 Welcome to today's lecture where we will be discussing the Rayleigh-Bernard convection problem. We will just continue from where we left off in the last class and just to make sure that we have everything in place what I have done is I have written down the equation which describes the perturbation V star talks about the functional dependency of y of the perturbation and V star is going to be governed by this sixth order ordinary differential equation. Now this particular ordinary differential equation was obtained by eliminating the u component of velocity, the pressure and the temperature. What we did was we linearized the governing equations about the steady state okay and then we converted the partial differential equations to ordinary differential equations and then we got this sixth order ordinary differential equation. Before the Navier-Stokes equations were only second order to begin with because we eliminated the other variables the u component of the velocity, the x component of the velocity and the temperature, the second order equation became a sixth order equation. This equation is subject to these boundary conditions. The boundary conditions are the same at both the surfaces at the lower wall and the upper wall because both of them are solid walls. We have the velocity being 0, this comes from the fact that the walls are impermeable and therefore there is no vertical component of velocity. The fact that you have a no slip boundary condition which tells you that the x component of velocity is 0, we use that to transform the boundary condition to the vertical component of velocity v. So and for this we use the continuity equation and we got that the first derivative of velocity in the y direction is 0. Remember capital D is the operator d by dy and this boundary condition came from the temperature boundary condition. So you see that the boundary conditions which were there for the temperature, the boundary conditions which were there for the x component of velocity all of that have been converted to boundary conditions on the vertical component of velocity v. Now what we like to do is take a close look at this equation and I want you to see that this is a linear homogeneous equation with homogeneous boundary conditions. What does this mean? This means that v star equal to 0 is always a solution to this problem. It satisfies the boundary conditions and it satisfies this differential equation. Now so in some sense this is like an eigenvalue problem. You have A x equals lambda x where A is my differential operator now, lambda is something like this quantity and x is my velocity vector. What we are interested in doing is we are interested in finding out conditions for which this equation has a nonzero solution. You know that v star equal to 0 is always a solution to this problem. Why? Because this is a homogeneous system for but can this have a nonzero solution. What is the question or under what conditions can this have a nonzero solution? Clearly the only parameters which are at your disposal are the temperature of the wall, the gap between the 2 plates, the properties of the fluid, the thermal diffusivity, the kinematic viscosity, the density dependency on temperature and so for some combination of these parameters, it is likely that this system will have a nonzero solution. Now if this system would have a nonzero solution, then for that group, for that combination of parameters, remember we are on the point of marginal stability or neutral stability. Why is that? Because we have derived this equation by imposing the condition that sigma the growth rate in time for the perturbation is 0. Now if you recall what we did in the last class, this equation was derived by imposing the condition that the real part of the growth rate and since the growth rate turned out to be real and not complex, the growth rate was 0. So if the growth rate is 0, you are on the point of marginal stability or just that transition point, the critical point where it is going to change from stable to unstable. So the point of neutral stability and the fact that this for a particular combination, this is going to have a nonzero solution is going to correspond to the point where you have the transition of the steady state from a stable steady state to an unstable steady state, okay. What I want to do is I want to point out here that this particular group of parameters leads me to seek a dimensionless number. Now I have unfortunately not made the equations dimensionless in the problem so far, okay and what I like to do is I like to now make this equation dimensionless and then I will possibly read what I have just told you to make my point very clear, okay. So remember v tilde the perturbation was given as v star of y times e power sigma t times sin alpha z, alpha x, sorry. Now since we have the same variable on both sides, we are not particularly worried about making the velocity dimensionless because we would choose a characteristic scale for the velocity and that would occur both on the left hand side and the right hand side and that is going to cancel off, okay. So imagine you had a characteristic velocity and you made this dimensionless, it would not really affect. But what we have here is this term here which possibly has some units, okay. So remember this term d, the capital D is given by d by dy. So this has units of reciprocal length, okay. The alpha here also has units of reciprocal length. For example, the alpha occurs as a wave number, okay. The wave number is such that it is the reciprocal of wavelength, okay. So alpha is something like 2 pi divided by lambda. So the point I am trying to make here is the alpha actually has units of reciprocal