 We were discussing about 2 pipes and working out some examples. Let us continue with another example. Let us say that you have a piping system or a single pipe in this example where a pitot tube is put. The fluid that is being handled in this system is air and the manometric fluid is water. The difference in the height of the 2 limbs of the manometer say is delta H which is given. The diameter of the pipe is 8 centimeter and the pressure upstream is 1 atmosphere approximately 101 kilopascal. Now what you are asked number 1, estimate the center line velocity, estimate the volume flow rate and the wall shear stress assuming smooth wall. Given dimension of delta H is 40 millimeter and a relationship between the center line velocity and the average velocity is given as 1 by 1 plus 1.33 square root of f where f is the friction factor. So this is the description of the problem. Now let us first look into the problem from a very basic consideration of the pitot tube that what you really can write regarding the difference in the properties between say 1 and 2. So the point 2 is supposed to be the so-called stagnation point and between 1 to 2 if one uses a pitot tube uses the well known expression of the pitot tube and that expression if you recall it does not account for any energy loss. It considers that the fluid is undergoing a reversible process without any energy loss and then it is possible to use the Bernoulli's equation along a streamline between points 1 to 2. So without any loss therefore it is like p1 by rho plus v1 square by 2 is equal to p2 by rho. This is ideal right and this is what equation is written for the pitot tube. Now somebody who is a very casual engineer will do a mistake in what? We will do a mistake in having these points 1 and 2 at some distance apart so that head loss between these 2 becomes important. If these 2 points are very close then that head loss may not be important but a very wrong approach of engineering may be to put them at some distance apart. I mean of course they will be at some distance apart but question is how to make that effect minimum because you are using an idealized equation where that effect is not present trying to predict whatever the velocity at the state 1 or at the point 1 using that but that will itself be erroneous if head losses are significant. Now between p1 and p2 of course you can relate that what changes by using the principle of manometry. So let us say that this height is h. So you can write the if you consider the same horizontal level you have the pressure at a is same as pressure at b that is you have p1 plus h rho of air into g plus delta h rho of water into g is equal to p2 plus h rho air g plus delta h rho air g. So from here you can when you write this p1 by rho this is rho of air or rho of water in this equation rho of air right. So to make use of this equation we just divided by rho of air but before that some terms will be cancelled out like h into rho a into g from both sides. So p1 by rho air minus p2 by rho air is equal to delta h into g into 1 minus rho water by rho air that means you have v1 square by 2g is equal to delta h g into rho water by rho air minus 1 sorry that g is okay. So from here you can find out what is v1 all other things are given. So you will be able to find out a value of v1 let me just tell you that what value you are expected to get out of this one. So you will get v1 as 25.5 meter per second. Now this is from an idealized analysis. The other important thing is that how this v1 is related to so this v1 is what this is velocity at a point and velocity at a point on the center line. So this is as good as the velocity at the center line one is a point on the center line. So we need not confuse between a point and the average over a section. So this is at a point we are writing okay. So because this emerge from the Bernoulli's equation between 2 points. Now this u center line and u average they are related by this equation. How this equation comes is a bit of a background information but it is quite simple that is this comes from the logarithmic law application of the logarithmic law for the velocity profile. So it is assumed for this problem that it is a turbulent flow over a smooth wall of a pipe and in the turbulent flow so the velocity profile is taken like this in this form okay. So here see u plus is what? u by some u reference this one and u reference is square root of tau wall by rho and what is y here? y is nothing but capital R- small r. y is the distance from the wall okay. So in this y is the distance from the wall not any arbitrary coordinate. So in the pipe coordinate system the cylindrical coordinate system this is the distance from the wall. So from this type of velocity profile you can find out the average velocity by integrating this over the section integral of u dA divided by the area. So and you can find out the velocity at the center line by putting small r equal to capital R okay and then I sorry small r equal to 0 that means y equal to capital R. Then you will get 2 expressions remember that these kappa and b are like sort of constants kappa is 0.4 or 0.41 may be to be a bit more accurate and b for a smooth pipe is close to 5. We have mentioned this earlier when we are discussing turbulent flows. Now with that if you find out the ratio you may relate that ratio with the friction factor. How you relate the ratio with the friction factor you know that tau wall is equal to the CF the friction factor is what? Tau wall by half rho u average square and