 Let us quickly revise what has been taught yesterday. Yesterday I started with special topics in computational heat conduction. There I talked about multi-solid heat conduction and non-linear heat conduction. Those are special cases of single solid heat conduction cases with constant thermal conductivity and thus there are some minor modifications you need to do in the code to solve those problems. Thereafter I went on to pure computational advection and heat transfer and then we went into computational heat convection. Convection is a combination of advection and diffusion. Thereafter I discussed for each of these cases I had taken some example problem which has some exact solution and I had shown you that how to, what is the, I started with finite volume discretization then I had shown you implementation detail solution algorithm and finally using all those code has been developed which is which are given to you for the lab session and using those codes results have been generated for the problems which we are showing yesterday. And thereafter I discussed finite volume discretization in the afternoon lecture session and towards the end I had started the solution of Navistoke equation on a staggered grid. I can understand I started this topic which is very important topic and maybe one of the most complex topic in this workshop on staggered grid yesterday evening why could understand that many of you who are looking into it probably for the first time it needs some revision. What I had planned is that I will begin this lecture where I will show you in detail basically I want to want you to get certain key ideas of this. So the first key idea is how staggered grid generated and how are the grid point locations. So let us suppose this is a control volume for simplicity and clarity I am just showing you 3 by 3 control volume and here I had mentioned that right now I am showing you 9 grid points. So this 9 grid points are used for pressure only pressure and temperature scalars and I said that the velocity grid point u velocity grid point are located at vertical phase centers and v grid points are located at horizontal phase centers. Now when we apply law of conservation of mass x momentum y momentum we have to take control volume where the corresponding grid point should lie at the centroid. So let us say if you want to apply law of conservation of mass then the pressure grid point should lie at the center. So what I will do is that I will show you one by one. So we start with this pressure grid point and let us look how the so for this pressure grid point we need a control volume where pressure is lying at the centroid. Now when pressure is lying at the centroid what I am trying to show is this control volume. Now in this control volume u velocity is lying here v velocity is lying here. Now if you want to name them then the representative control volume we have not noticed sorry this is the grid point for pressure. I had mentioned that the convention we follow in the positive phases we take the u this will be up this will be vp and then this will be uw this will be u sorry v south. Now where this is used? This used for applying law of conservation of mass is that clear? This is used for applying law of conservation of mass. How do you calculate mass flux? For this control volume let me denote this mass flux which is at the east face of this control volume this will be rho into normal velocity is u. So you do not need any interpolation note that. Similarly m north will be equal to rho into vp. Similarly you can calculate at the other phase. So note that by this type of staggering by this type of staggering we are avoiding any interpolation to calculate the mass flux at the phases of the control volume. Now let us take the u now we want to apply law of conservation of momentum x momentum. Let me do one thing y momentum and let me write energy also. So for x momentum equation I will take let us suppose this control volume. So this if I want to take a control volume such that this lies at the centroid then I have to take a control volume like this. So if I take a control volume where u velocity is lying at the centroid up if this is lying at the centroid then the corners are this v is that clear? If this is if you are having a uniform grid generation and if you want a control volume where this green square is at the centroid then this will be at the corner you are able to see. So now this will be v. Now let us try to what will be this this will be v what will be subscript here. Let me draw the pressure grid points also this will be the pressure grid point what will be the pressure here p subscript p this will be p subscript e because once you write this subscript for p then it is because all the reference which you take you take reference with respect to pressure. So if this is pp this is up then this will be v subscript will be p. If this is p subscript e this will be v subscript e any question on this? Sir could you please tell what is the advantage of writing in terms of p e and so on because when we write a program we are going to write using running index sir. This question is that what is the advantage of writing this subscript as in terms of letter why not write in terms of running