 So, good morning all, welcome to the next lecture. Today we will be revisiting the capillary instability problem through a separate analysis called the Rayleigh's work principle, okay. So, last class we had already looked at capillary instability and found that you know the configuration the cylindrical configuration of a jet is unstable for k greater than 1, you know stable for k greater than 1 and unstable for k less than 1. So, you had seen that you get a dispersion curve with a maxima close to k equal to 0.7. So, today we will revisit the problem, we will use something called as Rayleigh's work principle, it is a static argument to determine stability. When you do a linear stability analysis, you do something called as a dynamic analysis wherein you find out equations governing you know the variables the pressure and whatever variables are there in the problem, you get a dynamic equation how they are related in time and then you solve and get that dispersion equation that we had got in the last class. So, Rayleigh's work principle is a static argument wherein you look at the state that you are looking for the stability of say the cylindrical configuration of the jet and we disturb it like all cases where we analyze the stability and then we look at how the energy of the system has changed or you know the amount of work has been done on the system or done by the system. So, based on that we get the you know the criteria for stability of the system. So, I will just start with what Rayleigh's work principle states. So, for Rayleigh's work principle you first of all have a state of whose stability you are analyzing okay and the state is perturbed to a nearby okay and you find out whether work is required to be done on the system to take it to this state or whether the system does work and goes to that state okay. So, we determine the work done to take the system from you know the original state to a nearby state okay. Now if work is required to be done on the state to take it to a nearby state which means that you know it has gone to a higher energy state. So, which means this was a lower energy state and therefore this would be a stable state okay and if it is the other way where in the system you know while in this state particularly does work and goes to a neighboring state then since it has done work and gone to the neighboring state it has lost energy and it has gone to a lower energy state. So, this has to be a stable state compared to this. So, depending on the sign of you know the work done or basically the energy of the system determines whether it is stable or unstable. So, wherever it is at a lower state energy state would be the stable state okay. So, is that clear? So, I will just write it down if work is done by the system to move from state 1 which I would call this to move to state 2 which is a nearby perturbed state okay. So, work is done by the system which means it has lost energy and reached here. So, then state 1 will be unstable and similarly if work is you know required to be done on the system then device versa is true wherein the system would be stable okay. So, the problem that we are looking at today would be the problem of capillary instability wherein you have a jet of radius A. So, this I would call it is state 1 okay and what we are saying is that the total energy of the system right now it is an inviscid jet and it is stationary static. So, it is not even moving like the case which we had studied in the last class wherein we looked at a stationary thread of fluid and it was inviscid. So, the total energy of the system is basically dependent on the surface area which is there of the jet because it is when you have a jet and it tries to break up into drops it is basically trying to minimize the surface area or basically trying to minimize the surface energy of the system. So, the total energy of the system is determined by the surface area which is available for the jet okay. So, we have this as state 1 and we have the state 2 which would be a perturbed jet. So, we will right now just look at one wavelength it is infinite in the z direction. So, this is my r in the radial and z is the axial. So, it is an infinite jet. So, I am just looking right now at one wavelength of the jet okay. So, say I perturb it. So, this perturbation is about mean radius. So, this thing I will call it as r a cap. So, one thing you have to notice here is that this is r equal to a whereas I have mentioned here to be r as a cap that is important because of the geometry that we are looking at it is a cylindrical jet. So, when you started off with one wavelength of jet and you impose the perturbation which was of this you know cosine form the volume which is added you know here in the crest and the volume which is reduced in the trough are not equal because of you know it is a surface of revolution. You know if it was just a planar thing wherein it was symmetric then the thing which is added and subtracted are same but since it is a you know it is a volume of revolution the amount of volume that is getting added here is definitely more than what is being reduced by the trough of perturbation. Is that clear? It is clear? No. What I want you to see is that I initially had a jet which was of radius a and now on this jet I am imposing a perturbation which is of a cosine nature. So, what this means is that near the at this crest part the volume the you know the radius has increased and at the trough part the radius has decreased but if you try and calculate the volume which goes to the power of r square basically because it is a surface of revolution. So, you see that the amount of volume which