 Yeah, so welcome back to the final session today. We will continue with our discussion on the differential analysis and the final governing equation that we will be working right now with is the energy equation. So, we had talked about the energy equation, the formulation of the energy equation before the lunch break. So, what we are going to do is we are going to generate the differential form of the energy equation from our integral form under the simplifying assumptions that the flow will be treated as if it is inviscid in the sense that the work done by the viscous forces will be neglected altogether. We will not assume any sources in the flow and also we will neglect the body forces and finally the heat transfer into the CV would be assumed purely by conduction. So, with this let me go ahead. What we need to do now is we realize that this source term is going to be 0 because we have assumed that and we need to come up with some appropriate expressions for the heat transfer and the work done on the material contained in the CV. So, what I have shown here is a section of the control surface on the top here. So, let us focus on the top part right now and I have shown the unit outward normal for that elemental surface. I have also shown the heat flux vector which is again remember that we are going to treat this heat transfer only by conduction. So, this heat flux vector which is a heat flow per unit area is pointing in general in some other direction and it is pointing away from the surface. So, in order to calculate the conduction heat transfer into the control volume what we do is that we form a dot product of the heat flux vector with the unit outward normal that will basically provide us a component of the heat flux vector that is normal to the surface and that is the component that will actually carry out the heat transfer across the surface. Just like the normal component of the velocity was carrying the mass flow rate across the surface, the normal component of the heat flux vector will actually carry out the heat transfer across the surface. So, this q with two primes vector dotted with an area element with a vector specification using the unit normal outward normal and then integrate the entire thing over the control surface that is surrounding our control volume. However, since we are interested in the heat transfer into the control volume that is the way the balance statement was formed and since the heat flux vector was pointing away from the surface we add a negative sign to make sure that we are calculating the heat flow into the control volume. Now here let me point out that I could have taken this heat flux vector pointing in the opposite direction which is then what we want in some sense the heat flux going into the control volume in which case you need not have this negative sign obviously. However, keep in mind that in that situation the angle between the heat flux vector and the unit normal would be greater than 90 degree so that this dot product will anyway provide a negative sign. So, since we have chosen to do it in this fashion where the heat flux vector is going away from the control volume I am incorporating a negative sign and since the heat transfer is by conduction I employ the Fourier's law of heat conduction to represent the heat flux vector so that the heat flux vector is represented as minus the conductivity of the fluid material multiplied by the gradient of temperature. So, minus k times gradient of temperature is equal to the heat flux vector using the Fourier's law from heat conduction and then the minus and minus sign will cancel that is one thing we still have an area integral which we convert now into volume integral using our standard divergence theorem. So the vector function which is getting dotted with the unit vector here what we do with the vector function as usual we take the divergence of the vector function and integrate that over the control volume in the conversion of area integral into volume integral and that is what has been shown as the last part here. So finally, what we generate is an expression which is an integral expression for the net conduction heat transfer into the control volume on a integral basis in the sense that we are representing this as a volume integral which involves divergence of k multiplied by gradient of temperature that is integrated over the entire control volume. The reason we are again after integral representations over the control volume is because if you go back the two integrals on the left hand side of our balance statement have already been converted into a volume integral. So we want to generate volume integral type representation for both this q dot and w dot term. So that is as far as the q dot due to conduction is concerned as far as the work term is concerned since we have decided to neglect body forces and the viscous forces all together the only force that remains is the pressure force which will be basically acting on the surface of the control volume. Now again here I have shown an element of the control surface with a unit outward normal as usual pressure being a compressive stress it is acting in exactly the opposite direction as far as the unit outward normal is concerned. So pressure is acting normally but in a compressive manner on the control volume or the material contained within the control volume. The fluid velocity at the control surface is in general pointed in any direction. So therefore what we want is the work done here on the control volume because that is how in those terms this integral statement was formulated that we wanted heat transfer into the control volume due to conduction and we wanted the work done on the control volume. So since pressure acts on the control volume we are simply going to multiply pressure which has a negative sign because it is a compressive stress multiply that by the normal component of the velocity multiply that by the area element because what we are essentially generating is an expression such as force multiplied by velocity which will give us the rate of work done. So force here is