Lecture 10

[Music] [Music] uh this is lecture number 10 hello everyone welcome back and we'll continue the discussion that we had uh that we're having uh regarding the governing equations uh for fluid flows so in the last class we basically uh derived the momentum equations okay so what we did is basically you had conservation form so this is d r UI by DT plus d by d x j row UI UJ is equal to d i j dxj okay and plus row and bi right and we said that this is our time rate of change this is the temporal term uh this is the convection term okay uh this is the stress term which is related to the our viscous diffusion okay and finally we had this the body Force term okay and then we also related this uh t y j to the strin rate right and uh we basically part we found out is tal to 2 mu e and plus Lambda and uh that will be Delta i d u k by DX K right where EIG is the standard tensor symmetric part of the standard tensor in nothing but half DUI by dxj plus d j by DX I right so 2 i e will be DUI by DX Z plus d UJ by DX I right and we said that mu is the first coefficient of viscosity or the molecular viscosity as we know it and uh Lambda is uh is your bu viscosity right is the second coefficient of viscosity and if you write uh we also found that this Lambda + 2 mu by 3 okay this is called The Bu viscosity let's represent it using beta this is the coefficient of bulk viscosity bul viscosity okay so then uh we said that for normal uh flow conditions if the uh the flow rate is basically the speed is not very high it's not in Hypersonic region or or if the gas is let's say monatomic in that case uh the beta almost goes to zero and the Lambda becomes Min - 2 mu by 3 okay and this was called The Stokes hypothesis okay so this is exactly true for monoatomic gas because they have only the translation degree of freedom but uh when you trade the diatomic and triatomic gases then it becomes an hypothesis in that case also this Lambda + 23 or basically beta is nearly zero okay if beta is not zero then what happens that in most cases the Delta do u gr. u is equal to zero so the multiplication of this uh Lambda + 2 mu by 3 d. U goes to zero right which is the term that we have here okay Lambda multiplying by d k by DX which is d. U right so using this hypothesis you get this Lambda = to minus 2 mu by 3 right and if the flow is not Hypersonic you can use this hypothesis okay and for most flow condition it works okay but uh if the ch change is uh in the thermodynamics condition is very rapid and the system is not able to adjust to that thermodynamic change then you have to basically uh look for the variation in the bulk viscosity also and that usually happens at in very high speed flows like in Hypersonic flows or even high Mark number flows okay High supersonic flows right so if you use all those conditions okay there was another term here which was the P Delta I right uh so this is minus P Delta okay so this we had uh termed as the hydrostatic stress and these two are basically the detic state stress right so if you combine all of these then you'll get this final expression row UI U is equal to d p by dxi okay and plus you can uh write Sigma I by d x a and plus row b i where Sigma I is the detic part okay 2 mu e i z plus Lambda Delta i j d k by x k right so this is your final momentum equation you can express it in uh uh different forms okay and this is the differential form of course you can get the integral form also from the r transport theorem as we have got the uh the mass conservation equation right now let's uh uh try to write it in uh form that is called the non-conservation form okay so what do we do let's this is my a okay so let's take the left hand term this is row UI by DT plus del row UI J UI u j by Delux okay and let's try to uh basically use the chain rule okay so this will become Row D UI by DT plus UI d r by DT right this is the first term and the second term if I take the row UI together and UJ together okay so d by D XA row UI UJ okay and plus uh you will have row UI D UJ by dxj right this correct so what I'm trying to do I'm trying to use this uh Mass conservation relation to basically put it in a form where the row is outside the okay so uh if you collect the term so row UI and UJ D UI by dxj okay this is the first term this is the fourth term and plus UI Del by DT and plus Del by DX r u right we have this is the second and third right now this term identically goes to zero because of the mass conservation okay so then this will give you this Row D UI by DT plus u d UI byx Z right if you want to write in the vector form this will be Row D UI by DT or in more appropriately U by DT okay plus u do d u okay so remember this uh this is very very important to put a bracket there okay because this is not equal to U dodu okay so here uh we are putting the the Divergence of U with the D operator okay okay so that you have the expression u d by DX and then v d by d y and plus w d d z okay and depending on the direction of U you will have U or V or W here okay this person but this is different here you first taking the gradient of a vector right if you do d u here so then it's a it's basically a tensor and you're taking a DOT product of a vector U with that tensor so this is uh not the form that we have in the ne equations right so now let's put it back into a which is on the left side okay this is a so if you replace the left hand side with the expression that we had okay so then it will give you row by DT plus u d UI by dxj is equal to right hand side remains unaltered so just write let me YX plus row b or R GI okay this is the body Force so this is your the final form so this is called the non- conservation form okay and uh please understand that there's no difference if I call this B so there is no difference between a and b here okay the a is here and B is here so there's no difference between a and b okay both are the same thing it's only the expressing it in a different way but