Literature review RANS derivation and analysis

REYNOLDS DECOMPOSITION:

The Reynolds averaged Navier-stokes equations(RANS equation) are time-averaged equations of motion for fluid flow. The idea behind the equation is Reynolds decomposition whereby an instantaneous quantity is decomposed into its time-averaged and fluctuating quantities. For example, for a quantity `u` the decomposition would be.

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

The objective of this review is to apply the decomposition process to Navier Stokes(NS) equations and derive an expression for Reynolds\'s stress. The RANS equation is primarily used to approximate the turbulent flow.

To apply the Reynolds decomposition to the Navier-stokes equation, we need to substitute the Reynolds decomposition method by introducing time-averaged values and fluctuating values into the basic Navier-stokes equation.

OBJECTIVE:

In this project, we have to apply Reynolds decomposition to Navier Stokes equations and come up with the expression for Reynold\'s stress and also explain the Reynolds stress, turbulent viscosity, molecular viscosity.

Navier-Stokes Equation:

• Continuity Equation : `(del(u))/(del(x)) + (del(v))/(del(y)) = 0`
• Momentum Equation: `rho((del(u))/(del(t)) + u*(del(u))/(del(x)) + v*(del(u))/(del(y))) = -(delrho)/(delx) + mu*(del^2(u))/(del(y^2))`

Reynolds averaged Navier-stokes Derivation:

According to reynolds decomposition

`u(x,y,z,t) = overline(u(x,y,z,t)) + u\'(x,y,z,t)`--------------------------------------------(1)

`u` = instantaneous quantity`

`overlineu =` Time averaged quantity

`u\'` = Fluctuating quantity

if we integrate equation (1) we get

`1/t_i int_o^(t_i)u(x,y,z,t)delt = 1/t_i int_o^(t_i)overlineu(x,y,z,t)delt + 1/t_i int_o^(t_i)u\'(x,y,z,t)delt`

But we should know that the time integration of fluctuation terms will always leads to zero.

so from above equation `1/t_i int_o^(t_i)u\'(x,y,z,t)delt` = 0

The application of Reynolds Decomposition in N-S equations

Continuity Equation

`(del(u))/(del(x)) + (del(u))/(del(y)) = 0`

On applying reynolds decomposition

`(del(overlineu + u\'))/(del(x)) + (del(overlinev + v\'))/(del(y)) = 0`

on applying time integrating and averaging

`1/t_iint_0^(t_i)((del(overlineu + u\'))/(del(x))*dt + (del(overlinev + v\'))/(del(y))*dt) = 0`

`1/t_i{int_0^(t_i)(deloverlineu)/(delx)*dt+int_0^(t_i)(delu\')/(delx)*dt + int_0^(t_i)(deloverlinev)/(dely)*dt + int_0^(t_i)(delv\')/(dely)*dt = 0`

as we discussed before the time integration of fluctuationg terms will be zero the final equation is

`1/t_i{int_0^(t_i)(deloverlineu)/(delx)*dt + int_0^(t_i)(deloverlinev)/(dely)*dt } = 0`

on doing integration

`1/t_i{(deloverlineu)/(delx)*t_i+ (deloverlinev)/(dely)*t_i } = 0`

`(deloverlineu)/(delx)+ (deloverlinev)/(dely) = 0`

From above equation it is clear that Reynolds decomposition satisfies Continuity condition.

Momentum Equation:

`rho((del(u))/(del(t)) + u*(del(u))/(del(x)) + v*(del(u))/(del(y))) = -(delrho)/(delx) + mu*(del^2(u))/(del(y^2))`

On multiplying `1/rho` on both sides

(del(u))/(del(t)) + u*(del(u))/(del(x)) + v*(del(u))/(del(y)) = -1/rho(delrho)/(delx) + mu/rho*(del^2(u))/(del(y^2))

Now we are adding the x-component of velocity multiplied by continuity equation on LHS side for simplification purpose. In the above derivation contuinuity equation is satisfiedbthat means results of these adding terms will be zero hence the balance of the momentim equation is not disturbed.

`(del(u))/(del(t)) + u*(del(u))/(del(x)) + v*(del(u))/(del(y)) + u((del(u))/(del(x)) + (del(v))/(del(y)) ) = -1/rho(delrho)/(delx) + mu/rho*(del^2(u))/(del(y^2))`

After doing the simplification on application of product rule the equation changes to

`(del(u))/(del(t)) + 2u*(del(u))/(del(x)) + (del(u*v))/(del(y)) = -1/rho(delrho)/(delx) + mu/rho*(del^2(u))/(del(y^2))`

The term `2u*(del(u))/(del(x)) = (del(u^2))/(delx)`

`(del(u))/(del(t)) + (del(u^2))/(del(x)) + (del(u*v))/(del(y)) = -1/rho(delrho)/(delx) + mu/rho*(del^2(u))/(del(y^2))`