tau wall and u average are related by or you can relate tau wall with u r this u reference and you may relate u reference with u average and how you relate CF with f? So the CF is f by 4 right. So if you use this expression you use this velocity profile and for averaging also use the same thing relate that with the friction factor you will get this expression. This may be exactly derived by putting these numbers. So that is how this is there is not just a very magical thing just from the very basic understanding of the velocity profile and its averaging. Now let us say that you have this as a center line velocity. You are having a relationship between the center line velocity and the average velocity. If you know that then in one way you may straight away write the average velocity volume flow rate like that but only hindrance is you do not know the friction factor okay. So one of the ways again may be by the guess by the trial and error. So let us say that you make a trial say u average by u center line. See trial and error is not always hit and miss. It requires some intelligence and understanding of the problem. So if I give you a choice u average by u center line say 3 trials. One is 0.1, another is 0.5, another is 0.8. Out of these 3 which one you expect to be a better trial? 0.8 is expected to be a better trial in this case why? Because remember we are talking about a turbulent flow where the velocity profile is almost uniform. So there is not a great difference between the center line velocity and an average velocity. So of course the average velocity will be less than the center line velocity no doubt about it but how much it is less it depends on the skewness in the velocity profile. So for a turbulent flow it is almost uniform and therefore like if you have such choices may be 0.8 or 0.85 or 0.9 these types of gases are reasonable gases and so still if you guess 0.1 or 0.5 still okay but let us say you guess this as 2 then it is absolutely erratic because I mean that it does not matter whether it is laminar flow or turbulent flow whatever the center line velocity is always greater than the average velocity and so these types of basic considerations should be kept in mind anytime whenever you are having a iterated solution or a trial solution and putting a guess for that. So when you substitute this trial you will get a value of f from this equation straight in. Once you get the value of f from this equation then the question is that is this f what comes out from the relationship that you get from the Moody's diagram. Here remember that we are talking about a smooth wall. So hydraulically smooth pipe for that the friction factor should not be dependent on the wall roughness. So for a smooth wall it will depend on only the Reynolds number and for the hydraulically smooth pipe this is like 0.316 by Reynolds number to the power 0.25 otherwise one may directly read from the Moody's diagram the corresponding graphical graphically plotted values without looking into the function. So that means if you know f you will get a Reynolds number right and the Reynolds number is what? The Reynolds number is based on the average velocity right rho u average into d by mu. So once you get this f you will get a Reynolds number and that means you will get the average velocity. Put that average velocity back here and see that you get a new f. So in this way you iterate till you come to a convergence okay. So important consideration is that to have a distinction between the centerline velocity and the average velocity. The whole understanding is when you are applying the sort of Bernoulli's equation between the 2 points and neglecting losses you are talking about only velocities at the points. Whereas whenever the head losses are calculated those are based on Reynolds number which takes reference velocity as the average velocity over the section not velocity at a point that is the key concept that is used for solving this problem. So let me give you the answer I mean once it has converged then the remaining calculations are very straight forward and I need not repeat but let me just give you the answer. So the f is equal to the converged value of f is 0.0175. Then the q is 0.109 meter cube per second and tau wall is 1.23 Pascal. So this kinds of practical examples are important because in practice you have energy losses or head losses. Next what we will see is that we have till now discussed about the head loss but how is the head loss related to the energy of the fluid. That may be interesting to us because in our in the very beginning of our course when we are talking about inviscid flows we were discussing about the Bernoulli's equation and we found out later that the Bernoulli's equation sort of gives a mechanical energy balance for a system for a flowing fluid. Now therefore here we are seeing that even we might be tempted in using the Bernoulli's equation but because of certain losses that may not directly be applicable there might be certain errors. So these losses must have some relationship with the energy consideration in the pipe flow. So let us look into a bit more details of the energy considerations in pipe flow. The objective will be to figure out that how the head losses are related to the total energy balance. Let us say that you have a pipe of whatever section say circular pipe as an example and we are looking for the fluid in the pipe and trying to write an expression for the energy balance. So for any conservation equation we may start with the Reynolds transport