Lecture 09

[Music] hello everyone welcome back in the last class we derived the mass conservation equation using the renos transport theorem and uh we said that uh depending on what extensive quantity that you are going to use you can uh basically uh derive the mass conservation momentum conservation or the energy conservation equation so today we are going to derive the momentum conservation equation okay so if you use uh b equal to basically MV okay you can get b will be p and you can derive you can get this expression which will be uh similar to the mass conservation but you will have more forces or more stress terms in your momentum conservation right so that will give you row UI by DT why I here okay because momentum is a vector quantities you can derive it for individual velocities okay individual directions i j and k right mass and energy are scalar quantities okay you have just one equation right so this will be D by DX z r UI u z and then uh this is basically your uh this equally rate of change okay this one and this is your flux or the the rection okay how they are basically being transported through the surfaces okay and that should be equal to the forces okay there be some forces acting on the body okay that's the Newton's second law that this entire thing we'll see it shortly this represent basically the rate of change of momentum and that will equal to the forces okay so what are the forces that we have in fluid okay so we are writing in term which is basically the due to the stress and plus due to the body Force okay so I'm using G which is I mean which is typically use in the simulation which is the Gravity Force okay but it can be magnetic force or it can be any electric force or any force that you want right so if you are let's say doing the magnetic damic solution then it will have the magnetic force right now the most uh critical quantity here is the T J right if you want to write this expression uh in the vector form so that will be Del R U by DT plus d do r u u okay and which will be equal to to okay and plus row G okay here okay so this is how you write in the vector form and you many of you may be familiar with the vector form so that's why I wrote it for you but this uh does not I mean this is very trivial that you have uh rate of change is equal to the forces rate of change amount equal to forces but then what are the forces that we have in the fluid particle and that is where the derivation uh becomes very very important and this was basically given by navier and Stoke separately okay and that's why it's called devier nav Stoke equations so before that also people Tred to derive this expression but they did not consider the viscous forces right so uh that's why they are they are not the complete conservation law you can have you might have heard about the ier equation where you don't have the viscus term and uh that cannot be applied to any fluid or any realistic flow where you have got the viscosity okay now if you look at the variables that you have here okay as of now you have uh uh U V W okay but uh also row if it is compressible okay but also you have this tow term okay what is the tow this is a stress term they stress that the fluid is experiencing right so we need to find out the stress okay now the stress uh if you see in a fluid element okay if you have a three dimensional fluid element okay say this is your three dimensional fluid element okay this is your I okay this is your J and let's say this is your K okay so in that fluid element you can have forces which is let's say I this T is basically t i j it has got two uh indices okay so t i is I is basically in the direction uh uh I mean where the element is present and J is the direction in which the force is acting right so if you take let's say section here at a given I okay and then you apply the force in the same direction right then you'll get this so if I equal to 1 so then what you'll get is here is basically to 1 one right but in the same element you can apply the force in the upward Direction okay uh and if that direction is J or Y that will be to 1 2 right so it's I = to x j equal to Y and k equal to Z right and similarly you can apply in the Z direction that will be to 13 right similarly if you cut across let's say in the vertical Direction okay and then you can have basically to 2 1 to 22 and to 2 3 similarly if you cut in the the span wise or the in the third Direction which is the K Direction J Direction then you will have to 31 to 32 and to 3 3 so the to is now a matrix okay it's actually a tensor Matrix is not the correct term it's a it's a tensor okay to 1 one to one 2 to 1 3 although it looks like Matrix okay but there are certain rules which are applicable to tensor but not which are not applicable to the Matrix to 31 to 32 to 33 this is to 2 one okay now you can see that there are some uh uh in the diagonal which is to 2 1 to 1 1 to 2 2 and to 3 3 okay so these are