090 and see please calculate that quickly 00:30:43.27000:30:43.280 for me don't have to do this final 00:30:45.31000:30:45.320 calculation but write down all the terms 00:30:47.95000:30:47.960 for me in terms of how you would 00:30:49.27000:30:49.280 calculate the ln TD of that heat 00:30:51.61000:30:51.620 exchanger counter flow 00:31:19.17000:31:19.180 okay we do it as follows you don't need 00:31:21.88000:31:21.890 to do it physically temperature as a 00:31:26.95000:31:26.960 function of X there must be a hot stream 00:31:30.94000:31:30.950 in a cold stream so the hot stream 00:31:33.13000:31:33.140 obviously must be on top 00:31:34.90000:31:34.910 okay the hot stream always does 00:31:37.99000:31:38.000 something like that isn't it okay now I 00:31:40.99000:31:41.000 choose for no specific reason 00:31:44.52000:31:44.530 normally that as the inlet temperature 00:31:47.04000:31:47.050 the outside which is 80 come outside 00:31:52.84000:31:52.850 then it is 80 and then the outlet 00:31:55.66000:31:55.670 temperature there is equal to 40 you 00:31:58.00000:31:58.010 agree okay and then there must be a cold 00:32:01.63000:32:01.640 site okay now if that is the inlet 00:32:05.92000:32:05.930 temperature 80 and 40 the counter flow 00:32:11.05000:32:11.060 direction the counter flow direction 00:32:13.42000:32:13.430 then on the cold side what is the inlet 00:32:15.91000:32:15.920 temperature the inlet temperature on the 00:32:18.40000:32:18.410 cold side is equal to 20 and the outlet 00:32:21.40000:32:21.410 temperature on the cold side is equal to 00:32:23.32000:32:23.330 50 okay now you've got a counter flow 00:32:25.75000:32:25.760 heat exchanger you forced it almighty to 00:32:28.54000:32:28.550 be a counter flow heat exchanger 00:32:33.21000:32:33.220 does that make sense you happy lab okay 00:32:38.10000:32:38.110 so therefore we can L say LM TD is equal 00:32:45.76000:32:45.770 to this temperature difference which is 00:32:48.61000:32:48.620 equal to 30 conserve a temperature 00:32:53.83000:32:53.840 difference is 30 and that temperature 00:32:56.02000:32:56.030 difference is 20 okay so it is 30 minus 00:33:00.25000:33:00.260 20 divided by the limb of the T divided 00:33:04.65900:33:04.669 by 20 which is then equal to twenty four 00:33:10.84000:33:10.850 point seven degree Celsius 00:33:20.89000:33:20.90000:33:22.46000:33:22.470 now we need to get the correction factor 00:33:24.68000:33:24.690 if the correction factor is there is the 00:33:28.22000:33:28.230 two shall pause and the four to process 00:33:31.87900:33:31.889 as we have okay so based on that draw 00:33:35.72000:33:35.730 off which is the figure 11 point 18 B P 00:33:45.35000:33:45.360 is equal to T 2 minus T 1 divided by T 1 00:33:50.24000:33:50.250 minus T 1 okay so if we look at that 00:33:57.25900:33:57.269 then the shell is T 1 okay the shell in 00:34:04.27900:34:04.289 a temperature is equal to t 1 so that is 00:34:07.46000:34:07.470 equal to t 1 and that is equal to t 2 00:34:12.06900:34:12.079 degrees 00:34:14.62000:34:14.630 and in terms of the tubes what is going 00:34:18.22000:34:18.230 in is that temperature t-1 there and 00:34:21.55000:34:21.560 that is equal to t2 there you happy that 00:34:28.14000:34:28.150 okay so P is equal to 40 minus 80 00:34:38.46000:34:38.470 divided by 20 00:34:47.54000:34:47.550 41:20 - 80 and that is equal to 0.67 and 00:34:58.41000:34:58.420 R is equal to t1 minus t2 divided by 00:35:04.95000:35:04.960 small t2 minus small t1 which is then 00:35:09.78000:35:09.790 equal to 20 minus 50 divided by 40 minus 00:35:14.84900:35:14.859 80 which is equal to 0.75 00:35:25.10000:35:25.110 so based on that the friction factor is 