 Welcome back we were looking at the linear instability analysis of a plain liquid sheet this is essentially if you imagine a liquid sheet that is flat but has a finite thickness which is about we called it 2H exiting into an otherwise quiescent atmosphere. So we look at the density of the liquid sheet was rho 1 and it was moving with a velocity u 1 the density of the air outside if you want to call it the fluid outside was rho 2 on the top rho 3 on the bottom and u 2 and u 3 but essentially rho 2 and rho 3 will set equal to each other. So it is exactly the case of a plain liquid sheet exiting into atmosphere and one point we have to understand in all these linear instability analysis is that we are looking at the stability of an infinite liquid sheet. So there is no such thing as a nozzle from which the liquid sheet is exiting now you can do an instability analysis of what is called a semi infinite liquid sheet where you show the presence of a nozzle and then say they cannot you cannot have the possibility of disturbances upstream of that and you can only have the possibility of disturbances downstream of the nozzle. But we are not going to focus our attention we are only going to focus our attention today on the linear instability analysis of a plain liquid sheet exiting that of an infinite plain liquid sheet. So let us look at the schematic that we have like I said the sheet is of thickness 2H the density of the fluid is rho 1 and it is moving with a velocity u 1 uniform in the entire sheet and rho 3 we are going to set equal to rho 2. So we went through the process of writing the Euler's equations for both the fluids and then going through the perturbation process for the two fluids and we came as far as calculating the pressure field in the fluid above and the fluid below. The pressure field in the fluid above p 2 is given by this expression c 1 2 e power minus k y e power i k x times e power omega t I can write it slightly more compactly in the form that we are otherwise used to which is likewise p 3 and then so these are this is essentially the fluid above and the fluid below p 2 and p 3. As far as the liquid sheet itself is concerned we went through the process of eliminating one of the variables and using the normal mode assumption we got these equations 9 10 and 11. I can multiply equation 9 by i k and take 1 over omega d dy of equation 10 you will see why I am doing this it is essentially to get to the form that is in equation 11. So if I divide this by omega i k u 1 multiply by and multiply by i k. So let me do that equation 9 first becomes u 1 double prime equals minus 1 over rho 1 plus omega i k u 1 p 1 double prime. So i k u 1 double prime you have i k times i k that is minus k squared but with the other negative sign this becomes k squared over rho 1 I will call this equation 12. Now equation 10 is omega v 1 prime if I take derivative of this with respect to y I will call this equation 13 now using 11 you will find that this right hand side is 0 because of equation 11 what we have I will write this in a slightly different way but still recognizable. So essentially I have now an equation for p 1 double prime this is of the form equation 14 where lambda squared equals k squared omega over omega plus i k u 1 the rho 1 cancels out the solution to this call this 2 1 and 2 3 just for consistency of nomenclature. So I have this c 2 1 and c 2 3 are as yet undetermined and likewise you have these two constants in the pressures c 1 2 and c 1 3. So I have c 1 2 c 1 3 c 2 1 and c 2 3 so now I am going to go back to the schematic that we had the plain liquid sheet I will just copy this whole thing this is my unperturbed configuration if I look at the perturbations on this so I can have two kinds of perturbations on this and this is the fundamental difference between a plain liquid sheet and all of the analysis we had before this problem has two interfaces as opposed to one that has always been the case right. So what do we do with with the two interfaces I will call the above the interface on top as eta and the one on the bottom as eta bottom if I write the form of eta t and eta b so essentially y equal to eta b eta t of x comma t is the equation determining the of the top interface and y equals eta b x comma t now because of the linearity in the problem the this clearly has a functional form that is similar to the functional form that already has occurred in the pressures I will call this eta 0 t e power omega t plus i k x this is coming from the normal mode assumption. So now now that we know the interface shape we can write down the boundary conditions the boundary conditions at the t interface which is the top interface the first one like we have always written is the kinematic boundary condition I will write it from the direction of fluid 2 and from the direction of fluid 1 from the direction of fluid 2 what I know is that v 2 prime which is the y direction velocity in fluid 2 has to equal the partial derivative of eta t with respect to t this evaluated at y equal to h has to equal this from the fluid 1's point of view v 1 prime at y equal to h has to equal the partial derivative of eta t with respect to t plus u 1 partial derivative of eta t with respect to x and all these derivatives evaluated at the unperturbed free surface we have discussed this at least twice during the course of the previous lectures but I will reiterate one more point of one more point of view so to say that v 1 prime and v 2 prime in these two in this equation are the Eulerian velocities at the interface whereas the right hand side of each of these two equations is the same velocity at the same point written from a Lagrangian point of view so if I was a material particle on the interface and I am governed by the equation eta t of x comma t which means the rate of motion that I am subjected to is fixed therefore the material velocity at the interface say either in fluid 1 or fluid 2 has to equal the