Week 8: Literature review RANS derivation and analysis

Aim:

To Apply reynold's decomposition to the NS equations to get reynold's stress.

To understand terms of Reynold's stress.

Overview of turbulent viscosity and its difference between molecular viscosity.

Theory

Reynolds average Navier-Stokes equations(RANS)

In order to describe the behaviour of the turbulent flow Osborne Reynolds has came up with the concept of time averaging and decomposing the equations of the motion of the fliud flow. This method of decomposition is called Reynolds decomposition. Upon time averaging and decomposing we get 6 additional unknows along with the 4 variable unknows. These 6 unknows appear due to time averaging of momentum equations and because of this unknows we approach closure problem of turbulence. In order to get rid of these unknows there are certain models which can be used to model turbulence.

Continuity Equation

`(∂u)/(∂x)+(∂v)/(∂y)=0`

Momentum Equation

`ρ ⋅ ( (∂ u)/ (∂ t) + u ⋅ (∂ u)/ (∂ x) + v ⋅ (∂ u )/(∂ y) ) = −( ∂ p)/ (∂ x) + μ ⋅ (∂ ^(2)u )/(∂ y)^ 2`

Energy Equation

`(ρ_c ) * ((∂ T)/( ∂ t) + u ⋅( ∂ T)/( ∂ x) + v ⋅( ∂ T)/( ∂ y) ) = K ⋅ (∂ ^(2) T)/( ∂ x )^(2)+ μ⋅((∂u)/(∂y))^2`

Reynolds decomposition

It is a process of separating the flow variables like velocity into mean component(time averaged) and a fluctuating component. Due to the fluctuating velocity component, the fluid layers experience additional turbulent shear stress, known as Reynolds stresses. There is momentum exchange due to convective transport by the eddies which cause the faster-moving fluid layers to be decelerated and slower-moving layers to be accelerated. And in presence of temperature or concentration gradients, the eddy motions will also generate turbulent heat or species concentration fluxes.

Solution:

In order to take the time average, the momentary value is decomposed into 2 parts, i.e, the mean value and the fluctuating value.

In fluid dynamics and turbulence theory, Reynolds decomposition is a mathematical technique used to separate the expectation value of a quantity from its fluctuations. Here we split the the velocity component into a mean component and a fluctuating component for all three directional components.

The momentary velocity component is u, the time-averaged value is named `bar u` and the fluctuating velocity as u'. Hence the decomposition can mathematically be written as:

`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 )`

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

This is called as Reynolds Decomposition

Where `u(x,y,z)` is the instantanous velocity in at point (x,y,z) and at time t.

`bar(u)(x,y,z)` is the time averaged velocity in x direction. It is time averaged for a time period which is greater than the turbulent time scale.

`u′(x,y,z)`is the fluctuating component of velocity in x direction

Time averaged Velocity in x and y direction

`bar(u) ( x , y ) = 1 /(t_(Int)) ⋅ int_0^(t_(Int))u ( x , y , t ) ⋅ d t`

`bar(v) ( x , y ) = 1 /(t_(Int)) ⋅ int_0^(t_(Int))v( x , y , t ) ⋅ d t`

Also the time averaged value of the fluctuating component is always zero.

`1 /(t_(Int)) ⋅ int_0^(t_(Int))u' ( x , y , t ) ⋅ d t = 0`

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

Integrating both sides we get

`1 /(t_(Int)) ⋅ int_0^(t_(Int))u ( x , y , t ) ⋅ d t = 1 /(t_(Int)) ⋅ int_0^(t_(Int))bar(u) ( x , y , t ) ⋅ d t + 1 /(t_(Int)) ⋅ int_0^(t_(Int))u' ( x , y , t ) ⋅ d t`

and time averaging of a time averaged quantity results in the same value. So

`1 /(t_(Int)) ⋅ int_0^(t_(Int))bar(u) ( x , y , t ) ⋅ d t = bar(u)`

`bar(u) (x,y,t)=bar(u)(x,y,t) + 1 /(t_(Int)) ⋅ int_0^(t_(Int))u' ( x , y , t ) ⋅ d t`

