 Good morning we will continue our discussion of linear instability analysis and we will start by quickly recapping what we did the last time we wrote down a series of steps that we go through to start with an equilibrium solution and then we develop a linearized version of the governing equations about the equilibrium solution. So what this does is it helps us understand the behavior of a small perturbation around the equilibrium solution. Now if the equilibrium solution happens to be a constant like the case that we discussed the dispersion relation can be obtained in closed form if not one would have to solve for these eigenvalues essentially the dispersion relation gives us the eigenvalues at any point in time if the equilibrium solution is not a constant one would have to solve for those eigenvalues numerically. Say for example you look at if the I can apply the same exact methodology to look at the stability of the parabolic velocity profile in a round pipe when I take u equal to some 1 over r square over capital R square which is like my parabolic velocity profile it is not a constant in r it is not a constant in space essentially the mean the equilibrium solution is not a constant therefore what we end up with is a fourth order ordinary differential equation that and an eigenvalue problem involving this fourth order ordinary differential equation to which governs the growth or decay of the perturbation quantities that fourth order equation is very often called the over summer field equation. So essentially the over summer field equation is obtained through exactly the same process except in that case you would have to solve for the eigenvalue with the largest real part which is going to determine the growth of a given perturbation numerically whereas in this case we are able to obtain it in closed form meaning analytically because the mean flow is simply a constant. Now I do want to show that this is not something that is that came out of the way more fancy of a mathematician's mind this process of obtaining the most unstable wave number. So essentially if we go back to this dispersion relation it is some omega is a function of k omega is the growth rate also in mathematical terms you would call that the eigenvalue k is the wave number this growth rate as a function of the wave number is some algebraic equation in our case it turned out to be a polynomial but in general it could be any algebraic equation explicit in some instances even implicit but it is still it is a closed form solution that if I give you a k you can give me an omega and it could be not 1 omega or 2 omegas but many omegas if omega happens to be a transcendental equation if this equation happens to happens to be a transcendental equation but the one eigenvalue that I am interested in is the one with the largest real part for that k it is that one eigenvalue that is going to govern the growth of that disturbance. Now so from this we can find at one particular value I will call this k0 k0 is the what we would often call the most destructive wave number which means that for a given for a given disturbance of this wave number k0 the growth rate omega the corresponding growth rate omega 0 is the maximum of all the possible growth rates for all the other wave numbers okay which means because of the functional form e power omega t this particular wave number is going to cause is going to appear to grow the fast test in relation to all other wave numbers and this has been and has been validated experimentally in many situations you know like the example that we just solved you can show for yourself that when the water you can go back to the dispersion relation that we solved so if you replace the what we call fluid number one with air and the fluid number two with water and also said sigma to be equal to the surface tension of water you will find that the lambda 0 corresponding to k0 is on the order of about 10 centimeters 7 to 10 centimeters. So that is when air is blowing over a lake that is otherwise at rest and air is blowing at a velocity of 10 meters per second the ripples that we observe this theory predicts will be about 7 centimeters in wavelength this kind of a prediction can obviously be very easily validated in experiments and it has been validated okay. Now what we want to do is look at how this theory can be applied to sprays and atomization which is what we are after now. So if I take the first instance of an atomization problem the if I take a cylindrical jet ensuing out of a nozzle of radius r okay I am going to assume so this is now the case of an axis symmetric I am going to assume rho 1 and u1 are the density and velocity of the fluid outside rho 2 and u2 are the density and velocity of the fluid of the liquid itself. So imagine just a water faucet a faucet through which a jet of diameter 2 r is exiting into let us say air okay the air is at some velocity u1 and the fluid is at some other velocity u2. Now what we do actually end up considering is not this problem but the case of an infinite cylindrical jet so if I replace this with an idealized problem that is infinite in the z direction I can go through the exact same process that is step 1 write the governing equations I will use the subscript notation I like in our old notation we said capital quantities are the mean flow quantities and in this case ui happens to be a constant u1 and u2 both are constant in their domain of definition. So if I now simplify this equation you d ui dx is 0 because it is not a function of x and du dt is also 0 which implies just like we found in the other case du dx well in this case I should write not x but z because okay what we can show from this is if I write the same equation in the r direction what we find is that pi is a constant except if you know if you remember in the flat interface problem we found that because the interface curvature was 0 pi p1 is also equal to p2 so the pressure in both the fluids put together was a constant we will see here that because you have this interface having a curvature so if I this happens to be a round jet in cross section and the radius of that round jet is this capital R so what we find is that p2-p1 is equal to the sigma over capital R. So at the mean