/ News & Press / Video / Shell and Tube Heat Exchanger

Shell and Tube Heat Exchanger

WEBVTT
Kind: captions
Language: en

00:00:03.410
so here we have a shell and tube heat
00:00:07.030 00:00:07.040 exchanger it has one shell pass and
00:00:10.720 00:00:10.730 fifteen two passes so we'll write that
00:00:15.550 00:00:15.560 that the number of tube passes is equal
00:00:18.920 00:00:18.930 to 15 we have hot water on the tube side
00:00:22.810 00:00:22.820 so that's the hot fluid and it heats oil
00:00:26.570 00:00:26.580 on the shell side which is the cold
00:00:29.210 00:00:29.220 fluid so the length per pass of this
00:00:34.240 00:00:34.250 tube is three meters and it has a
00:00:39.290 00:00:39.300 diameter of 25 millimeters so our water
00:00:44.900 00:00:44.910 which is our hot fluid it enters so th I
00:00:49.430 00:00:49.440 is equal to 85 degrees C it's mass flow
00:00:55.490 00:00:55.500 rate is equal to 0.15 kilograms per
00:01:00.470 00:01:00.480 second and th out is equal to 20 degrees
00:01:05.780 00:01:05.790 Celsius so we have our cold Inlet
00:01:09.020 00:01:09.030 temperature which is the oil which is 5
00:01:12.380 00:01:12.390 degrees C and the outlet temperature is
00:01:16.730 00:01:16.740 30 degrees C and what we're looking for
00:01:20.179 00:01:20.189 is the average convective heat transfer
00:01:24.249 00:01:24.259 coefficient for the outer tube surface
00:01:26.989 00:01:26.999 and what we're given is this inner heat
00:01:31.880 00:01:31.890 transfer coefficient of 3600 watts per
00:01:37.249 00:01:37.259 meter squared K so how are we going to
00:01:40.819 00:01:40.829 find our H sub o well we're going to use
00:01:46.010 00:01:46.020 the overall heat transfer coefficient
00:01:48.620 00:01:48.630 which is U and we know that u a is equal
00:01:56.959 00:01:56.969 to the sum of the thermal resistances to
00:02:01.910 00:02:01.920 the minus 1 which is equal to 1 over H
00:02:07.240 00:02:07.250 outside times the area plus 1 over H
00:02:12.220 00:02:12.230 inside times the area raised to the my
00:02:16.309 00:02:16.319 this one so if we can find you a we can
00:02:20.059 00:02:20.069 find our h0 so we're going to use the
00:02:24.679 00:02:24.689 NTU effectiveness method because by
00:02:29.330 00:02:29.340 definition the NTU which is the number
00:02:34.160 00:02:34.170 of transfer units is equal to you a
00:02:38.949 00:02:38.959 which is what we're looking for
00:02:41.949 00:02:41.959 divided by the minimum heat capacity
00:02:46.129 00:02:46.139 rate to find NTU we consult a table
00:02:50.509 00:02:50.519 which gives NTU as a function of the
00:02:56.420 00:02:56.430 heat capacity rate ratio and the
00:03:03.099 00:03:03.109 effectiveness so the appropriate
00:03:06.099 00:03:06.109 equations are as follows let's start
00:03:10.429 00:03:10.439 with this C sub R this heat capacity
00:03:14.059 00:03:14.069 ratio and this is equal to the heat
00:03:18.020 00:03:18.030 capacity rate that's the minimum divided
00:03:21.589 00:03:21.599 by the one that's the maximum where this
00:03:24.890 00:03:24.900 heat capacity rate is equal to the mass
00:03:29.210 00:03:29.220 flow rate times the heat capacity so we
00:03:34.129 00:03:34.139 can find the C for the water because
00:03:38.059 00:03:38.069 that's equal to 0.15 kilograms per
00:03:43.460 00:03:43.470 second multiplied by its heat capacity
00:03:49.270 00:03:49.280 4184 joules per kilogram Kelvin and so
00:03:55.159 00:03:55.169 our heat capacity rate of the water is
00:03:57.879 00:03:57.889 six hundred and twenty seven point six
00:04:02.349 00:04:02.359 watts per Kelvin so how do we find the
00:04:06.709 00:04:06.719 heat capacity rate for the oil we don't
00:04:10.939 00:04:10.949 have a mass flow rate we don't have a
00:04:13.369 00:04:13.379 heat capacity but what we do have is
00:04:16.580 00:04:16.590 enough information to find the heat
00:04:21.399 00:04:21.409 transfer rate which is equal to a heat
00:04:25.250 00:04:25.260 capacity rate times the change in
00:04:28.279 00:04:28.289 temperature and
00:04:29.690 00:04:29.700 we have it for the water so this is
00:04:32.450 00:04:32.460 equal to six hundred and twenty seven
00:04:34.850 00:04:34.860 point six watts per K times the
00:04:39.560 00:04:39.570 difference in temperatures which is 85
00:04:44.770 00:04:44.780 minus 20 this is in degrees Celsius our
00:04:48.880 00:04:48.890 q dot is equal to 40 thousand seven
00:04:54.260 00:04:54.270 hundred and ninety four watts we assume
00:04:58.580 00:04:58.590 a well insulated system heat exchanger
