 So good morning all of you so yesterday we were looking at internal flows we started with basic introduction about the fundamentals of internal laminar internal flows. So I was talking about the growth of the laminar boundary layer the hydrodynamic boundary layer from the different walls of the duct and at some point downstream from the entrance the boundary layers merge and that is what is called the distance where they merge from the entrance of the duct is called the entry length okay the hydrodynamic entry length and beyond that you will find the entire region is dominated by viscous effects okay and also the velocity profiles if you draw somewhere downstream of the region to you will find that they take the shape of a parabolic velocity profile so how do we know that this is parabolic we will derive the from the basic equations we will show that it is a parabolic profile and apart from that if you also have an associated heat transfer either in form of uniform temperature or a uniform heat flux applied on to the walls of the duct so you can say that this is constant Q wall prime or T wall is constant in this case you will you will find a temperature profile so where the temperature is higher at the wall and is minimum at the center of the duct now the criteria for defining these three regions 1 2 3 that I have indicated here so in region 1 the velocity profile keeps changing along with the axial position okay so therefore this is a region where we cannot make much approximations in the navier stokes once it merges here so once the entrance length is reached as the two boundary layers if you plot the profile on from there onwards you will find that it is a parabolic profile and it is invariant of the location downstream so therefore for regions 2 1 3 you can put a criteria that du by dx is 0 that means the profiles are invariant of position downstream and this region is called as hydrodynamically fully developed region okay and when it comes to the temperature profiles however as you can see that when you apply a uniform heat flux or a wall temperature you keep on supplying energy and therefore the wall temperature keeps changing keeps increasing along with the location simultaneously the fluid which is receiving heat from the wall also keeps increasing in its temperature profile therefore you cannot put a similar criteria saying that dt by dx is 0 because dt by dx will keep on changing because of continuous heat addition therefore we have to look at a non-dimensional temperature ? which when defined in this particular fashion T – T wall by T mean – T wall so here unlike the external flows where you use the free stream temperature as the reference we have to use a new reference temperature which we call as a mean temperature or bulk temperature or the mixing cup temperature and we define the mean temperature we derived this yesterday based on that if you replace a non-uniform temperature with a constant or a uniform temperature which we call as T mean so the enthalpy of this and this should be the same okay because this is a flow process so therefore we talk in terms of enthalpy rather than internal energy so for that condition you get an expression for the mean temperature or bulk temperature which is nothing but mass weighted average temperature you see this this is the temperature and this is the mass so you are giving a mass weighted average at different locations radially and then you are summing or integrating them together okay so this is how you define a mean temperature and therefore condition on ? okay now the condition on ? will be now your wall temperature will be only a function of X and so is your mean temperature because you are averaging out along the radial direction so therefore if you look at the numerator so this T – T wall will have a certain variation along X which is exactly similar to the variation of T mean – T wall so therefore your ? as a whole does not have any variation with respect to X okay so therefore ? will be only a function of R right so therefore we can put a condition that if it has to be thermally fully developed then D ? by DX should be 0 so that means ? should be only a function of R okay so therefore if you plot the non-dimensional temperature profile it will look very similar to your velocity profile so where it will go 0 at the wall and you get some kind of a profile like this and this will remain invariant along the axial location okay so we also showed that if D ? by DX has to be 0 a consequently the heat transfer coefficient in region 3 okay where it is hydrodynamically and thermally fully developed has to be a constant irrespective of what thermal boundary condition that you use whether it is a constant wall temperature or constant heat flux okay so this is a very important condition that you get in internal flows which is unlike the external flows where your heat transfer coefficient is a local heat transfer coefficient and it keeps very continuously okay here once you get a fully developed profile both hydrodynamically and thermally the value of heat transfer coefficient has to be a constant so we will have to see now what is this constant value and it depends on the boundary condition okay so the next couple of lectures we will focus on only the region 3 which is the simplest