 Good morning, we will continue our discussion by starting to look at secondary atomization and we will see briefly what secondary atomization is, what role it plays and then we will come back and look at some simplified analysis to start with and you know what sort of techniques can be used to understand secondary atomization as a process okay, so we looked at several we looked at least two different analysis techniques that we can apply to primary atomization we looked at one in great detail called linear instability analysis, so this is where you have a bulk liquid coming out of the sprain nozzle and that bulk liquid is breaking up into drops and these drops are somehow determined by the instability structure that occurs on this liquid sheet, that was the premise on which we learnt the mathematical framework in which we can do this analysis right okay. Now we want to study the life of these drops what happens to these drops after they are formed, is there a possibility that they could break up further if so under what conditions and if after they break up and if they do break up what is the result of such break up process, so that is essentially what secondary atomization is, so like we said primary atomization is the process of the bulk liquid breaking up into drops, secondary atomization we will sort of loosely define anything that happens afterwards, so take a very simple example let us say I have the sprain nozzle and I have this liquid sheet and let us say there is a sheet and there is a drop that is formed it may be a drop it may be like a ring structure depending on what sort of break up happens, this ring is no longer in contact with the nozzle directly, so in other words if I try to spray a conducting liquid and I can measure the connectivity of each blob of liquid to it is to the parent nozzle, this blob that is formed which is let us say toroidal or semi-spherical whatever sort of quasi-spherical is no longer in contact with the nozzle, so the subsequent break up of these structures is what we will term secondary break up or secondary atomization, so there is a lot of secondary break up that happens in a spray as you can imagine because the drops that are formed from the primary break up process may not and in most instances are not spherical, but the final resulting drops are mostly spherical we will see why and how in a moment, so this is the sort of loose definition, so let us see where we can take this definition forward, so let us say what why do I want to study primary atomization, if I go back to that I will take a couple of very simple examples okay, here is an example of a rotary atomizer as we saw very briefly is a case where we have a disk, so this is I am going to just draw the outline of the disk say oh no this is the disk here, the disk is spinning and you can see there is a nice periodic structure in that sense of the liquid that is spewing out and this periodic structure is spaced as a mutally because of the serrations on the disk, so the primary break up process is essentially a result of a spilling liquid jet through those serrated edges and that spilling liquid jet breaking up due to capillary instability or due to some cross flow instability, so I have basically a cylindrical jet that is being squeezed out of this rotation of the circular disk and this cylindrical jet is also moving in an azimuthal sense which is similar to having an air across flow of air, so I have a cylindrical jet with a cross flow of air that is going to cause the break up of this liquid jet, we saw from primary instability from linear instability considerations that the break up of any liquid jet produces drops that are on the order of the size of the liquid jet itself, so if I have a round liquid jet undergoing capillary instability the size of the drops formed from that process will be on the order of the liquid jet diameter cannot be much smaller except if you have you know we saw in the shear instability with like a diesel injector but in a case like this the primary drops formed are very big, so for example on the right hand side here I have the drop size distribution that the manufacturer in this case Leda Bure claims to have measured in this rotary atomizer spray, the serrations in this particular case may be on the order of about 250 microns which means the primary drops that are formed are on the order of about 250 microns and if this is the distribution that was observed you can see that there are hardly any drops that are on the order of 250 microns, so even though the primary process caused drops that are on the order of 250 microns there are no drops in the real spray where measurable of that size, so the question then one has to ask is what happened in the middle that is where secondary atomization happened. Let us take one more example and see what we learn this is the size distribution measured in a real rain, so this we are now talking of the grandest of all sprays a very heavy downpour of rain, so if you take rain falling from the sky you have and if I had a way of sampling drop sizes in the rain and in fact these people Marshall and Palmer did that in this 1948 paper, they made measurements of the rain drops and showed that the number density which is now a probability density function of the drop sizes in a real rain follow a graph approximately like this, the point to note is that is this exponential tail where you have the ordinate and the abscissa is linear which means this is a case where the number density goes as e power – d over some d bar, we have seen this in our case in our studies of probability density functions and different amount of rainfall, so this is 1 mm per hour this is 5 mm per hour and 25 mm per hour, 25 mm per hour is like a huge downpour just to give you an idea right. So the in different instances the size distributions are different but qualitatively they are all very similar in their shapes, you can even go all the way back to