basically uh uh the stress in which uh the acting force in the same direction in which the element is present right so these are basically the normal stes T 1 one t 1 2 to 1 3 are basically the normal stes okay or the linear stes right whereas this t one2 t 13 or any other quantities basically what they have this tar one two so this is uh the direction is one but the force is acting in the uh the in the G direction or the second Direction so what it will do it will give you some sear right so that is basically the sear stess so these are basically the normal stresses and these are the shear stresses right all this to 1 2 to 2 1 to 3 1 to 2 2 3 and so on and so forth right and now remember this uh this tab is a symmetric tensor that means your tab one two will be equal to 2 1 similarly tow 3 1 = to to 1 3 and similarly tow 2 3 = to to 32 okay so this is a symmetric tensor but uh even if it is a symmetric tensor you have at least six unknown quantities which is to 1 one tow uh 22 tow 3 3 tow 1 2 tow 2 3 and to 3 1 right so how do you get this six stress quantities and then only you'll be able to basically solve this n equations right so let's try uh deriving this stress quantity okay now U if the element the a fluid element is at rest okay then also some stress is acting on this some force is acting on this and what is that force is the pressure Force right you can have pressure acting from normal Direction everywhere so it will be part of the normal stress right so the pressure is acting acting on this right if the fluid is in uh static state so this force is called the hydrostatic force or hydrostatic stress okay fourth per new area is called a stress right but when the fluid is in motion you will have additional stresses because of the motion of the fluid okay because you will have uh the velocity changing across layers okay so that will give you the different kind of stresses right so that that will be called as the deviatoric stress right so we'll basically divide our t i into two part one is the hydrostatic stress and other is the diator stress okay now I'm introducing uh atom just to represent in properly in your using tensor form is introducing a chronical Delta Delta I is chroner I hope I'm right okay Delta sometimes they don't use C okay so what is the property of this konal Delta so Delta I is equal to 1 if J is equal to 1 okay if J is equal to 1 otherwise it will be zero if J is not equal to 1 then this will become zero so the Delta I value will be zero sorry j is equal to I if J is equal to Y then it is 1 if J is not equal to I then it is 0o right so if you use this relation then you'll see that Delta is a the pressure will only act when I equal to J so then if you have you will if you write this term T let's say for only for hydrostatic okay then it will be only minus P because this will be Delta 1 1 it will be 0 this is 1 2 this is 1 3 0 0 minus p 0 0 0 minus P right so that uh as you can see that this is only oting the normal Direction because pressure always act normal to the surface so you can get this hydrostatic term so now we have basically found one term which is the hydrostatic term which is which acts even if the fluid in is in static State okay uh the more complicated part is the deviatoric part okay even in the deviatoric part you have two terms which are basically the normal stress or the linear stress as well as the uh the shear stress and we'll shortly derive those things okay now what we KN know from our experience or what we' discussed in the very first class uh is that uh in the fluid the stress is proportional to The Strain rate right so if you are simpleminded the flow is just 1 D so you can write t equal to Mu du by Dy okay that's what you have might have studied in your undergrad maybe the fluid mechanics course right but now uh we have a strain rate tensor in our case right because if the flow is threedimensional okay so if the flow is three dimensional let's say this is your strand rate okay this this value you know remember stren rate tensor I'm writing okay so you uh U will be known from your equation so this thing you know you don't know the tow but we know that T is related to somehow this standard ster that we have okay D X Y whatever okay now this standard tensor can be written in two parts okay one is D UI by DX Z Plus D J by dxi okay and plus half DUI by dxj minus D UJ by dxi right now if you closely look uh both the parts okay I'm not done anything I've just basically added this D by dxi and substracted that right so uh if you look observe it closely okay this is uh basically your symmetric part right so if you write your standard tens here okay so uh this is the symmetric part and this is the antisymmetric part okay let me complete this thought first anti symmetric or sometimes also called exus symmetric okay tensor and I I'm sure you know this quantity what is this this is basically the verticity right and this is your symmetric