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Thermodynamics - Steady Flow Energy Balance (1st Law), Heat Exchanger
WEBVTT Kind: captions Language: en
00:00:00.500 so this problem says that we have 00:00:03.24900:00:03.259 refrigerant-134a it's one well basically 00:00:06.86000:00:06.870 let's just say what the problem is doing 00:00:08.69000:00:08.700 so we have refrigerant-134a that's 00:00:11.53000:00:11.540 flowing through so it's flowing through 00:00:15.65000:00:15.660 a pipe that is basically being cooled by 00:00:21.14000:00:21.150 air so basically we have a heat 00:00:22.82000:00:22.830 exchanger and I'm gonna draw so this is 00:00:25.82000:00:25.830 let's just say that this is like a dr. 00:00:30.58900:00:30.599 something that has air flowing through 00:00:33.04900:00:33.059 it and then we have this pipe with the 00:00:38.78000:00:38.790 refrigerant flow it that goes through 00:00:43.04000:00:43.050 this duct or whatever it is so we have 00:00:49.79000:00:49.800 refrigerant flowing through the smaller 00:00:55.31000:00:55.320 pipe and then we have air flowing 00:00:57.97900:00:57.989 through this duct and so basically this 00:01:00.13900:01:00.149 refrigerant is flowing through here and 00:01:05.50000:01:05.510 as it flows through there it's being 00:01:07.94000:01:07.950 cooled by air so we have a heat 00:01:11.48000:01:11.490 exchanger and then it gives us some 00:01:14.30000:01:14.310 property data for the inlets and outlets 00:01:17.35900:01:17.369 of the air and the refrigerants I'm just 00:01:20.21000:01:20.220 gonna write that down and this is the 00:01:22.42900:01:22.439 part of the problem that I'm calling the 00:01:24.02000:01:24.030 set up and you want to do this like 00:01:27.17000:01:27.180 you'll want to set up all of your 00:01:28.55000:01:28.560 problems just so you have a really clear 00:01:30.77000:01:30.780 idea of what's going on and and you know 00:01:33.95000:01:33.960 how you're going to solve the problem so 00:01:37.06900:01:37.079 first of all it says that the 00:01:38.59000:01:38.600 refrigerant-134a at one mega Pascal and 00:01:44.63000:01:44.640 90 degrees C has to be cooled so that 00:01:47.06000:01:47.070 means that at the inlet let's just make 00:01:49.91000:01:49.920 this more clear so at the inlet the 00:01:53.17900:01:53.189 pressure of the refrigerant and I'm just 00:01:55.81900:01:55.829 gonna call it R for refrigerant so p1 R 00:02:00.24900:02:00.259 so one is for the inlet and then this 00:02:03.70900:02:03.719 refrigerant so if P is one mega Pascal 00:02:07.96900:02:07.979 and then t1 R is equal to 90 degrees 00:02:13.88000:02:13.890 and you can call these variables 00:02:16.97000:02:16.980 whatever you want just make sure it's 00:02:18.68000:02:18.690 something that makes sense otherwise the 00:02:23.06000:02:23.070 problem gets kind of confusing okay so 00:02:25.49000:02:25.500 we have the inlet for the refrigerant 00:02:27.50000:02:27.510 and it says that it's cooled to one mega 00:02:29.87000:02:29.880 Pascal in 30 degrees C so at the outlet 00:02:32.78000:02:32.790 we have P to R is equal to so the 00:02:37.73000:02:37.740 pressure doesn't change but the 00:02:39.08000:02:39.090 temperature is lower so T to R is equal 00:02:45.80000:02:45.810 to 27 no 30 degrees C okay and then it 00:02:52.61000:02:52.620 says the air enters at 100 kiloPascals 00:02:56.54000:02:56.550 so we have the air and I'm going to just 00:03:00.74000:03:00.750 call this P 1 air so I'm going to use 00:03:05.15000:03:05.160 consistent terminology so the air is at 00:03:10.16000:03:10.170 100 kiloPascals and the temperature is 00:03:16.37000:03:16.380 21 degrees so T 1 air is equal there 00:03:19.97000:03:19.980 sorry 27 degrees C so the air enters at 00:03:24.56000:03:24.570 100 kiloPascals and 27 degrees C it 00:03:28.52000:03:28.530 tells us the volume flow rate so we have 00:03:30.97900:03:30.989 the one air is equal to 600 meters cubed 00:03:38.93000:03:38.940 per minute so just don't let this 00:03:42.05000:03:42.060 confuse you the volume the volumetric 00:03:44.12000:03:44.130 flow rate there's an conserved property 00:03:46.64000:03:46.650 so the volumetric flow rate is likely 00:03:49.55000:03:49.560 going to change between the inlet and 00:03:53.72000:03:53.730 outlet of the air just because air is 00:03:56.09000:03:56.100 compressible so don't don't think that 00:04:00.38000:04:00.390 the volume volumetric flow rate is 600 00:04:03.56000:04:03.570 meters cubed per minute at the inlet for 00:04:06.92000:04:06.930 the air it's probably not equal to 600 00:04:10.91000:04:10.920 meters cubed per minute at the outlet 00:04:12.38000:04:12.390 okay and then it says that the air 00:04:14.96000:04:14.970 leaves at 95 kiloPascals so P 2a is 00:04:19.96900:04:19.979 equal to 100 kilopascals 00:04:23.71900:04:23.729 and I'm sorry 95 kiloPascals 00:04:27.82000:04:27.830 and sixty degrees C so T 2a is equal to 00:04:32.96000:04:32.970 sixty degrees C and that says determine 00:04:36.86000:04:36.870 the mass flow rate of the refrigerant so 00:04:39.68000:04:39.690 we're looking for the mass flow rate of 00:04:44.47000:04:44.480 the refrigerant and so that will bring 00:04:48.08000:04:48.090 us into our assumptions because we're 00:04:50.36000:04:50.370 going to assume that this is a steady 00:04:52.10000:04:52.110 flow problem her steady flow process so 00:04:56.87000:04:56.880 if you think about if you think about it 00:04:59.39000:04:59.400 the refrigerant and the air both have to 00:05:01.79000:05:01.800 be flowing as steady flow otherwise you 00:05:04.94000:05:04.950 would have air or refrigerant 00:05:06.38000:05:06.390 accumulating or disappearing from your 00:05:10.46000:05:10.470 from your system so say this is our 00:05:14.66000:05:14.670 system so basically if this were in 00:05:20.96000:05:20.970 startup or shutdown that might be the 00:05:23.03000:05:23.040 case but since it's not in startup or 00:05:25.40000:05:25.410 shutdown we're assuming it's been 00:05:27.65000:05:27.660 operating for a while we're going to 00:05:29.75000:05:29.760 assume that the mass is the mass of the 00:05:32.18000:05:32.190 refrigerant isn't changing in with time 00:05:34.91000:05:34.920 inside the side of control volume and 00:05:38.18000:05:38.190 