 Last time we were discussing about the integral forms of the conservation equations and as an example we looked into the integral form of the mass conservation and it is corresponding differential form also we revisited and we found out that it is possible to convert one form to the other. Now let us look into some more examples of the use of the integral form of the mass conservation equation. Let us say that you have a wedge shaped element like this with the axis oriented along x and y and the velocity field is a 2 dimensional velocity field is given by say u is given by this v is given by this one and 2 dimensional w is 0. Let us give some names to the faces of the elements here. The objective is to find out that what is the volume flow rate through AC? The dimensions are given let us say this is 1 meter or just one unit all are given in some consistent units say this is 2 these are given. So what we also are assuming that v0 is a constant it is not a function of any other variable that we are looking for. So how should we proceed with this problem? Let us say that we are interested to use the integral form of the mass conservation and that is one of the natural things that we should use here because there are the 3 faces across which fluid will enter and leave. So the net rate of transport should be given by the integral form of the mass conservation that is you have dm dt for the system plus this for the control volume plus integral of rho. The left hand side is 0 because no matter whatever system you are considering it is by definition of fixed mass. The right hand side because we are assuming that rho is not changing with time neither the control volume is changing with time. So what is our control volume? Let us say that this triangular shaped element is our control volume. So whenever we are making a control volume analysis it is important to identify what is the control volume that we are taking because you may take different control volumes. Of course for this problem this is an obvious choice of the control volume but there could be problems where there could be many different choices of the control volume. So the equations that you are writing should be pertinent to a particular control volume that you have chosen and that should be clearly mentioned. So this triangular shaped thing is a fixed control volume the volume of that is not changing with time. So it is not a deformable control volume neither the density is changing with time. So this term goes to 0. So what remains is this net term which is nothing but basically the net rate of outflow minus inflow of mass equal to 0. So if rho is a constant let us take a further simplification that rho is a constant then this will boil down to basically integral of v dot n dA over the control surface equal to 0. And when we say v here the relative velocity and absolute velocity are the same because it is a stationary control volume it is not a moving control volume. So this now we may break it up into 3 parts because the control volume has 3 different distinct oriented surfaces. So you can write this as the sum of the effects of AB, BC and AC. So for AB when you write what should be this corresponding expression. So let us identify that where is that AB so you are looking for this phase. You are interested to find out what is v dot n integral of that over the entire area. So one important thing is here n is not a variable it is like a line oriented along y axis. So n is a constant. So for AB what is n? So let us try to identify the what are the normal directions for the different edges here the edges are all straight lines. So they have unique normal directions. So let us say AB what is the n for that minus i then BC AC. So for AC let us say that this is the unit vector normal. So this will have its components let us say this angle is theta. So if the angle between the normal and the horizontal is theta then the angle between the edge and the vertical is also theta. That means you can say that n is definitely cos theta i plus sin theta j where you have tan theta equal to 1 by 2. So what will this imply? This implies that like you have identified the direction normals for all the phases only thing you require is the velocity. So the velocities for the phases for AB what is v? Let us write the in the same table let us try to write what is the velocity v cap. So for AB what is v cap? For AB x is 0. So u is not there there is some v okay. So it is minus v0 y by l j. For BC similarly you have y equal to 0. So it is v0 x by l i and for AC it is the real the sum of the 2 components because here x and y are both non-zero. So we are not writing it because it is like let us just write it as in general ui plus vj. Now to get the integral you have to keep in mind that if the velocity varies along that length then you have to integrate it over the length to get the total flow rate. When you come to the so let us start with AB you can of course evaluate this try to evaluate this but let us not do that bulwark. You see that the velocity along AB is like it is oriented along AB. So there is no normal component of that. So there is no net flux or influx or outflux of flow across AB because there is no normal component. So if you make a dot product of this that you can clearly make out and so that will not give rise to any net flow. Same might be true for BC. So the choice of the axis here has been such that if this is the velocity field then that is the case but the same is not the case for AC. So for AC how will you find out what is the total or the net rate of flow? So if something there is nothing that enters here there is nothing that enters or leaves here you expect that there is nothing enters or leaves here. How could you verify by doing the integral that it should be 0? No that is what like this is QAB