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Multiple Effect Evaporator - Mass and Enthalpy Balance

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00:00:01.420
in this tutorial we will consider mass
00:00:06.400 00:00:06.410 and enthalpy balances for a three effect
00:00:09.700 00:00:09.710 evaporator as we see here in this
00:00:14.080 00:00:14.090 schematic we have identified various
00:00:17.800 00:00:17.810 quantities related to the feed stream
00:00:21.399 00:00:21.409 the streams for vapors for the
00:00:26.800 00:00:26.810 concentrated product for our steam as
00:00:30.609 00:00:30.619 well as for condensate
00:00:32.589 00:00:32.599 so let's first write the mass balance so
00:00:37.950 00:00:37.960 for the feed the mass flow rate is m dot
00:00:42.460 00:00:42.470 F entering the bottom of the first
00:00:47.530 00:00:47.540 effect so m dot F equals the mass flow
00:00:52.660 00:00:52.670 rate of vapours leaving the first effect
00:00:56.260 00:00:56.270 m dot v1 note that those vapors are used
00:01:02.350 00:01:02.360 as heating medium for the second effect
00:01:04.810 00:01:04.820 and therefore they are leaving the
00:01:07.870 00:01:07.880 system later as a condensate so in our
00:01:12.910 00:01:12.920 mass balance we must account for the
00:01:15.849 00:01:15.859 mass of those vapors so we have m dot v1
00:01:19.950 00:01:19.960 similarly we have m dot v2 vapors coming
00:01:24.580 00:01:24.590 from the second effect n m dot v3
00:01:27.219 00:01:27.229 vapours coming from the third effect
00:01:29.200 00:01:29.210 plus of course the mass flow rate of the
00:01:32.679 00:01:32.689 concentrated product stream the solid
00:01:36.669 00:01:36.679 balance can be written as X F times m
00:01:41.289 00:01:41.299 dot F where XF is the solid fraction
00:01:45.429 00:01:45.439 which will be dimensionless so xf m dot
00:01:49.090 00:01:49.100 F tells us the amount of solid present
00:01:52.899 00:01:52.909 in the feed stream and that equals X P
00:01:57.219 00:01:57.229 the solid fraction in the concentrated
00:02:00.699 00:02:00.709 product stream times m dot P now for the
00:02:05.379 00:02:05.389 enthalpy balance we have to account for
00:02:09.120 00:02:09.130 enthalpy of each of the streams so the
00:02:14.050 00:02:14.060 enthalpy
00:02:14.970 00:02:14.980 entering with the feed is H F so we have
00:02:19.259 00:02:19.269 m dot F times H F plus m dot s times H V
00:02:26.250 00:02:26.260 s where H V s represents the enthalpy of
00:02:30.600 00:02:30.610 the saturated steam that we obtain from
00:02:33.990 00:02:34.000 steam tables that equals m dot v1 that's
00:02:40.050 00:02:40.060 the mass flow rate of the vapors coming
00:02:42.869 00:02:42.879 out from the first effect times H V 1 H
00:02:47.550 00:02:47.560 v1 is the enthalpy of the vapors from
00:02:50.330 00:02:50.340 first effect plus m dot F 1 which is the
00:02:55.890 00:02:55.900 amount of partially concentrated feed
00:02:59.610 00:02:59.620 coming out of Effect 1 times H F 1 where
00:03:05.729 00:03:05.739 H F 1 is the enthalpy of that pre
00:03:09.449 00:03:09.459 concentrated stream plus m dot s times h
00:03:14.520 00:03:14.530 c s where m dot s represents now the
00:03:17.970 00:03:17.980 mass flow rate of condensate leaving the
00:03:21.210 00:03:21.220 first effect times the enthalpy of that
00:03:24.809 00:03:24.819 condensate the second equation is for
00:03:28.949 00:03:28.959 the second effect of the evaporator for
00:03:32.580 00:03:32.590 which we have now our feed as m dot F 1
00:03:37.470 00:03:37.480 times H F 1 which represents now the
00:03:40.500 00:03:40.510 enthalpy of the feed stream into second
00:03:44.220 00:03:44.230 effect plus m dot V 1 times H V 1 note
00:03:49.559 00:03:49.569 that the vapors leaving from first
00:03:51.599 00:03:51.609 effect are now the heating medium for
00:03:54.720 00:03:54.730 the second effect that equals the mass
00:03:57.960 00:03:57.970 flow rate of vapours leaving the second
00:04:00.180 00:04:00.190 effect m dot v2 times the enthalpy of
00:04:04.080 00:04:04.090 those vapors H v2 plus m dot F 2 that is
00:04:10.500 00:04:10.510 the enthalpy with the mass flow rate of
00:04:12.869 00:04:12.879 this product stream leaving the second
00:04:16.409 00:04:16.419 effect times the enthalpy of that stream
00:04:21.439 00:04:21.449 HF 2 plus m dot v1
00:04:25.890 00:04:25.900 times h c1 the
00:04:28.679 00:04:28.689 enthalpy of that condensate so m dot v1
00:04:31.049 00:04:31.059 times HC 1 so our third equation is for
00:04:36.449 00:04:36.459 the third effect m dot F 2 which is the
00:04:40.889 00:04:40.899 mass flow rate of the feed that came
00:04:45.059 00:04:45.069 from the second effect and now entering
00:04:48.239 00:04:48.249 the third effect times its enthalpy H f2
00:04:53.749 00:04:53.759 plus m dot V 2 times H v2 representing
00:04:59.339 00:04:59.349 the enthalpy of the vapours that came
00:05:03.239 00:05:03.249 from the second effect and now being