the mass of the air isn't changing with 00:05:39.83000:05:39.840 time inside the control volume so what 00:05:43.16000:05:43.170 that means is that we're going to assume 00:05:46.97000:05:46.980 that the mass of the refrigerant in the 00:05:52.28000:05:52.290 mass flow rate in is equal to the mass 00:05:54.14000:05:54.150 flow rate of the refrigerant out so then 00:05:57.77000:05:57.780 this is just equal to the mass flow rate 00:05:59.48000:05:59.490 of the refrigerant which is what we're 00:06:02.15000:06:02.160 looking for and we can make the same 00:06:04.52000:06:04.530 assumption with the air so if we assume 00:06:06.47000:06:06.480 this is steady flow we're going to say 00:06:09.92000:06:09.930 that the mass flow rate of the air in is 00:06:13.10000:06:13.110 equal to the mass flow rate of the air 00:06:15.89000:06:15.900 out so we just have the mass flow rate 00:06:17.75000:06:17.760 of the air and this is where you don't 00:06:21.89000:06:21.900 want to get confused with those 00:06:23.27000:06:23.280 volumetric flow rate the mass is 00:06:26.15000:06:26.160 conserved like we know that mass is a 00:06:28.16000:06:28.170 conserved quantity so if this is steady 00:06:30.77000:06:30.780 flow that means that the change in the 00:06:34.43000:06:34.440 change of mass inside the system with 00:06:36.83000:06:36.840 time has to be equal to zero 00:06:38.96000:06:38.970 but 00:06:40.04000:06:40.050 the volumetric flow-rate like volume 00:06:42.32000:06:42.330 isn't a conserved quantity so the 00:06:45.32000:06:45.330 volumetric flow rate is going to change 00:06:47.45000:06:47.460 unless you make an assumption that 00:06:49.36900:06:49.379 you're dealing with an incompressible 00:06:51.08000:06:51.090 substance and air is definitely 00:06:53.37900:06:53.389 compressible so we can't make that 00:06:56.21000:06:56.220 assumption all right so we have some 00:06:59.83900:06:59.849 assumptions let's make a few more so 00:07:01.96900:07:01.979 basically what these assumptions we're 00:07:04.70000:07:04.710 assuming steady flow I'm also going to 00:07:09.95000:07:09.960 say so the air is ideal gas it's not a 00:07:13.96900:07:13.979 pretty low pressure and high temperature 00:07:15.49900:07:15.509 compared to the critical pressure and 00:07:18.08000:07:18.090 temperature for air so air is ideal and 00:07:22.18900:07:22.199 that means that we can assume constant 00:07:25.02900:07:25.039 specific heat and we're also going to 00:07:32.30000:07:32.310 assume well the change in potential 00:07:34.24900:07:34.259 energy is going to be zero I think 00:07:36.05000:07:36.060 that's pretty clear there's not going to 00:07:38.39000:07:38.400 be a large elevation change over this 00:07:40.33900:07:40.349 heat exchanger even if there is we 00:07:43.10000:07:43.110 haven't been given any data about it so 00:07:46.54000:07:46.550 it's we have to basically assume that 00:07:49.21900:07:49.229 zero most heat exchangers aren't going 00:07:52.79000:07:52.800 to have a huge elevation change but if 00:07:55.21900:07:55.229 you have one that happens to have a 00:07:56.65900:07:56.669 large elevation change you might need to 00:07:59.26900:07:59.279 consider the potential energy so just 00:08:01.01000:08:01.020 make sure you read the problem statement 00:08:02.80900:08:02.819 carefully or in real life when you get 00:08:05.08900:08:05.099 into your job if you're if you're 00:08:07.37000:08:07.380 analyzing a heat exchanger that's like 00:08:09.62000:08:09.630 several meters tall or something like 00:08:11.33000:08:11.340 that then you might want to consider the 00:08:14.17000:08:14.180 potential potential energy and also 00:08:18.95000:08:18.960 we're going to assume that the change in 00:08:21.86000:08:21.870 kinetic energy is zero that one's the 00:08:24.32000:08:24.330 low less obvious because we have flowing 00:08:26.39000:08:26.400 fluids in this like wall is the velocity 00:08:28.99900:08:29.009 changing well it could be but the thing 00:08:32.20900:08:32.219 with heat exchangers is the velocities 00:08:34.43000:08:34.440 they're usually pretty slow otherwise 00:08:36.70900:08:36.719 like if you think about it if the 00:08:38.56900:08:38.579 velocities through this heat exchanger 00:08:39.86000:08:39.870 were really fast there wouldn't be 00:08:42.31900:08:42.329 enough time for much heat transfer 00:08:44.90000:08:44.910 between the refrigerant and the air so 00:08:47.87000:08:47.880 we actually want these to move pretty 00:08:49.28000:08:49.290 slow so that there's time for maximum 00:08:52.69900:08:52.709 heat transfer to a 00:08:54.17000:08:54.180 and they're also going to be relatively 00:08:57.11000:08:57.120 unchanging and so we can assume that the 00:09:00.86000:09:00.870 kinetic energy is approximately zero if 00:09:05.06000:09:05.070 you're given and the other thing is 00:09:06.68000:09:06.690 we're not given any information in the 00:09:08.42000:09:08.430 problem about the velocities so even if 00:09:11.84000:09:11.850 we wanted to calculate the change in 00:09:13.43000:09:13.440 kinetic energy we don't have the 00:09:14.75000:09:14.760 information to do it so this is another 00:09:17.63000:09:17.640 thing to pay attention to you could 00:09:19.16000:09:19.170 potentially have a change in kinetic 00:09:21.05000:09:21.060 energy but you would need to know what 00:09:23.96000:09:23.970 the velocities are so if your problem 00:09:26.66000:09:26.670 gives you information about velocities 00:09:28.34000:09:28.350 that's something you might need to 00:09:30.35000:09:30.360 consider depending on what the 00:09:31.97000:09:31.980 velocities are or in real-life you're 00:09:34.58000:09:34.590 likely going to know what the velocities 00:09:38.63000:09:38.640 are of your fluid flowing through the 00:09:40.52000:09:40.530 heat exchanger so then you can get an 00:09:42.83000:09:42.840 idea of whether or not you need to 00:09:45.29000:09:45.300 consider the change in potential energy 00:09:48.37000:09:48.380 though there's no work so heat 00:09:51.32000:09:51.330 exchangers are passive devices and it 00:09:56.03000:09:56.040 also doesn't give us any information 00:09:57.20000:09:57.210 about heat transfer out of this out of 00:10:02.33000:10:02.340 the heat exchanger so basically when