Heat Exchanger Analysis 1

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Language: en

00:00:00.060
that's moving the chapter to dealing
00:00:02.629 00:00:02.639 with heat exchangers so this is a
00:00:05.690 00:00:05.700 introduction to heat exchangers there's
00:00:08.330 00:00:08.340 a lot of heat exchangers and maybe I
00:00:10.160 00:00:10.170 could just go through the whole host of
00:00:12.709 00:00:12.719 heat exchangers and different styles of
00:00:15.560 00:00:15.570 types will eventually do that but not on
00:00:17.599 00:00:17.609 the first day but here's a picture of
00:00:19.970 00:00:19.980 one that will analyze today it's a
00:00:22.970 00:00:22.980 concentric tube or double wall pipe or
00:00:27.410 00:00:27.420 pipe you know it's a pipe inside of a
00:00:29.779 00:00:29.789 pipe and the fluid inside the one pipe
00:00:33.110 00:00:33.120 runs maybe this way and it doesn't mix
00:00:35.479 00:00:35.489 with the fluid that's in the annulus
00:00:37.280 00:00:37.290 that may run like this and then go out
00:00:41.560 00:00:41.570 this is very similar this is a shell and
00:00:45.380 00:00:45.390 tube heat exchangers so you have a shell
00:00:47.479 00:00:47.489 side and then you have tubes and the
00:00:49.910 00:00:49.920 tubes you can see they come in and
00:00:52.180 00:00:52.190 they'll split in this manifold in two
00:00:55.400 00:00:55.410 different tubes flow it'll be a u-tube
00:00:58.389 00:00:58.399 sometimes they're straight through then
00:01:00.500 00:01:00.510 they have a manifold on the backend to
00:01:02.869 00:01:02.879 go in a straight run through the other
00:01:04.880 00:01:04.890 tubes different designs on shell and
00:01:07.250 00:01:07.260 tube heat exchangers and out to go so
00:01:09.770 00:01:09.780 fluid on the shell side and a fluid on
00:01:12.740 00:01:12.750 the tube side
00:01:13.640 00:01:13.650 this is automobile radiator so where are
00:01:17.510 00:01:17.520 two fluids the fluid with the liquid in
00:01:21.230 00:01:21.240 and out mainly water with some
00:01:24.260 00:01:24.270 antifreeze and then the air side goes
00:01:27.260 00:01:27.270 across so this is a cross flow heat
00:01:30.200 00:01:30.210 exchanger and then you have different
00:01:33.469 00:01:33.479 these are plate type heat exchangers
00:01:35.690 00:01:35.700 where you can actually these are
00:01:39.140 00:01:39.150 designed to take apart easily and
00:01:41.270 00:01:41.280 serviced easily and so you'll see a lot
00:01:43.819 00:01:43.829 of those there they're compromise
00:01:46.280 00:01:46.290 between the best efficiency and the best
00:01:48.469 00:01:48.479 service ability and the best reliability
00:01:51.230 00:01:51.240 and expandability you could long rods
00:01:55.760 00:01:55.770 bolt these together you could take the
00:01:58.130 00:01:58.140 apart and expand the heat exchanger if
00:02:00.350 00:02:00.360 you needed to increase its area make it
00:02:02.810 00:02:02.820 larger all right the fluid flow pattern
00:02:06.469 00:02:06.479 I have other illustrations to talk about
00:02:08.690 00:02:08.700 the fluid flow patterns in these hate
00:02:10.550 00:02:10.560 heat exchangers but not today
00:02:12.770 00:02:12.780 so let's go to the simplest heat
00:02:15.660 00:02:15.670 exchanger two concentric tube heat
00:02:18.600 00:02:18.610 exchanger and think of either parallel
00:02:21.660 00:02:21.670 flow or counter flow so normally we'd
00:02:25.260 00:02:25.270 sketch it something like this and say
00:02:28.260 00:02:28.270 have fluid that's on the hot fluid flows
00:02:32.430 00:02:32.440 this way and the cold fluid flows that
00:02:36.510 00:02:36.520 way and the hot fluid and the cold fluid
00:02:40.020 00:02:40.030 enter on the same side and exit on the
00:02:42.330 00:02:42.340 same side hence the flow are parallel
00:02:45.570 00:02:45.580 they're parallel you could maybe say
00:02:48.810 00:02:48.820 concurrent flow or some other
00:02:50.390 00:02:50.400 terminologies to say they're in the same
00:02:52.350 00:02:52.360 direction
00:02:53.090 00:02:53.100 how about counter flow and we'll have
00:02:57.240 00:02:57.250 the hot side come this way and the cold
00:03:01.350 00:03:01.360 side go that way and for introductory
00:03:06.150 00:03:06.160 purposes think about location a
00:03:10.380 00:03:10.390 coordinate system X runs from 0 to L the
00:03:14.520 00:03:14.530 length of those that heat exchanger and
00:03:17.610 00:03:17.620 we'll plot the temperature on the y axis
00:03:20.280 00:03:20.290 and think about the high temperature the
00:03:23.730 00:03:23.740 temperature hot coming in as the highest
00:03:26.010 00:03:26.020 temperature and the temperature of the
00:03:29.490 00:03:29.500 cold coming in would be the low
00:03:31.950 00:03:31.960 temperature and what do you think that
00:03:34.770 00:03:34.780 temperatures are going to do as it flows
00:03:37.410 00:03:37.420 through the heat exchanger you might
00:03:40.140 00:03:40.150 expect that the hot fluid to come down
00:03:43.350 00:03:43.360 some and exit at a lower temperature
00:03:46.590 00:03:46.600 temperature hot out and the cold fluid
00:03:50.760 00:03:50.770 could go up some temperature cold out
00:03:56.300 00:03:56.310 would it be reasonable for the cold to
00:03:59.880 00:03:59.890 somehow come out at a higher temperature
00:04:02.870 00:04:02.880 than the hot out no
00:04:10.050 00:04:10.060 No so this is not possible true it's
00:04:14.790 00:04:14.800 just not possible why is that not
00:04:16.409 00:04:16.419 possible well what's happening is is you
00:04:19.530 00:04:19.540 discretize this into little D X's and
00:04:22.170 00:04:22.180 for every little DX there's a little bit
00:04:24.330 00:04:24.340 of heat transfer a little bit of heat
00:04:25.950 00:04:25.960 transfer and if you had that case that I
00:04:27.960 00:04:27.970 just erased it would be like hold it
00:04:29.670 00:04:29.680 heat is coming out of a hot fluid which
00:04:31.740 00:04:31.750 is now cooler than the cold fluid and
00:04:33.570 00:04:33.580 flowing to the higher temperature cold
00:04:35.970 00:04:35.980 fluid that's not gonna work heat
00:04:38.460 00:04:38.470 transfer always goes from what's hot to
00:04:40.560 00:04:40.570 cold and things like this okay we're not
00:04:44.310 00:04:44.320 we don't think of a complicated system
00:04:47.070 00:04:47.080 to like air-conditioning system this is
00:04:50.250 00:04:50.260 just fluid exposed other fluids
00:04:54.390 00:04:54.400 separated by walls so convection and
00:04:56.940 00:04:56.950 conduction heat transfer all right when
00:05:00.990 00:05:01.000 might we expect if this was the case
00:05:03.840 00:05:03.850 right here and I asked you okay change
00:05:06.629 00:05:06.639 it up a little bit I start to increase
00:05:11.610 00:05:11.620 the mass flow rate of the hot fluid I
00:05:14.159 00:05:14.169 really start pumping more hot fluid
00:05:17.070 00:05:17.080 through the system how does it change
00:05:19.529 00:05:19.539 the red curve does it change the
00:05:22.440 00:05:22.450 temperature of the hot coming in if I
00:05:25.200 00:05:25.210 increase the mass flow rate of the hot
00:05:27.540 00:05:27.550 fluid how does that red curb the
00:05:31.860 00:05:31.870 temperature profile of the hot fluid
00:05:35.250 00:05:35.260 change as it goes through the heat
00:05:37.200 00:05:37.210 exchanger would it be either coming down
00:05:43.290 00:05:43.300 steeper or would it be shallower
00:05:48.710 00:05:48.720 shallower wouldn't it be shallower and
00:05:51.950 00:05:51.960 so the hot out would go up wouldn't it
00:05:56.840 00:05:56.850 the hot out okay what about the cold the
00:06:01.110 00:06:01.120 same temperature cold in what would the
00:06:04.140 00:06:04.150 cold temperature profile do would it
00:06:07.529 00:06:07.539 change would the cold outlet go up or
00:06:11.070 00:06:11.080 would it go down all I did was I did not
00:06:15.420 00:06:15.430 change the mass flow rate of the cold
00:06:17.219 00:06:17.229 fluid or the temperature coming into the
00:06:19.080 00:06:19.090 cold fluid all I did was increase the
00:06:22.080 00:06:22.090 master
00:06:22.570 00:06:22.580 right of the hot fluid would be a little
00:06:27.129 00:06:27.139 steeper so that the temperature of the
00:06:30.909 00:06:30.919 cold out actually went up a little bit
00:06:35.640 00:06:35.650 let's vote with your thumbs
00:06:38.100 00:06:38.110 so look okay alright so that's right now
00:06:42.369 00:06:42.379 you could play the reverse game and say
00:06:44.920 00:06:44.930 what happens if I change the mass flow
00:06:47.439 00:06:47.449 rate of the cold fluid and make those
00:06:49.420 00:06:49.430 sketches I really think that's helpful
00:06:51.369 00:06:51.379 let's do the same thing over here for
00:06:53.830 00:06:53.840 the counter current this is much more
00:06:56.320 00:06:56.330 popular and we'll see why why would you
00:06:58.689 00:06:58.699 design a heat exchanger to run in this
00:07:02.290 00:07:02.300 type of configuration well let's put the
00:07:05.409 00:07:05.419 same type of hot temperature whoops but
00:07:09.040 00:07:09.050 here what do I have on let's get
00:07:12.969 00:07:12.979 blackened in all right so over here I
00:07:16.540 00:07:16.550 have the temperature the hot fluid in
00:07:18.999 00:07:19.009 don't I and on this side I have the
00:07:22.749 00:07:22.759 temperature cold in so now describe the
00:07:27.519 00:07:27.529 temperature profile well you could have
00:07:30.279 00:07:30.289 a temperature profile that looks
00:07:31.959 00:07:31.969 something like this and something like
