 Welcome to the lecture where we are going to be primarily applying Reynolds transport theorem and derive the conservation of momentum, the Navier-Stokes equations which you are already familiar with okay and then we will talk about how you need boundary conditions to solve this and we will do it in a slightly different way from what you are already exposed to. Different way in the sense that the boundary may not be a straight line, the boundary could be curved and then how do you generalize the formulation of the boundary condition. So that is the strategy okay. So let us look at the conservation of momentum. So now n represents momentum and the corresponding intensive property like we saw yesterday eta is going to be the property per unit mass which is nothing but the velocity okay. Now what I am going to do is write down the Reynolds transport theorem and just substitute wherever we had eta velocity and we have the material derivative and remember now n is momentum okay and this is associated with the control mass and this is equal to the rate of change of momentum in the control volume. Like I said I am going to write this as my velocity vector v dv plus we have the efflux term across the control surface which is eta rho v dot da. All I have done is substituted wherever we had eta the velocity vector okay. We are going to convert this again to a volume integral and group these 2 terms okay that is just from calculus and since the control volume is fixed in time I move the time derivative inside like we did yesterday and so I have the control volume I have d by dt of rho v plus I am converting this also to a volume integral and I am going to write this as the divergence of rho v v okay and this entire thing I am going to integrate over the volume element. What I want to do now is to tell you that these terms on the right hand side are basically the inertial terms which you are familiar with in the Navier-Stokes equations normally when you write the Navier-Stokes equations you would write this on the left hand side okay. So these are the inertial terms the term on the left hand side represents the rate of change of momentum associated with the control mass. What we are going to do now is use the fact that from Newton's second law the rate of change of momentum associated with the control mass is going to be equal to the forces that are acting on the system okay. So what we want to do is we actually want to study rates of change of velocity at a fixed point so I want to get this in terms of the forces okay. So this is going to be written as the forces acting on the control mass which is occupying the control volume okay. So that is basically what the Lagrangian approach tells me that the forces are equal to the rate of change of momentum. And so now what we need to do is identify what are the different forces which are acting on the control volume the mass contained in the control volume and then we should have the Navier-Stokes equation since our objective is to derive the Navier-Stokes equation what we are going to do is and the Navier-Stokes equation talks about how the velocity pressure etc vary with space we are going to consider an infinitesimal control volume and to find out what the forces are that are acting on the control volume okay. So to keep things simple we are going to look at a rectangular coordinate system and let us say this is my x axis this is my y axis and this is my z axis okay and I am going to consider a cuboid this is the x direction and the two surfaces are going to be given by the surface on the left is at x the thickness here is dx so that the surface here is given by x plus dx okay this distance is dx or delta x maybe I should use delta x because then when I do the limit as delta x tends to 0 I can write this as dx just to be on the safe side okay. The lower surface is given by y equals y and the upper surface is going to be given by y equals y plus delta y okay similarly this surface here is going to have the z axis is coming outside the board the way I have drawn it so the surface of the back is given by z equals z and the surface in the front the front face is going to be given by z equals z plus delta z okay so the back face is at z equals z the front face is at z equals z plus delta z the two things on the sides right and left are given by x coordinates and this. Now the forces that are acting on the system there are basically two kinds of forces one which is the body force and then whatever is being exerted along the surface okay so let us do the easy one first which is the body force and the body force is going to be primarily because of gravity okay that is only one thing we are considering right now and that clearly is going to be what is the body force is acting on this control mass is going to be the density times the volume multiplied by the gravitational acceleration right so rho times g times delta x times delta y times delta z it is something bothering you show them everything is fine okay so that is the body force which is acting on the system. Now we are going to have surface forces acting on all the six surfaces and so let me just maybe write this thing this is at the back and this is at the front and this x plus delta x is at the right and this is on the left okay and y equals y is the bottom and this is at the top. So now as far as the viscous stresses are concerned the stresses we are acting on this I will just say for right now that the stress is a tensor and what does this mean it means I need to specify two things in order to completely specify my stress okay and normally the stress components are given by the symbol tau okay and let me just say that these are the two subscripts x and y what does this represent this represents the component of the tensor which is acting on a surface whose outward normal is in the x direction and which is in the direction of plus y okay. So this basically represents component surface whose normal is in the x direction and which is acting in the y direction okay. So we need to specify two things as opposed to a vector where you only specify the direction of the vector okay. So here I am specifying the normal to the surface as well as the direction. So since there are two things which are specifying okay this becomes a tensor we will see more of this in some detail later on when we do it more mathematically. But just for the sake of illustration physically because now we are going to get into some physical derivation for the Navier-Stokes equation. The other important thing which you should know is that there is a sign convention which you have to follow and the sign convention