 Welcome we will continue our discussion on sprays, but we will switch gears slightly we will start to look at the insides of some of these atomizers and nozzles. So, we are going for the next few lectures we will talk of design of atomizers and we will begin with design of a pressure swirl atomizer, but before that I want to provide a quick primer on the fluid mechanics of swirl we will see how swirl in general works. I will take two kinds of swirls and we will see what it looks like. So, if I take a container of liquid and spin it at some angular velocity omega about a point o. So, we are only going to look at two dimensional examples to start with that is sufficient to understand how things work. So, this is a container that is spinning and it is got some liquid in it this liquid at least at the wall is going to spin with the container and towards the middle it is not going to spin. So, this is one kind of swirl the other is where I take let us say an infinite infinitely big bath of liquid and insert a rod into this and spin it spin this rod at some velocity at some angular velocity omega. So, in one case I have liquid inside a container and the container is spinning. In the other case I have an infinite bath of liquid and the infinite bath of I immerse a rod into that bath of liquid and spin the rod at some angular velocity omega. I want to understand what the fluid mechanics is like in these two situations. I am going to look at inviscid flow for now governed by Euler's equations. So, if I write the Euler's equations in cylindrical polar coordinates I will write the whole thing this is our famed continuity equation. So, this is my theta momentum equation. So, what we find here is that I have three quantities u r which is the radial velocity u theta which is the angular velocity in the theta direction and p which is the pressure the fluid dynamic pressure in the in the hydro hydrostatic pressure in the fluid and that hydrostatic pressure is represented by capital P could be a function of r and theta. Now, I am going to simplify these equations from some very simply making some assumptions for assume axis symmetry. Now, the two kinds of flows I am interested in are these. So, in both these instances like in the first instance o is the axis the second instance this happens to be o and the axis passes through o perpendicular to the pointing along the direction of the angular velocity vector in both these instances. Now, so if I assume axis symmetry what do I have let us say u r is only a function of r u theta is only a function of r I will use the same symbols as there p of r. So, if I use these axis symmetry assumptions and scratch out terms that are not important if u r is only a function of r and I am also going to assume steady flow. So, things are spinning and we are well past the spin up phase that is the phase where the fluid was adjusting to a rotating container or a rotating rod and we are well past all of those transients and we are in a steady mode of operation. So, if I now go through scratch out terms and put in parenthesis why that term is being scratched out I will start with the first one. This is scratched out because of axis symmetry this is scratched out because of the transient nature not being present again this is due to axis symmetry this is due to transient nature not being present this is due to axis symmetry as well this is due to axis symmetry. So, I am going to take the simplified version of the continuity equation and see where it will take us says 1 over r d dr of r u r equal to 0. Now, if I simplify this I can first of all replace the partial derivatives with total derivatives because u r is only a function of r u r can only be of the form c 1 over r. So, if there is a radial velocity to this flow it can only be of the form c 1 over r c 1 is some arbitrary constant that is yet to be determined. Now, if I go to the theta momentum equation and see what that will tell us now before I go much further I am going to invoke the let us look at each case separately. If I take the solid body rotation case or if I take the I will call it the container case then r equal to 0 is contained inside the domain because of which the if u r is of the form c 1 over r then and this is the only admissible form for u r if r equal to 0 is contained inside the domain that amounts to an infinitely large radial velocity at the center line. So, therefore, the only admissible solution the only admissible solution c 1 equal to 0 which also implies that u r is identically 0 everywhere. Now, if I go to the r momentum equation what I have these are the only two surviving terms. So, if u r is equal to 0 then what I have is this further simplifies to rho u theta squared over r d p d r is equal to rho u theta squared over r. So, for the case where the axis is contained inside the container this is the only possibility and this should look physically familiar that this basically says pressure gradients balanced by centrifugal force. Now, if I take on the theta momentum and take the terms that I still have remaining u r d u theta d r plus u r u theta over r equal to 0 again if u r is equal to 0 like for the container all you get is an identity. So, as far as the Euler's equations are concerned you really cannot solve for u theta and p. So, you essentially have one equation which says d p d r is rho u theta squared over r and from there that is the only equation that you have and theta and p are both separate are two unknowns. So, in the inviscid case in the in an inviscid swirling flow you end up with one equation in two variables that you really cannot completely solve. So, there are two forms of. So, whatever be the form of u theta over r you will recover the pressure distribution associated with that that is set up by that centrifugal action that is essentially what this thing what that equation says. So, any form of u theta is a function of r that you impose that is that you impose on the flow field would give you the corresponding pressure field. So, if I look at what are all the possible physical forms of u theta over r and