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Single Effect Evaporator - Mass and Enthalpy Balance
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00:00:01.490 in this tutorial we will look at a heat 00:00:05.12000:00:05.130 and mass balance for a single effect 00:00:07.70000:00:07.710 evaporator let's look at the schematic 00:00:12.33900:00:12.349 so we have the various streams going 00:00:16.55000:00:16.560 into the evaporator the vapours leaving 00:00:19.84900:00:19.859 on the top we have steam coming on the 00:00:24.29000:00:24.300 side the feed entering at the bottom and 00:00:29.05000:00:29.060 the condensate leaving the system along 00:00:33.11000:00:33.120 with the concentrated product 00:00:35.77000:00:35.780 note that the tubes and the heating 00:00:39.52000:00:39.530 chest of this evaporator is called 00:00:43.34000:00:43.350 calendar iam now let's conduct a mass 00:00:48.61900:00:48.629 balance so for the feed we have m dot F 00:00:53.72000:00:53.730 which is the mass flow rate of the 00:00:56.95900:00:56.969 dilute feed in kilograms per second we 00:01:01.45900:01:01.469 have m dot s the mass flow rate of steam 00:01:06.23000:01:06.240 as kilograms per second we have m dot V 00:01:11.81000:01:11.820 as the mass flow rate of vapors exiting 00:01:17.12000:01:17.130 the evaporator in kilograms per second 00:01:21.19000:01:21.200 we have m dot P the mass flow rate of 00:01:26.02000:01:26.030 concentrated product in kilograms per 00:01:29.27000:01:29.280 second and we also have the solid 00:01:35.71900:01:35.729 fraction in the feed as XF which will be 00:01:40.85000:01:40.860 dimensionless 00:01:42.24900:01:42.259 similarly the solid fraction in the 00:01:46.58000:01:46.590 concentrated product is denoted by X P 00:01:50.60000:01:50.610 which will be dimensionless so we can 00:01:55.63900:01:55.649 write the mass balance by looking at 00:02:00.28000:02:00.290 what is entering the system and what is 00:02:03.88900:02:03.899 exiting the system so we have the feed 00:02:07.99900:02:08.009 entering the system and the mass flow 00:02:11.21000:02:11.220 rate is m dot F 00:02:13.32000:02:13.330 and that equals the mass flow rate of 00:02:16.77000:02:16.780 vapors leaving the system and m dot p 00:02:21.26000:02:21.270 the mass flow rate of concentrated 00:02:24.24000:02:24.250 product leaving the system note that in 00:02:28.59000:02:28.600 the evaporator steam and feed remain 00:02:32.58000:02:32.590 separated because this is an indirect 00:02:35.46000:02:35.470 way of heating when steam condenses it 00:02:39.33000:02:39.340 leaves the evaporator as a condensate so 00:02:44.04000:02:44.050 steam does not mix with the feed 00:02:46.32000:02:46.330 therefore we do not include the mass 00:02:49.77000:02:49.780 flow rate terms for either the steam or 00:02:52.59000:02:52.600 the condensate in the mass balance when 00:02:56.10000:02:56.110 we look at the solids balance we note 00:02:59.49000:02:59.500 that xf is the solid fraction in the 00:03:05.13000:03:05.140 feed and it is dimensionless 00:03:07.07000:03:07.080 so xf times m dot F will represent the 00:03:12.57000:03:12.580 amount of solids in the feed and that 00:03:14.90000:03:14.910 will equal X P which is the solid 00:03:20.04000:03:20.050 fraction in the concentrated product 00:03:22.92000:03:22.930 which is dimensionless times the mass 00:03:25.65000:03:25.660 flow rate of the product and that will 00:03:29.58000:03:29.590 represent the amount of solids present 00:03:31.65000:03:31.660 in the concentrated product stream so 00:03:35.31000:03:35.320 that is our equation 2 now for the 00:03:39.03000:03:39.040 enthalpy balance we write m dot F times 00:03:45.60000:03:45.610 HF where HF is the enthalpy of the 00:03:51.09000:03:51.100 entering feed in kilojoules per kilogram 00:03:55.46000:03:55.470 we also have enthalpy entering into the 00:03:59.69900:03:59.709 system with steam so we have m dot s the 00:04:04.86000:04:04.870 mass flow rate of steam times h vs where 00:04:10.02000:04:10.030 hv s is the enthalpy of saturated vapor 00:04:13.65000:04:13.660 at temperature TS that is the saturation 00:04:17.37000:04:17.380 temperature of steam and the units will 00:04:20.75900:04:20.769 be kilojoules per kilogram 00:04:23.37900:04:23.389 so the heat is entering the system 00:04:25.40900:04:25.419 either with the feed or with steam this 00:04:30.52000:04:30.530 is the only two ways that heat is 00:04:32.23000:04:32.240 entering this system and that equals the 00:04:36.96900:04:36.979 heat leaving with the vapors that is m 00:04:40.26900:04:40.279 dot V times H v1 where H v1 is the 00:04:48.36900:04:48.379 enthalpy of saturated vapors at 00:04:51.42000:04:51.430 temperature t1 t1 is the boiling point 00:04:55.95000:04:55.960 inside the evaporator that is maintained 00:05:00.01000:05:00.020 by certain vacuum conditions the units 00:05:04.42000:05:04.430 for H V 1 are kilojoules per kilogram so 00:05:09.42900:05:09.439 in addition to the heat leaving with the 00:05:11.89000:05:11.900 vapors we also have