corresponding velocity coming from the Eulerian description that is the physical essence of the kinematic boundary condition now if I write down the dynamic boundary conditions I will call these I have two kinematic boundary conditions at the top interface how many dynamic boundary conditions do I expect to have there is only going to be one because what that dynamic boundary condition amounts to is essentially a force balance on a tiny element so if I take if I take an element of fluid just like that and write down the force balance on that element of fluid that element of fluid is chosen to span both fluids but of an infinitesimal thickness so its own mass is nearly 0 which means the pressures acting on both sides and the net resultant forces should all balance out if I say force equals mass times acceleration the mass of the element itself is small then you are essentially looking at something that is where the forces are all equal since static equilibrium that is what we call the dynamic boundary condition so with that if I write down the pressure in fluid 2 minus the pressure in fluid 1 equals sigma times kappa we have already gone through this process of identifying what kappa would be if I know y minus eta t of x comma t equal to 0 is the equation of the surface kappa is given by this minus d square y d x square divided by 1 minus 1 plus d y d x the square raise to the power 3 half so if I simplify that knowing what the form of eta is so this is eta 0 t e power omega t plus i k x if I take derivative twice one derivative gives me i k times this another derivative gives me i k times i k which is minus k square so this becomes plus k square eta 0 t d y d x is essentially i k using the Taylor series expansion that we had I can clearly see that the minus 3 halves this term here is order epsilon squared so this simply becomes so this is the formula for the curvature so this is the formula for the curvature if I substitute that in the dynamic boundary condition I have p 2 minus p 1 evaluated at y equal to h is equal to sigma times k square eta 0 t e power we will actually do this just to sort of show what it looks like p 2 is known really speaking these are the prime quantities p 2 prime and p 1 prime because the mean pressure condition was where everything was at the same pressure so I can go get the forms for p 2 and p 1 p 2 is given by the c 1 2 e power minus k y times e power omega t plus i k x so this is c 1 2 e power minus k h times e power omega t plus i k x this is p 2 prime at y equal to h p 1 prime at y equal to h has these two c 2 1 and c 2 3 but this is p 1 double prime so we just have to multiply this by e power omega t plus i k x so this is simply the right hand the left hand side of this equation the right hand side says sigma k square eta 0 t e power omega t plus i k x so from here I have one equation that c 2 1 e power minus k h minus c 2 1 sin k h sorry c 1 2 minus c 2 3 cosine k h equals sigma k square eta 0 t likewise so I have this is my equation 17 which is I have 15 16 and 17 as the three boundary conditions at the t interface likewise if I write the boundary conditions at the bottom interface I will have the following first of all v 3 prime at y equal to minus h is equal to d eta b d t and v 2 prime at y equal to minus h equals d eta b d t and v 2 prime at y plus u 1 d eta b d x these are the kinematic boundary conditions at 18 19 just as we did the dynamic boundary condition part with the top interface I am going to implement this in some detail with the bottom interface if I go back I know what p 3 double prime looks like p 3 x comma y comma t is given by this whole thing if I evaluate so from this p 3 I can go to equation 7 as the equation that looks like equation 7 essentially because the fluid on top and bottom 1 and 2 are essentially the same the equation for the v velocity in fluid 3 will look exactly like equation 7 with the subscript 2 replaced by 3 so I know d v 3 prime d t equals minus 1 over rho 2 because rho 2 and rho 3 are the same and knowing p 3 prime is c 1 3 e power k y e power omega t plus i k x what we have we find is that knowing v 3 prime is also of the normal mode form so from here I find omega v 3 double prime is minus 1 over c 2 times c 1 3 k times e power k y this is the functional form this is the equation that gives you the functional form for v 3 double prime so if I use if I substitute that in the boundary condition 18 what we have it for 18 is v 3 prime evaluated at y equal to minus h equals d eta b d t the partial derivative of eta b with respect to t so if I evaluate v 3 prime at y equal to h I have minus 1 over rho 2 omega c 1 3 k e power plus k minus k h because y equals minus h e power omega t plus i k x equals omega eta 0 b e power omega t plus i k x notice how the exponentials cancel out in pretty much all these equations what we are left with is so this is 20 but comes from 18 I can likewise go through and do the same thing for 19 I am just showing you how you can you can take the solution that we have for the pressure and find the y direction velocity and use the kinematic boundary condition to come up with the expression relating the unknown constants and eta 0s eta 0 t and eta 0 b so now the last part is the dynamic boundary condition at the b interface and what we have here is p 2 minus p 3 equals sigma kappa evaluated at y equal to minus h you can find kappa to be the same k square eta 0 b e power omega t plus i k x up to order epsilon so let us quickly take stock what are the unknown constants that we are yet to find we have c 1 2 c 1 3 c 2 1 c 2 3 eta 0 b and eta 0 t we have six boundary conditions basically four kinematic boundary conditions and two dynamic boundary conditions in total when you go through this process though what you will it looks like a nice consistent Eigen value problem now mind you that all these boundary conditions yield only homogeneous equations if you look at equation 20 as an example c 