I 00:10:04.19000:10:04.200 say that we know there's heat transfer 00:10:05.90000:10:05.910 between the air and the refrigerant and 00:10:10.51000:10:10.520 so I mean that's pretty obvious but what 00:10:14.63000:10:14.640 we're interested in is both of them so 00:10:17.39000:10:17.400 our system is and so that our system is 00:10:23.32000:10:23.330 basically includes both pipes so it 00:10:28.19000:10:28.200 includes the pipe with the air and the 00:10:30.05000:10:30.060 pipe with the heat exchanger so if we 00:10:32.36000:10:32.370 assume that this is well insulated which 00:10:35.63000:10:35.640 is often a really good assumption with 00:10:39.11000:10:39.120 heat exchangers in real life they're 00:10:40.67000:10:40.680 often pretty well insulated we can 00:10:43.37000:10:43.380 assume that there's no heat transfer out 00:10:45.98000:10:45.990 of the system so there's no heat 00:10:48.83000:10:48.840 transfer here so Q is zero and this can 00:10:54.71000:10:54.720 be a little bit confusing because it's 00:10:56.45000:10:56.460 like well there is heat transfer here 00:10:59.06000:10:59.070 like from the refrigerant to the air but 00:11:02.44000:11:02.450 since that's inside our system we're not 00:11:05.81000:11:05.820 we don't need to consider that 00:11:07.61000:11:07.620 in our full energy balance so basically 00:11:10.46000:11:10.470 we're going to say that Q is equal to 00:11:12.98000:11:12.990 zero because they we're assuming that 00:11:15.65000:11:15.660 the heat exchanger is well insulated so 00:11:22.40000:11:22.410 basically in order to solve this problem 00:11:24.83000:11:24.840 we could theoretically solve the energy 00:11:27.95000:11:27.960 balance and get a value for Q but in 00:11:31.16000:11:31.170 this case we don't have enough 00:11:32.15000:11:32.160 information to do that we would need to 00:11:33.74000:11:33.750 know what both of the mass flow rates 00:11:36.77000:11:36.780 are we only know one really we're 00:11:40.25000:11:40.260 calculating the other one so we don't 00:11:42.62000:11:42.630 have enough information to calculate Q 00:11:45.77000:11:45.780 we weren't given Q so basically we're 00:11:48.35000:11:48.360 we're just going to assume that the heat 00:11:50.69000:11:50.700 exchanger is well insulated and there's 00:11:54.50000:11:54.510 no heat transfer to or from the heat 00:11:56.99000:11:57.000 exchanger so this means this adiabatic 00:11:59.98000:11:59.990 and so basically all adults put this - 00:12:04.10000:12:04.110 so all heat transfer occurs from the hot 00:12:11.09000:12:11.100 fluid to the cold fluid so basically all 00:12:16.61000:12:16.620 of we have heat transfer from the 00:12:19.52000:12:19.530 refrigerant to the air and we can just 00:12:22.91000:12:22.920 assume that all of the heat that's 00:12:24.77000:12:24.780 transferred out of the refrigerant is in 00:12:27.35000:12:27.360 the air like we don't have any losses 00:12:29.65000:12:29.660 outside of the system and we don't have 00:12:31.79000:12:31.800 any heat coming in so that actually 00:12:35.18000:12:35.190 greatly simplifies this analysis because 00:12:37.97000:12:37.980 we don't need to worry about Q in our 00:12:40.76000:12:40.770 energy balance so let's so I think that 00:12:45.11000:12:45.120 this is basically our problem set up we 00:12:49.67000:12:49.680 have our assumptions now let's write 00:12:52.19000:12:52.200 down the equations we need so we're 00:12:57.74000:12:57.750 going to need the first law energy 00:12:59.24000:12:59.250 balance and first steady flow and this 00:13:02.33000:13:02.340 problem is a little bit different from 00:13:03.94000:13:03.950 quite a few of the others I've been 00:13:05.90000:13:05.910 doing because with this one we have 00:13:07.55000:13:07.560 multiple inlets and outlets so basically 00:13:11.15000:13:11.160 if I redraw what I have going on here so 00:13:16.94000:13:16.950 we have an inlet here and inlet here 00:13:19.64000:13:19.650 outlet here outlet 00:13:21.29000:13:21.300 here so we have two inlets and two 00:13:23.66000:13:23.670 outlets so we need to consider our first 00:13:27.47000:13:27.480 law energy balance for so we need the 00:13:30.76900:13:30.779 steady flow equation that allows for 00:13:33.82900:13:33.839 multiple inlets and outlets and that was 00:13:36.88900:13:36.899 derived in the video that I did on the 00:13:40.50900:13:40.519 derivation for the for the first law 00:13:43.57900:13:43.589 energy balance and so go back and watch 00:13:47.84000:13:47.850 that if you want to see where this 00:13:49.69900:13:49.709 equation came from but for here I'm just 00:13:51.94900:13:51.959 going to write down the equation so we 00:13:54.31900:13:54.329 have Q minus W is equal to and then we 00:13:58.10000:13:58.110 have the sum and this is the mass out 00:14:02.92000:14:02.930 and these should be great this is and 00:14:09.07900:14:09.089 I'm gonna actually change this out to to 00:14:11.48000:14:11.490 just to simplify the writing so I'm just 00:14:14.18000:14:14.190 saying that well this is all of the out 00:14:18.13900:14:18.149 so let's just leave that like that but 00:14:20.03000:14:20.040 then I'm gonna call I'm just gonna call 00:14:24.34000:14:24.350 two here and one here so one is and two 00:14:28.69900:14:28.709 is out so this is H 2 plus V 2 squared 00:14:34.25000:14:34.260 over 2 plus G Z squared minus and then 00:14:40.43000:14:40.440 we have then then we sum all of our 00:14:42.76900:14:42.779 inlets this is MDOT and then I have H 2 00:14:48.47000:14:48.480 plus V 2 squared over 2 and actually 00:14:56.51000:14:56.520 these are 1 because this is the inlet so 00:14:59.03000:14:59.040 this is actually 1 1 and then this is 00:15:02.24000:15:02.250 over 2 plus gz1 all right so this is our 00:15:08.78000:15:08.790 general energy balance and as you 00:15:11.36000:15:11.370 probably guessed a lot of these terms 00:15:12.71000:15:12.720 are going to be 0 00:15:14.18000:15:14.190 I just wrote out the entire thing so 00:15:16.06900:15:16.079 that you know what equation to start 00:15:19.04000:15:19.050 with 00:15:19.55000:15:19.560 in fact when I'm solving these types of 00:15:21.35000:15:21.360 problems I always start with this 00:15:23.06000:15:23.070 equation and then I cancel out the terms 