So, welcome to the NPTEL lecture of cryogenic engineering and we are talking about gas separation and related to distillation column where the gas mixture gets separated. Just to take over view of the earlier lecture, in the earlier lecture we have seen the working of rectification column with the help of animation. If you remember that we had shown how the feed comes, how the gas components gets separated, the low boiling component, the high boiling component gets separated from a mixture. Then we have got a ponchan and separate macabre and thill and numerical techniques that are used to calculate the theoretical number of plates in the rectification column. So, every column consists of plates and a design of rectification column would consist of calculating the theoretical number of plates and there are several techniques depending on the complexities and assumptions and these are the three techniques ponchan and separate macabre and thill and numerical techniques of which we are going to talk about macapthill because this is a less general and is widely used for binary mixture and is valid at cryogenic temperature. So, it is a simpler technique of all three has been proved for cryogenic gas separation column. The major assumption is that the saturated vapor and saturated liquid enthalpies are independent of mole fraction. So, we assume here that the capital H and small h which is the enthalpy of the saturated vapor and saturated liquid are independent of molar composition. This is what we have seen in the earlier lecture in detail. In the earlier lecture the equation for operating lag in addition to what we just talked we also had derived the equations for operating line for stripping section and enriching section. The section above the feed is called enriching section of the column and the section below the feed is called a stripping section. We derive a relationship between the vapor fraction which goes up and the liquid molar fraction which comes down and this is what we call as operating line for stripping and enriching section. The locus of intersection of these two operating line denotes the feed conditions. So, feed comes at a particular molar composition with a particular quality and the intersection of these two operating line denotes the feed condition. The point of intersection of feed line or Q line this is what we call as quality line or a which denotes the wetness of the feed basically amount of liquid it brings to the total number of moles that is what we call as Q and the intersection of this Q line and the y is equal to x line gives the content of a component basically it gives the composition of the low boiling component in the mixture and it is called as XF. This particular lecture extending the same knowledge ahead in order to design a distillation column or a rectification column in terms of theoretical plate calculation. So, what we are going to do is we are going to talk about graphical solution for column design using Maccabre-Thiel method. So, Maccabre-Thiel method as we talked about is a simpler of all three and it is basically a graphical solution. You do not have to do lot of calculations over here the entire column can be represented on a graph and which helps us to calculate how many theoretical plates are required in the enriching section how many theoretical plates are required in the stripping section. And we can better understand this graphical technique using a tutorial which we will solve using an excel sheet and then we can see different parameters which affects this design of the column. So, what are these parameters? How do they affect the design of the column? This we can see all these three things we can see in this lecture. So, taking overview again in order to understand this method in details just see what we have earlier learned. So, if we plot the vapor and liquid mole fractions for a particular component for the operating line is shown over here. Let 45 degree diagonal this line represents y is equal to x line which is what is known to us. The desired purity and impurity of this component in top and bottom products are x d and x p respectively. We have talked about this x d denotes the purity requirement of a component in the condensate or d that is obtained from the top of the column while x b shows amount of that component in the product which is taken out from the boiler side. That means basically this length becomes the impurity for example, if you want to take oxygen from this this will denoting the amount of nitrogen existing in the oxygen while this will show amount of nitrogen in the condensate that is rich in nitrogen and it is taken from the top of the column. So, x d denotes the purity while x b denotes the impurity of the same component from the from b or the product which is taken from the boiler side.  This is the equation for the stripping line and this is what we had plot and we remember that this operating line intersects y is equal to x line at x is equal to x b because at this point this is the bottom most plate and therefore, when x m plus 1 becomes x b this equation reduced to y is equal to x or y m is equal to x p and therefore, we had found earlier we had derived earlier that if x m plus 1 is replaced as x p we will reduce to y is equal to x and this therefore, happens to be the point of intersection of the operating line for stripping section and y is equal to x line. The equation for operating line in enriching section is y n. So, m and n represent the stripping section and the enriching section respectively.  This equation represents the operating line for the enriching section which is that top part of the distillation column. So, this is operating line for the enriching section, this is operating line for the stripping section. Also to be seen from this that l by v or l n plus 1 by v m is the slope of this line while l n plus 1 by v n is the slope of operating line for the enriching section. So, these equations give the slopes of the two lines l n by v l n plus 1 by v and it also gives you the y intercept. That means, where these lines intersect the y axis. For example, for operating line for the enriching section we got d by v n into x d as the y intercept and therefore, we know this point and we know this point also. Joining these two points one can draw the operating line for the enriching section once we know the equation of the operating line for the enriching section. Similarly, for the stripping section we have got the slope of the line which is l a by v given by this equation and also we have got a y intercept which is minus b by v m into x b. So, what is this y intercept? Again it is a point of intersection of the operating line of enriching section with the y axis and which is what you can see in this graph that this point is going to be negative point negative side of the y axis and therefore, this point the y intercept can be located on the y axis at this point. Similarly, we know that this line intersect y is equal to x at x is equal to x b. Knowing these two points we could join this point and extend it further and in this way we will be able to draw the stripping operating line for the stripping section. Similarly, what I told you for the stripping section we can see the same thing same argument for the enriching section that the operating line for the enriching section intersects y is equal to x line at x is equal to x d. So, in this equation again if I put x n plus 1 is equal to x d for the top most plate from where d is taken out that means that the plate below the condenser this equation will get reduced to y is equal to x and therefore, we say that the operating line for enriching section intersects y is equal to x line at x is equal to x d. Now, let us assume that these two lines intersect at a point o and this denotes the feed condition.  