this is not coming from inviscid theory, but actually from viscous theory, but we will use it freely here essentially what you find is that u theta over r is of the form omega times r plus some gamma divided by 2 pi r. This is called solid body swirl and this is called potential swirl. So, if you look at any kind of a swirling flow remember that we are not restricted to one of these two forms except in the case of a viscous flow we will look at that later on, but as far as inviscid theory is concerned the only thing that you are allowed to conclude is that if you tell me what v theta is a function of r is there is a corresponding pressure field associated with that that means that can be calculated from the radial momentum equation. So, now let us go back to a pressure swirl atomizer and see what the basic construction of a pressure swirl atomizer looks like and see where this theory would come in handy. So, a typical pressure swirl atomizer is composed of let us say some n number of tangential inlets that bring fluid in these tangential inlets may be located at a distance d away from this center of this swirl chamber. So, I will just show one of the tangential inlets because the others are sort of hidden. So, this is a swirl chamber there is usually a contraction passage. So, all of this the rest of this nozzle is axis symmetric. So, this is the simplest kind of a pressure swirl atomizer that you will find. So, if I go back to this case and let us say if I was to characterize this is a cross sectional area this is some width w and some height h. So, if the depth of the depth of the slot is some h and w is the height w is the width of this passage and v is the velocity coming in then the total volume flow rate is v times w times h times n. n is the number of tangential slots v is the velocity coming in w and h are the width and height of the slots. So, if I look at what the potential what the possible velocity field here would look like inside and I am going to include this dimension r sub s as the radius of the swirl chamber. Now, if v is the velocity here one could imagine an invisible container between this point and the center line that is spinning at a linear velocity v or an angular velocity v over d. So, this part would have a velocity profile that looks like that. Now, if I look at what is happening outside here and if I draw a zoomed out picture past this point you essentially have to go all the way to the wall. So, this is the wall of the swirl chamber. Now, from this point where you have this tangential slot coming in to there it is like you can imagine an invisible rod that is spinning at an angular velocity v over d. So, as far as this part of the fluid is concerned it is like it is being driven by a rod of diameter 2 d or radius d that is spinning with an angular velocity v over d. So, this part of the velocity profile is going to look like a potential swirl. So, you have two parts you have this solid body rotation on the inside and a potential swirl kind of a flow on the outside. So, this part looks like v equal to r omega this part looks like or rather u theta equal to r omega this part looks like u theta equal to gamma over 2 pi r. So, in reality as you go close to the wall you also have like a boundary layer that is due to again viscous effects. So, we will sort of try to stay away from that for now we will look at viscous effects in an empirical fashion a little later on. So, now if you look at what this flow is doing total angular momentum coming in is m dot times v times d m dot equal to rho times q that is equal to rho times v times w times h times n the number of slots rho is the density of the fluid. So, this is how much angular momentum is being input into the swirl chamber. So, if I were to draw a control volume around this part if I do an angular momentum balance on this because we are looking at an inviscid flow the walls are not exerting any kind of a torque on the fluid. So, whatever angular momentum comes in is exiting from the bottom. So, if the total angular momentum coming in is of this form we will call this i dot in i dot out is equal to m dot times v at exit times d at exit over 2 if d at exit is that whatever is the angular velocity of the fluid coming in has to equal the angular momentum of the fluid coming in has to equal the angular momentum of the fluid exiting. So, if I equate the two this is only really true for a inviscid flow this is just the tangential component of the exit velocity. Now, if because the fluid is coming in tangentially into the swirl chamber it contains no axial momentum in the direction of the axis. But, for the fluid to exit out of this exit cross section it has to have an axial momentum. So, again if we do a simple control volume analysis one would have to imagine that this part of the wall is essentially exerting a downward force on the fluid causing it to generate an axial momentum. So, this axial momentum is now going to depend on the exit cross the mass flow rate divided by the exit cross sectional area divided by the density. So, if I look at the axial velocity I will call this u at exit this is m dot divided by rho a e the only problem here is a e is not equal to pi d e squared over 4 this is where the design process gets a little complicated a e is the exit area available for the fluid flow. And, because you have a swirling flow there is no guarantee that the entire cross sectional area is flooded with fluid flooded with the liquid. So, in order to understand this we will go back to our container rotating problem and see if we can make sense of it. So, why is we will go back to the solid body rotation. So, if I take a container filled with a fluid up to some height h and I spin this about an axis it is we have observed this in the past if not we can easily make these observations that this fluid meniscus is going to become deformed in the form of a parabola this is our static steady condition where does this come from this comes from say first of all u theta being equal to omega times r u theta of r being equal to and. So, if once I have this I have d p d