 So, welcome to the 25th lecture. We were dealing with gas separation and distillation column and this will be the last lecture to cover most of the aspect related to the gas separation and the distillation column design. So, in the earlier lecture, we have seen the importance of equilibrium curve, operating lines and Q line in the calculation of number of plates using graphical macapthal method. As I pointed to you earlier, what is most important is to know the equilibrium curve which is a given data for a given composition of the field and the pressure. What is important is to calculate the equations of the operating line, plot it properly and also the Q line and then do the staircase in business, staircase in methodology to calculate the number of theoretical plates both in the enriching section as well as in the stripping section. So, a tutorial problem was solved using an excel sheet. Now, when I did excel sheet, I have done lot of work behind that excel sheet because what I wanted to do was basically to see the effect of various parameters which is what you will see in this lecture. However, you will have to do the same thing on a graph paper and therefore, it needs really a good practice. So, please solve a problem. The same problem you can solve using a graph paper and calculate show all these curves, show all these lines and calculate the number of plates. So, a tutorial problem was solved using an excel sheet to understand the staircase in method and the concept of equilibrium liquid vapor mixture alright. The total number of theoretical plates given by the vertical line between the equilibrium and operating line. So, the number of theoretical plates in enriching section or in the stripping section is given by the number of vertical lines between the equilibrium line and operating line and we have seen what is the concept behind this because on the equilibrium line when we draw a horizontal line, we find the thermal equilibrium value of the vapor and the liquid while when we come vertically down, we come from the top most plate down steadily down to the lower and lower plates. The outline of this lecture is we will talk about concept of reflux. Now reflux is amount of liquid when the vapor goes to the top, the liquid comes down after condensation. So, this ratio of liquid which comes down to the vapor which goes up is called reflux and we will talk about that and then importantly, we will study various parameters using a tutorial. So, using the problem which we have used earlier, I will extend the same problem using an excel sheet to understand effect of various parameters on the number of theoretical plates that we had calculated and finally, I will be summarizing entire gas separation topic which I have covered during last 7 or 8 lectures alright. And then we got some assignment for you which I expect that all of you would solve by yourself and compare the results the solutions which were also given. Now, this is what our graphical methodology deals with what we have is operating line for the enriching section operating line for the stripping section and what we have is the equilibrium line. Also what we have got a purity on the top which is given by x d of the low boiling component and we got a impurity at the bottom or purity of high boiling component given as x b at this point and what we have is a diagonal line which represents y is equal to x line. Also we recall that these two points represent the top most and the bottom most plate in the column because on these points y is equal to x as we have found out. So, consider the operating lines for a column as shown in this figure. So, this basically represents a column when we want to use Macapthil methodology. Now, you can see that I have just drawn some schematic lines to show that the number of theoretical plates in this case are 5. How did we come to this conclusion? We can see vertical lines 1, 2, 3, 3 in the enriching section and 1 on 2 in the stripping section. So, what you can see here is the number of theoretical plates total are 5. Now, what you can see from this methodology? Same thing I drive this operating line in such a way that they come closer to y is equal to x line. So, if the OP line or the operating line approach closer to the diagonal which is y is equal to x line as shown in this figure what you can see here? If I draw similar curve over here the number of plates in the same case will be 1, 2, 3 and 4 only. So, we have come down from so many number of plates to a very small number. The number of theoretical plates are reduced to 4 in this case. What does this mean? This means again from understanding if the operating line move towards y is equal to x line the vertical line jumps are going to be very, very high. While if you go towards the equilibrium line these jumps are going to be smaller and smaller and therefore, the number of plates in this case very large by the number of plates when the operating line move towards y is equal to x line are going to be smaller and smaller. This is what we want to show from here. So, the extreme cases now if I want to talk about two extreme cases the extreme cases of the point of intersection operating line are one is the operating lines are very, very close to the equilibrium curve and this will happen when the point of intersection of the operating lines is on the equilibrium curve. When this is one of the extreme cases that when the operating lines intersect each other on the equilibrium curve and the other case will be the operating lines are lying on the diagonal itself that means they intersect on the diagonal which means that the operating lines are on the diagonal itself. So, you can see that in this case the operating lines are right on the diagonal while in other case the operating lines are very, very close to the equilibrium curve. These are the two extreme cases in this case as you know the number of plates are going to be very, very large as we just saw earlier while in this case when the operating lines intersect on the diagonal the number of plates are going to be minimum in this case. Also understand that the point of intersection of these lines cannot be beyond equilibrium section or cannot be below the y is