00:07:35.680 00:07:35.690 that so that the temperature of the hot
00:07:40.209 00:07:40.219 out is but well the temperature of the
00:07:43.689 00:07:43.699 hot fluid anywhere inside the heat
00:07:45.879 00:07:45.889 exchanger is always higher than the
00:07:49.390 00:07:49.400 local temperature of the cold fluid
00:07:52.029 00:07:52.039 right okay but as sketched what could
00:07:56.680 00:07:56.690 happen right here the temperature the
00:07:59.200 00:07:59.210 cold out if I did it very accurately he
00:08:06.610 00:08:06.620 could actually exceed the temperature of
00:08:09.339 00:08:09.349 the hot out couldn't it couldn't
00:08:13.570 00:08:13.580 yeah it could let me see
00:08:19.420 00:08:19.430 could you in the parallel flow could you
00:08:21.760 00:08:21.770 ever get the hot out lower than the cold
00:08:24.999 00:08:25.009 out no it's impossible but here it is
00:08:29.290 00:08:29.300 possible okay let me ask you a little
00:08:33.370 00:08:33.380 bit what happens if I make the same
00:08:36.850 00:08:36.860 temperature for the hot in and the same
00:08:40.089 00:08:40.099 temperature for the cold in but I make
00:08:42.250 00:08:42.260 the heat exchanger longer I increase its
00:08:46.060 00:08:46.070 length so it's like I have to say no put
00:08:49.030 00:08:49.040 the length over here and here is the
00:08:53.079 00:08:53.089 starting temperature temperature hot in
00:08:56.010 00:08:56.020 true but will what will the temperature
00:09:00.160 00:09:00.170 profile look like compared to the
00:09:02.620 00:09:02.630 shorter smaller heat exchanger now I
00:09:05.260 00:09:05.270 have a larger heat exchanger don't
00:09:08.199 00:09:08.209 change any flow rates the mass flow
00:09:11.980 00:09:11.990 rates the same fluid flow rates are the
00:09:14.079 00:09:14.089 same what would happen the temperature
00:09:19.060 00:09:19.070 difference would decrease right and so
00:09:22.360 00:09:22.370 maybe I know this is going to be a
00:09:24.880 00:09:24.890 complicated diagram but it made go like
00:09:28.180 00:09:28.190 this and the blue may go like this maybe
00:09:36.060 00:09:36.070 something like that did it actually
00:09:40.569 00:09:40.579 change the slope of the red line let's
00:09:44.350 00:09:44.360 say this was the original LED red line
00:09:46.690 00:09:46.700 and that's the second case red line that
00:09:49.060 00:09:49.070 it changed the slope of the red line
00:09:55.100 00:09:55.110 how about this that it changed the slope
00:09:57.259 00:09:57.269 of the blue line case one and case two
00:10:05.470 00:10:05.480 yeah and if you make it really long you
00:10:08.810 00:10:08.820 can see that if we would have just
00:10:10.750 00:10:10.760 extrapolated that blue line up maybe we
00:10:13.639 00:10:13.649 would got to a ridiculous case okay so
00:10:18.769 00:10:18.779 the slopes changed a little bit
00:10:20.960 00:10:20.970 didn't they both of them all right what
00:10:27.079 00:10:27.089 dictated the slope or there's a lot of
00:10:29.900 00:10:29.910 things that dictate to slope but I'm not
00:10:31.579 00:10:31.589 going to change the mass flow-rate and
00:10:33.350 00:10:33.360 I'm not going to change the specific
00:10:34.880 00:10:34.890 heats of either fluid streams for this
00:10:36.590 00:10:36.600 introduction right now for this part of
00:10:38.329 00:10:38.339 the discussion what what what what makes
00:10:41.720 00:10:41.730 it what makes this slope on the red line
00:10:44.269 00:10:44.279 and the slope on the blue line it's the
00:10:49.250 00:10:49.260 temperature difference between the red
00:10:51.019 00:10:51.029 line and the blue line what does the
00:10:52.759 00:10:52.769 temp local temperature differs between a
00:10:54.680 00:10:54.690 hot fluid and cold fluid and because
00:10:57.350 00:10:57.360 that dictates the rate of heat transfer
00:10:59.620 00:10:59.630 hence the rate of change of the hot it's
00:11:02.150 00:11:02.160 going to go down and the rate of change
00:11:04.220 00:11:04.230 of the cold that's going to go up in the
00:11:05.930 00:11:05.940 direction of flow that's pretty
00:11:11.180 00:11:11.190 complicated concept all right but let's
00:11:14.930 00:11:14.940 do this let's go back and play with it
00:11:17.300 00:11:17.310 where I have not changing the length but
00:11:20.689 00:11:20.699 I had the temperature of the hot fluid
00:11:23.000 00:11:23.010 coming in I'm not gonna change that the
00:11:26.120 00:11:26.130 temperatures are hot in I'm not gonna
00:11:28.250 00:11:28.260 change the cold in I'm not gonna change
00:11:31.639 00:11:31.649 anything with the cold fluid but what am
00:11:34.189 00:11:34.199 I gonna do to the hot fluid make it go
00:11:37.519 00:11:37.529 faster increase mass flow rate of the
00:11:40.790 00:11:40.800 hot fluid all right what will the red
00:11:43.819 00:11:43.829 curve do
00:11:49.870 00:11:49.880 it used to be here the red curve right
00:11:53.800 00:11:53.810 it used to be right here will it go up
00:11:57.470 00:11:57.480 or will it go down or will the red curve
00:12:00.079 00:12:00.089 stay the same so if I increase the mass
00:12:05.389 00:12:05.399 flow rate of the hot fluid the red curve
00:12:08.870 00:12:08.880 will go up stay the same or go down top
00:12:16.699 00:12:16.709 and you're exactly right so that's what
00:12:19.850 00:12:19.860 happens to the red curve okay what about
00:12:24.410 00:12:24.420 the blue curve it used to be like this
00:12:28.430 00:12:28.440 will the blue curve increase you know go
00:12:32.000 00:12:32.010 up a little bit steeper or not nothing
00:12:38.210 00:12:38.220 in the cold nothing in the cold it'll go
00:12:43.400 00:12:43.410 up and why would it go up a little bit
00:12:48.730 00:12:48.740 because the delta T is a little greater
00:12:52.220 00:12:52.230 to drive that heat transfer hence more
00:12:54.530 00:12:54.540 heat is going to get into the cold fluid
00:12:56.090 00:12:56.100 I didn't change the temperature of the
00:12:58.939 00:12:58.949 hot fluid all I did was change the mass
00:13:00.530 00:13:00.540 flow rate of the hot fluid but that was
00:13:04.189 00:13:04.199 enough to do it I'm not certain how many
00:13:08.420 00:13:08.430 of these I know that they're very
00:13:10.280 00:13:10.290 productive and you need to process these
00:13:12.199 00:13:12.209 type of questions in your mind I don't I
00:13:14.720 00:13:14.730 hope a lot of you are following along
00:13:16.730 00:13:16.740 but these could be very tricky did you
00:13:23.810 00:13:23.820 that's another good question so that now
00:13:26.550 00:13:26.560 we say what taste do I have when I would
00:13:29.190 00:13:29.200 expect the slopes to be pretty parallel
00:13:32.810 00:13:32.820 for the hot fluid in the cold food under
00:13:35.610 00:13:35.620 what conditions would that be somebody
00:13:38.460 00:13:38.470 would say well probably when the mass
00:13:40.620 00:13:40.630 flow rate of the hot fluid is pretty
00:13:42.630 00:13:42.640 close to the mass flow rate of the cold
00:13:44.250 00:13:44.260 fluid but it's not just the mass flow
00:13:48.240 00:13:48.250 rates it's the mass flow rate times a
00:13:50.850 00:13:50.860 specific key to the hot fluid and mass
00:13:53.130 00:13:53.140 flow rate times specific heat of the
00:13:54.750 00:13:54.760 cold fluid when the product of the mass
00:13:57.330 00:13:57.340 flow rates and the specific heats are
00:13:59.940 00:13:59.950 the same the slopes will be the same
00:14:04.580 00:14:04.590 that's it and so in this heat transfer
00:14:08.100 00:14:08.110 literature we introduce a cap C symbol
00:14:11.190 00:14:11.200 and the cap C is the mass flow rate
00:14:14.760 00:14:14.770 times the specific heat I'm not doing
00:14:17.880 00:14:17.890 phase change I'm not doing condensation
00:14:20.250 00:14:20.260 or vaporization yet we'll get there but
00:14:23.400 00:14:23.410 for introductory purposes it's just
00:14:25.470 00:14:25.480 single phase if it's a gas it stays a
00:14:28.110 00:14:28.120 gas if it's liquid it stays a liquid so
00:14:31.200 00:14:31.210 we have a constant specific heat
00:14:33.530 00:14:33.540 neglecting temperature dependent
00:14:35.370 00:14:35.380 specific heats and you have the mass
00:14:37.230 00:14:37.240 flow rate and you have the heat capacity
00:14:40.170 00:14:40.180 rate heat capacity rate all right new
00:14:46.920 00:14:46.930 term cap see it's used a lot and heat
00:14:49.200 00:14:49.210 exchanger literature and analysis and
00:14:52.020 00:14:52.030 methods what are the units both SI as
00:14:56.430 00:14:56.440 well as US customary system the units
00:14:58.920 00:14:58.930 for the heat capacity rate watts so
00:15:04.070 00:15:04.080 let's do it this way I'll scroll down
00:15:07.620 00:15:07.630 just a little bit so we have mass flow
00:15:10.290 00:15:10.300 rate specific heat let's work over here
00:15:12.750 00:15:12.760 in Si so we're gonna have kilograms per
00:15:15.570 00:15:15.580 second times kilojoules per kilogram
00:15:20.840 00:15:20.850 Kelvin or degree C kilograms cancel
00:15:25.760 00:15:25.770 I'll have forget the kilo I'll have
00:15:28.230 00:15:28.240 watts per degree C true all right mass
00:15:32.280 00:15:32.290 flow rate over in the USCS let's do
00:15:35.700 00:15:35.710 power
00:15:36.269 00:15:36.279 per minute or pounds per hour pounds per
00:15:39.960 00:15:39.970 something second and then specific heat
00:15:43.590 00:15:43.600 well that's how many BTUs per pound
00:15:47.960 00:15:47.970 degree F bit to use cancel so we'll have
00:15:52.290 00:15:52.300 BTUs per minute that's some rate of
00:15:55.079 00:15:55.089 transfer divided by degrees F it's
