Calculating the inlet height in a boundary layer control volume (Fluid Dynamics w/ Olivier Cleynen)

in this problem we’re taking a first look at boundary layer flow flow very close to surfaces in this problem we were in particular looking at air coming in and flowing over a flat plate and it’s coming in here from the left with a very nice smoothly uniformly distributed velocity 25 meters per second and it’s leaving towards the right and by the end of the plate it has now a very strange and complex velocity distribution that is not uniform anymore the question we’re trying to answer is what is the thickness of a control volume at the inlet that has the same mass flow as the thickness here has at the outlet it’s a bit of a strange question and so I’ll spend some time explaining what the question really means before I start answering it let me show you what the control volume could be to study this problem we could simply draw a rectangle on the bottom would be the plate and on top would be a straight line at the outlet you would have a velocity distribution that looks like this like so with progressively smaller arrows as you go down and then you would take this length here and we reproduce it here and this would be here the inlet velocity distribution there’s a problem with this control volume which is it doesn’t have a mass balance if you look at the mass flow exiting of the control volume here there’s deficit here this whole this whole velocity here is lacking if you went to integrate velocity with respect to area at the outlet you will not get the same number as in the inlet and so the question becomes where is this mass flow gone? you cannot create or destroy a mass in fluid mechanics and so the answer is the mass flow exits towards the top like this and with it it carries momentum it carries mass flow but we have no information about this mass flow on top so even though this is a correct control volume to study the problem it’s not a very useful one because we’re lacking information about what happens on top of the control volume. instead to study this problem we would draw a control volume that would look like this we will have a flat plate on the bottom and then the outlet here with the height Delta and then the inlet we would have less of a height so then when we look at this velocity distribution here at the outlet that looks like this like so and we take this maximum velocity here and we reproduce it here then we have here the same mass flow at inlet as we have at the outlet and for this to be correct then we have the top surface top surface must have kind of an unknown shape here that joins one side to the other woop should not auto correct this shape here like so in this control volume mass is conserved we have the same mass flow incoming as we have outgoing and there is no flow across the top surface of the control volume so the question we try to answer today is if this is known this is the height here this height is called Delta is the thickness of the boundary layer then what is the thickness at the inlet here? what is this here this height here which we call h1? what is this height yes? this is what if we try to find once we have this height here it will be a lot easier to calculate force inside this control volume which is the question in the next problem ok so let’s take a look at this let me push this up here and let’s now figure out what this height h1 is since we have a mass flow balance problem I can start by writing a mass balanced equation the mass of a balance equation in fluid mechanics for control volumes looks like this it says that 0 is the total amount of mass that is created or destroyed inside the control volume well it is the sum of two terms it is the change in time of mass inside the control volume and so this is here an integral over the whole volume with the whole control volume and to this I had a double integral over the control surface here and this is then an integral over an area I don’t like to remember this by heart because I miss it very often so I just fill in the terms by looking at the formula sheet every time and so for mas flow this is rho simply density in the middle and in the area it is Rho with a very annoying term which is V rel here dot in the relative velocity as a vector dot an N vector which is unit vector which is always pointing outwards by convention ok what does this equation become in our particular case well the whole first term here this whole term there here this is zero because the flow is steady and so even though this integral here of the total amount of mass inside the control volume this integral is not zero its change in time is 0 so this whole first term goes away what about the second term we’re going to split into two I’m going to say here it is minus for incoming so the integral over the incoming flow here of Rho V orthogonal here and I’m going to put an absolute value and DA this is for the incoming flow and then I have for the outgoing flow plus the integral whop this both of those should not be straight lines the integral out Rho V orthogonal da yeah so this is for the outgoing flow now let’s simplify it further and let’s write it out in a way that’s understandable the velocity incoming here is anyway orthogonal to the