 So, now what we what we intend to do is we will take a little digression and go that is we will go from go away from the main track and focus only on fluid mechanics for some time. That is as I said we will be going for conservation of mass and conservation of momentum before we go to conservation of mass and momentum and energy let us get acclimatized or let us get used to some terms of fluid mechanics because those are very essential for understanding. So, now let us see how do we go ahead with this. So, there are two approaches always one is differential approach another one is integral approach what is called as finite control volume approach, but here we do not intend to take the finite control volume approach. But let us just see what is the difference between the differential and the control volume approach. In the control volume approach this is the control volume approach which I think you would have applied this control volume approach extensively in case of thermodynamics and also in case of fluid mechanics where calculating the drag force and etcetera. But in the control volume approach what we get is only the integral parameters we are not worried about as you can see in this block if you carefully see inside the block there is a blade here inside the block what is happening I do not know all that I know is the free stream velocity and what is coming out I know, but I have no idea what is the velocity distribution, but what will I what am I looking for them I am looking for lift and drag forces on this blade without knowing the velocity distribution if I were to compute the lift and drag forces based on integral parameters that is what is called as control volume approach, but in a differential approach it is quite different that is you can see here complete streamlines have been plotted and I will be getting complete velocity profile and maybe if heat transfer is involved heat temperature profile let us not worry about temperature now it is fluid dynamics so velocity and pressure so I will have complete velocity profile using this velocity profile and pressure profile we will be computing the lift and the drag forces. So that is what is called as differential approach how do I proceed with the differential approach that is I will write I will take a small infinitesimal control volume that is again infinitesimal small control volume and in each control volume I will set up these equations of conservation of mass momentum and energy and do the bookkeeping and get the complete velocity velocity profile. So professor wants me to repeat that here we do not know the details exactly, but we will get to know the integral parameters, but here it is all about detailing we get to know the details why are details important you see here for a given profile blade there are some vertices form there are some flow separation form. So if you want to improve the design of the profile the only way to improve the design the profile is you will get the details if once you get the details then you will be able to improve the design of the profile that is why differential analysis will be more important. So now let us get to little abstract things control volume approach is generally used for getting the integral parameters like for engineering problems where in which you are not worried much about the velocity profiles, but you still want to get the net forces acting no not like that no it is not like only for lab scale experiments I just want to let us say I want to question ask this by professor Arun when will I use control volume approach now let us say I want to fix the launch sprinkler because launch sprinkler is going to rotate and it may fall off because of the rotational forces. Now I want to hold it I want to compute just the net rotational force and I want to hold that so that it does not fall off for that control volume approach is sufficient. So there are situations where in which control volume and we have solved these problems in control volume approach we have solved these problems in fluid mechanics. So the control volume approach is only going to give me the net forces which are acting it is not going to tell why this force is high or why this force is low or which is the which velocity is causing this force higher or lower for example drag or lift force. So differential approach helps me in improving the design but not to relegate the control volume approach that does not mean that control volume approach is not important most of the engineering problems we can solve by using the control volume approach. So now let us see let us get to a fluid particle if I take a fluid particle of course I have taken a fluid particle two dimensional form and it is I have taken in terms of square when a fluid particle moves from one location to another location what happens it can it is translating that is it is moving from one location to another here in this it is not undergoing any deformation you see I have this is the solid line is the fluid particle at an instant a t let us say in the next instant the fluid particle has translated into next location but notice that the fluid particle has not undergone any deformation that is its shape is preserved but in the next case let us say it is it has not moved but the fluid particle has undergone deformation but here I am mentioning that as linear deformation what does linear deformation mean it is only pulled or pushed its shape is more or less preserved that is it has not it has not undergone deformation or it has not like for example if it has undergone angular deformation this would be the case would have happened but it has been just stretched or in either x direction or in y direction if I call this as x and if I call this as y I have just stretched it so that is why it is linear deformation it can undergo rotation like translation it can just undergo rotation in rotation and translation there is no deformation but in angular deformation you see there is there is the