this is QBC this is QAC like if you evaluate by doing the dot products. Will the dot product automatically give it? It is expected that the dot product integral over that should automatically give it. So that you should check so that it gives you a confidence of how to calculate that because here I have given a special type of velocity field. So that actually this problem solution is not necessary. I mean I am trying to go through this through a formal route to give you an idea of what should be done in a case when the problem solution deserves that but here actually does not deserve. So this is a more intuitive case when nothing is entering and therefore is expected to leave. Now if by chance you get something which is leaving through this and nothing is entering so that will really violate the law of mass conservation. So that one has to be careful of. Now let us look into some other problem which is not as trivial as this one. So another example let us say there is a tank like this the velocity profile at the exit of the tank is through this pipe is given by this one and in terms of a local coordinate system this let us say that the local coordinate system is x1, y1 it is given as u1 is equal to some u0 into 1-y1 by y1 square by h square where y1 is the transverse coordinate and h is this height and the fluid entering here here the again in terms of the local coordinates you can specify it but that specification may not be necessary because it is given that it is a uniform velocity profile here with a velocity u infinity which is uniform and let us say this is h by 2 okay the directions of the axis are not given that means it is not given that what is this angle theta say theta 1 it is not given that what is this angle say theta 0 these are not given what is given it is given that the density is a constant you have to find out what is u0 given the width of this figure perpendicular to the plane where it is drawn is uniform okay so u infinity is given h is given these 2 things are given okay so how will you go about it again looks like a situation where mass conservation should be applied and if you want to apply the integral form of the mass conservation let us say that we consider a control volume so what should be a good choice of the control volume so a good choice of the control volume is something where across the surface we are totally confident about the velocity field so let us say that we make a choice of the control volume something like this so with this choice of the control volume you can write the law of mass conservation if you want to write that then like it is because rho is a constant again it will boil down to a case very similar to the previous one where eventually it will be integral of v dot n dA over the control surface equal to 0 the remaining terms will not be relevant so this is the only term that is relevant now out of the surfaces that you have only you have one inflow and one outflow surface across the other surfaces fluid is not flowing so those surfaces are not relevant so you may break it up into 2 integrals one for this inflow and another for the outflow okay so the first one is very straight forward let us do that so when you have the v dot n dA see this v is uniform over the area over which it is flowing for the inflow so you can take away this v out of the integral not only that the dot product also because n is also a constant here so v dot n entire thing you can take out of the integral so what it will become it will become v dot n into area of the face over which it is coming so let us call that this area of the face is here a naught and here the area of the face is say a1 so when you write v dot n you have to keep in mind that what is the direction of n along this surface so the direction of n is opposite to v right so v dot n will give the minus of the magnitude of v so for the area 0 or area o it will become this term will become what minus u infinity into h by 2 into the width let us say v is the width perpendicular to the plane of the figure the other area for a1 we cannot have the same consideration because the velocity varies over a1 so to see how the velocity varies with a1 we are or along a1 we are already given that with respect to the local transverse coordinate how it varies but n is a but you have to keep in mind that forget about that functional dependence on y1 u1 is going out of the area and what is n n is also oriented out of the area this means that if you just take the vector sense v dot n that will give you the magnitude of v into 1 because the dot products are in the same dot products of 2 vectors in the same sense that will be leading to that conclusion but because it is varying with y1 now you have to really do the integration so when you first have v dot n that will become u0 into 1-y1 square by 8 square that will be v dot n and dA dA is what say you take a small element on the axis y1 so this is a small element of the area so what is that small element of the area say at a height y1 from the center line see the coordinate system is from the center line so at a height y1 from the center line say you have taken a small area of width dy1 so the elemental area which is like the dA given symbolically it is dy1 into b if you integrate that from-h by 2 to h by 2 then that will represent that what happens for the area 1 so sum of these 2 should be equal to 0 clearly from the minus sign of the first term you can make that it is inflow and the plus sign of the second term means that it is outflow and it is possible to complete this integration in a very simple way we are not going into that just to save some time but important thing is that see this is the variation of u over the section so you can write it equivalently as integral of u dA in the fundamental form in a scalar form so this is