00:00:04.330 --> 00:00:08.510 In this problem we are asked to do an energy balance on a heat exchanger. 00:00:08.510 --> 00:00:15.090 We're given the feed which is superheated steam and were're told how much energy is 00:00:15.090 --> 00:00:18.840 removed from that steam to heat some reactor feed. 00:00:18.840 --> 00:00:24.020 We want to know what's the outlet temperature of that steam, what phases are present and 00:00:24.020 --> 00:00:32.480 then what's the enthalpy. 00:00:32.480 --> 00:00:37.690 So first thing we want to do is draw a diagram so we can input the information we already 00:00:37.690 --> 00:00:39.729 know about the system. 00:00:39.729 --> 00:00:44.630 So here we are showing the mas flow rate in, m dot, that's a constant so the same mass 00:00:44.630 --> 00:00:46.390 is leaving. 00:00:46.390 --> 00:00:54.350 We don't know the outlet temperature and the heat removed is - if we take as my system 00:00:54.350 --> 00:01:02.960 is the steam so I'm removing 450 kJ per second which corresponds to 450 kW. 00:01:02.960 --> 00:01:05.010 So what i want to do is the energy balance. 00:01:05.010 --> 00:01:13.740 It's a steady state system and so the energy balance is going to be steady state, we have 00:01:13.740 --> 00:01:19.040 energy flowing into the system, the inlet stream. 00:01:19.040 --> 00:01:25.640 Same mass flowrate out, enthalpy is going to be different and then we have a rate of 00:01:25.640 --> 00:01:26.710 heat removal. 00:01:26.710 --> 00:01:32.320 Of course in general we would have work but there is no work in this system. 00:01:32.320 --> 00:01:37.240 First thing we are going to do is look up inlet enthalpy using steam tables. 00:01:37.240 --> 00:01:43.230 So 200 C, 10 bar, go to superheated steam tables. 00:01:43.230 --> 00:01:48.360 And we know it's superheated, one because this pressure is less than saturated pressure 00:01:48.360 --> 00:01:51.140 and the problem statement states it out. 00:01:51.140 --> 00:01:53.420 And so the enthalpy I can just read directly. 00:01:53.420 --> 00:01:59.550 The enthalpy coming in, 2828 kJ per kg. 00:01:59.550 --> 00:02:05.100 I can solve this equation for the enthalpy out. 00:02:05.100 --> 00:02:10.140 So let me first rearrange the equation and the energy balance. 00:02:10.140 --> 00:02:17.319 So I rearranged the energy balance by taking these two terms to the other side of the equation 00:02:17.319 --> 00:02:21.470 and then I substituted in the values for Q dot and M dot. 00:02:21.470 --> 00:02:27.560 So H out is just H in minus 225. 00:02:27.560 --> 00:02:41.550 H in, something we just looked up in the steam tables, 2828 kJ per kg minus the 225 kJ per 00:02:41.550 --> 00:02:42.550 kg. 00:02:42.550 --> 00:02:47.780 And so the enthalpy out, in order to determine the temperature out we go back to the steam 00:02:47.780 --> 00:03:00.160 tables and we look at enthalpy at 10 bar for saturated vapor. 00:03:00.160 --> 00:03:02.709 And the enthalpy: 2777 kJ per kg. 00:03:02.709 --> 00:03:08.040 Which means, since we we're at the value less than that but we're at 10 bar, we must be 00:03:08.040 --> 00:03:09.900 in the two phase region. 00:03:09.900 --> 00:03:16.950 so that means we're at saturation conditions, we can calculate the quality, because the 00:03:16.950 --> 00:03:20.940 enthalpy is a mixture of so much vapor. 00:03:20.940 --> 00:03:29.270 So saturated vapor, so x is the fraction of vapor, 1-x is the fraction that's liquid. 00:03:29.270 --> 00:03:32.050 Enthalpy liquid saturation conditions. 00:03:32.050 --> 00:03:41.510 We know the total enthalpy, that's 2603 kJ/kg so I'm going to write down the values for 00:03:41.510 --> 00:03:45.300 saturated liquid, saturated vapor at 10 bar. 00:03:45.300 --> 00:03:49.010 So we can solve for x the quality. 00:03:49.010 --> 00:03:56.841 And is 0.91 so that means we know the temperature leaving, it's one of the things we wanted 00:03:56.841 --> 00:04:05.510 to determine and that's the saturation temperature, 10 bar 179.9, which, significant figures, 00:04:05.510 --> 00:04:06.510 180. 00:04:06.510 --> 00:04:16.849 We now know the quality so it's wet steam that's leaving. and it's enthalpy 2603 kJ/ 00:04:16.849 --> 00:04:19.099 00:04:19.099 --> 00:04:20.929 That's the information we are asked to determine.
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