equal to x line ok. The point of intersection cannot be above equilibrium and below diagonal lines because if you understand the methodology in which the way we drawn this operating line if this happens to be true then the law of conservation of mass gets violated alright. So, please put in your understanding the way we derived these equations if the operating line goes in this region above the equilibrium curve then the relationship between y and x will be violated the law of mass conservation in this case will get violated and therefore the operating line will always be between the equilibrium curve and diagonal curve the diagonal alright. So, therefore it is clear that the number of theoretical plates are maximum when the point of intersection on the equilibrium curve as I just said that this is the extreme case now and in this case for a given column for a given mixture the number of plates in this case going to be maximum because you can see the vertical lines are going to be very, very small over here I mean the number of vertical lines are going to be large in this case because the distance between the equilibrium curve and the operating line is very, very small alright while the number of theoretical plates will be minimum when the operating lines are intersecting or they are on the diagonal itself. Now, you can see here that the curve will have a big jumps. So, you can have big jumps and therefore the number of vertical lines in this case are going to be minimum if we do the staircase in business. So, the theoretical plates are going to be minimum when the point of intersection is on the diagonal line and this is very important to understand these are two extreme cases the actual case will be somewhere in between the actual case will be compromise between these two extreme cases. So, let us try to understand this concept of reflux ratio. In the earlier case we have seen the equation of an operating line for the enriching section given as y n is equal to n n plus 1 by v n x n plus 1 plus d by v n x d where l by v or l n plus 1 by v n is a ratio of liquid which is coming from the top plate to the vapour which is going up all right while d by v n is d is the moles per second leaving the condenser and v n again is the vapour which is going up while x d is the purity of the low boiling component. In the above section the l n plus v n the amount of liquid which is coming down from the top plate to the vapour which is moving up is called as internal reflux ratio or IR. This is the reflux ratio which denotes which indicates how much number of moles are coming down l n plus 1 to the number of moles which are going up for condensation v n which is going up all right because some of the liquid is taken out as d from this whatever condensation happens on the top some of the liquid is taken out as d remaining liquid will come down as l n plus 1. So, this is very important to understand. So, l n plus 1 by v n is called as internal reflux ratio while d by v n is called as external reflux ratio. So, what is coming out of the column is d this d value will have b and f also b which is coming out from the bottom or f which is coming in the field because it will all be governed by the law of conservation of mass all right. So, d by v n is called as external reflux ratio and l n plus 1 by v n is called as internal reflux ratio. Now, are they related if you remember my earlier expressions they are related and they are related by this expression l n plus 1 by v n is equal to 1 minus d by v n. So, one can understand from here that i r is equal to 1 minus e r what does it mean if i e r is equal to 0 i r is equal to 1 when is e r is equal to 0 when I do not take any d out. So, d by v n if i am equal to 0 that means I am not taking any condensate out from the condenser in that case we reach a case called i r is equal to 1 which means that l n plus 1 by v n is equal to 1. Let us see this cases now in this case if the e r or the external reflux ratio is 0 then internal reflux ratio becomes equal to 1. So, the column has 2 limiting reflux conditions which is I just showed to you by some mathematical formulation one condition is called as condition of total reflux what is the reflux ratio l n plus 1 by v n is called internal reflux ratio d by v n is called external reflux ratio. So, condition of total reflux is in this condition all the vapor of the top plate is condensed and is returned to liquid as column. So, whatever v goes up all this v will get condensed and it comes down in the column itself as l what does it mean it means that there is no d because all which is going up is coming down and this condition is called as total reflux condition. So, in this case l by v or l n plus 1 by v n is equal to 1 for enriching section and the moment we do l by v for as equal to 1 for the enriching section the l by v for the stripping section also would be equal to 1 because d has become equal to 0 in this case and by different calculation you will find that l by v for the stripping section also becomes equal to 1 making the operating lines to match with the diagonal as you can see here. I got both the lines operating line for the enriching section operating line for the stripping section both on the diagonal and therefore, the intersection can say we can say that is equal to x is equal to x a point they are basically intersecting on the same line. The column has minimum number of plates in this case as we have seen because if I start staircaseing you can see that I will get a minimum number of plates in this case. So, for i r is equal to 1 from the equation we get i r is equal to 1 minus e r we get e r is equal to 0 this is what I have told you earlier. So, moment I say e r equal to 0 I know d by v n also equal to 0 what does it mean d is equal to 0. Hence, in this condition when i r is equal to 1 you will get the product output in the column as 0. I can say the other way if I make d is equal to 0 if I decide that I do not want to take any condenser out if I make d is equal to 0 then my i r will be equal to 1 in this case alright when I make d is equal to 0 my i r is going to be 1 in this case which will approach again this condition. So, reverse is also true if I decide if I