 Hello, Amal Jyoti, college Kerala, you have a question please go ahead. So my question is that sir in the integral analysis 20, so it was given that the total energy balance for most of the cases for mass conservation and momentum conservation, we have written it as d by dt of a control volume or the property in a control volume is equal to the whatever property in minus property out plus energy source. So in that in this special case why did you consider the heat transfer and work transfer separately or can we do it in such a way that the heat transfer and work transfer can be included in the energy in and energy out so that it the equation becomes much similar to the energy balance equation. Why did we actually is there any anything common to the heat and work transfer so that we are not including it in energy in and energy out. Yes, so the question is related to the energy balance statement which is projected right now on the board. The question is if or why I should say the heat transfer and work transfer terms have been separately written out and whether they can be included in the E dot in that is energy in and energy out terms on the right hand side. Actually your question is really relevant and it is very correct and good question. The answer is that you are right there is no need to separately write out the Q dot and W dot term rather we can include them in the energy in and energy out. It is just that what I wanted to do was the energy in and energy out terms are specifically related to the fluid flow that is going in and out of the control volume. So as the fluid gets in and out of the control volume by virtue of its motion it carries with it certain energy in and out of the control volume whereas the energy so the heat transfer term is slightly different in the sense that it is considered to be a conduction type heat transfer which is happening because there exist temperature gradients within the flow. So that is the reason I wanted to separate it out from the energy terms that are associated with the fluid in and fluid out. Similarly, when it comes to the work term the work term is actually in an integral situation will usually include a shaft work in addition to the surface forces work as well. So in case of an integral analysis these are some special situations of shaft work and that is the reason I wanted to write them separately but in general you are absolutely right there is no reason to write this separately. However, when you calculate then E dot in and E dot out you have to explicitly include these terms as well. Okay sir thank you in and out. Shivaji University, Kolhapur you have a question please go ahead. Sir my question is is there any possibility of change in density with respect to time in incompressible flow? So the question is if there is a chance of change in density with respect to time in an incompressible flow? The answer is incompressible flow you can treat it as an unsteady flow as well. So in that case it is possible. However, if the incompressible flow is a constant density flow like what would happen in case of a liquid flow in that case there will be essentially no change of density with respect to time. Nirmai University you have a question please go ahead. Hello I have question in this slide kinematics 25. Here we have derived the relation between angular velocity and vorticity. So what is the significance of vorticity or how it differs from this angular velocity? So the question is on slide number 25 of kinematics where we discussed angular velocity and vorticity. Specifically what is the physical significance of vorticity and how it is related to angular velocity? Actually the relation is right there on the slide that the vorticity is 2 times the angular velocity. If you go back a couple of slides and if you follow the derivation of the way we have calculated the angular velocity you will realize that eventually it turns out to be one half of the curl of the velocity field which is what has been mentioned in slide number 25. And it is simply a nomenclature that the curl of the velocity field is called the vorticity. In principle on a physical basis both the vorticity and the angular velocity imply the same thing that is if either the vorticity or the angular velocity are non-zero we are dealing with what is called as a rotational flow which means that the fluid particle is spinning about its own axis as it moves. So physically speaking both vorticity and angular velocity imply spinning of a particle about its own axis as it is moving. Relations wise the relation is right on slide number 25 it simply so happens that vorticity is equal to 2 times the angular velocity. I have one more question in slide differential analysis 26 in this first equation is true for incompressible flow and second one we are deriving from first and in second equation it is said that density may not remain constant but in first equation density is constant so that is not clear. Yes so the question is on slide number 26 differential analysis what is getting pointed out and in some sense correctly is that the top equation is written for a constant density or incompressible flow whereas the bottom equation is written for inviscid flow. However it is been written that the density may not be constant so how is it possible is there some sort of an inconsistency. Actually I am sorry this is my fault I should have explained that the second equation which is at the bottom of the slide is not really coming from the top equation. In fact if you go back one slide which is slide number 25 the general equation that you see here which is for a viscous compressible flow from here which is the most general equation so far if you remove the viscous terms for the inviscid flow you will realize that you do get the equation on page number 26 the second one where the viscous terms are now missing. So in that sense the equation for the viscous inviscid flow which is now getting highlighted is actually following from the most general equation which is on 25 by removing the terms related to viscosity. So sorry about that