has been added in the crest is more than what gets reduced in a trough. So, basically when you give a perturbation of this form you have added mass into the system. So, to compensate and to equate the masses of the system you basically would have to have an a cap which is not same as a but less than a you know so that you can impose a perturbation and still conserve mass. So, the basic idea of choosing a cap which is not same as a is to impose conservation of mass. If you keep r as a and impose a perturbation of cosine form then the mass would never be conserved. You have to have a different a cap so that the mass remains conserved. So, like I said the amount of work which would be required to take the system from state 1 to state 2 would be dependent on the surface area of the two states you know. So, essentially all I have to do is to calculate the surface area of the jet for this particular state and surface area of the jet for this particular state and compare for what conditions you know the area would increase or decrease and thereby the energy would increase or decrease and therefore stability could be determined. So, I will call this form of perturbation as r a cap plus a small amplitude which is epsilon okay and this surface could be just r equal to a okay. So, now what I would do is calculate the area which is the curved surface area you know the curved surface area of the jet for this particular state and the volume of this particular state. So, let us find that so area for this particular case the state 1 would be just you know I am looking at the curved surface area. So, it is r d theta d z which has to be integrated theta goes from 0 to 2 pi and d z goes from 0 to lambda I am looking at one wavelength of the jet and r is a here. So, I can just replace r by a and I would get basically 2 pi a lambda just direct integration. So, this is the area which is there for my jet the curve surface area. So, I can say that you know the area per unit wavelength would be nothing but 2 pi a fine. Similarly, I would just calculate what my volume would be. So, this is the unit volume in the spherical coordinate system. So, I integrate r dr from 0 to r which is r equal to a and then theta from 0 to 2 pi and z grows goes from 0 to lambda okay. This should be directly the volume of a cylinder of radius a and length lambda. So, which you could easily get to be pi a square lambda. So, v by lambda or the volume per unit wavelength would be pi a square is this clear. So, now we do the same calculation for state 2 which is a perturbed jet. So, as we know for state 2 r is given by a cap this epsilon cos kz and if I am along the surface of the jet I can you know name this to be say ds which is you know along the arc length of the surface okay and ds and this would be dz and dr. So, if I zoom that part I would get ds, dr and dz, ds should come here yeah along the arc length. So, ds is going along the arc length and z is the axial direction and r is the radial direction. So, by Pythagoras theorem I can say that you know ds square would be dr square plus dz square and then dividing throughout by dz I would get ds by dz whole square 1 plus dr by dz or ds by dz could be written as 1 plus dr by dz to power half oh yeah thanks. So, this is important because while I am calculating the area of the perturbed jet I should be going along the arc length you know to get the curved surface area. So, I need ds by dz so that I can integrate directly in the z if I know ds by dz. So, this is my r and this is my ds by dz. So, ds by dz you know I can write I can get dr by dz from the equation of r here and then that would give me ds by dz to be you know 1 plus dr by dz only this term has z in it. So, it would be epsilon squared. So, cos differentiation would become sin I just substituted r in this and then got ds by dz and if I further use my binomial approximation for this I could get you know 1 plus half epsilon square k square sin square kz. So, any doubts till here? So, once I know ds by dz now I can calculate my you know the curved surface area for the perturbed jet using a. So, theta goes from 0 to 2 pi and dz z goes from 0 to lambda because it is again one wavelength. So, now we have to replace r by that form which is you know a cap plus epsilon cos kz. So, that gives us 0 to 2 pi 0 to lambda a cap plus epsilon cos kz and ds by dz would give me 1 plus epsilon square k square by 2 sin square kz. So, I can just multiply these guys and get a cap plus epsilon cos kz plus a cap epsilon square k square by 2 sin square kz plus an order of epsilon cube term integrated in d theta dz. Is this clear? So, I can directly integrate it in theta and write it as multiplied by 2 pi because none of these terms are a function of theta. So, that would give me directly 2 pi multiplying it and integrating the first term with respect to z it is a constant. So, I could directly write it as a cap times lambda a cap times lambda. The first term is a constant a cap. So, integrating it in z would directly give me lambda which are the limits lambda minus 0. The second term has cos kz in it. So, if I integrate the cos term it would become sin kz and sin kz if I put the limits the bottom limit would directly give me 0 because sin 0 is 0. The top limit if I put sin kz where z becomes lambda, lambda is nothing but 2 pi by k. So, 2 pi by k gets cancelled kk gets cancelled and I will just get sin 2 pi. So, even that becomes 0. So, essentially when you integrate you will see that and put the limits the cos term gets cancelled. So, what you will be left with is integral of this with respect to z and that should give you a cap epsilon square k square by 4. We