Literature review- RANS derivation and analysis

AIM:

To apply reynold's decomposition to the Navier-Stokes equations and come up with the expression for reynold's stress.

To explain briefly the meaning of reynold's stress.

To explain briefly about turbulent viscosity and its difference from molecular viscosity.

INTRODUCTION AND THEORY:

Navier-Stokes equations:

The Navier-Stokes are a set of differential equations that govern the motion of fluids. They represent the conservation of momentum.

The equations are given by:

Where,

u is the fluid velocity

p is the fluid pressure

ρ is the fluid density

μ is the fluid dynamic viscosity

The different terms correspond to the inertial forces (1), pressure forces (2), viscous forces (3), and the external forces applied to the fluid (4).

They are solved together with the continuity equation. Continuity equation represents the conservation of mass.

Continuity equation is given below:

`(∂rho)/(delt)` + `grad.(rhou)` = 0

and for incompressible fluids it is `grad.(rhou)` = 0.

If we solve the Navier-Stokes equations and the continuity equations, Numerically without making any approximations then we call this type of simulation as Direct Numerical Solution (DNS). But, solving these equations, particularly for turbulent flows with high reynolds numbers, using DNS takes a lot of time and effort and therefore is uneconomical and impractical especially for engineering applications. To make this economically viable, and still achieve the results close to the DNS solution, turbulence modelling has been developed.

One of the turbulence modelling approaches is Reynold's Averaged Navier- Stokes:

Reynold's Averaged Navier-Stokes equation:

RANS equations are time averaged equations of motion for fluid flow. These eqautions are obtained by decomposing the instantaneous quantities into time averaged and fluctuating quantities. As an example, the decomposition of a flow variable such as velocity, into mean and fluctuating term is shown below:

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

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

Where,

u(x,y,z,t) and v(x,y,z,t) are instantaeneous velocities at a point (x,y,z) in space and at a time t for a fluid flow and

`baru(x,y,z) and barv(x,y,z)` are the corresponding time averaged velocities upto a time which is much larger than the turbulent time scale.

`u^' (x,y,z,t) and v^'(x,y,z,t)` are the corresponding fluctuating velocities.

Time integral of the instantaneous velocities over the time interval from 0 to `t_("int")` are given by:

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

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

Now, let us time integrate the velocity decomposition equation for velocity u(x,y,z,t) over the time interval from 0 to `t_("int")`.

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

The time average of a time averaged quantity is the time averaged quantity itself. For example, if the velocity of a particle at 1st second is 1m/s and at 2nd second it is 2m/s and at 3rd second it is 3 m/s. So, the time average of the velocity will be `((1+2+3))/3 =2`m/s. But, if we again time average the time averaged velocity we get `((2+2+2))/3 =2`m/s, which is the time averaged quantity.

Therefore, the above velocity decomposition equation becomes

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

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

`rArr1/(t_"int").int_0^(t_"int")u^' (x,y,z,t)dt =0`

Thus, it can be concluded that the time average of a fluctuating quantity over a large time is zero.

REYNOLD'S DECOMPOSITION OF THE NAVIER- STOKES EQUATIONS FOR A BOUNDARY LAYER:

Let us decompose the governing equations for the boundary layer.

The governing equations for the boundary layer are as follows:

`(delu)/(dx) + (delv)/(dely) = 0`. This is the continuity equation.

`rho((delu)/(delt)+ u.(delu)/(delx) + v.(delu)/(dely)) = -(delp)/(delx) + mu.(del^2u)/(dely^2)`. This is the momentum equation in X-direction`

`rho_c((delT)/(delt)+ u.(delT)/(delx) + v.(delT)/(dely)) = K. (del^2T)/(delx^2) + mu.((delu)/(dely))^2`. This is the energy equation.