confusion but I hope this answers your question. Yes sir okay thank you. Jaipur engineering college go ahead if you have a question. Sir I want to ask that how you have applied Leibniz rule and divergence theorem in slide number 5. Yes so the question is related to slide number 5 on differential analysis specifically how the Leibniz rule is getting employed for the volume integral and how the divergence theorem is getting employed for the area integral. So if you see the Leibniz rule which was pointed out so the Leibniz rule is presently projected on the screen at the bottom of the screen. Obviously the way it was written here on the slide that you see right now is written for a one dimensional situation whereas what we have in that differential analysis slide is a three dimensional control volume but in principle what we are saying is that when we want to take this differentiation with respect to x inside the integral sign it will result in a partial differential sign which is one thing and then additionally two terms will show up if the limits on the integration are functions of x this is in one dimension. Now going back to the before we in fact go back this partial differentiation will be present when we go back to the differential analysis. However if you notice what we assumed in that differential analysis slide is that the control volume is fixed and because the control volume is fixed its limits which you can consider as the surrounding surface with respect to time which is what the differentiating variable is are actually 0 because the control volume chosen is a fixed control volume. So let me go there and then therefore point out that the only term that remains in that situation is the partial differentiation term. You can see here that the time derivative has been brought into the integral sign but then because the control volume is a stationary control volume its limits that is the integral limits are fixed with respect to time therefore only the partial differentiation remains. So that is the way the Leibniz rule has been employed for the first term. The divergence theorem simply says that the area integral gets converted into the volume integral with the function the vector function which is dotted with the unit outward normal and then integrated over the area surface area that vector function is converted into a divergence form. So the divergence of that vector function then is integrated over the control volume after converting the area integral into a volume integral. So that is precisely what has been done if you see the vector function here it is rho multiplied by vector v. Rho is just a scalar so it goes along with v and then the divergence of rho times v is then integrated over the entire control volume that is how the divergence theorem is applied for second term. In fact if you see that how these Leibniz rule and divergence theorem have been employed for later situations also when we went to momentum equations or momentum balances they are exactly the same. As long as we are dealing with a stationary control volume the limit on the integration as far as the volume integral is concerned is independent of time and therefore only that partial differentiation inside the integral remains there is no additional term as is earlier shown by the Leibniz rule. Similar thing for divergence theorem the vector function which is getting dotted with the unit outward normal is simply converted into its divergence and then that divergence of that vector function is then integrated over the entire control volume after employing the divergence theorem. So that is how the Leibniz rule and divergence theorem work. Thank you sir. Sir, I also want to ask in slide number 10 how have you written the equation number 1 x momentum equation in 2D. Yeah, so the question is on slide number 10 on differential analysis specifically how this top equation has been written. So if you go back to our balance statement so now we are talking about the balance for the linear momentum. So what we have is on the left hand side the rate of accumulation of linear momentum in the control volume on the right hand side we have rate of inflow minus rate of outflow and the rate of increase of linear momentum due to a source which we said the last term that is here the source term is due to the net force acting on the control volume or the material inside the control volume. So now with this if we go to slide number 10 it is simply each of those terms representing what I just described. On the left hand side what we have is the rate of accumulation term which is getting written down here. Right most is the net force term which is the source term in the x direction and the two middle terms are basically the terms that describe let me go back the rate of inflow minus the rate of outflow of the linear momentum in the x direction from the control volume. All you need to realize though is that as I was trying to point out in the lecture look at this particular slide right now here slide number 8. You will realize that there are two linear momentum coming in in the x direction one from the west phase and one from the south phase similarly there are two that are going out one from the east phase and one from the north phase and that is why because there are four linear momentum terms you end up generating two of these terms which represent the rate of inflow minus rate of outflow. If there were only two momentum terms you would generate only one of them that is all that is. So what I suggest is really that all the components required to perform the algebra are written down on slide number 8, 9 and all you need to do is the material from slide number 8 and 9 needs to be substituted in the balance equation and if you simplify the balance equation you will obtain this first equation on slide number 10. Santai Perrier-Vellore you have a question please go ahead. Sir good afternoon sir. I have a question in kinematics question 15 here the velocity getting changed to from V2 V plus dou V by dou x into del x here actually what is the physical significance