00:00:00.030 so let's do this example 100 liter of 00:00:03.37900:00:03.389 water per minute is to be cooled down 00:00:05.48000:00:05.490 from 90 to 65 degrees Celsius using cold 00:00:09.32000:00:09.330 water flux with an inner temperature of 00:00:11.72000:00:11.730 20 degrees with an heat exchanger with 00:00:14.57000:00:14.580 the area of 4 square meters and our task 00:00:17.81000:00:17.820 is to calculate the cold water flux 00:00:19.31000:00:19.320 needed if a counter current setup is 00:00:21.80000:00:21.810 chosen and the K value has been 00:00:25.18900:00:25.199 estimated to 1 kilo watt per square 00:00:27.58900:00:27.599 meter in Kelvin we also get some numbers 00:00:30.01900:00:30.029 here you could look this up in a 00:00:32.72000:00:32.730 handbook so on the hot side we have 970 00:00:37.49000:00:37.500 1 kilogram per cubic meter and 4190 1 00:00:40.63900:00:40.649 euro per kilogram Kelvin as the heat 00:00:43.25000:00:43.260 capacity and on the cold side well we 00:00:47.02900:00:47.039 actually don't know the temperature so 00:00:49.70000:00:49.710 you can't look that up but let's assume 00:00:53.36000:00:53.370 that is 996 kilogram per cubic meter 00:00:56.81000:00:56.820 density and 4175 as the heat capacity so 00:01:01.40000:01:01.410 what you would need to do if you do it 00:01:03.20000:01:03.210 is really really carefully is that you 00:01:05.42000:01:05.430 assume something as you see I'm a 00:01:07.91000:01:07.920 temperature range and then you look up 00:01:09.50000:01:09.510 the density and the heat capacity for 00:01:11.87000:01:11.880 that range and then you do your 00:01:14.48000:01:14.490 calculation and see does my result is 00:01:18.07000:01:18.080 that in coherent with the Assumption I 00:01:22.55000:01:22.560 did okay so we make energy balances so 00:01:28.60900:01:28.619 the energy that is being transferred and 00:01:32.06000:01:32.070 taken up respectively is the mass flux 00:01:34.34000:01:34.350 times the heat capacity times the 00:01:35.96000:01:35.970 temperature change in the respective 00:01:38.12000:01:38.130 flows and that must also be the same as 00:01:41.35900:01:41.369 the temperature that goes this sorry the 00:01:43.49000:01:43.500 heat flux that goes through the surface 00:01:45.98000:01:45.990 so K times a times the logarithmic mean 00:01:48.31900:01:48.329 temperature and not it's the logarithmic 00:01:51.13900:01:51.149 mean temperature because we have no 00:01:54.10900:01:54.119 phase changes so we have a nice study 00:01:56.88900:01:56.899 increase and decrease respectively of 00:02:00.35000:02:00.360 the temperature of the two mediums so 00:02:04.99900:02:05.009 with the numbers we have we can see that 00:02:07.73000:02:07.740 the hot medium is 100 00:02:10.30000:02:10.310 a minute so that zero point one cubic 00:02:13.42000:02:13.430 meter per minute times the density 00:02:15.40000:02:15.410 divided by sixty translated to kilograms 00:02:18.30900:02:18.319 per second and you have the change in 00:02:23.53000:02:23.540 temperature for the hot flow is 90 minus 00:02:27.13000:02:27.140 sixty five does twenty-five degrees 00:02:29.19900:02:29.209 and from that you can calculate the 00:02:31.99000:02:32.000 energy transferred so that's 00:02:34.24000:02:34.250 approximately 170 kilo watts so 00:02:37.93000:02:37.940 kilowatts per second when you do these 00:02:40.78000:02:40.790 calculations I recommend you to check 00:02:42.78900:02:42.799 the caf the unit's carefully because 00:02:46.00000:02:46.010 it's easy to do mistakes now we know 00:02:49.75000:02:49.760 everything now in the lost energy 00:02:54.46000:02:54.470 balance there Q equals K times a times 00:02:56.89000:02:56.900 Delta TL except the logarithmic mean 00:02:59.58900:02:59.599 temperature so we can calculate the 00:03:01.24000:03:01.250 logarithmic mean temperature as to Q 00:03:03.72900:03:03.739 divided by the overall heat transfer 00:03:04.93000:03:04.940 coefficient and area and we get that 00:03:08.62000:03:08.630 forty two point 39 and we know the 00:03:12.19000:03:12.200 formula for the log rating me 00:03:13.90000:03:13.910 temperature delta T 2 minus delta T 1 00:03:16.42000:03:16.430 divided by natural log of delta T 2 00:03:19.09000:03:19.100 minus natural logarithm delta T 1 and 00:03:22.53000:03:22.540 from that we get well first you have to 00:03:25.93000:03:25.940 realize that delta T 2 and that's 65 00:03:29.80000:03:29.810 minus 20 here it's it's important to 00:03:32.71000:03:32.720 realize it's a counter current setup so 00:03:35.89000:03:35.900 the coldest inflow and the coldest cold 00:03:39.72900:03:39.739 medium in flow should meet the coldest 00:03:42.78000:03:42.790 hot medium outflow so 65 minus 20 that's 00:03:48.25000:03:48.260 45 degrees on one side and then you can 00:03:50.65000:03:50.660 calculate the temperature difference on 00:03:52.72000:03:52.730 the other side and that becomes thirty 00:03:54.69900:03:54.709 nine point nine degrees Celsius we keep 00:03:56.94900:03:56.959 some extra data here just in case so we 00:04:01.50900:04:01.519 don't run into numerical problems and 00:04:05.58000:04:05.590 delta T one so that the temperature 00:04:08.80000:04:08.810 difference between the two flows on the 00:04:10.59900:04:10.609 other side that's 90 for the hot flow 00:04:13.75000:04:13.760 and the cold out we don't