Lecture 39 : Displacement and Momentum thickness

[Music] next we will try to go through some of the important concepts or important terminologies related to the boundary layer and these terminologies are displacement thickness and momentum thickness so first terminology that we will try to understand is displacement thickness Delta star what it is let us say that you have a flat plate but there is no growth of boundary layer so then what happens if you have uniform flow uniform flow will remain uniform and that means the plate is having no effect so if there was a stream line say there was a stream line like this this stream line would move parallel to the original direction so when you have uniform flow the stream lines are actually parallel to the flow direction so let us say this is one of such undisturbed stream lines now this is what that does not happen in reality in reality what happens in reality you have a growth of the boundary layer now let us say that the boundary layer is growing and the boundary layer is growing like this so because of the growth of the boundary layer what happens what will happen to the stream line the stream lines will originally be parallel to each other but when they come close to the plate will the stream line so thus so we have decided or we have come to a conclusion that the stream line cannot remain just like this so it has to get shifted or deflected question is it will be deflected upwards or downwards upwards so let us try to see that why it should get deflected upwards so let us say that because of the effect of the boundary layer the stream line has got deflected upwards so this same stream line which was undisturbed now it has got deflected upwards the reasoning is very straightforward if you consider two sections let us say that we consider one section here say section a and another section here say B if the flow is steady whatever is the flow rate through section a should be the same flow rate through section mean because see this black line this is a streamline so there cannot be any flow across it okay now if the same flow rate has to be maintained see this is the region where the flow is slowed down to compensate for that the stream line should be upward so that this extra flow Plus this flow becomes same as this flow right so the stream line has to move upward not downward question is how much does the stream line move upward let us say that we have to chosen such a reference that this is the Delta at the given X and let us say that this shift of the stream line is Delta star we want to find out what is this Delta star nature is not important for us this is just a rough sketch what is important is that whether it is shifted upwards or downwards because like we are not having any attention on what happens across the stream line because by definition of the stream line there is no flow across the stream line okay so let us write let us try to find out this Delta star what will be the basis by which we may find it out that the mass flow rate at the sections a and B must be the same so you must have m dot at the section a same as m dot at the section B so if the uniform velocity was u infinity at the section a it is Rho into u infinity into Delta into the width let us say the width is what B now the wit will get cancelled in both the sides so let us not just write the width for b4b you see there is one portion within the boundary layer so that is integral zero to Delta u dy where u is the velocity as a function of Y here outside the boundary layer you have u equal to u infinity so plus Rho integral of Delta to Delta plus Delta star into u infinity dy basically that into u infinity into Delta star so from here if we simplify one more step what we get you will get u infinity into Delta is equal to integral 0 to Delta u dy plus u infinity into Delta star this term you can also write as integral 0 to Delta u infinity dy all the same the whole idea is that we want to Club these two terms together so from here you can write Delta star is equal to 0 to Delta 1 minus u by u infinity dy by dividing all the terms by u infinity ok so this is known as displacement thickness so physically what it is indicating physically it is indicating may be a displacement or a shift in a streamline because of the existence of the boundary layer it may also be looked from a different angle let us say that we are trying to consider a case when this is a flat plate with boundary layer and retardation of the flow close to the wall and we consider an equivalent case where we do not consider the effect of the wall directly but what we do is we make a shift of the wall so the new location of the wall becomes say like this what should be the shape so that this problem with a boundary layer is equivalent to a problem with a uniform flow so this is a problem with a boundary layer this is a problem where we are trying to have the same flow rate but no boundary layer that means there will be a uniform flow so let us say that the uniform flow is like this with a velocity U infinity so here there is a growth of the boundary layer so you have a delta as a function of X here you don't consider the growth of the boundary layer but to avoid that analysis what you do you shift the plate a bit upwards because the effect of the boundary layer is it has a reduction in the flow rate so if you still want to use u infinity you make a shift of the plate somewhat so that your effective area of the flow gets reduced so that multiplied with u infinity should give the same as this one so let us say that this shift of the plate is a what is our constraint our constraint is that the flow rate should be the same this is a hypothetical uniform flow this is the DL boundary layer flow so here the velocity profile is like this this is U infinity so what is the flow rate here the flow rate here is integral of 0 to Delta u dy what is the flow rate here u infinity into Delta minus K of course I am just considering the volume flow rate per unit width of the plate so Rho into B that term I am NOT considering so if you equate these two what you will get from here is what is a a is nothing but integral 0 to Delta 1 minus u by u infinity dy so this is nothing but Delta star so what does it mean it means that the displacement thickness also may be looked at as a hypothetical displacement of the solid boundary so that the remaining flow within the length delta may be perceived as a hypothetical uniform flow still predicting the same correct flow rate okay so the advantage of this is that sometimes you do not want to analyze the details of the boundary layer but you just want to have a Brass estimate of the flow rate if you know what is