Lecture 11

[Music] hello everyone welcome back so we have finished the discussion on the deriving the momentum equations for the general flow which may be compressible incompressible so we have found out that those are uh basically uh we did not consider anything about the variation intensity and we also saw that uh you naturally basically get a mark number out of it and you can see how the pressure is or the density change related to the velocity change okay now we are going to go to derive the equations for truly incompass flow so what do I mean by that so what I'm saying is uh uh Mark number is equal to U by a okay so I'm talking about true incompressible flow which does not exist okay all flows have got some or other compressibility okay gases are more compressible than liquids uh but even liquid has got some compossibility but for all practical purposes you can ignore them and you can derive equations for truly incompressible flow that means your mark number basically tends to zero and that means your speed of sound TS to Infinity right so because the speed of sound is infinity okay uh in this kind of flows uh if you are assuming an in Inc fluid okay Inc of flow what will happen is that if you have any change anywhere let's say this is your computational domain okay and you make some change here okay just it here it will be immediately failed everywhere okay because your speed of sound is infinity okay on the other hand if the flow is compressible you will have finite speed of sound and it will travel in a direction in a given Direction uh and then it will go I mean the information will reach with that the speed of sound right so uh because of the Assumption of the flow to be incompressible uh your uh basically discre your scheme to basically uh solve an gradient uh or basically your physical system the way you basically uh emulate it to the numerical system changes right so before we go into this incompossible uh flow cfd let's try to see what are the governing equations for the incom flows okay so we already have the derive the comp equation okay so what you can do you can uh just uh use the assumptions that we have for the income s flow and then you can basically find out the Govern equations okay all the Govern equations so one thing is D do U equal to Z that we already derived using our mation law okay so we had d by DT plus d by D XJ low u z equal to Z and what we did D by DT Plus Row D UJ by dxj plus UJ D row by dxj equal to zero uh take these two together that become that gives you d by DT and what is this this is d. U Row D do U is equal to Z right and we said that for in a truly incal flow your D by DT goes to zero why because your change in density uh is zero okay the acceleration there cannot be any uh change in density following a streamline okay the total change in density has to be goes to zero and that gives us the relation which is Del do U equal to 0 okay and uh what is the I mean consequence of getting from this to this that we lose one dynamical equation right what is this d.u now it's just a ktic condition okay this is your dynamical equation okay you can solve for row here okay how the row is changing in time you can get that and because we assume that row to be constant uh we ultimately get this condition we say that the Divergence of velocity has to be zero at any point in your in a income cell flow right so this this we already got now the second thing is to get the momentum equation okay so let's get the momentum equation uh so momentum equation we'll use this condition and get this incompass momentum equation right so the momentum equation is d r UI by DT by Dy d x j r UI U uh will be minus d p by d x i plus d by d x a okay Sigma i j Sigma i j and plus r b okay now uh here this is General momentum that we have derived okay and now what we're driving is making a approximation that flow is truly incompressible so what happens to this equation and this was basically to Mu e i z and uh minus 2 mu by 3 uh that was basically Lambda okay and Delta i e KK right so this is the expression that we have been using okay so uh here I'm seeing LDA = to- 2 by 3 using the hypothesis so uh now make assumptions where we say that the flow is truly incompressible so the row can come out it does not matter whether row is inside or outside you can take it out so you can write it row du DT plus Row D by DX Z okay UI u z okay will be equal to minus DP by DX I okay and then you have this term d by DX x j Sigma i j uh plus r i okay now since your d. U equal to Z okay this term identically goes to zero right so try to get this term d by D XJ Sigma i j your stress term for an incompressible flow okay so that will be D by dxj to Mu J so that can be expanded here I'll write in expanded form okay mu or you can take the MU also out because see in the when we derive the compressible FL equation uh the MU is inside here because mu can also change in space for the compressible flow right uh because mu is a function of temperature okay in the composable flow the temperature changes and because of that your viscosity may change so you might have different viscosity at different locations okay in the case of incompressible flow uh your uh temperature is if the temperature is not changing you can have incompressible flow with the buenus approximation which will probably talk later where you might uh your density might be changing because of let's say CH you are adding some energy you adding some heat right so in that case you may have to consider the variation in viscosity uh with the temperature but you need not use the compressible equations for that uh those kind of flows right you may use the compressible but since the change in uh density is due to the temperature and not due to the pressure so those are not truly compressible flows right so you can basically add some uh sour in the in your momentum equations and uh you can have