this is not like a vector form so it is like just like because eventually with the dot product it has become a scalar you have to keep in mind this is nothing but the component of velocity which is normal to the area here fortunately all component is normal to the area there is no component which is cross that so this you may express through some quantity which is called as average velocity so average velocity is this divided by the area that means if this velocity was uniform but the same flow rate was there see if it was uniform then that uniform velocity times the area will give you the flow rate that is what the first term has told us and if it is not uniform obviously you have to integrate it to get the flow so if the flow rate were uniform in a hypothetical case sorry if the velocity profile was uniform in a hypothetical case but the flow rate becoming still the same as it is in the real case then if you equate those 2 flow rates then that equivalent hypothetical uniform velocity this is called as average velocity so it is like a equivalent uniform velocity that would have prevailed across the section satisfying the same volume flow rate as it is there in the real case so therefore this is like as good as some u1 average into a1 and this is just like u0 into a0 so it is just like a1 v1 equal to a2 v2 you have to keep in mind again that what is v1 and what is v2 these are very important again I am repeating these are fundamentally the average velocities over the sections 1 and 2 see when we are writing here 1 and 2 or may be 0 and 1 or whatever subscripts there is a fundamental difference from what we wrote for using the Bernoulli's equation those we wrote for points 1 and 2 now we are writing for sections 1 and 2 so may be subscript wise they look very similar but meaning is entirely different when it is uniform it does not matter it is as good as writing for a point because velocity does not vary from one point to the other but you do not have anything called as area at a point so it is basically when you write a1 no matter in the context of what we write what we have written earlier for 2 points when you use the Bernoulli's equation we should keep in mind that there also a1 was for the section that contain the point 1 so it is not that area of a point 1 or something like that but there we use velocity at the point 1 the reason is that in the Bernoulli's equation we use velocity at a point so we had to link it with velocity at a point now it is like we are linking things through velocity over an area so if you complete this problem you will get what is u0 because all other things are known now let us look into some other examples where may be we look into different case may be unsteady case let us say similar problem where you have tank like this and there is a free surface here so in the previous problem we assume that there is no change in level of the surface in the tank but in reality that does not happen so that is something which is a bit hypothetical in reality it may happen that approximately the change in this level is 0 because this is such a large area that no matter whatever is entering and leaving it is not changing the height these types of tanks are called as constant head tanks so they maintain constant head because with respect to the inflow and outflow the change in height of this is so small because of maybe this is a large reservoir so the area is so large that it does not change any level but the more realistic version of the previous problem is that yes the level of the water will also change so let us say that the height of the free surface from the bottom of the tank is h just to simplify the situation we now go back to a case of a uniform velocity profile because we have already seen that if it is non-uniform that it is not a very difficult thing we have to just integrate the velocity profile over the section so with that understanding let us say that you have a situation like this let us say it is uniform so you have a velocity v1 here and let us say area of cross section a1 you have a velocity v2 area of cross section e2 when we were talking about the previous problem see we could get rid of the situation of change of height of the tank even in a real case by some approximation that yes the rate at which the water is entering is the same at which the water is leaving so it is not changing the height of the level of the tank but when both are entering that is not the case so here the only chance of the height of the tank or the level of the water in the tank not changing with time maybe only for the consideration that the area of cross section of the tank a is so large as compared to the others that the corresponding change in height is very small otherwise here there will always be a change in height smallness or largeness depends on the areas in the previous problem you could cleverly come up with a situation where there is 0 change in height by making sure that the rate at which it enters is exactly the same at which it leaves so it appear to be bit hypothetical but it is not that hypothetical if it is really doing that so if you have a reservoir and water is entering and living at the same rate why should it change the level of the reservoir it will not but here both are entering so here our objective is to find out how the height is changing with time given these velocities areas and the information that these velocities are uniform over the area okay so let us take a control volume again write the equation so dm dt for the system is equal to this one left hand side 0 right hand side the first term let us assume that the density is a constant let us say that is given okay so if the density is a constant it will come out of the integral but the volume is not a constant so the volume within the control volume is what