 Welcome back everyone. Let us get on with our discussion on the differential analysis. So the slide that is projected right now on the screen was the last one that we were discussing when we break. So what we have here is the so called cohesive equations which are the differential momentum equations and as you can see the leftmost side or the leftmost part of the equation is the non-conservative representation. The middle part of the equation is the conservative representation. These are essentially equivalent and the right hand side of the part is simply telling the net force acting on the fluid particle and the net force calculation has been done based on the slide number 18 where various types of forces that are acting on the fluid element are shown. So having summed this appropriately you obtain the Cauchy's equations of motion. So now what is left to complete this differential momentum analysis is that we would like to connect these stress components such as sigma xx tau yx etc to the rates of strain through the Newton's law of viscosity. Once we do that we will be done with this part. So let us talk about it. These are what are called as constitutive relations for Newtonian fluids and essentially these are the relations between stress components and strain rates. Remember we talked about this on the first day when we obtained an expression for the viscosity dynamic viscosity and it was done in a very special sort of a flow situation which was a simple parallel shear flow. It was only a one-dimensional flow where the only non-zero component of the velocity was the axial velocity u which was a function of only y and the only shear stress that was acting was in that case considered as only tau, technically it is tau yx if you really want to look at it carefully and there we said that tau is equal to mu times du by dy that was the expression that we used to describe a Newtonian fluid. Now we are talking about extending that analysis to a general multi-dimensional situation and that analysis was performed by Stokes way back in 1846. So the Stokes relations have been around for more than 150 years. What we are going to do is we are not going to get into each and every detail of the Stokes' analysis. The reason is because it is a fairly tedious process to work out the entire proof of the relationships between stress and strain rates. Usually that is something that we do in an advanced fluid dynamics course. In this present material what we are going to do is we are going to simply list those relations and we will utilize those relations in the Cauchy's equations to complete our analysis. So I am just listing here in the slide number 21 under what assumptions Stokes arrived at these constitutive relations. The first assumption was of course that the Newton's law of viscosity was assumed namely that the stress component is directly proportional to strain rates. The next assumption was that the fluid was considered to be isotropic which means that it was a directionally independent behavior assumed for the fluid. No matter what direction you are looking at the fluid would behave in the same manner and it turns out that most fluids in practice are do conform to such behavior. So it is not a very seriously restrictive assumption and also the fact that in absence of strain rates which means that in absence of a flow the stresses should reduce to a corresponding static fluid condition. So under these assumptions Stokes arrived at the relations and the first set of relations is projected on slide number 21 right now and these are essentially relations between the shear stresses and the corresponding strain rates. The constant of proportionality is of course the dynamic viscosity and so tau xy equal to tau yx which we showed a few slides earlier is equal to mu times dv dx plus du dy is one such relation and similarly for the remaining three. If you look at the first relation since we are now dealing with multidimensional situation this additional factor of dv dx is coming in. If we were dealing with only that simple parallel shear flow as we had discussed on day number 1 we would only have this du dy multiply that by the viscosity mu and you will obtain your shear stress. So now these are what are called as generalized Newtonian relations if you want which will take into account multidimensionality of the flow. So this is as far as the shear stresses are concerned remember there were 6 shear stresses of course occurring in pairs so there are only 3 independent numbers so to say and then there are 3 normal stresses as well. So it turns out that the normal stress in a moving fluid is getting 2 contributions one comes from the pressure as we would expect but there is a viscous contribution in general as well and therefore what I have written is that a normal stress in general is written as minus p remember that pressure is compressive so that minus sign is assigned to take care of that compressive nature of the pressure plus a viscous contribution. Now this viscous contribution is something that Stokes derived through a series of manipulations and we are not going to get into those details. Let me just try to establish an analogy between this viscous contribution and what we have in case of an elastic solid as the normal stress in an elastic solid. So those who are familiar with again solid mechanics or strength of materials in a reasonable detail will probably remember that there is a generalized Hooke's law which will be applicable for an elastic solid in multi-dimensions and that generalized Hooke's law for an elastic solid is written in this form as is shown right now on the board where the normal stress