due to pressure acting over an area element dA and that multiplied by the normal component of the velocity will give you the work done by this pressure on the control volume. So the normal component of the velocity as usual is obtained through the dot product of the fluid velocity which is in general oriented at some angle with respect to the unit outward normal and this entire area integral then is converted using the divergence theorem by realizing that we then calculate the divergence of the function p times v. So that is what is shown here and then integrate that over the entire control volume the minus sign is simply carried. So that is all that is what I suggest is that when we formulate such expressions go step by step and make sure that you understand each step in terms of what is getting done. So just like what I said we are interested here in rate of work done. Rate of work done on the control volume is force acting on the control volume multiplied by a velocity component along the force that is how our standard expression for the rate of work done is from mechanics and that is how this term has been calculated as I just explained a couple of minutes back. So p multiplied by dA gives you the force due to pressure minus sign because it is a compressive force acting on the control volume v suffix n is the velocity of the fluid along the direction of the force which happens to be along the unit normal direction in the case of pressure which is calculated as a dot product of the velocity of the fluid with the unit outward normal and then we convert this entire thing into an area sorry a volume integral using the divergence theorem. So that is how the process goes in steps. So now if you go back what we have is the source term was essentially 0 because we said that we will assume no source and we have formulated the expressions for q dot and w dot. So putting all those together everything now is an expression over a control volume integral over control volume. So we can all club together utilize the standard assumption that or standard argument I should say that for an arbitrary control volume the integrand integrated over the control volume will become 0 implies that the integrand itself is going to be 0. So in order to create such a statement you will take the two terms on the right hand side on to the left hand side combine all of them together under one integral sign so that the right hand side will be equal to 0 and then you realize that the integrand is simply what I have boxed in the second equation. The only change that I have done from this equation to the boxed equation is that I am assuming that the thermal conductivity for the fluid medium is a constant so that it comes out of this divergence operation and del dot del operating on temperature t will provide you the Laplacian of t or del squared of t. So k times del squared t is the net heat transfer due to conduction on a per unit volume basis into the control volume. Remember all differential equations that we are generating are on a per unit volume basis so this is no exception either. This divergence of e multiplied by velocity is the work term the rate of work done term on the control volume again on a per unit volume basis. The two terms on the right on the left hand side are basically the accumulation of energy in the control volume and the net rate of outflow of energy from the control volume everything again on a per unit volume basis that is what precisely our original balance statement said we have simply converted it into a differential form which by definition is always on a per unit volume basis. Now remember that we also neglected body forces which meant that there was no requirement to include the potential energy because the only body force of interest that we normally take is the force due to gravity which will result in a change in the potential energy but since that entire body force is neglected the total energy is simply the thermodynamic internal energy plus the specific kinetic energy. So this small e is then i which is the thermodynamic internal energy plus v squared over 2 which is specific kinetic energy. So this is what we have right now as our most general energy equation of course under all these assumptions for the total energy. Now there are a couple of standard manipulation that are done to reduce this equation in some other form so let us look at that. What is done is we take the continuity equation which is written in this square brackets in the conservative form as we say and multiply that by the specific total energy so e times the continuity equation since continuity equation is identically equal to 0 we are not really changing the value by adding sorry multiplying this by e then after multiplying by e what is done is the resulting expression is subtracted from the original equation that we obtained little bit earlier which is here in the box. So remember see what is going to happen nothing is going to change on the right hand side ok right hand side will still remain k times del square t minus divergence of p times v. On the other hand left hand side will actually change in the sense that the first term here will be e multiplied by partial derivative of rho with respect to t that will be subtracted from partial derivative with respect to t of rho e. So you open up this and subtract this what is going to happen is what we will show on the white board in a second. So we will end up cancelling one term and what we will be left is only rho multiplied by partial derivative with respect to t of the total energy as far as the second term is concerned similar operation can be operation can be done and if you see this is our rho multiplied by the surviving term partial derivative of e with respect to time. The other term which is divergence of rho e v let me try to do this minus. So basically you will see that you want to get rid of this. So accordingly you split this divergence the term inside the divergence e times rho v is what will be the splitting. So e times divergence of rho v plus rho times divergence of e v that is what the