 So, welcome to today's lecture and what we will do is we will continue from where we left off okay and towards the end of the last lecture we had derived the normal stress boundary condition and the tangential stress boundary condition. Now specifically speaking what we have done is taken into account the fact that you could have a surface tension variation along the interface okay and there is a contribution of the surface tension variation along the interface in the tangential stress boundary condition. In the normal stress boundary condition it is only the surface tension which appears not the variation of the surface tension that is at every point on the interface the local value of the surface tension tells you what the difference is between the normal stresses whereas the difference in the tangential stresses is going to be given by the gradient of the surface tension okay. So, this is your this is in fact the most general form of the normal stress boundary condition and this is your tangential stress boundary condition. So, what I want to emphasize here is that T is the total stress which is given by – P plus P i plus tau. So, what you can do is you can actually substitute this back in and keep the pressure term separately and the shear stress term separately and then proceed okay. So, if you do that what you will get for the normal stress boundary condition is P tilde – P plus n dot tau – tau tilde dotted with n – gamma del dot n equals 0 okay. So, what you have suppose there is no liquid moving inside and you only have your spherical drop that then the tau term is 0 and the difference in the pressures is going to be balanced by the curvature term which is what you are used to from your surface tension courses earlier P1 – P2 equal to gamma divided by R or 2 gamma divided by R. Now, when it comes to the tangential stress boundary condition you could have a gradient of gamma can be induced by gradients of concentration or temperature okay and what we are interested in is for the Marangoni convection problem gradients of temperature. So, I am going to talk now about how this boundary condition manifests itself in the context of the Marangoni convection problem because once you know how this boundary condition translates to the Marangoni convection problem then you can solve because you know how to solve because this is the only new boundary condition which is coming into the picture okay. All the other boundary conditions you are familiar with for example in the Marangoni convection problem what do I have? I would have let us say solid wall here okay and this is my interface this is my gas liquid interface and let us say interface does not deform that is it remains flat okay there is no undulation of the interface just to keep life simple for right now. Then what are the boundary conditions you are going to use at the solid wall? You have temperature equals some fixed value t0 let us say you have the no slip boundary condition and the impermeable boundary condition for the velocities those are things which you know how to use okay then you have the no slip and the impermeable wall conditions. What about here? Since the interface is not deflecting the vertical component of the velocity is going to be 0 okay that is vertical component of velocity equals 0 then you have the heat loss boundary condition. What is the heat loss boundary condition? If this is a z axis boundary condition is –k dt by dz equals h times t-t ambient that is the other boundary condition at the interface okay and what is it normally used as a gas liquid interface? You normally say the shear stress is 0 when you have a gas liquid interface we normally say shear stress is 0 but we are going to modify it now because you have the shear stress equal to 0 when does that how does that arise from here that is a specific case of this general problem okay t is the shear stress exerted by the gas and that is 0 because it is inviscid normally you neglect the gradients in the surface tension and so that becomes 0 and what you are left with is t dot n equal to 0 and that is t dot t tilde dot n equal to 0 that is your 0 shear stress condition but now this guy is not 0 okay and so we want to see how this gets modified that is the idea okay. So now the tangential stress boundary condition is modified so let us look at how to calculate this tangential stress boundary condition we need to look at this surface tension gradient okay and this is the gradient along the surface. Let us look at our very specific problem that we have the specific problem I have is z is in this direction and let us say x is in this direction and y is into the board okay. So I mean I want to derive this tangential stress boundary condition for this problem here now what is grad as of gamma it is actually the gradient along the surface but if you go back to what I wrote last time it is the actual gradient minus n of n dot del that is the definition okay this is the normal component of the gradient I am subtracting that from the total gradient and I get the thing along the surface. Now what is the gradient operator it is i d by dx plus j d by dy plus k d by dz and what is n? n is the vector in the z direction that is k minus k and n dot del is k dot whatever the gradient so I am going to give me again d by dz. So I just wanted to show to you that of course since I kept my life simple n is just k but if you had a deformed interface then you have to go back to calculating n and then doing the dot product and all that in terms of the function f okay. So that is your grad as now I need to get grad as of gamma how does the surface tension vary but surface tension variation is induced by the temperature variation. So I am going to have to define something like gamma as gamma naught times 1 minus I am assuming that the surface tension varies linearly with temperature okay at t equal to t naught gamma equals gamma naught that is the surface tension value and this tells you the rate of variation the way I have defined the gamma t it is a positive quantity because surface tension is decreasing with temperature the decreasing part is included in minus sign okay. So remember the way I got things gamma t is positive I mean