 In this lecture we will discuss on the topic of principles of similarity and dimensional analysis. Let us first get a background or motivation of studying this topic. Let us say that we are trying to have an idea of the how to design an aircraft and we know that if you want to design an aircraft you want to have a clear idea of the lift and the drag forces as some of the fundamental entities to design it for the real system. At the same time we know that it is somewhat prohibitive to have many experiments on aircrafts of real sizes. So if you have an aircraft of real size and if you want to test it it is not only very expensive but also there are many other drawbacks associated with such experiments. So one might be interested to have a reduced model that is a model of aircraft of maybe a reduced size and then test it in a wind tunnel. So in a wind tunnel say keep the aircraft model and have a controlled flow of air relative velocity between the aircraft and the air and then from that if you measure the pressure distribution you may also measure other parameters say velocity distribution and so on and you will get a clear picture of what is the flow around the aircraft. The question is can you extrapolate this to the behaviour of the real aircraft that is a very big question. So the first question that we would like to answer is that given a study on the basis of a model of a different size how or whether you can extrapolate the results of those experiments to predict what is going to happen in reality in the real situation. So in the real situation whatever is the entity that is being used that is considered to be a prototype and the model is a version of a prototype a scaled version of the prototype. In this example the model is smaller than the prototype it is obvious because if you have a real aircraft that is quite large you want to reduce its size. It is not always true that models have to be smaller than the prototype. Sometimes the prototype itself may be inconveniently small and you might want to have a model a bit larger than that. Then will be that whether that scaling will affect your results or not that is one of the important things that you want to answer and if the scaling does not affect your prediction then the next question comes that how can you utilize the results from the experiments with the model to predict what is going to be for the prototype that is the first question. The second question is let us say that you are doing experiments with a model you may have many parameters which are influencing the results of your analysis or the results of your experiment. Now how could you reduce the number of parameters to a fewer but more effective ones may be more effective non-dimensional parameters. Why non-dimensional parameters are important because sometimes you may parameterize the result as a sole function of certain non-dimensional parameters. So as an example if you have say flow through a pipe you may have many experiments with different lengths different diameters or may be different densities of fluid different viscosities of fluid. But if you parameterize the result in terms of Reynolds number then if you keep the combinations of these such that the Reynolds number is unaltered the physical behavior is unaltered. That means we in some in such a case may reduce the parameters from say 4 parameters to an equivalent single non-dimensional parameter. So the big exercise or the big understanding is how can we make the parameterization of the experiments in terms of the reduced number of parameters using certain non-dimensional parameters. So these are the important questions that we would like to answer through the study of principles of similarity and dimensional analysis. Now when we say similarity what kind of similarity we look for the most intuitive form of term of similarity that appeals to us is a geometric similarity. So whenever we first studied about similarity in high school we only studied about geometric similarity. So if you have a figure and another figure which is geometrically similar you have basically ratios of the equivalent sides as identical or equivalent lengths as identical. So basically then you have a similarity in the length scales that is what we essentially found for geometric similarity. So geometric similarity is something which is intuitive that is if you want to study the flow past an aircraft you may make it small but geometrically you would always like to make it similar in terms of the actual big aircraft. So when we talk about similarity and the nature of similarities the first similarity that would come to our mind is geometric similarity okay and geometric similarity is the similarity in the geometry as obvious as that there is nothing more involved. The next type of similarity that we look for is known as kinematic similarity. So kinematic similarity again from the name it is clear it is similarity of motion. So maybe what it means is that if you have say 2 points p1 and p2 and say the velocity here say is u1 and the velocity here is u2 in a model in a prototype say you have a model like this where the equivalent point for p1 is p1 prime. Equivalent point for p2 is p2 prime then and say the velocities are u1 prime and u2 prime then u1 by u2 will be identical to u1 prime by u2 prime. So you will be basically having a sort of similar scale as the