00:05:05.369 00:05:05.379 used as heating medium in the third
00:05:07.439 00:05:07.449 effect that equals m dot V 3 times H v3
00:05:12.659 00:05:12.669 which represents the enthalpy leaving
00:05:15.389 00:05:15.399 with the vapors from the third effect
00:05:18.419 00:05:18.429 plus m dot P times HB 3 representing the
00:05:23.669 00:05:23.679 enthalpy leaving with the final
00:05:26.519 00:05:26.529 concentrated product stream plus m dot V
00:05:30.600 00:05:30.610 2 times HC 2 representing the enthalpy
00:05:34.679 00:05:34.689 leaving with the condensate from the
00:05:37.679 00:05:37.689 third effect so these three equations
00:05:41.329 00:05:41.339 for the enthalpy balances for first
00:05:44.759 00:05:44.769 second and third effect plus the solid
00:05:48.029 00:05:48.039 balance and the mass balance give us 5
00:05:51.089 00:05:51.099 equations so those 5 equations are
00:05:54.989 00:05:54.999 solved simultaneously to obtain the
00:05:59.009 00:05:59.019 unknowns in a given problem for heat
00:06:02.429 00:06:02.439 transfer areas we can use the following
00:06:06.839 00:06:06.849 equations for the three effects q1 the
00:06:11.189 00:06:11.199 rate of heat transfer in the first
00:06:12.989 00:06:12.999 effect will equal u 1 a 1 times TS minus
00:06:19.139 00:06:19.149 t1 equal m dot s h vs minus m dot s HC s
00:06:27.139 00:06:27.149 so the left hand side gives us the rate
00:06:31.379 00:06:31.389 of heat transfer across the tubes the
00:06:34.439 00:06:34.449 heating tubes in the first effect by
00:06:37.139 00:06:37.149 using the overall heat transfer
00:06:38.759 00:06:38.769 coefficient the area
00:06:41.270 00:06:41.280 and the temperature difference that is
00:06:43.250 00:06:43.260 between the temperature of the steam and
00:06:45.290 00:06:45.300 the boiling point inside the first
00:06:48.230 00:06:48.240 effect t1 and on the right hand side we
00:06:51.620 00:06:51.630 have the enthalpy associated with the
00:06:55.250 00:06:55.260 steam minus the enthalpy that leaves
00:06:58.370 00:06:58.380 with the condensate so that's the amount
00:07:00.560 00:07:00.570 of heat that gets transferred from steam
00:07:04.250 00:07:04.260 into the liquid stream second equation
00:07:10.460 00:07:10.470 for the second effect is q2 equals u to
00:07:13.909 00:07:13.919 a to t1 minus t2 equals m dot v1 h v1
00:07:21.760 00:07:21.770 minus m dot v1 HC 1 again this is
00:07:26.840 00:07:26.850 similar to our discussion for the first
00:07:31.310 00:07:31.320 equation the only difference here of
00:07:34.610 00:07:34.620 course we have used t1 minus t2 because
00:07:39.110 00:07:39.120 T 1 is the temperature of the vapors
00:07:41.900 00:07:41.910 coming in as heating medium and T 2 is
00:07:45.140 00:07:45.150 the boiling point temperature maintained
00:07:49.010 00:07:49.020 inside the second effect the third
00:07:52.520 00:07:52.530 equation for the third effect is Q 3
00:07:55.420 00:07:55.430 equals u 3 a 3 times T 2 minus T 3
00:08:00.940 00:08:00.950 equals m dot v 2h v2 minus m dot v2 HC 2
00:08:07.430 00:08:07.440 and again here T 2 minus T 3 represent a
00:08:12.350 00:08:12.360 temperature difference between the
00:08:15.040 00:08:15.050 heating medium which is the vapors
00:08:17.600 00:08:17.610 coming from the second effect at
00:08:20.659 00:08:20.669 temperature T 2 minus T 3 where T 3 is
00:08:24.440 00:08:24.450 the temperature maintained inside the
00:08:26.779 00:08:26.789 third effect we can obtain the Steam
00:08:30.680 00:08:30.690 economy as m dot v1 plus m dot v2 plus m
00:08:37.339 00:08:37.349 dot V 3 where these are the mass flow
00:08:40.820 00:08:40.830 rates of the vapor streams exiting from
00:08:45.170 00:08:45.180 each of the effects so that total amount
00:08:49.340 00:08:49.350 of vapors that are produced in the
00:08:51.320 00:08:51.330 triple effect evaporator
00:08:53.150 00:08:53.160 divided by
00:08:54.830 00:08:54.840 the mass flow rate of steam m dot s
00:08:57.850 00:08:57.860 typically the steam economy of a triple
00:09:01.340 00:09:01.350 effect evaporator is between two and
00:09:03.860 00:09:03.870 three now note that the units for these
00:09:07.250 00:09:07.260 various quantities that we have used we
00:09:10.940 00:09:10.950 can summarize them here m dot s m dot V
00:09:15.500 00:09:15.510 1 m dot V 2 m dot V 3 are in kilograms
00:09:19.220 00:09:19.230 per second and so are the M units 4 m
00:09:22.820 00:09:22.830 dot F m dot F 1 m dot F 2 m dot P they
00:09:27.920 00:09:27.930 are all kilograms per second as they are
00:09:30.829 00:09:30.839 mass flow rates XF and XP the solid
00:09:35.660 00:09:35.670 fractions are dimensionless and then the
00:09:39.230 00:09:39.240 enthalpy terms hv s hv 1 HB 2 h v3 hf h
00:09:45.410 00:09:45.420 f1 h f2 h p3 its CS h c1 and its c2 are
00:09:51.200 00:09:51.210 all in kilojoules per kilogram

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