 After that introduction to the basic concepts of mass transfer across interfaces in the last lecture, we resume our discussion on evaporation especially of water in its retardation. And in that context, we generalize upon our understanding of resistances, their analysis and concentrate in particular on diffusional resistance in the gas phase today. It is seen that satyl alcohol is or hexadecannol is effective in containing the evaporation losses without sacrificing on the dissolved oxygen concentration level in large water bodies. So, we talk a little bit further on this particular compound hexadecannol, we will quickly review in a couple of slides the performance of hexadecannol and some of the practical aspects which might appear surprising. Hexadecannol is very suitable in practice for retarding water evaporation. One advantage is that it spreads spontaneously and rapidly from solid beads. This could be kept in gauze covered wraps in surface and then we have the satyl alcohol or hexadecannol spreading rapidly over the surface. The surface pressure is about 40 dynes per centimeter and it effectively retards evaporation by about 50 percent. Notice when the reservoir surface if it is contaminated heavily then we might need to aid the spreading of hexadecannol by using kerosene. So, it is possible to dissolve hexadecannol in kerosene and then the solution can be made to spread rapidly. One may of course have to deal with issues like the VOCs emissions. Presumably the quantities of the spreading aid solvent would be miniscule so that should not in general matter. There is some interesting features about practical trials of hexadecannol which have been tried in the US sometime back. Other sterile alcohol suspensions were allowed to spread by using a perforated plastic hose and the spray application was chosen at the windward end of the reservoir so that the sterile alcohol could then propagate on the surface. Commercial satyl alcohol interestingly has only 45 percent purity. Next of it is meristil alcohol and sterile alcohol and this impure or mixed product has a greater resistance for retarding this evaporation of water of the order of about 14.7 into 10 raise to 4 centimeter square second per gram in Langmuir's terms compared to pure hexadecannol which offers a resistance of 5.3 into 10 raise to 4 centimeter square second per gram. There is another application of the effect of higher alcohols in reducing evaporation is found in nature this is a different one. Certain insects have wax cuticle of the order of a few microns in thickness and as the temperature rises about above about 30 degree centigrade this becomes permeable to water therefore the insect can lose water and cool itself. This has been documented by Biemann's work in 1955. Now let us move on to a bit of analysis as we go along we will take our starting point as an experimental observation. If we take small droplets of liquid or small spears of solid and consider the loss of this liquid or the solid into steel air experimentally one finds the rate of change of area of this droplet or solid spherical particle to be a constant. Let us refer to this as our equation 1 a rate of change of area is constant and for a spherical geometry a therefore is 4 pi a square where the small a is the radius of the droplet and for water this constant is about 3 into 10 raise to minus 5 centimetre square per second. For an iodine particle it is 5.4 into 10 raise to minus 4 centimetre square per second. Oh we generally have that because we want to conserve water mainly large water bodies as I said would encounter more losses than utilization 10 feet roughly per year that may be typical loss. The see the issue is the following that in general the petroleum based liquids they are for certain application and we would never choose to keep them in open conditions ok. So they would be always in storage tanks yet you may say that as we deplete or drain out this liquid the vapor will evaporate the issues there are while it may be possible to retard evaporation even there it might introduce impurities which are not desirable say for example if you have some of the higher molecular weight monolayers present in let us say a petroleum tank petrol tank we might not permit those as impurities in our fuel having said that it may always be possible to choose suitable monolayers which can permit this in this context I may elaborate a little bit further you probably all I have heard about blending of anhydrosethanol in petrol and diesel in many states of our country that is proposed and in some cases I think implemented also. So you have now another volatile component besides the petroleum ingredient alcohol and there are other issues they might be absorption of water vapor it is opposite of the losses of petroleum but they may be absorption of water which is a concern because if it happens in a storage tank the water would eventually settle but there is always a possibility of taking of that water along with petrol sometime that that would create problem in the engines. So one of the possible solutions would be that you safeguard against any separation of water which might occur if you have anhydrosethanol there will be a tendency for water to be absorbed by such a blend and if that is unavoidable then the only other thing you should do you could be able you would be able to do is make sure that that water does not separate if remains if it remains in dispersed form as very very tiny droplets or domains then you can achieve that purpose that could be done by using some suitable surfactants. What it means is now that you have these three key ingredients the hydrocarbon petroleum or diesel alcohol and surfactant if these components are present then we have a possibility of containing any Epsom water or any