velocity. Now if you do not want to consider the velocity as such but just have a more qualitative picture more qualitative picture but a more physical picture may be obtained from the concept of stream lines. So if you have kinematic similarity, similarity in motion means the stream lines which are there should also be geometrically similar because similarity of stream lines is an indicator of the similarity in the kinematics because stream lines relate to a visualization of the kinematics of the motion or the velocity vector. Now remember that when you have stream lines the contour of a body is also a stream line as we have discussed because through the contour of the body you do not have any penetrating flow. So it is also a stream line that means that if you want to have similarities in stream lines you must also have similarities in the contours of the bodies. That means for kinematic similarity, geometric similarity is a must. So that means whenever we say that there is a kinematic similarity that is prevailing implicitly we must understand that they are also geometrically similar. Stimultural restriction need not be imposed, kinematic similarity automatically ensures that just because the body of the surface, the contour of the body is itself a stream line. The third important concept regard to the similarity is the dynamic similarity. Dynamic similarity is the similarity in forces that means if there are 2 types of dominating forces which have certain ratio in the model the same ratio should be preserved in the prototype. That means let us say that you have inertia force and viscous force. So if you have these 2 as the important forces and competing forces then you must have the ratio of the inertia force and viscous force in the model same as the ratio of the inertia force by viscous force in the prototype or equivalently in this case that means Reynolds number in the model same as Reynolds number in the prototype okay. So again it may be inferred that dynamic similarity should imply kinematic similarity because if you do not have kinematic similarity then how can you have a dynamic similarity if you just think about inertia forces. So obviously it follows that dynamic similarity should have kinematic similarity and kinematic similarity in turn should have geometric similarity therefore it is as good as considering just the aspect of dynamic similarity. It will automatically ensure kinematic and geometric similarity. It is very easy to talk about this in theory but when you go to experiments it may be difficult to achieve these types of similarities as we are theoretically intending. We will look into certain examples to illustrate that but before that we have to discuss about one important thing which we did not explicitly mention when we talked about these similarities. So this discussion may give an indication that if you somehow have geometric kinematic and obvious dynamic similarity that means then you have essentially all types of similarities between the model and the prototype. So whatever experiment you do in the model you can extrapolate that to the behavior of the prototype that in a sense true but incomplete because the first and foremost requirement of a similarity is that the physics of the problem must be identical for model and the prototype. So let us say let us take an example say you have a pipe of diameter 10 meter. Now somebody says that I will have an experiment where I will have a geometrically similar thing with a diameter of 10 micron okay all other. So it is just as if a geometrically similar thing. Let us say by some way the Reynolds numbers are maintained to be the same. So velocities are adjusted in such a way that the Reynolds numbers are the same. So dynamic similarity is preserved and from the dynamic similarity it is possible to get a picture of the behavior in these 2 cases. So if one does an experiment with this maybe one is intending to extrapolate it for this case. It will be totally wrong because the physics of the problem has got changed. It has got changed in many ways. One of the most common way without looking into anything else is as you reduce the size surface tension effects become more and more important. So capillary effects will have a strong role to play in terms of dictating the dynamical behavior in this system where for this such a large pipe the capillary effects will not be that important. So physics of the problem has changed altogether whatever were the important physical aspects which were not important for which were not important for the larger diameter pipe for the much smaller size capillary it has become important. So no matter whether you maintain the Reynolds number to be the same you will come up with a wrong conclusion because in the small scale the Reynolds number is not maybe the dictating factor because inertia force is not important. So instead of going through the ratios of these certain forces or the non-dimensional numbers you have to first be ascertain whether this non-dimensional number is physically relevant for the physics which is occurring over that scale or not. So that is very very important. So you should not change from a scale to another scale for predicting the relative behavior between the model and the prototype in such a way that the physics of the problem changes altogether and that is one of the very important