theorem. So let us start with the Reynolds transport theorem where E is the total energy of the system. So we can write dE dt for the system where capital E is the total energy and small e is the energy per unit mass. So now let us make certain assumptions for simplification that let us assume that it is a steady situation. So the unsteady terms go away. So when you have this steady situation the next thing what you do, the next thing what you do is you try to write this expression for the dE dt of the system. So that it is what it gives the total rate of change of energy of the system and so if you have a system like this of whatever arbitrary configuration the total energy change of a system is something of fixed mass and identity and for that the total energy change is given by a particular form of the first law of thermodynamics. So you are having basically some interaction of heat and work. So you have let us say there is a heat transfer to the system say delta Q. There is a work done by the system say delta W. These are positive sign conventions that we will follow. So any heat transfer to the system we consider as positive. Any work done by the system that is energy flowing out of the system because of what we consider as positive. So let us say that some heat is transferred to the system and as an example some part of that is used to do work. The remaining will change the energy of the system. So you have delta Q-delta W is equal to the change in energy of the system. Obviously I am not writing these terms in a very formal way whenever in the next semester when you will be studying thermodynamics you will be studying in details of how to formally write all these terms. But we are just trying to make use of this and I am just trying to be at your level so that we can proceed further. Now when you write this energy keep in mind that this energy is the sum total of kinetic energy, potential energy and anything else other than kinetic and potential energy which is the function of the internal configuration which we call as internal energy. So let us just symbolically write it kinetic energy plus potential energy plus internal energy. In books of thermodynamics internal energy is given symbol of u as you have noticed maybe earlier but here because we already use u for velocity we use just i as a symbol for internal energy to avoid the confusion in the terminologies. Now this equation you can also write as a rate equation. So you can write Q dot-w dot is equal to dE dt of the system. So this we can write Q dot-w dot. Q dot-w dot of what? Q dot-w dot for a system but in the limit as delta t tends to 0 when it is derived this is same as this for the control volume as well. So this is as good as Q dot control volume-w dot control. What is the control volume? Say you have chosen some control volume which is across which some fluid enters in the pipe and it leaves the pipe and the boundary of the control volume is shown by this dotted line. Now let us concentrate on the right hand side. First of all this control volume is stationary so that the relative velocity and the absolute velocity they are the same. So this will be equal to integral of rho E. So integral of v dot n dA for the outflow it will be positive and the inflow it will be negative. So we can say that integral of rho UDA for outflow boundary – integral of rho E UDA for the inflow boundary. Because we have now lost the vector sense so we have put the proper algebraic sign to represent the vector sense. Now next is to split different terms based on like what are the important effects. So heat transfer so heat transfer of course there may be some heat transfer to the system or away from the system say you are heating the wall of the pipe is heated. So there may be a heat transfer from the surrounding to the system. If it is not heated then also there may be a heat transfer because of the temperature difference between the ambient from outside and the fluid that is there in the pipe and how such temperature difference may be created we will try to see. Now if you consider concentrate on the work done. So what is the work done here? By the fluid in the control volume. First of all you have the fluid let us consider the inflow the fluid is entering with the pressure P. So it is putting some energy to the control volume as it displaces some fluid and enters it. So what is the corresponding work done? See we have related this with the flow energy or flow work. So that same work we are referring here. So if you have let us say a small element of area here say dA. So the elemental work done is P in into dA into the displacement. Here we are writing the rate. So the rate of the displacement that is the velocity. So P in dA into u integral of that over the entire area sin plus or minus. You see this consistency of the first law of thermodynamics see any energy in the form of work if it is transferred from the inside of the system to the outside it is positive. Here the energy is being put into the system. So that is negative work in terms of the work. So that means you have q dot- so you have minus integral of P u dA for the in and for the outflow it will be plus. So that is the left hand side expression that we are having. Now the next is we are assuming that rho is a constant for this problem or for this discussion. So when rho is a constant we can take rho out of the integration in place of E what we can write. This is energy per unit mass. So first say if you write first kinetic energy u square by 2 potential