indices? You are right I could have alternately written this v as let us say v i plus 1 comma j finally because when we program it has to be done like that but traditional finite volume method for explanation as far as east neighbor, west neighbor, north neighbor which is just for understanding but when we do coding you note that all over what p means p means i comma j e capital E means i plus 1 capital N means capital N means j plus 1 capital S means j minus 1 capital N e north east means i plus 1 j plus 1. So for understanding we are writing neighbor with subscript east west north south in terms of neighbor that is what is commonly used in finite volume method for better understanding but note that in coding there is one to one correspondence between the subscript which are using here with the running indices. Now what will be the what will be the subscript for v here and here v s here and this will be v south east is that where the south east will be j minus 1 sorry j minus 1 i plus 1 south corresponds to j minus 1 east corresponds to i plus 1. Now let us take the this is for x momentum equation now let me do for similarly for now before I let me do this for y momentum equation y momentum I have to take from here. Now I take a control volume where this v p is lying at the centroid. So if you take this control volume now the pressure lies here let us first write pressure. So what will be the subscript of this pressure p the subscript of this pressure will be N if this is p this will be north neighbor. Now where you are sitting you are again here sitting at corners. Now once you know the p it is easy for you to write the if this is p p this will be u p if this is p n this will be u n and what will be this this will be u w because if this is p this will be the west neighbor and if this is north this will be north west is that clear. Now let us write the if you have energy equation in energy equation control volume used is pressure control volume and here again for pressure control volume the normal velocities are sitting at the face center. So here you have u p u w v p v s. Now I did all this just to give you a feel at step by step how you can understand you can take a small 3 by 3 control volume and from there you can draw the different control volume this is for applying law of conservation of mass continuity equation this is for applying law of conservation of x momentum law of conservation of y momentum law of conservation of energy. This was the first objective to give you a that how you can start from scratch and draw this grid points and note that I have written subscript with neighbors but finally you code you have to write in terms of running indices and there is one to one corresponds. Now the second thing which I want to highlight through this slide is mass flux calculation where mass flux comes mass flux comes in mass conservation anyway it will come in mass conservation but it also comes in all the transport equation what are transport equation x momentum y momentum energy equation where it comes this mass flux in the advection term. There is a the bulk motion of the fluid causes momentum transport and energy transport. So if you look into this you will realize that when you want to calculate mass flux in the law of conservation of mass and the law of conservation of energy like here m e is equals to rho u p m north phase center for let us say temperature this is rho v p. So here you do not have to do any interpolation but as soon as you go here come here if you want to calculate m e of for u control volume that what will how you will calculate this will be rho multiplied by do you have a u grid point here because mass flux here will be made up of u velocity for here for this case you had a u velocity sitting here for here you do not have u velocity sitting here. So if you do not have u velocity you have to do linear interpolation but what is the nearest neighbor u p n there will be one more neighbor that will be u e so this will be rho u p plus u e divided by 2. When you want to calculate mass flux on the north phase of u control volume now when you want to calculate v here you do not have a inverted triangle here but you have some neighbors what are the nearest neighbor these two neighbors okay. So this will be rho into nearest neighbor for v is this 2 so this will be v p plus v e divided by 2 is that clear similarly you can calculate for other cases if you go to rest here you want m e for v control volume how you will calculate this what are the nearest neighbor of u there is no u point here nearest neighbor at this this will be rho u n plus u p divided by 2 similarly if you want to calculate let us say here let me do it here because there is no space here let us say if I want to calculate m at south phase center of v then this will be rho I need v velocity here so I will use this and this neighbor so this will be v p plus v s divided by 2 so the message here is that for mass conservation and energy conservation when you want to calculate the mass flux you do not have to do any interpolation okay but in the momentum equations you have to do some interpolation to calculate the mass flux. So where you are avoiding interpolations all this we are doing to avoid some