splitting of the first term is minus e times divergence of rho v so that first and third will go away and only the second term will be left rho times divergence of e v so let me just use a two dimensional form for the sake of brevity. So divergence of rho times e times v if you expand it in the Cartesian form it will be d dx of rho times u rho times e times u plus d dy of rho times e times v then rho times divergence of was the second term. So divergence of rho v in our Cartesian form we can write as d dx of rho u plus d dy of rho v so that this is e times d dx of rho u plus e times d dy of rho v. So now we are going to subtract the equation number two from equation number one that is what we were up to. So therefore 1 minus 2 will give you see we have d dx of rho times e times u so we will split this as e multiplied by rho u so what we have is e times d dx of rho u and therefore 1 minus 2 will provide rho u times d dx of e as the first term. Similarly the second term we will split this as rho v times e we already have here e times d dy of rho v so therefore after subtraction what we will get is rho v times d dy of e. This is written since you realize that rho is a constant rho times I will write it in particularly in this form u d dx plus v d dy operating on e of course everything is getting done in a two dimensional form however I hope you realize what is going on here and in doing so we are generating rho times v dot gradient which is what the circled term is operating on e and that is what we are going to verify has been written on the slide which is this term rho times v dot gradient of e. So as you can see in fact this served as a good example of how to carry out some of these manipulations if we get confused suddenly for a vector operation you can always convert everything slowly and surely in the Cartesian form and carry out the manipulation in the Cartesian form and then once the Cartesian form manipulation is complete you can go back to the vector form. I really hope that you followed what I was up to rather than looking at the vector form directly which I have written on the slide I have done it in a Cartesian form and then in that process I have made sure that this term shows up. So the right hand side as I said remains exactly the way it is k times del square t minus divergence of p v which is on the right hand side. So now we have two terms rho multiplied by partial derivative with respect to time of the total energy plus rho times v dot del e. So let me go back to the whiteboard again this is what we have on the left hand side and that equals k del square t minus divergence of p times v on the right hand side. Now look at the left hand side I write this as rho times partial derivative with respect to time plus v dot grad operating on E equal to this right hand side but what is this? This is nothing but the substantial derivative partial derivative plus convective derivative operating on E. So this is nothing but rho times substantial derivative of E equal to the same right hand side. So the important point here is the following let me go back to my slide rho times substantial derivative of the total energy is what has been brought out on the left hand side. Since total energy is equal to the internal energy plus specific kinetic energy I am writing this explicitly as small e equal to i plus v squared over 2 and the right hand side remains the same. Compare the equation on slide number 32 which is projected here and my highlighter is sitting on it right now with equation on slide number 33 again boxed and where my highlighter is standing right now. I hope that you realize what we have done here. We had the equation in a conservative form shown on slide number 32 with this manipulation that we just worked out we have converted it into a non-conservative form. In fact if you remember we had done this conversion of conservative form into a non-conservative form for both the differential mass conservation equation or the continuity equation and later the momentum equation as well. And that is precisely what we have done. So this present set of manipulations was just to show that you can convert the energy equation also which was in conservative form as we obtained from our balance statements earlier to a non-conservative form. So now what is getting done here is that we have the equation presently on our screen in the non-conservative form for the total energy which involves or includes I should say both the thermodynamic internal energy and the specific kinetic energy. So what is done now is that a separate equation is generated only for the kinetic energy. In order to do that what we do is we take the momentum equation and dot with the velocity. Now remember that the momentum equation should be such that we are going to exclude the viscous forces that was part of our assumption. So if you want to choose the appropriate form of this momentum equation it will be minus or without the viscous terms which means that we are going to choose the Euler's equation. Is that fine? Furthermore we are going to neglect the body forces also as part of our assumptions. So the Euler's equation let me go back to it for a minute here. The Euler's equation is what we are going to choose minus the body force term which is the last term here on the right hand side. So all we do is we will take rho multiplied by the substantial derivative of velocity equal to minus the gradient of pressure and that will be our momentum equation because we are assuming an inviscid flow and we are assuming that we are not taking any gravity term or the body force terms. So now we do the following manipulation. We take the momentum equation and dot that with the velocity v and in doing so we seem to be obtaining something like this. So let us try to work this also from our Cartesian point of view. Before that let me write this boxed equation for the total energy in the Cartesian form at least using the Cartesian form for the last term on the right hand side. So let me rewrite that because this is something important and it may help in general how we go about