is it is a into h a is the cross sectional area of the tank and a is the height so this effectively boils down to what this effectively boils down to d dt of rho into d dt of a h area of cross section is a constant so this is as good as rho a dh dt see h is a function of time only and nothing else so this partial derivative becomes an ordinary derivative here and what about this term so it has now effect of the 2 areas 1 and 2 so because we have seen that formally what is the consequence of the dot product and all those things we will now try to write it directly without going through that route so for the surface 1 what will be-rho v1 a1 and for the surface 2-rho a2 v2 right so sum of this should be equal to 0 so from here you will get what is dh dt that is very straight forward again you can make an observation that if dh dt equal to 0 that is not a possibility because these 2 terms of the same sign cannot cancel that and only way it can cancel if v1 and v2 are of opposite sense that is if one is entering the other is leaving and then it again boils down to a1 v1 equal to a2 v2 so that is like a previous case that we had considered the other important thing that we might discuss in this context is that can we try to choose a different control volume let us say that we choose a control volume which is not the previous one but say the new dotted line that I am drawing fundamentally there is nothing wrong but it will not help us solving the problem why because now the control volume is cutting across some flow surfaces across which you do not know the velocity profile or you do not know how the velocity is so how will you find out this integral term you have to have those locations where at least you have an idea of how the velocity is varying or what is the velocity so the choice of the control volume is not that it is only something which is unique and you cannot choose anything else but if there are many alternatives you have to find out that what are the unknowns and knowns involved with that alternative and it is wise to choose a control volume which gives very easy situation by reducing the unknowns and the initial choice that we made is an obvious choice towards that let us take another example of similar type may be a bit different let us say that you have a conical tank just for a change radius capital R and height capital H there is a small hole at the bottom of the cone through which water is coming out with a velocity V e and because of this leaving of the water the height of the water in this conical tank is reducing maybe initially it was the full height capital H but because water is leaving at some instant of time say the height is like h small h so this small h is changing with time because water is continuously leaving let us say that it is leaving through the small hole with radius of small r both like this is a circular hole and cone is of course a circular cross section then you have to find out what is dh dt it is given that you may approximate V e by root 2 g small h it is given we have seen earlier that this is not actually a very correct estimation but it gives a sort of approximate situation under certain simplified assumptions we have discussed those assumptions in details when we were discussing about the Bernoulli's equation so this problem is fundamentally not very much different from the previous ones except that the geometry is such that you have a variable cross section area nothing bit more complex than that so let us try to use the conservation of mass here so the first term what it will be again assume rho is a constant if rho is a constant so the bad thing that we have approach is like this that we have tried to solve the problem without identifying the control volume I am trying to do it in the same way in which you are habituated to do just like straight away going to an equation and solving a problem now it is ridiculous we are trying to apply an equation for a control volume but we do not know what is the control volume and that perhaps we were trying to do so let us identify a control volume we will identify one type of control volume and I will leave it on you as an exercise to identify a different type of control volume and come to the same answer at the end and that will give you a good idea let us say that we identify a control volume like this let us say with respect to that control volume we are writing this term so with respect to this control volume when we are writing the term you have let us say that we neglect the density of the air which is there at the top of the water in comparison to that of the water it is not a very bad engineering assumption because the reason is that air is much much lighter than that of water right typically 1 by 1000 so in fact that is what we did also in our previous problems where if you have a tank may be in the tank some part is water but the remaining is air but when we wrote this term we did not write the term corresponding to air so that was an inherent assumption that we were making or keeping in mind but not explicitly stating so then like if you take rho as a constant and out of the integral what will become this integral of dv so that this term will become basically rho derivative of v with respect to time now the volume is the volume of the water so at this instant that we are considering the volume of the free surface is located here so the volume of the water is one third pi local r square so let us say let us give it a name say r1 one third pi r1 square h and from the semi vertical angle of the cone let us say theta you have tan theta is equal to r1 by h which is same as capital R by capital H so it is possible to write the whole volume in terms of small h so the whole volume becomes one third pi in place of r1 it is h tan theta so h cube tan square theta then