water which may be contaminating this mixture in advertently in small quantities it could be dispersed making a micro emulsion. Now water is now split as fun such fine domains which would be shorter than maybe quarter of wavelength of light it would never separate and unlike emulsions these micro emulsions are thermodynamically stable. So that is one solution which you can envisage for the opposite of retardation of evaporation if there is any water pickup then it is not allowed to create any problem it turns out that this strategy works very well and it actually improves the performance of the fuel as has been checked we have some research here conducted at IT Delhi and patented we shows that such some of such micro emulsion fuels can give you not just a better thermal efficiency but also contributes to the increase engine life because you allow for complete combustion at slightly lower temperature and certain mechanism which involves micro explosions of these water domains in the IC engine that allows you to practically prevent formation of particulates. Now why I am going into depth of this particular question is because the surfactant that would be very desirable here will also be able to retard the evaporation the question you started out with. So it is possible for certain applications to think of intelligently choosing a surfactant which would take care of many problems absorption of water evaporation losses and so on. So some of the dispersion concepts can be utilized in improving the basic product performance and also elevate the VOC's emissions. So but you have to do it carefully all right. So we have some idea now as to how this area would change with time for water and iodine how do we study experimentally such systems if evaporation is to be studied. You could think of may be very fine droplets if the droplets are about 1 micron in radius is about 1 micron then you might conveniently study this fine drop while it is falling through air while it is falling through air it would also lose the liquid through evaporation. We might be able to find out how evaporation occurs usually one finds a linear relation between the velocity of fall and the time elapsed. And we started out with rate of change of area with time equal to constant in terms of this radius we may be able to write analogously dA square by dt equal to another constant. Let us play with this a little bit. Other situation that one may think of is relatively larger drops say about a millimeter sized. In order to study how evaporation occurs for these larger drops now might not be able to have a free fall studied so easily because that will be rather quick. We might instead choose a micro balance span which is coated with carbon black and now such a drop about 1 millimeter size could be kept on this carbon black coating which would not allow it to spread and then you can observe how the weight loss occurs over time. You might be able to measure the radius also as a function of time. If you look at the loss of weight of such a drop of liquid then we could start with this weight being equal to 4 third pi a cube nearly spherical drop times the density of liquid that is the weight and dw by dt would be d by dt of a cube that will be approximately varying as d a cube by dt or as a square dA by dt that can be alternatively written as a dA square by dt. These are not equalities these are how they vary in proportion. So, this is also equal to a into dA square by dt but dA square by dt as we have seen just now the constant. So, dw by dt is equal to a and per area we may write dividing through by a this would go as 1 over a area will be proportional to a square a by a square is 1 by a. Sir, I was asking whether we are considering the effect of change in Gibbs free energy when the molecules are like evaporated from the surface. No, no, no see these are much larger drops right you are talking of over 1 millimeter size. So, now we are not going to worry about any change of surface tension or related quantities. So, and we are going mainly by experimental observations even in this argument when you substitute dA square by dt equal to a constant we are making use of extendable factor. These are two large drops for changes of surface tension of Gibbs free energy to occur. In any case even in those molecule level clusters having may be tens of molecules of water the surface tension would change only very little. But the main result now is that per area if you look at this rate of weight loss you see that now it goes inversely as the radius. So, the smaller the drop becomes greater is the rate of evaporation per area becomes very very large when the drop becomes very small. In the final moments of evaporation it might just appear in a flash or disappear in a flash. The same result is found for other liquids besides water for aniline, paracrysal and so on for a range of radii from half a micron to even about 2 millimeters. So, that goes on to prove that we are not definitely in that region where there may be other considerations. It is here that we argue that our equation 1 rate of change of area with time as being constant is consistent with the gas phase resistance R g being the rate controlling or the process being diffusion controlled. That is the only way you would envisage that this rate of change of area would be constant. There is some other resistance which is not dependent on what is happening inside the liquid that is controlling that is a gas phase resistance. So, we need to now get acquainted with bit of diffusional concentration. So, one could write from diffusion theory the following expression minus d w by dt where w is the small w is the weight of material evaporating is equal to 4 pi m d p a by RT where d is the diffusion coefficient in the gas phase m is the molecular