manipulating these equations and so on. I will maintain that k del square t term as it is as the net rate of heat transfer by conduction into the control volume that I want changed. On the other hand that minus divergence of pv I will write in the explicit Cartesian form. So if you expand it again I am using a two dimensional representation but if you expand this del dot pv in Cartesian form you will get d dx of p times u plus d dy of p times v and since this minus sign is sitting each of these terms will be minus. So this is our basic equation in the non-conservative form. Now let me look at this v dotted with momentum equation. So let me first write this as rho is a scalar so it can be operated or placed wherever you want minus v dot grad p. Rho times velocity vector v dotted with the substantial derivative of v is something that I am going to leave it to you show that this term turns out to be rho multiplied by substantial derivative of v squared over 2. So I will put a question mark here which means that we want to show this but I am going to leave it to you. On the other hand the right hand side which is somewhat simple which I will do is minus u times dp dx plus v times dp dy. So that is it. This is the second equation of interest and let me go back to my set of slides and I will point out that this is what we are achieving here. So I wrote that the left hand side turns out to be rho multiplied by the substantial derivative of the specific kinetic energy. So this manipulation you should be able to do. Again do it in Cartesian coordinate and see if you are getting it or not. It should be fairly doable. The right hand side I have explicitly written on the whiteboard in the form of a Cartesian component. Going back to the slide now what we do is we take the equation which is at the bottom and subtract it from this box equation. What will happen is you see that the top equation here has rho multiplied by the substantial derivative of i plus v squared over 2 from which you will subtract rho multiplied by the substantial derivative of v squared over 2. So what will be left is only rho times substantial derivative of i that will be left. k del square t will remain exactly the way it is on the right hand side whereas this del dot pv will actually cancel one term from here and if you want to go back to my whiteboard the term that I am going to highlight are the del dot pv term in the Cartesian form. And from these two we have to subtract these two. So please carry out this subtraction and you will see that what you are left with eventually is k del square t on the right hand side as it is minus p multiplied by the divergence of velocity. On the left hand side as we said what will be left out is only rho multiplied by the substantial derivative of the internal energy i that is what will be left out. And this is what is called as the differential thermal energy equation under the assumptions. Clearly our assumptions have been that we have neglected the viscous forces altogether, we have neglected body forces and what else we have neglected let me go back. We have no source in the flow and that is precisely what we have assumed. So under these assumptions we obtained first a total energy equation from the total energy equation. We subtracted something which is popularly called as the mechanical energy equation although I have not written down here explicitly. If you see fluid mechanics textbooks this rho multiplied by substantial derivative of the specific kinetic energy equal to whatever is called as the mechanical energy equation. So subtracting the mechanical energy equation from the total energy equation we generate what is called as the thermal energy equation. And further what we can do is we can invoke the definition of what is called as an enthalpy from thermodynamics. So enthalpy is basically the internal energy plus p multiplied by the specific volume or p divided by density. If you include or if you want to express the differential thermal energy equation using enthalpy you can easily convert the equation here to something called another form of differential thermal energy equation where what you end up generating is the substantial derivative of enthalpy. These are the final equations of our interest. Remember what we have done is we started as always with a balance statement in this case it was the energy balance. We made certain assumptions and employing certain mathematical tricks so to say we obtained our conservative form first. We did some more manipulations and obtained the non-conservative form for the total energy equation. From the non-conservative form of the total energy equation we subtracted the non-conservative form for the mechanical energy equation and we finally have obtained what is called as the thermal energy equation. Actually if you want to really take a few minutes and realize what is happening it will be very satisfying in the sense that what you see right now is a differential equation. So differential equations in our situations have been always obtained on a per unit volume basis. So this is no exception to that everything here is again a per unit volume basis. Just look at the form of this expression the thermal energy equation in particular that you are seeing. If you see it carefully you will realize that this looks exactly like the first law of thermodynamics as you remember from thermodynamics many of you have seen thermodynamics before. So in thermodynamics what we have is change in the energy expressed in terms of the heat transfer and the work transfer and that is precisely what this is. It is nothing but the first law of thermodynamics expressed for a fluid particle where I am now talking about the internal energy content of a given fluid particle. The reason because we are talking about a given fluid particle is since we are talking about a substantial derivative. So we are necessarily using a Lagrangian form. So as this fluid particle is moving in the fluid flow I am monitoring the change that it is experiencing in its internal energy. Those changes are occurring because there is a heat