interpolations I had mentioned in the beginning we are avoiding interpolation in the calculation of mass flux in the mass conservation and energy conservation we are also avoiding averaging in calculation of pressure in case of momentum equation what will be the pressure force here how you will calculate pressure force in x momentum it will be p p minus p e into delta y in this case it will be why momentum it will be p p minus p n into delta x the pressure force you want to calculate for this control volume and this are the phase center and at this phase center the pressure grid point is sitting so you do not need any interpolation so you are avoiding interpolation for mass flux in mass conservation and energy conservation and you are also avoiding interpolation for pressure in momentum equation is that clear any question on this in case of momentum equations while calculating the momentum flux okay since we have staggered the grid yeah we do not need any advection schemes no our question is we do not need advection scheme for x momentum equations since we have staggered the grid that is not that is not true even if we have staggered but the we have to make sure that we take the proper control volume where the proper variable is sitting at the centroid so for let us say x momentum we will take this control volume and in this control volume we have to make sure that the x momentum conserved law of conservation of x momentum is obeyed if you have to make sure that you need to have momentum flux entering from this phase this phase and leaving from this phase and this phase okay so you have to do all those which had been discussed earlier but it is just that the control volumes are different staggered for different conservation equation so the point is that here you may say that we are not obeying the conservation equation in this all the conservation equation in the same control volume that is true but this does not affect that final result we obey law of conservation of the different conservation law in different staggered control volumes you can see not in the same control volume that is true whenever you are conserving mass you are using momentum satisfying velocity whenever you are you are getting this point here the velocities which you are will be using finally the solution which you get you know that I had mentioned yesterday that this law of conservation of mass finally it becomes a pressure correction equation and in that you solve it iteratively and when you solve it iteratively in the last iteration when you get the correct result the velocities which satisfy this continuity equation in the last iteration they are in fact already satisfying the momentum equation finally what you end up is after that pressure correction iteration you get a velocity momentum satisfying velocity field which obeys the mass conservation. Now you have some question. Sir advexing scheme yeah whether we have talked about different advexing schemes advexing schemes we have to implement here or on the you have to use advexing schemes advexing schemes yeah but checkerboard distribution when it is when problem is identified suggestion is one is go for stagger grid with same up in scheme or we can avoid that problem with higher order schemes. No you cannot avoid by higher order scheme all the advexing schemes which I had discussed all those are applicable in this case this pressure velocity decoupling cannot be avoided by the it is not a problem of advexing scheme this velocity pressure decoupling is not a issue of advexing scheme that is completely independent okay so all the advexing schemes you have to use here. Sir I did not get the answer. Yeah. You did not get the answer I mean advexing schemes because we have the value of the value of the velocity of the pressure still we are using peak or central depressor. For which you are saying x momentum or energy? x momentum and y momentum. In x momentum y momentum you do not have u velocity here to calculate x momentum you do not have v velocity here to calculate the y momentum okay because in this control volume note that in this control volume the x momentum which is coming in you have to take here x momentum leaving x momentum entering from here x momentum leaving from here in this case it will be y momentum on all the four phase center. Only central you are using now. Can you repeat? I mean here you are showing only with central this thing. Central different. But we can use other schemes also you mean to. Okay note that do not get confused this mass flux I am what the way I am writing it is not advexing scheme do not call it as an advexing scheme because this is a driver I had mentioned that we do for passenger there is a second velocity which will be coming advected variable okay. When I was discussing finite volume method also at that time I highlighted this point and I mentioned that the normal velocity we can calculate by this averaging do not call this as advexing scheme. Because for advexing scheme you use the mass flux direction and then calculate or and magnitude as well to calculate the advected variable. There are two velocities coming into the advexing any other question? Okay now let me show you the formulation because yesterday I showed you some