weight p is the saturation vapor pressure and capital T is the absolute temperature. You see here that we have the diffusivity and a p by RT term. The vapor pressure by RT is the gas phase concentration that would be present at the interface between the droplet and air. So, you have those elements of what we thought the diffusion coefficient constituting part of the rate constant and p by RT constituting the concentration driving force. Far away from the surface of the drop the partial pressure of the vapor would be presumed as 0. You could derive this yourself from the basic diffusional concentration. I will leave that as part of your homework to think about. So, according to this if we look at the fractional weight loss we should be dividing the left hand side by w minus 1 by w d w by dt will be then proportional to the right hand side which contains a by a term containing a cube weight will contain volume and density volume will be proportional to a cube. So, when you divide by w the right hand side would scale as a by a cube or 1 by a square. So, minus d w by dt by w the fractional loss of weight will go as 1 by a square. If you take a water drop of initial radius of 1 millimeter at 18 degree centigrade in still dry air it would take about 11 minutes to disappear. We could write our equation 2 in terms of surface area. So, we have now 1 minus 1 by a d w by dt equal to whatever was there 4 pi m d p a by RT divided by 4 pi a square that simplifies to m d p by RT a 4 pi cancels 1 of the a is cancels. And therefore, we could write this as 1 by 4 pi a square that is a spherical surface area sphere surface area and the w is written in terms of density volume product. So, 4 third pi a cube is the volume 4 third pi is constant. So, taken out d by dt of a cube rho L that will be equal to m d p by RT a right or that could be simplified. When we simplify this will give you 3 a square rho L d a by dt 3 and 4 pi cancel a square cancels. And when you take this a on the other side we have this result minus a rho L minus a rho L d a by dt which could be alternatively written as minus rho L half d a square by dt or minus rho L by 8 pi d capital A by dt where capital is the area and that is equal to m d p by RT or minus d a by dt equal to 8 pi m d p by rho L RT. So, we have these 2 results we can write minus 1 by a d w by dt is equal to m d p by RT a or minus d a by dt equal to 8 pi m d p by rho L RT. When we look at the structure of the right hand side here for equation 4 we see that this equation 4 is in accord with the experimental equation 1 rate of change of area equal to constant and it would mean high rates of mass transfer per area at small a. This would be a situation corresponding to diffusion in a radial diffusion field. Sir capital P is a saturation diffusion. No, no, no. P is this P is same I think there is a bit of a inconsistent writing. This is the same P the saturated vapor pressure. Certain natural to be dependent upon the areas of the droplet. That is if you go down to extremely small droplets. So, one would see that if you really go down to that last phase where the evaporation is very rapid several things will come into picture. Apart from the a square in denominator there is also the increasing vapor pressure for smaller droplets. So, that will make that last part really rather quick. Now, the geometry requires that R g is reduced because the concentration falls off very rapidly away from the surface. When we compare our equation 4 against experiments the agreement is pretty good. For iodine we have minus dA by dt equal to 4.4 into 10 raise to minus 4 centimeter square per second calculated and the experimental is 5.4 into 10 raise to minus 4 centimeter square per second. This clearly establishes the importance of diffusion as the rate controlling step. If you apply the diffusion theory to very small droplets then it implies that minus 1 by capital A dW by dt varying as 1 over small a will actually approach infinity as the radius tends to 0. This result is a consequence of our assumption that R g is so large that R l can be neglected. That probably would not be true for very small droplets because for very small droplets R g would reduce very significantly. What is that? If you look at the length scale over which the gas phase diffusional resistance is operative as the droplet becomes smaller and smaller that length scale is also smaller. So, corresponding gas phase resistance in the gas film gets reduced as the droplet shrinks. That is what this is supposed to negate. So, we see here a catch we cannot say that R g alone will remain the controlling resistance. So, this is a consequence of our assumption in the entire history of evaporation when we start with the given droplet and it all ends in vapor. So, at some stage R l the liquid phase resistance will have to come into picture. Otherwise we have to deal with that kind of result minus 1 by A dW by dt becoming infinity So, the way to do that is bring back our idea that we have these resistances in series. So, when the droplet becomes very small we now have to bring in the liquid phase resistance and the way to add that resistance could be this minus 1 by A dW by dt as m p by R t A by d plus R l. See this if you do not have R l we have the same result as we had earlier m d p by R t A. We are merely writing this in that generic form rate is equal to driving force by resistance. The driving force is p by R t the saturated vapor pressure that is at the surface of the drop and A by d is the resistance in the gas phase. p by R t is the driving force in the gas phase for evaporation and A by d is the resistance in the gas phase. So, A by d is the gas phase resistance to that we must add the liquid phase resistance. So, that gives us equation 5 the right hand side is written