tasks for an experimental designer that one should not design an experiment which changes the physics altogether from what happens for the prototype. Now when you have these forces the ratio of these forces let us just look into certain examples where we consider the ratios of different forces and those will give certain non-dimensional numbers. Reynolds number is one which we have already learnt and referred to many times. Let us look into some other ones. Let us say that we want to find out a non-dimensional number. So some examples of non-dimensional numbers. These non-dimensional numbers are important because in terms of these you may reduce the number of parameters with respect to which you parameterize the results of your experiments. So let us say that we consider the ratio of the pressure force by inertia force as an example. Pressure force we will try to see what are the scales just in the same way as we did when we came up with an expression for the Reynolds number. So pressure force will be some pressure difference times an area right. So this times L square inertia force is first the mass that is rho into L cubed into acceleration. So acceleration is like u du dx so u square by L. So the ratio of these 2 becomes delta p by rho u square. Sometimes this is known as Euler number. Let us say we want to find out the ratio of inertia force by surface tension force. So inertia force by surface tension force so inertia force we have already seen. Surface tension force if sigma is the surface tension coefficient sigma into L. So this becomes rho u square L by sigma. This is known as Weber number. Let us say we want to find out inertia force by gravity force. So inertia force by gravity force will be rho L cube u square by L divided by gravity force is mg. So rho L cube g so that is u square by gl. So this is a non-dimensional number. Classically the square root of this one is considered to be one of the important non-dimensional numbers called as Froude number. That is just u by square root of gl okay. Then some more examples let us go through. Let us say we consider inertia force by elastic force. So rho L cube u square by L by elastic force. If you have modulus of elasticity as E then E into area E into L square right. So that is equal to what? U square divided by E by rho. What is E by rho? I mean what physical thing does it represent? It represents square root of E by rho represents what? The sonic velocity through the medium. So that is A that means this is nothing but u square by A square where A is the sonic velocity or sonic speed through the medium. And we know that sometimes the square root of this one which is the Mach number which is a very commonly used. The Mach number is the square root of this one that is u by A. Maybe another example let us say viscous force by surface tension force. So viscous force is what? If you just want to write it for a Newtonian fluid, mu into the velocity gradient is the shear stress. So mu u by L times L square, shear stress in the area and surface tension force is sigma into L right. So that is mu u by sigma. This is sometimes known as capillary number. So in this way many such non-dimensional numbers are possible. In fact hundreds of hundreds or hundreds of non-dimensional numbers are there depending on the ratios of different forces. But we have just introduced some of the more common ones which are which may be pertinent to an introductory level course. Now when we talk about similarity, we have to understand one thing that whether this similarity is going to be maintained for all cases and that means that can you predict the real behavior? So two questions we want to answer. Can you predict the real behavior without satisfying the similarity in certain cases? The inverse that is if we do not satisfy the similarity then what would be the consequence or is it possible that in all cases we satisfy the true similarity? And these types of interrelated questions we would try to answer through some examples or problems. Let us look into that. Let us say that we have one wind tunnel experiment, wind tunnel for drag force determination. The scale is 1 fourth scale and it is a test on an automobile with the length of the model as 1 meter, the density of the model and the density of the prototype they are. So this is when we say density of the model and prototype, these are not densities of the solids actually, these are densities of the fluids flowing around. So in a loose sense we lie density of model and prototype, does not mean the car density of the car. It is basically the density of the fluid that we are talking about the air. Then similarly the viscosities of the air conditions 2 into 10 to the power – 5 Pascal's second and the velocity of the prototype is 25 meter per second. The first part of the problem is calculate the velocity of the model. Second part is the drag force on the model is measured to be 600 Newton. Calculate the drag force on the prototype. Then the third part that experiments indicate that the range in which we are operating the drag coefficient is independent of Reynolds number and it is equal to CD for the model is 0.5 with a reference area as 0.19 meter square. Calculate the drag force on the prototype and then from that you find that why it is different from what is predicted in part b. The fourth part you find out what is the power required