in an elastic solid is expressed in terms of shear modulus times the strain remember in solids we have strains and no strain rates in fluids we have strain rates and then let us say another constant where there is this Poisson's ratio which I have shown as eta multiplied by the summation of all normal strains. So what ends up happening in a fluid is that essentially an analogous relation is obtained for this viscous contribution to the normal stress. The only difference being that of course the constants are different in a fluid but the main difference is that instead of strains as we have in solids we have strain rates. So with that analogy I will write the expression for a normal stress due to viscous action in a Newtonian fluid as 2 times mu times the strain rate in the x direction plus another constant which is denoted as lambda and which is popularly called as the second coefficient of viscosity multiplying the summation of the strain rates in all 3 directions. So if I go back and look at the expression for the normal stress in an elastic solid you can see that there is a clear analogy between the expression here and the expression for the fluid. So again please keep in mind that we are not deriving these relations from the first principles. Those exercises are available in advanced level fluid mechanics books. Normally we do those in an advanced level fluid mechanics class. At present this is sufficient for us to proceed because our purpose really is to get into the CFD side knowing the governing equations and to complete the derivation of governing equations we essentially need the expressions that will relate stresses to the strain rates. We have already seen the relations for shear stress and now is the set of relations for the normal stresses. So the normal stress finally in the x, y and z direction will be given in terms of these 3 relations which are called relations number 2 minus the pressure plus 2 times mu. Now the strain rate from our kinematics discussion yesterday if you will remember in the x direction is simply the x derivative of the x component of the velocity. Epsilon xx dot is simply du dx which is out here. Similarly epsilon yy dot is y derivative of y component of velocity and the strain rate in the z direction is z derivative of z component of the velocity. And if you remember obviously that the 3 strain rates add to what we call the divergence of velocity du dx plus dv dy plus dw dz. So therefore this normal stress in a Newtonian fluid is written as minus p plus 2 mu times du dx for the x direction plus lambda which is what is called as the second coefficient of viscosity multiplied by the divergence of velocity and analogously the 2 other normal viscous stresses are also added to minus p to obtain the normal stresses rather in a fluid. So the thing that we should note here is that if there is no flow and therefore no velocity gradients what remains is only sigma is equal to minus p. There are no shear stresses because all velocity components being 0 there will be no shear stress at all. So therefore this set of relations then automatically reduces to a static fluid condition where the only normal stress in a static fluid is equal to the negative of the pressure. So that way it is consistent. So what is done now is having noted remember noted we have not derived these relations having noted relation set number 1 and relation set number 2 which will relate the normal stresses to the strain rates and shear stresses to the strain rates. We go back and substitute those in these Cauchy's equations on the right hand side wherever you see either sigma xx or sigma zz or tau xz or tau zx or whatever else. This is again a fair amount of algebra nothing difficult but reasonable amount of algebra and if you want to perform it I would suggest that you go ahead and do it. However before that let us point out one more peculiarity of these relations. So relations 1 and 2 which we just looked at are what are called as the generalized Newton law of viscosity or also constitutive relations for Newtonian fluids. By these what we mean is that they are simply relating the stress components to the strain rates. Now going back to relations 2 which are for the normal stresses if you add these what you will get is sigma xx plus sigma yy plus sigma zz p, p, p and p so 3 times p with a minus sign. The divergence of velocity is common so it will be 3 lambda multiplied by divergence velocity and if you add these 2 or add these 3 2 mu is a constant du dx plus dv dy plus dw dz is another divergence of velocity. If you add these 3 relations what you get is sigma xx plus sigma yy plus sigma zz equal to minus 3p as I said the divergence of velocity which is essentially the addition of the strain rates will be multiplying a factor of 2 mu plus 3 lambda. Now it is convenient and many times a practice to define what is called as a mechanical pressure which is defined as the negative of the average of the 3 normal stresses. So the mechanical pressure by definition remember is sigma xx plus sigma yy plus sigma zz divided by 3 and take the negative of it. This is by definition an artificial quantity defined in case of fluids. So therefore with this definition we have the addition of these 3 normal stresses equal to minus 3 times that so called mechanical pressure which is denoted by p bar and if you substitute this on the left hand side here and divide the entire equation by 3 you will realize that this so called mechanical pressure is equal to the pressure as we know from thermodynamics which has what has been going on in the analysis plus this additional factor or minus additional whichever way you want to