energy gz and internal energy per unit mass i okay. So if you collect all the terms what you get at the end so you get q dot cv. Let us say that you take these terms of integral of P u dA to the right hand side. So if you take these terms to the right hand side you will see that it will club up with these ones u square by 2 gz i plus that you will have one P by rho. Because rho is there as a multiplier this is just P alone. To adjust with that you will have one rho multiplier outside. So in the bracket what will enter is P by rho. So you will have this equal to say if you take rho outside then integral of P by rho plus u square by 2 plus gz plus i into u dA same thing for the in flow. The next important thing is the integration of whatever integral appears in the 2 in flow and the out flow boundary terms. So when you write this integration you have to keep one thing in mind that you have to be careful whether the properties are varying over the cross section or not because these are integrals over the cross section. So let us assume that the pressure is not substantially varying over the cross section and that is in a way true that we have seen that the major pressure gradient is along the x direction. Then u will definitely vary with the cross section because you have a velocity profile it is not a uniform flow. The potential energy effect that also you may consider that the pipe diameter is not so large that there will be a great difference in potential energy effect. So certain terms of these you may very confidently take out of the integral assuming that those are constants. So like you can for example write say you divide q dot Cv by rho. So when you have P by rho or let us do one thing we will divide by rho in the next stage. Now see that if you take P by rho out of the integral then what you are left with in the integral is integral of u dA that is the volume flow rate that multiplied by the density is the mass flow rate. So we call it m dot okay. Next is you have rho by 2 integral of u cube dA. So this is remember we are writing for the outflows also let us give some names to this areas one for the inflow area and two for the outflow area. So this is P1 by rho this is integral over the section 1 and then similarly this is m dot gg. If you assume that the temperature is also not varying over the section then the internal energy you may assume to be a constant over the section. So plus say m dot i this is 2 right yes m dot gg this is 2 i2 then minus similar terms for 1. So minus P1 by rho m dot minus rho by 2 integral of u cube dA for the 1 minus m dot gg1 minus m dot i1 okay. Now let us say that we neglect this effect we do not neglect this effect of velocity variation outright but we somehow make up for our negligence. See if we do not consider this effect all together and say that the consider that the same velocity velocity same as the average velocity is there then one approximation to this term could be half m dot into u average square because this is what this is like kinetic energy because 1 rho into u into dA is like m dot and that into u square is like this one but this is erroneous why this is erroneous because this is not exactly same as this one because m dot is what rho into u average into A. So u average into u average square is u average cube that is not same as integral of u cube dA. So that is an error and that error has to be adjusted with a multiplying factor say alpha which we call as kinetic energy correction factor. So this alpha at the section 2 this may be different at different sections because velocity profiles may be different in general over different sections. So this alpha is known as kinetic energy correction factor. So what is this correction factor all about? This is the correction factor to correct the kinetic energy from a hypothetical consideration that it is based on the average velocity to the real kinetic energy that is there integrated over the cross section. So the kinetic energy correction factor will be what? So you have alpha into m dot is rho u average into A into half u average square. So u average cube is equal to rho by 2 integral of u cube dA. So you can now write an expression for alpha as integral of u cube dA this one. So if it is for a circular pipe, so this is as good as u by u average whole cube dA is 2 pi r dr by pi r square from 0 to i. So you can calculate the kinetic energy correction factor given the velocity profile. Now can you tell whether it will be more for laminar flow or turbulent flow? Laminar flow. Why it should be more for laminar flow? So it depends on the u by u average. So u by u average it is u deviates from u average significantly more for laminar flow. So you will have a more significant value of this one deviated from one. So if u was equal to u average throughout then kinetic energy correction factor would be 1. If it slightly deviates from u average then it will be very close to 1. But if it is largely deviating from u average say consider the fully developed laminar flow through a circular pipe. So u center line is 2 into u average. So you can see there is a large difference and that 2 factor will be there if you consider this kinetic energy correction factor. So it will be a large value. So I will leave it on you as an exercise that you calculate the kinetic energy correction factor for fully developed laminar flow through a circular pipe. Just substitute the velocity profile that u by u average equal to 2 into 1- small r square by capital R square and then just do the