of length and there is only one length scale in the system and that is the height or the gap between the 2 plates. So the length scale in the system is h. I am going to use this length scale and make my equations dimensionless, okay. And what does that mean? I am going to define alpha star as alpha multiplied by h. So this has units of reciprocal length and this is units of length, so this is dimensionless, okay. And I am going to write and if I want to make my equations dimensionless by scaling it with h, remember the domain of the problem now is going to go from 0 to 1. So now the domain of the problem extends from 0 to 1 instead of 0 to h, okay. And clearly what I am going to do is I am going to now write the alpha here in terms of my dimensionless wave number alpha star. I am going to write this as alpha star divided by capital H. The D squared is going to be the second derivative with respect to y and when I am making y dimensionless I am going to have an h coming out in the denominator, okay. So what I am going to do is I am going to define D star as d by dy star which is equal to 1 by h d by dy, sorry, is equal to h times d by dy, okay. This is a dimensionless derivative and this is what I get. So this is h times D. I am going to now, so remember this is the dimensionless derivative with respect to dimensionless coordinate and I am converting this to dimensional coordinate. I am going to substitute for D and alpha star in this equation here. So D I am going to write as D star divided by capital H, alpha I am going to write as alpha star divided by capital H, okay. And now I get D star-alpha star and this is all squared divided by H squared whole cube V star equals-beta g and alpha squared I like to write in terms of alpha again, alpha I like to write in terms of alpha star. So I am going to get alpha star squared divided by H squared, okay times Th-T naught divided by alpha T times nu times H times V star. So all I have done is converted that equation to a dimensionless form and remember now alpha star is dimensionless, okay because I have scaled it with H and D star is also dimensionless. I would have gotten this if I had scaled my equations made in dimensionless in the beginning. And all I am going to do is recognize that this is h to the power 6. Take h to the power 6 on this side and remember I have an H squared and an H here. So I have an H cube. So I have H power 6. When I take it to this side I get H cube. So I am going to write this as D star squared-alpha star squared whole cube V star equals-beta g Th-T naught times H cube divided by alpha T times nu times alpha star squared times V star, okay. I have D squared-alpha star L square star squared-alpha star squared whole cube operating on V star equal to this. I like to group these terms together and I want you to realize that beta in general is negative and Th-T0 is in general negative because the upper plate temperature is lower than the lower plate temperature. So this is negative and this is negative. So the quantity inside my brackets is a positive quantity, okay. And those of you who have done a course in heat transfer will remember that this is extremely similar to this number called the Grashof number which you have come across. Only thing is in the context of this problem this number is called the Rayleigh number. So I am going to define this Rayleigh number beta times g times Th-T0 times alpha T times nu times H cube, okay. And this is a dimensionless number just like you have the Reynolds number, the Prandtl number and the Schmidt number. And the whole idea of making this thing dimensionless was to get a dimensionless number. So we could talk things in terms of this dimensionless number, the Rayleigh number. So now I would like to recast this equation as d star squared minus alpha star squared whole cube equals minus of the Rayleigh number times alpha star squared v star, okay. And the boundary conditions to which it is subject to are v star equals dv star or d star v star equals d star squared minus alpha star squared whole squared equal to 0. All I have done is converted the boundary conditions also to the dimensionless form, okay. And this is straightforward because you just get a factor of capital H in the denominator and since that is not 0 this has to be 0. So now to go back to what I was saying, we are going to ask the question what is the Rayleigh number for which this has a nonzero solution. If I can find the Rayleigh number for which this has a nonzero solution then I will know that for that group of parameters equal to that number equal to that value I have the marginal stability or the transition from a stable steady state to an unstable steady state, okay. How do I go about solving this problem? Remember alpha star is a constant, okay. So what you really have is a sixth order equation with constant coefficients. Now you know how to solve this problem. So what you would normally do is you would seek a solution in the form of an exponential function. So the way you go about solving this problem is we seek the solution v star as, sorry there is a v star here, there is a v star there and there is a v star here, okay. There is a v star here and there is a v star there, okay. We seek v star as some function times some constant multiplied by an exponential function. You would substitute it in this equation and you would get a characteristic equation which would be a sixth order polynomial. You will get 6 roots which would give you 6 values for m and those would be your possible solutions to the homogeneous equation. Clearly the values of m would depend upon the Rayleigh number and the alpha star. Then what you would do is you would impose these boundary