the other term what will be this one how many flow boundaries are there in the control volume only one flow boundary only one exit boundary so what is that so in this boundary v and the normal vector are located oriented similarly so the dot product will give a positive term so it will be rho then what so v dot n v is uniform over the area let us assume that so v dot n will come out of the integral rho will come out of the integral and integral of d a will become a so it will become like rho v e into a right so it will become rho into v e into pi r square okay so then it is very straight forward these two together is 0 and you can differentiate v with respect to time it is like an ordinary derivative again because h is just a function of time so one third pi into 3 h square dh dt into tan square theta so if you substitute that here you can find out that what is dh dt at a given instant when the height is h small h okay. Now I will leave on you solution of the same problem but with the control volume choice like this say you have a control volume which is adapting itself with the movement of the water so this control volume is a moving or a deformable control volume in a way that if the water level is coming down this is also coming down with it okay so this type of control volume and then with respect to that type of control volume you make the same analysis and try to come up with the same answer okay. Let us work out another problem let us say that we have a rigid tank spherical tank which contains air and originally there was a valve located here which was preventing the air inside to go outside. Now the valve is opened and once the valve is opened the air will go out of this pipeline which is connected to the tank it is given that number one the state is uniform within the tank in tank and pipeline that means the properties are the same at a given instant of time everywhere that we are considering and number two is that the rate of mass flow out is proportional to the density at that instant okay this is given you have to find out that how the density is changing with time it is expected that the density will change with time because it is a rigid tank. Because it is a rigid tank if you take away air from it it is density will fall because now less mass is occupying the same volume so that you have to find out that how that density is changing with time again let us start with the choice of a control volume on which we want to have our analysis. So the control volume let us say that we choose a control volume like this if we choose a control volume like this it does not matter whether the valve is opened or closed there will always be a flow across it. So now if you are asked that what happens after the valve is open you should take it like I mean of course the pipeline does not end here so the pipeline continues beyond the valve. So you should take it like where the effect will be apparent when the valve is open but again in this particular case and only if our objective is to look for a mass conservation it makes no difference as such because even if you take your control volume like the surface which is to the right of the valve or to the left right at any instant whatever is the mass flow rate here the same is the mass flow rate along this pipeline in terms of the mass flow rate but that may not be the case if the density in the pipeline itself is a function of time. So if the density in the pipeline is a function of time and if there is a possibility of like see variable density cases are very typical cases. So for variable density cases it is not so trivial to say that like the mass flow rate is the same in all cases but here no matter how the density varies you can say that these two mass flow rates will be the same why yes area of cross section is same. So density is like I mean whatever it is like even if you forget about the density like just fundamentally think if you have a mass flow rate that goes out of this if the same mass flow rate does not go out of this where does that mass go is any mass like accumulated in between no velocity may be different if the density are different but if the density is uniform then that may be ruled out provided the time dependence of the density is not creating any big change. So if you see that like what happens within this pipeline in terms of the velocity or density something we do not have enough information really to talk about that but one important information we have that whatever happens individually to velocity density but the mass flow rate what goes through this is same as the mass flow rate that goes out because where otherwise it will go it cannot sit on the valve that mass. So we are assuming that there is no accumulation it is a steady flow system so if it is a steady flow like whatever enters here the same leaves here then like it does not matter really whether your control surface goes through these or through these or whatever. Now let us look into the different terms in this expression so if you look into the first term so here it is a rigid tank in the previous class we tried to solve a similar problem which is like a flexible balloon so there the volume could change with time that is maybe if it was a if it were a flexible balloon you could have expected that by taking away air out of it the balloon will try to shrink but here it is a rigid tank so it cannot respond to that it can only respond to that change by having a change in density inside but not it is change of it is own volume. So the volume of the tank is the density of the fluid in the tank is definitely varying with respect to time but the volume is not varying with time so that means you can the first thing that you can do is you can take the time derivative inside because the volume of the control volume is not a