which is followed is that if both the x and the y directions are positive then the stress component is taken to be positive okay. This could be one of the sign conventions you can use or if both of them are negative then the stress component is again positive. So basically what I am saying is if the tau if both x and y directions are positive or both are negative the stress component is positive whereas one of them is positive and the other is negative. So I am going to illustrate this it will become clear to you okay if one of them if one is positive and the other is negative the stress is negative okay. So we will just stick to this sign convention and then we will find out what are the forces which are acting on the system. So now clearly if I consider the phase on the right on this right surface I will have different forces which are acting okay. I will have 3 components for the stress okay. So let us look at the right surface this is at x equals x plus delta x okay maybe it is better to draw it slightly better way okay I am happy with that okay. Now this is the x direction right sorry this is the x direction x is normal to this. So I have a component which is going to be in the normal direction that is going to be the normal direction to the surface is the x direction and I can have the force which is acting on this resolved in 3 different directions x, y and z. So the normal component which is going to be in the x direction acting on the surface whose normal is also in the x direction I am going to give it a slightly different notation I am going to call it sigma okay and you will see why that is. So sigma xx is the normal component of the stress tensor. So what does this tell you? So whenever it is normal that means both the subscripts are the same because the direction of the stress is the same as the direction of the outward normal of the surface. So this is a normal component. Now what are the other components which are acting on this surface? Other components acting on this surface include tau xy because my surface is fixed okay it is tau xy and this is the stress component in the y direction but it is acting on the surface x plus delta x and tau xz is the stress component in the z direction. So the way I have drawn this tau xz is going to be in this direction okay and tau xy is going to be in this direction. So y is the vertical direction so I just stick to that and the way I have drawn these components all of these are positive because the direction of the component is in the upward the plus y direction the surface outward normal is also in the plus x direction so this is a positive component. This guy is also a positive component because the normal is in the plus x direction the direction of the stress component is also in the plus z direction. So like I said if both of them are positive then the stress is positive okay and the sigma xx is going to be a normal direction so just imagine something coming out of the board out of this thing so that is sigma xx. Now what we want to do is do a force balance right we have to find out all the forces so rather than talk about all the forces acting on the entire control surface what I am going to do is just talk about forces in the x direction okay. So I am just going to find what the forces are in the x direction and then by similarity we will get the forces in the y direction and the z direction. So let us just concentrate on what are the forces acting on the x direction clearly in the x direction the only component which is going to play a role is the sigma xx because this guy is acting in the y direction this component is acting in the z direction okay. So but what is going to happen is the other surfaces will have components in the x direction I need to take that into account okay. So now let us find out what are the forces acting on the other surfaces in the x direction and then we should be able to write get all the forces acting in the x direction. So on the upper face the okay the force acting in the x direction is going to be y because the upper face is outward normal is y the first subscript is y the direction is going to be x okay now yx and this is going to be evaluated at y plus delta y okay. But this is acting on what area element this is acting on an area element which is given by delta x delta z that is the area on which this component is working. So that total force is going to be delta x delta z this is on the upper surface and what about the force acting in the x direction at the lower face this is also going to be how yx divided evaluated at y multiplied by delta x delta z okay. I just I am going to indicate that the x direction is positive in this direction y is positive here. Okay if you want a positive component for this sorry you want a positive component for this the positive component is going to be acting to the right right because this is the one where x and y are both positive whereas here the negative component the component acting on the lower faces outward normal is in the negative x direction. So in order for me to have a positive tau the tau yx which is positive is going to be acting towards the left okay I am just going to indicate tau yx as being acting to the right here and tau yx as acting to the left here. So if you want to really look at the net force acting in the x direction from the upper and lower faces because the directions are opposite you are going to have to basically subtract the 2 okay and so the net forces force is going to be given by tau yx at y plus delta y minus tau yx evaluated at y the entire thing multiplied by delta x delta z okay. And this you can write as by just doing a Taylor series expansion you can approximate this as for a very small delta y you can write this as tau yx by dy times delta x times delta y times delta z okay. Now let us look at the contributions due to the other surfaces from the on the right face the force acting is going to be given by x is in the outward normal direction the direction is the normal force that we are looking at is going to be given by sigma xx okay times the area element which is delta y delta z and this is evaluated at x plus delta x on the left face the force acting is given by sigma xx evaluated at x delta y times delta z remember here the x component is the x direction is the outward normal here so your positive direction is going to be in the plus x direction here the outward normal is in the negative x direction so in order for the stress component to be positive you should be acting in the negative direction and that is what I am going to do now that sigma xx is acting this way here sigma xx is acting