So

`1 /(t_(Int)) ⋅ int_0^(t_(Int))u' ( x , y , t ) ⋅ d t = 0`

Reynolds Averaged NS Equation Derivation

Applying the Reynolds decomposition to the NS continuity equation

`(∂u)/(∂x)+(∂v)/(∂y)=0`

`(∂)/(∂x)(bar(u)+u')+(∂)/(∂y)(bar(v) +v')=0`

Integrate the entire equation and do a time averaging

`1 /(t_(Int))* int ∂ /(∂ x) ⋅ [ bar(u) + u ' ] d t + 1 /(t_(Int))* int ∂ /(∂ x) ⋅ [ bar(v) + v ' ] d t = 0`

`1 /(t_(Int))* int (∂ bar(u) )/(∂ x) d t+ 1 /(t_(Int))* int (∂ u') /(∂ x) d t + 1 /(t_(Int))* int (∂ bar(v) )/(∂ x) d t+ 1 /(t_(Int))* int (∂ v') /(∂ x) d t = 0`

Time integral of a fluctuating component is zero. So

`1 /(t_(Int))* int (∂ u') /(∂ x) d t =0 and 1 /(t_(Int))* int (∂ v') /(∂ x) d t = 0`

And `bar(u)` and `bar(v)` are not a function of time. So taking them out of integral sign yields

`(∂bar(u))/(∂ x) ⋅ 1/(t_(Int))∫ d t + (∂bar(v))/(∂ x) ⋅ 1/(t_(Int))∫ d t = 0`

Applying the limits of integration

`(∂bar(u))/(∂ x) ⋅ 1/(t_(Int)) int_0^(t_(Int)) d t + (∂bar(v))/(∂ x) ⋅ 1/(t_(Int)) int_0^(t_(Int)) d t = 0`

`(∂bar(u))/(∂x)⋅ t_(Int−0]/t_(Int) + (∂bar(v))/(∂x)⋅ t_(Int−0]/t_(Int) = 0`

So we get

`(∂bar(u))/(∂x) + (∂bar(v))/(∂y)= 0`

Which is Continuity Equation in terms of Time averaged variable

Momentum Equation:

`ρ((∂u)/(∂t)+u⋅(∂u)/(∂x)+v⋅(∂u)/(∂y))=−(dP)/(dx)+μ⋅(∂^2u)/(∂y)^2`

Dividing entire equation by Density `rho` and we have `μ /ρ = ν`. So eqaution becomes

`(∂u)/(∂t) + u ⋅ ((∂u)/(∂x)) + v ⋅ (∂u)/(∂y) = − 1/ ρ ⋅( ∂P)/(∂x) + ν ⋅ (∂^2u)/(∂y)^2`

Now add product of x component of velocity and continuity equation to LHS

`(∂u)/(∂t) + u ⋅ ((∂u)/(∂x)) + v ⋅ (∂u)/(∂y) + u⋅((∂u)/(∂x)+(∂v)/(∂y)) = − 1/ ρ ⋅( ∂P)/(∂x) + ν ⋅ (∂^2u)/(∂y)^2`

Applying product rule

`(∂ ( u v )) /(∂ y) = u ⋅ (∂ v)/ (∂ y) + v ⋅ (∂ u)/( ∂ y)`

`(∂u)/(∂t) +2. u ⋅ ((∂u)/(∂x)) + (∂ ( u v )) /(∂ y) = − 1/ ρ ⋅( ∂P)/(∂x) + ν ⋅ (∂^2u)/(∂y)^2`

We have `2. u ⋅ ((∂u)/(∂x)) = (∂ ( u ^2 ))/ (∂ x)`

`(∂u)/(∂t) + (∂ ( u ^2 ))/ (∂ x) + (∂ ( u v )) /(∂ y) = − 1/ ρ ⋅( ∂P)/(∂x) + ν ⋅ (∂^2u)/(∂y)^2`

Applying time integration and Reynold's decomposition,

`1/T* int_0^T(∂/(∂t)(baru+u')dt) + (∂ / (∂ x)(baru+u')^2dt) + (∂ /(∂ y)[(baru+u')(barv+v')]dt) = 1/T int_0^T− 1/ ρ ⋅( ∂)/(∂x)(barp+p')dt + ν ⋅ (∂^2(baru+u'))/(∂y)^2dt`