00:15:57.809 00:15:57.819 consistent isn't that consistent between
00:16:01.860 00:16:01.870 them so they do have a little funny
00:16:03.540 00:16:03.550 units
00:16:04.170 00:16:04.180 it's a beat to use per minute per degree
00:16:07.710 00:16:07.720 F or something like that yes well with
00:16:20.939 00:16:20.949 the advent of computers we can account
00:16:23.670 00:16:23.680 for their variable temperature dependent
00:16:26.069 00:16:26.079 properties as it flows through our
00:16:28.319 00:16:28.329 system and so we disparate eyes and you
00:16:30.929 00:16:30.939 can really do a good job of analysis but
00:16:33.480 00:16:33.490 for introductory purposes it's constant
00:16:36.889 00:16:36.899 some fluids are more susceptible their
00:16:40.949 00:16:40.959 properties change as a stronger function
00:16:43.170 00:16:43.180 of temperature so but we're gonna just
00:16:47.670 00:16:47.680 use an average temperature think about
00:16:50.220 00:16:50.230 water liquids or something like that
00:16:51.869 00:16:51.879 gases so that's an idea if if the flow
00:16:56.189 00:16:56.199 rate of the mass flow rate of the hot
00:16:58.290 00:16:58.300 fluid increases then the hot fluid curve
00:17:01.139 00:17:01.149 becomes flatter if the mass flow rate of
00:17:04.139 00:17:04.149 the cold fluid increases it becomes
00:17:07.289 00:17:07.299 flatter true that goes down and we
00:17:13.919 00:17:13.929 already explored increasing the length
00:17:15.929 00:17:15.939 so think about increasing mass flow
00:17:18.630 00:17:18.640 rates or increasing heat capacity rates
00:17:21.120 00:17:21.130 and then also increasing the area or
00:17:26.429 00:17:26.439 length here's two basic questions when
00:17:31.260 00:17:31.270 you're doing heat exchanger analysis
00:17:33.720 00:17:33.730 just basically the two questions are
00:17:36.029 00:17:36.039 you're gonna rate the heat exchanger or
00:17:38.490 00:17:38.500 you're gonna size the heat exchanger
00:17:41.190 00:17:41.200 you're gonna say for this heat exchanger
00:17:43.260 00:17:43.270 here it is think about it right here
00:17:45.180 00:17:45.190 here it is plop it down on somebody's
00:17:46.950 00:17:46.960 desk or there it is right there in the
00:17:49.230 00:17:49.240 shop
00:17:50.270 00:17:50.280 what will the rate of heat transfer be
00:17:52.580 00:17:52.590 with these two inlet fluids what will Q
00:17:56.450 00:17:56.460 be what will the rate of heat rate your
00:18:00.290 00:18:00.300 heat exchanger by predicting what it can
00:18:02.480 00:18:02.490 transfer that's the first question the
00:18:05.720 00:18:05.730 second question is is I know what I need
00:18:08.300 00:18:08.310 to transfer in my plan I know I need to
00:18:11.180 00:18:11.190 get diz many you know gallons per minute
00:18:13.670 00:18:13.680 of this type of substance with this
00:18:16.100 00:18:16.110 specific heat from this Inlet
00:18:17.870 00:18:17.880 temperature that outlet temperature
00:18:19.670 00:18:19.680 basically told me what I need to
00:18:21.620 00:18:21.630 transfer the rate of transfer needs to
00:18:24.860 00:18:24.870 be accomplished right the next question
00:18:28.520 00:18:28.530 is is how big should my heat exchanger
00:18:32.600 00:18:32.610 be to accomplish that goal
00:18:35.230 00:18:35.240 so you size your heat exchanger so what
00:18:39.080 00:18:39.090 you're gonna find is you'll find almost
00:18:40.850 00:18:40.860 like version a and version B of
00:18:44.680 00:18:44.690 equations in the heat exchanger and
00:18:48.070 00:18:48.080 chapter and they're gonna be like oh if
00:18:51.290 00:18:51.300 you want to rate it use the version a if
00:18:54.830 00:18:54.840 you want to size it version B maybe you
00:18:58.610 00:18:58.620 recall that different set of tables in
00:19:00.620 00:19:00.630 the heat transfer textbook well there's
00:19:04.010 00:19:04.020 two analysis methods that are out there
00:19:07.130 00:19:07.140 there's really more many more than just
00:19:09.350 00:19:09.360 two but to have survived and are still
00:19:11.480 00:19:11.490 in heat transfer textbooks one is called
00:19:14.180 00:19:14.190 the log mean temperature difference and
00:19:15.830 00:19:15.840 it's presented especially for
00:19:18.920 00:19:18.930 informational purposes in this class as
00:19:21.980 00:19:21.990 well as in most textbooks some textbooks
00:19:24.470 00:19:24.480 have relegated to an appendix and it may
00:19:27.620 00:19:27.630 eventually go away but it's the
00:19:30.620 00:19:30.630 effectiveness NTU method is the more
00:19:33.200 00:19:33.210 predominant method it is more flexible
00:19:35.870 00:19:35.880 method but when you get a more flexible
00:19:39.200 00:19:39.210 predominant method it's sometimes more
00:19:41.150 00:19:41.160 abstract and it's harder to understand
00:19:43.760 00:19:43.770 what's going on so in the log mean
00:19:47.210 00:19:47.220 temperature difference method you have
00:19:48.830 00:19:48.840 the rate of heat transfer is some
00:19:52.480 00:19:52.490 overall heat transfer coefficient in the
00:19:55.580 00:19:55.590 heat exchanger which you may recall
00:19:58.360 00:19:58.370 having to work with in the prerequisite
00:20:01.760 00:20:01.770 class the overall heat transfer Co
00:20:03.800 00:20:03.810 efficient
00:20:04.150 00:20:04.160 a lot of times we have a solid
00:20:08.400 00:20:08.410 separating a fluid from a fluid and we
00:20:12.520 00:20:12.530 think about having the temperature of
00:20:14.710 00:20:14.720 the fluid then we have a little
00:20:15.970 00:20:15.980 convective resistance little conduction
00:20:18.220 00:20:18.230 resistance and little convection
00:20:19.870 00:20:19.880 resistance so we'll have the 1 over H a
00:20:23.290 00:20:23.300 on that side the 1 over H a on that side
00:20:26.950 00:20:26.960 and an L over ka on that side and we
00:20:29.680 00:20:29.690 wrap them all together to get a 1 over
00:20:33.270 00:20:33.280 UA which is an overall heat transfer
00:20:36.340 00:20:36.350 coefficient has the same units like H
00:20:38.980 00:20:38.990 convection coefficient so we have that
00:20:42.460 00:20:42.470 big symbol I ran out of room sorry about
00:20:45.070 00:20:45.080 it let me scoot this down a little bit
00:20:46.750 00:20:46.760 all right so the in the log mean
00:20:49.780 00:20:49.790 temperature difference you have the rate
00:20:51.880 00:20:51.890 of heat transfer in that heat exchanger
00:20:53.830 00:20:53.840 is equal to you the overall heat
00:20:57.160 00:20:57.170 transfer coefficient times a that gives
00:20:59.650 00:20:59.660 me not only the physical size but the
00:21:01.810 00:21:01.820 thermal size of that heat exchanger
00:21:04.740 00:21:04.750 times an appropriate temperature
00:21:07.090 00:21:07.100 difference the log mean temperature
00:21:09.700 00:21:09.710 difference is the appropriate
00:21:12.010 00:21:12.020 temperature difference let's go back to
00:21:14.770 00:21:14.780 these plots right here
00:21:17.020 00:21:17.030 or maybe I'll make a new one so if I had
00:21:20.800 00:21:20.810 a heat exchanger and this works for
00:21:23.920 00:21:23.930 parallel flow counter flow cross flow
00:21:26.290 00:21:26.300 all kinds of heat exchangers you just
00:21:28.120 00:21:28.130 put a correction factor in or lookup a
00:21:30.400 00:21:30.410 core a table to get one of the
00:21:32.500 00:21:32.510 properties let's say the hot let me draw
00:21:35.800 00:21:35.810 it like this the hot fluid tea hot in
00:21:39.430 00:21:39.440 and what's down here tea cold in the
00:21:44.980 00:21:44.990 cold food goes like this the hot fluid
00:21:47.500 00:21:47.510 goes like that can you see that this
00:21:50.530 00:21:50.540 temperature difference throughout the
00:21:52.720 00:21:52.730 heat exchanger is roughly the same
00:21:55.410 00:21:55.420 that's the temperature difference that
00:21:58.210 00:21:58.220 this log mean would give you it would
00:22:01.090 00:22:01.100 give you that value all right but what
00:22:03.820 00:22:03.830 happens if the hot did something like
00:22:07.390 00:22:07.400 this and the blue did something like
00:22:10.360 00:22:10.370 that well what you would see that the
00:22:12.130 00:22:12.140 temperature difference over here is less
00:22:14.800 00:22:14.810 than the temperature difference in that
00:22:16.390 00:22:16.400 part of the heat exchanger
00:22:18.060 00:22:18.070 so it's like give me the appropriate
00:22:20.940 00:22:20.950 temperature difference that is
00:22:22.900 00:22:22.910 throughout the heat exchanger and the
00:22:24.880 00:22:24.890 log mean temperature difference does
00:22:26.770 00:22:26.780 exactly that not only will it work for
00:22:29.680 00:22:29.690 this counter flow but it works for the
00:22:33.250 00:22:33.260 more theoretical less practical parallel
00:22:37.090 00:22:37.100 flow where you can get some dramatic
00:22:41.220 00:22:41.230 differences because what does the
00:22:43.630 00:22:43.640 temperature difference look like over
00:22:45.640 00:22:45.650 here quite large what's the temperature
00:22:48.160 00:22:48.170 difference over there quite small and
00:22:51.330 00:22:51.340 it'll work in those cases as well so
00:22:55.210 00:22:55.220 what is that log mean temperature
00:22:56.470 00:22:56.480 difference well what you do is you say
00:22:58.660 00:22:58.670 tell me the temperature difference on
00:23:01.030 00:23:01.040 one side it really doesn't matter of
00:23:03.480 00:23:03.490 that heat exchange or delta T one
00:23:06.160 00:23:06.170 difference between the hot and the cold
00:23:08.170 00:23:08.180 on this side tell me the temperature
00:23:11.230 00:23:11.240 00:23:13.270 00:23:13.280 on the other side Delta P two and if