transfer into the particle which is given by this k times del square t term that is the heat transfer due to conduction into the fluid particle. Remember that the conduction heat transfer will occur whether it is a solid medium or whether it is a fluid medium as long as there exist temperature gradients there will always be a conduction heat transfer. So that is what this is and the final term on the right hand side is minus p times divergence of velocity should remind you of what is called as the p dv work from thermodynamics. This is actually no different from that. This is exactly the p dv work that you are used or at least many of you are used to seeing in thermodynamics. It is just that when it comes to fluid mechanics situation we are talking about everything on a rate basis and that del dot v or the divergence of velocity from kinematics is nothing but volumetric rate of strain and multiplied that by the pressure is essentially the p dv type work from thermodynamics. So finally what ends up happening is that you realize that you have essentially derived a special form of the first law of thermodynamics for a given fluid particle in this non-conservative expression. So same thing really in terms of the final equation is just that we have used enthalpy instead of internal energy. As the primary dependent variable on the left hand side that is about it. Other than that the basic interpretation from a physical point of view does not change at all. Here what we can say is that the p times dv or rather p dv work which was rather obvious in some sense in this expression is not as obvious in the second expression. Here what we see is that it turns up in the form of a substantial derivative of the pressure experienced by the particle. So here the way to interpret this is that as the particle is moving in the fluid essentially pressure forces are acting on it and assuming that so far we have not assumed that the situation is an incompressible one. So for a compressible flow the pressure forces are acting such that the volume of the fluid particle may reduce or expand. So it is undergoing a volumetric rate of strain because of which the pressure within the fluid particle will actually keep changing and that pressure change is represented through the substantial rate of change of pressure term. So fundamentally again there is no difference between the equation written out at the top which is a first law of thermodynamics statement for a fluid particle exactly the same thing for the bottom equation. Only thing is that we have chosen to write this in terms of enthalpy rather than internal energy. That is more or less it really we have completed the entire set of governing equations. So what I have done on the next slide is I have summarized these for a constant density, constant viscosity, constant thermal conductivity flows. So I have written these on this particular slide in vector form. Remember that we are dealing with constant density equation so that the continuity equation or the mass conservation equation simply reduces to divergence of velocity equal to 0. The Navier-Stokes equation is written both in non-conservative and conservative form. So if you want to choose the non-conservative expression you choose the leftmost which is this and the rightmost. If you want to use it in the conservative form you choose the middle part and the right part. Similarly the thermal energy equation is written for this situation which is constant density. So since it is a constant density situation and del dot v is equal to 0 going back this p times del dot v term will simply go away. And therefore what is left is only rho times the substantial derivative of internal energy which is our non-conservative statement equal to k del square t or the middle part and the right hand side will constitute the conservative expression. So these are written in vector form. What you should realize here is this is a system of equations which will be required to be solved if you are dealing with a constant density, constant viscosity, constant thermal conductivity flow where there is a heat transfer occurring through conduction. So how many equations are these really? This at the top is a scalar equation del dot v divergence of velocity is a scalar equation. It is only one. The momentum equations are three, one for each component of the velocity. Energy equation again is a scalar and here it is written in terms of the internal energy. So I have listed the unknowns which will be in general required to be found from this set of governing equations. These involve the three components of velocity or one vector velocity in general the pressure field, in general the temperature field and in general the internal energy field. So we have four equations and only three unknowns. So in order to complete the system what we say is that we express the internal energy in the form of a specific heat times the temperature where this specific heat will be essentially assumed to be known. So that is how we close this system so to say. Right now here I have written vector forms. Let me actually immediately show you the Cartesian 2D forms. So let me in fact project this for a second and then I would like to write this on the white board. Remember that this is for 2D form. Again for constant density, constant viscosity, constant thermal conductivity which is exactly what the previous set was. Only thing is now we are writing the Cartesian 2D form. So let me go to the white board and write the important equation the momentum equation. Of course let me write the continuity also in Cartesian form. So this is our continuity equation or the mass conservation equation for this constant density situation. I will write it in the non-conservative form in the x direction which explicitly written will read since I am doing a two dimensional formulation. Let me end it here and this is written in the x direction which will be then equal to minus dp dx plus mu times the Laplacian of the x component of the velocity. And for the purpose of completeness let me write this as well that there is a body force