00:35:27.77000:35:27.780 of the correction factor is equal to 0.9 00:35:30.95000:35:30.960 one 00:35:40.27000:35:40.280 okay now we've got the LM TD we've got 00:35:43.19000:35:43.200 the correction factor now we need to get 00:35:45.89000:35:45.900 you multiplied by the area we actually 00:35:48.14000:35:48.150 already have the surface area so we 00:35:50.87000:35:50.880 actually only need to get the overall 00:35:52.52000:35:52.530 heat transfer coefficient we can 00:35:54.26000:35:54.270 calculate it separately or we can 00:35:57.47000:35:57.480 calculate it as you multiplied by the 00:35:59.69000:35:59.700 area whichever we want to do remember 00:36:03.88000:36:03.890 one divided by UI is equal to one 00:36:10.91000:36:10.920 divided by the heat transfer coefficient 00:36:12.71000:36:12.720 on the inside multiplied by the area on 00:36:15.14000:36:15.150 the inside plus the fouling on the 00:36:18.47000:36:18.480 inside multiplied by the area plus the 00:36:21.65000:36:21.660 resistance of the tube to PI KL plus the 00:36:29.12000:36:29.130 fouling on the outside plus the flow 00:36:34.76000:36:34.770 resistance on the outside that is 00:36:38.63000:36:38.640 overall heat transfer coefficient 00:36:39.85900:36:39.869 correlation we've done that with a 00:36:42.08000:36:42.090 previous lecture 00:36:52.03000:36:52.040 okay the fowling at this takes for part 00:36:56.71000:36:56.720 a it is without any fouling the second 00:37:00.43000:37:00.440 part of the problem is with the fouling 00:37:02.85000:37:02.860 so firstly we can delete those two terms 00:37:08.04000:37:08.050 it has been given that it is a 00:37:11.05000:37:11.060 20-millimeter - but it is very very thin 00:37:13.66000:37:13.670 okay if it is thin then it means that 00:37:17.14000:37:17.150 that the diameter and that diameter is 00:37:19.33000:37:19.340 approximately the same the limb of one 00:37:22.15000:37:22.160 is equal to zero okay three four that 00:37:25.93000:37:25.940 term is also negligible okay and because 00:37:33.37000:37:33.380 it is thin that area and that area and 00:37:36.04000:37:36.050 that area is the same okay so in this 00:37:39.73000:37:39.740 specific case we can say 1 divided by u 00:37:42.40000:37:42.410 is equal to 1 divided by the heat 00:37:45.43000:37:45.440 transfer coefficient on the outside 00:37:47.16000:37:47.170 divided by 1 divided by the heat 00:37:49.66000:37:49.670 transfer coefficient on the inside plus 00:37:52.21000:37:52.220 1 divided by the transfer coefficient on 00:37:54.07000:37:54.080 the outside okay which is equal to 1 00:38:00.10000:38:00.11000:38:01.84000:38:01.850 on the inside which is equal to hundred 00:38:06.27000:38:06.280 160 plus 1 divided by the heat transfer 00:38:09.76000:38:09.770 coefficient on the outside which is 25 00:38:13.08000:38:13.090 so the overall heat transfer coefficient 00:38:16.30000:38:16.310 is going to be very close to the 00:38:19.87000:38:19.880 smallest deterrence or coefficient which 00:38:21.40000:38:21.410 is 25 and if we calculated it is going 00:38:24.49000:38:24.500 to be 20 1.6 watts per square meter 00:38:28.20000:38:28.210 Kelvin 00:38:35.53000:38:35.540 right so now we can calculate the heat 00:38:38.42000:38:38.430 transfer rate we can say the heat 00:38:40.01000:38:40.020 transfer rate is equal to u multiplied 00:38:42.34900:38:42.359 by the area multiplied by the correction 00:38:45.38000:38:45.390 factor multiplied by L MTD for a counter 00:38:50.69000:38:50.700 flow heat exchanger 00:39:14.42000:39:14.430 you is equal to 21.7 the surface area 00:39:20.22000:39:20.230 we've calculated as 3.77 00:39:24.58900:39:24.599 the correction factor is 0.9 1 and the 00:39:29.30900:39:29.319 