the very small sizes and you can see that there is a similar structure irrespective of the rate of rainfall. The point we are going to focus on is the fact that this slope is linear as you get towards larger and larger drop sizes, if I plot these if I plot this data in an axis like that essentially what I am looking at is a long tail just like that, so essentially this would be an e power – d over d bar type tail, so whether it is a rotary atomized spray or rain which are obviously coming from very different sources, the behavior towards the larger drops seems to be that you have this long exponential tail, now you can show I just took two examples that are very contrasting but you can take real sprays coming out of an air blast atomizer a simplex atomizer and many other different mechanisms of atomization and you will find that there is a long exponential tail that of the probability density as the drop size increases, this is a standard feature of most sprays. So the question then is like if all these are mechanically so different why are why is the measured drop size distribution showing such similar features, the answer to that is that when drops are produced by primary atomization processes they are relatively still unstable, so we really mean unstable in the sense of a linear instability analysis showing a positive growth rate and therefore they are likely to break up, so we will see in a moment under what conditions are they likely to break up and how and so therefore this is like I said the last part is just a hypothesis, it will be very careful when you listen to hypothesis that in most sprays secondary atomization processes are really responsible for drop sizes and their distributions much more than even the primary processes, so even if I did not do all the linear instability calculations but I just assumed that this spray nozzle is producing drops on the size of the liquid sheet or on the order of magnitude of the orifice diameter in the case of a diesel injector or on the order of magnitude of the liquid sheet in the case of either an air blast or a simplex atomizer but did a more thorough job of the secondary atomization processes modeling I might end up getting a better answer, so I might end up getting reasonably correct answer I should not say better, so that is the hypothesis under which we will learn secondary atomization and clearly I did not present this hypothesis when I was describing linear instability analysis because that is also an important process at least to know under what conditions jet is even likely to break up you cannot predict that without the help of a linear instability analysis. So we will come back to the rest of this a little later let me close this, so with that premise let us see why would a drop break up, so let us answer ask a question say I have formed a drop and I am going to assume an initial drop is spherical, let us say the drop is moving with the velocity Ud surrounded by an air flow that is at some velocity Ua, so we are going to ask the question why would the drop break up, if I assume for a moment well even before you ask the question why would the drop break up will answer the question what causes deformation, so break up is like an extreme event of a deformation process that has to initiate be initiated on the drop, so we will see what causes deformation of the drop clearly for that you need a stress you need a force right, so what causes the force as you can see I have a drag force the drag force on this drop is some drag coefficient times pi d squared over 4 let me keep the notation right, half rho I know you are going to say u squared but we have to be careful it is essentially a result of a relative velocity between the phases, so in this one dimensional sense the relative velocity is just Ud-Ua the squared this is the magnitude of the drag force this is not indicating the direction. Now does it matter if Ud is greater than Ua or Ua is greater than Ud not really because all I care about is the relative velocity I could have a stationary drop be impacted by sudden by a velocity by high speed air or I could have the drop be injected at a high velocity into quiescent air it does not matter, now this drag force if the drop was perfectly spherical and if the drop was let us say rigid that is now we are looking at a rigid sphere I can draw like a pressure profile around the drop something like that you can clearly see that the pressure on the front part would be larger than the pressure elsewhere etc, now if I have a perfect stokes drag then it would be a symmetric profile all around but anything that has a finite Reynolds number you are not going to see symmetric pressure distribution around the sphere. So now what does this pressure distribution do since we are now since we are allowing the drop to deform this drop is likely to deform in the direction of the exerted force sort of like that because the pressure force is now just trying to flatten out this into a pancake and if you let this process continue all the way forward it is going to break up now are there any other forces that are important yes the answer is we have the surface tension force so on one side we have the drag force and we have the surface tension force which is given by sigma times pi d if pi d is the circumference of this drop if I have to deform this drop from the black shape to the blue shape I have to increase the surface area and any increase in the surface area comes at the penalty of increased surface energy and that is worth done against surface tension force which is of this magnitude. So if I so if I allow these two forces to compete with each other which is what is actually happening let us see what happens if I have a competition between these two forces by the way this what is this row this row is that of the air let us be very clear about that so when these two forces are equal in magnitude are on the order of the same or on the order of the