Lecture 48 : Turbulence Models

Hello everyone, welcome back, we have been discussing about turbulent flows, we looked the RANS equations for the incompressible flow and compressible flow and we also started discussing about how to basically address the closure problem in the turbulent flows. So, the Reynolds stress that we get, how do you basically model them. So, we looked at the Boussinesq hypothesis briefly in the last class and we will basically elaborate it for all three dimensions in the present class and then we go to basically deriving turbulence models. So, this is the equations that we have derived last time for the compressible flows. Of course, when it is incompressible your RANS equations, this will go away and your mass conservation also becomes just del dot u bar equal to 0. But, I am doing it for a general flow, general compressible flow and incompressible is very easy if you can do it for the compressible flow. So, these are the terms that we had. So, we had this total stress instead of only the stress term, we had the total stress which includes the laminar stress as well as the turbulent stress. The laminar stress we already have the formula using the Newton's law and this can be written as 2 mu s i j star where I am using this s i j star to include this portion del u k by del x k in the diagonal. We have the turbulent flow which is given by minus rho bar u double prime u double prime tilde or in the case of incompressible flow it will be like rho u prime u prime bar. In the compressible flow, we have additional heat equation. So, that gives us the thermal conductivity term will solve. So, we have two terms one due to laminar that is the viscous diffusion and the other is the turbulent transport. So, this is the viscous diffusion of heat q j laminar and this is the turbulent transport of heat again this is a term which needs to be modeled. So, there are two terms which needs to be modeled I mean at least actually this is a six term here if you see there are six stresses and then you have a turbulent transport of heat term. So, how do we do that? We already discussed it in the last class that just the way the stress in a laminar flow basically is governed by the mean flow gradient. Same way we can think of that the turbulent stress or the Reynolds stress or the power stress basically is related to the mean flow gradient. So, that is a very bold approximation, but it works for most of the flows and the idea behind it is basically if you look at any simple shear flow. So, if you look at even the stresses due to the eddies they are somehow related to the flow gradient. So, if you can relate to the flow gradient the problem of having six unknown stress quantities basically three normal stresses and three independent shear stresses can be reduced to one unknown problem which is the eddy viscosity. So, what we said is for one due flow if we can write it here let us say if this is your flow I mean shear flow. So, you can write it as tau turbulent I mean it will have two parts one will be laminar part and other will be the turbulent part. So, laminar part of course is given as mu del u bar by del y and we are saying that your turbulent part also gives the stress as mu t del u bar by del y. Now, it is very important to remember that mu t is not the property of the fluid here mu is the viscosity which is the property of the fluid mu t is called the eddy viscosity or the turbulent viscosity this is not the property of the fluid, but rather it is the property of the flow. So, in fact if the flow has very high Reynolds number the mu t will be high if the flow has low Reynolds number means the velocity is small or in general the Reynolds number is small then the this quantity will be low and if the fluid laminar this quantity will be 0. Now, this is good enough for like one due flow how do you do it for three due flow. So, three due flow your stress terms I can do it here. So, your stress term as we have seen earlier that is be given by minus u prime u prime let me write in a in compressible only minus rho v prime v prime minus rho w prime w prime these are the normal stresses and then you will have minus rho u prime v prime minus rho u prime w prime minus rho v prime w prime these are the independent stresses and then we will have two terms here also. So, you can write it minus rho v prime u prime which is nothing but minus rho u prime v prime and here also you can have minus rho w prime u prime which is because of symmetry it will be equivalent to u prime w prime this quantity and this will be minus rho w prime v prime which is equivalent to this and if you use the definition that we had earlier using mu t. So, what we can write what is your strain tensor that will be this is 2 del u bar by del x 2 del u bar by del y and 2 del u bar by del z remember our earlier definitions we had derived it this kind of expressions while doing the deriving Navier Stokes equations and this will be nothing but del u bar by del x plus del v bar by del y and similarly it will be del u bar by del x plus del w bar by no. So, this will be d u by d y right. So, d u by d y plus d u by d x not del u by del x. So, del u bar by del y plus del u bar by del x and this will be del u bar by del z plus del w bar by remember