2:31 Good evening everyone, good evening to all of you, good evening to everyone, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening, good evening and chapter will be absorption we will complete the whole a2z right so today's target will be one shot target that we will complete our absorption in one shot right so first of all everyone will do this work you share the session ok you can put it on your status all the people who follow me ok everyone who studies
3:21 with me put the link of the session on your status ok you will put it on status before the class starts everyone will come Secondly, you can share the link in the group you are added to. And the longer your class lasts, you have to attend the class continuously. because you must have completed this series in 4 days so you must have understood the content of this series right so the longer you sit, the more additives you will get and the 10-10 lectures that we take to complete a chapter we are completing it within 2.5 hours. The chapter of 12 hours, 15
4:11 hours, 16 hours, we are completing it within 2-3 hours. So you are getting the whole content. And we will select some numericals, on the basis of which many applications will be cleared. And a lot of surrounding practice will be done here. So we will do this much work. Till then we will restart the session. so this is about myself. I will tell you about myself. Ok, the new students who are coming to this channel for the first time. So in case you want to know about me, I will tell you about my small achievements. Six times I have get qualified X experience in the HPCL Bangalore in 2019. I am
4:56 also selected in the HURL in 2021 and my M.Tech completed from IIT Guwahati. Now I will discuss 5 major subjects of Chemical Engineering. This time I have covered Mass Transfer Operation, IPC, Heat Transfer Operation, Engineering Mathematics and Thermodynamics. So whatever will go on my profile, you will get all the courses on plus. Along with this, here I told the kids to record thermodynamics. And here I have completed two of your small subjects and the second one has just started. One subject has been completed, which is plant design economics. I have completed it in just ten lectures. So whoever wants to see the students, they
5:41 can complete it in ten lectures. you can complete the entire plan design economics. Overall, there are 13 lectures. In the last two lectures, I have completed 100% PYQs. If you want to see the theory, then the entire theory part is ended in 10 lectures. After this, another subject of yours has just started, mechanical operation. So, those students who want to continue with mechanical operation, they will be running classes at 11 in the morning. Okay, you can do that. Whenever you come to Unacademy, you take plus and iconic, then this is the my referral code. Coming to Unacademy means that this referral code right This is the first session of day 5. We are taking two
6:32 subjects simultaneously on alternate day. On one day we do mass transfer operation and on next day we complete IPC. complete the chapter. So in one shot we gave you diffusion, in one shot I gave you mass transfer coefficient and today is the third day of mass transfer operation where we will again complete the big chapter absorption. So every day you are doing 2 number class. It can be 2 to 3 number or 4 number. So you are covering minimum 2 marks on a daily basis. So today's target is absorption. We will cover it in one shot. We will focus on the important topics in absorption. Like counter current absorption. Very very important topic. And the second topic is important. HTU and NTU concept. We will
7:21 we will focus on that and also will discuss some other things regarding this. So today's target is counter current absorption and the second topic we will study is HTU and NTU concept. Both are important topics. Both are highlighted. Both are frequent asked topics. We are going to discuss on these topics today. So, we will put the heading counter current absorption. So, write down the heading counter current absorption. No one percent of your time will be killed in any other thing. Here 100 percent of your time topic ki baat karenge aur 100% content aur crips content ki baat karenge right toh yaha pe either udhar ki cheeji hum log yaha pe bulkul bhi discuss nahi karenge
8:08 seeday seeday apne point par aake jaldi se baat karna hai start karte hain chalo write down the heading counter current absorption write down the heading counter current absorption 10-15 minute basic dekhne hai iske baad hum log thoda sa iske aage accelerate karte hain yes good evening sab logon ko good evening Harish Mansi Dhritik Great! The students who are serious are studying from home. So that the connectivity in the class is good. If you have a class, then you can revise a little chapter beforehand. So you will connect very well. So students do this, but some people come directly, so they have to work a little more. Let's start. Everyone is ready.
8:54 Everyone is ready. We will sit with a pen and paper. We will close the door. Lock the door and listen to the music with 1.5 to 2 hours concentration. We will revise the entire chapter. So, we are talking about counter current absorption. So, let's start with the basics. Make a column. Make a column. This is the column. We have a column. We will try to understand it very well. This is your absorption column.
9:41 This is the absorption column. We will understand it very well. we will understand each point very carefully. This is our absorption column, so what you should know is that this is its bottom and this is its top. This is the top section, we will call it top and this is the bottom section. We will represent the bottom with 1, so we will denote all the compositions and flow rates in the bottom with 1. Whatever will be in the top, whatever flow rates and compositions will be there, we will denote them with 2 naming. Here, the gas flow rate which we will represent with G, here your mole ratio will be Y1. Here, the gas flow rate will
10:30 be Gs and here the mole ratio will be Y2. Here the liquid flow rate will be Ls and capital X1 will be coming and the liquid will be coming out from here. This will be our Ls, solvent flow rate and here capital X1. Let me tell you again what is GS? It is the gas flow rate on solute free basis. on solid free basis so gs kmr it is the gas flow rate on solute free basis tick y1k it is the mole ratio okay here also our mole ratio is our gas flow rate here is going to be the same in
11:17 both the top and bottom places solvent flow rate is Ls which is on solid free basis and here when we talk about solid free basis then our flow rate will be constant so flow rate will be constant and your composition will be changing so how we will define Ls, Gs, X1, Y1 let's try to understand this quickly So, what will we write as GS? It is the carrier flow rate or gas flow rate on solute free basis. So, this is the carrier flow rate. And this is the gas flow rate on solute free basis. What is Y1? It is the mole ratio.