do d is equal to 0 I will get l by v is equal to 1 which means I have gone to a condition of total reflux. This is not a admissible case basically this is just a theoretical case nobody would like to have d is equal to 0 because the whole rectification column is to get some d at the top and b from the bottom is not it. Therefore, this is a very extreme case where if d equal to 0 then i is equal to 1 or when l by v is equal to 1 d is equal to 0 in this case continuing further it is clear that l n plus 1 is directly dependent on q d how are you getting l n plus 1 how do you get liquid? The liquid is coming because of condensation and how does condensate happen? How does condensation happen? This is because of the cooling effect q d that is available at low temperature at the top. Now, you see the most important cost of this column is going to be associated with this refrigeration effect or the cooling effect q d at a particular temperature. If you are talking about the mixture at one bar for nitrogen and oxygen the q d has to be coming at 77 Kelvin it has to be kilo watts 500 kilo watts which saw in the last problem in the last lecture that very high amount of the cost will be devolved to generate such a high cooling effect q d. If q d available is less for me if the condensation if the refrigeration effect is less in that case my l is going to get decreased and therefore, in that case my i r will also get decreased alright. So, if my condensation effect is less then my l by v also going to get less. So, if the value of q d increases it will increase i r also. So, depending if I got a high value of q d I will approach the value of 1 if I got a less value of q d then I will go away from 1 which means that I will go away from this column and I will try to move towards the equilibrium curve. So, cooling effect available q d is very very important parameter in order to decide whether we are close to the equilibrium curve or close to the diagonal right the cooling effect is the most costliest running cost in this case basically for the column operation. So, summarizing at high i r value what is the high i r value for the enriching section is 1 and what is the low value the lowest value will be when this operating line meet or intersect on the equilibrium curve. So, if I got a high i r value which is approaching 1 then we have reduced number of plates which we just saw then for a given d suppose I have fixed some value of d for high condensation load for a given d we got a high condensation load if I got a value i r value which is approaching 1 then I will have a high condensation load that means higher cooling effect is going to be required if I try to move towards the diagonal alright. If i r is equal to 1 the product output this is the extreme case the product output of the column is going to be 0 and this is what we call as condition of total reflux. So, i r is equal to 1 is a condition of total reflux in this case d is equal to 0 there is no output from the column and this is a very hypothetical theoretical case. Now, the other condition is condition of minimum reflux alright we had earlier condition of total reflux. Now, we have a condition of minimum reflux in which minimum l comes down not maximum l when the maximum l comes down from condensation we have got a condition of total reflux where l by v is equal to 1. However, if I got less number of l coming down because of whatever other reasons in that case I call this as a condition of less reflux or minimum reflux in extreme case. So, let us see what it is. So, I have got a condition of minimum reflux in this condition the amount of liquid to the column is decreased alright. So, liquid which is coming less with a decrease in vapor condensed and return to the top of the column. In the earlier case what was happening all the vapor which goes up will get condensed. So, l by v was equal to 1 in that case, but here now because of whatever reason I got less liquid coming out from the top and therefore, vapor going up is very high, but I am not able to condense all that into liquid or I am drawing more d from whatever condensate is happening on the top I am taking more d out. So, remaining liquid is coming down as l. So, this could be the two reasons because of which your l which is coming down from the condenser is going to get decreased. So, l by v in this case is going to be less than 1 much less than 1 for enriching section again correspondingly my l by v also would shift for the stripping section. So, this is the condition I am talking about and this is the extreme condition I am talking about when both of them meet on the equilibrium curve. So, for l by v less than 1 makes the operating lines to move closer to equilibrium curve the column has maximum number of plates in this case. So, the extreme condition this two operating lines are going to intersect on the equilibrium curve and in this case the number of plates are going to be maximum in this case all right. So, you can see that the number of plates in this curve are maximum because the gap between the equilibrium curve and operating line is very very small in this case. So, again for I r less than 1 from the equation I r equal to 1 I r equal to 1 minus E r we get E r more than 0 in this case. So, I will get when I r is less than 1 I will have some value of D because E r is more than 1. Hence in this condition unlike the earlier case there is a finite quantity of product output this case. So, D is not going to be 0 as it was in the earlier condition of total reflux. So, we know that l n plus 1 is directly dependent on Q D again. Therefore, as Q D decreases. So, this condition can arise because we have got a cooling effect at 77 Kelvin for example, is going to be very very less as I said the major cost is the cooling effect available at 77 Kelvin. So, if Q D decreases in this case your operating lines would shift towards the equilibrium curve. So, as Q D will decrease I r will decrease and therefore, your operating line will move towards the equilibrium curve meaning which that your number of columns are going to be very high in order to achieve the kind of purity which we want to achieve alright. It is also purity dependent because your point X