same magnitude then you have something interesting happening in term it is basically like a fight between equals drag force trying to deform this drop and surface tension force trying to resist the deformation okay so and when would that condition occur when this mathematical condition occurs and what you see on the left hand side is what we often called Weber number okay and this 8 over CD is a is another number for drag over as drag around a sphere if I plot CD CD happens to be a function of the Reynolds number of the Reynolds number around the flow and you know you can look up any standard textbook and you will see a graph that looks something like that for CD as a function of Reynolds number so essentially CD is like a number that comes from either empirical measurements or some source of some other source the point is this that really the value of CD is important but not as important as the idea that when this Weber number is is on the order of some critical Weber number okay so we will call this 8 over CD as some critical Weber number so when this when the real Weber number around my of the flow around the drop is on the order of the critical Weber number then I start to see deformation effects creep in if this surface tension force is much greater than the drag force essentially I have a very very rigid sphere that refuses to deform because its surface tension force is so high that it is able to keep the drop intact so it is almost like a perfectly spherical body moving through air on the other hand the drag force is very large in comparison to the surface tension force it is as though I do not even have a liquid drop it is just going to be pulverized it is as though I have a for lack of another way of saying it like I have a little mass of gas that is being impacted by some other stream of fluid and this mass of gas is just going to disintegrate without knowing any idea without having any effect of a restorative surface tension force okay these are the extremes of what the Weber number looks in the Weber number space if the Weber number is very large that means the surface tension force is insignificant in comparison to the aerodynamic force that is trying to deform the drop and if the Weber number is very small the surface tension force is so large that I would not have any kind of break up the transition between these two regimes happens at this critical value of the Weber number now very often people ask me the question you know if I write Weber number is rho u square d over sigma I have done a little bit of simplification u here is a relative velocity but for the remainder of this lecture in the next few we are going to assume that the drop is stationary so in other words if I impose a coordinate system that is moving with the drop at the velocity of ud then I all I care about is ua-ud so u is that relative velocity measured in the frame moving with the drop and if I write down the Weber number as rho u square d over sigma what is this row should it be rho a or rho l rho a or rho l the answer is it is it depends on what you want the Weber number to convey in this particular exposition I have I have shown Weber number as coming out of a competition between the aerodynamic drag force and surface tension force therefore you should this becomes rho a now if I was trying to in and this is typically what you will define Weber number as I mean whereas it is sort of a dimensionless number should come out of physical arguments like this and we will do that we will look at other such arguments in the next couple of we will look at other such dimensionless numbers that will fall out of other fall out of similar arguments if I take the next instance I will let me go back to my drop so we know now that Weber number is important and that this Weber number has to exceed some critical value and that critical value could be a function of the Reynolds number if Cd is a function of the Reynolds number right that critical value is a function of the Reynolds number right what is the Reynolds number in this particular case it would be rho a-u so I will use the u calculation so now I ask the same question okay I would not answer it but I will ask the question and then answer if I write Reynolds number as rho ud over mu is rho rho a or rho l and is mu mu a or mu l that depends on again what you want the Reynolds number to convey we will see a couple of instances if I want to talk of this Reynolds number I will make the distinction that this is that drop Reynolds number this is the Reynolds number of essentially the drag coefficient acting on a spherical drop so if I want to talk of this then red is rho a-u d over mu a because I am looking at the drag on a steel ball of diameter d that is what this sort of Cd versus Reynolds number comes so if I want to understand how this critical Weber number behaves as a function of the Reynolds number that Reynolds number is rho a-u d over mu a which is the viscosity of the air we will see another but for the breakup of the drop itself is the Reynolds number associated with the liquid viscosity important so in other words let us ask another question what role does liquid viscosity play in this whole process we saw the role of the liquid surface tension let us see what role does the viscosity play so if I take a drop and if I want to deform it slightly okay now these are all in the coordinate system moving with the drop so the drop is stationary and it is just deforming as this deformation process happens I have to set up a relative velocity field inside the drop so if every point in this drop is moving at exactly the same velocity there will be no deformation of the drop in the frame of reference fixed at this oh if I have to have deformation of this drop in the frame of reference fixed at some centre of mass of the drop let us say then different points in the drop have to be moving at different velocities something like that the moment I have different