deformation it is not just stretched or pulled it is undergoing in an angular form the shape is getting deformed so it can undergo translation or linear deformation rotation angular deformation all together or one by one it is like asking in each transfer it is will only conduction be there no not always it can be we can have conduction convection radiation most of the time all three are there so it is like that when a fluid particle moves from one location to another location that is from t0 to t0 plus delta t if we keep track of a fluid particle between two instance of time so it can have all four or only two or only three or only one it depends on situation to situation so now let us see how do we represent mathematically each of these motions and deformations because we are going to see velocity gradients even we wrote del u by del y but most of the times if we ask ourselves what is del u by del y what does it mean most of us will not be able to answer we will be able to answer if we understand these deformations and translations and rotations so let us see just one quick addition relative position of the fluid particle with respect to one another will change unlike a solid where you will not unless you have gone beyond the elastic limit in normal applications the relative position between the particles will remain the same major difference between fluid and a solid is that one particle with respect to another it need not maintain the same distance and the best example for this is even though mass flow rate remains the same in a nozzle the particles will be much much closer at the exit of the nozzle when the diameter becomes very small whereas the inlet diameter is larger they can be far apart so please remember this because acceleration of a fluid particle gets defined because of this concept. So let us take translation if I take translation I have a fluid particle which I have taken in the form of a square this is o and velocity in the x direction is u and the velocity in the y direction is v so if it has translated so it would have moved in this direction v into delta t that is velocity into time is the distance so here in the y direction and velocity in the x direction is u into delta t u into delta t so you have velocity in the x direction is u into delta t and velocity in the y direction as x into delta t so now let us see how does this happen in case of linear deformation I think translation is quite straightforward I do not think there is any confusion that is why I am not writing this on the y to board but I intend to write for linear deformation let me repeat for translation it has moved in the x direction u into delta t and in the y direction v into delta t so now let us take up linear deformation if I take linear deformation if I take the particle as so this is let us say o a b and c now let us say the velocity is u here and the velocity here again at b is also u okay now the velocity this is delta x this is delta x so what would be the velocity here what would be the velocity here by applying Fourier by applying Taylor's expansion series I can say that u at delta x is u plus we have done this like q x plus del q x by del x delta x so I am not going to tell again Taylor's series expansion I can write u plus del u by del x into delta x okay similarly here velocity would be u plus del u by del x into delta x plus del u by del x into delta x okay so if I ask myself how much my particle has got stretched the same particle I am redrawing o a b c how much it has undergone deformation if I if it has undergone deformation like this let us say if it has undergone deformation that is it has got stretched itself from a c it has shifted itself here it has shifted itself here it has shifted from here to here so what is this how much it has got stretched itself how much it has got stretched itself that is nothing but the velocity here and the velocity here has to be deducted if I deduct that I get del u that is I am writing this del u by del x into delta x and this has happened within a time frame of delta t so it has got stretched itself by del u by del x into delta x into delta t so this is the stretching so let me see let me show the same thing here in linear deformation so I had a fluid particle and it had it has undergone linear deformation in the x direction and how much it has got stretched is decided by del u by del x now I understand what is del u by del x what is del u by del x del u by del x means del u by del x means stretching stretching in x direction stretching in x direction so similarly let me write what will be let me write now just now I wrote del u by del x as stretching stretching in x direction x direction x direction now del v by del y similarly del v by del y will be stretching in y direction and del w by del z would be stretching in z direction so now I understand if the velocity variation in the same direction that is u is in the x direction if I have the same velocity in the same direction if there is a gradient that means that it is stretching in the x direction del u del v by del y means v is the velocity in the y direction that is stretching in the y direction del w w is the velocity in the z direction that is stretching in the z direction so mathematically it is a gradient but physically for a fluid particle it means stretching so velocity variation in the same direction means stretching in that direction that is what we need to keep in our minds okay now that is about the linear deformation let us now move on to what is called as angular deformation or before going to angular deformation yeah linear deformation that is what here it has been told I am telling this in little different form that is stretching in x direction I have just put limits but that does not matter what I am trying to say is that 1 upon delta v into d delta by d t is equal to limit as delta t times to 0 del u by del x del t by del t equal to del u by del x that is del u by del x means change in the volume because