D and X B have defined the purity requirements from your side. Summarizing at low I r values that is less than 1 we have increased number of plates which is what we see in this case. Also we have low condenser load in this case if the condenser load is less then we will shift the operating line will get shifted towards the equilibrium curve. And in this case we will get a finite quantity of product output D and B from the column this is what extreme condition would be. For the extreme case of high and low reflux ratio we have got plates minimum and high refrigeration cost. If I got I r value to be high equal to 1 I will get plate as minimum but it would involve a very high refrigeration cost while the plate will be maximum for low reflux ratio when I r is less quite less as compared to 1 the number of plates will be maximum but the refrigeration cost will be minimum. So, these are two extreme cases this case happen when both lines intersect on the diagonal this can happens when both line intersect on the equilibrium curve. Both these cases are not practical cases and therefore for practical purpose will be somewhere in between it is somewhere between the minimum and maximum reflux ratios. In practical cases the design of column is a compromise between the number of plates and refrigeration cost. I cannot have very high number of plate because my column height will increase in this case. At the same time I cannot spend lot of money on the refrigeration cost the cost of the refrigeration is very very important. So, I have to really compromise between number of plates and the refrigeration cost in this case. The typical value of reflux ratio that is L by V for the enriching section for distillation column are 1.15 or 1.35 times the minimum reflux ratio. So, what is important is find out minimum reflux ratio which will occur when both the operating lines intersect on the equilibrium curve and then multiply it by 1.15 to 1.35 times depending on again the cooling effect that is available with you and this could be the typical reflux ratio for a normal operation. For a cryogenic operation for nitrogen, oxygen, etcetera because the condensation load or the cooling load matters a lot the refrigeration cost is very important we keep it to a still a minimum level and a cryogenic condition we have typical value of 1.05 to 1.15 times the minimum reflux ratio in that case. So, that the cost of refrigeration is kept minimum in this case. It is important to note that the minimum reflux ratio is dependent on equilibrium curve. So, what is the equilibrium curve that decides what is my minimum reflux ratio? Equilibrium curve depends on what is your composition and what is the pressure of the feed. So, this will decide what is my minimum reflux ratio and corresponding to this minimum reflux ratio you can decide if you want to have a 15 percent more or 35 percent more depending on the kind of cooling effect that are going to be available to you throughout the execution throughout the running of this distillation column and this is going to be very important cost associated with this. Now, let us see different parameters which affect the reflux ratio. The slope of the operating line is a function of both L and V that is L by V is what it is called as reflux ratio which in turn depend on what is your D what is your Q D which is just talked about similarly what is your B and what is your Q B. So, this will decide in relation it is all related to even feed also alright. So, these are the parameters which are governed by law of conservation of mass law of conservation of energy this is also going to decide what is L by V of your column what is the reflux ratio of your column. So, it is important to note that any change in D or any change in B is going to alter other parameters which are going to alter the L by V because they are all related. So, it is important to note that any change in D alter other parameters as well as vice versa. So, if you change any other parameter if I change B, if I change Q D, if I change Q B all other parameters are related and therefore they are going to change and this will change the reflux ratio this is to confirm with the laws of conservation of mass and conservation of energy for the column as a whole alright. So, what we are going to do now is basically study the effect of different parameters on the theoretical plates of a given column and therefore, what I am calling this as a parametric study. So, we know that also in addition to all these parameters we know that the impurity of a column A in the bottom is given by X B and on the top it is X D. So, if I reduce the value of X B or if I increase the value of X D if I have a purity of 98 percent 99 percent requirement alright 97 percent requirement I will need to have more number of plates. Similarly, if I want to decrease the value of X B to 0.02 0.03 alright my number of plates would be required to be more because I will have to have more heat transfer interaction between the vapor which is going up and the liquid which is going to get coming down more and more condensation boiling should happen and therefore, this will result in more and more number of plates. So, the parameters which are affecting the design of column also is the your purity requirement X D and X B requirement they are also going to decide what are the number of theoretical plates that are required for a particular column design. So, reducing X B implies more purity for the bottom product hence more number of plates in the lower section. Similarly, the purity of component A in the top product is given by X D and if I increase the value of X D I will need more number of plates for the top column also. Again increasing X D would increase the number of plates in the top section. So, very important parameter X D and X B they will also determine what is the theoretical number of plates are required. Also what is the other parameter? The other parameter is the Q line what is the slope of this Q line? When Q is equal to 0 you know you got a horizontal line the slope of the Q line is 0 we got horizontal line and the slope of the line is infinite then it is parallel to Y axis and we could be anywhere in between when we got a