temperature as a passive scale scer and you can solve that problem right um so if the MU is not changing with temperature you can take the MU also out here okay because temperature is not bearing right so then we'll have two 2 e is what DUI by dxj plus d j by DX I okay and then you have - 2x3 and you have Delta i d. U right so this goes to zero anyways right for the incompass flow now we have this these terms remaining so mu d by DX Z and you use both the DUI by DX Z and plus d by DX z d u by d x i right so then this can be written as first part will be mu Del to UI by delx a² right and then second term you can basically interchange it you can take this D by DX a first and second will become D UJ by dxj okay what is this J is a D Index here okay so this is nothing but d by by dxi of d. U right so this will also go to zero and what is this part Del to u y by DX a² so this is nothing but because uh it will repeat in J so this will become this is basically nothing but Del to u y by D x² plus Del to u y by D y² or X1 squ X2 squ if you are comfortable with that d square you can it can be written as Mu d square UI right so the entire stress term that we had so complicated relation we basically get it to you can write this entire thing as new d s UI right so this is the final expression that you get okay uh if you want if you have the body Force you can use this body Force here this can also be expressed as D UI by DT plus d by d x z uy u will be minus d p by row okay x i okay plus uh so here it will mu okay this will be mu by row that becomes new d s UI plus row b right so this is p by row okay pressure per unit density okay so this is the quantity this is the definition of pressure in many of the solvers okay for example if you're using a an open form solver so the unit of p is not the Pascal okay uh but you will have this unit which is given by P by row okay that is that comes because of this definition and the other reason for treating not P but P by row is that anyway p is not a thermodynamic pressure in the incomp flow so that should be very clear to you that we cannot use P equal to row relation in the the incompressible flow okay it is not valid for incompressible flows truly incompressible flows okay because your uh your density is constant and temperature is constant you will not get the the pressure there the other uh physical aspect is that uh the pressure that you experience in the uh in an incomp flow is a mechanical pressure it's a purely mechanical pressure okay it is it is not the theramic pressure that you basically get in the compressible flow right so in fact that pressure is indeterminate in the Inc flow you cannot determine the absolute value of pressure in an incom flow what you can rather get is the pressure gradient so in an incomp flow if the flow is happening it does not matter what pressure you put here and what pressure you put here okay what matters is your what is your DP by DX okay or the pressure gradient so the pressure gradient uh is what determines how the flow is Will evolve and uh you can for the calculation purpose you can have some values of P1 and P2 right uh but those are not the absolute Valu so you should not be trying to relate it with the uh with a thermod pressure okay so now uh we got two relations here okay so the two relations are d. equal to Z uh and the other is your expression uh the momentum equation which is given here okay now what are the variables that you have that you want to solve for r a constant so you don't need that u b w and P in a three dimensional flow right but uh if you look at the Expressions okay so this is your first expression which uh which is coming from the uh the mass conservation and as we have been discussing this is a chematic condition this is not a dynamical equation okay but you have three dial equations for velocity okay for U V and W you can expand it and write in X momentum y momentum and Z momentum you can get the uh dynamical equation for the velocities right but you don't have the dynamical expression for the pressure so how do we determine minutes so that is a question right so the N equation do not have independent equation for pressure okay for compressible p third pressure as I told you and the mass conservation equation is the dynamical equation for density okay for incomp flow the p is a mechanical pressure okay so this is just a ktic constant or ktic condition on the velocity field we do not have dynamical equation so how do you pressure for Inc flows so what we'll try to do that if you do not have the equation for pressure we'll try to get an expression for pressure by using the relation which can use your city constant and your momentum equation so how do we do that okay so write your expression okay to minus x i by row plus new Del Square UI right if you write this in Vector form so sometimes you know uh I mean I can also work out work it out in a tensor form but sometimes it is easy to use the vector form because you can easily figure out for example if I say. U equal to Z okay you can easily uh figure that out right so what I'm going to do I'm goinging to write it is Vector form which is d. uu is equal to minus grad e by row plus new d s u okay so this is my uh equation okay and what is our goal we want to find out an equation for pressure okay this is what we want to get why because these are three equations for velocities in the momentum equation and there is another expression which is uh the ktic constant on velocity itself right so how do we get the pressure right what you could do you take a Divergence of a okay take the dience of the entire equation a so basically what you do you do D dot this entire thing right so what will happen you can write this D do D do uu will be equal to minus D do d p by row plus new d s okay D do d s u right uh okay let me write the first term also okay D dot d by DT right now we know D do equal to