tensor what it does it basically deforms the fluid deforms the fluid element right so this is what is important for your stress okay this will not cause you any stress in antic tensor uh because this is a verticity and that will give you the rotation right so the symmetric tensor is what is important for the derivation of uh stresses okay so I will write it as e okay and this will be basically half Omega but remember this uh this is not contributing to stress contributing to stress okay and this is antiemetic you can see it easily uh so if I can write it as a I or just if you write Omega i z right so that will be d i by d x zus d J by DX I right so if you do Omega j i that will become D J by dxi minus d i by DX right so that is basically minus Omega I right so you can see this is an anti symmetric tensor on the other hand if you write this e i j okay you you can exchange J and I it will remain the same that will become a e so this is a symmetric tensor right so you can write the symmetric tensor here uh okay so the E J can be written uh as basically d by DX here it will be half d by d y plus d v by D X because it is I = 1 g = 2 I = 1 and then J = 3 if you do that will be half d by d z plus d w by D X and simil you can write the entire uh standard tensor this will be DV by d y this will be DW by d z this quantity will be similar to this and you will have some more quantity here okay this will be similar to this okay then you have quantity which is this is I = 2 and gal to 3 right so that will be half d p by d z plus DW by d y and here we have Del W by d y okay and d v by d z okay and this would be half this is uh I = 3 and J = to 1 right so this will be half d w by D X and plus d by DJ okay please verify this that uh uh what we're doing is correct okay this will be probably half d by d y plus d DX let's see how many independent quantities that we have 1 2 3 because a symmetric tensor so this is four and this is five and this is six right so have six uh standard tensel okay that is present okay so we uh our goal is to basically use this standard tensor and derive the expression for the stress term okay now uh I am going through this derivation uh because it is uh important to understand why the pressure in the incompass flow is not a thermic pressure right so you might have read about this that uh uh in the case of uh Inc compal flow you don't use the equation of State okay and that's why you have to go through this pressure poision equation right in the compressive flow we use the equation of state and that is very very helpful because you can relate the pressure to the the density and the temperature okay so let's try to find out that why uh we cannot use the uh thermodynamic pressure uh in your in our incom flow for that we have to basically go through the derivation right so what we going to do is basically we are going to basically express our uh uh uh stress in terms of strength right so strength rate tensor is basically e k l and c i j KL so why I'm writing like this why ekl because IG is are the free index okay it cannot be basically used here so what I've done here this is a tensor of order two what is the order one tensor is basically a vector right and this is also a tensor of order two right so in order to basically use uh one tensor and get another tensor what you need is a transformation tensor and that has to be tensor of order four and uh uh why you need tensor of order four I mean to understand that you can just uh you might have read about this Transformers Matrix right so let's say have a vector okay which is b equal to a i + BJ + c k right and you want to transform it to another Vector W right so if you want to do that okay so you would require a matrix okay to transform this a BC into another value for another basis Vector right so this is what is your transformation matrix so if you have let's say this is t11 T12 t t13 T2 1 t22 t33 t31 t32 t33 so then if you want to get this first element let's call it capital A that will be equal to t11 a + T12 b + t13 C right so this is your first order tens and in order to get another first order tens from that okay you need a second order tensor right this is the tensor of order two right similarly you have a second order tensor which is your strandard tensor and we want to get this our stress tensor from that okay so we need a tensor of order four okay that is which is what is c i j k l right now it's a very uh complicated tens because it will have how many elements so in Vector you have it's a first order tens so your element will be three right this is a second order tens so that will be element what how many elements we had was so number of elements will be 3 square right the three the power order of the tensor if have fourth order tensor that will 3^ 4 that means you'll have 81 elements okay in this tensor and finding out all the1 elements is not easy okay for a given fluid right so we'll try to reduce it