12:05 In case, in the feed, we have A plus C mixture. What is A? A is carrier. And C is solute. this is the solute right here our beta solute and this is our carrier here our carrier is the solute right so our y1 will be the mole ratio so moles of c will be written by everyone it is defined as moles of c upon moles of a right so mole ratio solute to the carrier solute to the carrier so our a is our solute sorry did i say it in reverse ok take it in reverse like this take a carrier and solute of C. It is
12:50 not necessary that A and C will always be the same. It will be given in the question. According to that we have to take. Always read the question. Your carrier and solute can be changed. So C is the moles of C. Moles of C upon moles of A. so this is how y will be defined here gs is the carrier flow rate so gs is carrier flow rate on solute free basis and y1 is mole ratio similarly what is here again see this solvent is also solvent flow rate solvent flow
13:36 rate it is again solvent flow rate okay on solute free basis on solute free basis write all the terms once solid free basis here x which is defined how it is defined x which is defined how it is defined it is again defined x here your 2 will be x2 yes how x2 is defined it is the moles of moles of c upon our here son your top is solvent in solvent our b plus c is coming in solvent what is coming b plus c and this will be your moles of b So this is the solid to solvent or this is the solid to
14:23 carrier or solid to solvent. so we will write the mole ratio and the flow rate on solid free basis so all of you should know the basics like what is ls, gs, y, x here mole ratio will be given in mole fraction so we will convert it in mole ratio and in case you have total flow rate given then we will convert it to solid free basis so total flow rate will be given so we will convert it to solid free basis okay so in case you have to calculate gs so how will we calculate g1 in case you have here g1 is given gas flow rate so how will we calculate this 1 minus small y1 and small y1 is the mole fraction write one more thing here that
15:11 what is the relation between small y1 and capital y1 so small y is equal to In case you have to write, yes, you have to subtract capital Y, then small y divided by 1 minus small y and similarly you have to subtract capital X, then small x divided by 1 minus small x. Small x is written in the term of mole fraction. Again, small y is written in the term of mole fraction and So these are the basic parameters that you should know so that we will tell you the whole story on
16:00 these things. write down this first. These are the basic things that you should know because all the things are based on these. As soon as you complete the things, you will definitely say yes in the comment box. You will definitely say yes in the comment box, you will give a green signal, you can reply to the things
16:46 from your end. Okay, these are the basic things. I will do this. Now I will talk about operating line. I will talk
17:35 about operating line. I will draw an equation of operating line. I will draw an equation of operating line. I will draw an equation of operating line. I will draw an equation of operating line. Whenever we will find the equation of operating line for counter current flow, then keep in mind that where your variable is going to be. and the variable we will get from here is y and from here we will get x this is capital Y and this is capital X so these are two variables one is y and the other is x so
18:22 listen very carefully whenever we will take out the equation of operating line so in which equation will we always put the balance? in top or bottom? Where do we put the balance? Top or bottom? I have to put the equation of operating line, so will I put the balance in top or bottom section? If we want to calculate operating line equation then we can calculate it by applying valence But where will be easy when we will apply in top section Because this term is unknown This term is
19:07 unknown parameter This is unknown parameter If we apply in bottom then we have to calculate it Right. We have to do the calculation first. Then we can put it in the bottom. So we don't have to do its calculation. We will have x2 given. We will have y2 given. Okay. How much separation is happening. All these things will be known. So by going to a particular x. We will get y immediately. Right. By going to a particular x. We will get y immediately. That means there will be one line. You will have two variables. You will know x. You will know y. You will know y. You will know x. You will know x. Otherwise you will put it here. So x and y are your two variables. Along with that. and 3 variables will be there. So, because of this, we will always, whenever we have to find the equation of operating line,
19:54 we will always put the balance in the top section. So, we will always put the balance in the top section. So, we will put the balance in the top section. So, we will put the equation, operating line equation. So, here we will talk about operating line. operating line equation so here our gas flow rate is gs and here our y will be more y minus y2 absorption is happening is equal to ls ls and here what will happen to us, tell me, here what will be more, x will be more, right, x
20:41 minus x2 because solute is continuously being transferred to the bottom side so this solvent will be pure so as we go in downward direction the solute will start dissolving in the liquid so as we move towards the bottom the composition of x will increase So, as we move towards the bottom, the concentration of X will increase. Because X is a solute. Solute will dissolve in it. As we move upwards, it will be pure solvent. So, slowly your composition will increase. So, X will be more compared to X2. fine, very good, so from
21:28 here operating line equation will come capital Y is equal to capital Y is equal to Ls upon Gs Ls upon Gs and X minus X2 and plus E, your son, Y2 will be added here this will be Y2, so E will come out operating line equation this is known as the operating line equation, E will come out operating line equation this is called operating line equation this is known as operating line equation okay so this is your operating line where slope is how much see how much is your slope ls upon gs which is how is it positive slope what is our ls upon gs which is how is it son it is
22:15 positive okay how is it your positive right this is our positive slope is our ls upon gs but how is the slope positive one thing So this is the operating line equation for counter current absorption. For which? For counter current absorption. So if you get any numerical basic then you will make it from here. We ask you for basic and you make the compositions from any point. So for counter current absorption, how is your slope? Positive. This will be my co-current. So how will our slope be? Negative.
23:02 in case my co-current is negative, then slope will be negative. so this is the counter current absorption slope will be positive, operating line equation will be this, basic term will be this. very good fine done great now in
23:47 case write it down here, all the things are ok in case if I calculate minimum solvent how will I do it? if I have to calculate minimum solvent how will I do it? so next heading will be calculation of LS minimum write down the next heading calculation of LS minimum minimum solvent required how much will be required? if I have to calculate this how will I do it? If I have to calculate the minimum solvent, then how will it
24:37 be made? Listen very carefully. When a bigger figure is made, then we will make a bigger figure in our copy. Your figure should be big. If we talk about absorption, then the equilibrium curve that comes for absorption is like this. Composition was the largest for Y1. So I will do one thing, first I will draw an operating line. Then I draw the composition here.