00:15:25.75900:15:25.769 I don't need I feel like this is much 00:15:27.98000:15:27.990 easier than trying to write down an 00:15:30.86000:15:30.870 equation for a specific case and then 00:15:33.35000:15:33.360 figuring trying to figure out if it 00:15:34.87900:15:34.889 lies in your case because the equations 00:15:38.68900:15:38.699 that are written for specific cases they 00:15:41.17900:15:41.189 have assumptions built into them and if 00:15:45.41000:15:45.420 you just write down an equation for a 00:15:47.29900:15:47.309 specific case it might have an 00:15:49.54900:15:49.559 assumption that you can't make and it 00:15:51.22900:15:51.239 might be hard to figure that out just by 00:15:53.62900:15:53.639 looking at it so I like to start with 00:15:55.75900:15:55.769 this equation this is the general energy 00:15:57.65000:15:57.660 balance for steady flow so this is so if 00:16:02.03000:16:02.040 you don't have steady flow you'll need 00:16:04.90900:16:04.919 to be starting with a even more general 00:16:07.09900:16:07.109 equation than this one this one all 00:16:10.00900:16:10.019 we've assumed that all of the time 00:16:12.04900:16:12.059 derivatives are zero because we don't 00:16:14.23900:16:14.249 have anything changing with time in our 00:16:15.82900:16:15.839 system so this is the equation you'll 00:16:19.09900:16:19.109 start with for a steady flow for 00:16:24.31900:16:24.329 multiple inlets and outlets so now let's 00:16:26.59900:16:26.609 figure out which terms are zero we've 00:16:28.69900:16:28.709 already assumed the 00:16:30.19900:16:30.209 adiabatic there's no work and we've said 00:16:33.47000:16:33.480 that all of the change in potential and 00:16:35.74900:16:35.759 kinetic energies are zero so basically I 00:16:40.00900:16:40.019 will solve the terms are zero I also 00:16:42.10900:16:42.119 like to write down these equations and 00:16:44.09000:16:44.100 then my cancel out everything I don't 00:16:45.94900:16:45.959 need it it leaves you with a much 00:16:49.12900:16:49.139 simpler equation but let's just write 00:16:53.35900:16:53.369 down this so if I'm out x h2 and using 00:16:59.47900:16:59.489 to you on these was probably a bad idea 00:17:01.60900:17:01.619 I'm just gonna change these two out so 00:17:04.75900:17:04.769 these are out this is in and then same 00:17:07.93900:17:07.949 thing on these terms just because it was 00:17:11.62900:17:11.639 probably kind of confusing to write it 00:17:13.03900:17:13.049 like that 00:17:13.63900:17:13.649 so HL and then we have minus and then 00:17:18.52900:17:18.539 the sum of the inlets so this is in this 00:17:23.38900:17:23.399 one out and then H yeah so this is the 00:17:29.18000:17:29.190 equation the this is our energy balance 00:17:31.51900:17:31.529 that we can use for this particular case 00:17:34.61000:17:34.620 of this heat exchanger and the reason 00:17:38.06000:17:38.070 why I'd like to start with this equation 00:17:40.31000:17:40.320 is because if you just write down this 00:17:42.25900:17:42.269 equation which you can but if you just 00:17:45.28900:17:45.299 write down this equation you have to 00:17:48.30000:17:48.310 and think okay what is this assuming 00:17:49.89000:17:49.900 okay this is assuming those adiabatic 00:17:51.90000:17:51.910 there's no work the potential and 00:17:54.39000:17:54.400 kinetic energies are zero steady flow 00:17:56.72000:17:56.730 there's just a lot of assumptions built 00:17:59.07000:17:59.080 into this second equation whereas this 00:18:02.19000:18:02.200 first one the only assumption built into 00:18:05.55000:18:05.560 those first equations so far is that a 00:18:07.44000:18:07.450 steady flow so it's a much more general 00:18:11.16000:18:11.170 nonspecific equation to start with and 00:18:14.22000:18:14.230 if you start with it you're going to 00:18:16.92000:18:16.930 almost always get the right fine and 00:18:19.71000:18:19.720 you're always going you're almost always 00:18:21.33000:18:21.340 going to get the correct energy balance 00:18:23.16000:18:23.170 for your particular system all right now 00:18:27.21000:18:27.220 let's let's write in our actual inlets 00:18:32.49000:18:32.500 and outlets and I'm gonna use the 00:18:34.85000:18:34.860 terminology that I had before so just 00:18:39.69000:18:39.700 draw this again the hardest part with 00:18:42.36000:18:42.370 these with these heat exchangers is to 00:18:46.05000:18:46.060 just keep track of everything so the 00:18:48.24000:18:48.250 more little drawings and things you can 00:18:51.30000:18:51.310 have to kind of keep track of things the 00:18:53.43000:18:53.440 easier it's gonna be to get set up so 00:18:55.35000:18:55.360 I'm gonna call this so this entrance 00:18:58.71000:18:58.720 here is one refrigerant so that's in so 00:19:03.24000:19:03.250 one refrigerant this one's two 00:19:05.43000:19:05.440 refrigerant so the outlet refrigerant 00:19:07.29000:19:07.300 this one's one air and this one's two 00:19:10.86000:19:10.870 air so now let's go ahead and expand 00:19:15.00000:19:15.010 these summations for our specific 00:19:17.31000:19:17.320 problem so we know that we have two 00:19:19.80000:19:19.810 outlets so our first outlet is M a and 00:19:25.65000:19:25.660 remember that M two ay is equal to m1a 00:19:32.04000:19:32.050 and I'm just going to call this M a and 00:19:34.17000:19:34.180 the reason why is because we're assuming 00:19:35.88000:19:35.890 this a steady flow and the same thing 00:19:38.25000:19:38.260 with the refrigerant so M two R is equal 00:19:41.46000:19:41.470 to M one R and I'm just gonna call that 00:19:44.64000:19:44.650 M R so I'm just gonna have em a here and 00:19:48.51000:19:48.520 this is gonna be this is the outlet so 00:19:52.41000:19:52.420 the so the mass flow rate of the 00:19:56.67000:19:56.680 refrigerant at the outlet and then this 00:19:58.98000:19:58.990 is multiplied by the enthalpy at the 00:20:01.08000:20:01.090 outlet 00:20:02.19000:20:02.200 for A+ and then we have the refrigerants 00:20:05.78900:20:05.799 om R and then this is multiplied by the 00:20:08.84900:20:08.859 enthalpy of the refrigerant and then we 00:20:14.07000:20:14.080 have minus and then we want to sum our 00:20:17.34000:20:17.350 inlets so we have the mass flow rate of 00:20:20.77900:20:20.789 