w x is always 3 dimensional and there is what is the last term here this will be del v bar by del z plus del w bar by del y right. And what we are seeing that this is equal to this and multiply by mu t. So, again this term will be symmetric to this term the term here will be this is x and w. So, that will be similar to this term and the term that will appear here will be similar to this term right. So, now if you add this. So, if you use the business hypothesis this all the stress terms should be basically found by just knowing this mu t, because we know all this quantities all the gradients quantity is known by the momentum equation or mass conservation equation or basically equation right. Now, but there is one problem. So, if you add the trace here right. So, what is the trace here? So, the trace here will be minus rho u prime u prime bar plus v prime v prime bar plus w prime w prime bar right which is nothing but twice of the kinetic energy. So, basically minus 2 ray rho k bar which is the turbulence kinetic energy right. However, if you add these three terms what will happen you will this is written for the incompressible flow right. So, it is become 2 del u bar by del x plus del v bar by del y and del w bar by del z right that is nothing but del dot u and that will go to 0 right. So, there is a problem here right if you just use this definition and it is essentially saying that my turbulence kinetic energy is 0 which is not possible. If the flow is turbulent the flow has to have the kinetic energy and it has to somehow accounted here in this formulation that we have right. So, what we can do here is basically in order to make it physical. So, that when I add the trace it should give me the term which is minus 2 rho k bar what we will do my we will just basically be adding the quantity which is basically minus 2 by 3 rho k only in the diagonal term right. So, if you do that equation will look like let me here your equation will look like mu t here and this will become 2 del u bar by del x minus 2 by 3 rho k bar I am only writing the diagonal term you know all the other terms 2 del v bar by del y minus 2 by 3 rho k bar and then minus 2 del w bar by del z minus 2 by 3 rho k bar. This is to make it consistent we have other terms which are not written and if you add this 3 all this 3. So, this term will go to 0 because of the divergence condition and only what will remain is 3 into. So, this will be 3 into minus 2 by 3 rho k bar and this 3 3 will cancel out and that will give you the minus 2 rho k bar right. So, this is your final definition will can be written in a short form which is like this. So, this is your symmetric stress tensor this is written mu t as s i j bar is basically nothing but. So, s i j bar is del u bar del u i by del x j plus del u j by del x i half of that right. So, 2 of s i j is the tensor minus 2 by 3 rho k delta i j delta i j as we know the property the delta that it will become positive only or it will appear only when i is equal to j right. So, that is still only in the diagonal term right. So, we got the value of the Reynolds stress here for the incompressible flow and it is also consistent because if you take the trace of this you are getting the correct value. Now, the same thing you can think for the compressible flow also. So, it is very similar to the incompressible flow. Now, the definition of the Reynolds stress is basically the forward stress here and instead of s i j you will have s i j star because this term should also come in the diagonal. So, basically it will be minus 2 by 3 del dot u that will when you add all this minus 2 by 3 del by u. So, that will cancel out the divergence term which is appearing from the first equation right. So, let me write it here for the compressible flow what will be the difference there will be another term which is minus 2 by 3 del dot u minus 2 by 3 del dot u and minus 2 by 3 del dot u right. So, derivative term I mean this is basically to make sure that when you add all these things. So, if you add all the quantity on the trace you just talk about the trace for the compressible this is 2 del dot u minus 2 rho k and minus 2 del dot u again because of this term right. So, although the divergence is not 0 this will go to 0 that is essentially give you minus rho k tilde because this is the power of this right. So, again this is consistent for compressible flow also. So, whether it is compressible or incompressible we have reduced the closure problem to basically one single unknown which is mu t and for consistency we have used added minus 2 by 3 rho k in the diagonal term right. As you see that this is heavily modeled and we have used the physically consistent definitions, but again this is something which is not derived from the first principle right that is why it is called a model. So, that principle is called the Boussinesq hypothesis as given here or may be the Boussinesq model it is given around 1877 or also called Boussinesq eddy viscosity model because we are not basically solving for a stress we are solving for the eddy viscosity model. Now, our known is nothing but this eddy viscosity all other things are right. Now, as we saw earlier that we have another issue in the compressible flow that we do not know this term