Delta star then what you may say is that within this length this minus Delta star whatever is there the velocity is uniform so it is not that it is actually uniform it is a pseudo situation with which you are matching with the exact situation what you are not compromising with is the flow rate prediction so the suit of situation and the correct situation are giving the same flow rate that is the basis of this now although the mass flow rates over the sections a and B are the same but the momentum flow rates are not the same and we will see that what is the difference in the momentum flow rates over the sections a and B so let us write what is the momentum flux or maybe momentum rate of momentum transfer so the rate of momentum transfers at a what is this so you have what is the rate of mass transfer at a Rho u infinity into Delta that multiplied by u infinity is the rate of momentum transfer at a what is the mobile date of momentum transfer at B again one part within the boundary layer another part outside the boundary layer so whatever is within the boundary layer you have lo integral of u into u that is u square dy from zero to Delta plus it is basically Rho u infinity square into Delta star that is the top portion top portion outside the boundary layer velocity is U infinity so if you want to find out what is the difference between these two momentum flux momentum rate of momentum transfer at a and B the difference of these two so that you have Rho into u infinity square Delta minus integral 0 to Delta u square dy minus Rho u infinity square into Delta star is what Delta star is u infinity minus u by u infinity integral of that dy from 0 to Delta so one of the U infinities get cancelled out okay so when one of the U infinities get cancelled out the next step we can have a simplification let us go for that simplification so these becomes first of all these becomes minus Rho u infinity square Delta so there is one u infinity square Delta then there is minus one u infinity square Delta so this term gets cancelled out with the first term so what remains is Rho u u infinity minus u square dy from zero to delta u u infinity from the last term and u minus u square from the second term okay so now what you see that what this is not zero right therefore there is a difference now is this positive or negative this is positive because u infinity is greater than u so u into infinity is greater than u square so that means this is actually a representative of the very important fact that there is a momentum deficit at the section B so whatever was the rate of momentum transport at the section a the rate of momentum transport at the section B is somewhat less than that and this is indicator of how less it is so this you can write as Rho into u infinity square let me just taking u infinity square as a reference so u by u infinity into 1 minus u by u infinity dy just are not normalized or non-dimensional way of writing is the term which is there in the integral and see this term in the integral is a term that we have encountered in a momentum integral equation now wall by Rho u infinity square was DDX of this quantity and this we call as theta which is also given a symbol of momentum thickness so momentum thickness that is given by theta and what is the important is that this momentum thickness what it physically indicates it physically indicates the deficit in the rate of momentum transport across two sections because of the development of the boundary layer that is the physical implication of this but mathematically this is expressed just by this integral sometimes one also uses a third parameter which is known as the shape factor H which is given by Delta star by theta the ratio of these two why it is given a name of a shape factor is is somewhat obvious the obvious is that this ratio depends on the shape of the velocity profile solely on the shape of the velocity profile because if you see the integrals these integrals of course they themselves depend on the shape of the velocity profile but sort of even their ratio also therefore depends on the only on the shape of the velocity profile shape of the velocity profile means u by u infinity as a function of Y by Delta that is the shape of the velocity profile so once that shape is fixed this is also automatically fixed so that is why we call it a shape factor and if given a velocity profile it is possible to determine the shape factor now the important thing is that we have till now consider the boundary layer over a flat plate and that is one of the very simple cases but even in this very simple case we have not considered one thing that is the effect of turbulence we have considered that means implicitly that the boundary layer develops as a laminar boundary layer but in reality the boundary layer may have a transition towards turbulence and what is the reynolds number that is important here the reynolds number here is dependent the on the parameter the axial location X so you see that as you move along the plate this Reynolds increases beyond the critical Reynolds number you have the inertia forces dominating so much that a slight disturbance may trigger the onset of turbulence and then the laminar boundary layer changes its characteristic to a turbulent boundary layer may not be abruptly but at least over a given distance and the critical X or the critical Reynolds number at which this transition occurs for flow over a flat plate is roughly 5 into 10 to the power 5 so you see that it is the critical Reynolds number it depends on different geometries see for flow through flow through a circular pipe the critical Reynolds number was roughly of the order of 2000 they are the reference length was the diameter of the pipe here the reference length is the axial coordinate there the reference velocity was the average velocity here the reference velocity is the free stream velocity use infinity so the it's the critical Reynolds number is therefore not a magical number which is true for all cases like depending on the situation your reference length changes your reference velocity changes and the convention changes altogether now what happens to the growth of the boundary layer if you have such a transition to understand that let us work out a problem by which we will base Illustrated so just note down this problem that consider the turbulent boundary layer over a flat plate for which U by u infinity for which U by u infinity is given by Y by Delta to the power of 1 by 7 okay so this is the velocity profile for the turbulent boundary layer so when you say you would even body we are not writing it explicitly is what this is the average velocity this is not the instantaneous velocity