to overcome the drag force in the prototype. So these are the parts of the question. So we look into it one by one. So the physical scenario just try to get a picture of this that you have a you want to design a car and you want to have the car designed for a speed of 25 meter per second and then for that you are having model experiment where you are having the size of the model car 1 fourth of that of the prototype 1 and the wind conditions etc. at the same the length scale of the model is 1 meter. So what is the velocity of the model? So here inertia forces and viscous forces are important because this is like Reynolds number is strongly dictating it. So for the dynamic similarity you must have Reynolds number of the model same as Reynolds number of the prototype. So that means you have rho model u model we are calling v. So v model l model by mu model is equal to rho prototype v prototype l prototype by mu prototype. So the densities are the same the viscousities are the same. So vm is equal to vp into lp by lm. That is lp by lm 4. So this is 4 into 25 that is 100 meter per second. Next the drag force on the prototype see what is the important coefficient that what is the important relationship that should dictate the equivalence of the drag force it is the drag coefficient should be same as in the model and the prototype. So you must have cd of the model same as cd of the prototype. So you have the cd is what the drag force divided by half rho v square into the area that is l square is equal to fd of the prototype by half rho p vp square into lp square. So you have fd of the prototype what is the drag force on the prototype that is the drag force on the model into vp by vm square into lp by lm square densities get cancelled out. So what is vp by vm that is one-fourth right. So 1 by 16 and lp by lm is 4. So it will be what what it will be the what would be the drag force on the prototype. So these two get cancelled out. So this is the drag force on the model which is 600 Newton okay then let us consider the third part. Experiments indicate that for the range of Reynolds number in which one is operating for this case cd is independent of Reynolds number. We have seen that such ranges are there we have discussed about the physical situation under which it is like that. So then cd of the model is 0.5 area of the area of reference for that corresponding cd is 0.19 meter square. So you can calculate that what is the corresponding drag force on the model and corresponding drag force on the prototype. So if you calculate that let me just tell that what you get. So the drag force on the prototype that you get as of course cd of the prototype into half rho v prototype square into area of the prototype and you have the test on the area of the model. So you can write this as area of the model into what 16 right l. So area of the prototype by area of the model is l prototype square by l model square right. So it will be area of the model into lp by lm square right. Velocity of the prototype you are already given the drag force of the prototype is same as drag force of the sorry the drag coefficient of the prototype is same as drag coefficient of the model and that is obtained experimentally as 0.5. So from this if you calculate the drag force on the prototype this comes out to be 522.5 newton okay. Now interesting is not what is the exact calculation but why these 2 predictions are different. So the drag force prediction from part b is 600 newton from this one it is 522.5 newton. This is experimentally obtained so this has more authenticity because the drag coefficient for model and prototype same that you have used this velocity is known and this is just from the model area with the scale ratio. Now why you feel that this may be different. See the key is just try to use a common understanding that cd is independent of Reynolds number. That means you may have cd of the model equal to cd of the prototype without satisfying Reynolds number of the model same as Reynolds number of prototype right. So when you are assuring this and when you have in the range cd independent of Reynolds number maybe you could have achieved it with a different Reynolds number but still your prediction goes well because cd becomes independent of Reynolds number. So this is a case where without satisfying the so called dynamical similarity you are able to come up with a prediction by exploiting the physical behavior over that regime that cd is independent of Reynolds number okay. So these are critical tidbits of similarity not always like you blindly look into the similarity but also look into the context in which it is being applied. Then the fourth part of course that is very very obvious power required to overcome what is the power required to overcome the drag force? It is the drag force times the velocity. So that is 13062.5 watt that is the answer. Now let us try to work out another problem. There is a ship which is 100 meter long and it is expected to sail at 10 meter per second and its submerged area is 300 meter square. So this is the prototype. The model is 1 by 25 scale 1 is to 25. Find number 1 the model speed neglecting frictional effects. Second part is the drag force measured on the model is 60 Newton when tested in a towering tank. So towering tank is just like equivalent to a wind tunnel for a ship experiment. So there is