 I do not think I did thermodynamic consistency I mentioned it reverse osmosis and then boiling point elevation and freezing point depression okay we will discuss thermodynamic consistency first the idea is the following if you look at the Gibbs-Duhem equation looking at I will write the expression T of G excess by RT suppose I have to write this in capital letters G of G by G excess I mean ? G- ? G ideal mixing similar students have come if you are looking at vapor liquid equilibrium let us say in particular you are looking at a binary system the number of degrees of freedom is simply 2-2-2 2 phases so I can write this as D of G excess by RT first of all is equal to – H excess by RT squared dt plus P excess by RT dipping plus ? excess by RT is exactly log ? 1 Tx1-log ? 2 Tx2 remember this ? excess by RT ? – ? ideal this is simply log ? I and if I divide by the total number of moles dividing it for a closed system if I divide by the total number of moles and I simply get log ? 1 Dx1-log ? 2 Tx2 no plus sorry no minus here this is plus since there are 2 degrees of freedom there are either you can consider isothermal VLE in which case this term is 0 or this is 0 for isobaric in any case you get D of G excess by RT by Dx1 is equal to log ? 1 by ? 2 log ? 1-log ? 1 ? 2 Dx1 Dx2 is – Dx1 plus ? Tx1 I am sorry just ? where ? is either ? is H excess by RT squared – H excess by RT squared dt by Dx1 or it is V excess by RT Dp by Dx1 2 degrees of freedom so the degrees of freedom that I use either I do isothermal VLE in which case temperature is constant the other variable is X1 so if I treat T and X1 as variables in a 2 in a system with 2 degrees of freedom DG excess by DX1 will be this plus ? where ? is simply V excess by RT Dp DX1 see sometimes written as ? T this is usually written as ? P and this is ? T this means that constant temperature ? is V excess by RT DP by DX1 now if you integrate this over X1 the left hand side is simply DG excess by RT from X1 equal to 0 to 1 so you will simply get G excess by RT sorry from X1 equal to 0 to X1 equal to 1 this is equal to integral 0 to 1 log of ? 1 by ? 2 DX1 this left hand side is clearly 0 left hand side is simply G excess at X1 equal to 0 and X1 equal to 1 so it is clearly equal to 0 so you get the inclusion that log ? 1 by ? 2 DX1 integral 0 to 1 should be equal to ? – S so – 0 to 1 ? DX1 now this term is approximately 0 experimentally H excess by RT squared is very small and so is V excess by RT both these terms are usually small so you get the integral of this over DT DX1 to DX1 so because these that is why the symbol ? was used because this is approximately 0 this is a test of consistency if you have data experimental data on ? 1 for example you can calculate ? 1 from experimental measurements on pressure composition of the gas phase and compression of the liquid phase so this comes from experimental data if you have experimental data on ? 1 ? 2 you can plot log ? 1 by ? 2 versus X1 the area under the curve should be approximately 0 actually if you actually do the plotting you get log ? 1 by ? 2 versus X1 if you go to X1 going to 0 ? 2 is 1 you get log ? 1 infinity that will be the limit here so let us say log ? 1 infinity somewhere here and log ? 2 infinity here so you will get a curve like this this area there is a positive area and a negative area this area we will denote this area by P this area by N to indicate positive and negative areas so the experimental the integral test for thermodynamic consistency as I told you earlier people used to use the experimental data to verify the Gibbs Duhem equation now the Gibbs Duhem equation is so well established that the test for consistency of the data have you measured the data correctly you measure Y1 X1 etc if you have errors in the measurement then it would not satisfy the integral test the integral test says if P is the positive area and N is the negative area P P plus N divided by P plus N is algebraic quantity that is a negative area this is the positive area is less than a prescribed amount say 1% depending on the if then data is said to be consistent said to satisfy the integral test this test is called the integral test of consistency because you are taking an integral over the entire composition the original differential test is simply this of course this is almost never satisfied by the data simply because it is too hard to apply this is your Gibbs Duhem equation if you can verify at every point that this sum is 0 that is called the differential test usually differential test is satisfied to plus or minus 20% because your you have to plot log gamma and take the slope with respect to X1 these slopes are very difficult the scatter in the points will be so much very difficult for you to get an accurate slope so this differential test is to van is due to vanness in somebody but this is normally not used for want of accuracy for calculating slopes this possible there are instances of data there is a journal called journal of chemical engineering data this has a large number of data it will be it is exceptional cases where the differential test will be applied but the integral test is almost always applied this goes on the name of thermodynamic consistency and experimental data will always whenever you report experimental data you have to show that they satisfy the integral test otherwise they would not even accept it for publication this is one thing I omitted to discuss the other issues this is all this is for solvent-solvent systems and I have discussed it for binary you can actually discuss it in ternary and so on as far as you are concerned you just need to know what is thermodynamic consistency and how you test a binary system for data the other omission as I told you is reverse osmosis is far too important the process is also a nice plant in a place called Nareepa you this is in near Madurai it is a beautiful plant BHEL has constructed this plant what it involves is taking the there they using grid power not photovoltaic I am suggesting photovoltaic because grid power is also in short supply in India also drinking water supply is a need that is distributed there is no point in saying I will have 300 million tons of drinking water here in Chennai and then I will supply it to obscure