the air multiplied by the enthalpy of 00:20:26.82000:20:26.830 the air at the inlet plus the mass flow 00:20:30.57000:20:30.580 rate of the refrigerant multiplied by 00:20:33.11900:20:33.129 the enthalpy of the refrigerant at the 00:20:35.87900:20:35.889 inlet so now what I want to do is 00:20:38.87900:20:38.889 combine like terms so where I have this 00:20:41.62900:20:41.639 mass flow rate of air and mass flow rate 00:20:44.87900:20:44.889 of the refrigerant I want to just 00:20:46.64900:20:46.659 combine those to make this a little more 00:20:48.33000:20:48.340 concise the equation so zero so all I'm 00:20:52.83000:20:52.840 doing now I basically this is basically 00:20:55.04900:20:55.059 my equation all I'm doing now is moving 00:20:57.89900:20:57.909 some things around to make it a little 00:21:00.21000:21:00.220 bit easier to read so I have the mass 00:21:03.98900:21:03.999 flow rate of the air multiplied by the 00:21:08.19000:21:08.200 enthalpy of the air at the outlet most- 00:21:13.01900:21:13.029 the enthalpy of the air at the inlet 00:21:15.69000:21:15.700 plus the mass flow rate of the 00:21:20.07000:21:20.080 refrigerant multiplied by the enthalpy 00:21:22.91900:21:22.929 of the refrigerant at the outlet minus 00:21:25.83000:21:25.84000:21:28.20000:21:28.210 inlet okay so we have our energy balance 00:21:32.82000:21:32.830 and let's let's just take a step back 00:21:35.48900:21:35.499 and what we want to do at this point is 00:21:39.13900:21:39.149 determine what we're looking for and 00:21:42.11900:21:42.129 what we already have so first of all we 00:21:44.75900:21:44.769 have the Wii so if we go back up to our 00:21:48.74900:21:48.759 problem statement we have the mass flow 00:21:52.91900:21:52.929 we can't well we weren't given the mass 00:21:54.81000:21:54.820 flow rate of the air but we were given 00:21:56.51900:21:56.529 the volumetric flow rate at the inlet so 00:21:59.43000:21:59.440 we can calculate we can use that 00:22:01.28900:22:01.299 volumetric flow rate to calculate the 00:22:03.38900:22:03.399 mass flow rate at the inlet and then 00:22:05.78900:22:05.799 since we know the mass flow rate at the 00:22:07.56000:22:07.570 inlet is equal to the mass flow rate at 00:22:09.41900:22:09.429 the outlet we basically have the mass 00:22:11.78900:22:11.799 flow rate of the air so we're just going 00:22:13.91900:22:13.929 to calculate that and 00:22:15.75000:22:15.760 we're going to calculate that from the 00:22:18.33000:22:18.340 mass flow rate is equal to v1 so the 00:22:24.24000:22:24.250 volumetric flow rate at the inlet 00:22:26.43000:22:26.440 divided by the specific volume and since 00:22:29.64000:22:29.650 this is an ideal gas we can just 00:22:32.37000:22:32.380 calculate this volume from the ideal gas 00:22:34.53000:22:34.540 law so PV is equal to RT and we know 00:22:37.86000:22:37.870 that V is equal to RT over P so we have 00:22:42.69000:22:42.700 everything we need to calculate the mass 00:22:44.82000:22:44.830 flow rate of the air so basically we 00:22:47.67000:22:47.680 know that so if we go back down to this 00:22:49.50000:22:49.510 problem so this is known once we 00:22:53.40000:22:53.410 calculate it and I'll just rewrite that 00:22:54.93000:22:54.940 here so the mass flow rate of the air is 00:22:59.63000:22:59.640 equal to the volumetric flow rate and 00:23:06.45000:23:06.460 I'm going to actually be a little more 00:23:08.34000:23:08.350 specific so the mass flow rate of the 00:23:12.14000:23:12.150 air at the inlet is equal to the 00:23:15.57000:23:15.580 volumetric flow rate of the air at the 00:23:18.72000:23:18.730 inlet divided by the specific volume of 00:23:21.84000:23:21.850 the air at the inlet but we know that 00:23:25.53000:23:25.540 this is also equal to the mass flow rate 00:23:28.23000:23:28.240 of the air at the outlet so that's why 00:23:30.21000:23:30.220 that's just equal to the mass flow rate 00:23:32.34000:23:32.350 of the air and then since we assume that 00:23:36.30000:23:36.310 this is an ideal gas we can calculate 00:23:39.12000:23:39.130 the specific volume from the ideal gas 00:23:41.28000:23:41.290 law so the volume is just equal to RT 00:23:46.25000:23:46.260 over P we know the temperature pressure 00:23:49.26000:23:49.270 we can look up the gas constant so we 00:23:53.01000:23:53.020 have everything we need to calculate the 00:23:55.68000:23:55.690 mass flow rate of the air so we're going 00:23:58.26000:23:58.270 to basically assume that that's known 00:24:00.26000:24:00.270 this is what we're looking for so we 00:24:04.59000:24:04.600 need to figure out a way to get the 00:24:06.99000:24:07.000 enthalpies for these enthalpies of the 00:24:11.58000:24:11.590 refrigerant we're just going to look 00:24:12.69000:24:12.700 these up on the tables that's pretty 00:24:14.73000:24:14.740 easy because we're given all of the 00:24:16.77000:24:16.780 property data that we need to look those 00:24:18.45000:24:18.460 up and then for the enthalpies for the 00:24:21.51000:24:21.520 air we can do this a couple of different 00:24:24.27000:24:24.280 ways since this is an ideal gas we can 00:24:28.32000:24:28.330 we 00:24:29.10000:24:29.110 oh that the change in enthalpy is equal 00:24:31.32000:24:31.330 to C sub P t2 minus t1 so we can 00:24:36.00000:24:36.010 calculate this using a specific heat and 00:24:39.03000:24:39.040 then we would just look up the specific 00:24:40.98000:24:40.990 heat at the average temperature and we 00:24:43.95000:24:43.960 know T 2 and we know T 1 so we can 00:24:46.20000:24:46.210 calculate that the other way that we can 00:24:49.56000:24:49.570 do this is we can look up the enthalpies 00:24:55.31000:24:55.320 so look up H 2 a and H 1 a on the air 00:25:03.41900:25:03.429 table so there are two ways that we can 00:25:07.15900:25:07.169 calculate this I'm going to use the 00:25:10.95000:25:10.960 specific heat but just so you know you 00:25:14.25000:25:14.260 could also look up those values on the 00:25:16.28900:25:16.299 air table and you would get the same 00:25:18.45000:25:18.460 answer basically and if you want to 00:25:22.04900:25:22.059 verify that you can solve it both ways 00:25:23.73000:25:23.740 and just verify that you get the same 00:25:25.77000:25:25.780 answer