sometimes this becomes important otherwise you will get a dimensionless number which is negative and you are breaking your head as to what is happening okay. So now d gamma by d so what I am interested in is grad as of gamma is nothing but I d by dx of gamma plus j d by dy of gamma. So I am going to write this as d gamma by dt d gamma by d temperature multiplied by d temperature by dx okay. So d gamma by d temperature is nothing but minus gamma naught gamma t this is the slope of the curve. So this is minus gamma naught gamma t times I dt by dx plus j dt by dy okay. So all I have done is just use this chain rule d gamma by dx is d gamma by dt multiplied by dt by dx and we do not know what these are right that is something we need to find out and yeah. What about see I found out the gradient vector I am going to have to take the dot product with my tangential direction. Now there are 2 tangential directions one is along the x direction one is along the y direction okay. So I need to if I really want to use this equation I need to find what the component is along both the directions. So actually that is this actually 2 equations that is the point I am trying to make okay. So let us look at this guy t dot t minus t tilde dotted with n we will keep it simple okay we do not have to sit down and do the calculation because the interface is flat you guys can quickly tell me what the components are this guy is fluid at the top right. So that is inviscid. So this is not going to contribute okay I want to yeah that is right this is one and we are not looking at the normal stress boundary condition we look at the tangential stress boundary condition right. So this is actually tau t here is tau why because we are doing the tangential balance not the normal stress balance when we look at the normal stress balance only then the pressure comes into the picture okay. So I need to write this as tau minus tau tilde dotted n dot t what are the components here or what is tau minus tau tilde dotted n tau is 0 since we have gas which is inviscid what about tau tilde that is going to be given by I am interested in tau tilde dotted n okay and the n is in the z direction. So the 2 components we are actually going to participate we are going to contribute will be tau zx and tau zy okay. So I have tau zx and tau zy will be the 2 components okay clearly tau zx is acting in x direction and tau zy is acting in the y direction and so this particular term is again a vector tau minus tau tilde dotted n is again a vector having 2 components. So now I need to do the dot product with t which gives me the unit vector in x direction and the unit vector in the y direction okay. So now what I am saying is we take the balance in the x direction in the x direction what is it that is going to contribute tau zx and that is given by mu dw by dx plus du by dz okay that is the component which is occurring in x direction and this must be balanced by the component here acting in the x direction what I am saying is I am taking t in 2 directions t x direction and t in y direction okay the thing in the x direction is going to be the i term dt by dx. So this tau zx is already a minus sign so I am going to move this to this side and this is a gradient of gamma must be equal to minus gamma naught gamma t dt by dx that is my x component balance and similarly my y component balance is going to be mu times dw by dy plus dy by dz is this clear. So I am saying that what I have is a vectorial equation here before I take the dot product I am just saying that each component has to be 0 and so I am just saying the dot product in the x direction and the y direction. So now I need to make some simplifications I have assumed that the vertical component of velocity is 0 everywhere so dw by dx and dw by dy will be 0 and this interface we have assumed that there is no deflection of the interface see that is the reason I do not need to worry about the kinematic boundary condition okay and the normal stress boundary condition. So all I do is I just say that the vertical component of velocity is 0 for all x and y which means dw by dx and dw by dy will be 0 okay. So dw by dx equals dw by dy equals 0 if there is no deflection okay since interface is flat this tells you du by dz is 0 dv by dz is 0 sorry du by dz is equal to that du by dz is equal to that in case there is no Marangoni effect no surface tension temperature you get your 0 shear stress boundary condition okay so everything is fine. So what I need to do is write this equation and mu du by dz equals minus gamma naught gamma t dt by dx mu dv by dz minus gamma naught gamma t times dt by dx this is at the z equals d the interface. Now this is fine but I want you to recall what we did for the Rayleigh Binard problem the Rayleigh Binard problem you had basically a similar situation a stagnant liquid some temperature variation okay and you had your velocity components and if you go back what we did is we eliminated the pressure term we eliminated the velocity term and finally we ended up with only the 2 variables the vertical component of velocity and the temperature okay. So what I want to do is we are going to use the same approach as what we did for the Rayleigh Binard when it comes to solving in fact you guys are going to use the same approach when it comes to solving this equation I am just outlining what the procedure is. So I want to get rid of these components of velocities in terms of the vertical component of velocity and how do you do that to do that by using the equation of continuity right because that is the one which relates all these guys. So I am going to differentiate this with respect to x I am going to differentiate this with respect to I think there is some problem here rho t by dou y is it yeah that is good rho t by dou y I am happy then I differentiate this with respect to y then I get you know something having dv by dx something having dv by dy I go back to equation of continuity I have subtracted do something and get everything in terms of w and temperature okay and then finally I will have an equation for w and temperature only 2 variables like a hat for the Rayleigh Binard okay. So let us do this differentiate this with respect to x mu