25:24 The more you represent things with clarity in your copy, the more you will be able to understand it clearly. Our maximum composition was column Y1. So Y1 was maximum. After separation, Y2 was less. So this will be Y1. And Y2 will be Y1. After the separation. So this separation is fixed. How much separation you have to do? This separation is fixed. You have to remove 90%, 40%, 70%. So whatever you have to remove, this will be fixed. so this terminal point is fixed y1,y2 y1 will be the most in bottom and y2 will
26:09 be after separation Now see where will be x1 and x2? This is our pure solvent x2 and this is our x1. That means x1 will be more than x2. So the value of x2 will be more. The value of x2 will be less. The value of x1 will be more. So what will be our? X2 will be there and this will be our x1. This will be our x1. So which composition are you getting? You got this composition. Capital X1 comma capital Y1. You got this composition. Similarly, which composition did you get? capital X2 comma capital Y2 so this composition is capital X2 comma capital Y2 so now this point is of top or bottom? this point we
26:55 are taking from top or bottom? where is this point? is it of top or bottom? anyone here? where is this point of X1 Y1? of bottom right? x1,y1 is of bottom. So this is our point of bottom. And this point is of x2,y2. These are our top. So these are our compositions of top and this is our bottom. So bottom comes up and bottom is down. So don't get confused that this is our bottom. Here bottom is coming up and
27:41 here is our top. because top compositions are there because there is counter current right which is counter current because of this no sachin is not available because this part is not in the gate because of this what was its slope what was its slope what was your slope what was your slope ls upon gs Ls upon Gs will be the slope. So, you can see the slope is positive. How is this? It is a positive slope. So, how do you catch the positive slope? That both y and x should be increasing together. See, y is also increasing and x is also increasing. So, this is a positive slope. This is called a positive slope. In case y and x should be increasing together, this is called a
28:30 positive slope. Okay. Both y and x should be increasing together. This is called a positive slope. And any line is drawn like this, that is, your y and x should be means y is decreasing and x is increasing so this is called negative slope right this will be our positive slope so slope is positive ls upon gs I have made the compositions of top and bottom like this done everyone will make the figure now we will do one thing we will make one figure again in our copy one figure we will make again in the copy ok everyone will make one figure again in the copy everyone will make one figure again in the copy ok I will do the work in this you make it again
29:18 You make it again, I will work on it. Make a new figure in the copy again. Understand things better. Everyone will make a new figure again. So what was our composition? This was our x2, y2 and this composition was our x1, y1. Now listen very carefully. Now I am going to do the heading calculation of the LS minimum. Calculation of the LS minimum means Minimum solvent flow rate calculation. So what will we do? we have to find out what was the slope
30:04 of this line? Ls/gs Ls/gs was the slope of this line now I want Gs minimum I want Ls minimum that's why we can calculate the minimum solvent so to find out Ls minimum will this slope increase or decrease? will I decrease the slope of this line or increase the slope of this line? slow beta will decrease so i have to decrease the slope so to calculate the ls minimum first of all what i have to do is to calculate the ls minimum so where will you get the ls minimum calculation means where will you get the minimum solvent where your flow rate will be maximum where
30:53 you will have equilibrium fp so what i will do is i will move this line forward and how much will i take this minimum is up to equilibrium point so as i move this line forward up to equilibrium point because y is fixed now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can i go decreasing this now how much can So, I went to this point because we can't go beyond this point. So,
31:42 now what will be the slope of this? So, Ls/Gs will be the minimum slope. So whenever I have to calculate the minimum solvent, then which composition is going in your equilibrium? Now take x1 and take x1 double dash. So x1 double dash and y1 is also in bottom. And you will also get this composition in bottom. This also will be
32:27 in bottom. That means the bottom composition is in your equilibrium. okay bottom key compositions are both of equilibrium may hold a tabby up to solvent flow rate who got more up to minimum or just slope of the operating line or go up come in mom or slope of the operating line to have a minimum or a slope of the operating line minimum cover got job up to bottom key compositions equilibrium may hold a any key up to bottom key composition X1 and Y1 Equilibrium means the composition will pass from this equilibrium curve. So whatever composition we have to replace, we will replace the bottom composition. Why are we replacing the bottom composition? Because this figure tells
33:13 us that whenever we calculate the LS minimum, how far can this minimum go? up to this point, it can go up to this point only, here we will get L minimum. So, because of this whenever we set the equilibrium, we always set it at the bottom. So, from here you will solve by interchanging. So, first of all you have to find the Ls minimum. So, from here your Ls minimum will come. This is called Ls minimum upon Gs and Gs is fixed. So, GS will be fixed and LS minimum will be removed from here. This is the calculation of minimum solvent. Minimum solvent will be removed easily. But, in
34:01 case of operating solvent removal, So how will we get that? For that you have a condition that the operating solvent i.e. the optimum solvent flow rate will be 1.2 to 1.5 times of L minimum. This value will be given to you that from which factor will you multiply in L minimum. So whenever I talk about L optimum, so pay attention here whenever I talk about L optimum, for L optimum So for L Optimum I will multiply 1.2 to 1.5 times of L Minimum. In L Minimum I will multiply this factor. So from here L Optimum will be obtained. So when I multiply this positive factor,