both ways alright so it looks 00:25:29.90900:25:29.919 like we basically know what we're doing 00:25:32.90900:25:32.919 to solve this we're solving for the mass 00:25:35.01000:25:35.020 flow rate of the refrigerant we're going 00:25:37.40900:25:37.419 to look up the enthalpies of the 00:25:40.56000:25:40.570 refrigerant on the tables we're going to 00:25:43.94000:25:43.950 calculate the change in enthalpy of the 00:25:47.15900:25:47.169 air using a specific heat and we were 00:25:50.49000:25:50.500 gonna calculate the mass flow rate of 00:25:52.16900:25:52.179 the air from this relation so now let's 00:25:57.33000:25:57.340 get our data and like I said the hardest 00:26:00.90000:26:00.910 part about doing these heat exchanger 00:26:03.53900:26:03.549 problems is just keeping everything keep 00:26:05.61000:26:05.620 it's just keeping track of everything so 00:26:08.66900:26:08.679 just make sure you label everything and 00:26:11.07000:26:11.080 try and keep things really organized and 00:26:13.53000:26:13.540 you'll find that these problems are a 00:26:15.53900:26:15.549 lot easier so the way I'm gonna look up 00:26:18.24000:26:18.250 the data I'm gonna first do the 00:26:20.66900:26:20.679 refrigerant Inlet so r134a Inlet and at 00:26:27.36000:26:27.370 the inlet we have the p1 p1 R is equal 00:26:32.58000:26:32.590 to one mega Pascal and t1 R is equal to 00:26:39.90000:26:39.910 90 degrees C 00:26:43.52000:26:43.530 so basically we can determine from these 00:26:47.10000:26:47.110 values that this is in fact a 00:26:48.96000:26:48.970 superheated vapor so we need to get our 00:26:53.67000:26:53.680 enthalpy from the superheated vapor 00:26:55.23000:26:55.240 table so H r1 is equal to three twenty 00:26:59.97000:26:59.980 four point six six kilojoules per 00:27:04.53000:27:04.540 kilogram all right now let's do our 00:27:08.79000:27:08.800 outlet for the refrigerant so one r134a 00:27:12.59000:27:12.600 outlet we have the p2 r is equal to 1 00:27:20.18000:27:20.190 mega Pascal and t2 R is equal to 30 00:27:25.88000:27:25.890 degrees Celsius if we look up if we look 00:27:30.51000:27:30.520 up the saturation temperature this 00:27:33.15000:27:33.160 pressure you will see that this is a 00:27:34.80000:27:34.810 compressed liquid and since we don't 00:27:39.06000:27:39.070 actually have a compressed liquid data 00:27:41.28000:27:41.290 table for this temperature and pressure 00:27:43.53000:27:43.540 we're just going to assume that it's a 00:27:45.48000:27:45.490 saturated liquid so basically H are two 00:27:49.26000:27:49.270 touch like to you R is equal to the 00:27:54.24000:27:54.250 enthalpy of saturated liquid at 30 00:27:56.76000:27:56.770 degrees C and remember to look it up on 00:27:59.37000:27:59.380 the temperature table and not the 00:28:01.32000:28:01.330 pressure table so this is ninety three 00:28:03.99000:28:04.000 point five eight kilojoules per kilogram 00:28:10.52000:28:10.530 all right now let's do the air well I'll 00:28:14.91000:28:14.920 write down the data for the for the air 00:28:17.49000:28:17.500 and letting out light even though I'm 00:28:19.05000:28:19.060 using a specific heat but anyway the air 00:28:21.81000:28:21.820 inlet so we have the p1 a is equal to 00:28:26.46000:28:26.470 100 and naturally we don't even need the 00:28:29.22000:28:29.230 pressures to look things up on the air 00:28:30.90000:28:30.910 table we just need the temperature so T 00:28:33.39000:28:33.400 1 a is equal to 300 Kelvin this was 27 00:28:39.75000:28:39.760 degrees Celsius and so from the air 00:28:43.56000:28:43.570 table H 1 a is equal to 300 0.19 00:28:50.24000:28:50.250 kilojoules per kilogram and then the air 00:28:56.47000:28:56.480 outlet so t2a is equal to sixty degrees 00:29:04.60000:29:04.610 C which is 333 Kelvin H to a is equal to 00:29:14.77000:29:14.780 three three three point three three 00:29:19.38000:29:19.390 kilojoules per kilogram and then we also 00:29:24.10000:29:24.110 need so we have the data from the tables 00:29:27.28000:29:27.290 we also need the specific heat so like I 00:29:31.51000:29:31.520 said this is enthalpy data from the air 00:29:37.36000:29:37.370 table and you can either solve this 00:29:40.06000:29:40.070 problem using the data from the air 00:29:42.03900:29:42.049 table to get your change in enthalpy or 00:29:44.38000:29:44.390 you can use the specific heat and just 00:29:46.72000:29:46.730 calculate the change in enthalpy so I'm 00:29:49.65900:29:49.669 going to do it with a specific lunch 00:29:51.73000:29:51.740 like I guess I'm gonna just do it both 00:29:52.96000:29:52.970 ways so so a specific heat of air and 00:29:58.91900:29:58.929 I'm gonna use an average temperature of 00:30:01.45000:30:01.460 approximately 300 degrees C that's not 00:30:06.46000:30:06.470 the exact average well I guess actually 00:30:13.72000:30:13.730 I'm just using the temperature of the 00:30:15.70000:30:15.710 inlet but there's not a huge change 00:30:17.79900:30:17.809 between the temperatures and basically 00:30:21.52000:30:21.530 the reason why I did that was because I 00:30:24.12000:30:24.130 look at the data for air I have data for 00:30:32.32000:30:32.330 between 300 and 350 and I'm gonna be 00:30:40.84000:30:40.850 like the average of this is somewhere 00:30:42.88000:30:42.890 around 3:15 and the specific heat only 00:30:46.65900:30:46.669 varies between one point is zero zero 00:30:48.66900:30:48.679 five and one point is zero zero eight so 00:30:51.46000:30:51.470 basically I'm just using one point zero 00:30:55.21000:30:55.220 zero five kilojoules kilogram Kelvin 00:30:59.89000:30:59.900 although if you wanted to interpolate 00:31:01.78000:31:01.790 and get a more specific value there 00:31:05.64900:31:05.659 would be no problem in doing that I just 00:31:08.11000:31:08.120 think you would end up with probably 00:31:09.91000:31:09.920 about the same answer so let's also well 00:31:15.70000:31:15.710 I think this is all the data we need 00:31:17.11000:31:17.120 well we need our so we need our gas 00:31:19.18000:31:19.190 constant and we need the gas constant 00:31:23.56000:31:23.570 for air so that is zero point two eight 00:31:27.34000:31:27.350 seven zero kill a Pascal meter cube 00:31:34.29000:31:34.300 kilogram Kelvin okay