Now, everything depends on the feed line condition and the feed line condition is determined by the value of q namely we have got various conditions that if the feed is going to be saturated vapor where h f is equal to capital H in this case the q will be equal to 0 and if q is equal to 0 if you put this value q upon q minus 1 the slope also will be equal to 0 and therefore, this line is going to be horizontal or parallel to x axis alright. So, depending on the feed condition the slope of q line will change the next condition is saturated vapor when h f is equal to small h q is equal to 1 if I put q is equal to 1 in this equation the slope will be infinity therefore, this line is going to be parallel to y axis. Similarly, we have got a two phase where h f is going to be between capital H and small h we have seen all this in detail again in the earlier lecture the q therefore, would lie between 0 and 1 and the slope would be negative and if you got a slope negative the line could be in this direction between these two lines basically. Then we got a sub cooled liquid possibility and there we got a super cooled liquid possibility and therefore, in this case the slope will be positive and therefore, the slope could be in this direction the line could be in this direction. So, all these conditions which can exist for a feed line have been represented on this graph and this will decide because all these lines will intersect y is equal to x line at x is equal to x f and this will decide the number of plates in enriching section and number of plates in the stripping section. The point of intersection of feed line or q line and y is equal to x gives the content of the component a in the feed x f this is what we have seen. So, if we take these are the feed line the intersection of this line with y is equal to x will give you x f this intersection point is used to draw the feed line as shown. So, what I know I know the point of intersection I know the composition of the feed if I locate these two points I can join them and I can draw therefore, the q line or the feed line and this will decide the slope of this line or again this basically represent the condition of the feed. The variation of equilibrium vapor and liquid fraction for a particular component is called as equilibrium curve. If you remember my earlier lectures we have talked about this that the equilibrium curve exists for every composition for every mixture for a given composition and pressure. So, these are the points on which y and x vapor and liquid are in thermal equilibrium that means the temperatures are same and they remain in thermal equilibrium. So, equilibrium curve for a given composition for a given pressure is going to be a unique numbers and if we join all these points y is equal to x for various composition. So, you got a unique points and if you could join these points together for different temperatures on at each point on this curve y and x are in thermal equilibrium. So, if I know that if suppose this is oxygen and nitrogen at one bar then this point is at 77 Kelvin while this other point is going to be at 90 Kelvin and therefore, you can join these points across this temperature and at every point here the temperature will be in between 95 to let us say 77 and at each point for a given temperature y and x will be in thermal equilibrium. It means that at any point on this curve this is what I just said the vapor and liquid of this component are at same temperature they are in thermal equilibrium. Now, this condition exists on each plate if you remember again our earlier arguments in the third lecture of gas separation or something like that on each plate vapor and liquid are in thermal equilibrium because there is a heat transfer between the vapor which is going to the liquid and liquid which is coming from the top of the plate on every plate. Therefore, some vapor leaves and some liquid leaves this vapor and liquid on above this plate are going to be in thermal equilibrium. So, on each plate vapor and liquid are in thermal equilibrium therefore, the plate condition lies on the equilibrium line that is equilibrium curve gives the relationship between the liquid composition x n and vapor composition y n on the same plate. So, it relates y n and x n for every plate this is what it denoted by the equilibrium curve. Since the top and the bottom products have different boiling points for example, 77 Kelvin may be the product product had 95 Kelvin which is a oxygen. Since the top and the bottom products have different boiling points there is gradual variation of temperature across the length of the column. So, I said 95 should be 90 Kelvin which is oxygen 77 Kelvin which is nitrogen. So, you have got a 90 Kelvin at the bottom you got a 77 Kelvin on the top and at any point across the column at any point in the column at any plate in the column you have got a temperature in between these two values 90 to 77 Kelvin. So, when you come down from top to bottom you will have a temperature gradient from 77 Kelvin at the top to 90 Kelvin at the bottom. The operating line relates the variation of liquid and vapor mole fractions of a particular component across the length of the column. So, what is operating line? Operating line is basically giving you a relationship between y and x n plus 1 basically is a y n as a function of x n plus 1 and therefore, it relates the liquid and the vapor mole fractions of a particular component across the length that means at various temperatures. So, operating line gives a relationship between y n and x n plus 1 at various temperatures across the length of the column while the equilibrium curve is basically give you the relationship between y n and x n on every plate who are in thermal equilibrium with each other. These are the concepts for all these lines. So, let us now see all these things all the graphical representation with respect to the column. We are just showing you at top of the column where d is taken out, where condensate is taken out and we got some plates over there and let us concentrate on the n th plate for example. For this n th plate we got a relationship between y n and x n plus 1.  