with fluctuations so u by u infinity is y by Delta to the power 1 by 7 this is given for the turbulent boundary layer what you have to determine on the CF and CD that is the local skin friction drag coefficient and total drag coefficient on the plate for the flow over the plate if the entire boundary layer is turbulent and number two the same things CF and CD if the boundary layer undergoes our transition to turbulence at re critical equal to 5 into 10 to the power 5 okay then what is given given is that at the wall this is what is given at the wall we will see that why this is given at the wall this is from experiments this is actually a real data from experiments conducted by Blasius so this is from experiments and for laminar flow for laminar boundary layer it is given that CF is equal to 1.3 to 8 into reynolds number that is cdf like CD 1.3 to 8 into reynolds number to the power minus half CDF okay so with this problem statement let us try to work out this problem and this will give us some idea first of all this is the velocity profile which is again for turbulent flow it is not possible to determine the velocity profile exactly so this is just a curve fit of the experimental velocity profiles and therefore it is it is not a very accurate one the loss of accuracy is important at some place and let us see where what is the place so let us first write now wall by Rho u infinity square is equal to DDX of integral 0 to Delta u by u infinity into 1 minus u by u infinity dy so this part is a substitute d u by u infinity as a function of Y by Delta and then you will get this as something into D Delta DX that is fine substitute u by u infinity as Y by Delta to the power this now what about towel see if you want to calculate mu D u dy at the wall you see that that you cannot calculate using this velocity profile so this velocity profile is not exactly valid at the wall for calculating the wall shear stress and that is why the wall shear stress has to be separately determined from experimental conditions and that is what it has been experimentally determined see fluid mechanics is such a subject where you cannot independent go or grow with theory or experiment you have to somehow make a good combination of these for understanding the physical principles so towel by Rho u infinity square you will have 0.0225 into mu by u infinity into Delta to the power 1 by 4 so then it is straightforward you integrate this and get Delta as a function of X so if you integrate this you will get let me just tell you the answer quickly you will get Delta by X is equal to 0.37 into Reynolds number to the power minus one-fifth okay so once you get Delta by X it is easy to calculate the CF and CD just as we did for the previous cases so let me give you the answers at least so that you can verify later on so CF will be 0.05 eight into Reynolds number to the power minus one-fifth and CD is zero point zero seven two five into Reynolds number to the power minus one okay so these are the answers for the first part of the problem but this is not a very realistic representation why because we know that the boundary layer does not become turbulent from the very beginning the Reynolds number is based on the local X local X is initially very small so it is always initially laminar and then it becomes turbulent so the real picture may be something like this so you may have the flat plate in this way you have first a laminar boundary layer then it undergoes a transition to turbulent boundary layer let us say that this is the length x critical over which it is laminar and then beyond that it is turbulent so what the question will be that what is the length of your plate if the length is greater than X critical how do you find out X critical X critical is given by u infinity X critical by nu is 5 into 10 to the power 5 that is re critical so given u infinitive and always find out what is ex critical if your length of the plate is greater than the ex critical then in reality there is a great chance that the remaining flow is turbulent so your actual drag force actual is what if you say calculate the total drag force as an artefact or outcome of that the entire boundary layer is turbulent there is an error because of the presence of the laminar part so that error we have to nullify so what we can do we may consider that it is if drag force that is the drag force with from X equal to 0 to X equal to L considering it as turbulent fully turbulent then subtract the drag force from X equal to 0 to X equal to X critical for turbulent and add the drag drag force for that part for the laminar okay so as if the full thing was turbulent then you subtract the turbulent part from the initial and at the laminar part okay and you have to keep in mind that when you are writing the corresponding the CDs so this will be CD into half Rho u infinity square into B into L here it will be CD into half Rho u infinity square B into X critical and here also it will be the same CD into half Rho u infinity square B into X critical the difference is that in one case the CD is the laminar CD in another case is the turbulent CD so this is turbulent CD this is the turbulent CD and this is the laminar CD in a problem already it is given what is the expression for the laminar CD you have to be careful in place of L you have to now use the X critical as the reference length and then if you make a simplification this the net drag force is the F drag force actual by half Rho u infinity square into B into L it is the net CD and the answer of this will come out to be zero point zero seven two five Reynolds number to the power minus half minus one seven four zero by Reynolds number this is the answer to this so if you just substitute and make a simplification just check that whether you get this final expression so what this final expression says that there is a correction because of a part of the boundary layer being laminar and not the entire boundary layer being turbulent and this is the expression of the composite CD that takes care of the fact that a part is laminar and then it is turbulent so with this we stop our discussion today and in the next class we will start with the discussions on boundary layer where you have the effect of the pressure gradient till now we have considered the boundary layer without the effect of pressure gradient and boundary layer with the effect of the pressure gradient we will start discussing from the next class thank you

Channel: Advanced Fluid Mechanics - IITKGP

Views: 2,260

Keywords: Derivation of displacement and momentum thickness, Shape factor, Illustrative example.

Id: mwpciKNsEgY

Length: 28min 36sec (1716 seconds)

Published: Sat Mar 10 2018