r equal to rho u theta squared over r that comes from our radial momentum conservation. So, this is rho omega squared r squared over r which is rho omega squared r. So, if I integrate this p of r equal to half rho omega squared r squared plus a constant I will call p 0. Now, this is the case with no gravity included in this equation if I include gravity then this p 0 is minus rho g z and when I use that this I mean I can show this to come from the z momentum equation if we want. But for now we will just assume it is actually equal to p 0, but it is I mean p 0 can be a function of z if I have gravity included which is acting along the z direction from here I can calculate the an equation of an isobaric surface an isobar which is a constant pressure surface is given by half rho omega squared r squared minus rho g z equal to some pressure value and for atmospheric pressure I am going to set that equal to 0. So, if I simplify this what I have this is the equation of my parabola. So, if you take a set of parametric curves you know. So, if I take shapes of this z as a function of r for different values of this number omega squared over 2 g that is like a length scale right or I will write this. So, this now happens to be a dimensionless number which describes the competition between the rotating velocity omega rotational velocity omega and gravity g. So, as this dimensionless number becomes larger and larger what do these curves look like essentially if you know let us say that is equal to 1 you get a curve that looks like this going up to r equal to r as this number becomes larger and larger I am going to have to draw this one more time I will erase this part. So, as the as this number increases the fluid is going to be pushed more and more towards the wall the meniscus shape is going to look like a nearly flat meniscus at the bottom and it is going to rise up sharply near the wall. For a pressure swirl atomizer this is a gravity really has no role to play. So, in other words omega omega squared or r omega squared in relation to 2 g is such a large number that g is like practically being equal to 0. So, you are essentially in other words the spray the spray properties of this nozzle is independent of whether you are looking at the spray horizontally vertically horizontally vertically vertically upwards does not matter it does not matter what the component of g is you are going to get essentially what looks like the same spray. So, for that to happen this r omega squared over 2 g is a very large number. So, what looks like a parabola in our regular container spinning is now going to look like a thin film that is sticking to the walls of the container walls of this exit orifice. So, the liquid exiting through the pressure chamber here through the swirl chamber is going to be in that form and t is some film thickness. Now, what is it that is inside here the same stuff that is inside this parabola when it is spinning right at this point when I started the spin up process before I started the spin up process there was pure liquid. Now, as I spun the container up to some finite angular velocity omega the air which was outside here got dragged in here because of the pressure gradients and the net result of that is exactly the same as in this case except since there is no stabilizing force like gravity. The air is dragged essentially all the way into the swirl chamber and this is called the air core. If I look at where this comes from it is essentially coming from this isobaric surface. Now, being equal to r equal to some constant value r naught which is less than the diameter of the pipe itself. So, this part of the flow this part of the flow looks like a pipe a constant diameter pipe in which you have an inviscid swirling flow and that inviscid swirling flow is exiting into atmosphere and that atmospheric air is dragged in to this into the nozzle to create an air flow air core. So, this air core is primarily responsible for the cross sectional area a e not being equal to pi d e squared by 4 a e is the liquid exit area. So, this liquid exit area is now being is being less than this pi d e squared over 4 is what is responsible for which is actually is what is responsible for the diameters for these for this relation. In in in reality though this is a this is the reason precious swirl atomizers are very attractive because I can have a very large exit orifice d e and by appropriately changing the swirl geometry upstream I can cause the film thickness here to become thicker or thinner. So, if you want to go back by varying this number d d is the tangential offset of these slots by varying the tangential offset I can essentially create a higher amount of angular momentum entering the swirl chamber or lower as a consequence of that the swirl chamber the swirl chamber geometry then essentially dictates the value of a e the exit area available for the fluid flow. So, just to complete this discussion if I now take an isobaric surface including gravity or including some constant pressure p naught at the center or at the wall let us say. So, if I know the pressure at the wall then p of r is some p naught minus half omega squared r squared. So, if p naught is the absolute pressure absolute atmospheric pressure the pressure as a function of r is of this form. So, I can find a particular value of r where this is equal to 0. So, if the pressure at the center line is given I can find a radial cut off point at which the pressure is equal to the atmospheric pressure. So, essentially it is it is using statics to understand a dynamic system which is not always the right thing to do, but it gives us the concept of the radius of the air core as being that point at which the flow field the centrifugal pressure due to the flow field is balanced by the atmospheric pressure. So, you can sort of find that value at which the it is like a cut off point at which inside which the air core is stable and that is going to dictate the difference between these two is approximately equal to film thickness. We will stop here we will continue this discussion in the next class.