look at it. So it turns out that the thermodynamic pressure and the so called mechanical pressure the way it is defined is in general different because there is a factor here. If you look at it carefully the factor involves divergence of velocity. So for incompressible flow since we know that the divergence of velocity is identically equal to 0 this factor will be exactly 0 and in case of incompressible flow you will have the mechanical pressure equal to the thermodynamic pressure as we know. For general compressible flow this need not be the case. However for simplicity it was assumed by Stokes that this 2 mu over 3 plus lambda is simply equal to 0 thereby always getting rid of this factor and usually we do not then differentiate between this mechanical pressure and thermodynamic pressure. Remember there is some confusion always about this mechanical pressure and thermodynamic pressure. So the thermodynamic pressure as we know is essentially what we call is static pressure in fluid dynamics. The static pressure is something that you measure with the help of a probe if you are moving along with the fluid so that there is no relative velocity between the probe and the fluid. The mechanical pressure on the other hand is an artificially defined quantity which is defined as the minus of the average of these 3 normal stresses. It turns out that for incompressible flow both are identical so we do not have to really bother about it. For compressible flow where ideally this divergence of velocity is not really equal to 0 technically these 2 guys are different. However for simplicity Stokes assumed that this 2 thirds mu plus lambda is equal to 0 and thereby obtaining that the mechanical pressure is equal to thermodynamic pressure. So we do not have to really keep talking about these 2 different quantities all the time and also then as a result lambda is equal to minus to 2 thirds mu. Now it turns out that although apparently Stokes assumed this essentially not with lot of background theoretical background it turns out that even for most compressible situations this assumption actually is a really really good one. The only regions where it seems to fail is if you want to look at a structure of a shock wave for example as it occurs in case of a compressible flow. However other than such extremely isolated situations this Stokes's postulate is found to be very very good and therefore we never have to really talk about a difference between mechanical pressure and this thermodynamic pressure. We always talk about thermodynamic pressure and therefore with this knowledge what we do is as I told earlier take relations number 2 and relations number 1 substitute them in Stokes equation on the right hand side here perform a reasonable amount of algebra and finally in doing so you generate the so called Navier Stokes equations of motion. So what I have done here is that I have written down the Navier Stokes equation which is essentially a momentum equation for a Newtonian fluid remember that Cauchy's equations are the differential momentum equations written only in terms of the stress components we have not really discussed how these stress components are behaving so far. So the fundamental form of the momentum equations in terms of the stress components is what is called as the Cauchy's equations only when we include this constitutive relations set number 1 and set number 2 in the Cauchy's equations we call the resulting set Navier Stokes equation. So for the x direction then assuming a constant value of viscosity again for simplification purpose that is the only assumption by the way so far what we do is we obtain from the algebra that you perform on the Cauchy's equations for the x direction this Navier Stokes equation. So if you see again the left hand side is written in the non-conservative or the Lagrangian form the middle part is written in the conservative or the Eulerian part so it is up to you to choose either the non-conservative part or the conservative part the right hand side simply talks about the net force acting on the fluid element and what it shows is that there are essentially three types of forces that are acting on the fluid element one is given by the pressure gradient. So this is the net pressure gradient let us say in the x direction that is acting on the fluid the second and the third term on the right hand side are actually the viscous force is acting on the fluid element and we have not really assumed an incompressible flow and therefore the divergence of velocity remains here if we assume an incompressible flow this third term will go away and we will see that in a minute and remember that there was a body force term as well so what is left out of the body force is only that intensity which is essentially the body force per unit mass multiplied by the density so this is what is called popularly as the Navier-Stokes equation for a viscous with a constant viscosity compressible flow. So keep in mind that the Navier-Stokes equation is a momentum equation with the constitutive relations substituted in and the constitutive relations are derived only for a Newtonian fluid so the Navier-Stokes equation is a momentum equation for a Newtonian fluid I think there were a few questions asked earlier about non-Newtonian fluids as well. So it is quite likely that the non-Newtonian fluids will have essentially the same momentum equation up to here however the constitutive relations for the non-Newtonian fluids will not be these these are for Newtonian fluids if people have derived non-Newtonian fluid the constitutive relations which actually I am not aware of since I do not deal with those kinds of fluids