uh the a is called the conservation form and B is called a non-conservation form and the this is purely a cfd terminology it has called nothing to do with the fluid mechanics okay and uh when it is uh conjuration form so what why it is called conservation form is because you can use the row U variable inside your dydt okay or even the uh your flux okay uh the convection term okay so this row U becomes a conjured variable okay and we'll have some discussion about conj versus primitive variable but the advantage of using this Ru especially in compressive flow is that uh you you don't have to basically uh uh separate this row and U and that is very very useful when you uh do any kind of average ing or even uh the cfd calculation right so your Ru always comes with U or V or W or even the energy okay so um these terms are called the coner variables wherever if you use u v w and uh e okay then these are called the Primitive variables okay so uh in compressible solver coner variables are very very useful because you have the variation of density also along with other flow variation of other flow quantities right so you want to arest the uh the the variation in density and variation in velocity into the single uh variable which is the rowu variable okay so uh that is the reason it is called the conservation form because you have this row U entire row U uh in your uh Del by term okay and non-conservation form also preserves the same I mean property if you are solving it uh but it's only the uh your algorithm may become different if you want to solve in the learn conjuration form further we'll discuss about the Matrix form and you will see that uh when you write in The Matrix form basically when you want to solve a couple system of equations your mass conservation momentum conservation energy conservation all at the same time in that particular case uh your conservation form basically helps you in a great deal okay uh in solving all five together okay the another thing that you might want to check here is that if you really look this equation closer okay so what is this so this is mass what is this this is an acceleration okay and what is this one this is force right so this is basically uh your second law basically uh the mass into acceleration all everything is in per volume okay as we have been uh discussing that uh the volume changes continuously so that's why we do it per volume in the fluid flows so it gives you the uh the second law which is basically Mass into acceleration equal to net force right so that's that you can easily see when you cast it in the non- conjuration form right and once again I want to emphasize that when it is non- conservation it does not mean that uh uh it is uh not conserving some variable or some value in this case momentum but it's only to suggest that uh when you are casting in in a coupled form or couple system equations that uh this form is not suitable okay and this is equal to uh you can further cost this the Del UI by DT plus u DX is the substantial derivative or the total derivative and and that will give you Mass into acceleration I mean you can write this thing into the acceleration this is the Lan form okay so now let's go to the uh the isers equation okay in uh cfd many time uh you would be basically solving an ideal flow where you can put the viscosity to be zero right and uh we are able to solve problems without basically having any molecular viscosity because you have in most of the cfd solutions you have something called numerical viscosity and that's why without even putting uh the noly bond condition or ensuring that uh uh the flow velocity identically goes to zero in the wall you're still able to solve it right so what equations do you solve on this in these invit flows so basically what we do we ignore the viscus TS okay so if you ignore the visus TS okay remember in the true situation there is not no fluid which is inid okay a brief uh is discuss but you get uh some insights about the mechanism uh of basically balance between different forces when you solve it in the invit way another thing is that uh developing a cfd solver for is equation which is the also called the ined equation okay is easy uh the reason for that is the viscus force uh requires very cumbersome part of coding because you have uh so many visus terms available particular in the compressible part you have this DUI by DX Z plus d UJ by DX I andus 2x3 uh basically uh K by dxk so you have to calculate all the gradient terms everywhere okay and then uh put in a certain form and then you have to basically take the gradient of this entire quantity right and then it becomes little bit cumbersome in solving in terms of computational cost also it is more expensive right because uh you have to determine the VIS custom every time uh so in those cases you can use the is equation which is nothing but your uh momentum equation without the VIS custom okay this is basically Del by DX a row UI U uh is equal to uh you if you have some body Force you can use the body Force okay or it is just d p by dxi or use the body Force here right so this is your equation so the entire term that we had Delta y z by d x z goes to zero because this stress is due to the presence of the viscosity okay presence of the molecular viscosity so when you start writing a code many time you start writing uh first write the iser code okay then you add the viscus part and then you can include the effect of the Runar number variation or the viscosity variation right uh so this makes life very simple but remember if you're are simulating a truly uh I mean simulating a true flow there's no flow which is ined so no matter how large your anos number is okay you will still have boundary layer okay and uh so if you are solving to on a flat plate okay so you will have to have a bond layer okay and that you'll get when you solve the N Stoke equations okay if you solve using the iser equation okay because what you're