Week 8: Literature review - RANS derivation and analysis

AIM

* To derive Reynolds averaged navier stokes equation (RANS) by applying reynolds decomposition.

* To explain the reynolds stress term.

* To explain turbulent viscosity and how it is different from molecular viscosity.

INTRODUCTION

Reynolds averaged navier stokes equation is derived from the navier stokes equation forboundary layer, meaning the navier stokes equation used in the boundary layer region differes a bit from the navier stokes equation generally used. The NSE used in the boundary layerregion has some of the terms ignored. Continuity equation and momentum equation of navier stokes equation used in the boundary layer region is given below,

`(delu)/(delx) + (delv)/(dely) = 0` is the continuity equation for boundary layer region`

`rho*[(delu)/(delt) + u*(delu)/(delx) + v*(delu)/(dely)] = -(delp)/(delx) + mu*(del^2u)/(dely^2)` is the momentum equation for boundary layer region. Above mentioned these 2 equations are used to derive RANS`

Derivation of RANS mainly involves 2 steps.

1. To decompose the flow parameter into a average term and fluctuating term.

if we consider any flow parameter as f then applying reynolds decomosition means, `f = bar f + f prime`

where `f` is the flow parameter.

`barf` is the time average of flow parameter.

`fprime` is the fluctuating parameter.

2. To apply the decomposed flow parameter into the boundary layer NSE and derive the RANS.

so let us start with continuity equation which is given as, `(delu)/(delx) + (delv)/(dely) = 0`

in the above continuity equation the flow parameter to be decomposed are velocity in x and y dierction respectivley, i.e, u and v. Applying the decomposition we have,

`(del(bar u + u prime))/(delx) + (del(bar v + v prime))/(dely) = 0`

simplyfying, `(del(bar u))/(delx) + (del(u prime))/(delx) + (del(bar v))/(dely) + (del(v prime))/(dely) = 0`

applying the time averaging integral we have,

`1/t*int_0^t(del(bar u))/(delx) + 1/t*int_0^t (del(u prime))/(delx) + 1/t*int_0^t (del(bar v))/(dely) + 1/t*int_0^t (del(v prime))/(dely) = 0`

from the above the integral of fluctuating term is zero, this can be proved as shown below

suppose if we consider the velocity component along x-direction as the flow parameter, applying reynolds decomposition we have,

`u(x,y,z,t) = baru(x,y,z,) + uprime(x,y,z,t)`, applying the time averaged integral

`1/t*int_0^tu(x,y,z,t)*dt = 1/t*int_0^t bar u(x,y,z)*dt +1/tint_0^t uprime(x,y,z,t)*dt`

from the above `1/t int_0^tu(x,y,z,t)*dt = bar u(x,y,z)` also time averaged integral of `bar u(x,y,z)` is `bar u(x,y,z)`, hence the above equation can be written as

`bar u(x,y,z) = bar u(x,y,z) + 1/tint_0^tuprime(x,y,z,t)`

i.e, `1/t*int_0^t u prime(x,y,z,t) = 0`

so using the above we can express, `1/t*int_0^t(del(bar u))/(delx) + 1/t*int_0^t (del(u prime))/(delx) + 1/t*int_0^t (del(bar v))/(dely) + 1/t*int_0^t (del(v prime))/(dely) = 0` a`

`1/t*int_0^t(del(bar u))/(delx) + 1/t*int_0^t (del(bar v))/(dely) + = 0`

simplifying further, `(del(bar u))/(delx) * (t-0)/t + (del(barv))/(dely)*(t-0)/t = 0` and hence we have

`(del(bar u))/(delx) + (del(barv))/(delx) = 0`

* Now considering the momentum equation in the boundary layer region we have,

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

from this equation RANS is derived, so let us start by moving `rho` to R.H.S, we have

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

adding `u*((delu)/(delx) + (delv)/(dely))` on L.H.S we have,

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

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

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

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

To above simplified equation reynolds decomposition is applied and hence we have

`(del(bar u+ u prime))/(delt) + (del(baru + uprime)^2)/(delx) + (del((baru+uprime)*(barv+vprime)))/(dely) = -1/rho*(del(barp + pprime))/(delx) + nu*(del^2(baru + uprime))/(dely^2)`

applying time averaged integral and hence we have,

`1/t*int_0^t (del(bar u+ u prime))/(delt) + (del(baru + uprime)^2)/(delx) + (del((baru+uprime)*(barv+vprime)))/(dely) = 1/t*int_0^t-1/rho*(del(barp + pprime))/(delx) +1/t*int_0^t nu*(del^2(baru + uprime))/(dely^2)`

`1/t*int_0^t((del(baru))/(delt) + (del(uprime))/(delt) +(del(bar u)^2)/(delt) + (del(2*bar u*u prime))/(delt) + (del(uprime)^2)/(delt) + (del(bar u * bar v))/(dely) + (del(bar u*v prime))/(dely) + (del(u prime * bar v))/(dely) + (del(u prime*v prime))/(dely))*dt = 1/t*int_0^t-1/(rho)*(del(bar p))/(delx)*dt+ 1/t*int_0^t-1/(rho)*(del(p prime))/(delx)*dt + 1/t*int_0^tnu*(del^2(bar u))/(dely^2)*dt+1/t*int_0^tnu*(del^2(u prime))/(dely^2)*dt`

from the above equation the time averaged integral of fluctuating term becomes zero and hence the above equation can be written as,

`(del(bar u))/(delt) + (del(bar u)^2)/(delx) + (del(u prime)^2)/(delx) + (del(bar u*bar v))/(dely) + (del(u prime*v prime))/(dely) = -1/rho*(del(bar p))/(delx) + nu*(del^2(bar u))/(dely^2)`

above term can be further written as,

`(del(bar u))/(delt) + (del(bar u)^2)/(delx) + (del(bar u*bar v))/(dely) = -1/rho*(del(bar p))/(delx) + nu*(del^2(bar u))/(dely^2) - 1/t*int_0^t(del(uprime)^2)/(delx)-1/t*int_0^t(del(u prime*v prime))/(dely)`

from the above equation, `-1/t*int_0^t(del(u prime)^2)/(delx)` can be taken as zero since `u prime` which is the velocity fluctuation along x-direction inside the boundary layer is negligible.

but term `-1/t*int_0^t(del(u prime*vprime))/(dely)`cannot be approximated to zero, also this term has the unit N/m^2 which is the unit of stress and hence this term is called as reynold's stress.

* Now the reynolds stress is obtained the whole discussion squares down to finding a solution for the reynolds stress term. So the closure point comes into picture.

* There are certain models evolved which defines the closure equation. Spalart almaras model uses one equation to close the system, where as variants of k-ε and k-Ω models use 2 equations to close the problem.

* Main difference between turbulent viscosity and molecular viscosity is that, molecular viscosity is the property of the fluid whereas turbulent viscosity is the property of flow.

* Turbulent transfer of momentum by eddies giving rise to an internal fluid friction. It is greater for flows with more turbulence.

* Molecular viscosity also called as dynamic viscosity reffers to the internal friction between the layers of the fluid and hence causing resistance to the flow.