now let's we have 00:31:39.76000:31:39.770 so we have all of our data we have all 00:31:42.07000:31:42.080 of our equations let's all we have left 00:31:44.41000:31:44.420 now is to plug numbers into the 00:31:47.05000:31:47.060 equations and solve so I'm going to go 00:31:50.41000:31:50.420 back to I'm going to rewrite the 00:31:54.37000:31:54.380 equation I'm going to rewrite our energy 00:31:56.59000:31:56.600 balance that we're solving so m dot a 2 00:32:00.48000:32:00.490 a minus H 1 a plus m dot our H 2 R minus 00:32:10.78000:32:10.790 H 1 R and then what I want to do is well 00:32:23.83000:32:23.840 first of all I want to plug specific 00:32:25.51000:32:25.520 heat our specific heat relation into 00:32:28.06000:32:28.070 this equation so this is going to be CP 00:32:31.11000:32:31.120 t2 minus t1 and then I just want to 00:32:34.72000:32:34.730 solve this equation for the mass flow 00:32:37.63000:32:37.640 rate of the refrigerant because also 00:32:39.01000:32:39.020 we're actually interested in so if we do 00:32:41.44000:32:41.450 that we have negative MA and then H 2 a 00:32:46.59000:32:46.600 minus H 1 a is equal to the mass flow 00:32:51.79000:32:51.800 rate of the refrigerant and then we have 00:32:54.61000:32:54.620 h 2 R minus H 1 R and then I just want 00:33:01.42000:33:01.430 to multiply through this negative sign 00:33:03.01000:33:03.020 so I'm if I do that I get ma a dot H and 00:33:08.22000:33:08.230 I forgot to replace this with a specific 00:33:10.99000:33:11.000 heat so this should be C sub P t2 minus 00:33:16.09000:33:16.100 t1 so C sub P t2 minus t1 00:33:23.72000:33:23.730 is equal to m dot R and then I'm just 00:33:26.87000:33:26.880 going to swap the signs on these two 00:33:28.70000:33:28.710 enthalpies sigh of H 1 R minus H 2 R and 00:33:35.86000:33:35.870 then this means that the mass flow rate 00:33:39.40900:33:39.419 of the refrigerant is equal to the mass 00:33:42.83000:33:42.840 flow rate of the air multiplied by C so 00:33:45.79900:33:45.809 P t2 minus t1 over H 1 R minus H 2 R or 00:33:55.15900:33:55.169 if we didn't assume specific heat so 00:33:58.43000:33:58.440 let's say we just get data from the air 00:33:59.93000:33:59.940 table then the mass flow rate of the 00:34:02.62900:34:02.639 refrigerant is equal to the mass flow 00:34:05.12000:34:05.130 rate of the air multiplied by H 2 a 00:34:10.93000:34:10.940 minus H 1 a divided by H 1 R minus H 2 R 00:34:17.91900:34:17.929 so do you see how you can kind of solve 00:34:20.45000:34:20.460 this two different ways really all we're 00:34:23.14900:34:23.159 doing is trying to get the change in 00:34:24.58900:34:24.599 enthalpy you can get the change in 00:34:26.72000:34:26.730 enthalpy from table data if it's 00:34:28.30900:34:28.319 available or you can calculate it using 00:34:30.91900:34:30.929 like specific heat so now let's 00:34:34.52000:34:34.530 calculate the mass flow rate of the air 00:34:37.38900:34:37.399 so the mass flow rate of the air is 00:34:40.70000:34:40.710 equal to B 1 a over the specific volume 00:34:48.02000:34:48.030 and let's we need to calculate the 00:34:50.00000:34:50.010 specific volume so from the ideal gas 00:34:52.70000:34:52.710 law 00:34:53.14900:34:53.159 it's just RT over P and then and then we 00:35:00.10900:35:00.119 should have everything so R is zero 00:35:03.38000:35:03.390 point two eight seven zero kill a Pascal 00:35:08.90000:35:08.910 meter cubed kilogram Kelvin multiplied 00:35:14.39000:35:14.400 by three hundred Kelvin divided by 100 00:35:19.35900:35:19.369 kilo Pascal so then the kiloPascals 00:35:22.88000:35:22.890 cancel kelvins cancel we're left with 00:35:26.35900:35:26.369 meters cubed per kilogram so the 00:35:33.07900:35:33.089 specific volume is 0.8 00:35:37.27000:35:37.280 6-1 meters cubed per kilogram so that 00:35:42.44000:35:42.450 we're gonna plug this into here and we 00:35:44.59900:35:44.609 know the volumetric flow rate because it 00:35:46.84900:35:46.859 was given so the mass flow rate of the 00:35:50.72000:35:50.730 air is equal to 600 meters cube per 00:35:57.98000:35:57.990 minute divided by 0.8 6-1 meters cubed 00:36:05.09000:36:05.100 per kilogram which works out to 697 00:36:09.22000:36:09.230 kilogram per minute I'm gonna convert 00:36:12.44000:36:12.450 minute into second because that's a 00:36:13.97000:36:13.980 little more standard so one minute over 00:36:16.43000:36:16.440 60 seconds so then the mass flow rate of 00:36:20.42000:36:20.430 the air is equal to eleven point six 00:36:23.84000:36:23.850 kilograms per second alright so now we 00:36:27.71000:36:27.720 have everything we need to plug into 00:36:29.89000:36:29.900 either of these two equations and solve 00:36:32.63000:36:32.640 so we know the mass flow rate of the air 00:36:34.99000:36:35.000 we know the specific heat we know the 00:36:37.60900:36:37.619 temperatures we know that enthalpy of 00:36:40.19000:36:40.200 inlet and outlet of the refrigerant we 00:36:42.44000:36:42.450 also know the enthalpy of the inlet and 00:36:43.82000:36:43.830 outlet of the air and then just the 00:36:45.89000:36:45.900 refrigerant let's go down here and solve 00:36:49.04000:36:49.050 these so we have the mass flow rate of 00:36:52.55000:36:52.560 the refrigerant is equal to eleven point 00:36:56.18000:36:56.190 six kilograms per second multiplied by 00:37:00.71000:37:00.720 and this is the one with the specific 00:37:02.24000:37:02.250 heat so 0.05 kilojoules kilogram Kelvin 00:37:09.13000:37:09.140 multiplied by sixty minus 27 degrees C 00:37:13.40000:37:13.410 and we can do this because the Delta K 00:37:16.67000:37:16.680 and Delta C end up being the same so we 00:37:20.39000:37:20.400 did if we if we're just subtracting them 00:37:22.31000:37:22.320 to get the difference we don't need to 00:37:23.69000:37:23.700 convert to Kelvin so and then divided by 00:37:27.05000:37:27.060 three twenty four point six six 00:37:30.52000:37:30.530 kilojoules per kilogram minus ninety 00:37:35.71000:37:35.720 3.58 kilojoules per kilogram and just to 00:37:41.23000:37:41.240 specify the equation since it was off 00:37:44.00000:37:44.010 the screen the equation for this is the 00:37:48.32000:37:48.330 mass flow rate of the air multiplied by 00:37:51.02000:37:51.030 the 