can see that the integral of this would be just lambda by 2. You know sin square could be written as you know 1 minus cos 2 x by 2 and you will just see that the cos part again goes to 0 because it is cos just like this term. Only 1 by 2 would be left and then multiplied with these constants. So, you will directly get this term. So, this is the area of the perturbed jet which is nothing but 2 pi a cap lambda plus a cap epsilon square a cap k square lambda by 2. So, since this unperturbed was called a by lambda I will just call this guy a cap or something. So, this is the perturbed area a cap which you get like this. So, next we calculate the volume for the perturbed jet which is nothing but v cap integrating. So, you get the volume by integrating the volume element in the three coordinates. So, again you have to replace r by a cap plus epsilon cos kz. So, that would give me a cap plus epsilon cos kz whole square by 2. Replacing that here integrating this would give me r square by 2. So, that is directly r out there by 2 integrating 2 pi and 0 to lambda. Is this clear? Like integrating r dr would give me r square by 2 and r for the perturbed jet was this a plus epsilon cos kz. So, I just substituted instead of r a plus epsilon. This r is that r it is r here a cap plus epsilon cos kz. So, again this is none of them is a function of theta. So, you could directly integrate to get 2 pi. So, you will get pi integral 0 to lambda a cap square plus 2 epsilon cos kz plus epsilon square cos square kz. And again the cos term should go to 0 like we had got for the area case. And cos square could be again written as you know 1 plus cos 2x by 2. And then you will see that you know that cos 2x part again goes to 0 and then you are left with just 1 by 2 here. So, I will get pi times integrating a cap square would give me lambda. And this term would just give me epsilon square times lambda by 2 or v cap by lambda could directly be written as pi non square by 2. Now this is my perturbed volume per unit wavelength. And then what I had calculated previously for an unperturbed thing was was pi a square. So, the conservation of mass basically means that I have to equate both the volume because density is the same for the liquid even when you perturb it or in the unperturbed state. So, I could directly equate the volume and get a relation between you know how a cap should be dependent on a. Only once I get how a cap depends on a can I compare the areas because the area here is dependent on a and the area which was calculated for the perturbed jet depends on a cap. So, if we have to compare the areas you have to get a relation between a and a cap that comes from the conservation of mass which is equating both the volumes because the density is same. So, I have v cap here I have v here I just equate both of them to get a relation between a cap and a and then substitute that in the equation for area and then I can compare the area and get the conditions for what k would the system be unstable and for what k it would be stable. So, we will just do that. So, equating v and v cap to ensure conservation of mass we get pi a square is equal to pi a cap square plus. So, pi gets cancelled and then we get a cap equals a square minus epsilon square by 2. So, a cap would be a square minus epsilon square by 2 to the power half or I could just take a common out and then I could write 1 minus epsilon square by 2 a square and then again using the binomial you know equation approximation we get 1 minus epsilon square by 4 a square. So, I will just write down the a cap by lambda that we had got was 2 pi a cap plus epsilon square k square a cap pi by 2. Is it clear till here we just substitute the value of you know the expression for a cap that we have got into the expression for area of a perturbed jet. So, that we can calculate the area in terms of a and then compare it with the area of the unperturbed jet. So, what we will now do is just substitute a cap in this expression for the perturbed area and then compare it with what the area of the unperturbed jet was which is given by a by lambda. So, doing that gives us a cap by lambda is 2 pi a cap plus 1 square k square a cap 2. So, instead of a cap I have to replace it by a minus epsilon square by 4 a plus epsilon square k square pi by 2 into a cap which is a minus epsilon square 4 by a. So, expanding it we get 2 pi a minus 4 a squared yeah. So, it should be 4 a squared here and 4 a squared here correct. So, 2 pi a and then this would give me pi by 2 a square epsilon square minus this should give me pi 1 square k square a by 2 is that correct ok. So, 2 pi a minus is this correct yeah this looks correct yeah I did not carry the a square anyway somehow sorry yeah I can just write it as order of epsilon this one yeah ok. So, if I just write it as 2 pi a and take the epsilon square out you know and try to club this term together pi comes out let us say what else comes out nothing else 2 comes out. So, I am just left with 1 by a minus plus k square a is that correct plus higher order terms. So, I could just take a common so that you know I get something like this. So, for the time being we just do not worry about the higher order terms we just compare the area for the perturbed jet with the area of the unperturbed jet which is this and we see that you know 2 pi a was the only term which was there here and we have an additional term here which comes because of the induced perturbation that we had given to the jet ok. So, what we will do is we will just find what the change in area is for the two case that is just nothing but delta a ok we will find it to be per lambda to be a by lambda minus