LM TD is equal to twenty four point 00:39:32.06900:39:32.079 seven and that would give us a heat 00:39:36.63000:39:36.640 transfer rate of one thousand eight 00:39:38.76000:39:38.770 hundred and fifty watts for one point 00:39:42.78000:39:42.790 eight three kilowatts 00:39:55.16000:39:55.170 you happy with that okay let's do Part B 00:39:59.09000:39:59.100 or B is now with fouling of 0.2 palsy 00:40:05.34000:40:05.350 row so we need to recalculate the 00:40:10.17000:40:10.180 overall heat transfer coefficient so one 00:40:12.51000:40:12.520 divided by the overall heat transfer 00:40:14.04000:40:14.050 coefficient is equal to one divided by 00:40:17.46000:40:17.470 the heat transfer coefficient multiplied 00:40:19.29000:40:19.300 by the area fouling on the inside plus 00:40:24.18000:40:24.190 the learning of the diameter I chose 00:40:27.84000:40:27.850 divided by 2 pi KL plus the resistance 00:40:32.64000:40:32.650 on the outside plus 1 divided by the 00:40:37.14000:40:37.150 heat transfer coefficient on the outside 00:40:38.91000:40:38.920 multiplied by the area on outside 00:40:54.93900:40:54.949 okay again the tube is very thin so we 00:40:59.29900:40:59.309 can say that the resistance through the 00:41:01.03900:41:01.049 tube is negligible they didn't give us 00:41:03.52900:41:03.539 any fouling information on the inside 00:41:08.04900:41:08.059 for this special case where the area on 00:41:12.04900:41:12.059 the inside is equal to the area on the 00:41:14.23900:41:14.249 outside we can cancel all the areas and 00:41:21.78900:41:21.799 the result is one divided by the overall 00:41:24.91900:41:24.929 heat transfer coefficient is equal to 00:41:27.04900:41:27.059 one divided by the heat transfer 00:41:29.41900:41:29.429 coefficient on the inside plus the 00:41:32.17900:41:32.189 fouling resistance on the outside plus 00:41:35.65000:41:35.66000:41:38.23900:41:38.249 coefficient on the outside 00:41:47.28000:41:47.290 the transfer coefficient on the inside 00:41:51.21000:41:51.220 is equal to 160 the troweling resistance 00:41:56.95000:41:56.960 is equal to 0.2 'pl 0 6 plus 1 divided 00:42:03.82000:42:03.830 by the heat transfer coefficient on the 00:42:05.44000:42:05.450 outside which is equal to 25 and the 00:42:11.56000:42:11.570 result is if we calculate the overall 00:42:14.02000:42:14.030 heat transfer coefficient now is going 00:42:16.36000:42:16.370 to be 21.3 watts per square meter kelvin 00:42:25.11000:42:25.120 should it be lower than the other one 00:42:29.49000:42:29.500 obviously yes because of the fouling 00:42:32.58000:42:32.590 should decrease its efficiency we can 00:42:35.86000:42:35.870 see it is but it's not much not much so 00:42:40.03000:42:40.040 if you go and recalculate the overall 00:42:43.48000:42:43.490 heat transfer rate which is equal to UA 00:42:46.32000:42:46.330 x if x LM t-d-z which is now equal to 21 00:42:54.34000:42:54.350 point 3 multiplied by the surface area 00:42:57.94000:42:57.950 which is three point seven seven 00:43:00.48000:43:00.490 multiplied by the correction factor 0.9 00:43:03.58000:43:03.590 one multiply 5 and TD twenty four point 00:43:06.79000:43:06.800 seven and the result is the heat 00:43:12.88000:43:12.890 transfer rate of 1805 watts 00:43:18.99000:43:19.000 so if we could compare as it rolls right 00:43:21.51000:43:21.520 to that one we can see that they either 00:43:24.48000:43:24.490 decrease but at this stage it's not a 00:43:29.55000:43:29.560 lot after 10 years 00:43:34.19000:43:34.200 might be very significant any questions 00:43:38.58000:43:38.590 ladies and gentlemen if not thank you 00:43:41.19000:43:41.200 very much
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