On applying reynolds decomposition

`1/t_iint_0^(t^i){(del(overlineu + u\'))/(del(t))dt + (del(overlineu + u\')^2)/(del(x))dt + (del(((overlineu+u\')*(overlinev+v\')))/(del(y))dt)} = 1/t_iint_0^t_i{-1/rho(del((overlinerho+rho\'))/(delx)dt + mu/rho*(del^2(overlineu+u\'))/(del(y^2))dt)}`

On simplification above equation as we did in continuity equation we get

`(del(overlineu))/(del(t)) + (del(overlineu^2))/(del(x)) + (del(overlineu*overlinev))/(del(y)) = -1/rho(deloverlinerho)/(delx) + mu/rho*(del^2(overlineu))/(del(y^2))-1/t_iint_0^(t_i)(delu\'^2)/(delx)dt- 1/t_iint_0^(t_i)(del(u\'v\'))/(dely)dt`

The fluctuating terms in the right hand side is not equal to zero

`-1/t_iint_0^(t_i)(delu\'^2)/(delx)dt ne 0`

`1/t_iint_0^(t_i)(del(u\'v\'))/(dely)dt ne 0`

In general the average of the flutuating terms will be zero but when we consider the first terms the square of any value can\'t be negative so it cant be equal to zero. In the second term the multiplication of both component of velocities can\'t be predictable so we can\'t neglect that term. But when we are dealing with the terms along the Boundary layer the values along y-direction is more important than x-axis. Soo the first term deals with the square of the x- component velocity so it can be negligible because it shows a very minimum effect on the equation.

Therefore the final equation will be

`(del(overlineu))/(del(t)) + (del(overlineu^2))/(del(x)) + (del(overlineu*overlinev))/(del(y)) = -1/rho(deloverlinerho)/(delx) + mu/rho*(del^2(overlineu))/(del(y^2))- 1/t_iint_0^(t_i)(del(u\'v\'))/(dely)dt`

The above equation is rearranged to

`(del(overlineu))/(del(t)) + (del(overlineu^2))/(del(x)) + (del(overlineu*overlinev))/(del(y)) = -1/rho(deloverlinerho)/(delx) + 1/rhodel/(del(y)){mu*(del(overlineu))/(del(y))- rho/t_iint_0^(t_i)(u\'v\')dt}`

These are the inertial terms in the equation:

`(del(overlineu))/(del(t)) + (del(overlineu^2))/(del(x)) + (del(overlineu*overlinev))/(del(y))`

The pressure term: `-1/rho(deloverlinerho)/(delx)`

The molecular viscosity or viscous stress: `mu*(del(overlineu))/(del(y)`

The momentum diffusivity due to turbulence is also known as Reynold\'s stress: `rho/t_iint_0^(t_i)(u\'v\')dt`

Reynold\'s Stress: In the Fluid dynamics Reynold\'s stress is the component of the total stress tensor in a fluid obtained from the averaging operation over the N-S equation to account for turbulent fluctuations in fluid momentum.

Turbulent Viscosity (or) eddy viscosity: it is a coefficient that relates the average shear stress within a turbulent flow of water or air to the vertical gradient of velocity. The eddy viscosity depends on the fluid density and distance from the river bed or ground surface.

In a turbulent fluid, a linear interface between two different fluids breaks apart to form small-scale structures, also known as eddies. As these eddies grow and diminish in size (by energy cascades), they effectively alter the surface area of the interface between the fluids with different properties, thus altering the net transfer of momentum and scalar properties through the interfaces.

A computer performs a fluid dynamics simulation by numerical discretizing the Navier-Stokes equation over a domain that is split into grid boxes, as shown below.

(Computer simulation of the surface ocean, containing eddies)

The size of the grid boxes (also called resolution) determines the smallest scale of the eddies that the machine can resolve. In other words, if the grid size of your domain is 1 km, your machine won’t be able to capture the eddies smaller than 1 km. Which means that the net effective momentum transfer due to small eddies occurring within that box, is unknown to the machine. In order to estimate the net effective mixing at each of these grid boxes, we specify an effective viscosity and multiply it with the shear (or velocity gradient) of the fluid surrounding the grid box. This effective viscosity is called turbulent viscosity. The product of this turbulent viscosity and the shear gives us the net diffusive flux of fluid across the grid box. For momentum, that would be a diffusive momentum flux, and for temperature/density it would be a scalar flux.

How is it different from molecular viscosity?

• The Turbulent viscosity is an imaginary concept but molecular viscosity is a dynamic viscosity that will be present in all real fluids.
• The molecular velocity depends only on the properties of the fluid but the turbulent viscosity depends on the properties of the fluid, flow, adjacent boundary either fluid or solid.
• The Turbulent viscosity will only present in turbulent flow regions whereas molecular viscosity is present in all types of flows.