d square u dx dz equals minus gamma naught gamma t d square t by dx square I am going to add these guys and what do I get adding I get d by dz of du by dx plus dv by dy equals minus gamma naught gamma t d square t by dx square plus d square t by dy square and this from the equation of continuity is minus d w by dz so I get minus d square w by dz square equals that and which means mu multiplied by d square w by dz square equals gamma naught gamma t I am going to put temperature h is basically telling you from the horizontal direction or the surface x and y okay. So basically this tells me that the second derivative of the vertical component of velocity multiplied by the viscosity equals gamma naught gamma t times del h square t that is the boundary condition which is arising because of your surface tension dependent temperature dependence of extension. I want to specifically realize that in this boundary condition the velocity and the temperature are actually coupled okay the velocity and the temperature get coupled here. So basically velocity and temperature are coupled they go hand in hand then affect each other and it is to this boundary condition that this coupling is taking place because at the end of the day you have to be some interaction between these two it cannot be that they are decoupled velocity is doing whatever it wants temperature is doing whatever it wants because it is a net effect. So this is where whereas now if you went back to the Rayleigh-Bernard problem the coupling was through the differential equation because you had the gravity term which had temperature okay and you had the velocity term and if you go back to your equation you will see that it is a differential equation which had the coupling between the velocity and the temperature and the temperature and the velocity. In fact if you were to now go back okay so basically I want to tell you that this is the boundary condition which you are going to use at the top surface okay in addition to the other boundary conditions which are classical which you are comfortable with okay. So I just wanted to point out that when you solve your equation for the Marangoni stability problem or any Marangoni convection problem at the interface you will use this boundary condition in addition to other stuff. Of course I kept the interface flat if you keep the interface moving then you need to worry about how F changes what I want you to do and I am not going to solve the problem I am just going to tell you what exactly you have to do. So since this problem is so similar to the Rayleigh-Bernard problem you just have to mimic whatever we did for the Rayleigh-Bernard problem okay and the solution procedure for the onset of convection. You write down the Navier-Stokes equation the equation of continuity and the Navier-Stokes equation and energy balance. So what is the base state that you have? Base state is whose stability you are interested in finding out that is the one where the liquid does not move okay. So now another equation is U equals V equals W equals 0 that is the liquid stationary solution and what about temperature? Temperature is going to be linear but then the boundary condition in the top is slightly different remember it is not that the temperature of the top surface is fixed you need the boundary condition of minus you have to find the linear profile using the condition minus k dT dx equals h times T minus T ambient okay. Temperature is linear but upper boundary condition is okay. So this is what the base state is then you do the usual linearization okay. You linearize the Navier-Stokes equation what would you get? You would have the density the gravity term I am going to treat with density constant I am not going to I am not worried about including the effect of temperature variation of density in this Bernard-Gonig problem in the Rayleigh-Bernard problem I included that. So for all practical purposes my equations for the momentum are decoupled how the coupling occurs only through the gravity term okay. So point I am trying to make here is the velocity equations do not depend on temperature because I am assuming properties are constant okay the density is constant and so everywhere my velocity equations are independent but the velocity equations will affect the temperature equations it looks like it is a one way coupling is this clear okay. So that means temperature so you will get something like del power 4 w equals 0 that would be your fourth order equation okay. You can do it this way also if you go back to your Rayleigh-Bernard problem and you put the Rayleigh number equal to 0 because Rayleigh number remember contain that beta the coefficient of the density how it changes with temperature. So the density is not changing with temperature you just put that term equal to 0 that means the Rayleigh number is 0. So you can just go back to your Rayleigh-Bernard problem put Rayleigh number equal to 0 you will get this okay. The temperature equation will be del square theta equals some minus w something like that I have not derived the exact equation there may be some parameters here okay will be of this form this is the form because when you do the linearization you will have del square theta on one side okay the conduction term and then the inertial term will give you this minus w either yes it will be minus w because the slope is negative no this is the form of the equation not exact equation the point is theta is dependent upon the velocity it looks here that this is a one way coupling in the sense w does not depend on theta if you just look at the differential equations that is what it would appear it would appear that w is independent and theta depends on w but actually that is not true why because of this boundary condition there which actually relates the w and temperature okay. So that is basically what you have to do you have to remember that although it looks like they are independent they are actually coupled to each other and the coupling is through that boundary condition. So how do you go about solving this so you go about solving this the usual way which is del power 4 w equal to 0 and del square theta equal to minus w you assume you get a periodicity in the