that way okay and so the net force is going to be given by d by dx of sigma xx times delta x times delta y times delta z so taken care of the right and left face the top and bottom face you are going to take care of the front and back okay. So on the front face we have the force acting is going to be given by tau the first component is z because the normal is in the z direction looking at x direction x balance is what we are doing so tau zx okay but this is evaluated at z plus delta z multiplied by the area on which it is acting which is delta x delta y okay on the back face the force acting is tau xz evaluated at z multiplied by delta x delta y. So as far as the I got it wrong here right I got I should write this as zx as so the net force which is acting on the system because of these 2 faces the front and back face net force is going to be given by just subtracting the 2 because of the fact that the directions are different with the same convention that we are following and in fact if you had followed the other opposite convention again you are going to get the same thing okay we are going to get d by dz of tau zx times delta x times delta y times delta z. So clearly the delta x delta y and delta z is nothing but the volume element the volume of the infinitesimal cuboid that we just constructed what I want to do now is substitute these expressions for the forces. Now if you had an electric field and let us say your fluid had some kind of an electric dipole or you had a magnetic field then you would possibly have to include those forces okay right now we are just going to keep life simple we are not going to neglect all of those okay and it is likely that those forces are going to be more in the form of a body force. So I just wanted to mention that but then we are going to keep life simple right now and then we are just going to neglect all those forces and write out the sum of the forces so these net forces acting on the control mass in the control volume is going to be given by the sum of the body forces that we already wrote down and these 3 surface forces that we have okay. So I am going to be given by rho g multiplied by the volume element I am going to keep that common and I am going to write this as d by dx of sigma xx plus d by dy of tau yx plus and this is in x direction remember d by dz of tau zx multiplied by the volume element delta v okay. So these are the forces these are the surface forces that we have and this is the body force that we have and maybe what I need to do is be a bit more careful here I need to tell you that we are looking at the x component so I need to change this to the dx okay net forces acting on the control mass in the control volume in the x direction therefore. So now should be able to do the same thing in the y direction in the z direction so rather than go through the entire process we will just you know look at this and we should be in a position to actually write down the components for the other directions okay. But before doing that what I would like to do is simplify things a little bit so rather than talk about things with tensors we will have to keep things simple so let us go back to the Reynolds transport theorem and say that we are going to do conservation of momentum only in the x direction what that helps me do is instead of writing the velocity component in the vectorial form my eta I would write my eta as just vx okay so rather than do a general conservation of momentum I am going to do conservation of momentum in the x direction. So let us come back here and that is just to improve clarity okay and then we will definitely be doing things more tensorially but right now I do not want you guys to get tens looking at the tensors so we will just continue with this. So let us look at the momentum balance in the x direction okay so what this means is I already have the forces acting on the x direction so I can just write this rho gx plus d by dx of sigma xx plus d by dy of sigma sorry tau yx plus d by dz of tau zx okay multiplied by the volume element and this is the left hand side which I had and the right hand side from a Reynolds transport theorem is going to be triple integral over the control volume of d by dt of rho vx because I am looking at the specific property eta now is vx when I started the lecture today I just put eta as the velocity vector v but now I am doing only the momentum balance in the x direction. So I am going to restrict my substitute instead of eta vx okay plus divergence of rho vx into v that is what we have multiplied by dv okay so this is the above is from the Reynolds transport theorem where eta equals vx okay since we have an infinitesimal control volume a very small control volume we can basically say that this integral over the control volume can be written as this function multiplied by delta v okay since we are talking about a very small control volume so the right hand side can be simplified as since the control volume is infinitesimal we can write the right hand side as d by dt of rho vx plus divergence of rho vx v times delta v okay where delta v is nothing but delta x delta y delta z basically what I am trying to tell you is that this is some kind of an average thing inside the control volume okay and it is like very small integral fx dx can be written as fx times the delta x using a trapezoidal rule or a rectangular rule for doing the integration okay. So here we have this so now what I can do as you can see here what is happening is I have got rid of all my integral signs and since this is going to be valid for any arbitrary delta v so delta v cannot be 0 so I can actually remove the delta v and we get our equation that we want to do want to get which was objective which is the Navier Stokes equation in the x direction okay. So in the x direction the momentum equation is and I am going to now just flip the left hand side and the right hand side because then things would look very very familiar to you rho vx plus divergence of rho vx v okay equals d by dx of sigma xx plus d by dy of tau yx plus d by dz of tau zx plus rho g subscript x okay. I am going to focus on sigma xx sigma xx is my normal stress component okay and the way I have drawn sigma xx is it is positive when the x axis is the surface outward normal is in the positive x direction and the direction of the force is also in the positive x direction so if this is x plus delta x this force this is sigma xx so on this surface sigma xx is positive this way. Now supposing you have a fluid which is at rest okay then what is the force that is going to be acting on the surface it is basically the hydrodynamic pressure okay