conditions and try to find a non-zero combination of C1 for which you will be able to satisfy the boundary conditions, okay. So we do not want a situation where the constants C1 are all identically 0 because then you would get v star as 0. Remember we want v star to be non-zero. So the strategy is to try and solve this equation by seeking solution of this kind, getting a sixth order polynomial and then imposing the boundary conditions. Since you have 6 boundary conditions and you have 6 constants C1, C2, C3 maybe I should just write this as C1 e power m1 y, C2 e power m2 y. You want to impose the boundary conditions and then you would find that you have 6 equations in 6 unknowns. The 6 unknowns are the arbitrary constants Ci. We want a non-zero Ci and what you would do is you would formulate the problem as if it is a matrix problem, as if it is a matrix multiplying a vector. The vector contains my Ci and look for a solution for which you get a non-zero solution. Look for a condition for which you get a non-zero solution and from linear algebra you know that this is given by the determinant of the matrix being 0, okay. So let me just write this down. We seek V star as Ci e power mi y, okay. Substitute in the sixth order ODE and get 6 roots for the mi. That is I will have m1 m2 up to m6, okay and you know that V star is going to be given by C1 e power m1 y plus C2 e power m2 y plus etc. up to C6 e power m6 y. What we do is we use the 6 boundary conditions and recast these as a matrix which multiplies C1 C2 up to C6 equal to 0. The right hand side is going to be 0 because remember my boundary conditions are all homogeneous. These equations are going to be linear in the C1, C2, C3, C4 because my boundary conditions were linear, okay. So I can actually write this as a matrix multiplied by a vector and remember the elements here are going to be functions of depend on Rayleigh number and alpha star. So you would get a nonzero solution if the determinant of this matrix is 0. So as you can see there are 2 unknowns. One is the wave number, dimensionless wave number and the Rayleigh number. So what you would do is you would actually fix alpha star at some value and find the Rayleigh number for which the determinant is 0, okay. That is the idea and you would do this for different alpha stars. You will do it for different wave numbers and you would get different values of Rayleigh number and you would make a plot of this curve. So for every wave number you would know what the Rayleigh number is for which you have a nonzero solution, okay. Now you would have to possibly write a small computer code to actually do this problem in MATLAB. But what I like to do now is refer you to this book by Subramanian Chandrasekhar, Hydrodynamic Hydromagnetic Instabilities and the primary reason for me to refer you to this book is for you to see that how about 100 years back when scientists did not have access to the kind of software tools that you have access to, they were still able to solve these problems. So reading this particular section on the Rayleigh-Binard problem in Chandrasekhar, you will see how we use this arguments like symmetry, okay and tries to answer this question and that I have just talked about, how to go about finding a nonzero solution, okay. What I want to do is I want to tell you that we can still get some insight about this problem and get a feel for how this curve, how the Rayleigh number depends on the wave number by solving a hypothetical problem, okay. And so at the beginning when I started I wanted to tell you that I would told you that we would possibly get an analytical solution. So I am not going to disappoint you, I am going to still go about getting an analytical solution but then it will not be for the problem that we were talking about where we had the fluid between 2 solid walls, 2 rigid walls. Imagine now a hypothetical situation where you have a layer of fluid situation where let us say you have a layer of liquid which is suspended in the atmosphere. This is exactly what we had last time, only thing is in the sense that I have a liquid film which is extending from 0 to capital H or in dimensionless from 0 to 1. Instead of having a solid wall I have a gas-liquid interface here, I have a gas-liquid interface here. Remember at a gas-liquid interface the boundary condition is that of the shear stress being 0, okay. So at the gas-liquid interface we have the shear stress vanishing. I am going to keep my life simple and I am going to restrict myself to the case where the interface is going to remain flat, the interface does not deform, okay. So now since the interface does not deform I am going to still have V star equal to 0, okay. So since the interface does not deform V star equals 0 but now the shear stress is 0, you do not have a no-slip boundary condition where I put u as 0 but you would now have du star by dy equals 0, okay. So this is the 0 shear stress condition and remember what I want to do is I want to convert this condition on u to the condition on velocity V star. So what we do is we use the equation of continuity du star by dx plus dv star by dy equal to 0, that is my equation of continuity. I can use this condition only if I want to differentiate this with respect to y. So if I differentiate this with respect to y, I would get d square u star by dx dy plus d square v star, you know and really should be a bit more careful. I should just leave this as u and v by dy square equal to 0. So this is my 0 shear stress condition on the actual velocity and also on the perturbation and this is my equation of continuity and all I am trying to tell you is this condition of the 0 shear stress this is going to be true for