Lesson 7.4 The Heat Exchanger

WEBVTT
Kind: captions
Language: en

00:00:00.060
so this time we're gonna be talking
00:00:01.550 00:00:01.560 about heat exchangers now heat changes
00:00:05.420 00:00:05.430 are actually very similar to evaporators
00:00:08.210 00:00:08.220 and condensers in that all we have to
00:00:13.220 00:00:13.230 worry about is heat transfer so what
00:00:16.550 00:00:16.560 heat sugar does is literally move heat
00:00:19.070 00:00:19.080 from one substance to another so we're
00:00:22.250 00:00:22.260 gonna have a hot something substance
00:00:23.840 00:00:23.850 losing heat to a cooler substance which
00:00:26.330 00:00:26.340 gains it so let's say we have a pipe of
00:00:32.840 00:00:32.850 steam and then we it comes in contact
00:00:38.660 00:00:38.670 with another pipe of liquid nitrogen
00:00:44.170 00:00:44.180 now obviously steam is very very hot
00:00:46.880 00:00:46.890 whereas nut'n liquid nitrogen is very
00:00:49.190 00:00:49.200 very cold so he's going to want to go
00:00:51.590 00:00:51.600 from the water to the nitrogen it's
00:00:54.889 00:00:54.899 going to be a heat transfer here so it's
00:00:56.959 00:00:56.969 going to be a heat loss of the water is
00:00:59.779 00:00:59.789 going to be heat gain of nitrogen now
00:01:05.090 00:01:05.100 these losses are going to be equal the
00:01:07.100 00:01:07.110 opposite we're going to assume that
00:01:08.899 00:01:08.909 there's going to be no other ways for a
00:01:12.050 00:01:12.060 key to be exchanged like there's no vent
00:01:14.179 00:01:14.189 this these are well insulated they're
00:01:15.920 00:01:15.930 not going to leak go out or be absorbed
00:01:17.990 00:01:18.000 by the pipes so this heat transfer from
00:01:21.200 00:01:21.210 here to here is going to be equal the
00:01:23.810 00:01:23.820 opposite so positive here negative here
00:01:27.950 00:01:27.960 but they're still going to be have the
00:01:29.660 00:01:29.670 same absolute value so what we could say
00:01:32.030 00:01:32.040 is key n + e l0 conservation of energy
00:01:42.130 00:01:42.140 so Q in which in this case is nitrogen
00:01:50.240 00:01:50.250 because it's getting a Q it is going to
00:01:52.280 00:01:52.290 equal the mass flow rate of nitrogen
00:01:55.630 00:01:55.640 times h2 minus h1 oh these are of
00:02:02.330 00:02:02.340 mention then Q out what if I get these
00:02:08.600 00:02:08.610 little dots
00:02:12.800 00:02:12.810 h2o in this case not always this Q s
00:02:17.640 00:02:17.650 maybe h2o + 12 X 2 minus H 1 H 2 O now
00:02:29.030 00:02:29.040 notice we have six different numbers
00:02:31.649 00:02:31.659 here we have mass flow rate of nitrogen
00:02:34.740 00:02:34.750 mass flow rate of h2o these are two
00:02:37.380 00:02:37.390 different flow rates they are not
00:02:38.520 00:02:38.530 flowing in the same pipe so they can go
00:02:40.199 00:02:40.209 at different rates and then we have the
00:02:42.330 00:02:42.340 state one to state two both these
00:02:45.599 00:02:45.609 substances so we have six things to
00:02:48.509 00:02:48.519 worry about here so how would this look
00:02:51.809 00:02:51.819 in practice well let's say we have a big
00:02:58.920 00:02:58.930 ol heat exchangers this is a pipe cross
00:03:03.390 00:03:03.400 section of a pipe where this refrigerant
00:03:07.309 00:03:07.319 r134a going in and out but as it's going
00:03:11.910 00:03:11.920 through this part of the pipe it's
00:03:13.500 00:03:13.510 running over other pipes that are
00:03:17.069 00:03:17.079 running perpendicular so it's going into
00:03:19.110 00:03:19.120 the tape piece of paper of more pipes
00:03:22.500 00:03:22.510 and in this is cooling water
00:03:31.830 00:03:31.840 so we have gashes r134a running through
00:03:36.960 00:03:36.970 these pipes and it's going to be flowing
00:03:39.690 00:03:39.700 over cool what cool pipes that is water
00:03:42.930 00:03:42.940 flowing through it and on the other side
00:03:46.710 00:03:46.720 is going to come out as a liquid so uh