34:50 then the optimum slope will be more than this. the optimum slope will be more than this and the slope will be more than this and how much will be that? that will be this we have to operate on this and the corresponding composition of this will be x1 dash this is the reason why I wrote x1 double dash so the slope of this will be Ls upon Gs optimum This will be your optimum and operated. Whenever you get any question that you get the optimum flow rate or you get the actual composition that
35:37 how much will be going. So the actual composition will be this x1 dash which will be going out from the bottom. so this actual composition will be withdrawing from the bottom when it will be when you will operate on L optimum and how will L optimum come out first you have to find L minimum and to find L minimum which composition will you do in equilibrium whatever the bottom composition is you will do it in equilibrium so the bottom composition when you will do it in equilibrium then you will get L minimum and after finding L minimum you will multiply this factor in it you will get ALO and ALO you operate the system on that you will get the actual composition of the withdraw
36:25 actual composition you will get that how much is going out of your bottom so on this basic concept your gate questions come try to do something this is the concept that you have understood that how much composition is coming out of the bottom, you will calculate it, whatever you ask, you will calculate it, write it down, I have told you the concept of CRIPS, we will use this concept to practice some questions here. So, for which I have discussed this? Counter current absorption. So, we are discussing for counter current absorption
37:15 only. So, what is its rule? What is its assumption? Counter current absorption. Very nice. Let us read this concept properly and then we will try to apply this question in a little more detailed way. There is a question which has come in 2019. Try to do this. This question is for you
38:02 to practice in 2019. This question will clear your basic absorption. Everyone try to make
38:49 it. Your son has been asked this question in 2019. Everyone must have read the
39:36 question. counter current absorption tower 95% component a 95% composition remove 95% remove 95% remove mole ratio of inlet gas y1 composition 0.02 you more ratio catering up kept up in a carrier gas flow rate yeah Nikki GS up capacity the feed up the idea here's up capacity the feeder the idea a potential the equilibrium relation y is equal to 2x given a state line the year y and x are the mole ratio okay gas and liquid fit if that
40:23 tower operated twice the minimum solvent take a little given a In case your tower is operating 2 times of the minimum solvent flow rate. The mole ratio of A in exit liquid stream. So what will be the exit liquid stream? Here X1 is asked by you. So mole ratio of A in exit liquid stream is asked by you. So here GS will be there. And here Y2 will be there. And here LS will be there. And here X1 will be there. yeah so after put a guy what is the value of capital X 1 yeah also put a what is the value of capital X 1 yeah upset calculate car vara so capital
41:09 x1 is getting us to calculate this we have to calculate some people must be doing this that sir capital x1 what we will do is put these two in equilibrium and y is equal to 2x given y is equal to 2x given and y1 what do I do y1 is equal to 2 times of x1 and y1 is given 0.02 0.02 and divided by how much will come tell me two times divided by 2 times and is equal to x1 so is there any one in mind that 0.01 answer will come sir is there any one in mind that sir this answer should be there that y is equal to 2x is given to you y1 is equal to we will take these two in equilibrium and x1 will come out what is the mistake in this someone will tell me sir what is the
41:57 mistake in this x1 was asked to you x1 came out what is the mistake in this it may be that this can happen in mind because anything can happen what is the mistake in this sir anyone What is the mistake in this? Someone will tell me what is the mistake in this? What mistake did you make here? What mistake did you make here? What mistake did you make? Tell me this. This statement is wrong for equilibrium. I want you to say the right thing here. It is not taken out of this.
42:46 That means it is a conservative mind. It is not taken out of this. That means it has been taken out in some other way. You remember that. This maximum separation is not a fun thing. The right line that he has spoken here. First prime minister X. I am just trying to understand how you are thinking. Yes. Very good. Read this. In the question, it is written which Ix you want. You should say this. According to the question, they are asking if you roll on if first. If. If. If. If. If. If. If. If. If. If. If. If. If. If. If.
43:51 If. twice the minimum solvent. This is the actual asking of this question. When your tower operates 2 times of the minimum solvent, then what will be the exit stream of the mole ratio of A? In the question, it is asked that when tower operates twice, then how much will be the L-optimum
44:41 when it operates twice? when it operates twice. So from here, this two times is not operating, this has been removed at the minimum. Sir, here the two times that has been asked is not being justified, this has been removed at the minimum. When minimum is operated 2 times then what will be the value of x? First we have to calculate L minimum When we take L minimum 2 times then we will get L optimum When L optimum is operated then what will be the value of x? First we have to calculate L minimum your GS will be y1 minus y2 here you will get LS and for LS here we will get x1 minus x2 ok now how much will
45:29 be x2 see here it is written pure solvent x2 is the pure solvent so how much will be the value of x2 tell me x2 will be zero For pure solvent, our X2 will be 0. So for pure solvent, our X2 will be 0. So our GS is 50. And Y1 is 0.02. And what is the minus of Y2? Y2 is being removed by 95%. So what will be Y2? When I remove 95%, then Y2 will be 95% removed. So 0.05 multiplied by 0.02. So what will be this value? 0.02. double zero one check once again 0.001 right so this will be 0.001 which
46:15 equal to here will be your ls and ls i will say minimum when will i say ls minimum when your bottom composition will be in equilibrium so if bottom composition will be in equilibrium then how much will i keep in place of x1 y1 upon 2 y1 upon 2 and by putting value of y1 here, everyone will be able to tell me the LS minimum. How much is coming? How much is the LS minimum coming? 95. How much is the LS minimum coming? 95. Check it once. So it will come. How much will the LS optimum come? 2 times of this. So if we do its 2 times, then how much will it come? Tell me. 9280, 190. Check it once. 190 each and every one.