00:23:19.420 00:23:19.430 you're a student the first time you see
00:23:21.250 00:23:21.260 it you say well maybe I'll just add them
00:23:23.920 00:23:23.930 together and divide by two in many cases
00:23:26.650 00:23:26.660 you wouldn't be too far off but the
00:23:28.690 00:23:28.700 accurate log mean temperature difference
00:23:31.750 00:23:31.760 method would be delta T on one side
00:23:34.900 00:23:34.910 minus the delta T on the second side
00:23:36.940 00:23:36.950 divided by the natural log of the delta
00:23:40.360 00:23:40.370 T on one side divided by the delta T on
00:23:42.520 00:23:42.530 the second side we're gonna drive that
00:23:46.600 00:23:46.610 equation in a minute I'm giving you the
00:23:49.360 00:23:49.370 perspective before we get into the
00:23:50.950 00:23:50.960 brutal math all right the next method is
00:23:55.000 00:23:55.010 the effectiveness NTU method
00:23:57.870 00:23:57.880 effectiveness often is the epsilon or
00:24:01.180 00:24:01.190 some other Greek symbol for
00:24:02.940 00:24:02.950 effectiveness NTU is just like you know
00:24:07.270 00:24:07.280 LM TD log mean temperature difference is
00:24:11.010 00:24:11.020 NTU is the number of transfer units
00:24:16.230 00:24:16.240 number of transfer units so the number
00:24:21.490 00:24:21.500 of transfer units is just simply defined
00:24:23.950 00:24:23.960 as the you are forgot what was you
00:24:29.300 00:24:29.310 the overall heat transfer coefficient it
00:24:33.260 00:24:33.270 accounts for the fluid convection
00:24:36.050 00:24:36.060 resistance on both sides and the wall
00:24:39.110 00:24:39.120 typically conduction through it okay so
00:24:41.980 00:24:41.990 you a over see hey cap C Oh what was
00:24:48.500 00:24:48.510 that that was a product of mass flow
00:24:50.270 00:24:50.280 rate in specific heat what is cap see
00:24:52.760 00:24:52.770 the name of it heat capacity right
00:24:57.080 00:24:57.090 what are these funny units on it BTUs
00:25:00.110 00:25:00.120 per minute degree F or watts per degree
00:25:02.690 00:25:02.700 C and it's a product MC so it's the
00:25:06.200 00:25:06.210 minimum what seem in the minimum heat
00:25:11.150 00:25:11.160 capacity rate well I have two fluids I
00:25:13.580 00:25:13.590 have a hot fluid a cold fluid I can
00:25:15.290 00:25:15.300 calculate the heat capacity rates of
00:25:17.030 00:25:17.040 either one of those and I say which one
00:25:19.220 00:25:19.230 is the minimum and that's what is used
00:25:21.110 00:25:21.120 in the definition of the number of
00:25:22.700 00:25:22.710 transfer units and to use all right what
00:25:26.690 00:25:26.700 about the effectiveness well the
00:25:28.940 00:25:28.950 effectiveness is of definition Q Wow
00:25:33.290 00:25:33.300 what was Q used for what's the name of Q
00:25:37.870 00:25:37.880 Q is the the rate of heat transfer
00:25:41.210 00:25:41.220 occurring and my heat exchanger it's the
00:25:43.940 00:25:43.950 rate of heat transfer what would be the
00:25:45.320 00:25:45.330 typical SI unit watt or USCS BTUs per
00:25:53.360 00:25:53.370 hour BTUs per minute something it's a
00:25:55.550 00:25:55.560 rate of heat transfer right so the
00:25:58.370 00:25:58.380 effectiveness is the actual rate of heat
00:26:01.580 00:26:01.590 transfer being accomplished in that heat
00:26:03.560 00:26:03.570 exchanger divided by M ax I don't I keep
00:26:08.180 00:26:08.190 always confused what is ma X stands for
00:26:11.800 00:26:11.810 the maximum possible the maximum
00:26:15.770 00:26:15.780 possible and it takes a while to figure
00:26:19.250 00:26:19.260 this out but the let me give you the
00:26:22.730 00:26:22.740 equation then try to explain it the
00:26:24.350 00:26:24.360 maximum theoretical possible is if I
00:26:26.510 00:26:26.520 have the minimum heat capacity rate
00:26:29.420 00:26:29.430 times the maximum delta T ever possible
00:26:33.260 00:26:33.270 in this heat exchanger between the two
00:26:35.150 00:26:35.160 fluids the most extreme high high
00:26:37.190 00:26:37.200 temperature is the hot in is there a
00:26:39.620 00:26:39.630 higher temperature anywhere in the
00:26:41.300 00:26:41.310 system than the hot in
00:26:43.400 00:26:43.410 No and what's the lowest temperature
00:26:46.600 00:26:46.610 cold in so I got the most extreme
00:26:50.180 00:26:50.190 temperature that's physically being
00:26:52.250 00:26:52.260 realized times the minimum heat capacity
00:26:54.830 00:26:54.840 rate and that gives you the maximum rate
00:26:58.490 00:26:58.500 of heat transfer possible make that heat
00:27:03.110 00:27:03.120 exchanger as large as you want without
00:27:08.300 00:27:08.310 changing your fluid flow rates because
00:27:10.460 00:27:10.470 cement is based on comparing the hot
00:27:13.850 00:27:13.860 fluid in the cold fluid heat capacity
00:27:15.800 00:27:15.810 rates right
00:27:16.460 00:27:16.470 don't go changing seem in think about
00:27:18.500 00:27:18.510 conceptually changing the area let the
00:27:21.050 00:27:21.060 area go long long long huge long heat
00:27:24.740 00:27:24.750 exchanger right if we let the long long
00:27:28.340 00:27:28.350 long heat exchanger go with the high
00:27:30.950 00:27:30.960 mass flow rate of the hot fluid get rid
00:27:33.920 00:27:33.930 of this one here so this is the mass
00:27:37.460 00:27:37.470 flow F this is the mass flow rate of the
00:27:41.120 00:27:41.130 hot fluid it's very very high it's
00:27:43.580 00:27:43.590 nearly flat going across right the
00:27:46.310 00:27:46.320 temperature of the hot fluid out so I'm
00:27:48.200 00:27:48.210 a little bit less than the temperature
00:27:49.460 00:27:49.470 of the hot fluid in and I make this L go
00:27:53.330 00:27:53.340 very very long so we put a little how do
00:27:56.840 00:27:56.850 they draw that line like this saying
00:27:58.910 00:27:58.920 there's a break in it yeah right
00:28:04.450 00:28:04.460 something like that it's a break in it
00:28:06.920 00:28:06.930 so L is way out there what does the blue
00:28:11.690 00:28:11.700 fluid do will it go something like this
00:28:21.160 00:28:21.170 where what will the temperature of the
00:28:24.050 00:28:24.060 blue fluid do going out I have a
00:28:27.560 00:28:27.570 super-long it'll go to T hot in so you
00:28:33.590 00:28:33.600 could think about making it super long
00:28:35.480 00:28:35.490 and you would have the change in this
00:28:38.180 00:28:38.190 case can you tell that the heat capacity
00:28:40.160 00:28:40.170 rate of the cold fluid is less than heat
00:28:42.560 00:28:42.570 capacity rated the hot fluid we let the
00:28:44.720 00:28:44.730 mass flow rate the hot fluid go off to
00:28:46.340 00:28:46.350 infinity to achieve this so of course
00:28:48.590 00:28:48.600 the coke blue is less
00:28:52.110 00:28:52.120 so it's a little tricky there's a lot of
00:28:54.899 00:28:54.909 you know what if this what if that but
00:28:57.060 00:28:57.070 it's the minimum heat capacity rate
00:28:58.830 00:28:58.840 times the maximum temperature difference
00:29:01.200 00:29:01.210 anywhere the hot fluid in - the cold
00:29:04.289 00:29:04.299 fluid in that's the effectiveness NTU
00:29:06.600 00:29:06.610 method that's the more practical popular
00:29:10.110 00:29:10.120 implementable quicker method well I
00:29:15.840 00:29:15.850 thought about taking a blank page have
00:29:19.470 00:29:19.480 my notes and just start the derivation
00:29:22.250 00:29:22.260 and then I thought mmm I got that page
00:29:26.970 00:29:26.980 that page and that page the cover and I
00:29:32.669 00:29:32.679 think you would be asleep by that note
00:29:34.710 00:29:34.720 so let me say this it's in every heat
00:29:38.850 00:29:38.860 transfer textbook it is and it's in our
00:29:43.620 00:29:43.630 book - and the derivation is just what
00:29:46.980 00:29:46.990 they do in the book is they say and it
00:29:48.630 00:29:48.640 follows such that boom and they have two
00:29:50.940 00:29:50.950 lines and the work between those two
00:29:52.769 00:29:52.779 lines is about two or three pages of
00:29:54.690 00:29:54.700 work so anyway let me do this let me
00:29:57.060 00:29:57.070 kind of guide you through this so let's
00:30:00.990 00:30:01.000 start out we have to pick something to
00:30:02.820 00:30:02.830 analyze concentric tube it's easy
00:30:04.980 00:30:04.990 mathematically and easy to illustrate
00:30:07.080 00:30:07.090 counter flow because that's the
00:30:09.090 00:30:09.100 predominant way the fluids are actually
00:30:11.130 00:30:11.140 float through heat exchangers and we
00:30:13.230 00:30:13.240 want to derive for the temperature
00:30:15.659 00:30:15.669 distribution regardless if the hot
00:30:18.840 00:30:18.850 fluids flowing really fast or slow
00:30:21.380 00:30:21.390 regardless of the specific heat of the
00:30:24.330 00:30:24.340 hot fluid or cold fluid or the heat
00:30:26.460 00:30:26.470 capacity rates this is gonna work for
00:30:28.110 00:30:28.120 all those cases so you think for a
00:30:31.799 00:30:31.809 minute just cut a sketch a case put the
00:30:34.320 00:30:34.330 hot in over here the hot out cold in
00:30:36.870 00:30:36.880 cold out I didn't try to draw them to
00:30:39.299 00:30:39.309 have the same slope here's an expert
00:30:43.110 00:30:43.120 question which one of them just by
00:30:46.110 00:30:46.120 looking at the plot would have a slight
00:30:48.659 00:30:48.669 would have the lowest heat capacity rate
00:30:51.990 00:30:52.000 which one would have the highest heat
00:30:53.820 00:30:53.830 capacity rate this was a really hard
00:30:55.560 00:30:55.570 question it's a product a mass flow rate
