 Welcome to the third lecture of module 5. We are discussing distillation. So, before we proceed in this lecture, let us have a recap on our earlier lecture we had. In the last lecture, we have discussed deviation from ideality, deviation from ideal behavior. So, we discussed two cases. One is positive deviation, deviation from Raoult's law. We call minimum boiling azeotropes and second we discussed negative deviation from ideality, deviation from Raoult's law. We call it maximum boiling azeotropes and third we have discussed the enthalpy concentration diagram. Today we will discuss flash distillation. This flash distillation sometimes also called flash vaporization, jation or equilibrium distillation. This is equilibrium distillation, equilibrium distillation. Here we can see a schematic of the flash distillation column. This is a single stage operation, single stage operation where initially a liquid mixture, liquid mixture is pumped at higher pressure through a heater, which is over 100 here, heater and the feed is pumped. It raise the temperature of the feed and then there is a valve, this throttling valve. It is throttled mixture flows through the valve to the flash drum under reduced pressure. So, this causes the liquid to partially vaporize. As soon as it enters into the drum, the liquid and vapor separates. So, the vapor and liquid goes down, vapor goes off. As soon as the mixture enters the flash drum, vapor and liquid separate. Then vapor and liquid are allowed to come in equilibrium, vapor and liquid are allowed to come in equilibrium. So, the enough space is provided for the flash drum and liquid is provided. Then finally, the vapor and liquid are removed from the system. This flash drum is very common, particularly in petroleum industries, petroleum refining. So, in many cases, in distillation also, there sometimes we use the pre-flash. So, it reduces the load on separation. So, in this designing of flash drum, it is very important to provide enough space, so that the liquid and vapor separates, liquid and vapor separates. In general, the design of these sometimes it is of cyclone type. Now, we will consider a binary mixture in flash distillation and consider a mixture, which contains A and B. Now, flow rate, composition and enthalpy of feed. This is, we can write, F is the feed flow rate, F is the feed flow rate, X F and H F, which is given over here. Then, the flow rates, composition and enthalpy of distillate, which is given over here at the top, we can assume D, Y D, Y D and H D. The flow rate, composition and enthalpy of bottoms, that is, we can write W, X W and H W. So, with this nomenclature, if we do the material balance, first we will have the following assumptions. One is no heat loss to surrounding perfect mixing and ideal behavior of the vapor, ideal behavior of the vapor. Now, let us do the steady state material balance on this flash drum. So, if we do the overall material balance, total material balance, we can write F would be equal to D plus W equation 1. If we do the component balance, balance for A, we can write F X F is equal to D, Y D plus W X W. Similarly, we can write the energy balance equation. Energy balance equations, this is equation number 2. Energy balance equations would be F H F plus, this is H F plus Q would be equal to D H D plus W H W. So, this is equation 3. Now, from equation 2 and 1, we can write the D plus W X F would be equal to D Y D plus W X W. And from here, we can write D into X F minus Y D would be equal to D Y D plus W X W. And from here, we can write D into X F minus Y D would be equal to W X W minus X F, and from here we can write D into X F minus Y D would be equal to W X W minus X F, from which we can write minus W by D would be equal to Y D minus X F divided by W X F minus Y D divided by X W minus X F. So, this is equation 4. Now, using equation 3 and 1, we can write F H F plus Q would be equal to D H D plus W H W. This is equation 3. Now, if we substitute D by F minus from the total material balance F minus W H D plus W H W, and then we can write this would be equal to F H D minus W H D plus W H W. And we can write F H D minus W H D plus W H W, and then we can write this would be equal to F H D minus W H D plus W H W. And we can write F H F minus F H D plus Q would be equal to D H D plus W H W. So, we can write H F minus H D plus Q by F would be equal to W by F H W minus H D. And we can write W by F W by F multiplied by H D minus H W divided by W H D. So, we can write F would be equal to H D minus H F minus Q by F. So, this is equation number 5. We can also write this plus this. Now, similarly if we use energy balance equation F H F plus Q is equal to D H D plus W H W. Now, if we substitute D H D plus F minus D H W, and if we rearrange this equation in similar way, we can get D by F into H W minus H D plus F would be equal to H W minus H F plus Q by F. This is equation 6. So, if we divide equation 5 by equation 6, we would obtain minus W by D would be equal to W by F minus W by H D minus H F plus Q by F divided by H W minus H F plus Q by