function of time anymore and the next thing is that how this density changes with respect to time that does not change from one point to the other within the control volume why because it is given that is uniform state in the tank and in the pipeline that means the density may change with time but at a given instant of time the density is same everywhere within the control volume. So that means this is like an isolated term which does not depend on the volume so this you can take out of the integral so that means eventually it will become the volume of the tank into the rate of change of density with respect to time in the tank okay. So it is not so difficult to come up with this expression but we have to keep in mind that what are the important assumptions that are leading to this type of a simple proposition if the density was varying within the tank itself then we could not write it then we had to integrate it by keeping in mind that rho is a function of both position and time. Here the dependence on position we have frozen we have assumed that that is not there then for the next term so if you see that this now you have only one outflow boundary for the control surface when the valve is open the air is leaving here so you have an m dot exit which is the rate of mass flow rate out. So the rate of see this is eventually giving what this is giving a rate of mass flow rate outflow-inflow there is no inflows only outflow so this is as good as plus m dot e this is equal to 0 what is this m dot e? m dot e is proportional to the density so let us say that m dot e is equal to some k into rho so this rho is an instantaneous density that at that time whatever is the density in the system proportionate to that the mass is coming out so this you can write some k into rho where k is a constant so you can the next work is very easy all of you enjoy doing it very simple integration so d rho by rho is equal to minus k by v tan dt so you can integrate it from say time equal to t1 to time equal to t2 say the density changes from rho equal to rho1 to rho equal to rho2 so you can find out how the density changes with time okay so important is not the solution of the problem of course the solution is quite easy but solution is definitely having some importance but to my understanding the greater importance is what are the assumptions that are leading to the solution because in reality the problems are not so simple as many of these ones so these might not be very very common as our analysis equations because if you see these equations that we have written these are really on the basis of such simplified conditions or assumptions which might not prevail in practice so the final answer may be like interesting in terms of solving a particular problem but the reliability of the answer may not be so strong because you might have a strong variation of density within the tank itself but at least by having a simplified assumption it is giving us a fair idea of like how the analysis should involve the conservation of mass so when we have been discussing about the conservation of mass we have discussed certain types of problems what types of problems we have discussed we have discussed about one case when it is totally steady that is you have like the density first of all the density is not changing with time volume of the control volume is also not changing with time so that this term is not there so it is just a balance between the rate of outflow and inflow and physically that represents a condition where rate of outflow is equal to rate of inflow for that we have to keep 2 things in mind one is that what is the sense of the velocity velocity component normal to the area is it going is it opposite to the area vector or is it along the area vector the second point is is the velocity uniform over the cross section or is it non uniform accordingly we might need to integrate or not and when it comes to us unsteady case there are 2 types of possibilities one is the density is changing with time another is the volume of the control volume is changing with time and maybe a third case when both are changing with time but like we have perhaps not considered that case but that is like it is a combination of the cases that we have considered so all these cases have given us a form footing or understanding of how to use the integral forms of conservation equations what is the advantage of use of the integral form of conservation equation see when you are using an integral form the important advantage is that you are being abstracted from how things vary within the control volume you are only bothered about a gross manifestation in terms of what is entering what is leaving so what is happening inside you are just representing it in an integral sense or an overall sense you are not really representing it a point by point variation so we have to keep in mind here a very important thing what is the difference between an integral equation and a differential equation physically differential equation gives you the variation at a point whereas integral equation gives you the variation over a domain of course the domain is constituted of many such points but the differential equation is valid only at an identified point in the domain so the integral form when you are writing you may get back the differential form from that but at the same time you are not forced to track or to keep in mind that what is happening as a point by point variation so that differential nature of variation you may not be interested in so in such cases where you are not really interested in that it may be convenient to use the integral form so one has to keep in mind when should we use the integral form and when should we use the differential form it depends on the physical sense of the problem that we are trying to solve.