using some property of the fluid okay so that we can find out this C KL into something that we know for example we know the viscosity of the fluid and we also know that uh our uh stress in a very simple flow depends on the viscosity right so our goal is to find c i j KL okay uh in terms of let's say viscosity or related quantity but it should be uh uh a variable which does not have too many unknowns right like we have 81 unknowns here right so what we'll do we'll assume that the fluid is homogeneous and isotropic okay so what you mean homogeneous homogeneous means that is it is position independent so whatever quantity that we have it is independent of the uh the position right so that is homogeneous similarly uh it we saying that is AC Tropic that mean it is Direction independent that means if you rotate it okay across uh let's say any any B Vector the value will remain unchanged right so if you assume these two quantities you can write or represent this c i KL into basically uh three variables okay basically okay before we go there we can write it in terms of uh uh so we are basically trying to find a the scalar from the C KL okay because we want to reduce this 81 unknowns okay so this can be written the scalar s can be written as c i j k l and then it will multiply by uh you we require Vector because we need to basically contract and remove all these independent variables i j k and l okay BJ c k and DL right so these are the four vectors that we have okay which will each will contact with c k and ultimately give you this scalar right now if you assume that fluid is homogeneous and isotropic right so that means it will be rotation independent right so if you rotate it uh it will not change that means if you can take two vectors at a time you can let's say take a do B okay so uh this basically this these are vectors so that is nothing but a COS Theta the since the Theta is not changing right so what you can do you can write this in terms of the series of two vectors right so how many of such series that you can do I mean this we are doing assuming the conditions of isotropy right so that will become so we can write in a series of three vectors so this will be Alpha a i b i and this is CJ DJ okay so we took a together and CD together similarly you can do beta okay uh a I now you can take c i okay and then BJ and DJ right and similarly you can take the third one third combination okay that will be your gamma this is now we have taken AI Di and then basically BJ and CJ okay so now basically we reduce our 81 unknowns to basically three unknowns and of course we have to fix uh these vector quantity that is uh AI uh BJ I mean whatever the vectors that we have we will again make use of Kon Delta right so this can be written as Alpha AI so now it is B I want to write in terms of BJ right so that will be b i Delta I that means so I'm writing b i is equal to BJ Delta J right why because when J equals to I then only it will be valid so this can be written as BJ Delta a similarly you can see this is CJ now I want into CK right so you could actually because the repeated index so you can use this CK C CK DK so CK and this will be DL Delta KL right similarly if you use this uh beta so that will be a and C so this is AI and you want CK right CK Delta ik K right and then you have BJ and D L right so that will be Delta JL and plus gamma a i and you have D is L right so DL Delta i l and then you have BJ and c k Delta KL right so you can now compare this is basically c i j k l right so you can you can compare these values and this will give you uh c i k a i BJ CK DL so this a i BJ CK DL will get canceled out your c i j KL will become Alpha Delta i j Delta KL right and uh okay no so this is how we expressed it and now it has to be multiplied by your ekl right look at this expression this is your t i right so let's so then your entire thing will be c i k l e KL will be become Alpha Delta Delta k l e KL plus beta Delta ik K Delta i l e k l and plus Gamma Delta I Delta k l e e k l okay so before I go to the next uh uh slide you can just see this is Delta k e KL right so this can be written as EK K what is ekk is basically d u k by d x k right or in other words if you this is d. U or basically d u by D X plus d p by d y plus DW by d z right so this has become your Alpha Delta i j e k k right and uh similarly you can use the Expression here so this is i k and JL so you can reduce basically you can contract it okay so this will become beta Delta so if you contact it this is ik and that will give you Delta JL right Delta ik Delta G and this will be ekl right and similarly you have Gamma Delta I Delta KL and Delta ekl right so if you do the further construction you will have c i j k k l into e k l will be Alpha Delta i z e k k right plus beta you will you can contract it here both L and K will be contracted okay and what we'll get is basically e j plus gamma EJ I right this is the expression that we