region to start with where you can make approximations with respect to the velocity as well as the temperature profiles and we can derive the expression for it is just a value constant value for the two different boundary conditions one for the constant wall flux the other for constant wall temperature and now other thing is when you define Reynolds number in internal flows you have to be careful that you should use again another characteristic velocity called the mean velocity or the bulk velocity so once again what it means if you replace your velocity profile with a uniform variation all along the cross section okay the it should satisfy conservation of mass at that particular location so the mass flow across this particular cross section in this profile should be equal to uniform profile so if you do that you will get a definition for the mean velocity which is something like this which we saw yesterday and based on that you define your Reynolds number okay now for a circular cross section of course you know this is the diameter of the cross section but you can also extend this definition to non-circular cross sections right so if you have duct with a say rectangular or triangular or square cross section or some other cross section so you have to define your Reynolds number based on what is called the hydraulic diameter so you use hydraulic diameter which is defined as 4 times area by 3 meter okay so if you have a rectangular cross section with dimensions let me just so if I if I just tell you that this is your a and this is your B okay so accordingly this will be so your area now will be what this will be 4 times AB divided by 2 times A plus B right so this will be 2 AB by A plus B okay so this is how you have to replace your diameter with an equivalent hydraulic diameter and define your Reynolds number based on the hydraulic diameter so this is for the case of non-circular cross sections okay so now with sufficient understanding also we have classified the different regimes whether it is a laminar or a turbulent depending on the value of Reynolds number below a certain critical Reynolds number critical Reynolds number generally is about 2300 so below this value it is generally classified as laminar flows and above this generally considered as transition or turbulent although this value may vary in cases where you maintain a very uniform streamlined profile and your turbulent intensities are very low at the inlet and your valve is extremely smooth you can delay the transition to turbulence to as I has 8000 or sometimes even 10,000 okay experimentally it has been shown that you can delay the transition okay but this is a generic thumb rule okay so today what we will do is we will take up first the region 3 okay and in region 3 we will get the solution to the velocity and the temperature profile and now just before that I just want to write down the conservation equations for momentum and energy in region 3 okay so we will make certain approximations when it comes to region 3 so what are the approximation so why we are looking at region 3 is it is one of the simplest to start with because you can make several approximations so that your Navier Stokes equations becomes much simpler so you can get an exact solution to that so directly you found du by dx is equal to 0 therefore your velocity is a function of and you are now assuming although this is a three dimensional flow you are assuming that right now the profile is two dimensional that means you do not have any dependence of velocity on the ? direction so strictly speaking you have to write down the Navier Stokes equation in the cylindrical coordinate system because this is a duct with a circular cross section okay so but you are neglecting the variation of velocity and temperature with the third direction and also you are neglecting the variation with respect to the x direction so it is a function of only R okay and once you reach the fully developed condition therefore in region 3 all your inertial terms get knocked off because du by dx is 0 okay and V velocity is 0 correct there is no V velocity there is only axial velocity there is no radial velocity there is no azimuthal velocity so the inertial terms all disappear okay so there is no so you have your x momentum R momentum ? momentum so R momentum and ? momentum are not there right now only x momentum will be there and in x momentum on the left hand side you have du by dx 0 VR 0 V ? 0 so your entire inertial terms disappear so only you have your diffusion terms and your pressure gradient okay so therefore for this case your I will say VR and V ? these are 0 and also variation so this anyway I have said that use a function of R only okay so therefore the momentum will be only x momentum so you have – 1 by ? dp by dx this is your pressure gradient term on the right hand side plus you have your diffusion term once again in the diffusion terms the variation since your du by dx is 0 so your d square u by dx square will be 0 also the variation with the ? direction is also not there therefore your d square u by d ? square is also 0 okay so the only term what is the only term then which has to be included with respect to R okay the diffusion term due to the gradient with respect to R right so that will be in the cylindrical system