This is what a relationship between which is valid and it relates y n and x n plus 1. Revealing the operating line equation say for the top section it is clear that this equation relates x n plus 1 and y n for this component a. This is what I just said. So, let us see this is the n th plate and y n into v n is actually going to give you the or y n is the mole fraction of a component which is leaving this plate n th plate in the vapor condition while x n plus 1 is a molar fraction of the liquid which is coming from top plate above this n th plate. This is leaving l n plus 1 or x n plus 1 is the mole fraction of the liquid which is leaving the n plus 1 th plate. So, n th plate is receiving liquid from the plate above it and it is denoted by x n plus 1. The mole fraction of that component is denoted by x n plus 1 while y n is the mole fraction of the vapor of a component which is leaving this n th plate. So, all these curves the operating line for the enriching section stripping section q line and the equilibrium curve. So, all these curve and the lines are vital in calculation of the number of plates using mac apthil method and therefore, I wanted to give you the entire glimpse of graphical representation of the mac apthil. In view of this the slopes of the operating line and q line. So, what is this line basically? It has got slope it has got y intercepts associated with the physical condition. So, the slope of the operating line and q line the equilibrium curves and the purity requirements form the basis to determine the number of plates. So, all these parameter x b x d the slopes of operating line slope of q line equilibrium curves are all required in order that we calculate the theoretical number of plates for a given composition for a given separation yes separation problem. Now, let us start calculating or let us start understanding the procedure in which we calculate now the number of plates. The plate calculation method involves a stair casing method as explained below. So, we going to talk method which is normally also called as the stair casing method because we actually climb down the stair case in order to calculate the number of plates. The condensate d which is taken from the top and what is taken from the bottom is called b. What is taken from the top is low boiling component that is nitrogen in a mixture of nitrogen and oxygen. The condensate d collected at the top has a mole fraction of a component a that is nitrogen as x d. So, if I want d to have 98 percent of nitrogen it talks about the purity of nitrogen condensate which are received from the top and therefore, the value of x d is going to be 0.98 in that case. So, you got a liquid coming from the top whatever vapor goes up it gets condensed out of which I am going to take d out. I am going to take d moles out the remaining moles will come down whatever condensate happen a part of it is going to taken out and in order that always the l comes down in order that always we have got some l which is coming down remaining l is going to come down of the condensate. The liquid coming on to the top plate from the condenser has a mole fraction of x d. Hence point x d lies on the operating line as shown in the figure. So, operating line will have also have a point x d which is what we have proven for the top most plate. So, this plate receives the liquid as l n plus 1 for which lies the point x d. Now, let us come to this plate. Let us call this as n th plate which is a top most plate. As a result of whatever come from the top it has got its own vapor and liquid which is going to leave this plate alright. For this plate liquid and vapor are in thermal equilibrium. Therefore, the corresponding liquid fraction x n for this plate this is my n th plate now whatever liquid leaves whatever mole fraction leaves this plate is now will be called as x n whatever vapor leaves this place is going to be called as y n. And as we just talk this y n and x n are going to be in thermal equilibrium because now we are going to talk about this plate only. So, the corresponding liquid fraction what we receive from the top was x n plus 1 which has got a x d component. But for this x d now I would like to find as a result of the heat transfer between vapor and the liquid the vapor which is leaving at this place is going to be y n what is corresponding x n on this plate which is in thermal equilibrium alright. So, here I got a y n which is known to me because y n and x n plus 1 I know a relationship between which is operating on operating line. So, I would like to find out for this y n what is corresponding x n which is in thermal equilibrium. So, what do I do this equilibrium point that means for this y n the equilibrium point y n x n is going to be lying on the equilibrium curve which is what we just talk the equilibrium curve will show the equilibrium of y n and v n which exist on every plate. So, for this y n the corresponding x n can be found out by keeping y n constant that means I have to move horizontally keeping y n the same and the point of intersection of this horizontal line with the equilibrium curve is going to give me the condition which exist on this top most n th plate at the moment alright. So, this point is going to give me this equilibrium point is found by extending a horizontal line from x d to the equilibrium curve let this point be denoted by p. So, p denotes thermal equilibrium existing on this n th plate or which is a top most plate right now where y n and x n for this plate are in thermal equilibrium I hope you understood about this very important argument. Now, we have located y n v n for this top most plate and now let us go to the plate below this. Now, what was x n earlier which is existing on this plate if I am now my n th plate is now this my attention has come down from top most plate where I located this y n x n I will now come to the plate below and now this plate is going to receive x n plus 1 from the top that means this is the point this happens to be the x n plus 1 as far as my plate below is considered alright. So, the liquid from this plate flows this plate means top most plate flows over the wear whenever the liquid column goes above this wear this liquid will fall on the plate which is below it it goes over the wear and exchanges heat with the vapor coming from the top of the lower plate. So, now my attention has come to this lower plate for which this becomes x n plus 1. So, I have got a relationship between y n which is a vapor leaving at this point the mole fraction of the vapor which is leaving this point and I have got a relationship between this y n leaving and x n plus 1 now which is coming from top of this plate alright. So, we know that an operating line relates x n plus 1 with y n. So, now y n is the vapor leaving this plate and x n plus 1 is the point given by P which was existent earlier for the top most plate. So, if I want to find for this x n plus 1 corresponding y n which exist on this plate what I will do I will keep x n plus 1 constant and come down and the point where it intersect operating line because operating line gives a relationship between y n and x n plus 1. Now, I am going to come down keeping x n plus 1 constant point where it intersect the operating line for the enriching section is going to give me corresponding y n because it relates y n with x n plus 1 for this plate for this nth plate. Now, hence for this plate the corresponding vapor fraction lies on the operating line because it relates y n x n plus 1 for this plate. The vapor composition y n for this plate is found by extending a vertical line from P on to the operating line let this point be denoted by Q and therefore, I draw a vertical and I come down. So, can you understand from here that as soon as I have come down from this plate to this plate I derived a vertical line over here. Now, I do the similar argument now I have come here and I have