you can always go and substitute those in the Cauchy's equations here where we see these stress components and then obtain the corresponding momentum equations for the non-Newtonian fluids. But when we talk about Navier-Stokes equation we are necessarily implying a Newtonian fluid so it is a momentum equation developed for a Newtonian fluid the x direction equation is written out here immediately I am writing this in the general vector form let me in fact write this left hand side on on the whiteboard separately so that we know exactly what we are talking about and this is equal to the right hand side which was minus the pressure gradient plus mu times this turns out to be the Laplacian of the u velocity this third term on the right hand side is non-zero only for a compressible flow the reason is because this is nothing but the divergence of the velocity which is non-zero for a compressible flow plus the final term of the body force expressed in terms of the intensity on a per unit mass basis. So this rho times substantial derivative of u has been explicitly written in terms of the local acceleration and the convective acceleration. So get used to these expressions and it is up to you really which form to utilize presently what we are writing here is of course the non-conservative form but in fact what happens is many times the non-conservative form is easy to manipulate when it comes to some of the analytical solutions that we will work out later perhaps most likely tomorrow morning. However when it comes to the CFD applications especially for finite volume techniques the conservative forms of equation turn out to be more useful. So either way what I have done let me switch back to the document the slides what I have written here is on the left hand side the non-conservative form which I had just written explicitly on the whiteboard a minute back the middle part is essentially the same as the left part. However written in conservative form and the right hand side is again the various types of forces acting in the fluid which comes out to be pressure forces the viscous forces these two terms and the body force. Now again please remember that this is a differential equation so as with all differential equations this equation is also on a per unit volume basis. So in that sense this minus dp dx is to be interpreted as the net pressure gradient in the x direction on a per unit volume basis the next two terms together can be considered as the net viscous force on a per unit volume basis and rho times gx will be the net body force in the x direction all these are in the x direction of course on a per unit volume basis. Similarly the generalized vector form which is written analogously from the above equation is also on a per unit volume basis. So this so far is the so called Navier-Stokes equation for viscous compressible flow the only assumption in these forms has been that the viscosity has been assumed to be constant. So this form is actually little more general than what the mechanical engineering community will usually utilize. So let us just go to the form which is slightly simpler than the previous one and that is more used by the mechanical engineering community. So the mechanical engineering people normally will deal with a constant density or an incompressible type flow situation and if you go back to our previous slide the only difference then is going to be that this divergence of velocity term which appears in the case of a compressible flow situation will be identically equal to 0 if you are dealing with a constant viscosity or sorry constant density or incompressible flow situation. So that term is missing from the right hand side everything else is the same. So remember again that I had mentioned that this second term on the right hand side is the Laplacian of u velocity in the x direction equation. When it comes to the general vector form I have directly written this as del squared v where v is the vector. Now this del squared in our mathematical background on day number 1 has already been listed as the Laplacian which is given as partial derivative with respect to x the second derivative plus second partial derivative with respect to y plus second partial derivative with respect to z. So that del squared has already been listed earlier and we recognize that as the Laplacian operator. So the equation at the top which is boxed and with a tick sign here on the right is what we normally deal with when we come to most of the mechanical engineering applications. Clearly people who are dealing with aerospace type applications very high speed flows etc will usually deal with a viscous form of the compressible form of the Navier-Stokes equation. So this is for incompressible or as a special case you can say it is a constant density situation as well. There is a further simplification that you can obtain by treating the flow to be inviscid. So what we mean by inviscid is that we will simply say that the viscous forces are altogether absent from the flow and if that is the case the second term on the right hand side here will be essentially equal to 0 because there is no viscosity in the flow. Mu is essentially equal to 0 and then what you end up with is further simplified right hand side and the resulting equation is what is called as the Euler's equation. So the Euler's equation is something that we will use if we are dealing with a inviscid flow situation. Please note that here in the Euler's equation this density need not be a constant which is obviously seen through the conservative form. On the other hand I need not have written this density inside if it was a constant density situation