as m p by R t A by d plus R l. For very small drops the rate of evaporation tends to that in vacuum. For a drop of radius about 10 microns the correction due to this liquid phase resistance increases life time by less than 5 percent, but for about 10th of micron it has a 6 fold effect. So, near the end of this life of the drop the liquid phase resistance becomes a very significant one. On a different line maybe I should also tell you of that interesting example for demonstrating the gas phase resistance and liquid phase resistance in comparative sense. You could think of a situation I would need the visualizer please. If we consider a gas liquid interface and we think of transfer which is which is opposite to what we have been talking of from gas to liquid and we are not necessarily talking of opposite of evaporation that is condensation, but it may be absorption of some component A from an inert I into a liquid say B. Now you can think of a situation where we may have mixture of A and I. Then for making way up to the interface the A molecules will have to overcome the collisional barrier offered by the inert molecules as well as molecules of its own kind. So, there will be a resistance in the gas phase. In order to reach up to this interface that collisional barrier for A molecules to reach up to the surface will have to be scaled. Predominantly if inert is in excess then most of the intermolecular collisions will be A with I as far as A is concerned in its path up to the interface. So, the resistance in the gas phase will be highest when the concentration of A is lowest. So, very dilute gas phase mixture of A with inerts the resistance in the gas phase is likely to be significant. Now let us think of a slightly different situation. Supposing instead of the mixture we go to a situation where we have only pure A the gas is only pure A. Now the resistance that A molecules will have in reaching up to the interface will be the collisional barrier of molecules of the same kind. We may expect therefore, the gas phase resistance to be the minimum possible. The question then is if that is the situation does it mean that the resistance in the gas phase is 0? Sir, there will be some of vapor of liquid B as well. You are right if there will be if there is some vapor of B that will offer the resistance. Supposing we can neglect the presence of vapor in the gas phase then could we say that the gas phase resistance for pure A without inerts without vapor of B to be equal to 0. Maybe look at the time 0. At time 0 it is all pure A here gas phase is pure B here and the moment the gas has contacted the liquid the interface would become say saturated. So, the concentration in the liquid phase at the interface is the saturation concentration C s star. The moment the gases come in contact pure gases come in contact with pure liquid the interface is saturated. The liquid is pure B there is no A to begin with which means that at that time 0 the concentration of A everywhere inside the liquid is 0 C A in liquid is 0 everywhere except at the interface. At interface it is a finite saturation value which means if you are looking at the concentration profile you see now here that the concentration of A is becoming 0 within no distance from C A star it is dropping to 0 within no distance visualizing the interface as just the geometrical plane. If that is the case the concentration gradient at time 0 is infinity C A star minus 0 divided by 0 anything divided by 0 is infinity. So, this gradient is infinity if the concentration gradient at the surface at time 0 is infinity that means the flux is infinity because it is flux is proportional to gradient of concentration. So, if you look at the liquid alone on the left hand side liquid alone once C is that at time 0 the liquid has the potential on account of that infinite gradient to carry away infinite amount of gas per area per time. It is a capacity of the liquid to on account of that infinite gradient to carry the A. The question is can you even for as small time as let us say 100th of a picosecond have a capacity infinity met in practice. If you had infinite flux for even very very small time you would be able to transport infinite amount of A from gas into the liquid is that possible. We are assuming that the distance is a finite or 0. But the exact distance would be infinite initially at equal to 0 so that the net flux would be finite. No the exact distance is at the most equal to thickness of the interface which will be of the order of 25 nanometers. Even if you take 25 nanometers it will lead to fluxes which are practically very very large. And therefore, this argument of equating 25 nanometers to 0 and the gradient being infinity is a reasonable argument. So, the capacity of the liquid to carry infinite amount of gas per area per time at time 0 does not mean that it can be realized in practice. The question is why? Why is it not possible? So, because the concentration of the gas A at the interface is different from that in the bulk. So, they cannot be an infinite amount of flux at the equal. Let us put it this way that this C A star is the concentration of A the dissolved gas in the liquid. In the gas phase the concentration is some uniform concentration which will be much lower than this because gas is a rare or medium. So, maybe the gas phase concentration I can represent as C A g because pure gas it is constant right up to the interface at interface is corresponding to the C A star, but a lower value. There will be an equilibrium between liquid phase concentration C A star and C A g in the bulk of gas. But you are on correct line of thinking you are saying that the capacity is there to carry infinite flux, but can infinite flux be there provided by the gas and you are trying to say that it is not