two phase flow alright. So, the parameter Q is also very important because it will decide where it intersect here. So, number of plate in enriching section number of plate in stripping section can change depending on the slope of this Q line and therefore, it is very important that this parameter also has to be taken into account as a very important parameter to decide the number of plates both in the enriching as well as the stripping section. The parameter Q is a very important parameter in determine the slope of the Q line and approximate feed inlet condition and therefore, the number of theoretical plates this is very important parameters. So, let us see the effect of these parameters using the earlier problem which we have solved ok. So, in the earlier lecture we have solved the following tutorial and I will continue with the same tutorial in order to understand effect of all these parameters which we have been talking about. I will do this study again using the graphical method, but again I will use a excel worksheet. So, for example, excel worksheet is simpler for me, but for you it will be again you have to use a graph paper to understand effect of these parameters or also of course, you can develop the your own worksheet your own excel worksheet. So, you got the same problem and I will not go into details of this problem. We got the purity requirement given we got the feed composition given we have the Q value given also we got a value of B given and the Q D effect the condensation refrigerant effect available is also given. So, these are the parameters which we have been given the composition B feed line equation the Q and then we know that the purity requirements are 97 for the top nitrogen and the purity requirement for bottom is 0.05 nitrogen or 95 percent pure oxygen. And what we have to do now is to understand for the above problem calculate the number of theoretical plates for changes in value of Q D. I am going to change this value of cooling effect available at the top section of the condenser value of Q I am going to change I am going to change the value of x B and x D and correspondingly we will compare what happens if I increase the value of Q D what happens if I decrease the value of Q D to the number of theoretical plates that are required for the column design to attain a given purity. So, let us first see from the earlier section what we had seen was the operating line equation which we had was this the operating line for the stripping section we had for was this and we have got a Q line equation also as given with this based on this we have already constructed a graphical methodology we have already plotted the curves for the given requirement of x D and x B. Earlier lecture we have seen that the staircase in procedure we had done on the excel sheet and we have found that the number of total number of vertical lines were 9 in that case therefore, the total number of theoretical plate for the column are tabulated as below as what we had obtained earlier. So, in the enriching section I had 3 number of plates plus condenser stripping section the bottom section I had 6 number of plates plus boiler and this was the case when whatever was given in the problem as 500 kilowatt condensation cooling effect available and the purity requirement was 97 percent and 0.05 percent respectively for top and the bottom section. Now, let us take this as a tutorial we want to understand what are the effect of different parameters first parameter is let us see the effect of Q D which is the condensation cooling effect available on the top holding all other parameters as constant. So, let us I have just made excel sheet and we can see that the first column is whatever result we had obtained earlier for 500 kilowatt cooling effect available. So, what for that case we had got Q B is equal to this value L by V which is a reflux ratio internal reflux ratio as 0.77 L by V SS for the stripping section as 1.27 and the number of plates which we have just seen is 3 in enriching section and 6 in the bottom section or the stripping section. Now, let us go to the excel sheet back and let us see what happens if I change my Q D from 500 to a different value. So, let us see the excel worksheet. So, here in this case this is my Q D which is 500 kilowatt and now I will change it to the let us decrease it to 350 kilowatts 350,000 watts all other parameters will be will be changed now and therefore, again my slope has suddenly come down you see we have now gone to the less L by V less condensation load that is available. Let condensation effect that is available and therefore, my L by V for enriching section got decreased from 0.77 to 0.69 all right and correspondingly my stripping section also we change and now I will see what are the number of plates that could be calculated from this. So, again you can see that number of plates for the stripping section will be increased to 4. So, the total number of plates will be now be 6 plus 4 and this is a result if I decrease the cooling effect available from 500 kilowatts to 350 kilowatts and what is happening because of which my both the lines are shifting toward the equilibrium curve and now my total number of plates are 1, 2, 3 and 4 for the enriching section and the total number of plates for the stripping section are 8 over here. So, total number of plates are now 12 in this case. So, can I just compare this with earlier case. So, if I see now this is if I decrease the value of QD from 500 to 350 kilowatts my L by V goes from 0.77 to 0.60 that means it shifts to the equilibrium curve and therefore, the number of plates have increased from 9 to 12 as we know that as we shift toward the equilibrium curve the number of plates have increased and now let us go to the other case when I increase the cooling effect from 500 to a higher value again go to the graphical representation and now I will increase my cooling effect from 500 kilowatt to 900 kilowatt and suddenly you will find all the parameters have changed over here and my enriching section slope the operating line for the enriching section has gone to 0.88 that means the slope has increased alright. So, it has moved toward the diagonal now it has moved to the y is