I have considered stretching in the x direction only but if I have stretching in all three directions what will this change in the volume become it becomes change in the volume would become del u by del u by del u by del x plus del v by del y plus del w by del z that is stretching in x direction stretching in y direction stretching in z direction or deformation in x direction deformation in y direction deformation in y direction deformation in z direction okay so that is what we mean here rate of change of volume per unit volume due to the gradient okay either with del u by del x del v by del y del w by del z that is what is being stated here that is what is being stated here this is what is called as volumetric dilation rate. I will come back to this volumetric dilation rate when we come to what is the continuity equation or conservation of mass for now I would like you to remember that this del u by del x is stretching in x direction stretching in y and stretching in z I know I have repeated several times but this is intentional okay. So now let us worry about rotation and angular deformation before we go to angular deformation let me write what is and what is deformation or rotation. So now let us take you see here the particle I am going to draw this again but I just want to state this first you have velocity in the x direction is u and velocity in the y direction as v let me draw this that is I have taken again a fluid particle what are we up to we are trying to analyze we so far we analyzed the linear deformation and translation we are going to analyze now rotation and angular deformation. So I have taken again same particle o a b c o a b c I have a velocity in the x direction as u and the velocity in the y direction as v okay now what will be the velocity variation of v over the distance delta x which is the size of my particle fluid particle this is delta x what will be the variation again by Taylor's expansion series I can write this as v plus del v by del x into delta x okay. So if I were to write what is the distance over which this deformation has occurred that is same o a c b if I take that what will happen you have the deformation that is this distance this distance what is this distance equal to here I had a velocity v and here I have a velocity v plus del v by del x into delta x this will become del v by del x into delta x into delta t okay similarly I have velocity u here and the velocity here u is remember the velocity here as varied with y velocity variation here would be u plus del u by del y into delta y okay del u by del y into delta y so how much it has got deformed or rotated will be defined as what is this distance what would be this distance del u by del y into del delta y this has happened over a time period of delta t so that is delta t okay so now let us see what will now let us try to define what is called as rotation so here in rotation what we are going to say is that in rotation what we are going to say is that okay let me go to the document show that again come back and rewrite that so this is what we stated so next is if I take rotation so that is what is what is getting rotated o a is getting rotated to o a dash by an angle delta alpha so that is let me write that that is this angle is delta alpha so similarly this angle let us define that as delta beta okay so now if I go back again and if I have to compute what is this rotation what is this rotation of o a to o a prime is delta alpha so how can I quantify that that is omega o a that is the angular velocity is given by delta alpha by delta t for the limit delta t tending to 0 I think that makes sense that is angle upon change the angle upon change in time so that is limit of delta alpha by delta t within for delta t tending to 0 so what is this delta alpha delta alpha we have just now found delta v by del v by del x into delta x into delta t so delta alpha comes out to be delta v by del v by del x into delta t if I substitute that actually in a what here one more assumption I have made delta v what am I writing here tan delta alpha I am tan delta alpha is tan delta alpha is opposite upon adjacent but I am writing my delta alpha is very small so tan delta alpha approximates itself to delta alpha which is equal to let me write this on the white board because maybe too much I am telling so tan delta alpha tan delta alpha is approximated as delta alpha which is equal to opposite which is equal to opposite upon adjacent which is delta x that is del v by del x del v by del x into delta x into delta x into delta t upon delta x so delta x delta x cancelled out I get del v by del x into delta when I substitute that within the limit delta t vanishes and I am going to get omega oa as equal to del v by del x remember this omega oa is that is omega that is rotation of oa is del v by del x you see now variation of v which is in the y direction is with respect to x velocity variation of v in x direction not v in y direction v in y direction means it would have become stretching in y direction velocity variation of v in x direction is rotation. So, that is omega oa is del v by del x similarly omega o b can be worked out I am not going to work that work it out you can do it yourself all steps are here that is delta b delta b by delta t delta b is velocity gradient del u by del y. So, that is omega o b happens to be del u by del y, but now this fluid particle is rotating in which direction this fluid particle is rotating in z that is omega z it would become omega z that is I have x in this direction y in this direction omega in the z direction my fluid particle is rotating. So, omega z how do I quantify this omega z how do I quantify there is deformation there is rotation del v by del x it is rotating in what direction counter clockwise, but what about omega oa is rotating counter clockwise omega o b is rotating in clockwise. So, I have to take a notation if I take one as positive other becomes negative that is I am taking the notation counter