possible, but the arguments are not correct because now since there is no gradient in the concentration for the gas phase A we cannot argue that way. But we can we have to think of how would that infinite flux if it is to be carried into the liquid arrive at the interface. It has to come in terms of molecules of A reaching up to the surface. How would they reach you know how they would reach you know the strike rate of these A molecules on this surface that P by root 2 pi m k T kind of thing right. The thermal kinetic energy will permit molecules of A to move at certain velocities they can only come up to this surface at some representative velocity corresponding to that temperature. And since the temperature is finite that velocity is finite the strike rate is finite therefore the gas does not have more than a certain amount to be supplied per area per time. Although liquid has capacity to transfer if anything very much larger than what is actually present where to be made available. So, we have to now see that within the gas there is a self diffusion which is providing the barrier the collisional barrier is of like molecules. There is only a certain rate at which A can be permitted to impinge on the liquid surface. So, there is that initial resistance in the gas phase even for pure gas. Let us say R g 0 and let us put pure here some minimum resistance for pure gas at time 0 that has to be there that is a finite number you cannot transfer more than what arise here and what arise here is finite. So, this is one resistance. Let us look at a situation a little later I am basically reinforcing the same understanding in one step I added to gas phase. Gas phase resistance A by D the liquid phase resistance I am trying to demonstrate that in similar fashion how do you start with some minimum gas phase resistance. The minimum value being for pure gas in contact with the liquid and liquid being free in the beginning what is the liquid phase resistance. Let us think of that if this gradient is infinity that means the flux possible in the liquid is infinity which means resistance in the liquid is resistance in the liquid is 0 because once again flux or rate is driving force by resistance the flux is infinity driving force is finite that means resistance in the liquid is 0. So, you may say that R L 0 is 0. Because this flux is infinity. So, it should allow some of that particular A to transfer to volume, but as in the volume the computation is 0 that means it is not along. No, no, no, no, no, no, no, no, not that what we are saying is we are only looking at that initial moment time 0 what you are saying is what happens after time 0. If after time 0 the concentration remains 0 inside then it is true that liquid is not allowing A to get in, but we are still stuck at time 0. We are still not moved on we are still at time 0. At time 0 the gradient is infinity driving force is finite. So, driving force by resistance being equal to rate being infinity resistance must be 0. So, at initial moment there is only resistance in the gas phase presuming no resistance in the interface liquid phase resistance is 0 correct. Now, let us move on suppose we give some time what happens in the gas phase we still make the liquid non volatile we keep the liquid non volatile nothing of liquid evaporates. This concentration in the gas phase remains uniform pure gas concentration supply of gas may be the vessel in which the gas is there is large. So, that there is no depletion of C A G there is no reduction in pressure, but what happens now in the liquid phase as time progresses we do not express expect this gradient to remain infinite throughout at times little greater than 0 time greater than 0 may be the profile would have advanced up to here this would be now a new C A as a function of Z if this is the direction Z at some time t greater than 0 at time t greater than 0 the concentration inside the liquid in this depth would have become greater than 0. In the process the gradient has fallen from infinite to something finite still large, but it is much lower than infinity as time progresses this concentration is constant the profile would have shifted may be up to here and so on. So, that is the direction for increasing time as time increases the concentration would become non-zero over greater and greater depths as more and more liquid participates in this diffusional process more and more resistance in the liquid phase is being added gas phase resistance is constant at the same R G 0 for pure gas liquid phase is resistance rising from 0 to a greater and greater value. You may envisage that at some time the liquid phase resistance is so large compared to the gas phase resistance that for argument sec we may say that the gas phase resistance is only now 10 percent of the total. If R G 0 pure divided by R G 0 pure plus R L which is dependent on time if this is multiplied by 100 if that becomes 10 percent the liquid phase resistance must be 9 times the gas phase resistance. So, that the gas phase resistance only 10 percent of the total then we could say that for that time and greater time the gas phase resistance now as long as the gas is pure will be insignificant part of the total. The guess I want you to make and that is a dramatic one is for a typical system gas liquid system what time would that correspond to how much time will it take for the liquid phase resistance to become overwhelmingly large compared to the gas phase resistance. If we start with pure gas I will just answer that question and we will stop here it will be of the order of 10 picoseconds 10 raise to minus 11 seconds. That means it is such a fine time such a short time that the gas phase resistance becomes controlling beyond that time it is a liquid phase resistance all over. So, we will stop here.