know so we 00:04:15.97000:04:15.980 calculate that here delta T one was 00:04:18.82000:04:18.830 thirty nine point nine so we get ninety 00:04:21.09900:04:21.109 minus that nine point 00:04:22.06900:04:22.079 and that's fifty point one degrees 00:04:24.37900:04:24.389 Celsius so now we know the temperature 00:04:28.05900:04:28.069 so we can calculate the the mass flux 00:04:33.14000:04:33.150 now because now we also know how much 00:04:36.02000:04:36.030 the temperature increased for the cold 00:04:37.90900:04:37.919 medium so that's fifty point one minus 00:04:40.87900:04:40.889 the inflow temperature that's twenty 00:04:42.92000:04:42.930 that's thirty point one degrees Celsius 00:04:45.26000:04:45.270 and let's check that things look right 00:04:49.39900:04:49.409 here if you take the heat capacity of 00:04:53.54000:04:53.550 the code medium times the temperature 00:04:56.36000:04:56.370 change in that medium that's larger than 00:05:00.43900:05:00.449 the heat capacity of the hot medium 00:05:02.48000:05:02.490 times the temperature difference of the 00:05:04.39900:05:04.409 hot medium and that should result in 00:05:07.04000:05:07.050 mass flux W two that's less than W one 00:05:13.39000:05:13.400 so let's put in the numbers there so W - 00:05:17.86900:05:17.879 that must be Q divided by heat capacity 00:05:22.01000:05:22.020 and the temperature change and we get 00:05:24.70900:05:24.719 that as one point thirty five kilograms 00:05:27.83000:05:27.840 per second which is less than one point 00:05:29.83900:05:29.849 sixty two so it seems to be in good 00:05:32.60000:05:32.610 order and if we want to we can 00:05:34.18900:05:34.199 recalculate that as liters per minute so 00:05:37.24900:05:37.259 one point thirty five kilograms per 00:05:38.89900:05:38.909 second times sixty seconds divided by 00:05:42.40900:05:42.419 the density and you get zero point zero 00:05:44.83900:05:44.849 eighty one cubic meter per minute or 00:05:46.90900:05:46.919 eighty one liter per minute okay when 00:05:53.48000:05:53.490 you calculate heat exchangers the thing 00:05:58.12900:05:58.139 is that you need to know how much heat 00:06:00.20000:06:00.210 needs to be transferred what the for 00:06:02.18000:06:02.190 temperature should be and what the flow 00:06:04.45900:06:04.469 rate should be and the flow rate tells 00:06:10.36900:06:10.379 you the overall heat transfer 00:06:11.54000:06:11.550 coefficient and then you can calculate 00:06:13.67000:06:13.680 the area needed and once you have 00:06:17.08900:06:17.099 calculated area needed well then you 00:06:19.73000:06:19.740 have to think if you want to do this all 00:06:22.79000:06:22.800 the way okay if I know how big the area 00:06:26.54000:06:26.550 should be how big should then the heat 00:06:29.74900:06:29.759 exchanger be and that might influence 00:06:35.19900:06:35.209 the transaction area which in turn will 00:06:40.21900:06:40.229 influence the flow rate now the problem 00:06:44.42000:06:44.430 is you you have to do it right in here 00:06:48.26000:06:48.270 you have to guess a flow rate or you 00:06:51.13900:06:51.149 just set a flow rate you would like to 00:06:53.36000:06:53.370 have and then you do the calculations 00:06:55.27900:06:55.289 and then you do this assign whatever you 00:06:57.49900:06:57.509 do plate it a change here or whatever 00:07:00.26000:07:00.270 and then when you've done all the design 00:07:04.18900:07:04.199 choices you can calculate the flow rate 00:07:08.05900:07:08.069 and if the flow rate is different than 00:07:11.57000:07:11.580 the one you used before 00:07:12.70900:07:12.719 then you need to redo your calculations 00:07:14.62900:07:14.639 and continue and continue until it all 00:07:17.12000:07:17.130 fits if you have boiling well it all 00:07:27.07900:07:27.089 depends on how difficult equations are 00:07:29.65900:07:29.669 used we will use the natural convection 00:07:31.30900:07:31.319 boiling and then the first question we 00:07:34.37000:07:34.380 need to ask yourself is how much heat is 00:07:36.98000:07:36.990 need to be transferred and that is one 00:07:42.49900:07:42.509 thing and then it's the second thing is 00:07:44.32900:07:44.339 okay so how many watts per square meter 00:07:46.37000:07:46.380 will that be but not we're going to try 00:07:50.62900:07:50.639 to calculate the heat exchanger area and 00:07:52.76000:07:52.770 we need to know how much energy is being 00:07:55.18900:07:55.199 transferred per square meter hmm so 00:08:00.05000:08:00.060 guess something you guess how many watts 00:08:03.17000:08:03.180 per square meter you need to transfer 00:08:05.99000:08:06.000 and then you calculate the heat transfer 00:08:09.23000:08:09.240 coefficients using these equations and 00:08:11.44900:08:11.459 then you calculate the overall heat 00:08:13.67000:08:13.680 transfer coefficient and then you 00:08:15.17000:08:15.180 calculate area needed and then you 00:08:18.68000:08:18.690 calculate how many watts per square 00:08:19.90900:08:19.919 meter 00:08:20.48000:08:20.490 will that be and if you're lucky that is 00:08:23.77900:08:23.789 the same number as you guessed before 00:08:25.30900:08:25.319 and if you're unlucky just repeat and 00:08:28.27900:08:28.289 repeat and repeat until you find a 00:08:31.07000:08:31.080 solution
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