clockwise as positive. So, that means net rotation would be average of these two velocity that is del v by del v by del x minus del u by del y why did this minus come why did this minus come because of one is counter clockwise and this is clockwise and half has come because I have taken the average of the two. So, half of del v by del x minus del u by del y gives me rotation. Rotation in which direction? Similarly, I can write rotation in I think I will just write this that is omega z is equal to half of del v by del x minus del u by del y. So, similarly one can see note this velocity v is varying in x u is varying in y. Similarly, one can compute rotation in x direction as del w by del y minus what omega no this is omega x sorry I told omega y no this is right this is omega z that is along the z direction. So, it is right there is nothing wrong omega x equal to del v by del z and omega y equal to half of del u by del z minus del w by del x you see the velocity gradients are not in the same direction they are in the other direction. If the velocity variation is not in the same direction of the velocity, but in the other direction that means that it is either rotating or angular deformation. So, now let us see so let me take a recap we took linear deformation and we took into account translation and we now we have just done rotation. Now, let us see angular deformation before we go to angular deformation just for the sake of completion and this is what is called as vorticity omega z omega x omega y is what is called as vorticity vorticity is nothing but velocity gradients, but not in the same direction as the velocity, but in the other directions. So, omega equal to curl of v that is del cross v you can I think I am not going to do that you put the del cross v and figure out you are going to get that is what I am saying is you take i j k and del by del x del by del z u v w expand that determinant you are going to get the rotation x y and z. So, I am not going to do that because of paucity of time I think that is simple mathematics you can figure it out that is how the vectorial notation has come out here. So, that is curl v curl v gives the vorticity that is what that is the rotation vorticity by definition it is equal to twice the angular velocity or the rotation vector is the vorticity. Now one interesting situation in case if my fluid particle had indeed rotated like this then what would have happened what would be my gradients it will be what was my rotation if I put it would it was my rotation was del v by del x minus del v del u by del y why did I take del u by del y minus because it was clock way there in that case this would become positive and here by symmetry from the figure we see that as much as if these two angles are equal if the rotational angles are equal then the gradients are also going to be equal that means del u by del y will be equal to del v by del x. So, that is how we can figure out the rotation now let us see the angular deformation. So, what is angular deformation that is delta alpha plus delta beta that is the angular deformation. So, if I take the angular deformation the net angular deformation I am going to get as the notation we use for angular deformation is gamma dot that is gamma dot is equal to del u by del y plus del v by del x that is essentially stemming from the fact that delta gamma is equal to delta alpha plus delta beta that is the net angular deformation it is little different from rotation. So, this is the angular deformation now I guess you can relate what it is see what is this del u by del y we always use to write tau equal to minus mu into del u by del y that is because of strain angular deformation this is what is called as shearing strain this is what is called as shearing strain this is what is called as shearing strain. So, this is because of the angular deformation. So, let me just take a recap of whatever I have done quickly that is we said that just to take note of what we are up to we said that the fluid particle can undergo translation linear deformation rotation and angular deformation translation can be quantified by just u delta t and v delta t and linear deformation can be quantified as if it is in the x direction it is del u by del x del u by del x means stretching in the x direction and del v by del y will be stretching in the y direction and del w by del z will be stretching in the z direction. So, net stretching would be del u by del x plus del v by del z del v by del y and del w by del z this is stretching means what it is volume of my fluid particle is changing that is why it is called dilation dilation I think in medical analogy medical terms everyone will keep talking about dilation. So, that is this dilation that is swelling is there or not they say dilation. So, dilation is change in the volume that is it has got swollen or not. So, that is del u by del x plus del v by del y plus del w by del z that is the linear deformation. Now, angular deformation is the velocity gradient in the other direction u in y v in x w in y and w in y w in x is going to contribute for angular deformation. So, that is what we derived from fundamentals and we showed that rotation in z would be del v by del x and del u by del y. So, for shearing strain this is what is going to be very important for us more than rotation that is shearing strain is delta alpha plus delta beta what is delta alpha and delta beta this is delta alpha and this is delta beta that is the net shearing strain that is expressed in terms of velocity gradient that is del u by del y is delta alpha and del v by del x is delta beta as you can see this is nothing, but shear stress shear strain. So, shear stress will be given by viscosity into shear strain. So, that is how I get tau equal to minus del u by del y tau need not be always equal to del u by del y it is del u by del y only when I have delta alpha, but if I have delta beta my shearing strain would be del u by del y plus del v by del x then my