of partial derivative difference? Yes so the question is related to slide number 16 on kinematics where what we have shown is how the velocity is varying over this fluid particle over a distance delta x and delta y. So the specific question is as we go from point A to point B the velocity in y direction which is V is shown to change from V to V plus partial derivative with respect to x of V times delta x. So what is the physical significance is the question. See the only thing that you need to note here is that we are dealing with a two dimensional field. So if you are dealing with a two dimensional field let me go to the wide board for a for a second. For a two dimensional flow field both u and V velocities will be functions of x and y ok. In general both u and v will be functions of both x and y coordinates. So now going back to our slide what we are trying to show here is as we move from point A to point B we are simply moving in the x direction. So we are trying to find out how the variation of velocity V is along the x direction. Since velocity V is both function of x and y and since we are only moving in the x direction we are required to take the partial derivative of the velocity V with respect to x and then multiply by the delta x term so that the first order Taylor expansion is achieved. So that is the reason we employ partial derivative here. In fact if you note here the other component of the velocity which is u velocity or the x velocity that is also changing from A to B and in a similar manner it has been computed. The reason being that u is also a function of x and y in general in a two dimensional field. Since we are moving only in the x direction we will necessarily need to use partial differentiation with respect to x because essentially that y coordinate is considered to be constant as far as this x movement is concerned. Thank you sir. I have one more question differential analysis 18 when we are taking the momentum above O the shear stress tau xy along in y axis and the x axis corresponding momentum we are considering in differential analysis 19 tau xy minus tau yx it is not clear. So the question is on differential analysis slides number 18 and 19 and specifically the information provided on slide number 19 what are we really talking about here that is the question. So if you go back to slide number 18 which is what is projected right now what you will see is that I have shown all surface forces in terms of the stresses and the body forces that are acting in the x and y direction in our two dimensional fluid particle. So each phase if you see there will be a normal stress and then there will be a shear stress. Now what I have shown in the slide are the expressions for some of these stresses. So there are total of eight surface stresses out of those eight I have on purpose shown only four expressions the remaining four I was hoping that the participants will be able to work out. Addition to that what I have shown here is the body force which is passing through the center of mass. Now the slide number 19 simply says that if we want to take moments of all these forces about the point O what will happen. So since the normal stresses the four normal stresses and the body force are all passing through the point O there will be no moment arm or the moment arm for these forces will be 0 and that is why the moment due to the normal stresses and the body forces about the point O will be 0. The only moment which is non-zero is going to be because of the shear stresses. Now if you see the shear stresses they occur in pairs in the sense that you have a shear stress on the east side then you have a shear stress on the west side that forms an anti-clockwise couple there is a shear stress on the north face and a shear stress on the south face which form a clockwise couple. So if you want to evaluate a net counter clockwise or anti-clockwise couple you need to appropriately calculate the forces and multiply those by the moment arms for these shear stresses and that is what has been done here. So what you see on the top line in slide number 19 is the net counter clockwise moment that will be existing on our fluid particle because of the shear stresses and the net moment will come only because of the shear stresses other stresses will not contribute to the net moment that is what is getting pointed out at the top line. Thank you sir. Periyar Manny Ammai please go ahead if you have a question. Hello sir in the slide differential analysis 19 you have said earlier days that only in nanoscale dimensions we cannot apply the CFD approach but here in the middle of the slide in differential analysis 19 you have said if size is sinking to 0 means how this kind of products would happen since the CFD is not for nano or below 0 dimensions how this can happen sir since this kind of products may not happen. Okay so this is an interesting question the question is on slide number 19 which was being discussed a few minutes earlier and the specifically the question is that in this particular case we evaluate this net counter clockwise moment on the fluid particle and then evaluate or equate that to the moment of inertia multiplied by the angular acceleration and after some manipulations we bring the angular acceleration in the form as is shown right now here on the slide. So now the argument was that since delta x and delta y tend to 0 we say that this angular acceleration seems to tend to infinity. So the question that is getting asked is why are we talking about the particle shrinking to 0 whether we are then violating the continuum situation and the answer to that is that we are not really necessarily going down to the molecular level as what we would perhaps require if you are dealing with a nanoscale situation when we talk about delta x and delta y tending to 0 in this continuum situation it is in the calculus sense in the sense that we are not necessarily shrinking them to a point as the geometrical point that we talk about. We simply say that we are taking these arbitrarily to 0 arbitrarily small however keeping in mind