philosophy and then I said that if I show you the philosophy and then come to then I will show you the equation philosophy of pressure correction method. So the idea is that let us suppose you have a I am showing you only one control volume but in actual problem right now I showed you let us say nine control volume. But I am here I am showing you only one control volume and let us suppose the previous time level pressure is one atmosphere okay. So using the previous time level I am showing you in this grid point only but the neighboring grid point will also come in your calculation. So I mentioned yesterday that we take the pressures of the previous time level and if you follow a fully explicit method then you can calculate the advection and diffusion using the previous time level velocity and predict a velocity. Now you know that if this is a control volume for mass conservation then the velocity which you have predicted comes like this. Let us see whether you are getting what I am saying this control volume is mass conservation control volume okay. So let us suppose you obtain up star so this is pp okay. This pp is right now what one atmosphere so you predicted up star and multiplied by rho and surface area and you got this. You predicted u w star multiplied by rho and delta y you got this. So that way using the previous time level value of velocities and pressure you predicted normal velocities and multiplied by density and surface area and got this. So these are the predicted mass fluxes. Is it clear? I will show you equation also. I can switch back forth and show you what is the equation okay. What is the equation for this? So what is the equation? Let me start with so there is an original proposition. What is original proposition? Rho delta vp up to the power n plus 1 minus up to the power n divided by delta t. This is the finite volume discretization of unsteady term. Then the total advection term let me write it like this. This is equals to the pressure force acting in the x direction. pp minus pe into delta v plus diffusion we are using an explicit method. Advection and diffusion we will take at time level n. Now if you want to obtain correct velocity for the new time level you have to take this as at new time level. Is that clear? Okay now so this is the original proposition. Let me denote this as equation number 1. Now there is a predictor step. In predictor step what we do we write the same equation but the only difference is this pressure at n plus 1 we do not know. So what we will do? We will write this as pp n minus pe n into delta y plus dn. This reference side will be same. Rho delta vp up n plus 1 minus up n divided by delta t plus apn. Is there any mistake here? Up n plus 1 should be star. This should be star up star okay. So from this what we will get? Up star as a function of this advection diffusion will involve u velocity, v velocity at n pressure at n okay. So you got this. Is that clear? Is that clear? Similarly you can obtain vp star. Similarly you can obtain uw star. Similarly you can obtain vs star and multiplied by density and surface area and obtain this as 1 kg or 2 kg whatever it is. Is that clear? Okay. So here I am talking philosophy. In pen and paper I am showing you equations for that. One to one correspondence. Is that clear? Okay. Now what happens is philosophically if 4 kg per second is going out and 2 kg is going in so more mass is going out. So what if you have knob to control the pressure of this control volume. What you would like to do? You would like to more mass is going out. You want to decrease the mass input so that it is balanced. So you would like to decrease the pressure. When you want to decrease the pressure then what we have an idea which we call as the pressure correction. So let us suppose you want to have new pressure. New pressure is equals to old pressure plus pressure correction. Old pressure is 1 atmosphere. So the pressure correction will be in this case let us suppose we reduce the pressure and the final pressure is 0.5. So the reduction which we did was minus 0.5. So the pressure correction is minus. Note the word minus. Pressure correction is minus 0.5 and the mass imbalance is plus. Note the word plus. Plus 2 kg per second. So what is the relationship between the pressure correction and mass imbalance? Negative relationship. Okay. I will show you an expression where there will be a negative relationship. Okay but this derivation is little lengthy. Pressure correction is a function of mass flux correction. Now let us do the original proposition prediction. Now the way we are saying is that to get the correct value we have a predictor step plus character step. So if you want to get an equation for character step you have to subtract from predictor from original. If you subtract this to what will happen? You will get rho delta vp upn plus 1 minus up star divided by delta t. What will happen to this apn? This will cancel down. And on the right hand side you will get ppn plus 1 minus ppn minus of pen plus 1 minus ppn into delta 1. Yeah sorry. And this is what we call as what is this? This we call as ep prime. This we will call as ee prime and in fact this we call as up prime because the way we define is how we define upn