all x, du by dy is 0 for all x. So du by dy equals 0 for all x, therefore d square u by dx dy equals 0 and which means this condition implies that the second derivative of v with respect to y is 0. So this implies that d square v is 0. Remember for the rigid wall I got the first derivative equal to 0. Now I am getting the second derivative as being equal to 0 okay. So this hypothetical problem results in v star being 0, the second derivative being 0 and also the other boundary condition which is v star squared minus alpha star squared whole squared v star being 0. So this condition remember comes from my temperature equation. My temperature equation is such that my temperature is fixed. So my temperature perturbation is 0 at the lower wall, temperature perturbation is 0 at the upper wall and therefore that still remains the same and therefore this boundary condition remains the same. Only thing is the change from the solid rigid wall to a gas liquid interface has resulted in the change from the first derivative to the second derivative okay. Now what I would like to do is come back to the dimensionless form of the equation which is d star squared minus alpha star squared whole cube equals minus Rayleigh number times alpha star squared v star okay subject to v star equals and this is true at y equals 0 and h. I want to expand this and I am going to get d star power 4 of v star minus 2 alpha star squared d star squared v star plus alpha star power 4 v star equal to 0. Now since v star and the second derivative are both 0, this equation simplifies to this equation and it further simplifies to the fourth derivative being 0 because this is 0 and this is 0 okay and so what I have is the fourth derivative equal to 0. Now therefore what we have now is a problem wherein my v star is a function which satisfies all my even derivatives being 0, the function the even derivative of the second derivative and the fourth derivative are 0 okay and remember these are going to be 0 at 0 and 1. So these boundary conditions are satisfied at 0, 1 okay. I mean since I made a dimensionless I have got to write this at 0, 1 not at 0, h okay. So these are the dimensionless version. Now can you think of a function which actually has this property. So let me give you a clue. This is a trigonometric function and the function which satisfies this is the sinusoidal function sin n pi y. Clearly a n equals sin n pi y satisfies the boundary condition v star is 0. The second derivative gives me the same thing, something multiplied by sin. The fourth derivative gives me this. So automatically this function satisfies the boundary conditions but we want a function which satisfies not only the boundary conditions but also we satisfies the differential equation. So I want to make sure that this satisfies the differential equation okay. So this satisfies all 6 boundary conditions. So I do not worry about the boundary conditions but in order to satisfy my differential equation I am going to substitute this functional form into my differential equation here. And remember this is actually sin n pi y star because I made a dimensionless okay. When I do this I am going to differentiate it 2 times. I am going to get n square pi square with the negative value, negative sin. So I get minus n square pi square minus alpha star square whole cube. Instead of v star I would get, I have already substituted this. So I am going to get a n sin n pi y equals minus Rayleigh number times alpha star squared times a n sin n pi y. All I have done is substituted this functional form in my 6th order differential equation, differentiating this 2 times gives me n square pi square and this is whole cube and this is what I get. Clearly we want a non-zero a n sin n pi y okay. And I can take out this negative outside. I will get negative of this thing whole cube okay. And so this basically gives me the condition Rayleigh number is of the form n square pi square plus alpha square whole cube divided by alpha star squared okay. So when I was talking about the problem with the 6 constants and the determinant and I told you that you get an expression for Rayleigh number as a function of alpha star. We were not able to do it because it involved possibly writing a computer program. But this hypothetical problem gives me explicitly what the dependency is of Rayleigh number on alpha star squared okay. And clearly it also depends upon n. What we will do now is make a plot of this particular function. I am going to plot Rayleigh number versus alpha square. You realize that as alpha star tends to 0, this function is going to go to infinity is going to be unbounded because of the denominator. And so for low values of alpha star, this is going to be negligible compared to this. This is low and that is going to make my Rayleigh number go to infinity. For very large values of alpha star, this is going to dominate over this. I can neglect this. I will have alpha star per 6 divided by alpha star squared. So it goes as alpha star per 4 and again it is going to go to infinity. You also realize that this is all going to be positive. So I have a curve which is going to be confined only to the first quadrant and which has a shape of this kind okay. Point I am trying to make here is that this curve is going to have a shape of this kind. And I am going to draw this for different values of n. I will have a curve like this for n equal to 1. I would have a similar curve for n equals 2. Clearly the curve for n equals 2 is going to be having a magnitude or larger for Rayleigh number for every value of alpha squared because of the way the n is appearing here. So the point I am trying to make here is that the n equals 1 curve is the one