there are many other terms which we neglected in the compressible flow which were basically related to the viscous diffusion of the fluctuation system or the correlation of fluctuation term, but this is the turbulent transport basically this is the fluctuation which is u j double prime which is basically transporting the heat right. So, this is the term that needs to be found out. So, what we do is that it can also be written as you can take the c p inside and it becomes rho u j double prime h bar right. So, like we did the for the velocity gradient we will also relate it to the temperature gradient same way like we do the molecular diffusion of heat. So, how do you do the molecular diffusion of heat we have this k delta T by del x right. So, this is the power of this I am using tilde here. So, this is how you give the minus sign here or here lambda. So, let me use lambda here lambda is the thermal diffusivity or thermal conductivity. So, we will use the similar definition and we will say that our turbulent transport of heat is given by rho u j double prime h is given by a turbulent thermal conductivity which is acting on the temperature gradient right and that is how it is getting transported. So, this is by Reynolds analogy which relates the flow gradient or the velocity gradient to the temperature gradient right. Now, our unknown is what is the turbulent thermal conductivity right. So, if you recall while non-absorbing we had used the number called Prandtl number which was given by mu c p by k and we lambda like lambda here. So, lambda was mu c p by p r. So, this was molecular diffusivity by the thermal diffusivity we had used this definition. So, in the same spirit we will define a turbulent Prandtl number. So, which is given by p r t and that will be mu t viscosity c p by lambda t here. So, this is the eddy viscosity this is the turbulent thermal conductivity. So, that way you can now define or write our turbulent thermal conductivity as mu t c p by p r t right. Now, p r t is the turbulent Prandtl number which is often taken as 0.9 if the flow is not very high speed and like it is not like going to like high hypersonic regions. Then this definition works quite well especially in subsonic and low supersonic regions you can easily use this definition and now you can see that the only unknown we have still the mu t right all other things have been known. So, if you know the mu t c p is 1 for the gas and the turbulent Prandtl number is taken as 0.9 or around some that value and you can get the lambda t from lambda t you can calculate your rho u j double prime h bar and then you can close all the equations right. So, all the unknowns that we had earlier in this only these two were unknowns and then we have closed it by just using mu t right. We use the Boussinesq hypothesis as well as the Reynolds analysis. Now, all the models which basically relates the Reynolds stress or the published stress to the mean flow gradient come under the category of eddy viscosity model right. So, we are looking for finding out mu t right which is the eddy viscosity or you can write the kinematic eddy viscosity is mu t which will be given by mu t by rho. Now, all the models eddy viscosity model the one I think you might be using also k epsilon k omega s h t k omega even there are some non-linear viscosity eddy viscosity model also. There are many variations of k epsilon many variations of k omega they are all based on this eddy viscosity model and this works quite well for a wide variety of flow and it is much cheaper than the Reynolds stress model because Reynolds stress model if you want to derive the dynamical equation you need at least 6 dynamical equation right. So, for each of the stress term that will be very expensive. So, let us see how we can find out this mu t. So, again finding out eddy viscosity is not easy and you cannot basically derive a equation from the first principle to find this quantity, but you can make some dimensional argument. For example, if you look at this mu t this is has got the same dimension as mu which is nothing but meter square per second right. Now, I am looking so we have some scales in our mind from the eddy point of view and we want to use this scales to basically find out the eddy viscosity. So, what are the scales that we have for example, the length scale of the eddy or may be the velocity scale of the eddy. So, those kind of things that we know if you see this. So, in order to find this viscosity meter square per second we need to have two things one is the meter per second another is meter and that will become dimensionally consistent. So, we need a velocity scale for the eddy because it has a unit of meter per second and we also need a length scale for the eddy. So, you can write this mu t in terms of two scales which is u 0 and l 0 right. So, what is l 0 it is basically it is a typical eddy size. Now, eddies can be of many different sizes right. So, if you for example, if you see flow in a channel. So, you have a very large eddy all covering all the way from the top to bottom where you can have medium size eddies and then you can have very small eddies which are responsible for the dissipation. Recall our earlier basically picture of the eddy and we had also talked about