shear stress would be minus mu into all of this gradient that is del u by del y plus del v by del x. So, that is what has been shown here as the shearing strain. So, now, then we come to an important definition two more minutes before we sign off for coffee that is incompressible and compressible flow field for incompressible flow field you can see that you can see that here there is no change in the volume there is no distortion that is fluid may undergo translation distortion or rotation, but the net change in the volume net change in the volume should is 0 it can undergo translation it can undergo rotation it can undergo angular deformation it can yeah it can undergo angular deformation it can undergo linear deformation, but the net rate change in net change in the volume is 0. So, that is what is represented by this figure, but if there is a change in the volume which is represented by del u by del x plus del v by del y plus del w by del z is equal to 0. Of course, when angular deformation I am sorry angular deformation cannot occur because del u it is in my volume change change in volume is given by del u by del x plus del v by del y plus del w by del z. So, there is no angular deformation, but the all that we are saying is there is no change in the change in the volume can occur only because of del u by del x plus del v by del y and del w by del z. So, here change in the volume has occurred. So, that flow field is compressible flow field if there is a change we are going to revisit this you do not have to worry all that we are saying is if it is incompressible it is not going to undergo change in volume if it is compressible there is net change in volume. What is the next step after having introduced ourselves to the concepts in fluid mechanics that is translation rotation and angular deformation linear deformation all these things now we will come to the proper derivation of the conservation equation. These conservation equations have been derived in fluid mechanics in most curriculum hopefully at least the continuity equation would have been derived I am sure. Navier stokes in its full form I do not think it would have been derived in many universities many colleges, but here we are going to derive both these equations from first principles. So, what is this conservation of mass or continuity equation if I want to put it in words in fact this has been done even in thermodynamics where we would say that d m c v by d t is equal to m dot in minus m dot out I think most thermodynamics course is when before we do first law you would have used this equation conservation of mass or continuity. So, in thermo we would have done mass coming in if this is a control volume if mass is coming in mass is going out if they both are equal then we say mass is conserved at steady state d m c v by d t is equal to m dot in minus m dot out I am putting summation because there could be multiple streams coming in multiple streams going out so on and so forth. And there could be mass generation also, but in general this is the equation and it is steady state when the inlet rate is equal to the outgoing rate then you have m dot in equal to m dot out summation of mass coming in equal to mass going out. So, this is what we would use in react in fuels and combustion etcetera where fuel consisting of whatever hydrocarbon plus oxygen would give you carbon dioxide water so on and so forth we balance the mass as the first step. So, now what we are doing is essentially the same thing in a more classical form vectorial notation so on and so forth as we would do in fluid mechanics. So, the words statement remains the same and that is given here in blue time rate of change of mass of the coincident system is equal to time rate of change of mass of the contents of the control volume plus net rate of flow of mass through the control surface. What it means this d b by d b system by d t material derivative is equal to d by d t of an integral around the control volume rho b d v plus integral around the control surface rho b v dot n cap d a this quantity by the way this is come from Reynolds transport theorem we are not going to derive Reynolds transport theorem here essentially what it does is it relates something which is control volume which is changing its shape and having influx and outflow and generation is that sufficient I think I think that is more than enough yeah this is the equation of the so called Reynolds transport theorem it would be given in fact yesterday on Moodle there were lot of questions I answered some 11 or 12 questions on Moodle yesterday. So, those of you who have not visited the site please go many of the questions that you might be having have already been answered two people had asked for reference books for fluid mechanics and heat transfer. So, we are given some author names and Reynolds transport theorem has been stated derived very nicely in most of the books that we have mentioned. So, that forms the basis basically Reynolds transport theorem takes care of the change in the shape of the control surface in the derivation. So, let us keep that in mind and proceed. So, what we say is let us take a cube not going to write anything on the white board too much. So, this is the control volume which is of the shape of a rectangular parallelepiped the dimensions of the cube are d x in the x direction the light green ones are the coordinate directions x y and z unit vectors are i j and k. So, this is i this is j and this is k. So, what we are saying is there are six control surfaces. So, 1 2 3 4 the front and the back control surfaces. So, through each of these surface mass is going to come in. Therefore, mass flow happens through an area that is why it is a surface integral and this happens around the control surface. So, when we write this integral rho v dot n cap d a what it means it is nothing, but a more basic form of m dot is equal to rho a v or rho v times a rho a and what is that v that we talk about. So, in pipe flow