equal to x line and corresponding now number of plates in this case will be less because I am moving toward the enriching section this is what you can see from here. So, here if I draw the number of plates for the enriching section and the number of plates for the stripping section you can see that these two have decreased. So, my total number of plates have come down to 1, 2 and 3 for the enriching section and for the stripping section it has come down to 5 what does it mean if it means that if I increase the cooling load if I increase the cooling load keeping the D as same keeping the D as same my L by V has increased to 0.88 correspondingly my stripping section L by V has moved towards 1 it has decreased for the stripping section. However, it has increased for the enriching section what has happened because of which I have moved towards the diagonal I have moved towards the diagonal and therefore my number of plates have decreased to 5 plus 3 in this case. So, let us see the condition now when I increase my kilowatts or the cooling effect from 500 to 900 kilowatts. So, let us see my excel worksheet again. So, here you can see that I have increased my cooling effect from 500 kilowatts 900 kilowatts and correspondingly you will find change in the slopes for the enriching section in all these cases I am keeping my B, D and F as the same. So, law of conservation of mass remains the same you can see that my slope has increased from 0.77 for earlier case to 0.87 and correspondingly my stripping operating line slope has decreased towards 1 1.14 which means that I have moved from the equilibrium curve to the diagonal line which means that the number of plates in this case should decrease and if I continue on the staircase thing here you can see that the enriching section has got 1, 2 and 3 number of plates it should have 3 number of plates over here while for the stripping section I have got 1, 2, 3, 4 and 5 number of plates in the stripping section. So, you can see vertical lines have increased here in the 3 and 5. So, if I go back to my earlier case now I can see now that number of plates here will be 900 kilowatts corresponding to that L by V has 0.87 which is shifted towards 1 while L by V for the stripping section has 1.14 while you can see that the plates has increased from 3 plus 6 to 3 plus 5 in this case that means the total number of plates have decreased to 8 in this case has come to 9 in the earlier case with the given problem while it was 12 in a lower QD value. Now, let us go to the next problem where I would like to take an extreme case of very high cooling power if I have got a cooling power of let us say 9000 kilowatts correspondingly there will be lot of changes over here now. So, you can see I have increased my cooling load to 900000 watts and suddenly you will find that if you go to enriching section you can find my slope has come to 0.99. So, I am coming very very close to y is equal to x line in this case for the enriching section and also corresponding to that the stripping section slope also has come to 1, 1.01 what does it mean? It means that all the lines have shifted very close to y is equal to x and therefore I will have 3 number of plates in the enriching section and I will have 5 number of plates in the stripping section because you can see now my number of vertical lines 1, 2 in fact is 2.5, but I am completing it to 3 and in this case of the stripping section I got 1, 2, 3, 4 and 4.8, but I will say 5 number of plates in the stripping section. So, total number of plates have come down to 5 plus 3, 8. So, what does it show? It is a extreme case very high cooling power is available I have moved towards y is equal to x line the slopes have moved towards y is equal to x which is 1 and correspondingly if I now go back to comparison I can see now my number of you can see for 9000 kilowatts cooling effect available I have got slopes as 0.99 and 1.01 and now if I see number of plates it will be 5 plus 3, 8 number of plates which have got reduced as compared to previous case. So, in extreme case when I reach to 5 is equal to x line number of plates have got minimum in this case. Now, I can also see that if my condensation cooling load is decreased further to let us say 300 kilowatts then I move towards the equilibrium curve. So, if I again play with the excel sheet and if I now make the cooling load as 300 kilowatts corresponding to you can see now my number of plates are going to be drastically increase in this case. So, my slope has come down for the enriching section as 0.62 my slope has stripping section has increased to 1.51 that means I am going now away from the y is equal to x line going toward the equilibrium curve and correspondingly I can see now that number of plates in the enriching section has become equal to 4, 1, 2, 3 and 4 at the number of plates in the enriching section and correspondingly in the stripping section I have got 8 number of plates. So, total number of plates have increased to 12 and why did this happen? This happened because my cooling load availability got decreased to 300 kilowatts only. My slope decreased to 0.62 which went to very close value and the point of intercession has come very close to the equilibrium curve. As you remember this is a condition of minimum reflux in this case and therefore the number of plates in this case has increased. So, let us come back to now comparison curve and here if I add this 300 thing I have got total number of plates as 13 in this case which are very high as compared to 1. So, you can see that lessening the cooling load move towards equilibrium curve number of plates get decreased and if I increase the cooling load I move toward y is equal to the x line and the number of plates get decreased in this case. Now, let us go to the next parameter study which is we want to see the effect of x d and x b which is the purity level. What happens if I increase the value of x d or x b and corresponding to c? Let us see the excel sheet now to this. So, in this case I go back to my previous case where my cooling load was 500 kilowatts my purity was earlier in 0.97, but if I now increase the purity. So, I will increase my purity to 0.97 to 0.99 in this case. So, correspondingly