p and what is the direction of the hydrodynamic pressure it is in the direction of the inward normal so the hydrodynamic pressure acting on the surface is going to be basically because of molecules which come here and then they get reflected okay. So as far as the force on the molecule is concerned it is acting in the right direction towards the right but as far as the force on the surface is concerned it is acting towards the inward normal. So the reason why I have wrote this particular term as sigma xx is basically because I wanted to separate out the two effects one coming from the hydrostatic and one coming from the fluid flow. So the effect coming from hydrostatic in the normal direction is basically the pressure term and remember pressure acts even when fluid is not moving okay and is in the inward normal direction okay. So we are going to write sigma xx as –p plus tau xx this –p basically tells you that this pressure is in the inward normal direction and sigma xx is in the outward normal direction so basically what I am doing is the normal component I am resolving it in two directions one is in the when what you have when you have statics the pressure field and this is due to the flow okay. So I am going to substitute sigma xx now in terms of this and then you get what you are actually familiar with which is d by dt of rho vx plus divergence of rho vxv equals –dp by dx plus d by dx of tau xx plus d by dy of tau yx plus d by dz of tau zx okay. You can write a similar equation for all the three directions and of course I have missed out the body force term which is rho gx okay. We can similarly write for the y and z directions okay and you can you know write things in a very compact vectorial notation which is d by dt of rho v plus divergence of rho vv okay equals – gradient of p plus divergence of tau plus rho g okay. So this is the vectorial form of the Navier-Stokes equation. Now I just wanted to derive it this way so that I could illustrate to you that starting from the Reynolds transport theorem we use Newton's second law of motion only to take into account the rate of change of momentum associated with the control mass. So that is where we bring in the forces okay and the other terms the accumulation term and the convection term where you have the flow across the control surface those are what you are familiar with as inertial terms from your undergraduate class okay. So basically these are your surface forces the pressure is also a surface force but is the normal direction and tau represents all the components which is going to arise because of velocity whereas p will be present even when there is no flow even when you have a hydrostatic situation okay. I of course have jumped a little bit in the sense trying to put this thing together but we will take a look at how to evaluate divergence of rho vv divergence of tau in the next couple of days. What I want to do is just talk about the fact that what we have derived the equation of continuity and the Navier-Stokes equation is basically a differential equation okay and in order to solve these differential equations you need boundary conditions. So tomorrow what we are going to do is we are going to see how these boundary conditions can be put in a slightly general framework. Let me tell you what the motivation for that is. So I mean consider thin liquid jet coming from a tap okay. So if you go to the bathroom tomorrow or your sink and you just open the tap a little bit what is going to happen is water is going to fall right. So and clearly this water is going to be surrounded by air. What do you expect? I mean if this is the mouth of the faucet and this is the jet which is coming out clearly as the liquid is going to come down because of gravity the velocity is going to keep on increasing okay and if the velocity keeps on increasing the cross sectional area through which it is going to flow is going to keep on decreasing. Eventually the cross sectional area is going to be so small that you are going to have this particular jet break up into drops okay. So that is a problem which we are going to look at in some detail later on in the course how does the jet break up into drops. But what I want to talk about today is talk about the portion where the jet has not broken up into drops but the cross sectional area is changing okay. So the point is the jet coming out of the faucet of the tap okay is going to accelerate and the cross sectional area is going to decrease okay. Eventually it breaks into drops provided your flow rate is sufficiently low okay but tomorrow when you go to the bathroom and you are trying to fill your bucket you will possibly do this experiment and see does this indeed happen. But I want you to focus on before the break up the surface is not going to be a uniform is not going to be that of a uniform circular cylinder it is going to be of a cylinder whose radius is actually let us assume it is going to be circular whose radius is actually going to change okay. So what we want to do is we want to be able to apply boundary conditions to surfaces which are not going to be given by you know r equals constant where r is actually now a function of z okay. So and that is one of the things we are going to be doing in this course where your surface is not going to be given by lines where the coordinate is equal to a constant okay. So the surface of the jet is given by r equals some function of z. Let us say where z is the vertical direction the vertical direction is z. So normally you are used to applying boundary conditions to r equals a constant r0 because and the kind of boundary conditions that we are normally used to is applying things like continuity of tangential stress, continuity of normal stress if there is continuity of normal stress okay. So we would now like to find out how to go about applying boundary conditions here. So how does one apply boundary conditions on surfaces of this kind okay. So before that the reason is your coordinate system is r and z okay and we have derived our different stress components in terms of tau rz tau zz tau rr etc. But now the stress that is going to be acting on this surface is going to be in the normal component and in the tangential direction and the normal and the tangential directions are not in the classical r theta z because the surface is actually changing. So we need to be able to relate the shear stress and shear components in an arbitrary direction acting on an arbitrary plane to the classical things acting on the 2 classical directions r and z okay. So that is something we are going to see and then we will talk about the boundary conditions.