problems in fluid mechanics you would be given average velocity 2 meters per second 5 meters per second that is the average velocity given to us that is used for problem, but here what we are saying is v dot n represents the dot product of the velocity vector and the outward normal and this integration carried out over all the six control surfaces that is integral is essentially a summation. So, it is a concise form of a summation. So, this will take care of v dot n in flow out flow for a particular surface outward normal and the velocity need not be in the same direction. So, for a surface where just interrupt in fact, we do not have to go through this Reynolds transfer theorem. See professor had already told us that this conservation of mass this conservation of energy we have used e dot in minus e dot out plus e dot g equal to e dot s t. Again I can use the same equation here that is m dot in minus m dot out plus m dot g equal to m dot s t essentially that is what we are doing here m dot in if you see the control volume here m dot in is all rho u delta y delta z rho w delta z that is m dot in and getting out is m dot out and m dot g can be let us say a sprinkler is there in the middle of the control volume and m dot s t is the change of the mass with time that is essentially the density change. So, we do not have to go through Reynolds transport theorem if you are uncomfortable with this Reynolds transport theorem we can continue to use m dot in minus m dot out plus m dot g equal to m dot s t I think that will be very comfortable way of looking at this equation. So, what we are saying is when the velocity vector and the outward normal are in the same direction v dot n would come out to be like a positive quantity because of the angle. So, dot product is nothing but a vector a dot b vector is equal to magnitude of a times b times sin of theta. So, when the two are in the same direction you are going to have a positive quantity when they are in the op sorry cosine theta when the two are in the same direction you are going to have cos 0 which is 1 when the outward normal is in the opposite direction. So, in flow situation for let us say for this case flow is coming in outward normal is in the outward direction if I just show go on the whiteboard for those of you might get confused if let us take this cube this is the outward normal for this surface whatever be the direction if the velocity vector is in the same direction v dot n would be a positive quantity if instead of this we have flow going inwards we would get v dot n as a negative quantity. And this integration over the cross sectional sorry over the surface area which is what is given by this would our equation we said m dot in minus m dot out that plus and minus sin comes because of this dot product essentially when this is expanded it will give you rho v dot n dot rho v rho v n cap d a for surface 1 plus rho v n dot product for surface 2 so on and so forth. And when I will do this all the six terms if I write this v dot n sin will take care of itself and I will have three terms where the velocity and the unit normal are in the same direction that will have a positive sin three quantities where the velocity and the unit normal are in the opposite direction that will have a negative sin and that is how I will get three terms as positive three terms as negative. Now let us go with the derivation let us say for this I will do for one surface we can do for the other things very straight in a simple manner so then we can just do the derivation very quickly. For this left control surface the area associated is this area let me just go to the white board this control surface delta z is the thickness delta y is the width the area d a is equal to delta y and delta z flow is coming from left to right rho x direction velocity. So, this is going to be u if I expand that for this surface let me call this surface 1. So, this integral which was there I have written now for surface 1 is essentially rho u delta y delta z this is vector is one we are writing the magnitude. So, we are writing the magnitude so I can put this as one I will get rho u delta y delta z mass coming in now for the surface which is at a distance delta x this surface the right right phase of the cube which is at a distance delta x here for that surface we know again by Taylor series expansion. So, this second surface is here surface 2 it has the same dimensions but it is at a distance delta x away from it from the first surface. So, I will write for this mass flow rate in is here mass flow from this surface is nothing but rho u delta y delta z plus Taylor series expansion of this term is d by d x of rho u delta z. So, this is rho u delta y delta z the same function times delta x because that is the change in the x coordinate from that first location. So, that is what is written here rho u delta y delta z and this plus the derivative times delta x. So, I have written for the x direction flow I write the same thing for the y direction rho v delta x delta z is the area outward rho v delta x delta z plus the Taylor series term second term for the z direction I have two terms. Now, what I have to do is just put all these things together and which is what is done here d by d t around the control volume rho d v I will come to this term give me a minute this integral second term has been written. We have six terms for the second term and those six terms are essentially given in the previous set one term 1 2 3 4 5 6 6 terms are there and when we put that here we take the appropriate signs associated with them and we say when I do this Taylor series expansion of this term I will get rho u d y d z minus rho v d x d z minus rho w d x d y those will stay as it is plus rho u d y d z derivative of the this term. So, let me write one of these and then we will just put all of them together. So, I will get for I have written this is mass coming in this is the term for mass coming in this is the term for mass out of x