you can see that enriching section slope will change and everything will change in this case. So, my slope has increased to 0.78 now for enriching section the slope has increased to 0.78 corresponding stripping section also has got changed and here if I do the staircase in business again 1, 2, 3 and 4 I got 4 plates in enriching section and I got 6 in the stripping section. What does it mean? I have as soon as I increase the purity I went down from 6 plus 3 to 6 plus 4 my number of plates in the enriching section got increased from 3 to 4 in this case and this is the effect of basically change of purity or x d parameter from 0.97 to 0.99. So, you can again go back to my worksheet. So, here again earlier it was 0.97 3 plus 6 and now when I increase to 0.99 plates got increased to 4 plus 6. This is a change of purity for the x d. Similarly, we can see for x b case where earlier case 0.05 was a purity requirement we had a 3 plus 6 requirement and if I change my purity level now x b level to 0.02 for example or 0.0. So, we will let us change it to 0.02. So, my x b requirement has become very stringent on the bottom side on the stripping side and my x b has decreased from 0.05 to 0.02 corresponding my staircase is visible change and the number of lines required in the stripping section number of stripping plates required has increased to 8 now from 6 it has increased to 8 the slope has come to 0.76 the slope of stripping section has come to 1.28 in this case. So, we can see that number of plates in the stripping section has increased just because I made my purity demands from 0.05 to 0.02 or from 95 percent to 98 percent for oxygen purity alright this is as you know is a impurity for nitrogen in the bottom section which also means the purity for oxygen has become 98 percent in this case. So, let us go back to our slides and see the comparison. So, what has happened if I go from x b as 0.05 to 0.02 my number of plates have increased from 3 plus 6 to 3 plus 8 number of plates in the stripping section has increased from 6 to 8. Let us now go to the next parameter and let us see the effect of q what happens if my q parameter changes let us again go to the stripping section. So, here I can change my q value and let us go to original problem first and how will we change the value of q b q by changing 0.7 value to 0.8 value or something let us go to a very high value of 0.999 which is very high value of q and you can see that this is q is 0.99 that means everything is l basically everything is l or l by v is almost whatever coming is completely l that means my slope has become infinite q upon q minus 1 for the feed line has become infinite and you can see now this is my feed line over here vertical is the feed line and correspondingly I got different number of plates in the stripping section as well as the enriching section here. What we can see the q as soon as I change q to 0.99 my whatever input is happening is now is completely in l it is a liquid condition saturated liquid condition in this case l for the feed is going to be very high is equal to 1 whatever feed is coming is completely 100 percent is liquid in this case corresponding to this now I got a 6 plus 2 or 2 plus 6 as number of plates. Let us now change the l value to 0 if I change the value q value to be equal to 0 in that case. So, q was 0.7 plus and now I had q is equal to 1 which I wrote 0.99 in excel sheet my slope become infinite and I got number of plates as 2 plus 6 in this case. Now let us go to the value of q where q equal to 0 and when I make the value of q to be equal to 0 my number of plates become 6 plus 4 and this can again be seen from the excel worksheet. So, you can see that when I make the value of q is equal to 1 I got all the liquid coming in the thing the feed is coming as liquid if I make q is equal to 0 there is no liquid and the feed is completely vapor in this case I need more number of plates and if I put any other q as minus 0.2 0.3 for this value of slope I got number of plates to be in between these two extreme cases all other parameter remaining the same. So, maximum plates will happen when all the feed is coming in the vapor condition minimum number of plates when all the feed is coming in the liquid form or somewhere in between when you got a two phase that is coming in the as a feed in this case summarizing these parameters we have following inferences the number of theoretical plates decrease as the operating line approach close to the diagonal which we know the slope of the operating lines of enriching section decreases by increase in the value of d or reducing value of q d we know this now. Similarly, decreasing x b or increasing x d decreasing purity requirement of purity changes and if we increase this purity requirement it will increase the number of plates in the column the value of q is vital in determine the q line an approximate feed inlet condition and number of theoretical plates. So, q also dominates the number of plate calculations for this case with this summary of the different parameters I would like to summarize the entire gas separation chapter now this is what we have studied till now. So, if I want to summarize all the things which we have learned in last 7 or 8 lectures this is right from the beginning of from the work of separation different laws Routes law Dalton's law etcetera to the number of theoretical plates calculations and to see the effect of different parameters that affect the theoretical number of plates there are various other things also to be understood, but I cannot take and that that will go out of the scope of this particular lecture, but there are other things also to be understood that means how much high purity how to calculate number of plates at very very high purity requirements and thing like that, but this I will keep you for referring to the different books summarizing whatever I have taught to you in this last 7 or 8 lectures on gas separation are falling points in an ideal gas separation system all the processes are reversible and the work requirement is called as ideal work and this ideal work will be minimum for ideal conditions the equation for work requirement per mole of mixture to separate a mixture