for the top equation but that is fine. I can always resort to the non-conservative form when I want to represent it. So the key thing here is that this is essentially the set of governing equations that we will utilize and the second part in this course which will deal with finite volume methods of solving some of these equations will actually utilize the incompressible form of the governing equations. So what I suggest is that we get used to the continuity equation and the momentum equation or the Navier-Stokes equation as we say it now for the case of incompressible flow and what we will do is we will write these down explicitly again. We obtain these momentum equations as the Cauchy's equations and substituted the constitutive relations which were the Stokes's relations and finally now we have formulated these Navier-Stokes equations. The slide number 25 which is projected right now shows the Navier-Stokes equation for a constant viscosity but compressible flow. To make it applicable for incompressible flow all that we do is as I said this divergence of velocity is thrown out because that is identically equal to 0 for an incompressible flow and you obtain a subset of the previous equation. The inviscid flow is a further subset if you want where the entire viscous term is thrown out because there is no viscosity in the flow and you obtain what is called as the Euler's equation. Now as I was mentioning most of the mechanical engineering applications will be such that the flow is incompressible. So therefore the first equation which is written at the top is what will be mostly utilized perhaps even without including the body forces because many times it is the pressure forces and the viscous forces that are most important and body forces are many times not that critical. So it is likely that we are going to deal with this equation mostly when it comes to the CFD part next week. Therefore my suggestion would be to focus on this particular equation. So the way I would like to obtain this Bernoulli's equation is to start with the more general Euler's equation as what we had obtained in the previous slide here. So we have this inviscid flow governing equation or the Euler equation now written only using the non-conservation point of view. So the left hand side is essentially rho times substantial derivative of the velocity explicitly written as the local derivative and convective derivative. So we begin with inviscid assumption and further consider two dimensional steady constant density flow with gravitational force acting purely in the minus y direction. If you start and include these assumptions then the vector form which is at the top of the slide can be written in the independent Cartesian components in the two equations that have been written at the bottom of the slide equation 1 and 2 where we have the x direction. Remember it is a steady flow so that the local derivative will be non-existent. Remember that the gravitational force is only in the y direction that to negative so that the intensity which is minus g is taken in the y direction momentum equation and of course we have the constant density flow as we discussed. So density has been the equation has been divided all through by density otherwise the last term in the y momentum equation would be minus rho times g but now since it has been divided by rho it is only g. With this now let us say that we recall our equation of streamline which means that we are restricting changes in the x and y direction such that dx over u is equal to dy over v meaning that the slope dy over dx is restricted to be along the streamline in some sense because dy over dx will be then equal to v over u. So with this constraint in what we are going to do is we eliminate v in equation number 1. So what is v then v will be equal to u multiplied by dy divided by dx. We go back and we realize that here is v in equation number 1 so we eliminate that using the streamline equation. We eliminate likewise u in equation number 2. So here is equation number 2 on the previous slide and here is u. So we eliminate that u with v multiplied by dx divided by dy and if you do that you will realize that you obtain equations of this form. So what I am going to do is if you perform this algebra in fact the equations number 1 and 2 will get transformed into equations number 3 and 4 as are what seen here with the constraint that dx over u is equal to dy over v. And now if you add these 2 equations what you will realize is that you will add d dx of u squared over 2 dx. This should have been dy. I think there is a typo or let me work out that algebra post lunch. I think there is a typo here which I need to work out. So actually I will work out the entire algebra post lunch but what is going to happen is when you add these 2 you will obtain differentials of the magnitude of the velocity and since we are restricting our dx and dy to be along the streamline we are going to obtain these differentials along the streamline and then you integrate this equation along the streamline between any 2 points 1 and 2 on the streamline and you will obtain what is called as the Bernoulli equation. So usually what happens is the derivation of Bernoulli equation in standard textbooks many of them at least will be done directly along the streamline. So you will identify a fluid particle and you will employ the Newton's second law of motion to this fluid particle along the streamline and in doing so what will happen is that you will generate the Bernoulli equation. In this case what we are doing is we are starting from the more general inviscid flow governing equation which is the Euler equation writing that