47:05 Very good. Great. So L optimum is 95. Now in the question we have forgotten that when tower operates at L optimum, then we have to tell how much is the mole ratio going in solvent. Now they are asking this one. Now again you put this balance, so your gs will be y1 minus y2. Now operate on this, then tell us the composition. 190 and yes x1 minus, this will be 0 because it is pure. So now put it here, so this will be 50 and this will be 0.02 minus 0.001. and this will be 190x1 so from here you will
47:51 be calculating x1 will be 0.05 0.005 so 0.005 see this x1 which is double dash is less than that x1 which is double dash is less than that x1 double dash you will get maximum You will get maximum of 0.01. That will be less because composition is shifted in backward direction. So x1 is this dash. What we have taken out earlier was x1 double dash. That was maximum separation. but here you need optimum because you will not operate on maximum separation because there
48:38 required height of the column will be infinite because driving force is zero so required height will be infinite so we will not operate on minimum but on optimum when we operate on optimum then you will get x1' not x1'' so you can verify your answer with the help of x1'' so it should be less than x1'' right x1 double dash should be less than this point so everyone is clear once yes or no everyone understood say babu everyone will tell once in the comment box Sir, we have done this with the basic method. Now see, my purpose was not to come here and solve this question. I have to teach you something from this question. You have
49:27 to teach some things from this question. First of all, this was conventional way. Now I will tell you in advance how to solve such questions in a pinch. You practice those things a little bit, then many things will be easy for you. Many things will be easy for you, try to learn them. Tell me what you say. Do you have a good approach for this? Yes or no? I will arrange this question. I will try to tell you here. I will tell you another approach of this type of questions. How to solve this type of questions? We will be in learning part and
50:17 improving part. I want everyone to be in improving part. And understand all the things in a good way. In case I want to solve this question graphically. By method number 2. Everyone will put method number 2. We will try to understand very carefully. Alright. This was our equilibrium curve. Listen very carefully, you will understand one thing very well. Where there is something to understand, then you will understand things carefully. Where there is something to understand, you will understand carefully, there is no hurry. You will try to
51:04 understand things very carefully. Very nice. Great. What was the value of this? It was 0.02 and we had solved its value, so 0.001. right and we need what we have fixed x1 is our zero right this is our zero okay x2 is our what is our zero this value is our zero right and we have got this value this value what we have got this value so how much this value came x1 double dash is 0.01 and this is in equilibrium and this is straight line so we have equal composition and
51:50 we have this value now what i want is optimum value i want optimum value not minimum value so listen carefully what composition i want is minimum value what composition i want is optimum value I want this optimum line which I have drawn in the middle I want this composition What is my objective? To find out the discomposition What is my objective? To find out x1 dash Remember this What is my objective? To find out x1 dash Yes so my objective is to get x1 and x2 now listen to me, in the question what is being said? in the question it is being said that the minimum of
52:40 ls is 2 times of ls optimum yes or no This is what is being said in the question. We are going to use the same thing that was said in the question. So tell me, can I divide the LS minimum by GS? Yes, you can. And can I divide it by GS? Yes or no? No problem. I can divide it by GS. Now tell me, is the slope of the minimum operating line written? And is the slope of the optimum operating line written? Tell me. yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes
53:46 yes yes yes yes yes yes yes yes this this is I will take out the slope of this line and this line and multiply it by 2 by taking out the slope of this line. What is this? This is the minimum slope. So if I take out the slope of the minimum, I will multiply it by 2 and we will get this. change in y How much change in y is the slope? How much change in y is the slope? And how much change in x is the slope? We have this value 0 and we have
54:34 to subtract x1 dash So change in y upon change in x is this much So change in y is 0.02 minus 0.001 and divided by is equal to two times of is equal to two times of l minimum change in y is fixed, i need its slope change in y is fixed 0.02 minus 0.001 so here it will be 0.02 minus 0.001 divided by change in x change in x is going
55:25 to be this much change in x is going to be 0.01 and minus x1 dash so it will be 0.01 minus x1 dash one unknown and one equation is there so solve it easily see if the answer is coming or not check once x1 dash should be 0.00 Sorry, 0.01, yes, yes, yes, yes, yes. Here, change in x, see, here, I
56:15 am saying that only, it is a little stuck. Here, the slope will be, here, pay attention a little. Here, the change in x, because we are taking this line, this line, so the change in x, it will be taken so complete. So, 0.01 minus 1, yes. 0.01 minus 0. Now, your x1 dash is equal to 0.10. you can change in x completely so you can change in x completely so till here you have used the condition given in the next line and your
57:01 answer is here here you don't have to calculate the minimum, you have to calculate the minimum first and then the minimum and then the optimum and then the minimum and then the optimum and then the equation you don't have to do all that here slope, the condition given, if it is given 1.5 times then you will take 1.5 times if it is given 2 times then you will take 2 times just divide by gs and you will get the whole graph Whatever question you are solving here, it is completely with the help of graph. You have enjoyed the whole of your graphical analysis in this question. So with this type
57:50 of question we can solve the question. This is another way to solve this question very quickly. So this is one way. You can do it this way too. Whenever there is a question of this type, try to solve it with the shortest way. So solve it with the shortest way. time save ultimately so this was the basic absorption if you ask numerical then ask like this ok there is one more way to solve this what
58:36 do you say there is one more way to solve this there is one more way to solve this but that is a little big way by using this equilibrium relation ok y = 2x we can also use this equilibrium relation right wise culture to react to equilibrium relation and I scope a katana to regress a huge character look solve car second a canisko be Nicaragua okay is conicara a cookie both sorry method so jayenge a BSU go you buy school to your to X given a now is co-op huge character sector so I'm
59:21 Karna jacket why is called to 2x each characters yes yes extraction key to regress a very nice Kartika connected yes a extraction we curtail a equilibrium relation will use character a cigar sir Remember this method, this method is best for you, later everyone will do it with conventional method, with this you will save time. Next, so this was your heading, everyone will put the concept of the HTU and NTU. Write down the heading, concept of the HTU and NTU. Try to see the concept of HTU and NTU.
1:00:11 See, the atmosphere of weddings is going on in a strong way, you must be seeing a little bit of singing. Right, the wedding is going on, get educated quickly, everything will be done quickly. Sir, add me to the group. Look, I have added a lot of children to the group, but the children have gone out of the group. Children have gone out of the group by cheating. First he said, sir, add us, I will stay live and when I added it live, now they are not coming to study. So the boys have cheated.