Z and these are the linear operator so basically you can uh interchange the position so you can write it Del by DT D do U plus d do D do uu will be minus what is this this is Del Square P by row will be you can here also you can write Del s d you can interchange Del squ operator and the D operator because Del squ is a linear operator and the diers operator you can so this will go to zero and this will also go to zero right so what you'll get is minus d s p is equal to Row D do D uu or you can take the row on the left also it does not matter okay a constant okay but let's because we have been following that [Music] okay okay so what we got from this so This is called the pressure poison equation and why we had to get this equation is because we knew the uh the kmetic instr in velocity but we do not have any expression for the pressure and we need to basically uh get the pressure so we use this condition okay this uh called as X so we used X in a and got this new equation so now this equation the momentum equation as well as the pr po equation together can give you the solution for UV W and P so why it is called pressure poison equation is because this what is the form of this equation this is basically d s f = 5 right that is the form which is the equation of the the I mean the class of Po equation right and please please note that you need a special technique to solve this kind of equations because uh when you try to solve this kind of problem these are the class of electric problems and uh uh basically you will get uh basically equations uh which are in implicit form and you need a inversion Matrix that becomes very very uh expensive so you want to use some iterative algorithms so uh in that case you have some librar which are developed for solving this kind of poison equation or La equation right but this is not uh easy to solve compared to what we are we have been using for the the convection equation okay so we'll come to this kind of problems also this is an elliptic equation okay when we come to the classification of equation uh we'll get back to this that why it is challenging to solve or computationally expensive to solve these kind of eliptic equations okay so we got this pressure poison equation okay which is given here okay and the same thing and then now that you have the pressure poon equation okay uh you can basically make of the pressure pois equation to determine the correct values of velocity as well as the pressure so the solution is done typically in two ways okay so the first step is the solve the moment equation the momentum you already know uh to obtain the the velocity field for a known pressure so you can take a pressure okay uh guess a pressure value and solve the momentum equation and get a velocity field right using the momentum equation okay but it will not give you equal to zero okay in this the first step right you need to correct it okay so to correct it what you'll do you'll use the velocity okay so this is a PR POS equation for example in 2D okay here U is basically d by DX okay and similarly u y is d by d y and so on so forth right so this is the equation PR equation for the 2D flow and so once you know the velocity uh values from a uh pressure which is your initial pressure you calculate recalculate the pressure using your pressure poision equation and get a new pressure right but when you do this it will make sure that Divergence goes to zero right because while deriving this we have ensured that del. U goes to zero right so uh this is the twostep press and now you have an updated pressor use this new pressor to calculate the velocity again and then uh use the pressure pois again to ensure that diver is zero and this process gets repeated okay so this is how you solve your income flow we'll not go into the detail of solving the incompressible flow equations uh in this course we'll be mostly using the the compressible uh flow solver okay uh because we want to go and uh deal with the problems which involve socks and strong discontinuities okay uh the incomp flow cannot give you that kind of solution right an incomp flow solver uh so final thing uh before you conclude the class today is a brief discussion between the conservation and non-conservation form right so uh if you recall we had used this uh del row U by DT plus d by d x a to u u so this was your conservative conservation form okay which was minus DP by dxi plus d by d x a sigma I okay and plus RI this is the conservation form and you can uh take this uh row out and use the equation of continuity to get this non-conservative form okay okay something like thisal to- d p by DX i x a okay this is the non conservation form okay okay and um uh this can also be written as duy by DT okay the L expression okay and we had discussed that there's no difference between this form and this form okay uh this has been obtained using uh directly from your conservation law and this has been obtained by making further consideration that your D do goes to zero and then we obtain this non-conservation form and we argued that this is basically Mass into acceleration which gives you the force right uh so if the flow is continuous and flow is smooth you can use either of these forms okay to solve the problem but if the flow has got socks okay or strong discontinuities then you will have trouble solving using the uh if you use the non-conservation form and why is that so what happens across a sock for example if this is a sock here okay so will have basically you can have the change in density okay so your density increases so row2 is much much greater than Row one uh your velocity decreases okay uh U1 is much much less than U2 and so on and so forth right so your the Primitive variables go strong change across these strong discontinuities like socks and all right but your flux okay so which is row U okay so this uh you can use smoothly despite your row decreasing