term again on a per unit volume basis. So our hope is that in this first week everyone is will get sufficiently used to deal with such equations. And we will do a few more manipulations of these equations tomorrow. This all these equations in some exact solutions form. But right now this is what you should note down. Again going back to my slide what I did on the white board was I simply wrote this row multiplied by substantial derivative of u explicitly everything else is already there. Two dimensions so u and v components and finally the energy equation. So this was written in Cartesian form. It is useful to know some of the other forms of the coordinate systems also. So in particular I am going to simply mention the same set of equations that we wrote in Cartesian forms earlier in this cylindrical coordinates. So the cylindrical coordinates are the radial distance, an azimuthal angle theta and an axial distance z. And corresponding to r theta z we have the three velocity components which I have chosen to represent using v suffix r, v suffix theta and v suffix z. Some useful relations from coordinate geometry are listed here. So the x coordinate in Cartesian system is equal to r multiplied by cosine theta, y is r multiplied by sine theta and z is same as z. The unit vectors in the radial direction and the theta direction for the plane polar coordinate system are written in terms of their counterparts in Cartesian coordinates. So i hat and j hat are our standard Cartesian unit vectors and i r and i theta are the unit vectors in radial and tangential or azimuthal direction. What I have also listed is an expression for the Laplacian and the expression for the substantial derivative directly without really deriving these. Some of the or more or less all of the standard good fluid mechanics textbooks will actually list the expressions for the equations that we have been talking about in cylindrical coordinates as well. And that is what I have done in this particular slide number 38. So here I have the same set of governing equations namely for constant density, constant viscosity and constant thermal conductivity flow written out in cylindrical coordinates. And typically what ends up happening is that these are very very long expressions usually. So I have chosen to use this del square for brevity. If you see the del squared operation is quite long, I wanted to avoid the avoid these equations being very very long and cluttered up. But you know the expression for del squared the Laplacian operator in cylindrical coordinate system which is given on the previous slide. Similarly, the expression for the substantial derivative is provided here. Using these we can we can utilize these equations also. What I am going to do tomorrow is that we will actually solve some of the so called exact solutions problem where solutions based on these governing equations are found. And one example I will take which will be in cylindrical polar coordinates where we will use some of these governing equations. Otherwise at the moment you can treat these as for your reference whenever you require to use a cylindrical coordinate system under the assumption of constant density, constant viscosity, constant thermal conductivity flows you can utilize this. So that basically ends the discussion on the differential analysis. Fundamentally what I want you to realize and understand is that the governing equations have now been derived in the differential form as well. The momentum equations which we are calling as the Navier-Stokes equations are derived for the situation of constant viscosity but you also have the form available with you in your notes for a compressible flow as well. In general what we are going to deal with will be equations for incompressible situation or constant density situation but for your reference these compressible flow equations are available as part of the derivation. The energy equation has been derived only under certain restrictive assumptions let us say. The specific important assumption that we have utilized is that the viscous forces are all together absent. So because of that there is one important term that comes in the energy equation called the viscous dissipation term which is actually going to be absent. However for most mechanical engineering applications which involve low speed flows this is a fairly standard assumption to neglect the viscous dissipation term and that is the framework within which we will work in this particular workshop. So what we will do is we will take certain simplified situation because the purpose is just to show how this non-dimensionalization is done and the purpose of doing this non-dimensionalization is to bring about the non-dimensional parameters of importance once you have chosen a certain set of governing equations. So the idea is that there is a certain flow that you are analyzing based on your knowledge or intuition or whatever you have decided to choose a certain set of governing equations. From the manipulation of these governing equations we will be able to automatically find out which governing parameters of non-dimensional nature are important for the specific fluid flow situation. So the standard way to proceed in this non-dimensionalization is that we define non-dimensional coordinates. So we choose here Cartesian system for the purpose of illustration. A non-dimensional x coordinate is defined as x coordinate divided by some characteristic length of the problem. So for example if you are dealing with a flow inside a pipe we know that the diameter of the pipe is taken as the characteristic length. If there is a external flow such as flow over an airfoil for example, the standard characteristic length is the chord length of the airfoil and so on. If there is an external flow over a bluff body like a cylinder let us say or a sphere again the diameter of the cylinder or the sphere is taken as the characteristic length. So each problem typically has some characteristic length and we choose to non-dimensionalize the x and y coordinates with this characteristic length. So going back to