that still we are not breaching the continuum limit so the idea is simply to put up an argument that if you start reducing the denominator to a smaller and smaller value it seems to say that the angular acceleration will keep on increasing and the argument was that this is physically not possible because there is no special case that we are putting up here it is just a general discussion and to simply come up with the conclusion then that the numerator also must be essentially then very very close to 0 in order for the angular acceleration to be finite we use the argument of delta x and delta y tending to 0. You do not please consider that we are breaching the continuum limit we are still within the continuum limit by no means when we say that we take the delta x and delta y to 0 we are shrinking it to a real geometrical point that is never the case in the case of fluid mechanics. It is just that we are taking it arbitrarily small however we are still within the continuum limit the idea was that just to point out if you take it arbitrarily small alpha will be unnecessarily large which does not make any sense on a physical basis that is all. Okay thank you. PVG Pune. Pune Mijerthi Guru please go ahead if you have a question. My question is regarding the slide number 2 on differential analysis that is the balance statement for an elemental control volume the last term on the right hand side rate of increase of mass is there any illustration where the rate of increase of mass will take place that is question number 1. Question number 2 can you please explain the concept of conservative and non-conservative equations for differential analysis. Thank you. Yeah so there are two questions one is related to the general balance statement that has been mentioned on couple of times actually in integral analysis also and here as well. So the question is specifically with respect to the source term and if there is any rate of increase of mass because of a source in a practical situation. So in fact this is the general statement that has been written and I have mentioned a few times that the rate of increase of mass due to a source is essentially not possible unless you have a nuclear type of reaction happening in your flow field where you are actually either destroying or creating mass through the Einstein's relation of E equal to mc square such situations are not really to be encountered in general fluid mechanics. So for mass balance this term will be automatically 0 in general. The second question was related to conservative form and non-conservative form. So let me go ahead one more slide and point out slide number 3 here. The conservative form for example right now we are talking about conservation of mass here what you see right now on your screen at the bottom is the conservative form of the mass balance equation or the continuity equation. I have mentioned a few times that the conservative form you obtain when you employ a balance equation such as what we have been employing on a control volume basically you obtain a relation which says that there is a rate of accumulation and that rate of accumulation is because of the quantity that comes in is different from the quantity that goes out etc. So when you employ such balance equations on control volumes you will generate a conservative form of the equation. On the other hand if you decide to employ a Lagrangian form of approach wherein you will be then following a given fluid particle then inherently you will generate a non-conservative form of governing equation. So let me take you to slide number 6 and right at the bottom of slide number 6 we have obtained the continuity equation using this so called Lagrangian form by following a given fluid particle which always has a fixed amount of mass and then essentially saying that the substantial derivative of the mass content for the given fluid particle is going to be identically 0 and in doing so we inherently generate the non-conservative form of the governing equation. So this is the general guideline if you employ a balance statement you will generate the conservative forms if you employ a Lagrangian approach you will generate a non-conservative form. R K college Rajkot please go ahead if you have a question. Yes sir good afternoon sir in slide number 7 differential analysis I have two questions first is for stream function u is considered positive and v is considered negative what happens if the direction is vice versa if we take v as positive and u as negative what is the reason and my second question is city potential function since the stream function is defined only for two dimension and phi is defined for three dimension. So there are two questions first question is on the stream function as we have defined in slide number 7 on the differential analysis and the question is we define the x velocity as plus y derivative of psi whereas v as minus x derivative of psi so why is that this has been done to be honest with you I think it can be done either way so it does not matter whether you define u as minus d psi d y and then v as plus d psi d x typically the convention that people in fluid mechanics have followed is that u has been defined as d psi d y and v has been defined as minus d psi d x the negative sign whether it is on v or whether it is on u is immaterial the reason is because you take this couple of expressions and plug them back into the continuity equation you will satisfy the continuity equation unconditionally whether you define u as plus d psi d y or minus d psi d y and then vice versa. So in that sense there is no hard and fast rule that this has to be plus and this has to be minus is just that I think some accepted conventions are being followed here that most of the fluid mechanics literature seems to define the u and v velocities in this fashion. The second question was related to something called velocity potential and we have not talked about velocity potential because we have not really dealt with irrotational flows in any detail. So velocity potential is a concept that is normally utilized when