00:30:58.860 00:30:58.870 times specific heat so which one of them
00:31:01.830 00:31:01.840 has the highest C CAP C
00:31:04.960 00:31:04.970 the cold the cold does very good the
00:31:09.909 00:31:09.919 cold food because it doesn't change as
00:31:11.950 00:31:11.960 much in the temperature through the heat
00:31:14.560 00:31:14.570 exchanger so it's a larger heat capacity
00:31:17.799 00:31:17.809 rate you think about higher mass flow
00:31:20.739 00:31:20.749 rate higher mass flow rate specific heat
00:31:22.930 00:31:22.940 it's really that thermal property okay
00:31:24.789 00:31:24.799 we talked about a we're going to talk
00:31:27.279 00:31:27.289 about every little DX so if I'm going to
00:31:30.070 00:31:30.080 talk about every little DX and I need
00:31:31.869 00:31:31.879 the really the DA that goes with it the
00:31:34.299 00:31:34.309 perimeter times the D X makes sense so
00:31:36.249 00:31:36.259 even if you don't have a perimeter
00:31:37.330 00:31:37.340 introduce a perimeter in a concentric
00:31:39.820 00:31:39.830 tube it makes sense to have a perimeter
00:31:41.879 00:31:41.889 but we're talk about the amount of heat
00:31:44.680 00:31:44.690 the small D Q transferred at every
00:31:48.219 00:31:48.229 little DX okay so notice mathematically
00:31:52.779 00:31:52.789 what am I writing D H T DX the rate of
00:31:56.739 00:31:56.749 change of the hot fluid temperature as I
00:31:59.259 00:31:59.269 move in the positive direction of change
00:32:02.109 00:32:02.119 of X do you see this what I'm trying to
00:32:04.180 00:32:04.190 say right here is guess what it's
00:32:08.019 00:32:08.029 positive don't overthink it the fluid is
00:32:12.519 00:32:12.529 flowing in the negative x direction but
00:32:15.519 00:32:15.529 when I plot the temperature T H is a
00:32:18.580 00:32:18.590 function of X the slope is positive
00:32:22.979 00:32:22.989 likewise the change of the cold
00:32:26.080 00:32:26.090 temperature with respect to X is also
00:32:28.570 00:32:28.580 positive they both have a positive and
00:32:30.789 00:32:30.799 I'm going to be working in the same X X
00:32:33.369 00:32:33.379 goes from 0 to L all right so the heat
00:32:38.049 00:32:38.059 capacity rates with the units watts per
00:32:40.299 00:32:40.309 degree C or as shown for the cold and
00:32:42.339 00:32:42.349 the hot fluid and I look and I say what
00:32:44.889 00:32:44.899 is my rate equation to predict what this
00:32:47.619 00:32:47.629 term is DQ because that's what controls
00:32:50.469 00:32:50.479 how the temperatures change isn't it if
00:32:53.259 00:32:53.269 it's I can get a lot of heat transfer
00:32:56.580 00:32:56.590 for the same delta T the temperature the
00:33:00.669 00:33:00.679 hot in the temperature the cold will
00:33:02.049 00:33:02.059 more rapidly change isn't this my rate
00:33:04.719 00:33:04.729 equation
00:33:05.379 00:33:05.389 it's a u a DT and U is overall heat
00:33:10.119 00:33:10.129 transfer coefficient also we know just
00:33:13.509 00:33:13.519 if I look only at the cold fluid the
00:33:16.599 00:33:16.609 amount of heat into the cold fluid
00:33:18.790 00:33:18.800 makes the temperature the cold fluid go
00:33:20.890 00:33:20.900 up by what heat capacity rate times the
00:33:25.030 00:33:25.040 change in the cold fluid temperature
00:33:27.390 00:33:27.400 does this equation look sense makes
00:33:29.950 00:33:29.960 sense this one came from our rate
00:33:32.770 00:33:32.780 equation it's basically our description
00:33:35.590 00:33:35.600 of rates of heat transfer it's a
00:33:37.960 00:33:37.970 combination a convection conduction
00:33:39.550 00:33:39.560 convection combination of Newton's four
00:33:43.390 00:33:43.400 EA's Newton's law what is this second
00:33:47.710 00:33:47.720 one it's not Newton's it's not for yays
00:33:50.860 00:33:50.870 law what is it you just landed on the
00:33:55.900 00:33:55.910 sheet of paper I don't know it looked
00:33:57.610 00:33:57.620 good it looked like it was sensible to
00:33:59.380 00:33:59.390 write down no no no what's the basis for
00:34:02.680 00:34:02.690 it genius
00:34:06.180 00:34:06.190 if I give them a little a plus a little
00:34:08.950 00:34:08.960 star you know walk around and let
00:34:10.690 00:34:10.700 kindergarten put a little star on the
00:34:12.250 00:34:12.260 paper that's all you have to do to make
00:34:13.780 00:34:13.790 me happy Ben the same thing in thermal
00:34:16.060 00:34:16.070 one thermo two heat transfer now thermal
00:34:18.130 00:34:18.140 systems design professor the answer to
00:34:22.180 00:34:22.190 the question is the first law of
00:34:24.159 00:34:24.169 thermodynamics you're right isn't it it
00:34:30.820 00:34:30.830 is and guess what
00:34:32.500 00:34:32.510 for those that want to show off now
00:34:34.560 00:34:34.570 let's take a look at D Q and what its
00:34:37.270 00:34:37.280 effect on the hot fluid stream first law
00:34:41.440 00:34:41.450 thermodynamics conservation of energy I
00:34:43.600 00:34:43.610 transfer it heat in to the system and
00:34:45.880 00:34:45.890 there's now going to be increase in the
00:34:48.130 00:34:48.140 temperature that flowing fluid all right
00:34:50.110 00:34:50.120 we combine now this equation we get rid
00:34:54.520 00:34:54.530 of the del Q's we put the rate on both
00:34:56.710 00:34:56.720 sides here here now I have two equations
00:35:00.010 00:35:00.020 I want to combine them if you never did
00:35:04.870 00:35:04.880 this before you will say I don't know
00:35:07.270 00:35:07.280 which way you're going well trust me
00:35:09.610 00:35:09.620 let's start driving to Houston but I got
00:35:11.710 00:35:11.720 to first get on UTSA Boulevard and I got
00:35:14.170 00:35:14.180 to get on i-10 and I got a turn here
00:35:16.090 00:35:16.100 right it's like somebody else did this
00:35:18.550 00:35:18.560 before and so you combine them this way
00:35:21.960 00:35:21.970 algebraically it's just algebra and you
00:35:24.190 00:35:24.200 get this equation and it's starting to
00:35:27.010 00:35:27.020 look like changes in temperature
00:35:30.130 00:35:30.140 difference proportional to chain
00:35:32.650 00:35:32.660 change in location DX I'm getting a
00:35:35.620 00:35:35.630 differential equation and then I recall
00:35:38.589 00:35:38.599 at X equal to zero the DTD C is equal to
00:35:42.760 00:35:42.770 D H out minus D cold in I could do it
00:35:46.059 00:35:46.069 and have done it for the parallel flow
00:35:48.279 00:35:48.289 let's do it for the counter flow it's a
00:35:49.960 00:35:49.970 little more tricky with the counter flow
00:35:51.370 00:35:51.380 conceptually the parallel flow is
00:35:53.140 00:35:53.150 probably what most texts do anyway but
00:35:56.260 00:35:56.270 you then integrate from X equal to zero
00:35:59.380 00:35:59.390 to arbitrary X you know the initial
00:36:02.410 00:36:02.420 condition at X equal to zero and you get
00:36:04.960 00:36:04.970 this equation look at the temperature
00:36:07.420 00:36:07.430 hot minus temperature cold at that
00:36:09.370 00:36:09.380 location at that location X in our
00:36:12.190 00:36:12.200 counter flow concentric tube heat
00:36:14.620 00:36:14.630 exchanger is equal to th out minus T
00:36:18.609 00:36:18.619 cold in over at X equal to zero times
00:36:22.510 00:36:22.520 and I couldn't fit it on one line the
00:36:24.520 00:36:24.530 exponential of 1 over CH minus 1 over C
00:36:27.819 00:36:27.829 cold all times u P X it's a function of
00:36:30.760 00:36:30.770 X it's an exponential function of X you
00:36:36.069 00:36:36.079 can play a game you could explore the
00:36:37.900 00:36:37.910 solution this is tricky I'm sorry but
00:36:39.940 00:36:39.950 this is just what heat exchanger
00:36:41.440 00:36:41.450 analysis is if the heat capacity rate of
00:36:44.289 00:36:44.299 the cold fluid is greater than heat
00:36:45.760 00:36:45.770 capacity ready to cold hot fluid then
00:36:47.680 00:36:47.690 look at it the temperature difference
00:36:49.960 00:36:49.970 the local temperature difference equal
00:36:51.579 00:36:51.589 to a theta initial at the knot and
00:36:54.220 00:36:54.230 coming in X exponent of something that's
00:36:58.029 00:36:58.039 positive that's the key positive times X
00:37:01.390 00:37:01.400 so what's going to happen to the
00:37:03.220 00:37:03.230 temperature differences I move in
00:37:04.930 00:37:04.940 increasing X it opens up doesn't it it
00:37:10.059 00:37:10.069 opens up and then if you say okay what
00:37:13.569 00:37:13.579 about the hot fluid being greater than
00:37:15.640 00:37:15.650 the cold fluid then if I look up here
00:37:19.920 00:37:19.930 that term is now negative I have e to
00:37:26.319 00:37:26.329 the negative X what does that do with
00:37:28.900 00:37:28.910 increasing X and theta decreases as I
00:37:32.529 00:37:32.539 move it gives me the boat the same that
00:37:35.650 00:37:35.660 gives me the right shapes doesn't it in
00:37:38.740 00:37:38.750 the limit as C cold and C hot equal each
00:37:42.339 00:37:42.349 other I have e to
00:37:44.539 00:37:44.549 the 0x it's constant temperature
00:37:49.130 00:37:49.140 difference through the heat exchanger I
00:37:52.299 00:37:52.309 didn't lose you yet now that we have the
00:37:57.109 00:37:57.119 temperature difference anywhere in the
00:38:00.079 00:38:00.089 heat exchanger is a function of X we can
00:38:02.089 00:38:02.099 derive the log mean temperature method
00:38:05.449 00:38:05.459 okay so you just say well don't stop at
00:38:09.169 00:38:09.179 X take it all the way to L so if you
00:38:11.989 00:38:11.999 evaluate the same equation we had it all