plus 1 is equals to up star plus up prime. ppn plus 1 is equals to ppn plus p prime. Is that clear? This plus this is equals to this. Greater path character is the final value. We are using some symbol as you know that we use for we get some velocity which will not obey the continuity equation. So we will use subscript star just to denote. It is just a symbol to denote. And star is the predicted value. So this is the correct value which we should get. Actually this strictly speaking this upn plus 1 is actually new value of up star. Yeah. What means the path for ppn plus 1? Yeah. What means the path for ppn plus 1? There we do not introduce. Yeah I can understand what you are trying to emphasize. But actually in the first iteration it is n. But in the subsequent iteration it becomes star. Yeah you can say that it is true. Because even this u star may be up. In the first iteration this is pp to the power n. But in subsequent iteration it becomes pp star. That is true. Okay. So is that clear? Now from here what you can get? From here you can get? Yeah. Okay let me I have not written the heading let me write down. What is this? This step is corrector step. Is that okay? Corrector step is this is 1. This is 2. Corrector step is 1 minus 2. Okay. Now here what we get is what we got is up prime as a function of p prime. This up prime if you multiply by density and surface area what it will become for a mass conservation I am talking. I am talking for mass conservation. If you in the mass conservation this up is sitting at the face of the pressure control volume. Correct? In case of pressure control volume where is this up? Up is here. Okay. So if you multiply this by rho so rho up prime will be can I write this as mp prime mass flux is the product of density and normal velocity. So this becomes this becomes function of p prime. So here I am showing an equation of mass flux correction as a function of pressure correction. Is that clear? The p prime you write that is the pp and pe. I had written actually it should be I am not writing the subscript here also I had not written the subscript neighbors will be involved but I do not want to write it as a big expression. So that way I had written only p prime but implicitly I mean that there will be I am not writing any subscript but there will be many subscript coming. I want it to be little short and general. Here also I am not writing subscript intentionally. There will be lot of subscript which will be coming for this velocities when you are predicting this u star also. Okay. So this is the equation to predict velocity from the velocity of the previous time. This is to calculate the mass flux correction as a function of pressure correction. We need one more equation. What is that equation? Pressure correction as a function of mass in panel. That from where you will get that? You will get that from continuity equation. So let us use continuity equation. This continuity equation will convert into pressure correction equation p prime equation. Okay. And how it gets converted? What is the discrete form of up n plus 1 minus uw n plus 1 into delta y plus rho vp n plus 1 minus vs n plus 1 into delta x should be equal to 0. Now what we do here is this if you look into this equation. Okay. In this equation I can write this up n plus 1. How I can write this up n plus 1? This up n plus 1 let me write it here is equals to up star plus delta t divided by rho p prime minus p p prime divided by delta x into delta y. Is this correct? Sorry not from here. This is coming from this. Sorry. This is coming from here. This delta vp is what? This is delta x into delta y. Is it okay? So this is pp prime. This is p e prime multiplied by delta y. But when you divide by volume you will get one delta x in the denominator. Up n plus 1 is equals to up star. This delta t divided by rho delta vp will become delta x delta y. So this delta y will be divided by delta x delta y and then you get this. And then you substitute this into this. So similarly you will expression for uw, similarly for vvvp and similarly for vs. Once you substitute it and then do algebraic manipulation finally you get an expression like this. Is this correct? So from here you got an equation un plus 1 as a function of up star and up prime from here. Similarly you can get for vp star, uw star, vs star okay. And then you substitute and then do algebraic manipulation you will get an equation like this. And then you study I had mentioned that this discretization is very much similar to study state heat conduction diffusion term. And this is very much similar. So this is like a del square p by del x square. This is del sorry del square p prime by del x square. This is okay let me do in there. So the equation is something like this. Del square p prime by del x square plus del square p prime by del y square. That square is equals to minus of summation of m star. This is what? This is the Poisson equation. And here the equation which you have got is p prime as a function of mass imbalance. So we have got three equations. What are those three equations? Mass flux prediction. From here you can predict mass flux. This is for mass flux correction as a function of pressure correction. And this is pressure correction as