we deal with irrotational flows and those are also called as potential flows as some of you will obviously know. So in case of velocity potential you will recall that the velocity vector is defined as gradient of the velocity potential and there also there is a debate whether it should be given a negative sign or not. So let me just switch to wide board for a second. So velocity potential which is phi is a scalar function again and the velocity vector is defined as the gradient of this scalar function and as you correctly pointed out that the stream function is valid only for a two dimensional flow whereas the velocity potential can be defined for a three dimensional flow as well. The reason that we do not really have a stream function for three dimensional flow is because for a three dimensional flow then we cannot really identically satisfy that continuity equation with any kind of definition that the stream function may have. So only in two dimensions that continuity equation which is divergence of velocity equal to 0 can be identically satisfied by the stream function definition. So stream function is indeed restricted to 2D velocity potential can be used for 3D as well. G. H. Risoni College Nagpur please go ahead if you have a question. In differential analysis slide number 28 you are given del u by del x equal to del y by del x of u square by 2 so please explain and in differential analysis 14 sign connection. So there are two questions one is on differential analysis 28 where we were deriving that Bernoulli equation. How this partial derivative with respect to x of u squared over 2 times the x has been obtained. In fact I actually worked out the entire derivation for this so let me try to point out what we were doing in there. So this term here which now you see on your screen let me write it again on the white board is u multiplied by partial derivative of u with respect to x. This is simply written as d dx of u squared over 2 that is all that is. So if I want to bring this u inside the derivative I really need to make this u squared over 2 because let us see what happens now. u squared over 2 if you want to differentiate with respect to x this 2 is a common 1 half is a common fact or constant which will bring out d dx of u squared which is 1 half using the chain rule 2 sorry 2 u du dx and then that 2 and 2 will cancel and you will get your u multiplied by du dx which was what you started with. So this is the only manipulation that has been done that I want to bring this u here inside the derivative sign and when I want to bring that it will have to multiply this u and also divide by 2. So it has to be a d dx of u squared over 2. So you employ the chain rule when you are differentiating d dx of u squared over 2 and then you will get back your original form. So that is what has been done nothing more and your second question was related to 14 slide number 14 sign convention I thought you said sign convention. So let me go to the next slide. So what we said is that we will consider a stress component positive if both the surface normal for the surface over which the stress component is acting and the stress component itself both these guys point together either in the minus coordinate direction or plus coordinate direction and that is how we decide whether a stress component is positive or not. So for example let me draw something on the white board. So let me have a rectangular control volume or a fluid element if you want. Let me look at the left vertical face and the right vertical face. So if I show a stress component pointing in negative x direction on the left face and if I show a stress component obviously these are normal stress components pointing in the positive x direction for the right face. Now if I want to show the surface normal for the right face it is the unit surface normal that is also pointing in the positive x direction and so is the stress component. So therefore since both the stress component and the unit normal are pointing together in the positive x direction we call this a positive stress component. Similarly on the left face now the stress component is pointing in the negative x direction and the unit outward normal will also point out in the negative x direction. So again since both n hat which is a unit normal and sigma are together pointing in the negative x direction we call this a positive stress. See the reason why this convention is getting followed is if you are familiar with strength of materials or solid mechanics if you see these two stresses right now the way they are shown they act as tensile stresses on the on the material. So a tensile stress will be considered as positive in general from the solid mechanics or applied mechanics background and in order to preserve that convention we are also doing the same thing here. So I hope that answers your question. Okay thank you sir. Shastra University, Tanjavur please go ahead if you have a question. For calculating volumetric strain rate in the numerator new volume has taken as for two dimensional flow then new length into new rate. But it is newly formed that system is not actually it is not rectangular shape. So how can we take a new length into new rate that is first question and second question I have some about this pressure while taking pressure for static fluid and for dynamic fluid I have some doubts regarding for taking in momentum equation for dynamic fluid actually pressure should be will be in all direction but it will be cancelled automatically why should we take in component wise that is one more doubt these two are my doubts. The first question is related to calculation of the volumetric strain rate where the volumetric strain rate has been calculated as the new volume minus the old volume divided by the old volume and then divide that by the time interval over which this is happening. What is getting pointed out is that the new volume is not going to be rectangular so why are we treating this as a rectangular volume when we calculate it as a new volume. The point is correct actually what you are pointing