00:05:01.430 00:05:01.440 should be well insulated so the amount
00:05:04.520 00:05:04.530 of the heat transfer provided by the
00:05:06.740 00:05:06.750 water is the amount gained by the oil so
00:05:10.820 00:05:10.830 that means our Q de is going to be equal
00:05:14.180 00:05:14.190 to the heat capacity rate of the oil
00:05:17.950 00:05:17.960 times the change in temperature of that
00:05:21.530 00:05:21.540 oil which is thirty minus five degrees
00:05:26.090 00:05:26.100 Celsius and so that heat capacity rate
00:05:30.290 00:05:30.300 is equal to sixteen thirty-two
00:05:35.620 00:05:35.630 watts per K so looking at this it's
00:05:41.240 00:05:41.250 clear that this one is the minimum heat
00:05:43.970 00:05:43.980 capacity rate this one is the maximum
00:05:48.380 00:05:48.390 heat capacity rate and so from that we
00:05:52.430 00:05:52.440 can find that ratio that we need which
00:05:55.760 00:05:55.770 is equal to 0.38 so the next thing we're
00:06:01.460 00:06:01.470 going to need to find is this epsilon
00:06:04.450 00:06:04.460 which is the effectiveness and that's
00:06:07.760 00:06:07.770 equal to the heat transfer rate divided
00:06:11.690 00:06:11.700 by the maximum heat transfer rate so we
00:06:15.830 00:06:15.840 have our Q dot now we have to find our Q
00:06:20.480 00:06:20.490 max and that's equal to the minimum heat
00:06:25.130 00:06:25.140 capacity rate times the inlet
00:06:29.270 00:06:29.280 temperature of the hot stream minus the
00:06:33.410 00:06:33.420 inlet temperature of the Cold Stream and
00:06:37.160 00:06:37.170 so when we write that out that's going
00:06:41.540 00:06:41.550 to be equal
00:06:43.309 00:06:43.319 - six hundred and twenty seven point six
00:06:47.769 00:06:47.779 watts per K Kelvin eighty-five degrees
00:06:53.859 00:06:53.869 minus five degrees and our Q Max is
00:06:57.700 00:06:57.710 equal to fifty thousand two hundred and
00:07:02.269 00:07:02.279 eight watts so if we divide the Q that
00:07:05.899 00:07:05.909 we found here by our Q max we find that
00:07:12.019 00:07:12.029 our effectiveness is equal to zero point
00:07:16.909 00:07:16.919 eight one and now what we do is we put
00:07:21.170 00:07:21.180 everything back into these messy
00:07:24.079 00:07:24.089 equations to solve for our big E and
00:07:27.579 00:07:27.589 from that to solve from our NTU so
00:07:32.320 00:07:32.330 calculating this we find that our Big E
00:07:35.510 00:07:35.520 is equal to one point zero five and our
00:07:41.179 00:07:41.189 NTU is equal to six point eight five and
00:07:46.839 00:07:46.849 now we can go back to use that to find
00:07:50.869 00:07:50.879 our you a so are you a is equal to this
00:07:56.659 00:07:56.669 NTU six point eight five times our
00:08:02.170 00:08:02.180 minimum heat capacity rate six hundred
00:08:06.110 00:08:06.120 and twenty seven point six watts per
00:08:09.529 00:08:09.539 Kelvin and that equals four thousand two
00:08:13.309 00:08:13.319 hundred and ninety eight point two watts
00:08:18.249 00:08:18.259 per K and so now we go back to where we
00:08:23.869 00:08:23.879 use that to find right here our outside
00:08:31.329 00:08:31.339 convective heat transfer coefficient so
00:08:37.159 00:08:37.169 we can write that one over u a is equal
00:08:42.199 00:08:42.209 to one over the outside heat transfer
00:08:45.559 00:08:45.569 coefficient times a plus one over the
00:08:48.800 00:08:48.810 inside heat transfer coefficient times a
00:08:51.769 00:08:51.779 and the area's cancel out so we're left
00:08:55.939 00:08:55.949 with one over
00:08:57.040 00:08:57.050 you is equal to 1 over H o plus 1 over H
00:09:02.019 00:09:02.029 I now we're going to calculate that area
00:09:06.360 00:09:06.370 using the 0.025 meter diameter and we
00:09:11.650 00:09:11.660 come up with a total area of 3.5 3
00:09:16.150 00:09:16.160 meters squared so from there we can get
00:09:19.990 00:09:20.000 our U which is equal to the 42 98.2
00:09:24.100 00:09:24.110 watts per K divided by 3.5 3 meters
00:09:28.269 00:09:28.279 squared and our overall heat transfer
00:09:31.560 00:09:31.570 coefficient is 1216 watts per meter
00:09:36.009 00:09:36.019 squared K so our 1 over U is equal to
00:09:40.300 00:09:40.310 zero point zero zero 8 meter squared
00:09:43.960 00:09:43.970 Kelvin per watts which leaves us with an
00:09:47.710 00:09:47.720 outside heat transfer coefficient of
00:09:53.490 00:09:53.500 1836 point six watts per meter squared
00:09:57.759 00:09:57.769 okay

Sales phone