inlet of the flow so this V orthogonal here it’s going to be V in one so I can just write it away so the integral over the inlet flow of Rho 1 V 1 hey yeah plus the integral over the outgoing flow of Rho 2 V 2 dA like so and this is still equal to zero now let’s work out step by step what we have in here we have on the incoming flow v1 is perfectly uniform because if you look down here at the bottom left the v1 is just capital u it is 25 meters per second everywhere so the integral of constant Rho constant V vs area it’s just Rho V area yes here it is Rho 1 V 1 a 1 like so Plus at the outlet at the other the velocity is not uniform it has this very weird distribution it is the integral of here Rho 2 V 2. V 2 is the strange velocity distribution which if I gather up my equation over there is expressed it has u Y over Delta to the power of 1 over 7 to the power 1 over 6 dA let’s keep this going and now let’s split up a the Inlet what this term a 1 is because this is very interesting now we have minus Rho 1 V 1, V 1 is just u so and a 1 is 2 components it has on the width side on the Z side it has length L 2 and on the height side this vertical side it has the length we’re looking for which is length H 1 and on the right side on the outlets let’s work it out - Rho is constant and gets out of the integral Rho 2, u is also constant and can also be removed from this and we’re left with the integral here over the whole area of Y over Delta to the power 1 over 6 okay so now let’s take it one step further and leave this here as such for the inlet minus Rho 1 u 1 ya to H 1 perhaps let me just stop for a second and explain again what we’re trying to do here the unknown in this equation is H 1 we’ve got all of the other terms and this is equal to 0 so we know it’s down there on the other side we’re only looking for h1 and we’re formulating this mass balance equation as such we’re saying basically the inlet has the same mass flow as the outlet what is the height h1 so that the inlet has exactly the same mass flow as the outlet so h1 is the unknown in this whole equation Let me continue now at the bottom with this I say this is Rho 2 u2 the integral over the whole area is done again in two dimensions one is going to be the dimension across the plate this is going to be the dimension L2 nothing changes according to Z so everything is just multiplied by L 2 and then I have the integral here of let’s split this into two components so that it’s useful Delta to the power minus 1 over 6 this is just a number and Y to the power 1 over 6 this is the coordinate and then we’re left here with the coordinates that is the vertical direction that is away from the plate and this is dy like so from where to where do we do this integral well where are the limits well we start at zero and Y is equal to 0 the height of the plate here and then we move up along the plate until we reach the height Delta like so now there are a few things that we can simplify and remove from equation we try to move this up again we have zero on the other side so a few terms that are the same one in both parts of the equation can be removed one is the density Rho Rho 1 Rho 2 are the same U 1 and u are the same thing and I also have here L 2 which is the same on both sides so if I divide everything by those three terms they just get away just go away so I still here have 0, 0 is equal to minus I’m just left with h1, our unknown plus here the integral from 0 to Delta of Y to the power of minus 1 over 6 Delta to the power minus 1 over 6 y to the power of 1 over 6 dy it seems a bit scary it’s actually not that complicated Delta to the power minus 1 over 6 this is just one number it’s not affected by Y and so we leave it alone and then Y to the power 1 over 6 is integrated exactly the same way as if we had Y to the power 2 and so Y to the power 2 would become 1 over 3 (so 1 over 2 plus 1) of Y to the power 2 plus 1 that’s the same thing applies here we have here minus of h1 at the inlet plus here Delta to the power minus 1 over 6 not affected by Y and then we have 1 over one 6th plus one here of Y to the power ¹⁄₆ plus 1 over here this is integrated between 0 and Delta like so all right so just time now to just group up the numbers because we have the solution already 0 is equal to minus H 1 plus let’s see what we have here we have still Delta to the power minus 1 over 6… … …and this is evaluated from 0 to delta and so now we can get finally what we want to express h1 is equal to … … … here we have it this is the result that we’re looking for and so now we can just put in numbers into this we have h1 is equal to Delta, Delta is 2 centimeters so it is 0.02 meters multiplied by 6 over 7 and this is equal to (I put this into the calculator before) 1.71 4 times 10 to the power minus 2 meters and we can just write it like so h1 is equal to one point seven one four centimeters okay so this is the final result again what we’re calculating here is the height if I go back here the height of the inlet control volume that is so that it has exactly the same mass flow as the outlet over here and so this is how you use the mass balance equation to calculate and adjust your control volume so that you have exactly the same mass flow outgoing in as the mass flow incoming. .