have got okay this is nothing but t okay now you can further use the property of the Symmetry because this t i equal to t z i right so if you use this property of symmetry uh you can show that your beta is equal to gamma right you can uh write an expression here and you'll find that this is beta equal to gamma so you can write this expression t i is equal to Alpha Delta I e k k please go back and show this that beta is equal to gamma and plus beta into 2 e i okay now you can uh basically uh we know that our St is proportional to The Strain rate right in the case of fluid in the solid we have discussed this is proportional St is proportional to The Strain right so if you use this identity so this is your standard tensor e so then your beta becomes your viscosity okay and then Alpha is also a viscosity okay which is called which is denoted as Lambda so this is called the first coefficient of viscosity of viscosity and this one is the second coefficient and this is related to the uh bulk viscosity and it will only act when your dilation is not zero so basically when L do U is not zero which you can see here also because this is related to only ekk so let's write it here Lambda Delta i z ekk plus mu 2 mu e i right this is your t i right so we related our stress to the standard tensor right now we want to find out how the Lambda is related to Mu right so what we'll do now is that uh take this trace of this uh tensor stress uh tensor so we'll add this to 1 1 to 1 2 to 2 2 to 3 3 and basically divide by three right so this is to what is to 1 one Delta this is E11 right which is basically d by DX right see I equal to J right plus 2 mu E11 right which is again the uh this is uh d. just one second okay so the T 1 one if you do so this is Delta will become I will become one J will become one so this is Lambda and then this will be present the ekk nothing but D do U okay and plus 2 mu E11 which is basically just d by DX okay similarly you can have t22 which is Lambda d. U okay I'm missing the vector signs so please be careful 2 mu d v by d y okay and T 33 is equal to Lambda D do U plus 2 mu d w by DJ okay okay uh so remember we have missed one quantity here which is basically your uh so this is only the dec part so you can have you must have the uh the pressure part also so that will be minus P Delta Z right so it will be there everywhere or what you can do you can write it as just only this is T deviatoric will be this okay this is T de here and you have hydrostatic part also so that will be this will minus P minus p andus P if you add three all this three and divide by three okay take an average of that so this will be T 1 1 plus T 1 2 T 22 + to 3 3 by 3 will become uh minus P okay plus Lambda + 2 mu by 3+ Lambda Delta d. U + 2 mu by 3 and that is also del. U because d d x plus d d y plus d d z that is also d. U okay and we Define this as a mechanical pressure okay we call it pme so minus PM is equal to minus p + Lambda + 2 mu by 3 D do U okay now this is an interesting uh term so this is the mechanical pressure and what this is the thermodynamic pressure so this for this P you can use p P equal to row r t but for this this is the mechanical pressure you cannot use P equal to r t relation because that is not the thermodynamic pressure there a thermodynamic pressure plus you have something which is coming due to the the Divergence due to the volumetric change in the the fluid right now it so happens that in uh I mean this P mechanical depends on the basically your uh the degrees of freedom right so if the degrees of freedom are let's say only three that means only the translation degrees of freedom there so energy can be very uh I mean transferred only in the the energy mode or only the translational then this P mechanical and P will be the same right but if you have the vibrational component also or rotational component also that means the degrees of freedom are higher in that case your P mechanical and P will not be same and your volumetric uh stren uh will play a role okay in determining the P mechanical okay so that's uh one thing uh but another thing is that let's say if you have a system which is changing very fast you're changing the theramic state of the system very very fast then the system has to adjust to get the equilibrium but sometimes the response time of the system is not similar to the time which is at which you are changing the system so in that cases also P mechanical will not be equal to P but uh those are situations which are very rare situations okay for example if you have a balloon which is basically changing the state very very fast you are expanding and Contracting very very fast okay then uh your P mechanical will not be equal to the P but in most cases okay uh in our normal situations okay uh first of all if the translation motion is only important translation mode of energy is the only important and other thing is uh if the the change in the system is