flow condition the capital p2 which is the pressure inside the liquid is constant inside the liquid p1 is the pressure in the fluid outside that is in itself constant but p2-p1 has to be equal to this sigma over R because of the surface tension pressure that occurs at the interface so let us just make sure we understand this understand the physics of how this comes about if I take a piece of that some delta theta in diameter in angle for take a piece of the jet and if p1 happens to be the pressure p2 happens to be the pressure here and p1 happens to be the pressure outside there is a surface tension force that acts at the cut section so if I cut the fluid at some point here and here there is a surface tension force that acts in either direction and this surface tension force has a component in this direction and a component in this direction what you will find is that this p2-p1 has to be sufficient to balance the normal component of these forces the horizontal the vertical component as shown in this figure will cancel out okay and from that you get this curvature dependence so if I take this angle the sign and the cosine essentially if I say that the normal component of this force which is this force is a function of this delta theta and therefore you start to get these curvature effects so if I take the perturbation quantities if I write u as the u bar i plus ui vector u is the total velocity field ui is the mean velocity field mean or I would like to call it the equilibrium velocity field and this is my perturbation component if I go through the same process that I went through before and linearize these equations I will keep these in the vector notation just to sort of make things a little easy for us from continuity equation what we find is that del dot ui is equal to 0 this says that the perturbation velocity field is also divergence free meaning the perturbation velocity field has to instantaneously obey this incompressibility condition essentially del dot u being 0 is coming from the fact that the fluid is incompressible and so if I these are now the linearized equations that I am writing I do not want to go through the same process of showing you how I substitute the mean plus perturbation into the full governing equations take out all the order epsilon squared terms keep only the order epsilon terms and that is when you get these linearized equations okay so I am skipping those two steps in the interest of time to just show you the equations that you get from this linearization first equation is del dot u equal to 0 and the second equation so if I take the total pressure to be equal to this pi plus little pi just like we wrote in the previous case okay so if I now take I will mark these now as equations 1 and 2 if I take divergence of equation 2 and if I use equation 1 what we end up getting is this equation del dot grad of pi the divergence of gradient of the pressure in each of the fluids is 0 which can also be written as del squared pi equal to 0 now del squared is a linear operator so I will just sort of write this out in open form this is our equation that we get from just combining the system of linear equations that we have now I want to point out one thing here the that will reinforce some of the concepts that we discussed in the earlier linear instability analysis at this stage we introduced what we call the normal mode expansion correct we said we will write pi as e power omega t plus ikz plus im theta okay in the previous case we did not have this im theta in this case I am I have two spatial variables over which the perturbation can vary in the z direction as well as the the theta direction we will see what this means in just a moment but essentially I am assuming that my perturbation has two spatial variable forms two spatial variables involved and one time variable and if I do that I have this and of course I have the pre multiplier is pre p prime i of r is what we what we said would be the eigen function in the r direction now this is what we called the normal mode expansion okay what I want to show you is that really speaking you do not even need to think of that as an assumption the normal mode expansion is not an assumption that we make I will show you how if I look at equation 3 so we want to answer where does this come from okay we in the previous analysis I showed this to you as though it was an assumption okay but I want to show you that it is really not an assumption that there is a fairly simple procedure to obtain it okay. So if I look at equation 3 is the Laplace equation in three dimensions okay so this is if I replace this p i with capital T for temperature it is standard heat conduction equation in three in cylindrical polar coordinates steady state heat conduction equation in cylindrical polar coordinates how would you solve that equation you have to first identify which are the homogeneous directions and which is the direction in which you have you may have some sort of an inhomogeneity right in this particular instance z is a coordinate in the vertical direction so the z coordinate basically is infinite the theta coordinate is a periodic coordinate okay so clearly those are the two directions in which the so the coordinate has homogeneous boundary condition homogeneous or periodic really I should be saying periodic boundary conditions right so if I simply do a separation of variable solution on equation 3 so if I say p i of r theta z and time is equal to r of r, t times theta of I am going to write this is capital theta of theta, t times capital Z of z, t in fact we do not even need the t in every one of these I will show you in a moment that even that is not required time some capital T of t if I take this assumption which is essentially coming from separation of variables okay and I introduce that into the into the governing equation okay what I will end up seeing here is so if I divide by capital R I will write this first term also in this notation of in this prime notation indicating differentiation with respect to its own argument now if I the standard procedure of solving you know a three-dimensional heat conduction equation if I divide by r theta z and t what you end up seeing is this now if I non-dimensionalize the r