some artificial tank which as it simulates the sea or something like that and there the model is tested that is the towering tank at model speed. So from that estimate the total drag force on the prototype considering frictional effects also. There are some other data given for the problem but we will come into that one by one. First let us broadly look into the at least the first part of the problem. Second part there is some extra data that is given. Now before going into the problem let us try to have an understanding of the problem. So when there is a ship now what are the important resistance effects which are there? So one of the important resistance is of course the frictional resistance is there the viscous effect. The other important resistance is called as wave making resistance. So because of formation of the water waves and there it is a sort of a gravity dependent phenomena. So that wave making experiment resistance and there may be a third resistance which is because of formation of local eddies and so on but that is usually much much negligible as compared to the other 2. So here you have 2 types of important resistances. One is the resistance because of the wave or the wave making resistance and the other is the frictional resistance. So when you consider let us say that when you consider the wave making resistance. So if you consider the wave making resistance then what are the important forces which will be important inertia force and gravity force. So then the similarity will be determined by the Froude number ratio of the inertia force and gravity force. So if you consider that similarity then you have V model by square root of gl model is equal to V prototype by square root of gl prototype. This is for that is considering the wave making resistance. So Froude number of the model same as Froude number of the prototype. Also if you consider the viscous resistances then Reynolds number of the model and Reynolds number of the prototype they should be identical. So V model into L model by the kinematic viscosity of the model is equal to V prototype L prototype by kinematic viscosity of the prototype. So from here what you get V model by V prototype is equal to square root of L model by L prototype G is not changing. So then and from here what you get V model by V prototype is equal to L model by L prototype. Still you are having the same water with same kinematic viscosity. Now can you do an experiment where you satisfy both. That means if you want to satisfy this you must satisfy Lm by Lp is equal to square root of Lm by L right. But they are not one. One is to one model is no model right. One is to one then model is a prototype right. So this you cannot satisfy. These are very interesting situations that you come up with the important non-dimensional numbers you see and you cannot satisfy. So you have to come to a compromise that which one will you satisfy. Let us say that you satisfy this because I mean you may satisfy only one of these for your similarity. So let us say you give it a priority and that is what is considered in the first part. That is you find out a model velocity neglecting the frictional effects. So if you neglect the frictional effect then this is the solely dominating factor for the similarity because then Reynolds number is not important if the frictional effects are not important. So if you consider that then the velocity of the model it comes out to be 2 meter per second okay. Now that second part of the problem and for that some extra information is given from experimental data and let us just note down those that extra information. So what are the extra things? The Reynolds number of the model is 8 into 10 to the power 6. The Reynolds number for the prototype is 10 to the power 9 and you can clearly see that these two Reynolds numbers are different because you cannot simultaneously satisfy these two that is what we have seen. And from the experimental data of CD versus Reynolds number from this the CD is CD for the model is 0.003 this is the experimental data. So what we are writing is whatever has been obtained from experiments considering these calculations have been made with a consideration of the density as 1000 kg per meter cube for water. So now the question is what is the frictional drag force on the model? See when you consider this CD, this CD is a representative of what? This CD is a representative of the frictional drag right. It is not a representative of the wave drag. There are 2 drags. So out of the total drag force you can isolate the frictional drag. So how you isolate the frictional drag? So frictional drag is the CD frictional times 1.5 rho V square times the area right. So if you substitute these values, so let us see that what is the frictional drag force on the model? CD for the model rho m Vm square m. So if you substitute all the values, all these values are given actually. You will get 2.88 Newton okay. So with this one, so if you so you know the length scale so from the area of the prototype you can calculate the area of the model. By the square of the length they will vary. These are already known, density known CD also, this CD important is this frictional CD that has been calculated. That is what you have to keep in mind. So this is the frictional drag force. So what is the wave making drag force on the model? The total is 60 Newton, so 