Literature review : RANS derivation and analysis

Aim: The aim here is to derive the reynold's stress expression by performing reynold decomposition on the NS equation.if we take any standard book on boundary layer theory we see that the following equations are valid. The below derivation is derived for 2 dimensional equation

equations that govern the boundary layer are

`(delu)/(delx) + (delv)/(dely) + (delw)/(delz) = 0` --------Continuity equation`

`rho((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely) + w(delu)/(delz))= - (delp)/(delx) + mu((del^2u)/(delx^2) + (del^2u)/(dely^2) + (del^2u)/(delz^2)) =>`Momentum equation along X-direction`

`rho_c((delT)/(delt) + u(delT)/(delx) + v(delT)/(dely)) = k(del^2T)/(delx^2) + mu((delu)/(dely))^2 => Energy equation`

all the above equations are valid inside the boundary layer, we see that some of the terms are missing because inside the boundary layer we can neglect some terms based upon the order of magnitude analysis. in this case we will have a final set of equation which you can simulate boundary layer on any type of geometry.

but so far we have used only the turbulence model. So whenever we simulate a shock flow problem or a conjugate problem we use a turbulent model.

If we solve the above 3 equation it is completely capable of giving solutions for laminar and turbulent problems. ny doing that we can say that we are performing the direct numerical analysis(DNS)

Why do we need turbulence model

here in the above image shown if we take a velocity probe and put it in the turbulent region to measure the velocity as a function of timethen we get a profile like this

here the velocity variation is continuously fluctuating and if we measure the frequency of fluctuation and inverse it we can calculate something called as turbulent time step and this time step is very small and it occurs for a small distance.

so if by taking the governing equation and integrating it iover that small time step and is we have a high number of computational cells then we don't need a turbulence model, but this approach is quite impractical for engineering problems simulations, and that is why we use the turbulence model so that we can still capture the effect of turbulence by using a courser grid and a larger time step

now we are going to take the governing equation and integrate it over atime which is much larger then he turbulent time scale , so this is called the averaging process.

then we apply the reynold's decomposition and convert this original set of equation into a form called as Reynold Average Navier Stokes equation(RANS)

we than use this equations along with the turbulence model to simulate the turbulent flow , in this process we end up with some unknown terms which have to be modelled.

the first step is to split the velocity component or any other flow variable into the mean component and a fluctuating component this is called the reynold decomposition

`u(x,y,z,t) = bar u(x,y,z) + u'(x,y,z,t)`

`v(x,y,z,t) = bar v(x,y,z) + v'(x,y,z,t)`

`ubrace(w(x,y,z,t))_("1st ") = ubrace(bar w(x,y,z))_("2nd") + ubrace(w'(x,y,z,t))_("3rd")`

1st is the instantenous velocity at any space and time

2nd is the mean component function of space and 3rd is the fluctuating component function of time and space

`baru(x,y) = 1/t_"int"int_0 ^(t_"int") u(x,y,t) dt`

`barv(x,y) = 1/t_"int"int_0 ^(t_"int") v(x,y,t) dt`

here after integrating the velocity component we get the averaged velocity which is independent of time, another interesting property is that when we integrating the fluctuations its always going to be zero
let us derive why this is the case

`u(x,y,z,t) = bar u(x,y,z) + u'(x,y,z,t)`

this is reynold decomposition and let us say that i time average this

`1/t_"int"int_0^(t_"int")u(x,y,z,t) dt =1/t_"int"int_0^(t_"int") bar u(x,y,z)dt + 1/t_"int"int_0^(t_"int")u'(x,y,z,t)dt`

now the 2nd term in this the averaged quantity now if we time average the averaged quantity we get an average quantity

so,

`baru(x,y,z) = baru(x,y,z) + 1/t_"int"int_0^(t_"int")u'(x,y,z,t) =0`

so here the 3rd term which is the fluctuating term will become zero

so the idea is when we integrate for a longer time step the fluctuation tends to become zero, and this rule can be applied to other quantities as well

now applying this integral rule to the continuity equation

`(delu)/(delx) + (delv)/(dely) = 0`

applying decomposition

`del/(delx)(baru +u') + del/(dely)(barv + v') =0`

integrating the above w.r.t.time

`1/t_"int"intdel/(delx)(baru +u')dt +1/t_"int"int del/(dely)(barv + v')dt =0`

`1/t_"int"int(delbaru)/(delx)dt + cancel(1/t_"int"int(delu')/(delx)dt) + 1/t_"int"int(delbarv)/(dely)dt + cancel(1/t_"int"int(delv')/(dely)dt) =0`