nu into 1 by R d by dr into R du by dr right so this is all in cylindrical system this is not Cartesian Cartesian you do not have 1 by R so if you write the Laplacian operator in cylindrical coordinate system you will get 1 by R d by dr into R okay and the shear stress in the radial direction is du by ? into du by dr okay that ? divided by ? is your kinematic viscosity here so this should be equal to 0 so this is a big approximation okay so that is why much simpler to look at region 3 because you are not including any of the inertial terms okay only pressure gradient term balances your viscous forces okay so this is the very significant terms which are important in internal flows in external flows for flat plate you can say that dp by dx is 0 because your free stream velocity is invariant of X but here you have only these two terms and they have to balance each other so you cannot say simply the dp by dx is 0 in fact the pressure gradient is driving the flow here okay so that is why internal flows are pressure driven flows okay so therefore they incur a considerable pressure drop to drive flow internally whereas external flows even if you use a fan or a blower for convection so you do not have any pressure drop or very negligible whereas an internal flow internal flows pressure drop is a very important parameter because to drive the flow you need a pressure drop okay so let us write down the energy equation so the complete energy equation is your u dt by dx plus v dt by dr in the cylindrical coordinate system plus okay so I call this as vr v ? by r dt by d ? this is your inertial terms so I am writing everything under steady state income steady state incompressible these are the other approximation incompressible flow so this should be equal to the diffusion terms that is 1 by r d by dr into r into k dt by dr this is your diffusion term along the radial coordinate so you have 1 by r d by dr r into qr qr is k into dt by dr okay plus you have d by dx of k dt by dx plus 1 by r square d by d ? into k into dt by d ? so we are also neglecting what viscous dissipation you have included everything all the terms except the viscous dissipation now once again for this case you can directly say since you do not have vr and v ? so therefore the convective terms involving vr and v ? 0 right and we have also said that variation of both velocity and temperature with respect to the circumferential direction is 0 correct so therefore you have only one convective term involving the axial velocity which is balancing your diffusion in the radial direction as well as diffusion in the axial direction this is your axial conduction this is your radial conduction but once again we will make an approximation that compared to your radial diffusion or radial conduction your axial conduction is very minor because if you look at the profiles of temperature the gradient is dominant along the r direction therefore the diffusion will be much pronounced in the radial direction so compared to the radial conduction we will also neglect the axial conduction term so finally you are left with you dt by dx is equal to 1 by r so 1 by r d by dr r k dt by dr so this is your energy equation in region 3 okay so now since we have written down the equations we will go ahead and start solving them so I will erase this okay so let us call this as equation number 1 and let us start with the momentum solution first see also when you write down the y momentum equation it comes out that your dp by dy is approximately 0 because all the other terms are negligible right so therefore p is invariant of y it is only a function of x okay so this is the conclusion as far as your pressure is concerned it is a function of x okay so if you write the y momentum all the inertial terms are 0 all the diffusion terms are 0 so that will leave with dp by dy is 0 okay therefore the pressure will be only a function of x therefore you have pressure gradient only along the x direction which balances the diffusion term diffusion of velocity in the radial direction so with that why is that required because now this is a constant term because if this has to balance this this is a function of r this is a function of x so they both have to be constants correct then only this can balance this so once this is a constant that means we can directly integrate it out so this is the simplification okay so therefore since p is a function of x only okay so minus 1 by rho dp by dx should be a constant right since you know this is a function of x this is a function of r they can balance only if they are they are constant therefore we can integrate one integrating one with respect to r so we get if you integrate once r du by dr will be you have to now tell which one correct but dp by dx also has to be a constant because this is a function of p is a function of x and this is a derivative with respect to x this is a derivative with respect to r and u is a function of r so these two can be equal only if they are equal to a constant there is a only possibility okay so this will be nu by rho rho rho cancels this will be 1 by 1 by nu dp by dx so therefore I can replace the partial derivative with an ordinary differential because u is a function of only r and p is a function of x here so I can say this is r so there will be an