sorry row increasing and U decreasing across the sock you can have uh this conserved variables which is rowu which can have a smooth transition across the so right so that's why uh using this conservative form where you are treating and you together is useful uh if you have flow have socks and that's why this form is basically used in compressible cfd okay the other advantage of using uh this form is that let's try to write uh basically your expression uh using uh this form and so so if you have door DX or it can be DT also so the left side I'm writing the conservation form and right side I'm using the non conation form okay right so this is both are uh equal I've just used the chain rule okay but let's try to do uh solve this problem on the left side and right side separately right so if you're using let's say this is your grid okay I I + 1 I -1 i - 2 I + 2 and so on so forth right so if you use Simple defining scheme whatever you have studyed in your basic math course let's say use the central differing scheme so you can write row u i I + 1 row u i -1 by 2 Delta X right and here if you solve so here have row okay UI + 1 - UI - 1 by 2 deltax + U okay and row I + 1 minus row i - 1 by 2 Delta X okay so the question is what is the value of row you should be using and what is the value you should be using whether it is I + 1 or IUS one or I what is the value that you want to use here you can use here okay similarly for you what is the value that is suitable for this right so you will see uh that we will quickly run into trouble when you use this uh non-conservation form and the left side is not equal to the right side in the disc form okay in here it is fine in the continuous form it is fine when the disc I is form this left is not equal to right uh and you have I mean when use I + 1 or IUS one I you may get different values the the last advantage that I want to discuss is that this if you have like boundary to boundary okay this is let's say I = to 1 to I = to n right so if you express it here so you can have basically row U uh 2 minus ru you zero I mean you can have a ghost cell here okay this is z and this is n+1 okay outside the boundary okay if you have a periodic boundary then you can n = to 0 and n + 1 = to 1 right so that will be 2 Delta X plus you will have row u3 minus row U1 by basically 2 Delta X and so on so forth right so you can basically uh then you'll have row u4 minus R U2 by 2 Delta X and so for so on and so forth right so if you add all of this okay so here for example this row U2 goes to zero right similarly in the next term we'll have row u 1 going to zero and row3 going to zero and so on and so forth right so all all of these terms will basically collapse and what you will have is basically in this case row1 + 1 minus r u 0 by uh I mean 2 Delta right so whatever flux you have getting in and getting out you can calculate that this can be better understood if you let's say use the uh the oneid difference okay and not the central difference okay so if you do that let let me work that out so now what we're doing we're using the one-sided difference okay yeah xal to DX plus u DX okay this is basically row u i + 1 minus r u i by Delta X right so that will be basically R U2 minus r U1 by Delta X Plus row u3 minus r U2 by Delta X okay plus row U 4 minus r u3 by Delta X and it will keep on going till you have row u n minus r u n minus1 by Delta X right so you can see that here your row U2 is getting cancelled okay row u3 is getting cancelled and similarly row u4 and everything will get cancelled okay here r u n minus one will get canceled and what you'll get is basically row u n minus row U1 by Delta X right so if this is your grid starting from 1 to n right so you can see that all these values inside have vanished right so what you have so this is your inflow so this is influx and this is your outflux right so out flux is row u n and this is row U1 right so only the difference between the two is what you are getting here right which is expected also your net change is basically whatever is you getting in and difference between the what is getting out and what is getting in right and this is because your of your discon so this is called the telescopic collapse of flux okay so this will happen only when you use the conservation form okay if you use the non-conservation form so if you use this expression so let's say I'm using I here row i d i + UI + 1 - u y by Delta X okay plus uh uy row I + 1 minus row I by Delta X okay in the next case it will be row I + 1 u i + 2 - u i + 1 by Delta X Plus UI + 1 row I + 1 I + 2 row i - 1 + 1 by Delta X and so on and so forth right it will go till the end so here because uh I mean here I've used the point this is I this is I + 1 and this is IUS one so while calc derivative I use the value I here for row and U okay when you calculating the derivative at that point but you can use i+ one also as you have discussed earlier right so that was one issue that what value you should select for these different variables the other issue is that when you sum this up this will not lead to the telescopic collaps of flux okay so you'll have some differences okay between the value that you're getting here and when you're calculating the value here right so that is another disadvantage of using the non- conjuration form so uh with that we'll finish the class today and uh we'll be using the conservation form for our cfd calculation because we dealing with the compressible flow cfd and we'll write all these expression in The Matrix form but before that there is another conservation that is important that we need to deal with that's the energy conservation okay so in the next class we'll finish the energy conservation okay then we'll cast all these equations in a nice way uh in a matrix form and then we start solving uh those problems okay flow problems thank you very much [Music] [Music] [Music]