located my y n and now I would like to find out what is x n for this plate that means y n and x n are going to be in thermal equilibrium and if I want to locate that point what I will do I will again for this y n I would like to locate the point on equilibrium curve and the point where this line horizontal line will intersect the equilibrium curve this will locate y n and x n for this plate because y n and x n will now be in thermal equilibrium and by drawing a horizontal I will get the location like the way we had done earlier for P. I will get one more point which basically depict the y n and v n which are in thermal equilibrium on this plate having done that I will again go vertically down now to go to the next plate. So, this stair casing will continue and this is what I will show you now the stair casing should be continued till the stairs cross the point O. So, how long I will do like this I will do depending on the curves of this two operating line till I cross my intersection point O and this as I said number of vertical line would then denote the plate requirement. So, here for example, 1 2 and may be 2 and half. So, I can round it to 3. So, may be I can say that 3 plates in fact I could say 2.5 plates, but 2.5 plates cannot exist therefore, I will say 3 plates are required in the enriching section. What is important is have you understood this why do I go horizontally why do I come vertically please understand that and please that is the most important physics behind all this problem. So, the stair casing should continue till the stair cross the point O that is the intersection of operating line as shown in the figure. It is clear that every horizontal line gives the condition of liquid vapor on the same plate which are in thermal equilibrium while every vertical line gives the vapor conditions for the plate below the earlier plate this is what we just talked about. It means that every vertical line indicates the need for the plate in the enriching section till the stair casing crosses the point O. That means we will go on doing this stair casing till the time we hit point O and this are the number of plates required in order to get particular purity depending on the slopes of operating line and the queue line. The same exercise could be done for lower section with x d as the desired impurity of the component A in the bottom. So, I will do the same thing I will go vertical over here look at the point of equilibrium go down here look at y n corresponding to y at look at x n plus 1 on thermal equilibrium and go down here. So, these are the number of plates vertical lines in the stripping section these are the number of plates I would get in the enriching section. This will be more clear when I take a tutorial, but what is important is you have to understand why do I go this horizontal and vertical what is this business all about basically that is the most important thing having understood that the problem solving remains only kind of algebra. As mentioned earlier each vertical line indicating a plate the total number of vertical lines in top and bottom section together with boiler and condenser surfaces give the total number of theoretical plates required. So, these are the plates in the column in addition to that we have got a condenser surface and you have got a boiler surface complete all these things will give you theoretical plate requirement. So, I can say 3 plates plus condenser as a requirement enriching section similarly 3 plates in the bottom the stripping section plus boiler is the requirement for the stripping section from the adjacent hypothetical figure the total number of vertical lines are 4 these are the just the number of point if I see the vertical lines. However, you have to see that this is a half line also therefore, we have to take the integral of that basically you have to take the round it off. Hence, the total number of theoretical plates can be tabulated as from this technique the top is 3 plus 1 that is 3 number of plates plus condenser plate bottom is 3 number of plates plus the boiler plate. So, these are theoretically calculated number of plates by this technique by this graphical technique given by Macapthil in the top and the bottom or the enriching section and stripping section all right I think this is the theory behind the calculation of number of plates from the adjacent figure it is important to note that during the heat exchange process that is along the vertical line the liquid composition x is constant. So, when you come down x remains constant when you go horizontal y remains constant similarly along the horizontal line the vapor composition y remains constant also note that in moving from top to bottom the Macapthil diagram starts with a horizontal line here on the top while it is in the vertical line why does this happen this happens because the liquid will come from top to bottom this is because the liquid flows downwards and it represented by a vertical line and therefore, we do not have a liquid going down from here because there is no plate below however this plate receives from the top. So, this is the argument behind we go horizontal here we go vertical at the end with this background now let us do the entire column mass balance and energy balance which is required in every problem. In the earlier lecture we have balanced mole and enthalpy for top bottom and mid section respectively, but consider the column as a whole the following equations hold true what is coming is F what is leaving is D what is leaving is B what are the energies the condensation that the cooling effect the refrigeration effect required is Q D and whatever energy is given for baller is Q B these are all energies these are all mass which is entering and leaving the column. So, simple mass balance is F is equal to B plus D whatever feed number of moles B and D are leaving multiplying the above equation with the mole fractions of a particular component suppose I want to do balance for only nitrogen or only oxygen corresponding to that I have to multiply by the mole fraction of that component X F into F let us say for nitrogen then X B B into X plus X D D is giving you mole balance for this component. Similarly, for the entire column the enthalpy or the energy balance can be written as what is entering and what is leaving. So, Q B is entering H F F is entering Q D is the energy which is taken out. So, the condensation happens and D and B are leaving. So, corresponding that enthalpy is H D D and H B B. So, if I do the mass energy balance these are in is equal to out this is what equation and therefore, all the arguments regarding the plate calculation would be clear in the tutorial. We have to consider this energy balance we have to consider this mass balance having calculated these things we have to now understand the operating line equations Q line equations equilibrium curves and then go for the tutorial which is explained from the next slide from where we will understand how to calculate the theoretical number of plates alright. So, now, let us come to the tutorial consider a rectification column for nitrogen and oxygen separation plant operating at one atmosphere determine the number of theoretical plates required to yield 97 percent nitrogen at top and 95 percent oxygen at the bottom