this picture of the energy content in the eddies. So, the large eddies take the eddy from the mean flow and basically the lowest size eddy basically is responsible for the dissipation. So, this is for the dissipation and here it takes the energy from mean flow and call it production. In between they have a cascade of eddies largest eddy will transport to the next size and next size and next size and so on and so forth. So, it is not easy to find one typical eddy size you have. So, many eddies present in your flow right. So, which eddy size should you take? So, that is one problem. Second is how do you find the turbulent velocity right. So, one approach could be like take the u tau as you if you recall in the boundary layer flow we have been using this scale which is very close to the. So, we had this tau w wall shear stress will be rho u tau square. So, you can define a scale of velocity which is tau w by rho right, but then this scale is valid only near the wall your turbulent flow need not be confined only close to the wall right. You can take the free stream velocity also, but that will also not give you the I mean that is the mean flow. So, that cannot be used for the turbulent velocity scale right. So, what is the best velocity scale you can think of again go to the dimensional argument. Let us say that somehow you know the turbulence kinetic energy k this is the turbulence kinetic energy which we have been writing as half u prime u prime w prime g prime plus w prime w prime right. So, if you know the turbulence kinetic energy which is unit is meter square per second square. So, take the square root of that a square root of k will become a velocity scale. In most of the turbulence model you use the velocity scale which is the square root of the turbulence kinetic energy again for that we need to find out what is the turbulence kinetic energy right. So, the models like k epsilon k omega or s h t k omega or any variant of this basically try to solve a dynamical equation for the turbulence kinetic energy. Then you need another equation because we need to find out the length scale also. So, the other equation will give you the length scale and that is how you find the eddy viscosity using these two scales right. So, before going to those advanced models let us look at what kind of models we have been we have seen in c f d and this is from very early days of development of RANS where we started with very empirical formulation basically no equation for turbulence kinetic energy or the length scale or anything, but use some empirical relation. The first is like you will have tried with constant eddy viscosity. So, just by the information that eddies contain energy and the viscosity is basically higher I mean the mixing is much much higher than the normal viscosity laminar viscosity. So, you can take the viscosity is like let us say 10 times of the laminar viscosity or 50 times of the laminar viscosity and see how the performance is. Second is you can use this l 0 geometrically that means that let us say that if you have a channel flow right. So, your eddy is confined it cannot be bigger than this right. So, you can use some fraction of that length based on the geometry you can use that as your length scale and then you can use this u 0 from the mean flow gradient. So, you know the mean flow gradient somehow you can use the mean flow gradient to get this u 0. So, basically you are not solving any additional equation only the four range equation that you have you are going to solve them and so and you get the value of u 0 and l 0 and you that way you can get the nu t right. So, this is called the 0 equation model. Then there are half equation models that means that instead of solving a p d. So, typically the models the equation that we get may be seeing the continuity equation mass conservation equation momentum conservation they all solve p d right. So, instead of solving a p d we can solve an o d right. So, it is easy to solve. So, this is kind of equation model half equation model one example is the johnson king model then we have one equation model. So, typically in these kind of models we specify the length scale geometrically. So, l 0 is basically given by the geometry by the size if this is a free shear flow may be the wake width or the jet width all those kind of width you can take boundary layer thickness those kind of things you can take for your length scale and then, but you need an equation for the u 0. So, this kind of models are called the one equation model and then we have two equation models. So, we have to basically get the equations for both the length scale as well as velocity scale. So, we are not using any geometrical parameter and the problem with the geometrical parameter is that if the flow is complex right. So, then this length scale is changing everywhere right let us say you have a configuration which is going like this something like this you are basically your width is changing everywhere the flow is changing everywhere if you use a geometric scale let us say you define it based on this length this may not work here this may not work here and so on and so forth right. So, geometrical characterization of the length scale is not suitable for complex flow right. So, let us