00:38:13.939 00:38:13.949 the way to L isn't this the temperature
00:38:15.799 00:38:15.809 hot n minus cold out equal to what was
00:38:20.959 00:38:20.969 that X equal to 0 times e to the minus
00:38:23.419 00:38:23.429 everything here I put P times L that's a
00:38:26.349 00:38:26.359 so I have you a I went the whole area
00:38:30.620 00:38:30.630 now you just do the algebra and you
00:38:35.239 00:38:35.249 recall that the C sub H can be replaced
00:38:38.120 00:38:38.130 by Q over the temperature difference of
00:38:41.299 00:38:41.309 the temperature change experienced by
00:38:43.279 00:38:43.289 the hot fluid as well as the heat
00:38:44.719 00:38:44.729 capacity rate of cold is Q divided by
00:38:47.449 00:38:47.459 the temperature change of the cold fluid
00:38:49.759 00:38:49.769 you put that in you're doing some
00:38:52.400 00:38:52.410 algebra and lo and behold you get Q is
00:38:55.789 00:38:55.799 equal to UA times a big blob and you
00:39:00.529 00:39:00.539 call that the log mean temperature
00:39:02.329 00:39:02.339 difference so that big log mean
00:39:06.650 00:39:06.660 temperature difference can be read as
00:39:08.359 00:39:08.369 talked about the temperature difference
00:39:10.999 00:39:11.009 on one side the temperature difference
00:39:13.640 00:39:13.650 on the other side and it doesn't matter
00:39:17.120 00:39:17.130 which one you put it can be DT 1 minus
00:39:20.509 00:39:20.519 DT 2 divided by natural log of V T 1
00:39:23.179 00:39:23.189 over DT 2 or DT 2 minus D T 1 does it
00:39:27.039 00:39:27.049 these equations it's like hold it which
00:39:29.839 00:39:29.849 one is right the one on the left or the
00:39:31.609 00:39:31.619 one on the right they're the same how is
00:39:35.239 00:39:35.249 that true do this I have the natural log
00:39:37.939 00:39:37.949 of a over B how is that related to the
00:39:40.370 00:39:40.380 natural log of B over a how is that
00:39:44.749 00:39:44.759 related how is the natural log of a over
00:39:47.299 00:39:47.309 B related to the natural log of B over a
00:39:51.490 00:39:51.500 it's a negative sign and so what happens
00:39:55.670 00:39:55.680 is you can flip these in the denominator
00:39:59.269 00:39:59.279 just put the negative sign and then you
00:40:02.319 00:40:02.329 change up on the top that's all so there
00:40:08.720 00:40:08.730 you go that's a log mean temperature
00:40:10.039 00:40:10.049 difference it works great if you know
00:40:13.309 00:40:13.319 the temperatures of the fluids in and
00:40:15.730 00:40:15.740 out it's more challenging it leads to an
00:40:19.279 00:40:19.289 iterative solution or approach when you
00:40:22.039 00:40:22.049 don't know the outlet temperatures but
00:40:23.870 00:40:23.880 you're asked to calculate them you kind
00:40:26.180 00:40:26.190 of have to guess what is the outlet
00:40:28.220 00:40:28.230 temperatures so that I can get an
00:40:29.569 00:40:29.579 estimate of the delta T log mean to get
00:40:31.430 00:40:31.440 an estimate acute get an estimate of
00:40:33.170 00:40:33.180 outlet temperatures they go back and
00:40:34.819 00:40:34.829 correct and it's a leads to an iterative
00:40:37.490 00:40:37.500 solution it's not bad it's accurate it's
00:40:41.990 00:40:42.000 very accurate the next method is the
00:40:45.829 00:40:45.839 effectiveness NTU method we had a brief
00:40:49.099 00:40:49.109 introduction well what you can do is you
00:40:52.039 00:40:52.049 say okay the genius that thought this up
00:40:54.349 00:40:54.359 was truly a genius I don't know who it
00:40:56.329 00:40:56.339 was originally somewhere I should
00:40:58.549 00:40:58.559 research it right and then we should put
00:41:00.200 00:41:00.210 we should replace it and put their name
00:41:01.990 00:41:02.000 the Joneses method right and honor Jones
00:41:05.660 00:41:05.670 that whoever did it so we we know all of
00:41:10.160 00:41:10.170 our heat capacity rates at this point in
00:41:13.789 00:41:13.799 the analysis it's easier if you just
00:41:15.799 00:41:15.809 pick one to be the dominant heat
00:41:18.049 00:41:18.059 capacity right the other would be though
00:41:19.460 00:41:19.470 but if you did this twice and you
00:41:21.859 00:41:21.869 switched it and said no don't let cold
00:41:23.720 00:41:23.730 be the greater than the hot let the hot
00:41:25.730 00:41:25.740 be greater than the cold you'll get the
00:41:27.620 00:41:27.630 same exact result but it's easier to
00:41:30.799 00:41:30.809 just pick one and say this is the
00:41:32.749 00:41:32.759 relative difference between them okay so
00:41:34.940 00:41:34.950 we're just picking that one so the
00:41:37.249 00:41:37.259 effectiveness is Q max Q over Q max so
00:41:41.990 00:41:42.000 the minimum in this case in general it's
00:41:44.480 00:41:44.490 the minimum of the cold and hot but for
00:41:46.279 00:41:46.289 this case it's going to be the hot fluid
00:41:48.589 00:41:48.599 is the minimum heat capacity rate hence
00:41:51.440 00:41:51.450 the definition of men to use as UA over
00:41:54.049 00:41:54.059 si men in this case it would be UA over
00:41:57.019 00:41:57.029 CH then we have Q max it's typically in
00:42:02.210 00:42:02.220 general seem n times this temperature
00:42:05.180 00:42:05.190 difference now it's going to be C H
00:42:06.890 00:42:06.900 times that temperature difference and
00:42:09.589 00:42:09.599 then the actual rate of heat transfer
00:42:11.410 00:42:11.420 unraveling effectiveness will be the
00:42:13.819 00:42:13.829 effectiveness times the heat capacity
00:42:16.130 00:42:16.140 rate of the hot fluid times the
00:42:17.990 00:42:18.000 temperature difference is this exciting
00:42:21.620 00:42:21.630 I see one big yawn are we following this
00:42:29.020 00:42:29.030 would have been helpful if I would have
00:42:31.309 00:42:31.319 pretended that I didn't have my notes
00:42:33.140 00:42:33.150 and then press you that I'm able to
00:42:34.670 00:42:34.680 drive it would take 20 minutes longer
00:42:37.990 00:42:38.000 okay let's slug through professor get
00:42:41.180 00:42:41.190 the pill take the pain you know so we
00:42:44.750 00:42:44.760 had the hot fluid out as the cold flu
00:42:46.490 00:42:46.500 the hot fluid in - that how much heat is
00:42:49.280 00:42:49.290 transferred the rate of heat transfer
00:42:50.599 00:42:50.609 divided by the CH so the hot fluid out
00:42:56.150 00:42:56.160 is equal to what is Q replace it right
00:42:58.970 00:42:58.980 here put that in there to see HS cancel
00:43:01.270 00:43:01.280 kind of like a little bingo whoo look at
00:43:03.859 00:43:03.869 that and we can say well delta T 2 is T
00:43:08.809 00:43:08.819 hot out I'm sorry T hot out minus T cold
00:43:12.530 00:43:12.540 in and I can replace what is T cold in
00:43:19.450 00:43:19.460 right here I just put that into this one
00:43:22.760 00:43:22.770 into here do the algebra and I get delta
00:43:25.910 00:43:25.920 T 2 on one of the sides is T hot n minus
00:43:29.300 00:43:29.310 T cold n times 1 minus epsilon the
00:43:32.329 00:43:32.339 effectiveness this what I just did here
00:43:36.559 00:43:36.569 needs to be done again focusing on the
00:43:38.960 00:43:38.970 cold and you get delta T 1 is equal to T
00:43:43.160 00:43:43.170 ha T and minus T cold out without any
00:43:45.290 00:43:45.300 effectiveness ok algebra on those now
00:43:50.980 00:43:50.990 I'm sorry this one gets delta T 1 after
00:43:54.770 00:43:54.780 you substitute for T cold out right here
00:43:59.470 00:43:59.480 yet the substitute you get this long
00:44:02.599 00:44:02.609 equation for Delta T 1 you recall the
00:44:05.930 00:44:05.940 log mean temperature difference stick in
00:44:08.000 00:44:08.010 delta T ones delta T 2 delta T 1 delta T
00:44:11.270 00:44:11.280 2 they expand you look for things to
00:44:15.050 00:44:15.060 cancel you only find that to cancel you
00:44:17.960 00:44:17.970 now know that
00:44:19.040 00:44:19.050 q is equal to UA log mean temperature
00:44:20.930 00:44:20.940 difference so you have the whole log
00:44:25.130 00:44:25.140 mean temperature difference put in I
00:44:26.870 00:44:26.880 know that Q Max
00:44:28.700 00:44:28.710 / C min times T hot n minus T cold in
00:44:33.410 00:44:33.420 that all in blue is just multiplied by
00:44:36.170 00:44:36.180 unity I just multiplied by 1 that way I
00:44:40.820 00:44:40.830 can get the definition of the number of
00:44:44.540 00:44:44.550 transfer units I can cancel the
00:44:49.850 00:44:49.860 effectiveness times Q Max is equal to Q
00:44:52.850 00:44:52.860 so all those three red lines cancel and
00:44:55.010 00:44:55.020 these Delta T's one of them set cancel
00:44:57.650 00:44:57.660 and here is the final result the number
00:45:00.380 00:45:00.390 of transfer units is equal to 1 over 1
00:45:04.310 00:45:04.320 minus C R the ratio of heat capacity
00:45:07.310 00:45:07.320 rates minimum over maximum times the
00:45:10.100 00:45:10.110 natural log 1 minus the effectiveness
00:45:12.320 00:45:12.330 ratio of heat capacity rates divided by
00:45:14.750 00:45:14.760 1 minus and you put a little star there
00:45:16.940 00:45:16.950 and you say we did it so if I'm
00:45:20.750 00:45:20.760 interested in knowing if I need the size
00:45:23.540 00:45:23.550 that I use that first equation because
00:45:26.060 00:45:26.070 NTU the area is embedded in NTU if I
00:45:30.470 00:45:30.480 need to size it you can rewrite that
00:45:32.690 00:45:32.700 same equation with effectiveness on one
00:45:35.060 00:45:35.070 side that's the second equation and then
00:45:38.060 00:45:38.070 everything else is on the other side so
