Week 8: Literature review - RANS derivation and analysis

ASSIGNMENT NAME: - Literature Review - RANS Derivation and Analysis

Objective: - To apply Reynolds decomposition to the NS equation to get the RANS equation

What is turbulence?

In Fluid Dynamics, turbulence or turbulent flow is fluid motion characterized by chaotic changes in pressure and flow velocity. It is in contrast to a laminar flow, which occurs when a fluid flows in parallel layers, with no disruption between those layers.

Why we need turbulence modelling:

Most flows occurring in nature and in engineering applications are turbulent in nature. It is difficult to define turbulence. The time scale associated with turbulence is very small. We can solve turbulence flows using the Navier stroke equation but we need small time step and number of computational cells will be large in order capture turbulence physics. The process is complicated and indefinite time. And because of that we have turbulence model can capture turbulence physical with coarse grid and large time steps one such turbulence model called as Reynold average Navier stroke model

Turbulence modelling approach

• RANS (Reynold Average Navier strokes)
• LES (Large eddy simulation)

RANS (Reynold Average Navier strokes): -

The objective of the turbulence models for the RANS equations is to compute the Reynolds stresses, which can be done by three main categories of RANS-based turbulence models:

• Linear eddy viscosity models
• Nonlinear eddy viscosity models
• Reynolds stress model (RSM)

LES (Large eddy simulation): -

• In LES, the smallest scales of turbulence are spatially filtered out while the largest, most energy containing scales are resolved directly.
• LES modelling offers increased range of applicability and increased fidelity of the solution but all of this comes with an increased computational cost due to the time step requirements, as we can no longer consider the flow steady, and increased mesh resolution required to capture more details of the flow.

RANS Derivation

We have continuity equation

`∂u/∂x+∂v/∂y=0`

X- momentum equation:

`ρ(∂u/dt+u ∂u/∂x+v ∂v/∂y)=-∂p/∂x+μ (∂^2 u)/〖∂y〗^2`

Energy Equations

`ρc(∂T/dt+u ∂T/∂x+v ∂T/∂y)=-K (∂^2 T)/〖∂x〗^2 +μ (∂^2 u)/〖∂y〗^2`

Derivation Reynold averages Navier stroke Equation

The flow parameters are split into the mean and fluctuating components.

Example

The mean time average velocity is the function of space while fluctuating velocity is function of both time and space

Consider

`u ̅=mean time average components`

`u^'=fluctuating compoents`

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

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

Apply Reynold decomposition continuity equation

`∂/∂x (u ̅+u^' )+∂/∂y (v ̅+v^' )=0`

`1/t ∫〖(∂/∂x (u ̅+u^' )+∂/∂y (v ̅+v^' ))=0〗`

`1/t ∫〖(∂u ̅)/∂x dt+1/t ∫∂u'/∂x+1/t ∫(∂v ̅)/∂y dt+1/t ∫∂v'/∂t=0〗`

Here

`1/t ∫(∂u^')/∂x=0`

`1/t ∫∂v'/∂t=0`

`1/t ∫(∂u ̅)/∂x dt+1/t ∫(∂v ̅)/∂y dt=0`

`1/t (∂u ̅)/∂x ∫dt+1/t (∂v ̅)/∂y ∫dt=0`

`1/t (∂u ̅)/∂x ∫_0^tdt+1/t (∂v ̅)/∂y ∫_0^tdt`

After simplification we get

`(∂u ̅)/∂x+(∂v ̅)/∂y=0`

Momentum equation

`ρ(∂u/dt+u ∂u/∂x+v ∂v/∂y)=-∂p/∂x+μ (∂^2 u)/〖∂y〗^2`

Adding the product of X components of velocity and continuity equation to left side of equation