with n constituents is you know this formula which we have derived alright. So, I will not go this is all sigma depending on number of n components from j is equal to 1 to n y j log of 1 by y j this is what will determine the work requirement per mole of mixture the work per mole of mixture is always less than work per mole of its constituents for any mixture we have understood this by various problems and tutorials also. Also W i m divided by n m is maximum the ideal work per mole is maximum when the percentage composition of all its ingredients are equal from mathematics we can understand this W i by n m is going to be maximum in this case. If number of components number of phases and degrees of freedom of a mixture in thermal equilibrium are denoted by C P and F then the gives physical gives F is equal to C minus P plus 2 and again we have understood this with various tutorials and problems Dalton's law relates partial pressure of non-reacting ideal gases Raoult's law relates the vapor pressure with the liquid mole fraction of a component in a mixture is a very important Raoult's law because of which we could solve a lot of problems also. The Gibbs Dalton's law is an application of Dalton's law to the vapor above the liquid phase in fact these 3 laws you know make the basic foundation for the separation column is not it. The Murphrey efficiency if you remember Murphrey efficiency for a plate is defined as the ratio of actual change in mole fraction to the maximum possible change that can happen on each plate alright. So, Murphrey efficiency can be different for each plate also. So, this is actual change of mole fraction to the maximum possible change that can happen if ideal heat transfer happens in a plate. Macapthil method is less general and is widely used for binary mixture at cryogenic temperature and this is what we studied. The major assumptions in this method is that the liquid and vapor enthalpies are independent of mole fraction and this was a reasonable reasonably good assumptions in order that we understand the variations of different parameters if you are otherwise have to calculate the enthalpies for different composition for every plate alright. So, this is the major assumption which facilitated the calculations and made it very simple. The equations of operating line for enriching and stripping sections are these and this you know this very well now. The locus of intersection of these operating lines denote the feed condition and it is given as y is equal to q upon q minus 1 x plus x f upon 1 minus q and q upon q minus is the slope of the feed line. The point of intersection of feed line or q line and y is equal to x gives the content of a component in feed which is x f alright. So, x f denotes the composition of low boiling component in the feed in this case. So, this was what our Macapthil graphical method and the equation of the feed line is this and different q conditions can be understood and we had understood for saturated vapor your q is equal to 0 the slope is equal to 0 for saturated liquid q is equal to 1 and slope is infinite in that case and you got a two phase flow where the q lies between 0 and 1 slope is going to be negative and you got a sub cooled and super cooled condition also. For all this condition we got different lines and this is what is going to determine the number of plates in enriching and stripping sections on top part and bottom part over here. Equilibrium curves gives the relationship between the liquid composition x n and vapor composition y n on the same plate is a very important. This is what facilitated us to go horizontal and vertical on when we wanted to calculate the number of plates in the enriching and stripping section. The operating line relate the variation of liquid x n plus 1 and vapor y n because we got a y n as a function of x n plus 1 these are the mole fraction liquid and vapor mole fraction of a particular component across the length of the column. The operating line gives the operating parameter for every plate the equilibrium curves gives the operating parameter for each plate and this is across the plate because it is x n plus 1 and y n. The plate calculation is a stair casing method which involves locating equilibrium conditions on equilibrium line and operating line alright. So, this is also very well known. In a McAfthel diagram each horizontal line gives the condition of a liquid vapor on the same plate which are in thermal equilibrium. Each vertical line gives the vapor condition for the plate with respect to liquid that leaves the earlier plate which is x n plus 1 and this is y n v n this v n for the plate which is below the top. The total number of vertical lines in top and bottom section together with boiler and condenser surfaces is a total number of theoretical plates required. At the high IR value we have reduced number of plates, high condenser loads, product output is 0 which is IR equal to 1 in that case and at low IR value we have increased number of plates, low condenser loads, finite quantity of product output. This is what we have learnt earlier I is equal to 1 and low IR value what happens. The number of theoretical plate decreases as the operating line approach close to diagonal. This is what we study from different problems tutorials which we took that when the operating lines are going to a diagonal number of plates will decrease when we reach the IR equal to 1 value. So, high IR value we are very close to diagonal, lower value we are close to equilibrium line which results in increased number of plates and lower condenser loads. Again the purity requirement decrease in x b and increase in x d will increase the number of plates in the column. So, purity requirement also dominates the number of plates in the column. The perforated plates are sensitive to vapor flow rates. With this now let us come to see the assignments and please go through the assignment. The problem has been in line with what we have done earlier and for which we have got various parameters please go through this line and also calculate maximum and minimum number of plates in this case using a graphical sheet. The answers to this are given please check your answers and thank you very much.