doing you're ignoring the viscosity so then you will not have the one layer and whatever your free stream will come here here the free stream will start I mean maybe start above the bond layer right so this is from the oiler equation this is from the sorry this is from the this is from theer equation and here uh in the case of Oiler equation you do not say that U equal to Z at the wall right here U will be zero in the N because this discuss viscous forces included and the viscous forces will ensure that uh the molecules stick to the the wall right in in this case U does not go to zero in theer equation the V will still go to zero okay so V is in this direction the vertical velocity that means the no penetration will be insured but no sleep will not be insured okay so what we'll have is the sleep bation in the flow so this is one mistake many uh students do they solve the uh equation but then uh they say that they have used the No Sleep bonding condition which is not correct okay if you're using the equation it will be a sleep bonding condition okay so now that we have theer equation uh we can further simplify it let's write it for 1D flow okay and uh uh let's say that there is no I mean flow study okay then Miss D by D ter goes to zero and then body forces are negligable okay I mean this is reasonable assumption in the fluid flow I mean the mostly your body force is the gravity force you might go to MSD and all where the magnetic force intive forces are also important but generally for the flow I mean uh uh which is uh flowing in homogeneous X Direction okay uh you can ignore the body forces here okay so this is row b equal to Z so then it will become d by DX remember this is 1D flow so UI will be u j this is row u² is equal to minus d p by DX right all other term goes to zero right so here you can see that your uh uh uh change in density okay so because this is uh 1D flow so the D by DX can be written d by DX and because this is a study also okay okay minus DP by DX so you can Rel you can see that uh uh you can relate the change in pressure to the uh your change in velocity right so this can further be simplified you can write it row U duu is equal to minus DP right why we doing this we are trying to see uh where the the compressibility uh becomes important okay um so uh remember we had discussed that if the change is pressure is high uh then you can use the compressibility relation to identify if your density change is also important right and then you can call this flow is compressible right so you can actually further cast it for an isentropic flow because that is also an reasonable assumption as we have discussed earlier that uh if the flow does not have boundary layer then you can assume it to be entropic because you're not adding any heat and there is no dissipation wh dissipation and all so in that case you'll have for the atropic flow you have we have derived the the equation uh for the speed of sound okay expression for the speed of sound a² equal to DP by DX okay for the isentropic flow okay DP by D okay so the DP uh in this let's call this expression X you put in X then become row u d u is equal to Min - a² d r right so now what I'll do I'll multiply with youu and divide with u this is u² and I'll take a squ on this a square du by U take minus on this side will be D by right so what is this U by a we have discussed it earlier U by is equal to Mark number so minus m² du by U is equal to D by row right see what we got here is the uh the relation between your change in density per density with the change velocity per un velocity right and you can see that it depends on your mark number not only depends but it strongly depends on the mark number because this is the the proportionality constant is basically M squ right so if m is less than 1 okay which is the case of subsonic flow so in that case your um change in densities this is d by D row or the change in density is very very small okay okay you can just call it changeing density small with change in velocity compar to the change in velocity right when Mark is equal to 1 okay which is the Sonic flow your change in density is basically equal to the change in velocity right if you change the velocity by certain amount the density will also proportionally change by the same amount right is small here in this is equal to change in velocity and when m is m is greater than one then it is supersonic flow so in that case your change in density is much greater than the change in velocity so that is why you can now understand that when you um go to basically high and High Mark number your density change becomes very important and uh you cannot ignore the compressibility okay in fact I mean the general thumb rule is when the mark number is greater than3 okay the change in density which is d by row uh is uh basically more than 5% okay the the the change inity is more than 5% and because of that you cannot ignore the uh compressibility and you have to basically uh use the compressible N equations okay now let's uh just uh uh finish uh with one final thought okay which is uh basically the incom flow and we'll start uh more discuss on incom flow in the next class but uh just to uh give you an idea that in the case of income flow your your d. U is basically zero okay that means basically D UK K by dxk equal to Z and that actually simplifies a lot of things okay in the case of income flow your viscous uh stress term becomes very very simple okay but then you also lose something in the sense that you can no longer use P equal to R relation okay why because your density is constant your temperature is constant so you will get a constant value right so your pressure is not the thermodynamic pressure but instead it is the mechanical pressure that we derived in the last class okay so uh uh we'll look at the incompossible flu equations in the next class and then we'll uh continue from there okay thank you very much [Music] [Music] [Music]