direction. So, for the x direction therefore, I will say this quantity I told you because the unit normal is in the opposite direction to the flow this will have a minus sign associated with it. So, I will have minus the integral rho v dot n cap d a over the control surface when I write for surface 1 and 2 would be minus rho u delta y delta z that minus sign came because of this opposite direction plus rho u delta y delta z plus v by d x of rho u delta y delta z times delta x. I will just copied the second term as it is this is going to cancel off for the pair of surfaces perpendicular to the x direction I have only one term left that is d by d x of rho u times d u delta x delta y delta z represents the volume associated with this control volume. So, by this logic I can get terms for each of those two surfaces. So, this was for the first surface minus rho v delta x delta delta z coupled with these two would be for the second surface y direction z direction would be this and these two all the terms would get cancelled out except the derivative terms. So, I am left with three terms here second term number two, three and four are left over terms from this second integration the first one is nothing but d by d t of rho delta v what is this? This is that volume dilation term that was told change in the dimensions of the control volume leading to the change in the mass of the control volume. So, either it can be change in the volume or it can be a change in the density because we are not talking of solid mechanics we are talking of fluid mechanics because mass is rho times v the control volume volume is delta v the fluid in the control volume has a density rho either of these can change if I apply a lot of pressure compressibility effect will cause the change in the pressure leading to a change in the mass. So, this rho is going to also be a variable that is why we have to take the rho inside the derivative of this expression. So, rate of change of mass remember e dot in minus e dot out plus e dot generated is e dot stored this is that term e dot stored term change in the energy content of the control volume same thing change in the mass of the control volume. So, very simply I can write this as integral d by d t of rho delta x delta y delta z. So, once I put these two together I am left with this four terms which form what we call as the continuity equation delta v which is given by delta x delta y delta z is not equal to 0. So, I can divide through by that and this form of the equation that you see d rho by d t plus d by d x of rho u plus d by d x of rho v plus d by d 0 of rho w equal to 0 this is the general form of the conservation of mass or the continuity equation. So, let me just go finish this and then we will come back to this again what is this telling me I am not yet over we have not seen this in a more comfortable form d rho by d t represents what change of density we we into this. So, we interpret it as a change of density, but essentially it is the compressibility effect compressibility effect is there not only because of this time term, but is also there in a space term how is this is a product of two variables rho u rho v rho w. So, if I apply product rule and expand I will get six terms here. So, the total continuity equation will have seven terms and then what I do is I couple these terms together where density is involved as the dependent variable. So, d rho by d x d rho by d y d rho by d z these terms are coupled together. Let me write instead of talking like this human was derived as d rho by d t plus d by d x of rho u plus d by d y of rho v plus d by d z of rho w equal to 0. All I am now doing is expanding this d rho by d t plus rho d u by d x plus rho d v by d y plus rho d w by d z. I have taken one term out of each of the three derivatives then I will write u d rho by d x plus v d rho by d y plus w d rho by d z equal to 0. Now what I am saying is this term and this term combining these two instead of taking this together separately I will call this as material derivative or substantial derivative of density. So, d rho by d t plus rho times d u by d x plus d v by d y plus d w by d z equal to this is what we call as the continuity equation. And when I have incompressible flow situation this d rho by density remains constant. So, all these terms here which density is a variable that goes off to 0. I get rho times d u by d x plus d v by d y plus d w by d z equal to 0. And this is what most of us will use in class for fluid mechanics this is what most of us would have used I will just write this once more d rho by d t and I will come back to this material derivative do not think it is just left like that rho times d u by d x plus d v by d y plus d w by d z equal to 0. When the fluid is incompressible the flow is incompressible we can have both. So, flow can be the air for example, is a compressible fluid, but the flow when it happens at low speed low Mach number the flow of a compressible fluid by nature the fluid can be compressed can have changes in density, but the flow in that situation need not be a compressible flow it can be incompressible flow of air. So, please keep that in mind liquid for example, is an incompressible substance. So, it will not be like a compressible flow situation ever. So, even if we have air if it flows at low speeds we can treat the flow as an incompressible flow situation even though the fluid is the compressible fluid. So, for situations where density changes are not there I can throw this out and therefore, divide through by rho my famous continuity equation is what is given now. This is what I think most of us are familiar with in fact, we will do it in two dimensions. So, this will also go to 0 I will get d u by d x plus d v by d y equal to 0 all of us remember appreciate use abuse this equation. Now, what is this material or substantial derivative we wrote this as follows. So, this operator just like you have curl and divergence this is also an operator which is