the slide in a similar manner the velocity components are non-dimensionalized using some characteristic velocity for the problem. So the characteristic velocity for the problem can be if you are talking about an internal flow in a pipe the bulk velocity or the mass average velocity. If it is an external flow then it can be typically a free stream velocity and so on. So some characteristic velocity will always be available and we will non-dimensionalize the velocities using that characteristic value. Similarly the pressure is non-dimensionalized. What I mean by non-dimensionalized is that this p star then is the ratio of p which is our pressure divided by the characteristic value of the pressure so that this ratio is non-dimensional. So all these starred quantities that you see are non-dimensionalized. x over x, x over l is a length scale over length scale, u over uc is a velocity scale over velocity scale and so on. When it comes to the temperature usually the temperature non-dimensionalization is done in terms of the ratio of certain temperature differences. So the denominator here for the temperature non-dimensionalization is a temperature difference between typically a surface temperature value and a characteristic temperature value. So what I mean by surface temperature value is let us say again you have a flow inside a pipe and you have some heat transfer going from the pipe into the flow. So those who are familiar with heat transfer will realize that we talk about a constant temperature boundary condition or a constant heat flux boundary condition corresponding to which there is some temperature on the surface. So that temperature on the surface minus again some characteristic value of the temperature which could be a bulk mean temperature for internal flows or a free stream temperature for external flows. So the characteristic difference is on the numerator and T minus the characteristic value of the temperature is on the numerator where T is any general temperature within the flow field. With this you will see what happens to the governing equations. So we choose the continuity equation for constant density or incompressible flow which is du dx plus dv dy and what we do is we employ the chain rule to replace this d dx in terms of the d dx star times dx star dx. So this is the standard chain rule application. So let me work this out on the whiteboard. So what we have is u star equal to 2 divided by the characteristic velocity and x star equal to x divided by the characteristic length. So this u then will be simply u star multiplied by uc and this differential dx star over dx will be simply equal to 1 over lc because star is equal to x over lc. So therefore the term du dx will get non-dimensionalized as uc which is a constant value. It is a characteristic constant value divided by lc. So this term and the derivative from dx star dx what is left is du star dx star. Going back to my slide you will realize that exactly the same factor uc over lc will come from the second term also so that the non-dimensional form of the energy equation after cancelling this uc over lc is essentially the same as what it was to begin with. Only thing is that all terms have been replaced with their non-dimensional counterparts. So there is nothing significant happening in the non-dimensionalization of this continuity equation. If you want to look at the momentum equation there is a little bit of algebra involved and what you do is you just go through this step by step. In fact on these slides I have worked out the complete algebra that is going to be required in this non-dimensionalization. So if you work out the entire algebra you will realize that the original momentum equation which was assumed to be steady flow by the way so that I think I forgot to mention that we will assume steady flow for the purpose of illustration. That is the reason that time derivative term in the momentum equation is missing here and remember that I have divided through by the density at the top of this equation. So then mu divided by rho is replaced as the kinematic viscosity where my highlighter is standing right now. Also going back a couple of slides just for the yeah here this is what we have normally the Laplacian of u. So this Laplacian of u we write this for the purpose of convenience of non-dimensionalization as an internal derivative with respect to x and then an outer derivative with respect to x. It is exactly the same thing but it is written for convenience of the procedure of non-dimensionalization and that is how the term on the right hand side here is written similar term for the y derivative as well. Having done this then you can actually perform the non-dimensionalization. For example let us see this d dx is replaced as d dx star dx star dx. One u is replaced as u star times u c other u is replaced as u star times u c other u is replaced as u star times u c and so on. So if you actually simplify this equation it just looks little bit cluttered but there is nothing really in here that is very troublesome. All that you need to do is collect all these u c's and this dx star over dx will give you one over lc so that you will realize that u c squared over lc is the common term that will come out from the left hand side and similarly some things will come out from the right hand side. In doing so eventually you will realize that there is a non-dimensional number or ratio that seems to be multiplying the non-dimensional pressure gradient. So the number that is multiplying the non-dimensional pressure gradient is this characteristic pressure divided by rho times u c squared. The non-dimensional ratio or the non-dimensional number that seems to multiply the viscous terms seems to be this mu which is the kinematic viscosity divided by u c times lc. So we will come to these numbers in a minute. This was done for the y momentum equation exactly the same thing we can do for sorry x momentum equation it was done. We can do the same thing for y momentum equation and we can choose to include this body force as minus g in the y momentum equation and if you carry out