which is talking about the period requirement feed stream is 50 percent this is X F feed stream is 50 percent nitrogen and 50 percent oxygen mole fraction of liquid in feed stream is 0.7 this is talking about quality Q 0.7 mole liquid per mole mixture the desired flow rate at the bottom which is B is 20 mole per second and the heat removed in the condenser at the top is 500 kilowatt. So, various parameters are given we have got a purity requirement given we have got a X F requirement given we have got a quality given we have got a desired flow rates from the bottom and the top have been given we have to calculate and we have got a condenser cooling effect also given as 500 kilowatt what you have to calculate is theoretical number of plates for this. So, let us convert that into what is given working pressure one atmosphere mixture is of nitrogen oxygen feed stream 50 percent nitrogen 50 percent oxygen bottom flow rates 20 moles per second which is B feed line is 0.7 as Q. So, we know in fact the slope of the line for the above mixture the requirement of nitrogen is X D that is 97 percent which is X D requirement of oxygen is 95 percent and therefore, X B happens to be amount of nitrogen in this oxygen which is 0.05. So, this happens to be kind of impurity for oxygen. So, calculate total number of theoretical plates for this problem. So, how do we do start with the molar balance F is equal to B plus D we have got a mole balance for particular component also let us say for nitrogen. So, for nitrogen we have got a 50 percent nitrogen in the feed which is X F 0.5 X B is the oxygen requirement in the bottom which is 0.05 this is amount of nitrogen in oxygen which is B 0.05 and X D is the amount of nitrogen in condensate which is 0.970 actually talks about purity of nitrogen condensate to be taken from the top and B is equal to 20. So, F is equal to 20 plus D from this equation. Similarly, I have got a expression for the mole balance of nitrogen putting the values over here solving these two equations what we get is F is equal to 39.14 moles per second D is equal to 19.14 moles second. So, F is coming as 39 moles what is coming out is 19 as condensate what is coming on both from the bottom is 20 moles per second. So, we have got F B and D values. Now, what we do is energy balance. So, enthalpy balance calculate the value of Q B because Q D is given to us as 500 kilowatts we know we have to calculate enthalpies for these values calculate enthalpies for the feed depending on the composition and get the value of Q B. So, fraction of stream in feed is Q which is known to us enthalpy H F is equal to Q H plus 1 minus Q H which is coming from this quality expression. Now, let us find out enthalpy let us locate the point as 0.5 on the bottom 50 percent nitrogen 50 percent oxygen corresponding to this point find out enthalpy capital H and small h. So, small h is 1084 and capital H is 6992 and this will not change and therefore, we will assume that the capital H and small h are now constant. So, related to this if we calculate H F putting these values over here H F is equal to 2856. So, point lies somewhere in between it is a two phase flow is not it because it is called liquid content anyway that is been given corresponding to that corresponding to 0.7 liquid which is a Q we got a H F as enthalpy of the feed putting these values we got a Q D as 500 kilowatt H D 1084 H B is same irrespective of the composition H F is this F B D we know putting these values over here what you get Q B is 430.6 kilowatt ok. So, we know Q B we know Q D now. So, we want to ultimately find out the operating line for enriching section operating line for stripping section and Q line. So, what is operating line for enriching section for which I should know the slope of the line I should know the y component of this line. For calculating the operating line equation I should know D by V D by V given by this equation because D by would give me L by V if you know they are related D by V is related to L by V. So, these are the different values we have just calculated putting those values over here I get D by V equal to 0.226 once I get D by V I know L by V or L n plus 1 by V n is equal to 1 minus D by V n and therefore, L n plus by V n is 0.773. Now, this is nothing but the slope of the operating line for the enriching section. So, L by V is known for the enriching section as 0.773. So, if I put these values and if I get D by V value put in over here I got a equation for x D is 0.971 put all those over here L by V is 0.773 x n plus 1 D by V x D this is my enriching section operating line equation y is equal to 0.773 x n plus 1 point plus 0.22 slope of this line is 0.773 y intercept is 0.22 this line intersect y is equal to x at x is equal to x D I know both these points y intercept and x is equal to x D I can draw graphically the operating line for the enriching section having done this now I would like to do the same thing with the stripping section. So, operating line for stripping section for which I will have to calculate B by V m now put all those values which we have just calculated Q B enthalpies B etcetera put these values over here calculate B by V m B by V has come out here corresponding to that find out L m by V m which is nothing but the slope of the stripping section these are all calculations now what is important is to calculate B by V once you get B by V calculate L by V. So, now I know the slope of the stripping section line operating line for the stripping section I will put these values as I have done for the earlier values putting all these values what I get equation for the operating line of the stripping section. So, this has got a slope of 1.274 and y intercept of minus 0.013. So, again I could draw stripping section line operating line for the stripping section the third is a equation for the feed line the feed is given by this formula the Q is given by this formula I got all other variables available with me the equation for the feed line is Q upon Q minus 1 x plus these putting these values over here I get y is equal to minus 2.34 x plus 1.67 that means the slope of the feed line or the Q line is minus 2.34 this gives you the direction of the Q line well this is the y intercept of the Q line and again you know that this line intercept y is equal to x at x is equal to x f x f is equal to 0.5 if you remember. So, I have got all the three lines. So, in summary now before I go to the graphical solution now I need to know what are my equations. So, operating line for enriching section is given by this equation slope 0.773 y intercept 0.22 operating line for stripping section is given by these slope 1.274 y intercept is minus 0.013 and Q line y is equal to minus 2.34 plus 1.67. So, I now know the equation of Q line also with this now I am ready to go for my stair casing business provided I know the equilibrium curve for this particular composition and pressure also. So, the stair casing procedure is shown on excel sheet to have a better understanding of the method now I do not want to draw it on here, but I will now go to the excel sheet you will have to use your graph paper to calculate this because it cannot be done on paper. So, please have a graph paper use the graph paper plot these lines and then you would be able to do the stair casing