00:45:40.190 00:45:40.200 if I know the area and I want to know
00:45:42.110 00:45:42.120 the actual amount of heat transfer then
00:45:44.510 00:45:44.520 I would get the effectiveness and then
00:45:46.280 00:45:46.290 use that to get the actual Q so use this
00:45:49.310 00:45:49.320 second equation to rate and use the
00:45:53.210 00:45:53.220 first equation to size the heat
00:45:55.610 00:45:55.620 exchanger they're the same equation
00:45:58.810 00:45:58.820 they're the same equation they're
00:46:01.070 00:46:01.080 mathematically the same alright do I
00:46:04.730 00:46:04.740 have enough time to solve the problem
00:46:07.660 00:46:07.670 water enters a counterflow concentric
00:46:11.030 00:46:11.040 tube heat exchanger at 150 pound mass
00:46:14.270 00:46:14.280 per minute so our heat exchanger is
00:46:18.350 00:46:18.360 gonna be like this and I'm gonna divide
00:46:21.200 00:46:21.210 it like that let's just do the hot fluid
00:46:24.560 00:46:24.570 on one side and we'll put the cold flu
00:46:28.599 00:46:28.609 on the other side and the water enters
00:46:32.709 00:46:32.719 it's going from sixty degrees to one
00:46:36.370 00:46:36.380 hundred and forty is the water the cold
00:46:38.499 00:46:38.509 fluid or the hot fluid the temperature
00:46:41.739 00:46:41.749 of the cold in is sixty degrees F naught
00:46:46.179 00:46:46.189 to C and it goes out temperature cold
00:46:49.029 00:46:49.039 out is one hundred and forty degrees F
00:46:51.579 00:46:51.589 Troop it's the cold fluid waters the
00:46:53.949 00:46:53.959 cold fluid it's so we have the mass flow
00:46:57.429 00:46:57.439 rate of the cold fluid is one hundred
00:46:59.410 00:46:59.420 and fifty pounds per minute they didn't
00:47:02.769 00:47:02.779 tell me the specific heat of water it's
00:47:09.459 00:47:09.469 about one pound
00:47:10.390 00:47:10.400 BTU per pound degree F that's specific
00:47:13.719 00:47:13.729 heat of water you would get that out of
00:47:15.609 00:47:15.619 the table and her just remember it
00:47:17.499 00:47:17.509 what's doing the heating so the water is
00:47:21.150 00:47:21.160 the colder fluid oil is the hot fluid it
00:47:27.039 00:47:27.049 comes in temperature hot in of two
00:47:30.699 00:47:30.709 hundred and forty degrees F and they
00:47:33.189 00:47:33.199 tell you the temperature the hot out is
00:47:35.709 00:47:35.719 equal to 80 degrees F notice that the
00:47:40.779 00:47:40.789 oil comes out lower temperature than the
00:47:43.029 00:47:43.039 water goes out makes sense yeah and the
00:47:50.169 00:47:50.179 oil has a specific heat let me put it
00:47:53.049 00:47:53.059 over here a specific heat of the hot
00:47:55.989 00:47:55.999 fluid is point four eight beat to use
00:47:59.319 00:47:59.329 per pound mass degree F the overall heat
00:48:03.669 00:48:03.679 transfer coefficient is fifty two point
00:48:05.559 00:48:05.569 eight BTU per hour per foot squared
00:48:08.109 00:48:08.119 degrees what information did that just
00:48:10.390 00:48:10.400 give us their cap you that's right
00:48:13.329 00:48:13.339 fifty two point a etc what's the
00:48:18.099 00:48:18.109 question
00:48:18.839 00:48:18.849 determine the heat transfer area what
00:48:23.559 00:48:23.569 are we trying to do are we trying to
00:48:25.509 00:48:25.519 rate it or size it size it yeah
00:48:28.809 00:48:28.819 determine the heat transfer area we're
00:48:31.150 00:48:31.160 gonna use the effectiveness NTU method
00:48:32.979 00:48:32.989 how are we going to solve this problem
00:48:39.680 00:48:39.690 think in general that I'm going to
00:48:42.359 00:48:42.369 eventually get NTU equal to some
00:48:46.170 00:48:46.180 function of effectiveness and see our
00:48:49.190 00:48:49.200 ratio of heat capacity rates once I get
00:48:52.829 00:48:52.839 the number of transfer units then I
00:48:55.109 00:48:55.119 recall that the number of transfer units
00:48:57.120 00:48:57.130 is UA
00:48:57.809 00:48:57.819 over C min and so the area is the number
00:49:01.890 00:49:01.900 of transfer units times C min divided by
00:49:06.569 00:49:06.579 U so I got to get the number of transfer
00:49:10.019 00:49:10.029 units then unravel it it's like it's a
00:49:13.259 00:49:13.269 dimensionless ratio isn't it is it NTU
00:49:17.039 00:49:17.049 dimensionless it's purely dimensionless
00:49:19.559 00:49:19.569 yep so okay well I need to get the
00:49:23.279 00:49:23.289 effectiveness well how am I gonna get
00:49:25.470 00:49:25.480 the effectiveness well actually can I
00:49:27.930 00:49:27.940 calculate Q actual what is the rate of
00:49:30.660 00:49:30.670 heat transfer that occurs in this heat
00:49:33.059 00:49:33.069 exchanger sure you would say either from
00:49:39.569 00:49:39.579 the perspective of the cold fluid which
00:49:42.180 00:49:42.190 has a heat capacity rate of 150 beat to
00:49:47.130 00:49:47.140 use per minute degree F or the hot fluid
00:49:53.880 00:49:53.890 they don't give us a capacity rate of
00:49:56.370 00:49:56.380 the hot fluid do they but can they is is
00:49:59.160 00:49:59.170 this equal to the cold fluid times the
00:50:01.499 00:50:01.509 temperature of the cold out minus
00:50:04.559 00:50:04.569 temperature of the cold in its
00:50:10.200 00:50:10.210 conservation of energy the cold fluid
00:50:12.299 00:50:12.309 underwent this change mass flow rate of
00:50:14.789 00:50:14.799 the colds given heat the specific heat
00:50:17.519 00:50:17.529 is cold is given so we calculate the Q
00:50:20.099 00:50:20.109 and there's 12,000 BTUs per minute
00:50:24.980 00:50:24.990 well there's 12,000 BTUs per minute
00:50:28.019 00:50:28.029 getting into the water where is it
00:50:29.549 00:50:29.559 coming from no oil and if it's coming
00:50:33.180 00:50:33.190 from the oil then what we say is what
00:50:35.549 00:50:35.559 comes out of the oil goes into the water
00:50:37.920 00:50:37.930 so it's 12,000 so the Q is equal to the
00:50:41.400 00:50:41.410 C of the hot times the temperature of
00:50:43.740 00:50:43.750 the hot n minus temperature of the hot
00:50:46.920 00:50:46.930 out true I do a little check I just
00:50:50.039 00:50:50.049 calculated the Q
00:50:51.790 00:50:51.800 have the two temperatures of the hot
00:50:53.260 00:50:53.270 fluid I can get the heat capacity rate
00:50:56.020 00:50:56.030 of the hot fluid the heat capacity rate
00:50:58.900 00:50:58.910 of the hot fluid is calculated to be 75
00:51:02.850 00:51:02.860 BTUs per minute degree F do a quick
00:51:08.020 00:51:08.030 comparison which one is the minimum so C
00:51:15.070 00:51:15.080 min is of the hot and it's 75 BTU per
00:51:20.580 00:51:20.590 minute degree F I'm gonna need this C
00:51:23.920 00:51:23.930 sub R what is C sub R the ratio of
00:51:27.400 00:51:27.410 minimum to maximum so what's the minimum
00:51:31.090 00:51:31.100 75 what's the maximum 150 it comes out
00:51:34.330 00:51:34.340 to 0.5 Oh have C sub R is 1/2
00:51:38.710 00:51:38.720 okay now I know Q what I really need and
00:51:47.920 00:51:47.930 I just calculated C sub R right here and
00:51:51.280 00:51:51.290 I if I get the effectiveness then I can
00:51:54.100 00:51:54.110 just plug it into my equation to
00:51:55.750 00:51:55.760 calculate the number of transfer units
00:51:57.520 00:51:57.530 well what equation am I going to plug
00:51:58.900 00:51:58.910 into either version a or version B which
00:52:05.260 00:52:05.270 equation yeah we're going to be we want
00:52:12.490 00:52:12.500 to get n to use so we're gonna stick it
00:52:15.160 00:52:15.170 into that second version B I said bees
00:52:18.400 00:52:18.410 that that equation right there that's
00:52:21.310 00:52:21.320 our roadmap okay so I need to get the
00:52:24.640 00:52:24.650 effectiveness but how do I get
00:52:26.260 00:52:26.270 defectiveness I need to get the maximum
00:52:28.300 00:52:28.310 possible
00:52:28.990 00:52:29.000 what is that C min times temperature hot
00:52:32.170 00:52:32.180 n minus temperature cold in and so the
00:52:36.640 00:52:36.650 maximum rate of heat transfer is 13 500
00:52:42.180 00:52:42.190 BTUs per min so the effectiveness is
00:52:47.020 00:52:47.030 what I actually get divided by the
00:52:49.690 00:52:49.700 maximum it's 12,000 divided by 13 500 it
00:52:55.690 00:52:55.700 comes in at 0.8 a
00:53:00.590 00:53:00.600 eight-nine 89% close to 90% a little bit
00:53:07.230 00:53:07.240 under 90% we know the effectiveness the
00:53:10.920 00:53:10.930 NT use we use that equation the NTU is
00:53:13.920 00:53:13.930 equal to and I'll just rewrite it
00:53:15.440 00:53:15.450 natural log of 1 minus the effectiveness
00:53:19.110 00:53:19.120 times the ratio of heat capacity rates
00:53:22.260 00:53:22.270 divided by 1 minus effectiveness divided
00:53:24.630 00:53:24.640 by 1 minus C sub R when you do that the
00:53:28.320 00:53:28.330 number of transfer units comes in at 3.2
00:53:31.460 00:53:31.470 1 8 9
00:53:33.950 00:53:33.960 are there any units with that not it's
00:53:38.430 00:53:38.440 dimensionless but when we unravel it up
00:53:41.940 00:53:41.950 here to get the area we'll multiply it
00:53:46.710 00:53:46.720 by C min we'll divide by cap you the
00:53:49.560 00:53:49.570 overall heat transfer coefficient and we
00:53:51.780 00:53:51.790 calculate the area to be 274 square feet
00:53:59.180 00:53:59.190 it's a large area does that look good
00:54:04.640 00:54:04.650 guess what I like to do we solve a
00:54:06.840 00:54:06.850 problem but then we turn it around and
00:54:08.790 00:54:08.800 change it up just a little bit and so
00:54:11.820 00:54:11.830 I'm ready to ask you for a change so the