r2 by 2 right here r2 by 2 plus a C1 okay so now if I integrate it again this will be 1 by nu so my rr cancels here so this will be dp by dx which is a constant and if I integrate it again this will become r2 by 4 plus this is C1 by r if I integrate that that will be C1 ln of r plus C2 okay so this is my solution now I have to find the constant C1 and C2 how do I find the constant boundary conditions okay so what are the boundary condition so u at r equal to r0 so what is u at r equal to r0 r goes to r0 0 and what is the other boundary condition in the case of external flows I have you we can give a boundary condition where your u going to very large why why to infinity then your u becomes u infinity whereas here you do not have anything like that exactly so at r equal to 0 u should be finite now directly you can look at this equation if r goes to 0 this goes to infinity so in order to make u finite C1 has to be 0 okay so directly this gives you C1 is 0 so you can use u of r u at r equal to r0 equal to 0 to calculate C2 and if you do that your final velocity profile comes out as okay so this is your velocity profile and you can directly see that for a given pressure gradient dp by dx for a fixed pressure gradient the variation with respect to r is parabolic okay so at r equal to r0 this will be 1 minus 1 0 and at r equal to 0 this will be 1 by 4 minus 1 by 4 new dp by dx into r0 square okay so now we want to see what is the maximum velocity so because when we draw the profile we draw it like this and we know that we are plotting it with a maximum velocity at r equal to 0 but let us show that the maximum velocity or center line velocity is maximum velocity okay so how do we show that how do we show the location of maximum velocity yeah so you take the derivative du with respect to dr equate it to 0 and you will find that r comes out as 0 so that means the center line is the inflection point okay so there could be a maximum or minimum of course there has to be a maximum so this is the region where so u max is equal to u center line okay the center line velocity is the maximum velocity and the value of the maximum velocity here is 1 by 4 nu dp by dx r0 square this is the value of maximum velocity so now we will go ahead and calculate the other the other is the mean velocity which is 2 by r0 square integral 0 to r0 u of r r dr okay so if you substitute the velocity profile and integrate it r0 square minus 1 by 4 nu dp by dx these are all constant terms which can be pulled out outside the integral 0 to r0 1 minus r by r0 the whole square into r dr okay so if you integrate it out you will get minus 1 by 2 nu dp by dx into r0 square by 4 so which can be simplified to u m is minus 1 by 8 nu dp by dx into r0 square so therefore you see u m is a it is now a fixed value in the fully developed region because dp by dx is a constant so everything is constant so therefore u m is a constant value in the fully developed region okay so this is your expression for u m we also found out the expression for uc therefore you can also write your ur u of r as this entire thing is nothing but uc okay and you can see from this expression that your centerline velocity is actually twice of the mean velocity right so therefore you can express your u as twice of u m into 1 minus r by r0 the whole square which is also u centerline 1 minus r by r0 the whole square okay so this is this is how your general velocity profile is written as a variation in terms of r is it clear since you have already done this I am not spending too much of time you know you can just go and look into any fluid mechanics textbook and we will find the derivation so any questions so far so you can clearly see that at centerline you have r equal to r0 you have your centerline and from there you have a parabolic velocity profile distribution okay so therefore we will find out other integral quantities the important integral quantity will define an internal flow is called the friction factor okay in external flows we define the walsh stress and from there we calculated a non dimensional skin friction coefficient CF okay the same way we can also define a skin friction coefficient okay but generally people refer to as we will show that we will show that the equivalent of a skin friction coefficient is your fanning friction factor okay so there are two different friction factors which are defined so one which is more popular is your Darcy or your moody friction fact so named after these two people who have defined it and it is defined as f is equal to ?p by a fro u m square l by d because here your l by d is also an important factor so as far as your pressure drop is concerned because dp by dx is nothing but ?p by l so that has to be included as it is okay so this is how your moody's friction factor is defined the other friction factor is also called fanning's friction factor and he defines it as so that is ?p by twice rho u m square l by d so you see the basic form is the same it is only the factor here they use a factor of half and here fanning uses a factor of 2 and you will see why fanning defined it this way is that for the fully developed internal flows okay so the fanning friction factor exactly comes out to be the same as skin friction coefficient okay so therefore you defined it in a