00:00:03.449 --> 00:00:10.370 In this screen cast I am going to describe the operation of a single effect evaporator. 00:00:10.370 --> 00:00:17.720 Which is used to remove water from an aqueous solution in order to concentrate the solution 00:00:17.720 --> 00:00:24.340 this may be a solution that contains sugar, sodium chloride, orange juice, and the final 00:00:24.340 --> 00:00:27.670 objective is this concentrated solution. 00:00:27.670 --> 00:00:33.190 So what we are going to look at is a schematic representation of this single effect evaporator. 00:00:33.190 --> 00:00:40.900 So in this schematic representation we have a feed coming in which is a liquid and we 00:00:40.900 --> 00:00:50.280 are going to remove vapor and therefore end up with a concentrated liquid leaving the 00:00:50.280 --> 00:00:51.970 system. 00:00:51.970 --> 00:00:55.630 So we will label this feed, F in kg/hr coming in. 00:00:55.630 --> 00:01:03.309 And it is going to be coming in with a concentration of the solute, the solid that are dissolved, 00:01:03.309 --> 00:01:07.049 a mass fraction xF. 00:01:07.049 --> 00:01:11.520 Vapor is going to be pure solvent, which in general most of the systems we are looking 00:01:11.520 --> 00:01:12.520 at,this is water. 00:01:12.520 --> 00:01:16.719 So that means that he mole fraction of the solute is zero there. 00:01:16.719 --> 00:01:18.979 And then we have the liquid. 00:01:18.979 --> 00:01:22.329 So this is the concentrated liquid. 00:01:22.329 --> 00:01:28.709 This is our objective is to create this concentrated liquid. 00:01:28.709 --> 00:01:30.909 And so we have mass fraction of the liquid. 00:01:30.909 --> 00:01:36.630 The mass fraction of the liquid is greater than the mass fraction in the feed. 00:01:36.630 --> 00:01:43.460 A couple of things to note, remember this is a dilute liquid, feed temperature is less 00:01:43.460 --> 00:01:47.289 than the temperature in the evaporator. 00:01:47.289 --> 00:01:52.219 We are going to call this temperature T of the evaporator (TE) and we are going to assume 00:01:52.219 --> 00:01:53.289 this is well mixed. 00:01:53.289 --> 00:01:55.630 And this is important. 00:01:55.630 --> 00:02:01.500 Assuming this is well mixed, the temperature here, the temperature here, are also the evaporator 00:02:01.500 --> 00:02:09.090 temperature and this is because this is mixed in the evaporator sufficiently such that we 00:02:09.090 --> 00:02:11.160 have no gradients in temperature. 00:02:11.160 --> 00:02:17.269 So the pressure in the evaporator is going to determine what the evaporation temperature 00:02:17.269 --> 00:02:19.170 is for the system. 00:02:19.170 --> 00:02:25.810 The other aspect of this single effect evaporator is that we have steam coming in to cause the 00:02:25.810 --> 00:02:31.670 evaporations so we are going to have steam, which is causing the evaporation so we will 00:02:31.670 --> 00:02:38.269 have steam flow, comes into the evaporator and moves through the same flow rate as steam 00:02:38.269 --> 00:02:40.420 then leaves the evaporator. 00:02:40.420 --> 00:02:43.459 So, also kg/hr. 00:02:43.459 --> 00:02:48.360 Temperature of the steam (TS) and we are going to assume this is saturated steam. 00:02:48.360 --> 00:02:54.060 Quality of 1 and here we are assuming all the steam is condense so then the liquid is 00:02:54.060 --> 00:02:57.520 dripping out of the exchanger tube. 00:02:57.520 --> 00:03:00.849 So this is at the same temperature. 00:03:00.849 --> 00:03:06.260 So let's just look at some of the governing equations for this system. 00:03:06.260 --> 00:03:12.280 One would be a mass balance and we will do the mass balance on the overall feed. 00:03:12.280 --> 00:03:16.780 So this feed coming in is split into two streams. 00:03:16.780 --> 00:03:18.280 Liquid plus vapor. 00:03:18.280 --> 00:03:22.250 And then we can do a mass balance on the solute, the solid. 00:03:22.250 --> 00:03:26.829 So the mass fraction in the feed times the flow rate. 00:03:26.829 --> 00:03:34.079 The only place these solids leave are in the liquid phase, the mass fraction of the liquid 00:03:34.079 --> 00:03:39.329 times the mass flow rate of the liquid that is leaving. 00:03:39.329 --> 00:03:42.269 So this is our mass balances for the system. 00:03:42.269 --> 00:03:48.860 One of the energy balances that is important for this single effect evaporator is the heat 00:03:48.860 --> 00:03:50.439 transfer. 00:03:50.439 --> 00:03:58.140 So the heat transfer rate, this is kJ per hr, equal to an overall heat transfer coefficient. 00:03:58.140 --> 00:04:04.519 So this heat transfer coefficient is taking into account all the heat transfer both sides 00:04:04.519 --> 00:04:07.840 of the heat transfer tubing. 00:04:07.840 --> 00:04:19.910 So this is kJ/m^2 per hour per degree K. Heat transfer area which would be in square meters. 00:04:19.910 --> 00:04:27.780 And then the temperature difference so the TS, which of course is higher than the TE. 00:04:27.780 --> 00:04:31.330 Bigger this temperature difference of course the faster the heat transfer. 00:04:31.330 --> 00:04:34.330 And so these are the basic equations. 00:04:34.330 --> 00:04:41.070 We now also have to do enthalpy balances on the streams that we will look at in a subsequent 00:04:41.070 --> 00:04:43.190 screen cast. 00:04:43.190 --> 00:04:49.880 There are more efficient ways to carry out evaporation for example a triple effect evaporator 00:04:49.880 --> 00:04:56.320 to have three of these in series, that saves a significant cost on the steam but adds a 00:04:56.320 --> 00:05:03.800 cost of the construction and so it is a relative tradeoff, it is a tradeoff of the cost of, 00:05:03.800 --> 00:05:08.630 if steam is cheap, then this might be the better approach.
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