60-2.88 that is the wave making drag force. So that is 57.12 Newton. What is the wave making drag force on the prototype? If you have the wave making drag force on the model, how do you calculate it? Yes, for the wave making drag force you must have CD of the wave making drag for the model and prototype to be same. So that means the wave making drag for the model by half rho V model square into L model square is equal to the wave making drag for the prototype by half rho V prototype square into L prototype square right. So from here you can get that the wave making drag on the prototype, if you calculate it comes out to be 8.92 into 10 to the power 5 Newton. I am just giving these numbers because you can calculate this at your laser time and see. So now what is the frictional drag on the prototype? What is the frictional drag on the prototype? See your objective is to estimate the drag force considering frictional effects as well. So this is the wave making resistance only. So what is the frictional drag on the prototype? So CD of the prototype into half rho V prototype square into the area of the prototype right. So this CD, this is the frictional CD that means which one, this one. So if you substitute that that will come out to be 0.225 into 10 to the power 5 Newton. Some of these calculations may be bit erroneous but just I am outlining the procedure. So just concentrate more on the procedure. So then so this is frictional CD that we have to keep in mind and then the total drag force on the prototype is the sum of the wave making plus the frictional one okay. So what you see here is that you get a velocity by neglecting frictional effects using that you use the similarity in terms of the drag coefficient where you consider the wave effects because this velocity was calculated by considering the wave effects. On the top of that from the experimental data whatever you get you utilize that for calculating the frictional coefficient, the frictional resistance and then add those together to get the total resistance that is the whole idea okay. Now we will come to the other important part of this discussion of the dimensional analysis that we have seen certain non-dimensional numbers but these non-dimensional numbers how will we know that what are the important non-dimensional numbers for a particular problem. If you have a particular number of variables and that is given by something known as Buckingham pi theorem. So let us try to look into that. So what we are looking for that if you have many variables for a system how you reduce those dimensional variables into some equivalent functional relationship between non-dimensional numbers that this theorem tries to highlight how to do that. So it says that if there are m physical parameters and n fundamental dimensions then the functional relationship between all these may be written in terms of m-n number of non-dimensional parameters or these are sometimes known as pi terms that was the terminology used by Buckingham when he introduced it. So this is as good as having a functional relationship between some non-dimensional numbers pi 1, pi 2 up to pi m-n that is equal to 0. So you are reducing m number of physical parameters to m-n number of dimensionless parameters. How you do that the best way in which we will understand is through an example. So let us take an example to understand that how we reduce this example. Let us say that you want to estimate the pressure drop in a pipe of a given length l. What is important for us is the pressure drop divided by the length because we know that dp dx is physically what is the important parameter. It is a dependent parameter, it is a function of which variables. So if you have a pipe of say diameter d in which a fluid is flowing then what are the parameters on which this should depend? The diameter of the pipe that is 1 then what fluid properties? Viscosity, density, average velocity, the surface roughness, average surface roughness. So this, so first of all you must have a physical idea of the problem. You cannot do a mathematical exercise without having a physical idea of what are the important parameters. So we have identified the physical parameters. So this is the sort of a dependent variable and these are the independent variables. Now we want to see that what type of functional relationship should hold true for that. So for that we will write the dimensions of these parameters. So what is the dimension of this one? l. What is the dimension of viscosity? m l to the power – 1, t to the power – 1, viscosity, sorry density, m l to the power – 3, velocity, l t to the power – 1, epsilon, l. What is the dimension of delta p by l? So delta p is what? Delta p is Newton per meter square. Newton is mass into acceleration. So m l t to the power – 2 that is the, so that divided by this one that is the pressure and then divided by l, right. So this is m l to the power – 2, t to the power – 2, right. So now let us see that how many number of fundamental dimensions are there. So number of fundamental dimensions, what, 3, m, l and t and how many, so that is equal to n for the Buckingham pi theorem and how many number of variables are there, 1, 2, 3, 4, 5 and 6. So number of pi terms or dimensionless terms is 6-3 that is 3. So we will just show that how to find out these pi terms. So first to find out the pi terms, you have to