which will ultimately give you the number of plates in stripping section and number of plates in the enriching section. So, let us see the excel sheet now having done the equations of the stripping section and the enriching section now let us see this excel sheet which I have prepared with my students and this excel sheet actually will give you the number of plates that could be there for the enriching section and the number of the plates that we have in the stripping section. Now, please try to understand this excel sheet and I expect that you also would be able to develop such excel sheet now I can change any parameters over here and I can get calculations done in the excel sheet itself I will not go through all the details, but I will just show you for this problem that what we had was xf is equal to 0.5 alright. So, we had this quantity as given the amount of nitrogen in the feed as xf is 0.5 xb is basically the purity of the bottom product which is 0.05 this is the amount of nitrogen in oxygen that you get at the bottom product and xd is the purity of the nitrogen that you get from the top alright. At the same time what we have is a bottom flow rate is 20 moles per second then what we have also calculated as we did earlier is enthalpy capital H enthalpy 6992 and small h enthalpy of the liquid as 1084 and this will not change with the composition that is what our assumption has been also the value of q is 0.7 the data what we have also is qd that is 500 kilowatt is condensation heat load that is that the amount of heat to be removed on the top enthalpy of the liquid hd and hb which is independent of composition therefore is same as 1084 joules per mole while enthalpy of the feed is 2856 which we have calculated in the earlier calculations. Having done this when I write these values I automatically got the calculations done and the values of d and f what you get from the top and what we get as f which we have also calculated earlier automatically comes over here in this excel sheet alright. Having done this I got the value of qb to be calculated as 430 kilowatts it is the amount of heat to be given at the boilers at the bottom then based on this I get the values of the slope and y intercept for enriching section and for stripping section also for the q line. So, we got three lines the operating line for the enriching section has a slope of 0.77 which we have calculated it has got a y intercept of 0.22 then stripping section the operating line for stripping section we have got a slope of 1.27 and it is y intercept is very close to the 00 or the origin which is at minus 0.01 as far as the q line is concerned the feed line is concerned it has got a slope which is negative minus 2.33 line and this line has a y intercept of 1.67 alright. Having done this now let us come to see what we do when we plot this macapthil graph is a graphical representation and therefore what we can see on the right side now is a graphical representation of this. So, concentrate on this now you can see these lines which are already plot and this is the operating line for the enriching section this is the operating line for the stripping section how did I find these lines you can see that this is the purity of the nitrogen which is 97 percent as we had seen here. So, I will I will locate this point on this y is equal to x line. So, what we have is a three line here y is equal to x line then we are going to operating line for the enriching section we are going to operating line for the stripping section and we have got the third line which is the q line feed line alright. So, let us see how we have plotted these lines. So, enriching section has first point which I could locate on this curve as 0.97. So, I as we know that y is equal to x for the top most plate and this represent top most plate this is the y is equal to x line operating line for the enriching section intersect y is equal to x at 0.97. So, this is my first point also for this the y intercept is 0.22 for the enriching line. So, other point is 0.22 and when I join these two points I get the operating line for the enriching section. So, this is my operating line for the enriching section as shown in this graph. Now, let us see the operating line for the stripping section the operating line for the stripping section also has intersection at y is equal to x which is at 0.05 because depending on the purity of the bottom product or the impurity of the bottom product the amount of nitrogen it has which is 0.05. So, this is my first point here and its y intercept is minus 0.01 which could be at this point. So, if I join these two points I will get the operating line for the stripping section. So, this is my operating line for the stripping section this is my operating line for the enriching sections I have plotted these lines also what we have is a q line the q line is basically this and it has got two points which we know one point is intersection of this operating line for the stripping section operating line for the enriching section. So, I get one point from here and what is the other point this feed line as we know feed line intersects y is equal to x at x is equal to x f which is 0.5 alright. So, this amount of nitrogen in the feed we have got this point therefore and we have got this intersection point if we join these two what we get is a feed line or a q line. The fourth line also is very important rather curve which is a very equilibrium curve and the data is known to us this data is known for one atmosphere for such a mixture I got y at x known to me from distribution coefficient and thing like that this data is a standard data I will know this equilibrium points and therefore I can plot this curve on this line. So, my capital section now start with graphical process first and most important part is that we have plotted operating line for the enriching section operating line for the stripping section feed line or the q line and the equilibrium curve alright. Having done all these things on the graph then what we do is to start the stair casing process which is a very simple graphical process, but the most important aspect of this is to calculate this slope for the enriching sections stripping section line all these points first we have to calculate and then get these lines one can do this on a simple graph sheet also I have done it for excel sheet because tomorrow I can change these parameters and immediately I will get change of parameters as far as the slopes and y intercept everything are concerned. Now, in order to do stair casing in a graphical format I have developed my own technique. So, here you can see that as soon as I have got a enriching section I have got a stripping section here the enriching section start with the first horizontal line from here we know this I will first draw the horizontal line which I do from here alright. So, this is what I get the corresponding liquid for the top most plate which I draw a horizontal line and on equilibrium curve now where I get corresponding liquid point now I will come down. So, stair casing method starts I will come vertical down which I come here and this is where I come on operating line for this enriching section I get this point now I go horizontally again as I have explained earlier now I go down again and here you can say that I have crossed the enriching section and you can see that the verticals are 1 2 and 3 which represents number of plates in the enriching section because this is the point where my stripping section starts and we have crossed this point this point sometimes could be 0.5 plate or something it could be representation because this point can come on this side in that case we will say 2.5 plates, but then we take the entire integer. So, we can take the next rounded off and we can say 3 plates in this case I can see that there could be around 3 plates in the enriching section followed this is just a stair casing technique I am talking about the theory behind this we have already discussed. Now I start from this point where my stripping section starts. So, the first point has to be vertical.  