00:03:54.770 00:03:54.780 let's see we have to give you a few a
00:03:58.350 00:03:58.360 few notes we have for the refrigerant we
00:04:01.979 00:04:01.989 know that it has a mass flow rate of one
00:04:05.130 00:04:05.140 kilogram per second
00:04:07.699 00:04:07.709 we know that at state one as it comes in
00:04:15.080 00:04:15.090 has a pressure of one mega Pascal we
00:04:23.280 00:04:23.290 know that has a temperature of 60
00:04:25.200 00:04:25.210 degrees Celsius
00:04:27.440 00:04:27.450 no one stake to it has a pressure of 880
00:04:37.880 00:04:37.890 7.6 kPa we know has a temperature of 35
00:04:45.690 00:04:45.700 degrees Celsius and we also know that it
00:04:48.900 00:04:48.910 is a liquid and for water
00:05:02.879 00:05:02.889 so you know he won it's been equal e to
00:05:09.039 00:05:09.049 which equals to mega pascals
00:05:12.219 00:05:12.229 that's something I gave you pressure
00:05:13.749 00:05:13.759 doesn't have to stay the same but since
00:05:17.709 00:05:17.719 it's got to stay water throughout it's a
00:05:19.719 00:05:19.729 good assumption let me know it goes from
00:05:22.269 00:05:22.279 20 degrees Celsius to 40 degrees Celsius
00:05:31.409 00:05:31.419 and what this problem is asking is I'm a
00:05:34.209 00:05:34.219 mass flow rate of water right so we're
00:05:40.809 00:05:40.819 going to use this equation here in plus
00:05:43.299 00:05:43.309 Q out equals 0 conservation of energy so
00:05:47.169 00:05:47.179 Q in that's going to be water so he wins
00:05:51.219 00:05:51.229 is going to equal mass flow rate of
00:05:52.989 00:05:52.999 water times h2 water minus h1 of water
00:06:02.279 00:06:02.289 plus mass flow rate of the refrigerant
00:06:08.759 00:06:08.769 times h2 minus h1 so we need no no all
00:06:19.239 00:06:19.249 these enthalpies fairly straightforward
00:06:22.749 00:06:22.759 looking the trap in the tables we know
00:06:26.639 00:06:26.649 each one of the refrigerant at 1
00:06:30.069 00:06:30.079 megapascal and 60 degrees Celsius is 400
00:06:37.059 00:06:37.069 for 440 1.89 kilojoules per kilogram and
00:06:45.489 00:06:45.499 we know that h2 is a liquid at least at
00:06:50.409 00:06:50.419 this state of pressure and temperature
00:06:52.569 00:06:52.579 it's a it's somewhere along saturation
00:06:56.949 00:06:56.959 line so we have we know that it says
00:07:00.609 00:07:00.619 when it's liquid we just take the liquid
00:07:02.889 00:07:02.899 value from the table which is 249 point
00:07:06.819 00:07:06.829 10 kilojoules per kilogram
00:07:12.170 00:07:12.180 ah for water and to megapascals 20
00:07:15.920 00:07:15.930 Celsius that's super cool you go through
00:07:18.050 00:07:18.060 the table and we know that it's enthalpy
00:07:24.610 00:07:24.620 is a three point eight - coming in at
00:07:30.770 00:07:30.780 four degrees it's one sixty seven point
00:07:34.880 00:07:34.890 two nine so like like for the last video
00:07:38.540 00:07:38.550 we know we noticed a gain in the water
00:07:41.660 00:07:41.670 these heats coming into the water we
00:07:43.670 00:07:43.680 notice a loss in the refrigerant B's is
00:07:46.910 00:07:46.920 cooling down so mass flow rate of water
00:07:56.860 00:07:56.870 is going to equal ms flow rate the
00:08:01.940 00:08:01.950 refrigerant times h2 minus h1 or h2
00:08:12.400 00:08:12.410 minus h1 refrigerant water plug and chug
00:08:20.030 00:08:20.040 two equals one program plus second times
00:08:36.240 00:08:36.250 right equals zero so there's actually
00:08:39.569 00:08:39.579 going to be a negative side of it
00:08:40.949 00:08:40.959 because we're going to have to subtract
00:08:42.649 00:08:42.659 over this over so we can divide void
00:08:49.619 00:08:49.629 equal to 4 9.1 - 4 4 1.9 all over one
00:09:02.129 00:09:02.139 six seven point two nine - a three point
00:09:07.050 00:09:07.060 eight two negative sign which equals two
00:09:14.910 00:09:14.920 point three one kilograms assign
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