in Cartesian form and then we are going to manipulate the 2 Cartesian forms to get to the Bernoulli equation. Let me try to do the algebra right away on the whiteboard. So what we have is that the Cartesian forms in the x and y direction of the Euler equation. So the streamline equation as we know is dx over u equal to dy over v. So therefore from here if I rearrange v will be equal to u times dy over dx and also if I want to rearrange in terms of u, u will be equal to v times dx over dy. So let me now try to eliminate v in the first equation and u in the second equation. That is what I had written. So let me just make sure that that is what is going on. So what I have done here now is from the top equation here I have multiplied the entire top equation by dx and that is how I get the second equation. Is that fine? Now you see this u times d dx of u term. This is exactly same as partial derivative with respect to x of u squared over 2. Similarly the second term here is partial derivative with respect to y of u squared over 2 dy equal to minus 1 over rho dp dx times delta x okay. So this is what I should have obtained. Let me see what I have done on the slide. So this is correct. Equation number 3 is indeed correct. Let me try to see what the equation 2 will turn into. So this was our original equation number 2. Now we are going to eliminate u with the help of that streamline equation okay. So far so good. Now multiply this by dy and therefore you will have v partial derivative of v with respect to x dx plus v partial derivative of v with respect to y dy equal to minus 1 over rho partial derivative of p with respect to y dy minus g times dy and as before the first term v times partial derivative of v with respect to x is written as partial derivative with respect to x of v squared over 2 dx. Second term is partial derivative of y with respect sorry of v squared over 2 dy equal to minus 1 over rho dp dy dy minus g dy. So nothing changes on the right hand side. So this is what the second equation should have changed to. So let me go back to my slide. So equation number 4 as has been written out here is also actually correct. So now what is done is we add equation number 3 and 4. So if we add equation number 3 and 4 let us see what happens. And now you collect like terms. On the right hand side the first term here is essentially the differential of pressure. So equal to minus 1 over rho dp minus g times dy. On the left hand side what we have then is differential of u squared over 2 plus differential of v squared over 2 as the two terms. So which combine into differential of the magnitude squared over 2. And this is equal to minus 1 over rho dp minus g dy. And that is precisely what was obtained. And because we are restricting ourselves to utilizing this dx over u equal to dy over v what we will say is that we are necessarily obtaining this differential equation along a streamline. And therefore if you want to now integrate this it has to be integrated along the streamline between any two points 1 and 2. So now integrating this is no problem. You will realize that integrate d of v squared over 2 integrated between 1 and 2 will simply give you the differences in the squares of the velocities between 2 and 1. So on the left hand side you will get v2 squared minus v1 squared over 2. On the right hand side this will generate minus of p2 minus p1 over rho and minus g times y2 minus y1. If you rearrange this what you will get is p over rho plus v squared over 2 plus gy is essentially a constant which is nothing but the Bernoulli equation. So the reason why I wanted to do this small derivation is because many times this way of deriving the Bernoulli equation is not outlined in many of the standard fluid mechanics books. The way we have gone about it, you remember we start from the Euler equation in the vector form which is our basic governing equation for the inviscid flow. Then we write this under the assumptions of two dimensional steady constant density flow with gravity acting only in the negative y direction. So we write the explicit x and y forms under these assumptions. Then employ dx and dy to be restricted such that they are along the streamline using the equation of the streamline and then eliminate v in equation 1, eliminate 2 in equation 2 and then do a little bit of manipulation to come up with a differential form of the Bernoulli equation which then we integrate along the streamline between point 1 and 2 and obtain our standard Bernoulli equation. So remember that you know Bernoulli equation is something that most of us have already seen always seen. In fact Bernoulli equation is something that is routinely used in most of the undergraduate fluid mechanics books many times incorrectly. So the point is to be noted here that the Bernoulli equation is valid only for steady inviscid constant density flow along a streamline at least the way it has been derived in here. So keep this in mind that you know this Bernoulli equation can be derived in this fashion as well. This was just a side exercise. Technically this has nothing to do with the CFD course that we will be really pursuing. But I thought that since we are on equations of fluid mechanics some of these side issues which are slightly interesting should also be pointed out so that some of you will go back to your fluid mechanics teaching and perhaps you can think of incorporating this way of deriving a Bernoulli equation as well rather than starting directly from a streamline with a fluid particle traveling along the streamline. So I really hope that this is something that is understandable as how we have gone about it. What I want to do now