convenient form where you can replace the skin friction coefficient which you are used to in external flows and that will become equal to your friction factor in the internal flows okay so this can also be written as minus dp by dx into d by twice rho u m square right because ?p by l is nothing but minus dp by dx this is a pressure gradient so since this is pressure gradient keeps so your pressure changes is decreasing therefore if you take a pressure gradient will be negative therefore you have to put a negative sign here okay so this is the way that you write it and you can now substitute for dp by dx from this expression in terms of so your let me call this as equation number 2 from equation number 2 you can write dp by dx in terms of u m correct and you can substitute that here so you will find that some of the terms u m terms will cancel off so if you substitute that this will be 8 ? into u m divided by r 0 square you have 2 ? u m square okay so this should come out as 8 u by ? u m r 0 okay so the u m u m cancels here and okay so this is d here so d is twice r 0 right and now you can define Reynolds number okay so this is ? u m and this you can write as d 0 by 2 okay so this will become 16 divided by Reynolds number similarly if you had substituted here for the moody friction factor this expression will come out as 64 by Reynolds number okay so that is the difference the factor is the difference and now if you define your skin friction coefficient CF for internal flows as ? by ? u m square okay and your ? is nothing but minus ? du by dr at r equal to r 0 divided by ? u m square so you can calculate you have the velocity profile you can calculate the derivative with respect to r and you have to calculate derivative at r equal to r 0 and you substitute it so you will get an expression for CF which also comes out to be exactly identical to 16 by R E D so that is why the way the fanning friction factor was defined both fanning friction factor and skin friction coefficient become identical okay so therefore for the fully developed case your CF is equal to your F let me call this as F fanning here to differentiate it from F Darcy so this is a another important result that you should remember okay now if you plot the friction factor I think all of you know how it looks so this is called the Moody's chart right I hope all of you remember the Moody's chart in fluid mechanics that is a in internal flows there is a single most important thing that you might have learnt okay if you plot F as a function of R E D okay now remember if you plot on a log log scale now this is this is a hyperbola right you have F say F Darcy F into R E D is equal to 64 as a rectangular hyperbola so if you plot on a log log scale this becomes linear okay so on a log log scale if you plot a log log scale so this will look with your increasing Reynolds number the friction factor comes down right so this is your relationship F is equal to 64 by R E D and this is valid for certain Reynolds number till what critical Reynolds number you can say 2300 so once critical Reynolds number exceeds the flow becomes turbulent and the variation will not be as straightforward as this it becomes completely different and in the turbulent region you will find it is not a single line but you will have several multiple lines like this okay so what these multiple lines indicate is the friction factor is not only a function of Reynolds number but it is also function of the roughness okay so you plot what is called a non-dimensional roughness that is roughness height divided by the diameter of the tube so with increasing roughness your friction factor increases so this is completely turbulent whereas in the laminar flow you find a very linearly decreasing profile if you plot it on a log log scale so this is called the Darcy or Moody chart from which you can pick up for a corresponding value of Reynolds number what is the exact value of the friction factor okay I think all of you are familiar with this in the case of non-circular ducts okay this is valid for a circular duct okay if you go to non-circular ducts so your friction factor F becomes 64 by some factor Phi times R E it is not 64 by R E 64 by R E is only for circular ducts because we have seen that we have substituted all this relation for circular ducts but generally for non-circular ducts you have an additional factor Phi and this Phi will be a function of the aspect ratio so if you have a non-circular cross section okay you have aspect ratio A over B okay so it is a function of this aspect ratio suppose you take a rectangular cross section you can plot Phi as a function of A by B so says A by B going from 0 to say 0.5 and on the top scale say from 0.5 to 1 so you have a variation something like this corresponding to this and you have a variation corresponding to A by B of 0.5 to 1 so you have typically variation of Phi for a smaller range of A by B 0 to 0.5 you have one variation for A by B varying between 0.5 and 1 so at the limiting value 1 F becomes 1 basically okay so then it becomes equal to a circular cross a square cross section okay so then you have the value of Phi at somewhat approaching to 1 when you are A by B approaches 1 okay so therefore you have to know this factor and then you have to plug it for a non-circular cross