00:37:51.47000:37:51.480 a big heat multiplied by t2 minus t1 00:37:54.32000:37:54.330 over H 1 R minus H 2 R so this is the 00:37:59.63000:37:59.640 equation that I'm solving here so now 00:38:03.41000:38:03.420 let's make sure our units work out so 00:38:05.21000:38:05.220 the kilojoules cancel this kilogram 00:38:09.41000:38:09.420 cancels and the Kelvin cancels and we're 00:38:12.74000:38:12.750 left with kilograms per second which is 00:38:15.05000:38:15.060 what we want so this works out to one 00:38:17.57000:38:17.580 point six six kilograms per second now 00:38:24.34900:38:24.359 let's just verify that we could have 00:38:25.82000:38:25.830 also solved this with the equation using 00:38:27.94000:38:27.950 the specific heats of the air our sorry 00:38:30.80000:38:30.810 the enthalpies of the air from the air 00:38:33.47000:38:33.480 table so that equation was the mass flow 00:38:37.31000:38:37.320 rate of the air x h2 a minus h1 a 00:38:44.38000:38:44.390 divided by H 1 R minus H 2 R and so I 00:38:53.87000:38:53.880 don't think I'm going to go through and 00:38:55.22000:38:55.230 plug all of those values in again well 00:38:58.04000:38:58.050 actually maybe I will so this is equal 00:39:00.32000:39:00.330 to eleven point six kilograms per second 00:39:07.30000:39:07.310 x and then h2 the enthalpy of the air at 00:39:12.62000:39:12.630 that outlet was three three three point 00:39:15.23000:39:15.240 three three kilojoules per kilogram 00:39:19.39000:39:19.400 minus 300 point one nine kilojoules per 00:39:24.56000:39:24.570 kilogram divided by three twenty four 00:39:29.18000:39:29.190 point six six kilojoules per kilogram 00:39:33.91000:39:33.920 minus ninety three point five eight 00:39:38.05000:39:38.060 kilojoules per kilogram so let's just 00:39:43.09900:39:43.109 make sure the unit's cancel out again so 00:39:45.85900:39:45.869 this these kilograms all cancel the 00:39:50.12000:39:50.130 kilojoules all cancel and we're left 00:39:52.22000:39:52.230 with kilograms per second so this works 00:39:55.16000:39:55.170 out to one point six six kilograms per 00:39:59.14000:39:59.150 second so we got the same value with 00:40:03.05000:40:03.060 both methods 00:40:04.50000:40:04.510 basically what I want you to see is like 00:40:07.20000:40:07.210 if we go back to our energy balance so 00:40:10.83000:40:10.840 this basically this is what we end up 00:40:15.06000:40:15.070 with for our energy balance and we're 00:40:18.99000:40:19.000 often left with a scenario where we need 00:40:21.60000:40:21.610 to calculate Delta H or Delta U and 00:40:26.87000:40:26.880 basically there are different ways that 00:40:29.58000:40:29.590 we can do that we can look them up on 00:40:31.14000:40:31.150 the tables if the tables are available 00:40:32.58000:40:32.590 or weak if their tables aren't available 00:40:35.25000:40:35.260 we can calculate those using specific 00:40:38.79000:40:38.800 heats so however you do it is fine just 00:40:43.17000:40:43.180 be really clear about what you're doing 00:40:45.95000:40:45.960 so basically the way we did this problem 00:40:48.50000:40:48.510 was first of all we wrote down all of 00:40:51.75000:40:51.760 the information from the problem and our 00:40:54.30000:40:54.310 setup figured out exactly what we were 00:40:57.09000:40:57.100 solving for what was going on then we 00:40:59.64000:40:59.650 made some assumptions and the 00:41:01.56000:41:01.570 assumptions are really important because 00:41:03.03000:41:03.040 they tell you basically what equation 00:41:06.84000:41:06.850 you need to use I mean we know that we 00:41:08.97000:41:08.980 need to use this steady flow energy 00:41:11.40000:41:11.410 balance equation but what if you don't 00:41:14.19000:41:14.200 know and I didn't get the whole thing I 00:41:16.98000:41:16.990 mean this is a huge equation we don't 00:41:19.62000:41:19.630 want to solve this entire thing by hand 00:41:21.78000:41:21.790 plus we don't have all of the 00:41:23.76000:41:23.770 information to even solve this entire 00:41:25.71000:41:25.720 equation so basically we need to make 00:41:28.98000:41:28.990 assumptions so that we can make this 00:41:31.02000:41:31.030 equation solvable and also simplify it 00:41:34.62000:41:34.630 so we so we can get a reason reasonable 00:41:37.62000:41:37.630 answer fairly quickly so that's why we 00:41:41.37000:41:41.380 need to make assumptions so we know what 00:41:43.41000:41:43.420 we can cancel out of our equations and 00:41:45.62000:41:45.630 get an equation that's solvable 00:41:48.15000:41:48.160 with a lot of engineering problems if 00:41:50.19000:41:50.200 you don't make assumptions the only way 00:41:52.23000:41:52.240 to solve your equations is with a 00:41:54.18000:41:54.190 numerical with a numerical solution and 00:41:58.68000:41:58.690 that's not very convenient if you just 00:42:01.29000:42:01.300 want a really quick answer or something 00:42:03.29000:42:03.300 so anyway we did the equations and then 00:42:07.20000:42:07.210 we looked up data on the tables and we 00:42:10.32000:42:10.330 solved this problem and these problems 00:42:14.70000:42:14.710 from here on are going to start getting 00:42:17.94000:42:17.950 a little more complicated and actually 00:42:21.27000:42:21.280 complicated as a wrong word they're just 00:42:22.80000:42:22.810 gonna start having more things that you 00:42:25.02000:42:25.030 need to do especially when we get into 00:42:26.46000:42:26.470 the second law you're gonna have to 00:42:27.83900:42:27.849 start solving for entropy as well as 00:42:31.89000:42:31.900 some of the other things we've been 00:42:33.56900:42:33.579 solving for so doing these problems that 00:42:37.02000:42:37.030 are really ordered manner like I've done 00:42:38.97000:42:38.980 here is going to really help because 00:42:42.69000:42:42.700 these problems are gonna get really 00:42:44.19000:42:44.200 confusing really fast if you're not 00:42:46.47000:42:46.480 approaching them in a methodical manner 00:42:49.68000:42:49.690 and so this is how I like to do them and 00:42:52.31900:42:52.329 this is how I recommend doing them 00:42:54.15000:42:54.160 anyway I hope this was helpful thanks 00:42:56.91000:42:56.920 for watching
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