points inside the drop moving at different velocities I automatically I am implying that there is a velocity gradient field which means you are going to create up a stress field viscous stress field so let us see if we can estimate that viscous stress field so if I again take a diamond drop of diameter D the viscous stress field inside the liquid drop is what I am now thinking of is what I am now trying to model so that goes as mu the liquid viscosity times any sort of a velocity gradient that you can set up inside the drop okay so let us take the extreme case where this you now we are looking at a stationary drop impacted by a velocity you right so if this you is the velocity of the farther most far farthest point on the drop and I will take the extreme case where my o is at rest okay so over a distance r radius of the drop the velocity is going from some u to almost 0 so the viscous stress is going to be on the order of that we are only estimating the order of magnitude okay so just to like if I put one mm drop in a relative velocity field let us say 10 meters per second if I put plug in the numbers here that will give me an order of magnitude of the stress that is expected okay since we are dealing with D for diameter in the previous example values the same D because we are only establishing an order of magnitude that factor of 2 really does not matter so if this is the stress field and the same and the surface tension field surface tension pressure that is resisting this deformation okay is sigma over D okay now both notice both of these have units of stress force per unit area I can compare the stresses or I can compare the areas it does not really matter I mean I can compare the forces it does not really matter so all I will do is multiplied by like a pi D squared over 4 or if I just multiplied by D squared okay on both sides and it does not really change anything that I am about to do which is to when the viscous stress is on the order of the stress the restorative surface tension stress okay or let me be clear if I am comparing this viscous stress to the surface tension stress what I am trying to say is that the surface tension stress is essentially destabilizing the drop because of the oscillations that it is creating the viscous stress is trying to damp out the oscillations of the drop so this particular comparison if you notice is only intended to show what are the if I have a surface tension pressure that is trying to somehow destabilize the drop okay like for example if I take a really type instability so just a cylindrical jet of liquid cylindrical column of liquid infinite column of liquid in air we saw from linear instability calculations that that column of liquid is unstable it would rather break up into drops of a certain size spaced at a certain distance lambda that lambda comes from the linear the point of maximum growth rate so that is a process where surface tension is driving the growth of the instability that is the cause of the instability and that surface tension driver driving force if that is much greater than the viscous damping force then you essentially have a break up process if you if the viscous damping stresses are going to be much greater than the surface tension driving force then you are going to change the rate of the growth and eventually could even stabilize the process we will see what happens okay so essentially if I write this comparison at this point I want to sort of bring in all the different dimensionless arguments that you could make so you have this other number called the capillary number which comes from a competition between the viscous stresses that arise in the fluid during the course of deformation if I do not have a gradient field set up inside the drop a velocity gradient field I do not have any viscous stresses so if the entire drop is translating I do not have any viscous stresses inside the drop it is only that during this deformation process that I have gradient set up inside the drop because of and because of the deformation process and on the right hand side here I am looking at the surface tension forces that could be causing that destabilizing. So if I say on the other hand the viscous stresses inside okay remember are proportional to the gradients in other words they are absolutely 0 unless there is a gradient field in the gradient field is what is actually causing the gradient field in our instance it is the aerodynamic force so I have this half rho a u squared which is the pressure force so if I look at the comparison between the viscous stresses and the aerodynamic stresses what do I have I will ignore the factor half because I am only interested in orders of magnitude notice kind of the weirdness of the u of the see we are forming dimensionless groups but you never think of a rho a u d over mu l until you write this comparison of until you use physical argument so Reynolds number with a mu a in the denominator is not really expressing the competition between the viscous stresses inside the drop and the aerodynamic stresses that are causing that deformation you never be able to establish the competition between these two between these two physical forces now so I have this this kind of Reynolds number this capillary number and the Weber number that we formed already so let us just write those down now as you can tell this is a very weird Reynolds number right so what people find it instead of dealing with the row and mu coming from the different fluids people find it useful to define another parameter called the owners over the number which is basically square root of the Weber number divided by Reynolds number which if I do the calculation I have rho a square root rho a u square root d divided by square root sigma times the Reynolds number is mu a mu l mu a u d the u cancels out so what I have is a mu l divided by sorry see the point of some of these like derivations of dimensionless groupings without the use