infinite direction which is x okay and seek solutions of the form e power i alpha x then convert it to an ordinary differential equation the z direction okay and then you will get solutions and you use the condition that I mean there will be some arbitrary constants coming you want to get a non-zero solution to the system of equations to find the answer of this thing. The other important thing which I missed out is there is a dimensionless number which is going to come when you solve this problem and the dimensionless number which comes is going to come through this boundary condition here this dimensionless number which comes is I mean when you make the equation dimensionless and when you solve this dimensionless number is called Marangoni number okay and that will be the dimensionless number which comes on the right hand side. So for example this equation the tangential stress condition on being made dimensionless MA the Marangoni number okay MA is the Marangoni number and clearly here again which has a role to play in the sense it tries to damp out the convection if the gamma t is very large the surface tension dependence on temperature is large then that guy is going to overcome the viscous damping. So for sufficiently large values of Marangoni number you expect to see convection okay for yeah and this is the equivalent of your Rayleigh number that you have. In Rayleigh number you have the beta term which was how the density dependent on temperature here you have the surface tension dependency on temperature and clearly because surface tension is a interface property it is going to occur only in the boundary condition okay whereas the gravity term is a bulk property is occurring in the differential equation. So this Marangoni number above Marangoni number critical we expect convection okay because Marangoni number is 0 when gamma t is 0 there is going to be nothing going on it is going to be just sitting as it is the liquid. So your job now is to find this critical Marangoni number so how do you go about doing that to find Marangoni number critical we have to seek W as W star of z times e power sigma t plus i beta x i beta x this is x no rx here yeah and theta as theta star of z times e power sigma t plus i beta x that is periodic solutions in the x direction which is infinity growing linearly in time exponentially in time okay and this is my z dependency. I am going to substitute this in these equations which come by linearization I mean you have done this linearization before so many times so just go back to the Rayleigh-Bernard problems same thing you make a dimension less like you did earlier and you proceed okay and what we will do is we are looking for neutral stability right so the point where things are just going to go from stable to unstable for the onset of convection. So here again you can prove that sigma has to be real sigma is not complex okay so we are going to find the point of transition from stable to unstable by putting sigma equal to 0 just like we did for the Rayleigh-Bernard problem okay. So here sigma is real and so put sigma equal to 0 to get the onset of convection clearly this Marangoni number critical will depend upon the heat transfer coefficient because there is an extra parameter which is coming into the picture. So this is a heat transfer coefficient so you will get a Biot number kind of thing when you make a dimensionless HD by K. So for different values of Biot number you will get different Marangoni number curves and again you will have a critical wave number at which the convection is going to start and you can find out what the wave number is. So the analysis is exactly the same as whatever we have done for Rayleigh-Bernard. In fact I think maybe these days with things like Mathematica if you can get the ordinary differential equation you can possibly plug it in to Mathematica and get your solution. Get the condition for which the determinant is 0 and plot Marangoni number versus the wave number for which you get a nonzero solution okay and this will be for different Biot numbers. So that is what I want you to do in fact I want you to actually calculate this particular thing. I want you to calculate Marangoni number versus beta. Beta remember is a wave number okay you will get some curve like this and this Copperilla curve is for a fixed Biot number, Biot number is fixed there will be 2 dimensionless parameters in the problem. One is the Marangoni number one is the Biot number so for a fixed Biot number you will find the thing. So if the Biot number is different you will get one more curve you get a family of curves and you need to find out what this is and so this is the critical Marangoni number okay. So your job is to solve these 2 equations by substituting this form for the solution just like we did earlier okay and since we have done it earlier I am not repeating it. So you guys just do it get and since it is a linear equation your solution is going to be in the form of sin hyperbolic beta x z sin cos hyperbolic beta z things like that. Then once you have the solutions for theta and this put the boundary conditions find a non-zero solution by putting the determinant equal to 0. The determinant equal to 0 for the determinant of a matrix which has Marangoni number beta and Biot number that matrix will have all these 3 parameters you understand. So you fix Biot number then there are 2 parameters remaining Marangoni number and beta. So what different betas find Marangoni number for which the determinant is 0 get this curve get that minimum okay. So that is what you have to do and that will tell you this thing but I think the important point I want to emphasize this is boundary condition because that I think is the new thing here and you should be able to include this boundary condition and you should be able to solve. You can generalize this for systems where there is a temperature variation and where there is a concentration variation when there is a liquid-liquid layer 2 layers of liquids so many things can be done once you understand how this boundary condition has to be formulated okay. So that is as far as Marangoni convection is concerned so tomorrow we will solve some other problem okay thanks.