a function of mass imbalance. So now let us go to the slide. So I had already told you that this 2 kg, 1 kg per second is obtained from which equation? This equation. Now this pressure correction minus 0.5 where you get? 1 atmosphere to 0.5 the pressure correction is minus 0.5. That you get from where? Pressure correction as a function of mass imbalance. Mass imbalance I am not saying any equation because that is very easy to suggest an addition. This is obtained from this equation. So that minus 0.5 is obtained from this equation. I forgot to, you can see here there is a negative sign. So there is a negative relationship which I mentioned earlier. Mass imbalance is plus 2 kg. So the pressure correction is minus 0.5. Pressure correction is you have negative sign here. So that relationship is what I am discussing here philosophically. There also you can see it. Now how we calculate minus 0.5? I had shown you an equation to calculate as a function of this 4 minus 2 kg per second. Now with the new pressure value what we do is that we solve one equation. Which is this equation? So this 2 kg per second and 1 kg per second should change. There should be some positive or some negative thing. Some addition or subtraction it should occur. What will be that? That will be mass flux correction. Right now this 2 kg per second and 1 kg are mass flux prediction. Now we need some mass flux correction as a function of pressure correction. And what is the equation for this? This is the equation. Mass flux correction as a function of pressure correction. So what is the mass flux correction you will get here? For the north and east phase it will be for negative phases it is plus 2. And for north and south it is minus 1. And the equation from where it is obtained is this. Mass flux correction as a function of pressure correction. Once you get that what is the next thing? Again you will get new mass imbalance. And from the new mass imbalance there will be new pressure correction. What is the new pressure correction? Plus. Why it is plus? Because the mass imbalance is minus. Minus how much? 4 kg per second. 2 minus 6. Mass imbalance is minus 4 kg. Mass imbalance is 1 plus 1, 2 minus 3 plus 3. So this is minus 4 let me repeat again. So due to this pressure correction of minus 0.5 from this equation you will get from pressure correction you will get mass flux correction. That mass flux correction is minus 1 for the north and east phases and plus 2 for west and south phases correct using that equation. Is that clear? Now in this you can calculate the mass imbalance. And I had shown you what is the equation to obtain pressure correction as a function of mass imbalance. Using that equation the pressure correction as a function of mass imbalance comes out to be plus 1.5 atmosphere. So the pressure correction is plus mass flux mass imbalance is minus 4 kg per second. Is that clear? So I showed you full expression but this is I had shown you symbolically how things happen ok. So we predict mass flux then we calculate mass imbalance then we have equation for pressure correction as negative of mass imbalance from there you can know that minus 0.5. Then we have equation for mass flux correction as a function of pressure correction from where let us suppose you got 2 kg and minus 1 kg. And this so this is your the first picture is your first iteration this is second iteration then you will have third picture third iteration fourth iteration note that I am showing you this for one control volume. What you will have if you have million points it will happen for million control volumes. Any question on this? This problem you have taken unsteady problem correct. Suppose if it is a steady steady problem. Yeah. Still you will approach in a pseudo time pseudo transient problem or how do you. The question is that I am using as unsteady state approach and if the problem is steady state then this approach is called as a pseudo transient approach. There is another approach which I am not discussing here because this is more general that is a steady state approach. I think this question is then a steady state approach whether we go we solve the problem in this manner the answer is yes. Without pseudo transient also can we solve? Yeah. Without pseudo transient also this. That means in only one step we will get the answer. You know not one step. No one cycle of iteration. No no there will be two loops anyway in that case also. The only difference is that in case instead of time loop there you have iteration loop but that will be there anyway loops will be there also. If you know that the problem if you are if you are sure that the problem is a steady state problem and if you follow a steady state approach it takes less computational time. Yeah yeah. And ultimately that decides which one you should follow. Sir I think as I understand actually you are not doing corrections individually for each control volume we are simultaneously solving the correction for mass pressure first and then the mass simultaneously I think because the way the equations are. No it is not simultaneous we do one by one. Each control volume separately. Yes each control volume one by one. Okay but that involves say corrections for other cells also. Neighbor yeah