variable using capital R then essentially what I have is that each of these terms is only of this term is only a function of z this term is only a function of theta and if the sum of three functions that are each of a particular independent variable have to all add up to 0 the only way they can add up to 0 is if each of them is a constant the standard solution by separation of variables so from there you get the equation that z equals minus k squared if I say theta double prime over theta equals some minus m squared and I get do this by separating out the r part so if this part is equal to some minus k squared then the solution for this so essentially if I start out so the assumption of periodicity in the z direction and theta direction comes from taking the system of equations that we have which is equations one and two eliminating some of the unknowns in favor of the others so like for example in this case I eliminated u i vector which is u v w velocities in favor of p so I end up getting one homogeneous equation in terms of the pressure and after I have that which I have written out in full form as equation three identifying which are the periodic directions once I identify which are the periodic directions I can then solve this as though I am solving this problem by separation of variables and this is and what I end up with is what we call the normal mode assumption in the past in the last class so essentially in a mathematically this is a strum level problem that yields orthogonal Eigen functions in k is in the z direction and orthogonal Eigen functions in the theta direction okay and essentially this normal mode assumption is just coming from the fact that after eliminating these equations you get a strum level problem okay so now let us just so it comes from the fact that we are dealing with a strum level problem this is the answer okay in the previous case it was a strum level problem in two dimensions x and y if we did this in the previous case we would have simply gotten del square p i equal to 0 and there del square p i would have simply been the partial derivative of p i with respect to x twice partial second derivative plus del square p del x del y squared equal to 0 so you just have been del square p del x squared plus del square p del y squared equal to 0 you identify that x is the periodic direction in that problem y happens to be the depth direction so it is the it is not the periodic direction and so you get a sine function e power i k x in the x direction and corresponding to each e power i k x you get either e power minus k y or plus k y depending on which of the two fluids you are in the standard solution to Laplace equation in two dimensions okay so now let us take this forward if I take this normal mode assumption of the form 4 and substitute into 3 what do I get now p prime is only a function of r full building yeah this p prime we said is a function of r and so essentially this equation will call this equation 5 really should be written as an ordinary differential equation so just to be precise is an ordinary differential equation in p prime as it turns out this is called the modified Bessel equation and the solutions of this are the general solution okay this c i 1 and c i 2 are actually four constants you have c 1 1 for i equal to 1 and i equal to 2 c 1 1 1 2 2 1 2 2 now when i equal to 1 what we do want to know is that the solution is bounded inside i equal to 1 so just to be clear i equal to 1 is the outside fluid the graph of i m k r for i m x as a function of x looks something like this for different values of m this happens to be m equal to 0 and this is qualitatively how any m greater than 0 looks okay so as r becomes large k r becomes large and i m of k r increases indefinitely as r becomes large so all we know is that c 1 1 has to be equal to 0 it is coming from the solution not just being bounded but the solution needing to disappear towards 0 likewise when i equal to 2 this happens to be the case with the inside fluid now we use the fact that the solution has to be bounded inside the liquid jet inside the fluid so just to be clear if i draw a graph of k m of x as a function of x this is a functional form so at x for as x becomes smaller and smaller the value of k m of x becomes larger and larger this is called the k m is called the modified Bessel function of the second kind and i m is the modified Bessel function of the first kind so i m has this kind of a property k m has this kind of a property so when i equal to 2 if k m k r happens to as r becomes small k m k r becomes unbounded which means that c 2 2 has to be equal to 0 so from here i can write the full pressure in the outside fluid now once i find p 1 and p 2 what i also know is that u i has to be composed of the same normal modes as p i so if i you mean you can this is again this is also not an assumption i can go back to the original equation number 2 which said d dt so if i take the solution for p i that i have substitute in this equation essentially take the gradient of p i and said term wise how these have the parts of u i corresponding to time and spatial variation have to equal you end up getting exactly this so this is a characteristic of any linear problem that the response of the linear system is always the same as the response in the forcing function so what we want to do is use that fact and say u i is also is also of the same functional form in r theta z and time now so essentially that gives us an analytical solution for all of u i and p i and then finally like we said we have two sets of boundary conditions one is called the kinematic boundary condition that says d dt plus u i d dz acting on the interface is equal to u i dot er this is my way of saying that this is the radial component of velocity that i am concerned about on this side also eta is some perturbation just like of the same exact form as the one we have before and then we have the dynamic boundary condition p 2 minus p 1 now these are the differences in the perturbation pressures alone and these are functions of the principle radii of curvature okay we will continue with this discussion in the next class where we will start from here and work our way to the dispersion relation for a cylindrical jet.