The fluctuating terms are zero, hence, the equation reduces to,

`(∂baru)/(∂t) + (∂ ( baru ^2 ))/ (∂ x) + (∂ ( bar(u v) )) /(∂ y) = − 1/ ρ ⋅( ∂barP)/(∂x) + ν ⋅ (∂^2baru)/(∂y)^2 - 1/ T * int_0^T [ ∂/ (∂ y) ( u ' v ' ) + (∂ ^2u ' ) /(∂ x) ] ⋅ d t`

The above equation is similar to the original momentum equation except it it is in time averaged form and we have two additional terms on the RHS.

The additional time-integration terms in the RHS are not equal to zero, however, to simplify the model, certain important assumptions are to be applied

The velocity gradients in the y direction will be much more stronger than the gradients in the x direction beacuse of the boundary layer. So fluctuation in y direction will also be lower. Then we can ignore the below term

`-1/ T * int_0^T (∂ ^2u ' ) /(∂ x) d t`

Now simpliflying the RHS, it becomes

`− 1/ ρ ⋅( ∂barP)/(∂x) + 1 /ρ ∂/ (∂ y) [ μ ⋅ (∂baru) /(∂ y) + ρ/T int_0^T( u ' v ' ) d t ]`

So equation is

`(∂baru)/(∂t) + (∂ ( baru ^2 ))/ (∂ x) + (∂ ( bar(u v) )) /(∂ y) = − 1/ ρ ⋅( ∂barP)/(∂x) + 1 /ρ ∂/ (∂ y) [ μ ⋅ (∂baru) /(∂ y) + ρ/T int_0^T( u ' v ' ) d t ]`

and this is RANS Equation

In the above equation,

`(∂baru)/(∂t) + (∂ ( baru ^2 ))/ (∂ x) + (∂ ( bar(u v) )) /(∂ y)` are the inertial terms,

`− 1/ ρ ⋅( ∂barP)/(∂x)` is the pressure term,

`μ ⋅ (∂baru) /(∂ y)` is the molecular viscosity,

`ρ/T int_0^T( u ' v ' ) d t` is the momentum diffusivity because of turbulence,

`∂/ (∂ y) [μ ⋅ (∂baru) /(∂ y)]` is the viscous stress term,

and `∂/ (∂ y) [ρ/T int_0^T( u ' v ' ) d t]` is the Reynold's stress term.

What is Reynold's stress:

Reynolds stress is the component of the total stress tensor in a fluid obtained from the averaging operation over the Navier-stokes equation to account for turbulent fluctuations in fluid momentum.

Turbulent viscosity:

In a turbulent fluid, a linear interface between two different fluids breaks apart to form small-scale structures, also known as eddies, 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. This effective viscosity is called turbulent viscosity.

The coefficient that relates the average shear stress and rate of velocity gradient in the turbulent flow is called turbulent viscosity. The turbulent transfer of momentum by eddies give rise to internal fluid friction on a large scale. It is also known as eddy viscosity. These are enormous in nature and to capture these to extreme details is near impossible, if done computationally, it is very costly. Thus we devise a way to capture net effective mixing of eddies using viscosity models. Turbulent viscosity increases the shear rate and affects the velocity and heat transfer.

Molecular viscosity:

It is a fluid property and is the same as viscosity. Its coefficient is the same as dynamic viscosity. Molecular viscosity is the transport of mass motion momentum solely by random motions of individual molecules not moving together in coherent groups. Molecular viscosity is analogous in laminar flow to eddy viscosity in a turbulent flow. It can also be constructed like some sort of internal friction between two layers of fluid developed while flowing.

Turbulent viscosity is a turbulent transfer of momentum by eddies giving rise to internal fluid friction which is analogous to the action of molecular viscosity in laminar flow but takes place in much larger space.

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

Turbulent viscosity is also known as eddy viscosity. It is due to the velocity fluctuations in a turbulent flow at high Reynolds number. It depends on the flow, it is not a fluid property. However, molecular viscosity is a fluid property. Molecular viscosity is same as the dynamic viscosity. Molecular viscosity purely depends on the fluid type and the temperature whereas the Turbulent viscosity depends also on flow.