00:00:00.620 in this simple heat exchanger example 00:00:04.10000:00:04.110 the hardest part is understanding the 00:00:05.84000:00:05.850 Givens and the drawing we are told that 00:00:09.70900:00:09.719 a hot a stream which is the stream over 00:00:16.07000:00:16.080 here is used to heat cold a which is the 00:00:20.54000:00:20.550 stream over here in a heat exchanger and 00:00:22.82000:00:22.830 we are given this drawing so the first 00:00:26.12000:00:26.130 thing that we need to understand is that 00:00:29.14000:00:29.150 we are expected to know that the heat 00:00:31.91000:00:31.920 exchanger will be partitioned in this 00:00:33.59000:00:33.600 way so that there is a hot side and a 00:00:38.86900:00:38.879 cold side and that these parts do not 00:00:42.11000:00:42.120 communicate material between one another 00:00:44.60000:00:44.610 but that there will be an energy 00:00:48.25000:00:48.260 transfer between these two sections we 00:00:55.70000:00:55.710 are told that we may assume that CPS are 00:00:58.79000:00:58.800 constant independent of the state of the 00:01:01.72900:01:01.739 a and we are told that the flow of the 00:01:07.25000:01:07.260 hottie is 50 percent of the flow of the 00:01:10.52000:01:10.530 cold e so let's give ourselves some some 00:01:16.13000:01:16.140 names for these things let's say this is 00:01:18.98000:01:18.990 a B C and D and let's start writing down 00:01:26.46900:01:26.479 the equations that we have so this 00:01:30.46000:01:30.470 problem does not require us to know the 00:01:34.81900:01:34.829 value of the CP of a since we are 00:01:39.02000:01:39.030 exchanging materials between one another 00:01:41.26000:01:41.270 we don't have to go and look that up we 00:01:45.38000:01:45.390 can relatively easily see that there 00:01:48.59000:01:48.600 will be two different ways of 00:01:51.02000:01:51.030 approaching this problem the first way 00:01:55.71900:01:55.729 we will simply do a total energy balance 00:01:59.63000:01:59.640 and that will mean that we have in 00:02:09.02000:02:09.030 equals art 00:02:10.80000:02:10.810 and we will therefore have the enthalpy 00:02:14.78000:02:14.790 in a plus the enthalpy in C equal to the 00:02:28.38000:02:28.390 out streams QB + QD for all four of 00:02:37.05000:02:37.060 these we may write a equation of the 00:02:41.72900:02:41.739 form Qi is equal to MI CP bar I and then 00:02:50.72900:02:50.739 we can use the temperature of the stream 00:02:55.13000:02:55.140 minus some arbitrary reference now we 00:03:01.11000:03:01.120 are told that the rate at which these 00:03:06.71000:03:06.720 cold air streams is fed is twice the 00:03:10.71000:03:10.720 rate at which the hot air flows and we 00:03:14.69900:03:14.709 can assume that the flow rate throughout 00:03:18.56900:03:18.579 each section is constant so a is going 00:03:22.56000:03:22.570 to be equal to B a is going to be equal 00:03:26.64000:03:26.650 to twice C and C is going to be equal to 00:03:30.15000:03:30.160 D so if we say that in mass terms we 00:03:35.61000:03:35.620 will say M a is equal to M B M D is 00:03:43.89000:03:43.900 equal to M C and M a equal to twice M C 00:03:53.59900:03:53.609 now we can reason that since the CP is 00:03:56.55000:03:56.560 are all the same they will end up 00:03:59.47900:03:59.489 dividing out of all of these terms so 00:04:04.41000:04:04.420 we've got four Q terms they all contain 00:04:07.14000:04:07.150 the same CP term if we if we were to 00:04:10.08000:04:10.090 substitute them in they would all cancel 00:04:11.88000:04:11.890 out we can also see that all of the 00:04:17.37000:04:17.380 masses will end up being written in 00:04:19.58900:04:19.599 terms of M C 00:04:22.28000:04:22.290 in other words for a and B we will write 00:04:27.37000:04:27.380 to MC and for D and C we will write just 00:04:33.44000:04:33.450 MC so let's do that substitution clearly 00:04:38.81000:04:38.820 see the CPS are going to cancel out of 00:04:42.74000:04:42.750 these equations and that all of these 00:04:46.66000:04:46.670 masses will end up becoming in terms of 00:04:51.20000:04:51.210 MC once we've done that substitution we 00:04:58.97000:04:58.980 find that MC also cancels out of these 00:05:03.17000:05:03.180 equations multiplying out makes it clear 00:05:08.12000:05:08.130 that we have the same number of TRS on 00:05:12.11000:05:12.120 both sides of those equation so this one 00:05:15.62000:05:15.630 cancels that one and this one cancels 00:05:18.65000:05:18.660 that one we are left with a simple 00:05:22.34000:05:22.350 equation this is a satisfying result as 00:05:26.93000:05:26.940 we can interpret this as saying the 00:05:30.04000:05:30.050 temperature of B this final temperature 00:05:33.35000:05:33.360 will be wherever that started plus the 00:05:37.37000:05:37.380 energy transferred from the hot stream 00:05:40.61000:05:40.620 which is flowing at half the rate of the 00:05:43.94000:05:43.950 cold stream another way of solving this 00:05:49.79000:05:49.800 problem is to reason through the uptake 00:05:54.56000:05:54.570 and exchange of energy so we can say 00:05:57.28000:05:57.290 that Q is the energy transferred from 00:06:10.15000:06:10.160 the hot stream into the cold stream and 00:06:14.68000:06:14.690 so that must be equal on both sides to 00:06:19.33000:06:19.340 on the one hand it must be equal to what 00:06:23.09000:06:23.100 we previously defined as Q C minus Q D 00:06:29.12000:06:29.130 and it must also be equal to 00:06:34.64000:06:34.650 q b- q a since both of these terms are 00:06:45.24000:06:45.250 differences in the same stream it 00:06:49.26000:06:49.270 becomes a lot easier to write down the 00:06:51.81000:06:51.820 energy difference there are similar 00:06:57.27000:06:57.280 cancellations as before and we are a 00:07:02.61000:07:02.620 small set of manipulations away from the 00:07:05.31000:07:05.320 same answer we find unsurprisingly the 00:07:10.11000:07:10.120 same relationship and just for 00:07:13.23000:07:13.240 completeness when you plug in all those 00:07:15.42000:07:15.430 numbers you get 220 degrees Celsius
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