look at some of these models. So, this is a big picture of the turbulence models. So, we are now into the eddy viscosity model, but as we told that we can another approach is the stress model is more expensive because we use six models for six different stresses in the eddy viscosity model the only thing that we need to find out is mu t, but then it depends on length scale and the velocity scale that you need to find out. So, first is basically the linear model. So, broadly speaking let us say that your stress is linearly dependent on your velocity gradient. So, then it is a linear model and if it is non-linear dependent that means that you have additional terms here instead of just not only mu t, but let us say mu t square or some other things then it becomes a non-linear model. So, this is to basically make the model more and more complex. So, that you can address more complex flow problems. So, we have the zero equation model for example, the Prandtl's machine life model we will discuss that in the next class. Then we have the Baldwin-Lomax model which is also the zero equation model. There are many more models I am not going to discuss here there is a C. B. K Smith model. Then there is a one equation model like a Spilett-Alameda's model is a one equation model very popular model for external dynamics flow. Let us say you want to simulate the flow over a wing you can use the Spilett-Alameda's model, but this model does not use a length scale only the velocity scale. So, this may not be suitable for flows which are very sensitive to the length scale for example, flow in turbine blades like in the thermo machine. Then we have the two equation models. So, broadly the most popular models are k epsilon and k omega and then there are further categories to that we have the standard k epsilon, realizable k epsilon, R N G k epsilon. So, once the k epsilon model has been derived there have been very many variation in the k epsilon over time and people basically try to fit the coefficients or the definitions of the transport equation especially for the turbulent dissipation rate which is epsilon here. So, you can see here k and epsilon, but ultimately what we are trying to find out is the L naught and U naught. We will use this parameter of kinetic energy and the dissipation rate to get this U naught and L naught. So, U naught is fine you can just take k to the power half and using the dimensional analysis we can also find the L naught from epsilon and k. So, these are different models which have been derived over time. Then Cuyamoga model, so is a famous ones are basically Wilcox k omega and then Mentor's SST k omega and then Wilcox also has many variations. So, if you go to the NASA turbulent site you can see different models and how they when they have been proposed. If you are going to use these models in your for your simulation make sure that you know that which model you are exactly using and what are the parameters that you are using. Then we have non-linear models, so this is of course will be more expensive, but this is suitable for some kind of non-equilibrium flows. So, for example Q B k epsilon is one model, E A R S M which is another model. Then you have V 2 omega model which is a relatively new model, but this is becoming very popular recently. So, these are the models apart from these you also have the transitional models which have not included in this. So, the point here is that these models are good for fully turbulent flows, but there are reasons for example if you look at the typical boundary layer profile the flow is laminar here and then turbulent here. This is laminar, this is turbulent, in between the flow is transitional. So, how do you find out the transition region and how to model this part. So, that is also a matter of research, this is very common in most of the flows sites where there will be some region which is transitional. The transition can be natural transition, the bypass transition, it can be separation induced transition. There will be separation in your flow that can also become transitional. So, then there are models which basically account for the transition regime also. So, there are two models which are most popular, one is the basically four equation k omega s s t model. In this apart from this two equation of k omega s s t you have two more equations for one is for intermittency gamma, another is the momentum thickness based nano small r e theta. The second is the laminar kinetic energy equation, laminar kinetic energy model l k model which has not only the turbulent kinetic energy k t, but also k l. You also solve for the laminar kinetic energy because as I have been saying that in the transition regime the flow is neither laminar nor turbulent, it is in between. Flow is trying to become turbulent, but if you keep the sufficient it may become, it may go back to laminar also. So, we will start with some very simple models like zero equation model and then we go to the two equation model and then we will look at the k epsilon k omega s s t k omega and that is how we will finish this discussion about the turbulence model. We will see you in the next class. Thank you very much. Keywords- Zero equation model, k epsilon k omega s t k omega, k epsilon k omega s t k