00:54:16.530 00:54:16.540 same wording up here instead of saying
00:54:20.070 00:54:20.080 something about the the sizing it we say
00:54:27.690 00:54:27.700 no what's we're going to do is the water
00:54:30.090 00:54:30.100 flow rate it's gonna change it used to
00:54:32.970 00:54:32.980 be 150 used to be 150 pounds per minute
00:54:36.870 00:54:36.880 we're now going to change it to 110
00:54:40.050 00:54:40.060 pounds per minute what did the mass flow
00:54:43.080 00:54:43.090 rate of the water do went down now right
00:54:50.220 00:54:50.230 away the temperature of the water going
00:54:52.860 00:54:52.870 out we have to relax the temperature of
00:54:56.010 00:54:56.020 the oil going out we have to relax
00:54:58.560 00:54:58.570 because hey that we just changed an
00:55:01.290 00:55:01.300 input in the real system I change one of
00:55:03.630 00:55:03.640 those the mass flow rate of the cold
00:55:05.220 00:55:05.230 fluid
00:55:07.579 00:55:07.589 the area of the size has not changed so
00:55:10.819 00:55:10.829 I'm gonna stay with my area to be 274
00:55:15.049 00:55:15.059 foot square
00:55:15.859 00:55:15.869 this is really just like you might think
00:55:17.509 00:55:17.519 hey I sized that piece of equipment and
00:55:19.670 00:55:19.680 you guys in the plant you just changed
00:55:21.799 00:55:21.809 the flow rates of course now our outlet
00:55:25.459 00:55:25.469 temperatures are different than what I
00:55:26.989 00:55:26.999 told you they would be because you
00:55:29.299 00:55:29.309 changed my Inlet flow rates you've got
00:55:31.099 00:55:31.109 to put it back to where I told you right
00:55:33.400 00:55:33.410 something like that you can see this
00:55:35.749 00:55:35.759 dialog playing out well before I go
00:55:38.509 00:55:38.519 through the brute force calculation of
00:55:40.670 00:55:40.680 what is the new exit water temperature
00:55:43.069 00:55:43.079 the temperature of the cold fluid outlet
00:55:47.449 00:55:47.459 I want to ask you a few questions don't
00:55:51.439 00:55:51.449 get psyched out but these are the type
00:55:55.130 00:55:55.140 of questions so it's really hard to
00:55:56.749 00:55:56.759 answer given that problem just set it up
00:56:00.170 00:56:00.180 all we did was same size we got to size
00:56:02.749 00:56:02.759 now somebody's reduced the flow rate of
00:56:05.179 00:56:05.189 the cold fluid the water through the
00:56:07.130 00:56:07.140 system will cue change will the
00:56:11.569 00:56:11.579 temperature the cold on the outlet
00:56:13.130 00:56:13.140 change will the temperature of the hot
00:56:14.929 00:56:14.939 on outlet change will the effectiveness
00:56:17.299 00:56:17.309 change will the number transfer units
00:56:19.099 00:56:19.109 change and we'll see ratio change you're
00:56:21.439 00:56:21.449 getting from very physical to abstract
00:56:23.419 00:56:23.429 aren't we let's handle the Q first huh
00:56:29.630 00:56:29.640 or which one is easiest the temperature
00:56:32.630 00:56:32.640 the cold fluid out is probably the
00:56:34.429 00:56:34.439 easiest it used to come in at 60 and go
00:56:39.650 00:56:39.660 out at 140 but this was now changed to
00:56:42.799 00:56:42.809 one one oh what happens to the outlet
00:56:46.459 00:56:46.469 temperature of the cold fluid it was 140
00:56:49.549 00:56:49.559 is it going to go up is it going to be
00:56:51.499 00:56:51.509 now 150 is it going to go down or is it
00:56:55.609 00:56:55.619 going to remain the same the temperature
00:56:57.589 00:56:57.599 of the cold fluid on the outlet
00:57:05.010 00:57:05.020 you say up up up up all right so let me
00:57:13.510 00:57:13.520 see if I have the numbers here it'll go
00:57:19.120 00:57:19.130 up to 164 degrees F all right now that
00:57:29.080 00:57:29.090 we have that one correctly answered
00:57:30.790 00:57:30.800 let's take a look at this one the
00:57:33.040 00:57:33.050 temperature the hot fluid on the outlet
00:57:34.840 00:57:34.850 it was 80 it was 80 how about if I do
00:57:40.000 00:57:40.010 this and so I want to really get
00:57:44.140 00:57:44.150 everybody to answer this question so I'm
00:57:46.960 00:57:46.970 gonna pause and walk around well maybe
00:57:52.360 00:57:52.370 this wasn't the easiest question isn't
00:57:54.280 00:57:54.290 this not possibly a little easier let's
00:57:58.600 00:57:58.610 go solve this problem so the water went
00:58:04.030 00:58:04.040 from 60 to 140 but now it went from 60
00:58:09.460 00:58:09.470 to 100 64 degrees F this is the new case
00:58:13.000 00:58:13.010 because of the lower flow rate what did
00:58:15.460 00:58:15.470 the queue change because isn't Q equal
00:58:21.220 00:58:21.230 to the heat capacity rate of the cold
00:58:23.230 00:58:23.240 fluid times the temperature change of
00:58:25.120 00:58:25.130 the cold fluid and isn't the temperature
00:58:27.070 00:58:27.080 change greater than it was before isn't
00:58:30.220 00:58:30.230 it also Q is equal to the heat capacity
00:58:34.060 00:58:34.070 flow rate at hot fluid times the
00:58:35.590 00:58:35.600 temperature change it hot fluid now that
00:58:38.200 00:58:38.210 it came in 80 and it used to go out I'm
00:58:42.130 00:58:42.140 sorry it came in 240 and it used to go
00:58:45.280 00:58:45.290 out 80 it's still coming in 240 so this
00:58:51.040 00:58:51.050 has to go lower than 80 did the heat
00:58:55.000 00:58:55.010 capacity rate of the cold fluid change
00:58:58.050 00:58:58.060 going down so what happened to the
00:59:05.170 00:59:05.180 actual Q this went up yeah but this went
00:59:09.520 00:59:09.530 down more
00:59:13.349 00:59:13.359 and so you almost have to slug through
00:59:15.479 00:59:15.489 the numbers instead of getting 12,000
00:59:17.759 00:59:17.769 beat to use per minute you only got 11
00:59:20.959 00:59:20.969 457b tears per minute and if you only
00:59:26.430 00:59:26.440 get 11 500 instead of 1200 for 12,000
00:59:31.049 00:59:31.059 what happens here it comes out at 87
00:59:38.660 00:59:38.670 degrees F alright alright how about the
00:59:45.569 00:59:45.579 effectiveness it used to be almost 90%
00:59:50.539 00:59:50.549 well you have to work with these things
00:59:53.009 00:59:53.019 and it goes down to about 85% and then
00:59:57.359 00:59:57.369 what about the number of transfer units
00:59:59.130 00:59:59.140 it was 3.2 - well what's the definition
01:00:02.339 01:00:02.349 a number of transfer units you a over C
01:00:06.029 01:00:06.039 min did the length of the heat exchange
01:00:08.519 01:00:08.529 of change or the area change no that you
01:00:11.729 01:00:11.739 change no same you how about the minimum
01:00:14.910 01:00:14.920 heat capacity well the minimum was
01:00:16.709 01:00:16.719 either from the cold fluid or the hot
01:00:19.949 01:00:19.959 fluid the hot fluid last time gave us
01:00:23.039 01:00:23.049 the minimum true it was 75 the cold
01:00:26.279 01:00:26.289 fluid was 150 but with the new flow rate
01:00:30.059 01:00:30.069 it's a hundred and ten which one's the
01:00:34.349 01:00:34.359 minimum it's still the hot so guess what
01:00:39.420 01:00:39.430 did not change the number of transfer
01:00:44.039 01:00:44.049 units did not change and then what about
01:00:48.150 01:00:48.160 the ratio of heat capacity rates
01:00:50.099 01:00:50.109 it was 0.5 0.5 was the 75 over 150 but
01:00:57.390 01:00:57.400 now it's 75 over 1 100 what's happened
01:01:00.689 01:01:00.699 to the heat capacity rate nudged up a
01:01:05.699 01:01:05.709 little bit hidden it so it increased
01:01:10.240 01:01:10.250 here it is on plots because these are
01:01:12.790 01:01:12.800 the equations we were using and they're
01:01:14.740 01:01:14.750 pretty abstract but here is in a plot we
01:01:19.150 01:01:19.160 had the first case in light blue where
01:01:21.310 01:01:21.320 it was three point two two for the
01:01:23.290 01:01:23.300 number of transfer units and the ratio
01:01:26.110 01:01:26.120 of men to maximum was 0.5 so there's the
01:01:29.590 01:01:29.600 curve of 0.5 it's right where that blue
01:01:31.930 01:01:31.940 dot is we read off defectiveness it's
01:01:35.410 01:01:35.420 right around 90% a little bit low lower
01:01:37.660 01:01:37.670 than 90% 89% right we changed it what
01:01:42.850 01:01:42.860 happened we slowed down the cold fluid
01:01:46.650 01:01:46.660 the ratio of heat capacity rates went 1
01:01:50.380 01:01:50.390 110 whoops 75 over 1 110 which goes a
01:01:55.960 01:01:55.970 little bit higher almost point 7 well
01:01:57.850 01:01:57.860 they don't have 0.7 so you interpolate
01:02:00.010 01:02:00.020 between the line of 0.5 and point 7 5 so
01:02:03.460 01:02:03.470 there's my interpolation for the heat
01:02:06.460 01:02:06.470 capacity rate of 0.68 the same number of
01:02:09.850 01:02:09.860 transfer units what the defectiveness
01:02:11.560 01:02:11.570 dude dropped a little so you can see it
01:02:15.310 01:02:15.320 on plots as well as work it out in
01:02:18.400 01:02:18.410 equations okay I think I'm done with my
01:02:22.450 01:02:22.460 time sorry I didn't slug through that
01:02:24.700 01:02:24.710 one numerically but I gave you the
01:02:27.310 01:02:27.320 answers did I not so the answer is it's
01:02:32.140 01:02:32.150 that's the change that's the change it
01:02:35.740 01:02:35.750 goes from 80 to 87 degrees F goes from
01:02:40.390 01:02:40.400 89 to 9 85 percent stays the same no
01:02:44.770 01:02:44.780 change and this one changes from point 5
01:02:49.630 01:02:49.640 to what point six what does I say eight
01:02:53.260 01:02:53.270 point six eight point six eight thank
01:02:56.860 01:02:56.870 you very much
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