F. If we look into the H x y diagram. So, this points D which is H D and Y D divided by H W minus H F plus Q by F. If we look into the H x y diagram. So, this points D which is H D and Y D divided by and here W which is H W x W. This point represent H W x W and the point which is given over here is H F plus Q by F and this is corresponding to x F. All these three points 1, 2, 3, 4, and 3 say this is M and this is say W. So, W M D are in the same straight line and minus W by D and Y D minus x F divided by x W minus x F plus Q by F plus Q by F plus Q by F minus these are the slopes of the operating line slopes of the operating line. Then with this slope if we plot this line then the line connecting to this say U and V then U and V if the effluent streams are in equilibrium U V will be the slope of tie line. Now, we will do the another rearrangement of the material balance equations and we define the fraction vaporized of the feed fraction vaporized of the feed which is defined as F small f is equal to D by F. Now, from the material balance equation 1 which is F equal to D plus W if we divide both side by F it will be 1 is equal to D by F plus W by F which we can write F plus W by F and so from this we can write W by F would be 1 minus F. So, this is equation 7. So, now for the component material balance we can re-written as component material balance in terms of fraction vaporized of the feed fraction vaporized we can write X F would be equal to D by F Y D plus W by F X W. So, which we can write F Y D plus 1 minus F X W. So, this is equation 8. Now, if we write this equation we can in this equation we have 2 unknowns Y D and X W 2 unknowns. So, there should be a relations between these 2 which will give the relations between Y D and X W. So, that we can able to solve. So, this is provided by the equilibrium curve or relative volatility equation. The relative volatility as we have discussed for a binary system we can write alpha AB would be equal to Y A by X A divided by Y B by X B where Y B would be 1 minus Y A and X B would be 1 minus X A. So, if we rearrange these equations it will give Y would be equal to alpha X divided by 1 plus alpha minus 1 X. So, this is equation 9. Let us consider an example. A mixture of a liquid containing 50 mole percent n-hexane and 50 mole percent n-heptane is subjected to flush distillation at a total pressure of 1 atmosphere and at 40 degree centigrade to vaporize 50 mole percent of the feed. The relative volatility of n-hexane in the mixture is 2.36. The vapor pressure for this system can be obtained from the following Antoine equations which this problem we have already discussed in our first lecture. So, ln P v is equal to A minus B by C plus T where P v in millimeter H G and T in degree centigrade. The boiling point and the constants A B and C are given in the following table. N-hexane is this and the value of N-hexane which is not here N-heptane is 98.5. This is 15.587, 2911.32 and 226.65. Now, we have to calculate the compositions of the vapor and liquid leaving the flush chamber considering an equilibrium stage. The information which are given the feed contains 50 mole percent normal hexane and normal heptane. So, we have X F is 0.5 and D is 50 mole percent is vaporized. D is 50 mole. If we assume feed F is 100 mole as the basis, then D is 50 mole and W would be 50 mole. So, minus W by D would be minus 50 by 50. So, it would be minus 1. Now, this is the value at X F which is at 0.5 and then with the slope of minus 1, we will draw the line, the equilibrium data which are obtained from the equations which is given Y is equal to alpha X divided by 1 plus alpha minus 1 X. So, we can write this is 2.36 X divided by 1 plus 2.36 minus 1 X which would be 2.36 X divided by 1 plus 1.36 X. So, with this we can obtain the value of X and Y. So, like if we take 0, this will be 0. If we take 0.076, this value is 0.163 and if it is 0.191, then it will be 0.37, 0.341, this value would be 0.55 and 0.51, this would be 0.71. So, this is the value of X and Y divided by 1.1, 0.70, it would be 0.85 and this is 1 at 1. So, with this we can plot the equilibrium line. This is 45 degree diagonal and this is the equilibrium line. So, this is XY diagram. Now, in the XY diagram with the slope we can obtain this values and this gives a value of Y d which is about 0.65 mole fraction and X w we can calculate from this point is around X w around 0.375 mole fraction. Now, from the Antoine equations as we have discussed before since we have equilibrium line and we have the vapor pressure data at different temperature we can plot the bubble point and dew point curve. So, this is the plot which is shown over here. So, first we have to draw the bubble point curve and then from the equilibrium points equilibrium XY plot draw a vertical line and then from the intersections between the two between the from the intersection section between the vertical line and the bubble point obtain the temperature. So, the temperature is around in this case is around T is about 83 degree C. So, thank you for your attention.