Reynold's decomposition for the continuity equation:

`(del(baru(x,y,z)+u^'(x,y,z,t)))/(dx) + (del(barv(x,y,z)+v^'(x,y,z,t)))/(dely) = 0`

By time integrating the above equation over the time interval 0 to`t_("int")` we get

`1/(t_("int")).int_0^(t_("int"))((del(baru(x,y,z)+u^'(x,y,z,t)))/(dx))dt + 1/(t_("int")).int_0^(t_("int"))((del(barv(x,y,z)+v^'(x,y,z,t)))/(dely))dt = 0`

`rArr1/(t_("int")).int_0^(t_("int"))(del(baru(x,y,z)))/(delx)dt+1/(t_("int")).int_0^(t_("int"))(del(u^'(x,y,z,t)))/(dx)dt + 1/(t_("int")).int_0^(t_("int"))(del(barv(x,y,z)))/(dely)dt+ 1/(t_("i nt")).int_0^(t_("int"))(del(v^'(x,y,z,t)))/(dy)dt = 0`

As the time averaged velocities have zero time averaged value, we get

`1/(t_("int")).int_0^(t_("int"))(del(baru(x,y,z)))/(delx)dt+cancel(1/(t_("int")).int_0^(t_("int"))(del(u^'(x,y,z,t)))/(dx)dt) + 1/(t_("int")).int_0^(t_("int"))(del(barv(x,y,z)))/(dely)dt+ cancel(1/(t_("int")).int_0^(t_("int"))(del(v^'(x,y,z,t)))/(dy)dt) = 0`

`rArr1/(t_("int")).int_0^(t_("int"))(del(baru(x,y,z)))/(delx)dt+1/(t_("int")).int_0^(t_("int"))(del(barv(x,y,z)))/(dely)dt = 0`

`rArr(del(baru(x,y,z)))/(delx)1/(t_("int")).int_0^(t_("int"))dt+(del(barv(x,y,z)))/(dely)1/(t_("int")).int_0^(t_("int"))dt = 0`

`rArr(del(baru(x,y,z)))/(delx)1/cancel(t_("int")).cancel((t_("int")-0))+(del(barv(x,y,z)))/(dely)1/cancel(t_("int")).cancel((t_("int")-0)) = 0`

`rArr(del(baru(x,y,z)))/(delx)+(del(barv(x,y,z)))/(dely) = 0`

Which is the continuity equation in terms of time averaged velocities.

Reynold's decomposition for the momentum equation:

For simplicity let us use the following variables with different notations without changing its meaning:

`u(x,y,z,t)=u, baru(x,y,z)= baru, u^'(x,y,z,t)=u^' , v(x,y,z,t)=v, barv(x,y,z)= barv, v^'(x,y,z,t)=v^', p(x,y,z,t)=p, barp(x,y,z)= barp, p^'(x,y,z,t)=p^'`

Rearranging the mommentum equation to a suitable form to make life easy:

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

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

writing `mu/rho = nu` we get,

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

adding `u.((delu)/(delx) + (delv)/(dely))` on LHS of the above equation. As the value of `u.((delu)/(delx) + (delv)/(dely))` is zero so the above equation will remain the same.

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

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

We know that `u.(delv)/(dely) +v.(delu)/(dely) = (del(u.v))/(dely)` and `2.u.(delu)/(delx)= (delu^2)/(delx)`

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

Now, Applying the Reynold's decomposition to the above simplified form of momentum equation.

`rArr(del(baru+u^'))/(delt)+(del(baru+(u^'))^2)/(delx) + ((del((baru+u^').(barv+v^')))/(dely)) = -1/rho.((del(barp+p^'))/(delx)) + nu.((del^2(baru+u^'))/(dely^2))`

By time integrating the above equation, we get

`rArr1/(t_"int").int_0^(t_"int")(del(baru+u^'))/(delt)dt+1/(t_"int").int_0^(t_"int")(del(baru+(u^'))^2)/(delx)dt + 1/(t_"int").int_0^(t_"int")((del((baru+u^').(barv+v^')))/(dely))dt = 1/(t_"int").int_0^(t_"int")(-1/rho.(del(barp+p^'))/(delx))dt +`

`1/(t_"int").int_0^(t_"int")(nu.((del^2(baru+u^'))/(dely^2)))dt`

`rArr1/(t_"int").int_0^(t_"int")(del(baru))/(delt)dt+cancel(1/(t_"int").int_0^(t_"int")(delu^')/(delt)dt)+1/(t_"int").int_0^(t_"int")(del(baru^2))/(delx)dt+1/(t_"int").int_0^(t_"int")(del(u^')^2)/(delx)dt+cancel(1/(t_"int").int_0^(t_"int")(del(2.baru.u^'))/(delx)dt )+1/(t_"int").int_0^(t_"int")(del(baru.barv))/(dely)dt +`

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

`1/(t_"int").int_0^(t_"int")(nu.((del^2(baru))/(dely^2)))dt+cancel(1/(t_"int").int_0^(t_"int")(nu.((del^2(u^'))/(dely^2)))dt)`

The value of above terms which have been cancelled is zero as we know that the time average of the fluctuating terms is zero.