heat leaving with 00:05:14.89000:05:14.900 with concentrated product stream and 00:05:17.85000:05:17.860 that would be denoted by m dot P times h 00:05:22.32900:05:22.339 p1 where H p1 is the enthalpy of 00:05:26.55000:05:26.560 concentrated product stream in 00:05:28.80900:05:28.819 kilojoules per kilogram we also have 00:05:32.80000:05:32.810 heat leaving the system with the 00:05:35.64900:05:35.659 condensate this is the condensate that 00:05:38.95000:05:38.960 formed on the outside of the tubes as 00:05:42.20900:05:42.219 steam condensed into condensate and the 00:05:46.74900:05:46.759 enthalpy content will be m dot s that is 00:05:51.63900:05:51.649 the mass flow rate of the steam times HC 00:05:56.07900:05:56.089 s which is the enthalpy of the 00:05:59.01900:05:59.029 condensate in kilojoules per kilogram 00:06:02.07000:06:02.080 again note that the heating is done in 00:06:05.76900:06:05.779 an indirect manner so all the mass of 00:06:09.07000:06:09.080 the steam entering this heating chamber 00:06:12.18900:06:12.199 leaves as condensate after it condenses 00:06:16.62900:06:16.639 into liquid water so the mass flow rate 00:06:20.11000:06:20.120 of the steam m dot s is the same as the 00:06:23.95000:06:23.960 mass flow rate of the condensate as we 00:06:27.27900:06:27.289 see on our schematic so let's see how we 00:06:30.73000:06:30.740 can determine each term in our enthalpy 00:06:35.52900:06:35.539 balance 00:06:36.82000:06:36.830 so we have the enthalpy of the entering 00:06:39.96900:06:39.979 stream H F equals CP f which is the 00:06:45.15900:06:45.169 specific heat of the feed times T F 00:06:50.64900:06:50.659 which is the temperature of the feed 00:06:53.79000:06:53.800 minus 0 degree C we base all our 00:06:57.99900:06:58.009 calculations for enthalpy with a 00:07:00.61000:07:00.620 reference temperature of 0 degree C note 00:07:05.05000:07:05.060 that enthalpy is always relative rather 00:07:08.20000:07:08.210 than absolute value so H F then the 00:07:13.60000:07:13.610 units will be kilojoules per kilogram 00:07:16.46900:07:16.479 since we have the specific heat 00:07:21.11900:07:21.129 multiplied by the temperature we are 00:07:25.02900:07:25.039 going to assume that we are using 00:07:27.24000:07:27.250 saturated steam then H V s is obtained 00:07:32.98000:07:32.990 from steam tables depending upon the 00:07:36.76000:07:36.770 saturation pressure or saturation 00:07:39.45900:07:39.469 temperature of the steam each V 1 is 00:07:44.14000:07:44.150 obtained also from the steam tables as 00:07:47.87900:07:47.88900:07:51.36000:07:51.370 temperature T 1 H P 1 is the enthalpy of 00:07:57.43000:07:57.440 the concentrated product stream so we 00:08:01.71900:08:01.729 use the specific heat of the 00:08:04.60000:08:04.610 concentrated product CPP and multiply 00:08:08.52900:08:08.539 that with t 1 where T 1 is the 00:08:12.18900:08:12.199 temperature maintained inside the 00:08:14.55900:08:14.569 evaporator - 0 degree C where 0 degree C 00:08:20.92000:08:20.930 of course is the reference temperature H 00:08:23.68000:08:23.690 CS is obtained from steam tables as 00:08:28.86000:08:28.870 enthalpy of saturated liquid at 00:08:32.69900:08:32.709 temperature TS now we are going to 00:08:36.31000:08:36.320 assume that the condensate leaves the 00:08:39.85000:08:39.860 evaporator at temperature TS if the 00:08:44.41000:08:44.420 condensate leaves at a lower temperature 00:08:47.17000:08:47.180 than TS 00:08:49.51900:08:49.529 then we will need to determine that 00:08:52.61000:08:52.620 additional sensible heat leaving with 00:08:55.48900:08:55.499 the condensate the area for heat 00:08:58.85000:08:58.860 transfer in the calendar can be obtained 00:09:03.30900:09:03.319 with the following equation Q the rate 00:09:08.56900:09:08.579 of heat transfer across the tubes will 00:09:13.12900:09:13.139 equal u times a times TS minus t1 equals 00:09:19.91000:09:19.920 m dot s hv s minus m dot s HC s where u 00:09:28.16000:09:28.170 is the overall heat transfer coefficient 00:09:30.53000:09:30.540 in watts per square meter kelvin a is 00:09:35.26900:09:35.279 the area in square meters and of course 00:09:39.17000:09:39.180 q is the rate of heat transfer in watts 00:09:42.13900:09:42.149 so this equation can be used to 00:09:46.61000:09:46.620 determine the area required to 00:09:49.81900:09:49.829 accomplish a certain amount of heat 00:09:52.87900:09:52.889 transfer for concentrating liquid feed 00:09:56.25900:09:56.269 the performance of the evaporator is 00:09:59.41000:09:59.420 expressed using a term called steam 00:10:02.44900:10:02.459 economy which equals m dot V divided by 00:10:07.85000:10:07.860 m dot s where m dot V is the mass flow 00:10:12.25900:10:12.269 rate of vapors and m dot s is the mass 00:10:15.53000:10:15.540 flow rate of steam entering the 00:10:17.84000:10:17.850 evaporator a typical value of steam 00:10:20.56900:10:20.579 economy for a single effect evaporator 00:10:23.60000:10:23.610 is close to 1
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