select some variables known as repeating variables. What are the repeating variables? You must have certain variables. So number of repeating variables is same as the number of fundamental dimensions. That means here you have 3 number of repeating variables. How to choose the repeating variables? There are certain important things. First of all, out of the 3 repeating variables you choose, none of those should be the dependent variable that is the first thing. None of those should be dimensionless and none of those should be having same dimension and maybe the most important thing is collectively they should contain all the dimensions. That is your objective is to select 3 variables out of these in such a way that none of these are dimensionless, none of these are of the same dimension and when they are considered collectively they will contain all the dimensions. So you have a choice say let us consider rho, v and d as an example. So if you consider rho, v and d, so rho contains m, l, v contains l and t. So it is good enough to consider rho, v and d they together contain m, l, t all and none of these are of same dimension and none of these are dimensionless. Now in this way you could have many such possible combinations out of these ones and that you are free to choose. So if you choose this, then what we do? The first pi term is written as say rho to the power x into v to the power y into d to the power z into one of the variables say delta p by l. Similarly, pi to you will have rho to the power x, v to the power y say let us call this x1, y1, z1 say x2, y2, z2 into the other remaining variables. Mu and pi3 as rho to the power x3, v to the power y3, d to the power z3 into epsilon. Then how would you calculate pi1, pi2 and pi3? So you have to keep in mind that this is dimensionless. So that means let us take an example. Let us calculate the pi2 just to show you as an example. So rho to the power x2 is what? m to the power x2 into l to the power –3 x2, then v to the power –y2. So v to the power –y2 is l to the power –y2 into t to the power –y2. This is not minus, I have confused it, this was the over bar, right, sorry. So this was as the bar, v bar, right. So m to the power x2 into l to the power –3 x2, then l to the power y2 into t to the power –y2 that is for the d to the power y2 and then d to the power z2, so that is l to the power z2 into mu. What is mu? Mu is m l to the power –1, t to the power –1. So this should be dimensionless. That means from this what you get? x2 plus 1 is equal to 0 for m for l –3 x2 plus y2 –1 equal to 0 for t –y2 –1 equal to 0, right, which, oh, plus g2, oh sorry, sorry, in the second term you have the g2, right. This is plus g2, second term because of that l, okay. So from this you can calculate x2 equal to –1, y2 equal to –1 and what is g2? g2 is also –1. So what you get as pi1, sorry, pi2 rho v bar mu by rho v bar d, right. You can easily recognize that it is 1 by the Reynolds number. See when you get a dimensionless number, now what dimensionless number you will use in practice it depends on the convention. So you may use 1 by pi2 also and that is what we actually use. So in this way you can calculate pi1, pi3, what will be pi3? Can you tell? Like this is just by common sense you can say, epsilon by d, other terms will go away. So pi3 will be epsilon by d and pi1, so if you write say delta p by l, so pi1, can you calculate it? It will be, of course, there will be some v square by 2g type of thing that will come there because delta p is there. So you will have a v square by 2g which has a unit of what, unit of length and there is also a d. So it may be non-dimensionalized by d, so this will be an important non-dimensional parameter but then you have to see that you also have a delta p by l. So how that delta p by l combines with that, I am leaving it on you as an exercise, we do not have much time left for this lecture but the exercise that I leave on you is, at the end you have to show that this boils down to that hf which is expressed as delta p by l is nothing but hf rho g by l, from that hf will be a function of what, Reynolds number and epsilon by d or rather you may better way in which you write that delta p by l or even hf form is fine. This small f is a non-dimensional number, this into l by d into v square by 2g that form. So it is a non-dimensional function of Reynolds number and epsilon by d times this one. So from here you can get actually what is pi1 that is hf is delta p by rho g and delta p by rho g into d by l into 2g by v square or its inverse that you will get or basically you may get the inverse of this one. So this is a form which is the Darcy-Weisbach equation. See that you get this equation not in the exact value type, in the form of an exact value but you get the functional dependence. It reduces your number of experiments and makes you to come up with a dimensionless parameter and the choice of the dimensionless parameter is it depends on the physical situation. If pi2 is a dimensionless parameter and pi3 may be pi2 into pi3 is a dimensionless parameter, pi2 to the power half is a dimensionless parameter. So important is you have 3 independent dimensional parameters, you may make many other dimensionless parameters using this one. So the dimensionless parameters that you make it depends on your physical situation. Let us stop this lecture here. Thank you.