So, you can see that 1 2 3 4 5 6 or we can take 7, but I will say let us say 6 as vertical line because we are very close to this point of intersection. So, you can see now this is my Macapthil graph graphical representation and we have done stair casing technique from enriching section from the top and we found that there could be 3 verticals and therefore 3 plates in the enriching sections and we have got 1 2 3 4 5 6 as the plates in the stripping section meaning which that we have got in all 3 plates in the enriching section and 6 plates in the stripping section alright. So, theory of this has already been explained to you this is the way the graphical technique for the Macapthil method will be used. With this now let us go back to our original presentation from the excel sheet it is clear that the total number of vertical lines are 9. Therefore, the total number of theoretical plates for this column can be tabulated as given below the enriching section we have 3 plates and 1 condenser while for the stripping section we have got 6 plates and the boiler at the bottom meaning which we have got 9 plates total approximately this is all approximate and we have got a condenser and boiler to add up to this alright. So, this is the way one should work on a Macapthil diagram Macapthil technique to calculate number of plates for the enriching section and for the stripping section. Everything understood how to draw this excel sheet how to draw this on the graph paper how to calculate number of plates in enriching section and stripping section. Now, let us see different parameters which affects the number of plate and that is the most important thing. So, let us see the number of plates how do they vary with different parameters. In general the equation for an operating line is given as y is equal to l by v x and k this is what we know. So, definitely l by v plays a very important role the slope of the line plays a very important role it is clear that the slope of an operating line is a function of l and v for an enriching section the slope can be reduced. So, if I want to change the slope I can change the slope by either changing l or by v for enriching section what we have is a condenser this condenser will decide depending on the available cooling effect or depending on the value of d because d is the number of moles in the condenser taken out. So, how do I change the value of l the slope can be reduced by decreasing l this is possible by drawing out more liquid from condenser that is increasing d. So, I have kept the cooling effect the same that means I get condensate some condensate of which now I am taking more d out instead of taking some quantity I am taking more d out that means less liquid will go down now. So, l is going to decrease and when l decreases l by v also would decrease the other thing is instead of taking not more that d I will do what I do reducing the condenser heat load that means I have got less cooling effect available. So, whatever vapor comes on the top not I do not have available cooling effect I have got less cooling effect and therefore, depending on q d depending on the available cooling effect 300 kilowatt 400 kilowatt I will decrease the condensate overall condensate only that means the value of l will decrease. So, now let us see the effect of how do these things happen on the excel sheet let us come back to the excel sheet to understand this. So, we will see this effect and similar this thing can be understood for the stripping section also. So, in stripping section the slope is reduced by increasing v. So, I can control the slope now depending on what do I do on the boiler. So, I got a q v is a heat which is supplied to the boiler the slope can be reduced by increasing v. So, l by v can be decreased by increasing the denominator by increasing the v and how do I do this thing this is possible by drawing out less liquid from the boiler that is by decreasing v. So, I can take less v and therefore, more v is going to go back I will not use all the liquid which is coming down, but I am going to take less liquid and therefore, remaining liquid will get vaporized and therefore, v will increased. So, this is one of the ways the other ways by increasing the boiler heat q v or either supply more heat and therefore, v will increase more and more liquid will get vaporized. Therefore, as a result of this l by v will decrease. So, depending on what we just saw the excel sheet I am just summarizing here what did we see. We saw that if I my d earlier was 19.15 corresponding to this my l by v was 0.77 and number of plates therefore, were 4 and 6 4 in enriching section 6 in stripping section. If I increase the value of d which is what we just saw if I increase the d amount of condensate taken from the condenser my l by v is going to decrease. So, 0.77 got decreased to 0.66 as a result of l by v decrease the line shifted towards the equilibrium section because of the slope change the line operating line shifted towards the equilibrium line equilibrium curve which resulted in increase number of plates. So, now I need 4 plus 8 that means I need 12 number of plates instead of 10 plates over here. So, this was the effect which you can see what happens with the decrease in the slope what happens in the decrease in the slope of the enriching section. So, as l by v decreased the operating line shifted towards the equilibrium curves which resulted in the increase of number of plates. So, simple fact if your lines slope shift towards the equilibrium curve the number of plates would increase. The second parameter which we just saw was q d if I decrease q d from 500 to 300 that means if I decrease the condensation cooling effect which is available here if I decrease 500 from 500 to 300 my l by v decrease from 0.77 to 0.62 substantial decrease this also shifted the lines towards the equilibrium curve and which resulted in increasing number of plates from 10 to 13. So, here I have to put three additional plates look one each additional plate is a very important design change is a very important investment that has to be done on a design of column basically. So, this has to be understood very well. So, decrease d resulted in increase number of plates stripping section I can decrease b this is what we say I can decrease l by v by decreasing b. So, what I did I decrease b from 20 to 10 l by v decrease from 1.27 to 1.13 the number of plates now the slope has decreased for the stripping section number of plates change from 10 