villages never get there. So the whole idea is to have a series of plants and one of the biggest problems in reverse osmosis the membrane let me just describe the thermodynamic principle then I will describe the plant for you let us say you have a membrane as usual thermodynamics is do not have to be practical you can make drawings just to indicate this is a membrane the process so I have here I want H2O plus salt here this is component to this is component one this is just pure water ideally what I would like a seawater to come in for me to take out pure water typically the TDS the total dissolved solids and water that you drink is about 300 400 very acceptable for portable water but TDS in reverse osmosis water will be about 60 70 very low less than 100 and if water has very low TDS it is said to be very soft and it is an excellent solvent so very low TDS water cannot be passed through mild steel pipes and the curious thing about Naripayur is the following the state government wanted a this is a backward village and they wanted drinking water supply for this so they put up an experimental unit I think it is quite a large unit it is almost 1 million litres per day and they set up this plant BHEL set up this plant the reverse osmosis plant so what you do is take brain the process is something like this you have to take brain from seawater you will have a pump that pumps this out this is C in this brain has passed through a lot of processing the final unit is a membrane unit you should have brought you know if I have left my stick in this I have photographs of this membrane I will show you the membrane unit essentially consists of tubes because you need very large surface area each of these there are tubes like this inside each of the whole thing is a cylindrical tube of about one foot diameter the one you have here in the hostel is about one foot which in which there is a large number of tubes made of membrane semi permeable membrane the semi permeable membrane is such that it will let water through and no salt through so if I am doing an equilibrium between these two phases for at thermodynamic equilibrium you have mu 2 on that this is a this is beta alpha is equal to mu 2 beta for the moment let me assume that in fact the pressures will be different let us assume the pressure here is P and the pressure here is P prime the temperature is T mu 2 alpha is the chemical potential of water so you I have to write the chemical potential water in the mixture so it will be at T and P plus RT ln ? 2 X 2 that is the chemical potential model for water in solution on the right hand side it is pure water so it is mu 2 pure I do not need alpha here sorry this is liquid mu 2 liquid pure if you like this is also mu 2 liquid pure but this is at T and P prime in fact if the pressures were the same these two will cancel you will get log ? 2 X 2 has to be equal to 0 so ? 2 X 2 has to be equal to 1 so X 2 will be 1 which means essentially what will happen is if they were at the same pressure all the water will come this side it will keep diluting the solution till it gets more and more dilute and all the water will be on this side you actually want it the other way around the equilibrium process will cause water to come back in here so you need a pressure difference P – P prime to drive the water this way that can be calculated from here you get log ? 2 I will complete that picture in a minute log ? 2 X 2 is equal to this minus this right is equal to mu 2 liquid pure at T P prime – mu 2 liquid pure at T and P you know ? mu 2 by ? P is simply the specific volume of the pure substance and water is practically incompressible so this is simply V 2 liquid pure V 2 liquid into P prime – P this is written as V 2 liquid into ? with a – sign because P – P prime is called the osmotic pressure so this three lines it is a definition P – P prime is ? and ? is called the osmotic pressure yeah now this is all right now this one is RT here you can get an idea in practice you have to take ? 2 into account but for the moment let us just assume for sufficiently dilute solutions and see what is not really sufficiently dilute you have to actually do the calculations with ? 2 dilute solutions your log X 2 log ? 2 X 2 is approximately log X 2 because X 2 is approximately 1 X 1 goes to 0 X 2 goes to 1 tending to 1 not exactly the limit so this ? 2 becomes 1 when X 2 goes to 1 so you get approximately log X 2 which again is equal to 1 – X 1 log I am sorry log of 1 – X 1 which is – X 1 which is approximately – X 1 this is equal to – V 2 liquid into ? by RT so your ? the osmotic pressure is RT by V 2 liquid into X 1 to give you an idea of the order of magnitude let us calculate this in atmospheres R is for this system ET approximately ET temperature let us take room temperature by V 2 liquid is 18 x X 1 to mole fraction of salt how much does this come to 940 this is about 33,000 300 x X 1 your mole fraction of salt is 0.1 you are talking of 130 atmospheres actually typically for salt water for sea water the osmotic pressure is about 75 atmospheres so the engineering challenges you have a semi permeable membrane this membrane by its very structure for example the two kinds of membranes say the polymer membranes are made by DuPont so actually the first manufacturer of these membranes was Monsanto now the largest manufacturer is DuPont in the many other companies many are hidden DuPonts they are all subsidiaries and so on and then there is cellulose acetate membrane I forget the polymeric membranes exact name there is a polymeric one there is a cellulose acetate one this is non biodegradable this is degradable this is biodegradable cellulose acetate the membrane has cannot have pinholes if you have pinholes you will get all the salt water coming through so you are talking of a membrane that does not permit salt through but permits water the size difference after all is quite small between sodium chloride and water so you are talking of membrane that has it is like a molecular filter it filters out the sodium chloride sense the water through but it also has to withstand in a pressure of about 70 atmospheres just typically x1 will be about you can calculate x1 70 by 1300 7 by 130 what 0.25% is for x1 is 0.05 if it is sodium chloride you are