`(delbaru)/(delx)1/t_"int"intdt + (delbarv)/(dely)1/t_"int"intdt = 0`

now applying the limits to integration

`(delbaru)/(delx)1/t_"int"int_0^(t_"int")dt + (delbarv)/(dely)1/t_"int"int_0^(t_"int")dt = 0`

`(delbaru)/(delx)[[(t_"int")-0]]/t_"int" + (delbarv)/(dely)[[(t_"int")-0]]/t_"int" =0`

`(delbaru)/(delx) + (delbarv)/(dely) =0`

so the time averaged velocity component satisfy the continuity equation

`rho((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = -(delp)/(delx) + mu(del^2u)/(dely^2)`

`((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = 1/(rho)((-delp)/(delx)) + mu/(rho)(del^2u)/(dely^2)`

replacing `mu/rho`as it kinematic viscosity

`((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = 1/(rho)((-delp)/(delx)) + gamma(del^2u)/(dely^2)`

now on LHS we will add a term that is x component of velocity multiplied by the continuity equation

we are doing this to convert the above equation to a much more suitable form.

`((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) +ubrace( u((delu)/(delx) + (delv)/(dely))) = 1/(rho)((-delp)/(delx)) + gamma(del^2u)/(dely^2)`

the underlined term in the above equation is = 0 so it doesn't change anything, now rearranging the terms

`(delu)/(delt)+2u(delu)/(delx)+del/(dely)(uv)= -1/rho(delp)/(delx)+gamma(del^2u)/(dely^2)`

`(delu)/(delt)+del/(delx)(u^2)+del/(dely)(uv)=-1/rho(delp)/(delx)+gamma(del^2u)/(dely^2)`

now we are ready to apply the time integration and also the reynold decomposition

`1/t_"int"{int_0^(t_"int")del/(delt)(baru+u') + del/(delx)(baru+u')^2+del/(dely)[(baru+u')(barv+v')]} = 1/t_"int" (((-1))/rho)int_0^(t_"int")[(del(barp+p'))/(delx)+gamma(del^2)/(dely^2)(baru+u')]dt`

now expanding the equation

`1/t_"int"int_0^(t_"int")(delbaru)/(delt)dt+cancel((delu')/(delt)dt)+(delbaru^2)/(delx)dt+(delu'^2)/(delx)dt+cancel(del/(delx)2baru.u'dt)+del/(dely)(baru.barv)dt+cancel(del/(dely)(baru.v')dt)+cancel(del/(dely)(u'.barv)dt)+del/(dely)(u'v')dt = 1/t_"int"int(-1)/(rho)(delbarp)/(delx)dt-cancel(1/rho(delp')/(delx)dx)+gamma(del^2baru)/(dely^2)dt+cancel(gamma(del^2u')/(dely^2)dt)`

now cleaning the equation a bit we get

`1/t_"int"int_0^(t_"int"){(delbaru)/(delt)+(delbaru^2)/(delx)+del/(dely)(baru.barv)}dt = 1/t_"int"int_0^(t_"int"){-1/rho(delbarp)/(delx)+gamma^2(del^2u)/(dely^2)-del/(delx)(u'^2)-del/(dely)(u'v')}dt`

so here the initial terms are same as that of momentum equation the difference is that they were instantaneous quantities and here they are time averaged

now we will make an assumption in the boundary layer, the layer is very small especially when you compare it to the plate, the argument that we can make is the gradient in the velocity component are going to be much more stronger in the y direction compare to the x direction because because in that small space the values for the velocity component changes a lot anlog the y axis , but along the x axis it changes are not that much and hence we are going to neglect it

`-1/t_"int"int_0^(t_"int")del/(delx)(u'^2)dt=>` neglecting this in the RHS

this is true only inside a boundary layer

so the RHS of the equation becomes

`=1/t_"int"int_0^(t_"int"){-1/rho(delbarp)/(delx)+1/rho(del)/(dely)[mu(delbaru)/(dely)-(u'v')]}dt`

now here the term `mu(delbaru)/(dely) =>` is the molecular viscosity term

and `1/t_"int"int_0^(t_"int")(u'v')dt =>`is the momentum diffusity because of turbulence

now here we can say that the diffusion of momentum takes place in 2 ways 1 is through molecular viscosity and the other is through turbulence motion