out is that everything is happening simultaneously as we have repeatedly said. So all these four effects the translation rotation deformation and the two types of deformation that is they are all happening simultaneously as the fluid particle moves from one location to the other. However for the purpose of analysis what we assume is that since that delta t is very very small these are typically very small in terms of the magnitudes and that is why to a first order accurate expression we calculate those volumetric strain rates by simply assuming that the stretching is such that the new volume is also approximately rectangular otherwise you are absolutely right that it is not really remaining a complete rectangle as time progresses. But these are all results which are accurate to the first order accuracy as we say. However the first order accuracy in case of continuum fluid mechanics has been found to be sufficiently good and that is the reason we accept these results and move on with those. So however technically speaking you are I would say right. Sir, second question is regarding pressure while taking momentum in momentum equation for calculating momentum in for one dimensional flow we have taken v1 v1 plus m1 v1 like that we have taken for momentum calculation momentum in. So in that we have taken momentum due to pressure also p1 v1. So while taking that I have p1 a1 sorry p1 a1 my doubt is pressure acts in all direction if pressure is at position and it will be same in all direction it does not have components in all components if you take any component it will be same. But while calculating why we are taking for x direction and y direction and z direction specifically that is my doubts just I want more clarifications on pressure itself. See I do not know if you are referring to any specific slide from any specific topic the question is related to the pressure and it is variation. So let me try to just answer it in little general terms the pressure is equal is what is getting pointed however it is equal only at a given point. So the Pascal's law as you are quoting here will say that the pressure is equal in all directions but it is only at a given point. So as long as you are dealing with a fluid domain in general the pressure will vary from point to point as you go from one location to the next and that is the reason we have this gradient of pressure as the net force acting due to the pressure differences from point to point when it comes to a moving fluid that is all really to it at a given point I understand that it is equal in all directions. However in general from point to point in a continuum fluid domain the pressure will vary and because of which there is a gradient of pressure which will act as an accelerating force for a fluid particle. So therefore we end up taking gradient of pressure as the net force due to pressure in the momentum equations. Ok sir still I have some clarification because if you take pressure at one point it will be in all directions it will be cancelled as its own but while calculating for momentum we are taking in one component x component and if you are calculating for x momentum we are taking in one direction and similarly in y direction and similarly in z direction but it will be cancelled automatically. So what is the requirement of taking it for as a component wise that is my doubt. Can you please point out any specific slide with which I can have this discussion? I have general doubts sir. Because see the if you are referring to let us say an integral analysis where we are dealing with a with an enclosed completely closed control volume and if it is given as part of the problem statement that you can assume that the pressure is completely uniform in the entire domain then what is going to happen is the pressure contribution from all sides on the control volume will simply cancel each other out and there will be no net effect of the pressure force. So that is that is fine. However in general the pressure is not really constant and it does change from point to point and that is the reason we normally want to include that in our in our analysis especially when it comes to differential analysis we want to include that in terms of the gradient of the pressure which will be acting as an accelerating force for a fluid particle. But in if you are referring to any specific integral control volume type analysis we have simply assumed in some cases that the pressure is uniform for a completely enclosed control volume and in which case it will cancel cancel its contribution. As far as momentum flow etcetera are concerned pressure has nothing to do with momentum flow is always calculated as the mass flow rate multiplied by the velocity of the flow. So there the pressure does not come into picture at all the pressure will come in only in the calculation of the net forces that are acting on a fluid particle or equivalently on a control volume. So there is no connection between momentum flow in and out of a control volume and and the pressure. Yeah Nitte Meenakshi please go ahead if you have a question. The relation established between stress component and strain rate in the slide 21. So by assuming isotropic fluids so will it be the same relations will it be applicable to unsteady flow field that is first question and the second question how easily we can define the stream function without any formulae and please can you elaborate more on isotropic fluids thank you with the example. So the question is at least one of the questions is on these Stokes relations or the constitutive relations and the question was about isotropic nature of the fluids what is meant by that. So isotropic nature of the fluid simply means that the fluid will exhibit a stress and strain rate relationship identically in all possible directions. So technically you can say that this is an assumption it is a simplifying assumption that has been used in the Stokes analysis and even though we have not really discussed the Stokes analysis completely in detail I wanted to simply mention it. It so happens that most fluids