integration. Now you see that this kinetic energy correction factor if you put let us see that what equation you will get at the end. So now let us divide all the terms by m dot okay. So if you divide all the terms by m dot you have q dot by m dot is equal to phi 2 by rho plus say alpha into u2 square by 2 because you have already divided by m dot which is rho into u bar. Then plus g is e2 internal energy term we will just write separately minus p1 by rho plus alpha u1 square by 2 plus g is e1 plus internal energy 2- internal energy 1 right. So we may just rearrange it a little bit to write that you have p1 by rho plus alpha u1 square by 2 plus g is e1 is equal to p2 by rho plus alpha u2 square by 2 plus g is e2 plus internal energy 2- internal energy 1- this one right okay. So many times when you say alpha maybe it is better to write alpha 1 and alpha 2. Now if you consider these alphas as 1 this will look like a modified Bernoulli's equation that here you have the total mechanical energy at 1, here you have the mechanical energy at 2 and you have a term here the correction term. This correction term if it is 0 then it is just like the Bernoulli's equation that you have studied earlier. So sometimes this is known as modified Bernoulli's equation. Again that is a very wrong concept. This has nothing to do with the Bernoulli's equation except the form because Bernoulli's equation you are writing between 2 points. Here you are writing the equation between 2 sections 1 and 2. So be very very careful and this is a very important misconception that people have. Many times you see that in some of the industrial applications even the kinetic energy correction factor is omitted and then still it works. It works beautifully because many of the engineering flows are so turbulent that kinetic energy correction factor is very close to 1. That means considering that or not considering that does not matter but it is a matter of negligence or understanding. So if you understand that it has to be there but for a highly turbulent flow you neglect it that is one thing. But a very bad thing is you do not know or do not understand that this has to be there. So that misconception has to be avoided. So do not take it as a modified Bernoulli's equation. Better we just call it as an energy equation which looks like a modified form of Bernoulli's equation with the terms in the Bernoulli's equation adjusted with something. So what is this now? We will now concentrate on the physical meaning of this. So internal energy 2-internal energy 1. What is this? So you have basically let us say that you have so at the section 2 say at the section 1 the fluid has entered. At the section 2 let us say that heat transfer is 0. Let us say that you have insulated the wall of the pipe so that there is no heat transfer across the control volume. So then what do you expect? That term to be positive or negative. This is pure physical understanding. Do not try to go for any mathematics to describe it. Think about this. You have because of viscous effects the relative motion between various fluid layers. It is as if like you are rubbing one of your palms with the other. So you have the frictional resistance because of relative motion between various fluid layers and because of that frictional to overcome that frictional resistance what will happen? There will be some work that is necessary but that work is not a useful work. So the entire work is dissipated and where it is dissipated? It is dissipated in the form of intermolecular form of energy. So the fluid gets overheated. So it increases the temperature of the fluid. So because of viscous action whatever work is necessary to overcome that that is eventually manifested in the form of an increased temperature. This is known as viscous dissipation. So whenever you will be studying maybe heat transfer later on you will go through the greater details of what is viscous dissipation but it is very important to have a qualitative understanding of it. That you have velocity gradients between fluid layers and there is a relative motion between the fluid layers to overcome that resistance. Some energy has to be spent or some work needs to be done but that work is not manifested in the form of useful work. So that what it only does is it increase the temperature of the system through the internal energy rise. So we can conclude that this term is always positive. Now you may say that I will have a heat transfer which is more than this one. But see spontaneously that effect is not going to be there. In a limiting case what may happen? See when the system is overheated you have a higher temperature than the surrounding. So you will have a heat transfer. So this spontaneous heat transfer is what? This spontaneous heat transfer itself is negative. So here you have to remember that this was put with the sign convention that heat transfer to the system is positive. Here the system is getting overheated. So there will be heat transfer from that to the surroundings. So that itself will become negative. So the sum total of that will be positive. That means what you can say that this represents the total mechanical energy at section 1. This represents the total mechanical energy at section 2. This is a positive term. That means the total mechanical energy at section 2 is less than total mechanical energy at section 1. So there is some loss of energy and that is manifested in the form of the head loss that we have seen. So this expressed in the form of head that is if you divide all the