`∂u/dt+u ∂u/∂x+v ∂v/∂y+u(∂u/∂x+∂v/∂y)=-1/ρ ∂p/∂x+ϑ (∂^2 u)/〖∂y〗^2`

`∂u/dt+2u ∂u/∂x+u ∂u/∂x+v ∂v/∂y+u ∂v/∂y=-1/ρ ∂p/∂x+ϑ (∂^2 u)/〖∂y〗^2`

Applying product rule

`∂u/dt+2u ∂u/∂x+(∂(uv))/∂y=-1/ρ ∂p/∂x+ϑ (∂^2 u)/〖∂y〗^2`

`∂u/dt+(〖∂(u〗^2))/∂x+(∂(uv))/∂y=-1/ρ ∂p/∂x+ϑ (∂^2 u)/〖∂y〗^2`

Apply Reynold decomposition Momentum equation

`1/t ∫_0^t((∂(u ̅+u^'))/∂t+〖∂(u ̅+u^')〗^2/∂x+(∂((u ̅+u^' )(v ̅+v^' )))/∂y) dt=1/t ∫_0^t(-1/ρ (∂(p ̅+p^'))/∂x+ϑ (∂^2 (u ̅+u^'))/〖∂y〗^2 ) dt`

`1/t ∫_0^t((∂u ̅)/∂t dt+〖∂u〗^'/∂t dt+〖∂(u ̅)〗^2/∂x dt+〖∂(u^')〗^2/∂x dt+∂(u ̅u^' )/∂y dt+∂(u ̅v ̅ )/∂y dt+∂(u ̅v^' )/∂y dt+∂(v ̅u^' )/∂y dt+∂(u'v')/∂y dt) =1/t ∫_0^t(-1/ρ (∂p ̅)/∂x dt-1/ρ (∂p^')/∂x dt+ϑ (∂^2 u ̅)/〖∂y〗^2 dt+ϑ (∂^2 u^')/〖∂y〗^2 dt)`

Cut term is equal to zero

Rearranging the above question

`(∂u ̅)/∂t+〖∂(u ̅)〗^2/∂x+∂(u ̅v ̅ )/∂y=-1/ρ (∂p ̅)/∂x+ϑ (∂^2 u ̅)/〖∂y〗^2 -1/t ∫_0^t〖〖∂(u^')〗^2/∂x dt-1/t ∫_0^t〖∂(u'v')/∂y dt〗〗`

`(∂u ̅)/∂t+〖∂(u ̅)〗^2/∂x+∂(u ̅v ̅ )/∂y=-1/ρ (∂p ̅)/∂x-1/ρ ∂/dy ((μ∂u ̅)/dy-ρ/t ∫_0^t(u^' v^' )dt)`

Reynold average Navier strokes equation can be written as

So, the Reynold stress is

Reynold stress is,

Reynolds stress=

`=-ρ/t ∫_0^t(u^' v^' )dt`

`=-ρu^' v^'`

Stress is combination of laminar and turbulent stress.

So,

`T=T_laminar+T_turbulent`

What is turbulent viscosity? How is it different from molecular viscosity?

The coefficient that relates the average shear stress and rate of velocity Gradient in the turbulent flows is called turbulent viscosity. The viscosity in the turbulent flow is non homogenous and it varies in space.

The turbulent transfer of momentum by eddies creates internal fluid friction it is fundamental idea of how we define viscosity in turbulent flow

`T_(turbulent )= -ρ(u'v') ̅`

`T_(turbulent )= μ_t (∂u ̅)/dy`

Here,

`μ_t=turbulent viscosity`

Molecular viscosity is the same value as dynamic viscosity. Molecular viscosity is the transport of mass motion momentum by the random motion of individual molecules not moving together in coherent groups

Molecular viscosity is analogous in laminar flow to eddy viscosity in turbulent flow

The molecular viscosity is a macroscopic fluid property, you can think of it as what distinguish honey and water and the turbulent viscosity is a flow the increased momentum transport through the flow field due to turbulent velocity fluctuation

Molecular viscosity is a property of the fluid, the turbulent viscosity is the property of the flow

NOTE:PLEASE REFER WORD FILE ( Word file was accurate notation of all derviative equation)