neighboring it is coupled between the neighbors but not all the neighbors because we have a sparse matrix. Yeah but yeah. We are solving simultaneous equations every time so can means are we constructing the equation for all the volumes put together at a time and then make a correction or. Okay let me put this question in a slightly different way I think the way he is asking is I am solving let us say pressure correction Poisson equation. We solve it iteratively we start with an initial value of pressure correction and we solve it iteratively and it is a sparse matrix so we are using an iterative method of solving. Okay so the way you solve let us say a diffusion equation with a source term iteratively that is the way we are solving it. Corrections simultaneously for different cells is involved say different volumes are involved here. Yeah yeah. So obviously unless I know the correction for other volume I cannot find out here. Correct. Are we not doing the correction simultaneously for all the. Yeah simultaneously we are doing when I later on discuss the implementation detail that will come out more clearly we are doing simultaneously you are right because what is pointing is out that I do one iteration of pressure correction then I get an updated value of this okay then I solve it again so it is not that this is a constant value then I solve it it converges I update this after every iteration that is true. Sorry is it that in this case then for correction pressure correction also we need a boundary conditions. Yeah we will come to that we will need because as this is an elliptic double derivative in X and Y direction you will need pressure correction boundary condition as well however if you are using a staggered grid that can be avoided but for collocated grid you need this. Anyway I will discuss the pressure correction boundary condition towards the end of this lecture. So I hope now you got more clear idea of what you have from yesterday whatever you have understood. Now let us go to quickly I will just revise whatever I thought the equations so in semi explicit method this is an original proposition then this is the predictor step then this is the difference between the two which we call as the corrector step and then this becomes the velocity correction this becomes the pressure correction. Similarly for v velocity these are the equations and for continuity this is the way I had shown earlier that for continuity use the velocities of the new time levels and then use substitute u p n plus 1 v p n plus 1 as a function of u star and p prime u star v star and once you substitute you get this right now here I am showing you the algebraic form of the equation. So the point here to note which he was pointing out is that we saw we do one iteration of pressure correction and then after doing one iteration we get new value of this u p star because as I mentioned in my previous slide I emphasize that this u p n plus 1 you do not get in actually one iteration this is actually u p star new value as a function of old value. So once the mass conjunction occurs then this new value of u p star becomes u n plus 1 because this is an iterative method of solution this you do not get in one iteration many times we when we do formulation we fall off symbols because here I cannot write it as u p star but I can also not write it as u p n plus 1 because so I use one of the symbol and then I clarify that this is not u p n plus 1 this is actually u p star new value and this become n plus 1 only when it is the continuity equation in the last iteration. In the implicit method the difference is that we take the advection and diffusion at the new time level. Now you may be having question that why we are calling that method as a semi explicit because we have to take pressure implicit if we had take if we could have obtained correct solution just think what do you mean by a fully explicit means let us suppose you take pressure at old time level and let us assume that it gives you the correct u velocities and v velocities then and if so those two equations are sufficient to obtain the solution because those velocity needs to obey the mass conservation. So fully explicit is not possible in case of new stroke equation at least you have to take pressure implicit however if you want to do fully implicit also that is also not possible because this breaks the philosophy with which we are solving the set of algebraic equation note that we want to solve here algebraic equation where the coefficient is sparse. So the idea is that we want to solve algebraic equation where we should get an equation where the value at one point should be function of its sum of the neighbor not all other points all the neighbors. So like if you have yellow circle 25 interior grid points your equation should not be that the equation for one point is a function of 24 neighbors because if the coefficient matrix is dense then if you follow this iterative method it is very costly. Yes it is an explicit we use in semi implicit also you we do not use stability criteria in semi implicit method but in semi you know in any explicit method we have to use stability criteria. Sir but semi implicit also we use sir. Semi implicit we do not use stability