term by g then it is expressed in the form of unit of length or head. That is nothing but the head loss that we have calculated for the pipe flow problems. Now consider one important thing. If we ask you that what is the work done by the fluid to overcome the wall shear stress. The shear stress at the wall. What will be that? Yes. What should be the work done to overcome that? Keep in mind one thing. How do you calculate the work? Some force or here the rate of work. So some force multiplied with some velocity. So what where we are concentrating at the wall? At the wall there is some shear force. What is the velocity of the fluid relative to the wall? 0. That means there is no work done to overcome wall shear stress at the wall. Again this is a very very important thing because these are loose wrong concepts that because there is wall shear stress there is some work done to overcome wall shear stress and that is why fluid is losing energy and all those blah blah things people give as explanations. But you have to be very very particular in this. Where is the work if the velocity is 0? So there is no work done to overcome the shear stress at the wall. That work is 0. Only whatever is the work internally that is manifested in the form of this internal energy change. But not a sort of a useful work by the displacement at the wall obviously because it is a no slip boundary condition. So from this analysis what we understand is that we can cast the energy consideration between 2 sections 1 and 2 as p1 by rho g plus alpha into e1 square by 2g plus z1 is equal to p2 by rho g plus alpha 2 into u2 square by 2g. Remember all these are average velocities plus z2 plus head loss which is a positive term. This is sometimes known as modified Bernoulli's equation or better to say this is an energy equation modified mechanical energy equation which still represents the conservation but considering a loss. Now what could be these losses? What could be the sources of these losses? One of the sources of the losses is because of the viscous effects that we have already discussed that is the head loss hf. We could characterize it that how it is different for laminar flows, turbulent flows and so on. But there also could be other types of losses and other types of losses are possible because of other changes present in a piping system. For example you have a small pipe. Now that small pipe is getting changed to a larger diameter. Now this is something where there is a loss. Why there is a loss? See of course the diameter change is there but why diameter change induces a loss. So if you consider the streamlines like this the streamlines because of the sudden change in cross section will be having their curvature in this way. So locally what will happen? There will be eddies formed in this way. These eddies do not participate in contributing to the energy of the main flow. So whatever energy is there associated with the rotation of these eddies that is a loss so far as the main flow transmitter is concerned. So this is also a loss. This loss was not taken into account for calculating the Hf. Similar things may be there for like if you have valves in a piping system because those are creating some resistances. If you have a valve. So if the valve is totally closed the fluid cannot flow. If the valve is partially closed and partially open the fluid may flow. So obviously there may be other forms of resistances and those losses because of the other forms of resistances are known as minor losses in a piping system. So we will now look into minor losses in a piping system. Again this minor loss is a misnomer because sometimes so when there is something minor there is something major. So what is this major? Major is this head loss due to the friction this we call as major loss. But in many practical considerations minor loss becomes much much greater than the major loss. So it is the name major and minor should not be confused in a literal sense. So just because originally the loss considerations were there from the pipe friction considerations and in one way it is major because in a pipeline if there is nothing else at least the physical resistance because of fluid friction is there over a length. You may not have a valve you may not have a sudden change in cross section but the length of the pipe itself is present. So the reason of naming the major losses this loss will be there. Other losses minor losses are losses they may be there or may not be there depending on what fittings are there in a piping system. But if they are there sometimes they are much much more important than the major loss. So the relative importance need not be misunderstood. So we will look into some examples for which give rise to the minor losses. So the first example is flow through sudden expansion. So here what is important is we are able to see that there is some loss and let us say that the diameters of the smaller and the larger pipes are d1 and d2. We are interested to find out the loss because of this flow through sudden expansion. And when we find the loss due to sudden expansion we isolate the effect of the loss due to the length of the pipe. So we only consider the loss due to this sudden expansion effect not the length effect. So what we do we just take a control volume and try to apply the Reynolds transport theorem momentum conservation. So let us say that we take this section as section A and take this section as section B. So for this control volume if we want to write the Reynolds transport theorem. So