which is lowest okay. And what we want to do is we want you to realize that along this curve remember the growth rate is 0 because that was the condition which we had imposed on getting the solution to the equation when we were simplifying. So the growth rate of the curve, sigma, the growth rate of the disturbances is 0 along this curve okay. So let me try to tell you what we have done. I started solving a real problem of a fluid between 2 rigid walls and I said look I have to do it numerically and you have to do it numerically in the sense you have to solve this problem numerically and that is going to be one of the assignments. But I do not want to give up, I want to explain to you something about the physics of this problem and to allow me to do that I have gone to this hypothetical problem where my boundary condition now has changed from a no-slip boundary condition to a 0 shear stress boundary condition. This helps me get all even derivative 0, helps me guess a solution to my 6 order equation. The solution I have guessed is the sinusoidal curve okay and then I have substituted the sinusoidal curve and got into a situation where my boundary conditions are satisfied and I just want to make sure the differential equation is satisfied and that tells me how the Rayleigh number depends on alpha star squared. What we want to do is find out for n equals 1, find the point at which this minimum occurs and I am going to leave this as an exercise for you to calculate the value of alpha star at which you have the minimum and the corresponding value of the Rayleigh number. But I am going to tell you what the corresponding value of the Rayleigh number is and this value of the Rayleigh number is 657 approximately and this is remember for the hypothetical problem. So far we have not bothered about the alpha, the wave number of the disturbance. It would appear, we have not told you what this wave number is okay. We have just done the problem for a general arbitrary wave number and that is how this analysis has gone. Now how do I know or how does the system decide what the wave number is of the disturbance? That is the question and by looking at this picture we can get some insight. When you give a perturbation to a system, this perturbation is going to be composed of different wave numbers. It is going to be an arbitrary perturbation. You cannot in a real situation impose a perturbation of a fixed wave number in general. So you would give an arbitrary perturbation and this arbitrary perturbation can be decomposed using something like a Fourier series analysis into different modes. So those different modes corresponds to the sin alpha x. So when I have an arbitrary perturbation with the function of x, I would then say this has some component sin alpha 1x, some component sin alpha 2x. This is like decomposing a vector into a set of basis vector components okay. Now what we are asking is I want to find out how does each wave number behave? How does the disturbance in each wave number behave? Clearly I have told you along this curve sigma is 0 which means this curve is dividing the space into 2 regions. On one side of the curve sigma is going to be negative and on the other side sigma is going to be positive. Now we can use common sense to figure out where sigma is going to be negative. Clearly for low values of Rayleigh number, this corresponds to low values of the temperature gradient or the temperature difference. I expect the system to be stable. So if the system is going to be stable, this corresponds to sigma being negative and the region above this curve is going to correspond to sigma being positive. Now this curve therefore divides this into 2 portions where sigma is negative, 1 where sigma is greater than 0. Now if the Rayleigh number is sufficiently small, it turns out that the growth rate sigma is negative for all the wave numbers which means if I give an arbitrary disturbance, if I want to decompose it into different wave numbers into components of different wave numbers, each of them is going to have sigma negative. So in this region the system is stable. So for Rayleigh number less than 657 and all wave numbers have sigma negative. So system is stable. For Rayleigh number which is just greater than 657, so just slightly above this line, you see that if I give an arbitrary disturbance, I am going to resolve it into again different wave numbers. Wave numbers on this side are stable. But the wave numbers here are unstable. So you would therefore see in your real system the disturbance which grows and that is going to correspond to a disturbance which has a wave number closer to the value where this minimum is occurring. So the disturbance which grows corresponds to the alpha star of the minimum, the minimum value of alpha star. And this one remember is the one which is going to have the fastest growth rate because close to this sigma is 0. So if you have a Rayleigh number given by this value corresponding to this alpha star, this is farthest from this boundary. So the sigma value is going to be largest. As you come closer to this boundary, sigma is going to be 0. And so this disturbance with this wave number is going to be the one which is going to be farthest. So what I am saying is the one disturbance which is going to manifest itself in a real situation is going to correspond to the one with this alpha star. And this is the one which the system is going to decide by itself. So the non-linear interactions of the system tell you that the wave number which you are going to observe when it becomes unstable is going to be given by this alpha star. We will see more of this in the next class where I will discuss this in more detail. Thank you.