1 3 is my unknown constant eta 0 b is one of the unknown one of the unknowns in the list that we had just here under likewise every one of our equations would be and like for example here c 1 2 c 2 1 c 2 3 and eta 0 t so every one of these equations would yield you a homogeneous equation every one of the boundary conditions would yield the homogeneous equation in these and six unknowns that are shown here but the problem is that you will find that when you go through this process the rank of the matrix is less than even five so essentially you have a six by six matrix if I write this out in the form of I can where x vector is basically I can write these six equations in this form what we will find is that we the in order for me to set the determinant to 0 I do need one more condition because some of the equations give you the same they are essentially the same equation so will not you will not be able to eliminate all these equations to get you to the characteristic equation the condition that is used is basically assigning a value to this ratio now let us quickly look at what this ratio means if I take two menisci and if one of this meniscus has a perturbation of that form I have two possibilities for the bottom one it is in phase or exactly out of phase so this number can either be plus or minus one so depending on whether you choose this to be equal to plus one or whether you choose that equal to minus one you get two different characteristic equations arising from setting this determinant equal to 0 so these are of this form but determinant of a equal to 0 is the characteristic equation so whether you choose this ratio to be equal to plus one or whether you choose it equal to minus one gives you two separate characteristic equations that is essentially if you go back what is the characteristic equation characteristic equation is some f of k right so you have one equation for eta 0 b over eta 0 t equal to one which are called sinuous perturbations sinuous mode would be a better way to say it and you would get another equation this is called the varicose so for a given wavelength or for a given wave number for a given wave number I can get one growth rate if the disturbance was in the sinuous mode and I can get another growth rate if the disturbance was in the varicose mode so for a given flow situation I need to know what disturbance that I am looking at so I can plot my dispersion diagram which is omega versus k so I want to ultimately identify the value of k for which omega is a maximum and for me to get to that condition I need to know which mode it is that I am looking at so typically what you could do is that you could plot a dispersion diagram for both the sinuous mode and the varicose mode and choose the one k that has the maximum growth rate of all the modes possible of both the sinuous and the varicose modes so in fact this is kind of nice if you go back and look at what this actually physically means the if I assume the sheet is of the if I assume the sheets to be composed of parallel menisci initially in the unperturbed condition the sinuous mode says that both of them are in phase which is like saying that the sheet is going to if I look at the top meniscus and the bottom meniscus as being the top and bottom surfaces of my palm essentially they are both in phase sort of like that so this is like a flag flapping in a breeze for example would be the case of a of a sinuous mode except for a rigid object you cannot have a varicose mode but since you have you are dealing with a liquid sheet the liquid sheet would go into a mode where it thickens in at one point and thins at another point completely symmetric about the half way axis about the x equal to about the y equal to 0 position. So there are situations where we see that a sheet thickens and thins which in the varicose mode there are situations where the sheet is more of a flapping nature. So depending on the flow condition you may get one kind of break up or the other now if you go back and look at the analysis that we have that we have performed there is nothing in this analysis that says that the liquid sheet that you have a liquid sheet exiting into air they are just fluids of some densities I could have a gaseous sheet. So like a sheet of gas that is let us say entering and otherwise quiescent liquid. Now you can imagine how if I did that the varicose mode would essentially mean bubbles forming the minisci sort of being anti symmetric to each other or minisci being mirror images about the y equal to 0 axis would amount to sort of like a bubble becoming trapped. So if you did this analysis you would find that the situation where rho 2 is less than rho 1 and rho 1 is less than rho 2 and rho 3 would give you the varicose mode showing the dominant behavior whereas if you had a liquid sheet exiting into air generally speaking you would find the sinuous mode dominating. So let us look at some results so if I completed this analysis what I will get is a dispersion relation that looks like this with this alpha being equal to 2 H in our analysis this is from this paper by Ibrahim and Akpan from the Akta Mechanica. This u 0 in their nomenclature is our u 2 sorry our u 1 the rho liquid is what we called rho 1 and rho g is what we call rho 2. Now if I so this is essentially a quadratic in omega so that I can solve just like the previous situation I can write omega is minus tan hyperbolic I can essentially I can write the dispersion relation in closed form where I will get two roots for a given value of k. Now this is for a sinuous mode so somewhere along you will see if you go back to this reference they would have assumed that eta 0 t or eta 0 b over eta 0 t is equal to 1. Now if I look at the results from if I look at an extension of this now all thus far we have only looked at inviscid analysis we will at the beginning of the next class start off with including the effect of viscosity liquid viscosity and see what we find.