nothing, but d by d t of that quantity plus u d by d x of that quantity plus v d by d y of that quantity plus w d by d z of that what is this telling this quantity this derivative this material or substantial derivative is very very important and it is unique to our subject of fluid mechanics because it involves two things I would say one is the variation of this quantity this quantity is given by an empty bracket right now whatever be the quantity. So, that is why if we go back to our document where Reynolds transport theorem was first put this quantity is just a variable b some variable b if we have this material derivative that we wrote is nothing, but the change of that quantity with respect to time plus some u times change of that quantity with respect to x so on and so forth these three terms which are there all of us agree that this is a different form this three together are of one form this is the time variation of that quantity change in that quantity with respect to time that is easy to understand how does this quantity b vary with respect to time that is something which all of us can appreciate velocity changes with respect to time is an acceleration. So, that is easy, but what is this this is u d x by this what does that mean change in that quantity with respect to position. So, this is the remember professor taught us the fluid particle can change its position also solid if I have if I have a solid the solid will just move and if I push this all particles associated with the solid would move together. Whereas, in a fluid what will happen one particle will move little away from its datum it will move a little away from another particle. So, the relative position of the particles are going to change with respect to one another. So, spatial change associated is given by these terms these three terms contribute to that and they are married or coupled with their appropriate velocities in that direction u times d u by I mean sorry u times change in that that quantity with respect to x. So, this quantity has this thing has moved, but it has moved because of what because of the x component of the velocity the change in the position in the x direction is because of the u velocity. So, that variation is taken here and this is unique to fluid mechanics because of the nature of the composition of a fluid where the particles can move with respect to one another and change their relative position these terms are going to be there all the time and this quantity occurs frequently. So, we call this as the material derivative and remember I had told in I made a statement that acceleration of the fluid particle this is what we are saying in solid mechanics we say velocity of a fluid particle velocity changes with respect to time all of us have done problems in 12 standard where a ball is dropped and so on and so forth we calculate the acceleration of the object. So, d v by d t velocity at two different times are given change in the velocity by the time elapsed that is acceleration. So, when the velocity is constant acceleration was equal to 0 this is what we said, but now in fluid mechanics if I re if I put v in this bracket even in steady flow what do I mean by steady flow velocity is not changing with respect to time we can have the velocity acceleration that is what is exemplified by this material derivative business steady state all of us know nothing changes with respect to time. So, this term will go to 0, so this is steady state, but change because of the relative position need not be equal to 0 and we gave a classic example of nozzle thermodynamics all of us have studied nozzles diffusers nozzle is a device which is used to accelerate a flow flow goes from a high cross section area to a low one if you take a control volume all of us have analyzed nozzle I do not want to go into the details mass coming in mass going out when we will say m dot i is equal to m dot e what it means masses it is a steady state situation, but all of us know that velocity at inlet and velocity at outlet v e is much greater than v i. So, even in steady flow we say it is a steady flow situation definitely we can say looking at this that nozzles function is to cause acceleration. So, acceleration is positive that is because of d v by d v by d t is d v by d t plus these terms d u by d x plus d v by d x plus v times d v by d y plus w times d v by d z these are all vectors let me put this so that we do not get confused. These are the x y z components of the velocity this is the velocity vector this is 0 in steady flow because we say nothing changes with respect to time, but this quantity is definitely not 0 in case of a nozzle therefore, acceleration is always there. So, something very unique to fluid mechanics is this and material derivative is what is going to explain this is given very nicely in the textbook by Fox and McDonald in fact, I did not know this concept for a very long time until I studied it from there. So, it is given very nicely there the derivation of fluid particle acceleration is also given nicely there. So, this is my continuity equation so if I want to write the general form of the continuity equation it is this one and put this together in a concise form material derivative of density plus rho times del dot e equal to 0 in steady flow this can be taken as 0 incompressible flow also this can be taken as 0 either with this term will go off incompressible flow or even incompressible flow this is 0 rho gets divided you get del dot e equal to 0 that is the commonly given form of the continuity equation. Now, conservation of mass is very easy to implement understand etcetera if you know the velocity field we can write some expressions and get the continuity equation and it is one of the most easiest steps easiest things to do in fluid mechanics. In fact, d u by d x plus d v by d y equal to 0 if I know one of the velocity variation I can get the other one by just substituting in this equation and trying to get the other velocity.