is I want to go back and revisit our energy equation which if you remember we touched upon it in the integral form. However at that instant what we did was we simply mentioned what was going to be the integral form of the energy equation and we said that we are going to look at this integral energy equation little more in detail when we actually use it to convert it to a differential form. So now that we are talking about differential equations let me utilize the energy equation in the integral form and convert the integral form of the energy equation into a differential form. If you remember now what we have done is for the mass conservation which is the continuity equation in differential form and the Navier-Stokes equations which we have just completed talking about we derived those both from converting the integral form into a differential form and also from a balance statement that that is how we have gone about. In the case of energy equation I am going to follow only the conversion of the integral form to the differential form approach. You can look into any heat transfer book a standard heat transfer book where convective heat transfer is described in good detail and there more often than not you will see the derivation of the energy equation using the balance approach. So for example, if you look at Professor Sukhatme's heat transfer book or the heat transfer book by Incropera and do it these are couple of standard heat transfer textbooks that we use. You will see the balance approach to obtain the energy equation on a differential basis. What we are going to do here is we will start with the energy equation that we had obtained on integral basis and using our standard Leibniz rule and divergence theorem we will convert that eventually into a differential form. So if you recall then the energy equation written for the total energy which was capital E and the specific total energy that is on a per unit mass basis was small e. If you remember from our integral analysis discussion on day one it was in the form that exists right now at the top of the slide. Now what we want to do is we want to utilize our standard Leibniz rule and divergence theorems to convert the integrals on the left hand side to volume integrals. The first one is already a volume integral but with the Leibniz rule what we are doing is we are bringing the time derivative inside and since we are using a stationary control volume it becomes a partial derivative. The area integral on the left hand side the second term is converted into a volume integral using the divergence theorem as we have done two previous times for the mass conservation and the momentum equations. The right hand side terms if you recall from our energy equation discussion on the integral basis two days back they involve a source term, a heat transfer term because of conduction and a work transfer term. So, right now I am leaving them as they are in the form that we talked about. What we will do is that we will derive an energy equation under a certain set of assumptions. This certain set of assumptions is not necessarily extremely restrictive what happens in case of typical flow situations which are relatively low speed the viscous work is actually typically much smaller in comparison with the heat transfer and the work done by the pressure forces. Remember that this is an assumption that is valid only for low speed flow as is outlined on the slide here. Low speed flows are typical in mechanical engineering if you are dealing with high speed flows such as what the aerospace people may encounter and that too if the high speed flow is going past a solid surface the gradients of velocities are very large within the boundary layers that are formed on the surface. In those cases the viscous work cannot be really neglected but however for the purpose of our course rather we will assume that we are dealing with a relatively low speed flow and therefore we will neglect the work done by the viscous forces all together. In other words what we are saying is that let us treat as if the flow is behaving in an inviscid manner which is an assumption under which we will derive things. Let us assume that there is no source of energy in the flow remember that the energy source term came because of some sort of a chemical type reaction perhaps occurring and we will assume that there is no such thing happening in our flow. We will neglect body forces as well so when we neglect body forces the work done by the body forces either in the form of a potential energy term in the total energy expression or explicitly calculated as part of the W dot term will be neglected all together. So when we say such things that we will neglect this we will neglect that technically what that really means is that those terms are much smaller in comparison with other terms. So it is something like an order of magnitude analysis that is built into this derivation earlier we said that the work done by the viscous forces will be neglected. The neglecting of that work will be happening because we are essentially implying that that work is much less than the other form of the work. Similarly body forces will be neglected in the sense that they are going to be much smaller in comparison with other forces which are essentially the pressure forces now if we are going to neglect viscous terms all together. And furthermore last part is that let the heat transfer into the CV be purely by conduction so this Q dot which is the heat transfer into the control volume let it be purely by conduction. So these are the assumptions with which we will begin actually what I want to do is I will in fact stop right now.