1:00:56 Then the grade exam will deceive them. Take care of yourself. First he said add me. I added everyone. I said all of you are honest boys. I added. I am working hard at night. I am giving you 5 questions on a daily basis. After this class, I do exercise daily for you for an hour. And on a daily basis, I will tell you I put an effort from my end daily because I thought these guys are working hard so I should also give them an hour. So after the class ended, I am giving them a PDF of 5 questions daily. Now what the boys did, they got added. They messaged us and said that these guys are honest. Now what are they doing? I don't know what they are doing, they must be roaming around in winters. Right, so this is the
1:01:45 problem. I don't have any problem in adding. I will add and give you 5 to 10 questions. But the problem is that you don't cheat tomorrow. Right, right. So don't cheat tomorrow. Yes, be honest. I take all the classes of all the students tomorrow. Okay. Okay, they will add. Message me. Join the telegram group. Right. Join the telegram group. After joining this, message me. I will add. Sir, I am commenting while taking the rounds. you are not too busy right? the one who
1:02:31 takes the rounds will leave you. I think he is too busy. His concentration is somewhere else. His concentration is not here. Right? His concentration is somewhere else. So take the rounds first. First make sure. then you do the class whatever you do first do it for sure there it is not sure that your rounds are not complete son so after the class first do it for sure there so write down the heading concept of the HTU and NTU yes the kids who are coming regularly I will add them those who are
1:03:18 not coming let's talk to them tomorrow they will add me, Patel brother will message me by joining the telegram group Right on the heading, Concept of the HTU and NTU. Listen very carefully. You will understand very well. The concept of HTU and NTU. Based on this, I will ask you a good question. What is the concept of HTU and NTU actually? Here we calculate the height. When we calculate the height, then ht and ntu come in between. So by the multiplication of ntu and htu, we calculate the height of the column. So whenever we calculate the height of the column, then z is equal to the
1:04:07 multiple ways of calculating the height. Like to get the flux, you were taking the liquid side of the driving force, so your flux was coming from there. and you were taking driving force of gas side and you were multiplying coefficient then also your flux used to come out similarly here also you are calculating the height of column so you can calculate the height of column on basis of liquid and you can calculate the height of column on basis of gases so multiple ways are also here so here we will write first of all HTU HTUG into NTUG okay
1:05:09 z=HTUOL*NTUOL what is this sir? lets try to understand this what is this? whatever height calculation i have done this is liquid based now what is HTUG and NTUG this is the height of transfer unit based on gas side number of transfer unit based on gas side height of transfer unit is written as G this means this is the height corresponding to the individual mass transfer coefficient this is the number of transfer unit corresponding to the individual driving force you understand individual
1:05:55 right? y minus yi this is individual Similarly, what is this on the basis of the overall mass transfer coefficient? This is the overall driving force on the basis of y minus y star. What is this on the basis of y minus y star? What is this on the basis of koy? Similarly, this is in the liquid phase. Individual and individual. This is the overall and this is the overall. This is your overall and this is your individual. Write it down. Now, how is HTU defined and how is NTU defined? We will try to understand that on this side. we will understand it very carefully. So our HTU is, H T U G, this is
1:06:42 our definition, G S upon K C divided by individual mass transfer coefficient small k y into interfacial area. Canity U G, how we will write it, see very carefully, D Y upon dy/y-y_i this is the individual value y-y_i now the limit I will set is lower value to the upper limit lower value to the higher value because the positive quantity will come from here so lower value to higher value so higher value will be in the bottom and lower value will be in
1:07:27 the top so I will use y2 I will go from Y2 to Y1 because now this positive value will come out of your NTU otherwise you have put lower to higher sorry you have put Y1 to Y2 so its value will come out negative then also it is not a problem NTU will always be positive So in magnitude, there will be no changes, only minus and plus sign will change. So always keep that positive because the number of transfer units can be negative. So here in case you are not getting it, your answer will not be wrong. If 2.3 comes, then it will be 2.3 only. There will be only sign changes. So you have understood that NTUG will never be negative otherwise your RLOB will be positive. Otherwise, to play the calculation safe side game, then lower to upper limit. So this is
1:08:15 how we will calculate NTUG. Next one. When we talk about overall, pay attention to this point. Try to understand this point very well. In case if I talk about OG, then it will convert to KYKOG KOY and this YI will convert to YI*. yi will be converted into yi* and koy will be converted into koy and all the rest will remain as AC whenever we write htuog, gs/ac = koy*a and whenever
1:09:01 I write ntuog, the limit will remain the same, yi* will be converted into yi* because the overall driving force y-y* is for absorption So, pay attention to this, I am writing for absorption. Okay, for whom I have written this? For absorption. Write down. For whom I have written this? For absorption. Now I will write this also for absorption Listen very carefully, here the students get a little confused that Sir, this is written somewhere yi*-y and somewhere yi-y So whenever we talk about absorption, the figure you saw in
1:09:51 absorption was yi-y So write yi-y, in case you have written the gas side of stripping, then write yi-y For stripping operation, you have written gas side, then you have to write yi-y because then the composition will decrease here. So, pay attention that both things are there here. In case you are writing gas side for absorption, then you have to write like this. If you are writing gas side for stripping, then this composition will change. Are you getting it? Yes or no? Now, I will write liquid side for absorption. So, now how will I write NTUOL? So how will H T U L be defined? This will be defined as
1:10:36 L S upon AC. This is the thing which we have to clear. And here we will get K X into A and here we will get N T U L. Now pay attention everyone. Here pay a little attention. D X into. Now see. Now I am writing for absorption. So for absorption, I will write xi minus x. Now when you solve the limit, a negative sign will come out from here. This means that the limit I have to apply is higher to lower. Then ultimately your