of practical importance seem to follow this isotropic behavior quite reasonably and therefore for most fluids these Stokes relations are found to be quite applicable. So most of the practical fluids that you can think of including all gases and more or less all types of fluids do follow this quite well. So that is all about the Stokes relations and isotropic nature of the fluid. Can you please repeat your question on stream function. The relations which we established will it be applicable to unsteady flow feed. So that is the first question which I asked and the other one is how we can easily define the stream function without any formula. Yeah so the question is about stream function and the first part of the question is whether it is applicable for unsteady flows. It is actually it is in fact unsteady it is applicable for unsteady flows as well. Let me go quickly to the whiteboard. So in general really I should have probably mentioned this in my slides. In general the stream function is defined for a two-dimensional unsteady flow as well. So it is just that we did not choose to talk about unsteady flows but in general stream function is indeed defined for an unsteady flow. Even though it is defined for unsteady flow the relations that we have for the velocity components namely u is equal to u is d psi dy and v is equal to minus d psi dx will still remain good whether the flow is unsteady or not. So the answer is yes it is applicable for unsteady flows and the relations still remain exactly the same. You also asked whether we would know stream functions without any specific formula and actually to be honest with you the concept of stream function is fairly restrictive. We do not really utilize the concept of stream function nowadays. It is an old concept and for some specially derived flow type situations you can actually have a close form relation for a stream function. So the idea is that in general the stream function is a useful concept when it comes to plotting the CFD results in terms of stream line plots and so on. So that is all the utilities in my opinion in today's world otherwise it is a really old concept and somewhat restrictive as well. R C Patel, Shirpur if you have a question please go ahead. What is the mean for order density function for 0 for irrotational flow? Then why? It is only used for application for irrotational flow only. It is applicable for other flow also. See the so the question is about the concept of vorticity and if vorticity is 0 we call this as irrotational flow. So what we have done here is we have formulated the expression for the rotational motion of a fluid particle. So what I mean by rotational motion is that as the particle is moving it is spinning about its own axis. So if it is happening in a fluid then we say that the angular velocity is non-zero and by the virtue of the relations that we derived which are projected right now on slide number 25 here we simply establish the relation between this so called vorticity which is defined as the curl of the velocity and the angular velocity. So in that sense both vorticity and angular velocity imply the same thing namely physically spinning of a fluid particle and if the fluid particle for whatever reason let us say is not spinning about its own axis as it is moving in a flow we simply say that it is experiencing 0 angular velocity or 0 vorticity and corresponding to that we say that the flow is irrotational. So there are some classes of flows where the action of viscosity is very very low because of which the velocity gradients are very low and if the velocity gradients are essentially non-existent from the expressions that have been listed at the top you may be able to figure out that the angular velocity will be exactly 0. So in such situations which are typically applicable in inviscid flows you will see that the flow can be treated as irrotational which simply means that a particle that is moving in an irrotational flow does not spin about its own axis and that is what has been shown in the sketch at the right bottom. Can I give an angle please go ahead if you have a question. So I have a doubt in that slide number 22 in a differential analysis here normal stresses sigma which we are going to be providing that minus p plus sigma xs viscous can you please elaborate it. Yes so the question is on slide number 22 on differential analysis specifically on the normal stresses that we talked about and I have mentioned that the normal stress in general will be composed of a pressure contribution and a viscous contribution as well. See in general this is true in a fluid. So normally what we are used to assume is that the normal stress in a fluid or on a fluid particle if you want to look at it that way is simply equal to pressure and pressure being compressive we are going to employ that minus sign for it. However in general what is found is that there is a non-zero contribution from the action of viscosity as well on the normal stresses or in normal stresses. So there is a component that gets added because of viscosity in normal stresses. It may not be that obvious to follow that the viscosity is actually contributing to a normal stress but it does and although to be very honest with you we have not gone into serious details of the derivation of the stokes relations. I have simply listed here what are those and that is what I would like to really limit my answer to that at this point you just note that there is a contribution from the viscous effects as well to the normal stresses and I have simply listed the expressions that exist for the normal viscous stress by coating an analogy with the normal stress that occurs in an elastic solid which follows the Hooke's law. Vijay T. I. Mumbai if you have a question please go ahead. Sir how do you physically interpret stream function? So the question is on how to physically interpret a stream function. There are two ways you can interpret the stream function. One of which was explained in one of the slides. Let me try to pull up that slide. So one of the ways you can interpret this stream