to 8 number of plates where 10 now it has come to 8 only that means the stripping section slope if you decrease the stripping section slope the stripping line moves towards the y is equal to x line it goes away from the equilibrium section which results in the decrease of the plate in case of stripping section. Similarly, what I could do is increase the value of q b that means increase the amount of it to be given which will increase v and therefore l by v decrease. So, if I increase the q b from 0.32 to 630 my l by v decreases from 1.27 to 1.19 again this moves the curves towards the y is equal to x line away from equilibrium curve and therefore number of plates get reduced from 10 to 8. So, please understand the what happens if I do l by v change in the enriching section and if I do l by v change in a stripping section this is what it shows extending parameters study from the excel sheet it is clear that the minimum possible slope for operating line for stripping section is 1 which is when it approaches the y is equal to x line for stripping section while the maximum possible slope for operating line is going to be enriching section is going to be 1. So, for operating line for enriching section it will have a maximum slope to be equal to 1 while for operating line because if I reduce the slope it will come the minimum value that can come is 1. So, here it is minimum value for the stripping section here it is going to be maximum value for the enriching section for the slope of operating line as equal to 1 suppose the slope is 1 for both minimum for this and maximum for this both the lines the operating line will get merged to y is equal to x which is diagonal and in this case as this line approach to a y is equal to x is diagonal the column will have minimum number of plates this is just a hypothetical situation basically l by v is equal to 1 that means you are not getting any product whatever v is going up everything gets condensed but just to understand that if I do not want any product and if I just want to operate the column in a steady state manner for both the operating line will get merged to y is equal to x as a diagonal the column has minimum number of plates and that is what I am saying that if this operating line move towards the diagonal which is y is equal to x line your number of plates will go on reducing while if your operating line start moving towards equilibrium curve which is away from the diagonal the number of plate will start increasing this is a simple principle to understand the slope business in distillation column design we know that the impurity of the component a and the bottom product is given as x b so x b and x d also decide the number of plates if your impurity are demanding your number of plates are going to be very very important so reducing x b that means more purity for the bottom product therefore more number of plates for the lower section so x b is to be lowered or x d to be increased which is 0.97 to 0.99 also will have more number of plates similarly the purity of a component a in the top product is given by x d again increasing x d would increase number of plates in the top section so increasing x d or decreasing x b would result in more number of plates these two effects can be clearly seen in the excel sheet so this is again a very important decision that will change number of plates which we can see again from the excel sheet the third is a q line the content of the component a in the feed stream is given by x f this parameter is very important in determine the slope of the q line and an approximate feed inlet position the x f also govern the position of the point o which is point of interest in the operating line the slope of q line is also vital in determination of a number of plates which is very important so how does this q affect the number of plates we can see here from the excel sheet that if I got q is equal to 0 for which slope is 0 when slope is 0 that means parallel to x axis the number of plates are 11 5 plus 6 if I got q is equal to 1 that means slope is infinity it is parallel to y axis that means what does it mean only liquid is coming 3 plus 6 number of plates have got reduced from 11 to 8 so slope is 0 slope is infinity q is equal to 0 or q is equal to 1 when q is equal to 0 it is only saturated vapor coming if q is equal to 1 that means the entire feed is in liquid condition you got a plates which are maximum 6 plus 5 when there is no liquid saturated vapor is coming we got 11 plates we got reduced to 6 plus 3 9 and if I got a feed in between 0.7 slope of minus 2.3 we got number of plates as 10 1.2 as slope slope of 6 q is equal to 1.2 you got number of plates as 9 so it tells you slope of q line value of q slope of q line also determines are very important to determine the number of plates having understood this I would like to give you a problem or assignments to solve in a similar line basically I would expect that you also develop your excel sheet in order to understand this problem and solve this problem I have given the answers over here which is for cross checking for you calculate number of plates and see if it tallies with this.


What's going on guys and gals it's a Chicago trader here with episode three of the fundamental series I hope you guys are staying safe with the crazy market that we're in right now If you haven't had a chance to watch the first two videos they're posted on the website and they kind of build off of each other So if you haven't watched them go watch them. This one's going to be a little bit shorter It's on something a little bit more niche and I'm actually on vacation right now. So Let's get into it. There should be a PDF With the rights offering and registration restrictions So if we can scroll down to the first one, so What is the rights offering? essentially with the rights Existing shareholders you have the right to buy a specified number of new securities from the issuer Set a specified price and There's a certain period of time that you have that you have to participate in it or not So you can see each unit consists of one share of series five convertible preferred stock and you get 200 warrants So that's with a single unit. So how much is a unit? It's a thousand bucks. There's 10,000 total So that's $10 million they raised through this rights offering So preferred stock had a conversion price of $5 So what happens is you know these and convertibles and you know some of the other ways that they raised money They can make amendments to and modify some of the You know details of it. So if we come to the next one, which is filed on the 8th of January It's the same rights offering to just modify it. So notice right here. There's now 300 warrants You know, it's still raising the same amount of money So they close it at 970 the first day after it ran you can see that has this you know It says the float here is 2.5 million roughly, you know These will be a little bit off as you know that the flow can slowly and gradually increase after you know filing. So, I mean, it's pretty simple What was it close at 970 you know times 2 point let's just say six for simplicity divided by three so you get you know about eight eight and a half million dollars right around there


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