1:11:22 positive will come out. so higher, we have to keep lower above so lower will be x2 and here will be x1 so here x1 to x2 has to go see here this is higher limit and this is lower limit for absorption I am talking about counter current so this is our x1 and this is x2 because a negative sign will come from here and because of this I have to reverse the limit this will be our lower value and this will be our higher value so ultimately your result will be positive in case you are not able to do it then it is possible that your result will be negative and you can make it positive later but keep on understanding this way only Now, whenever I talk about overall, this Kx will be converted
1:12:09 to Kox and this Xi will be converted to Xi*. Actually, it will not be Xi but it will be converted to X* and this will also be converted to Y*. So, pay attention to the composition I have written for absorption. Note this thing and write. These are the all relations for absorption. I have written this for absorption. And this gas side is written in absorption and this liquid side is written in absorption. Whose side is
1:12:55 written? This liquid side is written. And this gas side is written. Now this composition will change whenever I talk about stripping. The things will remain the same, just x minus xi will be there, here yi minus yi will be there, this much will be your change and nothing else will happen. So whenever I will write for stripping, these four formulas come separately. and for absorption, 4 formulas are given. What are the changes? Not 4, but 8 are written. Sorry, 2, 2, 4 and 4 are written. 8 are written and 8 are the formulas for stripping. So, 16 formulas are made from here. 16 formulas are made from here. All 16 of 16
1:13:41 I have written in plus. All 16 of 16 I have written in plus. You should understand everything. So, K, O, Y will be in case of gas, this will be in case of overall, this will be in case of overall, this will be in case of individual and all are for absorption. In case of stripping, just change the driving force. So, these 16 formulas are derived from here. These 8 are written and only 8 more are derived for stripping. You will go crazy. In case you sit to remember, you will go crazy. they
1:14:26 can never remember it, they just have to understand it and there is a lot of confusion in this you just have to understand that for absorption there is gas phase and liquid phase and for stripping there is gas phase and liquid phase you just have to know this statement, see how it goes it goes like this, you have to understand from here whenever you have a problem in composition, make this diagram make this diagram and connect it with this now let me take a quiz, you will tell me for whom it is written Tell me, you will tell me for whom it is written. I will write
1:15:15 a formula here. What do you say? I write one formula out of 16 here. Will you tell me? Tell me. Yes, the limit will be the same, you just have to understand that carefully. I will write one here and give it to you. You will tell me for whom it is written.
1:16:03 Tell me, it will be clear from here. Should I write? Let's write. Tell me, for whom it is written? First of all, what is written here? NTUG, NTUOL, what is written here? Whose formula is written here? First of all, you tell me this. Then you define it. Yes, what is written here? NTUOL. Okay. And for whom is it written? NTUOL is written, but for whom
1:16:52 is it written? For absorption or stripping? NTUOL is written for stripping or for absorption? This is written for stripping. NTUOL is written for stripping. Which side is written? Gas side or liquid side? Liquid side. So see what will happen in stripping?
1:17:38 X* will be more. Right? See it once again. See it once again. For stripping or for absorption? See it once again. see it once again see it once again make a figure see it will shine x* minus x in the case of driving force absorption we used to make this diagram in the case of
1:18:24 stripping yes we used to make this in absorption in absorption x* used to be more are you stuck? in the case of stripping right what will happen x* is in the case of absorption and here we have X so this X*-X is written this is the absorption case written in liquid side NTUOL is there no problem in this but this is the absorption
1:19:11 case absorption in liquid side absorption in liquid side absorption in liquid side is written yes absorption in liquid side written in the last one yes Karthika do you understand? now see we used to make it for absorption do you remember? this was our y and this was our y* and
1:20:00 y goes with our hypothetical x* so x*-x is written here absorption is written for liquid side give me one more you are getting stuck give me one more yes or no say it carefully you have studied it right master? come here then give
1:20:45 me one more one more once more come on tell me come on tell me Think and answer in the comment box. You will get 2 numbers. Think and answer. Think and answer. You will get 2 numbers. Always make it. See something like this. This will be yi and this will be yi. So yi will be more than
1:21:31 yi. Yi is greater than yi. This means stripping. This formula is written for stripping. yi minus yi is written for stripping. So, this is confirmed as a stripping case. Where did I write this? Sir, it is written on the gas side. What is NTUG? right
1:22:22 Suppose there is a problem in setting the limit. If this is a stripping case, then you know that this will be more than x2. This will be more because the case of stripping will be going here. And this will be less. So y1 will be less and y2 will be more. y2 will be more. From here one negative will come out. Because it is minus x. So one negative will come out from here. That means I have to set the limit upside down. That means I have to keep the value higher here. So the higher count will be y2. If you have a stepping then y2 will be more so here you do y2. You change the limit. This will be y2 and this will be y1. so you have confirmed the limit this way now it is perfect with the limit so y2 will be this and y1 will be this because y1 is less so when you put
1:23:11 the lower limit here this will be lower and this will be higher so 1 minus sign will come because of the limit and 1 minus sign will come from here so minus minus will be positive so in this way you have checked the limit are you getting it? yes or no? Yes, for absorption, y minus y is i, so it will be absorption. So always keep this in mind, that how does this vary? In its case, all these formulas come. In its case, all these formulas come. I think that now so much discussion has been done, now it is clear. Is the concept clear to everyone? Yes or no? Will you