a cap by lambda. So, that would give us 2 pi a gets cancelled. So, this is 2 pi a a by lambda minus 2 pi a plus epsilon square pi by 2 a into k square a square minus 1 ok which is any doubts. So, this term gets cancelled and what we are left is nothing but epsilon square pi by 2 a into 1 minus k square a square I took the minus here 1 minus k square is that correct I mean you could take it anyway you could have subtracted this minus this or this minus this does not matter. So, this is my change in area per unit wavelength. So, if I try and solve for the case where delta a is 0 you know which would be the condition wherein the perturbation that I am giving is not changing the area. So, that is like the onset of instability you know where you know on one side it is positive on one side it is negative and at that particular k it is 0. So, that is like the point which you had got in the linear stability analysis as k going to 1 in the non-dimensional frame. So, here if you put delta a to be 0 implies that you know my k a will be 1 because this term is non-zero. So, 1 minus k square a square should be 0 which means k a should be 1 or I will call this a particular k which is called the k c or the cut off wave number or the critical wave number. So, that k c is nothing but 1 by a. So, you should see that you know the k c which is 1 by a is exactly same as what we had got through the linear stability analysis. You know remember the linear stability analysis you had the growth rate which was sigma squared or sigma whatever k c k you had a maxima and then you had a cut off after which it was stable below which it is unstable. So, this was k c which was 1 by a. So, you get the same information by not doing any complicated you know linear stability analysis just by energy arguments from Rayleigh's work principle. So, this is the motivation to do a Rayleigh work principle analysis of the system. So, what we will now do is we will look at k on either side of k c. So, k greater than k c which means if my k is greater than k c this the negative term becomes dominant or basically my delta a would be less than 0. So, if delta a is less than 0 which means the perturbed jet has lower area is that clear if my k is greater than k c. So, if I put k c which was 1 by a is that clear if k becomes greater than k c k c was the thing which was making both the terms equal and getting it cancelled. If my k is greater than k c then this term would be dominant or I will get my area to be negative which is less than 0 okay and if k is less than k c my delta a would be greater than 0. So, what does this tell me about the stability of the jet if the perturbed jet has lower area compared to what it was it means it has gone to a lower state which means the jet was jet has gone to a lower state when I perturbed it. A correct it has more area because it is negative. So, it has more area because this term is negative this term is negative. So, the delta is negative which means this has more area. So, if the perturbed jet has more area which means it has gone to a higher energy system energy state you know. So, it is gone to a higher energy state by perturbation that means where it was stable the previous state was stable because it was at a lower energy state okay. So, the jet is stable and when k is less than k c and you have delta a greater than 0 it is the reverse of what this would say. So, my perturbed jet would be having lesser area and thereby energy therefore jet is unstable. So, this is evident from the dispersion curve also you have positive growth rates for k less than k c. So, k less than k c the jet is unstable therefore it has positive growth rates whereas for jet for k greater than k c you have negative growth rates which means that you know the jet is stable. Yeah that is all for the class here thanks. Any doubts? Delta a equal to 0 gives you neutral stability yeah. Yeah. So, but so in that curve I mean from the linear stability analysis. Yeah. k equal to 0 is also a neutral stability point but through this we do not see that. Yeah that is true but k equal to 0 is not a it is not a perturbation which would conserve mass in any case. You cannot give a perturbation k equal to 0 and have the mass conserved yeah sigma square yeah. But you would get the same thing if you plot you would get the same form of perturbation. k goes to infinity then a will go to a hat for it anyways that should No it would be the jet would always be r equal to a cap plus epsilon if k goes to infinity k goes to 0 like it is k z right. So, you would always have the r which is larger than the thing. So, you cannot actually conserve. So, that is why k equal to 0 is not considered as a perturbation I mean whatever analysis whatever values of growth rate you get you cannot actually conserve mass by that perturbation. So, it is not done anything apart from k equal to 0 is considered fine. And then linear probability we took the perturbation of the interface as a plus some delta right. Yeah yeah. So, there we took a and not a. That was because it was infinite. So, it does not matter much. So, it should be infinite right just be just considering. No when you are doing an analysis per unit wavelength I think it matters for the mass to be conserved. Whereas, in that I think the continuity equation and all those the equations which are there for conservation of mass would take care of the way the velocity perturbations change when you impose that perturbation. So, that the mass remains conserved because you are solving for those mass conservation equation which is the continuity equation along with even when you flow. Yeah it is a flow and then you yeah.