section with a different aspect ratio and that will give you the corresponding friction factor so anyway these are some additional details which you can find in textbooks so next what we are what we are going to do is take the heat transfer problem so very quickly in the another 5 10 minutes and see how we can find the temperature profiles in the thermally fully developed region okay now we know how the velocity profile looks so now in region 3 temperature solution so once you get the velocity solution for velocity profile naturally the next part will be to get the solution for the temperature and we have already I have already told you that the heat transfer coefficient is a constant in region 3 so we have to know what is the value of this constant so therefore first we look at a boundary condition which will be the uniform wall flux that is q wall double prime is a constant okay now we have already defined my D by DX of ? and ? is T – T wall by T – Tm – T wall this is equal to 0 for thermally fully developed case so we will expand upon this term okay so if you integrate if you differentiate it by parts okay how so what do you get so this is Vdu – Udv by V2 right so if you differentiate and expand it you can just try I will give you some time you should be getting an expression like this so all of you please try expanding this differential here okay so maybe you can complete it as a homework but you should be able to get very straightforward you should be able to get this now the important fact is when you apply the uniform heat flux boundary condition from the Newton's law of cooling you can say that q wall double prime is H into ? T T wall – T mean so since your wall flux is uniform therefore DQ by DX will be 0 right there is no variation with respect to X and therefore your this will give that DT wall by DX should be equal to DT mean by DX very important conclusion as it because coming from the fact that you have a uniform heat flux so therefore if you look at the profiles if you substitute let us call this as now number one this is number two if you substitute this conclusion that is DT wall by DX should be equal to DTM by DX what happens these two terms will cancel out so therefore you directly get the fact also that DT by DX should be equal to DT wall by DX which in fact it is equal to DTM by DX okay so the slopes of all the three profile like you are any temperature at any location radial location your wall temperature and your mean temperature they are identical and it can be true only so how should the profile look if you plot your temperature on the Y axis and X if I want to show temperature at any radial location wall temperature and mean temperature and if the slopes have to be the same it has to be exactly parallel all the three have to be parallel and in fact we will see that this has to be just linear okay so you can say then this is your T wall okay this is may be T at some radial location and say this is your T mean so this is a function of X this is a function of R and X this is a function of X but all of them in the region three will be linear okay and the slopes are identical okay so this is a very important conclusion which characterizes a fully developed flow with a constant heat flux boundary condition so therefore if you apply this fact to the energy equation okay so what was the energy equation that we had written we had written if you u by ? DT by DX oh okay there is one more thing I forgot row CP on the left hand side of the convective terms in the when you wrote the energy equation that should be row CP okay on the right hand side you have K okay so I am taking row CP by K which is 1 by A DT by DX should be equal to 1 by R D by DR R DT by DR now the fact I am going to use here is my DT by DX is exactly identical to DTM by DX so these are all just a constant okay so this can this can be through only if they are all constant because T is a function of both R and X if this slope has to be equal to this and this this has to be equal to X DT wall by DX the slope has to be a constant value and therefore it is a linear variation okay so this is an important fact you can just replace this as DT mean by DX now this is a constant so therefore you can see that you can now integrate this equation with respect to R just like the way you integrated the velocity profile because now this term is a constant and therefore this does not become a partial this can be written as an ordinary differential equation okay so I will just give you the result of the integration and we will stop there so if you integrate this equation you will get solution for T as a function of DT or DTM by DX 1 by A DT by DX and you have to also substitute the parabolic velocity profile which you obtained from the fully developed hydrodynamic solution so then you integrate it with respect to R so you get so this is your temperature profile so we will stop here for today so tomorrow we will start continue from here apply the boundary conditions and calculate the constants and also similar to calculating integral quantities for velocity like skin friction coefficient and friction factor we will calculate the heat flux and the heat flux anyways a constant will calculate the heat transfer coefficient and we will know what exactly the constant value is