1:23:56 write everything? Tell me brother, now there is no confusion in this. Yes sir, all clear now? Great. Done, done, done, done. Very nice. Can I ask you a couple of questions? Do you want to ask some questions? All of you try to solve this one. Very nice. This is a very confusing question. Because it is a subject of mass transfer. If you don't do it correctly, you will get into trouble. Do this. This is a question from 1988. I was
1:24:42 not born till then. It is a question from 1988. It is a very old question. Do it but it is a good question. It is a question of old age. This question is old age gold, that's why I brought it. Is the screen getting a little cut? Wait wait wait one
1:25:29 minute. Now it's okay, now it's okay, right? Alright, now it's okay. Try it, great. What answer? Till now, the answer should come. Till now, the answer should come. What? Try it. What should come? Answer it, you guys. I
1:26:18 want you to answer. Then I will get confidence. Otherwise, if you will keep on studying, you will not be able to do it, then how will the work go? I will be more happy. I have done it on
1:27:04 my time. Give me ABCD. Why are you typing so big? Give me the answer in ABCD. B is coming, right? Okay. So Priya, Chitej, Swamya. Swamya, the answer is wrong. Shubham, Prince Goswami. Is there anyone else? So see. Continuous. counter current observer operating a very dilute concentration the volumetric gas flow rate volumetric gas flow rate come across
1:27:49 me mehra G upon a k value of k given mass transfer coefficient is proportional to G to power 0.8 any key mass transfer coefficient up cup proportional to yes kg FK y you know G to power 0.8 the gas flow rate is doubled Now gas flow rate mol per unit time per unit empty area. The gas is doubled what will be the ratio of the height of the transfer unit. Means G2 = 2 times of G1. So
1:28:35 how will the height vary? Asked you. One minute Arjun Jani. I have given one minute. And it is a question of one minute. So you know what is the height of the transfer unit? H T U G. Is the mass transfer coefficient given overall? ok, give me any number, no problem H T U G you know this is your G S upon AC divided by mass transfer coefficient KY into A this is done so you must be understanding from here that brother this is your H
1:29:26 T U G it is directly proportional to G S divided by KY and this is directly proportional to mass transfer coefficient is varying like this so put here G2 power 0.8 so means how height of transfer unit is varying how H, T, U, G is varying proportional to tell me G2 power 1 is here G2 power 8 is here so it will be G2 power 0.2 and G is 2 times so how much will be H, T, U, G 2 to power 0.2 So, HtuG2 upon HtuG1 will be 2 to the power 2 times
1:30:11 0.2. Match it. Yes, do it. Plus 0.2. Very nice. So, this was a small question but it was a correct question. There were chances of it being wrong. In case any child did not pay attention to it. In case he did not take this Gs, he did inversely proportional to Ky, then your answer will be wrong, the second match will be done. In case inversely proportional to this, you
1:30:59 did not consider G, then it will be a mess. You have to consider G too. So how much will be the value of G2 by G1? 2. 2 to power 0. Is it clear? Everyone will tell me once. Sir, I couldn't understand the language in the first reading. Language is the same as you practice, we give the test series because of this. The advantage of the test series is that you read a lot of questions. When you read a lot of questions, you understand what the matter is. Sir, G2 by G1, how is it G2 by G1? When G2 is 2 times G1, then it will be 2 times 0.2.
1:31:50 This is G2 power 1 and here it is 0.8 so 1 minus 0.8 means 0.2 of positive. So this was the question. Right Arjun? Next one. Everyone will add headings. Ntu calculations. Write down the heading Ntu calculations. Let us understand about Ntu calculations that how many different types of Ntu can we calculate. Calculation of the Ntu. How many ways can
1:32:39 we calculate Ntu? we understand it in short, it is important too and we understand it well. Everyone is ready? Everyone already knows what is absorption factor? What is the absorption factor? Do you know? Everyone knows stripping factor? So here we will write ntq calculation when in bracket a = 1, see very well understand, you will understand it
1:33:24 completely in the table and when a not equal to 1 A is the absorption factor, it is the ratio of the slope of the operating line to the equilibrium line. When A = 1 and A = 1, we are going to talk about that. So what is NTU? We have to do NTU calculation for two cases. One case is A = 1. This is the case number 1. And the second case is when A = 1. This is the case number 1. So write case number 1 on one side.
1:34:11 and write case 2 on one side. See, it will be in this slide. Case 2 is in one slide. So, what is case 2? a0 = 1. Now, a = 1. It means, a = 1 means that both the operating lines will be parallel to each other. Equilibrium line. suppose this is the equilibrium line or this is the operating line both will be parallel in the case of h equal to 1. so what will be the driving force here will be constant not zero. driving force will be constant in the case of h equal to 1, driving force
1:34:57 will be constant this is our y and this is our x so y is corresponding to y1 equilibrium this is y1 so y1 equilibrium this is y2 so y2 equilibrium you will get this so means the driving force is constant you can find it anywhere you will get constant driving force fine? great if you think of driving force vertically then you will enjoy more if you think vertically then you will enjoy more you will understand better So, the driving force on y axis is constant. For this, I have to do the calculation
1:35:44 of NTU. Let me explain you a little bit. I will explain you the basic of NTU. I will just talk about NTU. I had just taught you dy/y-y. We had written this as y-y. Put the limit from y2 to y1. y2 to y1 so this is higher value this is lower value this is positive value understand it very carefully Now when A = 1 is the case then what will be the driving force? This will be constant. This driving force will be constant because of constant because it is a parallel line. So when the driving force will be constant then E will come out. Now when E will come out then what will be
1:36:30 left inside? dy. And when I will do the integration of dy, y and when I will put a limit on y, then y1-y2. It means, understand very carefully. it means what happened NTU NTU OG how can I write better in case of H equal to 1 how can I write Y1 minus Y2 this overall change has happened divided by now see listen very carefully now what has come out here your Y minus Y equilibrium now whether you put it in the bottom or in the top whether you put it in the bottom or in the top you will get the same result but what will I do here I will write this in the top of the equation for easy
1:37:16 calculation. I will write this as y2-y2* I will write this. I will not read the bottom because I will never write it. I will never write this because I will never
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