not very very fast then your P mechanical may be equal to p and in that case you can write this Lambda + 2 mu by 3 d. U is equal to 0 right now here two things here either del. equal to 0 or you have Lambda + 2 by 2 mu by 3 equal to 0 so this is uh anyway valid for your incompressible flow so for the incompressible flow it will anyway go to zero okay uh in other cases you get this situation Lambda + 2 3 = to 0 that gives you basically Lambda = to - 2 mu by 3 so we are relating your second coefficient of viscosity for the with the first coent viscosity and uh uh so this is called The Stokes hypothesis and why it is hypothesis because you cannot prove it okay it can be proven uh if it is the only mode of energy your translational mode but in other cases it's an hypothesis but it works well for most of the situations uh there are two reasons one is that either the del. is very very small okay and then also it will go to zero th this entire term will go to zero Lambda plus 23 okay or your Lambda plus 2 through is very very small okay however if you go to a very high speed flow okay let's say the Hypersonic cases and all in that those cases this term becomes very important and the hypo Strokes hypothesis may not be uh valid because there the vibrational mode of energy becomes very very important uh uh the molecules start to dissociate you will have ion formation and in those cases you cannot use this relation of Lambda = to- 23 and you have to basically uh look for the variation of the V viscosity okay so uh that is mostly in the situation where value of the hyponic flows and that's why have to basically do special treatment of The Bu now let's finish uh this thought I mean for the normal situation so this we have already derived Lambda and mu okay so Lambda + 2 by 3 equal to 0 so if you use all these terms so I use this t y z is equal to okay uh minus P Delta a and plus Sigma a okay this is the deviatoric term and this is your PR term so if you use that expression we had this tar y z by delx Z right so that will become if you use this star form so this is minus Del by DX z p Delta a remember this uh uh this will only work when I equal to Z and because our direction is in I here right this is I'm doing for row I so this will become minus d p by delx I right and similarly you have this Sigma which will come here and sigma I've written in the in the this form this is nothing but mu and e+ 2 e minus okay this is uh 2 e mu and plus Lambda Delta i j e KK right this is ekk Delta UK by delx K Lambda is equal to - 2x3 mu and E is basically d by DX Z half d by DX plus d by dxi so this completes our derivation so that means we can calculate the stress term using the velocity gradient that we have present so now the number of variables have reduced okay uh the only thing that you need to know is now the viscosity right which is this quantity okay the viscosity may vary with the temperature uh as it happens in the uh case of compressive flow and we'll have some laws to basically how to calculate the viscosity the temperature varation with the temperature for incompass flow the viscosity can be assumed to be constant because the temperature fluctuation is not there temperature varation is not there and for the incomp flow you can take this uh uh quantity to be zero 2x3 Delta is a UK K dxk because this this quantity goes to zero okay so at the end let's derive for the incomp flow incomp flow you have row equal to constant that's one quantity okay and D UK K by DX K is equal to Del equal to Z okay so if you work that out so then you'll have Del Sigma is a by delx so this is our Sigma right so partial Sigma i j by dxj will be D by dxj okay and DUI by DX J plus d u j by DX I okay and uh uh if you work that out okay and you'll take the all three components okay so you'll get this Del U by D x² plus Del to u y by Del y² and plus Del to UI by D x² d z squ right so which is nothing but d s UI okay so and recall that you need to basically make sure that del. U goes to zero when you're deriving this okay so those components will cancel out and this is the expression that you get and this becomes even simpler for the income sub flow uh there is no second coefficient of viscosity and that's why all the terms is - 2x3 Delta a ekk goes to zero okay and it becomes very very simple it is a lling term okay and that is the reason as will uh study later that in the case of incomp flow it is more like a elliptic equation because of this presence of this lapian term where in the case of compos flow it's more like an hyperbolic system okay so with that uh we will uh uh include our thought okay in the next class we'll look at some of the non-conservation form and we'll also derive the energy equation and uh this is this was a quite uh extensive derivation so I would request you to please go back and uh uh basically work on this okay and uh we'll uh meet in the next class thank you very much [Music] [Music] [Music]