criteria in explicit we use in implicit we do not use as we do in earlier cases. So in the solution methodology in semi implicit is also very much similar the difference is that this advection and diffusion at they are at n plus one time level ok this is for u velocity this is for v velocity to do the prediction what we do is that we take the pressure at time level n ok. Then in this case we get a set of algebraic equations like if there are 25 points you will get 25 equation and in each equation there will be not more than 5 unknown because you can have maximum 4 neighbors plus central points of 5 unknown. So this x momentum prediction y momentum prediction in the semi explicit method you got an equation where let us suppose you have 25 equation in each equation there was one unknown but here in this case this x momentum equation and y momentum equation prediction velocity prediction they are iterative in nature. So you have one loop for the convergence of this second loop for the convergence of this and the third loop for the convergence of where is in and over and above this three are inner loop over and above you are anyway you have a loop for study steady state convergence in a semi explicit method you only have one inner loop and one outer loop inner loop for applying law of conservation of mall that is pressure pressure correction loop and the outer loop is for study state convergence. So how you will get the equation for correction you just subtract these two this minus this will be what if you subtract the two you get an equation like this now if you convert this into algebraic form you will get an equation like this. Now this equation has a problem the problem is that the velocity correction is a function of the velocity correction of the neighboring point and this velocity correction in function of the neighbor point are function of pressure correction. So if you use this equation the problem with this equation is that the coefficient matrix which you get will be very dense as I said you will get if there are 25 grid points you will get equations where in each equation the value at a particular point will be function of 24 neighbors and in CFD we do not have 24 25 points we have millions of points. So we have millions of equations. So just compare let us suppose if you have millions of equations and each equation is a function of four neighbors versus each equation is a function of million neighbors just think about the computational cost. So you will see in all many CFD books that till this step this is fully implicit and between this step and this step we are doing something due to which it becomes semi-implicit. You see this statement in let us say book by Patankar that this equation to calculate the velocity correction of a particular point it involves velocity correction of the neighboring point and this velocity correction will have the pressure correction of further neighboring point this will make the equation coefficient matrix dense. Now we want to avoid that because it will be computational cost will increase a lot. So we drop this term drop this one is when you drop this term this will be equal to this this is this is that clear any question as you done in semi-explicit method here also what we can know when discretizing only for those terms we can give nth level previous level then it get cancelled the while subtracting getting pressure correction velocity correction equation that term will get automatically cancelled this term will get automatically cancelled your thing this term how it can get cancelled I am not actually from here to get this this is a very lot of step I am not showing you all those details step because this APn plus 1 minus AP star actually involves neighbor so the neighboring velocity corrections will come up so it is a big equation it will not never cancelled stop. This is a big expression I am just writing because this consist of small a that small a consist of that is small is the momentum flux which here it consist of mass flux the normal velocity surface areas and so on. So it will never cancelled. So out of this semi-implicit semi-explicit so which one is going to converge past how can we say that or is it problem specific. We will come back to little later let us first complete what are the different method then we will discuss the comparison between the different method so in the similarly for V velocities we can get so note that here what I am basically showing is the although it is a velocity but the velocity sits in the continuity equation I can also call this as mass flux correction for the east phase as a function of pressure correction. And this is I can call it as a mass flux correction of the north phase center of the pressure control volume as a function of pressure correction. Then from this actually this is Vp n plus 1 minus Vp star. You can write Vp n plus 1 is equals to Vp star plus this term and then you can substitute into this continuity equation. What is this? This is Vp n plus 1 is equals to Vp star plus this is Vp prime. And once you substitute this you get an equation which I have shown you earlier for semi-explicit also. This is the pressure correction equation obtained from discrete form of continuity equation.