of Buckingham pi theorem which is what you typically learn in the under grad fluid mechanics is very important you would never be able to physically cast these kinds of groups without using physical arguments like this so another fallout would be never say I will divide the numerator and denominator by rho a and replace this with a kinematic viscosity of sorts because they are coming from different they are they are of different fluids okay. So now let us see this is the role that viscosity plays so I want to look at this dimensionless group to tell me how important is liquid viscosity this is very small that means the effect of liquid viscosity itself would be is negligible whereas for the same Weber number if I have a high onus organ number the effect of the liquid viscosity is now important. Now if I use simple Buckingham pi theorem arguments I will show you what else might happen what do we do in Buckingham pi theorem we essentially start out by making a list of all the parameters that are relevant okay I do not think I missed out anything so if I have 1, 2, 3, 4, 5, 6, 7 parameters I can form n-3 which is 7-3 or 4 dimensionless groups have written 3 of them here 1, 2, 3 onus organ number is not the 4th dimensionless group because it was formed out of Weber number and this definition of Reynolds number so I am looking for a 4th group which in this case turns out to be the density ratio rho A over rho L. So in other words I cannot form a rho A over rho L by any combination of these 3 groups primarily because rho L does not even occur anywhere in here. So what does this tell us so we have now identified the 4 dimensionless groups that are important now I could have gone through a similar physical process to identify this kind of a group the way to do that is what is the role of the liquid inertia in relation to the air inertia. So if I say the liquid is some density rho L its mass is d cubed proportional to d cubed and its velocity is u so rho u rho L times the mass is like times the let us say essentially is a measure of the liquid phase inertia we want to look at this is essentially going to tell us we want to look at the competition between liquid phase inertia and gas phase inertia and this dimensionless group is one way to find that. Now nothing in our arguments was restricted to drops in air if you notice rho A I could be looking at the break up of bubbles in an infinite medium of liquid for all this argument is concerned so it really does not matter. So another way to think of this is I can take a combination of the owner's order number that I have defined here this definition and this definition of rho bar and through some appropriate pairings of OH and rho bar define another OH bar which would be mu L divided by square root rho L d sigma which if you simply have just be OH times square root of rho bar. So once I have four dimensionless groups I can find many many more from just multiplicative and divisive combinations of all of these of the basic four but which ones do I pay attention to physically you cannot get to that using Buckingham pie theorem you will need the physical arguments that we used before if you go back we used forces to draw our analogies another way to do the same thing is using times so I could use what is called a timescale analysis to see what is it that the if the time what are the different orders of magnitudes of the different timescales and from there draw comparisons to which ones would be comparable which ones would be relevant so we would not do that today I just want to I will probably come back to that later on when we do need to talk of secondary break up but for now I have identified these groups so what can we learn from that the start of any understanding is experiments for the most part right so let us see what experiments have showed in this particular instance what you have here on this axis it is not clear is the owner's organ number and this is the Weber number that we define what this is a this is some very detailed data from all these different sources that professor GM faith and his co-workers put together into this regime diagram what is a regime diagram if I have Weber number and owner's organ number and if I identify a certain range of owner's organ numbers and Weber numbers what do you expect would happen in that range okay so in different parts of this graph as you can see there is different physics happening we will see individually what each of those are so if I have very low Weber number and very high owner's organ number will take the easier ones to understand then we will come back to the rest so I have I will draw a schematic of a drop being hit with air at some velocity you this is some diameter little d liquid viscosity mu l sigma etc we have defined the owner's organ number if I have a very high owner's organ number on this order of magnitude and very low Weber number it means that there is the liquid viscosity is so high that it is practically preventing any deformation of the drop as you can see they say that it will move but there is hardly any this break hardly any breakup expected even hardly any deformation expected less than 5% we will go to the other extreme up here very low owner's organ number and very high Weber numbers this is almost like our like a blob of gas that is going to be impacted that surface tension and viscosity play hardly any role that is where you get this thing called the shear breakup where I have a drop and I have tiny fragments being shed from the sides of the drop so it is almost like I have flow around a spherical surface and that flow is just pinching off tiny drops this is like our diesel spray as you can tell in this kind of a mode of this kind of breakup the range of drop sizes expected from the breakup process are going to be much smaller than the parent drop itself and in the rest of the regime you have other physics happening that will come back and discuss in the next class.