talking of x1 is moles of sodium chloride into what is the molecular weight of sodium chloride in a CLS what 38.5 okay let us say approximately 60 if you are calculating moles if M is the molality it means it is moles per 1000 grams of solvent by 1000 by 18 x1 is simply M by this sorry this is correct M by 60 now this is moles per 1000 grams of solvent so plus M that is all so if this is 0.05 this is about 55 0.05 so M is what M is approximately 55 x 0.05 0.275 now 2.75 grams per 1000 grams of solvent or into 60 give me 120 plus 45 165 grams per 1000 grams of solvent roughly 15% by weight so C water is that composition so you get about 70 atmospheres for this and for 70 and just this is about 70 atmospheres for C water for brackish water that you take from the ground the value comes to about 20 atmospheres so reverse osmosis plants will differ because you have to have a membrane that is mechanically supported it is usually supported with metallic or ceramic supports which will essentially still allow this water to come through will the membrane should not tear it should not have pinholes you have maintenance problem continuously right now we do not have a membrane manufacturing unit here you must know that these membrane units are also used in dialysis we have actually done a piece of work here just mention that to you because it is an interesting problem for chemical engineers on a large scale is much harder to try but let me explain this the membrane units are like this I will get you a picture of this I will this is each of them is called a lumen and each of these is millimeters in thickness I mean these each lumen is only about 1 millimeter in radius so you are talking a very thin membranes which are kept all over the place there will be thousands of them in one bundle and all of these will be inside a shell so what will happen is you will have a tube like this let us say this is the inlet this is the outlet you have a series of tubes through which liquid will pass the shell side you will get out the water alone will come out this is seawater pure water will be taken out here this will be concentrated brain now this unit is there in your hostel what you will see is the outside shell the ceramic shell water that comes out is concentrated brain in your case you are actually using you are not bringing in seawater you are using brackish water so we have a bore well that picks up the water that is at something like 800 TDS then goes through this unit this will come out much higher concentration you can calculate by mass balance this will come out at about 40 the problem with reverse osmosis is it would not give you it insists on giving you a certain quality of water we will get 50 60 TDS although you are satisfied with 300 the membranes do not make that compromise the membranes will give you 60 but let me come back here this membrane the question is the same thing happens in the case of dialysis artificial kidney in the artificial kidney this is your blood plus urea blood containing urea here you get pure blood out here you get urea out which is a waste in this case you do not have this problem these are the lumens inside each lumen looks like this it is called a lumen and each lumen is about 0.5 0.2 mm diameter 0.2 mm varies the main advantage of a small large number of small diameter tubes is surface area you get that is what gives you the productivity so the urea is removed and wasted urea is a larger molecule than any of the most of these chlorides so what they actually do is they use a dialysate outside the solution outside the pipe will be identical with blood except for its urea they do not need the blood consists of also proteins proteins won't come through anyway so you can leave the proteins out but all the other components you will match the composition of the individual blood so they have this with labeled with the patient's name because it is reused and the outside solution is prepared exactly to the same specifications as this blood here except for the urea so the only thing that will come out of this thing is urea and there is a rule that blood should not touch any of the other parts so the whole thing is done in PVC this thing which is known to be biocompatible a special PVC material is used for the piping and when the blood is taken out it goes directly into this there is a peristaltic pump the way you pump this is the pipeline the tube goes through this the tube from the patient will go through this but there is a cam arrangement as this rotates every time it will pinch the my drawing is terrible of course these are two solid surfaces as it pinches it as it rotates it will pinch it and send out a pulse so these pumps these peristaltic pumps are used for this no contact of the blood with any other okay so the reverse osmosis is a very important part and I think for desalination it is going to be the way to do it because one of the most reversible ways of you can calculate the work here the work here is simply the volumetric flow times the pressure difference right W s dot is simply 5 times the volumetric flow V dot of flow and you can calculate that work you will find that is much smaller than the comparable processes of evaporation for evaporation you have to supply 540 calories per gram and for freezing you have to still supply 80 calories per gram in addition to that for freezing you have a unit you have the Carnot efficiency because you have to you are essentially talking about heat pump so if you take that into account you will come back to 540 but with multiple effect evaporation you can do with about 50 to 60 calories per gram here it is even less because the osmotic pressure you are talking of flow per gram you are talking of 118 this is all you have to do is take 1 gram of the flow which is 1 cc into pi 70 atmospheres so 70 into 70 cc atmospheres 70 cc atmospheres will come to what less than 2 calories as compared to 50 calories if you have a multiple effect of operator the factor of 10 so in principle this should be much less but there are hidden costs when you make comparisons like this you have to take the energy for making the membrane you are paying there right here it is less but you have to take the membrane manufacture plus this together to find out what the