`rArr1/(t_"int").int_0^(t_"int")(del(baru))/(delt)dt+1/(t_"int").int_0^(t_"int")(del(baru^2))/(delx)dt+1/(t_"int").int_0^(t_"int")(del(u^')^2)/(delx)dt+1/(t_"int").int_0^(t_"int")(del(baru.barv))/(dely)dt +1/(t_"int").int_0^(t_"int")(del(u^'.v^'))/(dely)dt = 1/(t_"int").int_0^(t_"int")(-1/rho.(delbarp)/(delx))dt +`

`1/(t_"int").int_0^(t_"int")(nu.((del^2(baru))/(dely^2)))dt`

`rArr1/cancel((t_"int")).(del(baru))/(delt).cancel((t_"int"-0))+1/cancel(t_"int").(del(baru^2))/(delx).cancel((t_"int"-0))+1/(t_"int").int_0^(t_"int")(del(u^')^2)/(delx)dt+1/cancel(t_"int").(del(baru.barv))/(dely).cancel((t_"int"-0)) +1/(t_"int").int_0^(t_"int")(del(u^'.v^'))/(dely)dt =`

`1/cancel(t_"int").(-1/rho.(delbarp)/(delx)).cancel((t_"int"-0)) +1/cancel(t_"int").(nu.((del^2(baru))/(dely^2))).cancel((t_"int"-0))`

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

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

Assuming in the boundary layer, the gradients of the velocity components in the x- direction are much smaller when compared to the velocity gradients in the y- direction.

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

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

Simplifying the terms on the RHS, we get

`rArrubrace((del(baru))/(delt)+(del(baru^2))/(delx)+(del(baru.barv))/(dely))_("inertia terms")= ubrace((-1/rho.(delbarp)/(delx)))_("pressure term")+1/rho.del/(dely)(ubrace((mu.(del(baru))/(dely)))_("viscosity term")-ubrace(rho/(t_"int").int_0^(t_"int")(u^'.v^')dt)_("momentum diffusivity due to turbulence "))`

The above equation is the reynold's decomposition of momentum equation at the boundary layer.

The Reynold's stress is given by,`1/(t_"int").int_0^(t_"int")(del(u^'.v^'))/(dely)dt`

The understanding of reynold's stress:

Reynolds stress is the component of the total stress in a fluid obtained by the reynold's decomposition of the Navier- Stokes equations. The reynold's stress at the boundary layer is given by:

Reynold's stress =`1/(t_"int").int_0^(t_"int")(del(u^'.v^'))/(dely)dt`

It can also be said that the mean forces per unit area imposed on the mean flow by turbulent fluctuations.

Thus, to solve the RANS equation is to mainly obtain the value of Reynold's stress. Many models have been developed to obtain th evalue of Reynold's stress. The most popular ones are RNG k-`epsi` and k-`omega`sst and they depend upon the requirement of the solution.

Understanding Turbulent viscosity and its difference from molecular viscosity:

In a laminar flow, the diffusion is performed only by molecular viscosity. But, if the flow becomes turbulent, the molecular viscosity is not enough to account for the total diffusion of the fluid motion. Thus, the term called turbulent viscosity is added to the molecular viscosity to account for the momentumm diffusion of the fluid flow in this case. Thus, the molecular viscosity accounts for the momentum diffusivity due to turbulence.

The turbulent viscosity inside a boundary layer is given by,

`rho/(t_"int").int_0^(t_"int")(u^'.v^')dt`

The major difference between the molecular viscosity and the turbulent viscosity is that the molecular viscosity is the resistance of an adjacent layer of a fluid when a fluid layer moves forward and this can be related to the shear stress applied but the turbulent viscosity is a virtual viscosity, and not a physical concept.