the procedure similar to that we will generate the same two numbers that you saw multiplying the non-dimensional pressure gradient and non-dimensional viscous term. Additionally we will generate one more term because we are including this g in the y momentum equation. And now I have written down here what these numbers are actually many of you will be familiar with these numbers. So this characteristic pressure divided by rho multiplied by the square of the characteristic velocity is what is called as an Euler number. Characteristic velocity multiplied by characteristic length divided by the kinematic viscosity is popularly what is called as the Reynolds number and this what you see as the last term here is essentially 1 over a Froude number square. So the Froude number is the characteristic velocity divided by square root of g times the characteristic length. So what has been done here is that through these non-dimensionalization processes what we are able to see is once we choose a certain set of governing equations immediately the corresponding non-dimensional parameters of the interest for the particular problem that you are analyzing are coming out through this non-dimensionalization. Many of you will be familiar with this Buckingham's pi theorem for example. In Buckingham's pi theorem we perform essentially a similar operation what we do is we list all sorts of parameters that we think are important for a problem and then using the pi theorem you find out which non-dimensional parameters of interest are of significance. So this procedure of non-dimensionalizing the governing equations is an alternative way of finding the important non-dimensional parameters of interest for the problem. So if by any chance we had decided that there was no requirement of including this body force in terms of gravity we will say that the Froude number will not generate at all as part of the non-dimensionalization because there will be no g term at all in the governing equations. In that sense we will say that we have decided that gravity is not really influencing the problem at hand so the only two non-dimensional parameters of interest in that case would be the Euler number and the Reynolds number. So typically always more or less you will have pressure gradient and the viscous terms present in your analysis so that the Reynolds number and in certain high speed situations the Euler number is also an important parameter. Usually for mechanical engineering applications it is the Reynolds number that is the most important parameter. The energy equation is written next and the energy equation is actually slightly more cumbersome because if you realize the non-dimensional temperature is defined in terms of the ratio of two temperature differences so appropriately you have to carry out the non-dimensionalization. I actually have worked out everything here and what I will suggest is that just go through this algebra we can come back to this tomorrow morning but right now what I will say is that if you carry out this algebra you will realize that the parameter of interest having chosen the form of the energy equation in this particular form. So what we have is an energy term because of the inflow and outflow of flow and the conduction heat transfer only what you will realize is that the product of Reynolds number and Prandtl number turns out to be the non-dimensional parameter of interest. So as perhaps many of you might know this is I think termed as the Peclet number, the product of Reynolds and Prandtl number if I remember my heat transfer correctly let me verify this. So the audience here is saying that that is correct. So rather than writing this as a product of Reynolds number and Prandtl number I can write this as the Peclet number directly which is essentially a number which describes the strength of convection to the strength of diffusion. In this particular situation of the energy equation the strength of heat convection to the strength of heat diffusion which is happen heat diffusion is happening because of the conduction otherwise the flow of the fluid is carrying out the heat convection. So that heat convection to the heat diffusion will be given by the Peclet number. So the purpose of these non-dimensionalization is that you have decided to analyze a certain fluid flow situation, you end up choosing a certain set of governing equations and from those governing equations you can immediately figure out which non-dimensional parameters of interest are going to be your governing parameters of interest at least as far as the physical parameters are concerned. There may be some geometric parameters as well which anyway are taken in through these non-dimensional lengths and so on. But other than that as far as what sorts of forces are important in your situation at least for the fluid mechanics part you will realize that you will end up generating numbers such as this Reynolds number, Froude number and Euler number. From the CFD point of view it is sometimes useful to solve the governing equations in a numerical manner in their non-dimensional form. For some standard prototypical situations like say flow in a channel or flow past a circular cylinder or things like those which are fairly common occurrences you can actually employ the equations in the non-dimensional form. What ends up happening in that case is that if you are thinking about flow across a cylinder then non-dimensional solution will provide flow across all cylinders. As long as you are able to replicate the Reynolds number let us say you are essentially in a position to generate data for a 5 meter diameter cylinder or 5 centimeter diameter cylinder. So, in situations where the geometries are fairly straightforward sometimes the CFD analyst choose to utilize this non-dimensional form and that is precisely one of the reasons why we wanted to introduce this also. But the main purpose of non-dimensionalization is to bring about the governing parameters of interest. So, with this we are at the end of our differential analysis. So, with this we will stop for today. Thanks for attending.