function is that when we have a stream line the value of the stream function turns out to be a constant for a given stream line. So psi equal to constant will provide you a stream line field. So if there are various stream lines in a stream line field each stream line will have one value of the stream function associated with it that is one way of looking at stream function. The other way is that if you let me draw a sketch on the whiteboard. So let us say that we are looking at a pair of stream lines for which the value of stream function is psi 1 and psi 2 and let us assume that psi 2 is greater than psi 1. So it can be shown that psi 2 minus psi 1 is actually equal to the volume flow rate between these two stream lines on a per unit depth basis. So this is a very standard derivation which you can find in one of the text books that I have mentioned. But since you wanted to know the physical interpretation of stream function this is the most relevant interpretation from a physical basis that the difference between the values of stream function for two different stream lines will give you the volume flow rate that is flowing between these two stream lines on essentially a per unit depth basis because by definition we are always talking about two dimensions when we talk about stream function. Sir in differential analysis slide number 23 what is the significance of lambda? Yeah so the question is on slide number 23 differential analysis what is the significance of lambda? The second coefficient of viscosity. So let me answer this in a slightly hazy manner in the sense that we have not really talked about the complete detail of the Stokes's development in which the relations between stresses and strain rates are established. However, if you actually follow the complete treatment in one of these advanced fluid mechanics books you will see that the normal viscous stress that a Newtonian fluid will experience will actually contain two constants. So one of the constants is usually determined as the dynamic viscosity. So that is what is getting highlighted here as mu the other constant is usually called as the second coefficient of viscosity and that is what is given the symbol lambda. So unfortunately since we have not really completely talked about how these Stokes's relations are derived I am not in a position to talk about these in a serious detail. However, at this point what you can note is that the normal viscous stress in a fluid when it is expressed in terms of the strain rates will contain two constants in the expression. One of them which is called lambda here is what is called as the second coefficient of viscosity that is all I will right now say and what I will do is at the end of the course I will put up a list of some of these advanced textbooks in which you can actually try to follow the entire Stokes's development and there you will be able to see where this set of couple of constants will appear. I want to know what is the difference between the thermodynamic pressure and mechanical pressure and how we equate equal what is the difference between thermodynamic pressure and mechanical pressure. Yeah, so the question is about what is meant by the thermodynamic pressure and what is meant by mechanical pressure. So thermodynamic pressure is what we call a static pressure in fluid mechanics terminology and static pressure or thermodynamic pressure is something that you measure if you move with the flow so that the sensor with which you are measuring the pressure does not have any relative velocity with the flow. If this is what is happening then we call it as a static pressure. So in case of thermodynamics you talk about usually closed systems let us say a gas inside a container that is stationary and if you include a sensor in such a system there will be an interaction of large number of gas molecules with the sensor and they will transfer their momentum on to the sensor which will be eventually interpreted by the sensor in terms of a pressure. So this is what we call a static pressure and since we are dealing with a static fluid in the case of the closed system as I just talked about in fluid mechanics. In order to replicate this static nature what we do in case of a moving fluid is that we have to move along with the fluid at the local speed of the fluid so that there is no relative velocity between the sensor and the fluid so that the sensor essentially thinks that it is measuring a static pressure. So that is what we call as either a static pressure or thermodynamic pressure. Mechanical pressure as I explained earlier in the lecture is an artificially defined term which is simply as the slide projected right now shows is negative of the average of the 3 normal stresses and in general it is different by a factor that is shown on the third equation here. However, for the case of incompressible flow we have divergence of velocity equal to 0 or the 3 linear strain rates added together will identically be equal to 0. So in the case of incompressible flow there is no difference between that so called thermodynamic pressure and mechanical pressure. If you are dealing with strongly compressible flows and those also involving shock waves and such then you actually have to perhaps deal with these two separately even then typically using this Tuxes postulate which essentially assume that the thermodynamic pressure is equal to mechanical pressure you do not see any serious errors in the analysis. So for all practical purposes the pressure that we know from thermodynamics is what you can treat as the mechanical pressure only in case of strongly compressible flow you may have to deal with something different but usually these situations are very very scarce. So normally you do not have to really bother about it I just wanted to point out if you want to read a slightly advanced textbook you will normally come across such terminologies and at least I thought I should point out to you what is meant by thermodynamic pressure and what is meant by mechanical pressure. I will stop this question and answer session right now.