the resultant force along the x direction if it is x. I am just writing the final simplified form because we have discussed about such problems many times. So we assume it is a steady flow and VB-VS. What are the forces acting on the control volume? So you have a pressure force on the 2 ends. So when you have the pressure force on the 2 ends see also there is a shear force here at the wall. But we are neglecting that effect because that effect is already considered in the major loss. So that does not mean that that effect is not there. We are isolating that effect from the minor loss effect. So the pressure force so what is the pressure distribution at 1? Let us at A. So we have a section 1 which is say for here upstream and the section say 2 which is at a downstream. Let us consider that the velocity profiles are approximately uniform at 1 and 2. When is it possible? It is possible when it is a very highly turbulent flow. So let us assume that the flow is highly turbulent so that the velocity profiles are almost uniform that is the kinetic energy correction factor is not important. Now when you have say you want to write the force due to pressure here. See it is important to note that the pressure is acting here in this way and this is let us assume that this pressure and this pressure is not greatly different and that will not be greatly different because these kinetic energies are not greatly different and this length we are not considering so large that there will be a huge loss of head because of the friction. So here you have the pressure at this one roughly same as P1 if P is not located far away from the section A. If it is quite close then that is possible. So what we are considering is that in addition to the pressure acting over this part the same pressure also acts over the upper and the lower parts. This is an assumption. So and that assumption is well justified because the change in pressure from this one is not felt so easily by this one because this is just a small recirculating region. So the change in pressure is felt only when the proper bounding streamlines are making it feel. So here it is just a local recirculation this does not understand so easily that what would be the change in pressure from this to the subsequent section. So you have P1 so the entire pressure here is like P1 and the area is like A2 so that you have to understand this. It is not the area of one but the total area over which as if this P1 is acting so it is basically P1 into area of the section A area of the section A and B are the same. So this minus P2 into A2 is equal to rho Q now VB and VA so VB is like V2 that is fine. What about VA? See the velocities here are not contributing to the energy so VA is VA average is like roughly it is like taken as V1 average again it is an approximation. So what it considers is that the average velocity is like this is the part of the section where the velocity effects are important and this part of the section just has a velocity which is the same as the average velocity as this one and this is totally distributed uniformly. So that is what is so there are lots of approximations involved but many of these approximations are not so bad if the flow is highly turbulent. Now that is number 1. Number 2 is you can write the difference so practically you are considering this as like section 1 and this as like section 2. Section 1 is this part this small part. So if you now write the energy equation like you can write P1 by rho plus V1 average square by 2 is equal to P2 by rho. So kinetic energy collection factor we are not considering because we are assuming it close to 1 plus V2 square by 2 plus head loss. Also the points are so located we are neglecting the change in potential energy. So the head loss we can calculate. So divide by g to call it a unit of head. Head loss is P1 minus P2 by rho g plus V1 square minus V2 square by 2 g. Now P1 minus P2 you can write in place of Q you can write a2 V2. So that a2 gets cancelled from the 2 sides so you get this as V2 into V2 minus V1. So this is since Q equal to a2 V2 which is same as a1 V1. So that is from this equation. So then you can substitute that hl equal to in place of P1 minus P2 you will have V2 into V2 minus V1 by so there is a rho here that there is a rho here right rho into this one. So rho gets cancelled out then plus V1 square minus V2 square by 2 g. So if you simplify this it will be V1 minus V2 whole square by 2 g you can clearly see that just one step. So what we can get from this one is a very interesting thing that I mean this shows that it has to be always a positive thing. So the head loss is positive and it is a function of the difference in the average velocities over the 2 sections. One special case is that let us say this is a small pipe entering into a very large reservoir. So you have a small pipe like this it is entering into a large reservoir. So what is happening is fluid is exiting from the pipe to a reservoir and then that is a special case of this one with D2 by D1 very large and then what does it become? This becomes approximately V1 square by 2 g this is known as exit loss. So exit loss is very important in engineering because it signifies the exit of the fluid from a pipeline to a reservoir. So it is a special case where the ratios of these 2 sizes are grossly different otherwise this is the formula straight away you can use. So this is one example of a minor loss in the next class we will look into some other examples of a minor loss. Let us stop here. Thank you.