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Literature review RANS derivation and analysis

Objective: To apply Reynold\'s decomposition to the NS equations to get the RANS equations. 1. Understanding of the terms Reynold\'s stress 2. What is turbulent viscosity? How is it different…

GAURAV KHARWADE
updated on 13 Feb 2020

Objective: To apply Reynold\'s decomposition to the NS equations to get the RANS equations.
1. Understanding of the terms Reynold\'s stress
2. What is turbulent viscosity? How is it different from molecular viscosity?

Theory:

What is Turbulence?

$⋆$ $***$ Turbulence is looked at as Unsteady, irregular (aperiodic) motion in which transport quantities (mass, momentum, scalar species) fluctuated in space and time. It is in contrast to laminar flows, which occurs when a fluid flows in parallel layers, with no disruption between those layers.

$⋆$ $***$ Turbulence is caused by excessive kinetic energy in parts of a fluid flow, which overcomes the damping effect of the fluid\'s viscosity. For this reason, turbulence is commonly realized in low viscosity fluids.

$\"\"$

Properties associated with TURBULENCE:

1. Irregularity: Irregular and random in nature, requires statistical methods to resolve.
2. Diffusivity: Rapid mixing of heat, mass, momentum, etc. one of the most important properties.
3. Large Re: Turbulence always occurs at a high Re number. It often results from the instability of laminar flow as Re becomes too large.
4. 3D vorticity fluctuation: Rotational and 3D, characterized by a high level of fluctuating vorticity. Vorticity dynamics play a central role in the analysis.
5. Dissipation: Turbulence is always dissipative. Viscous shear stresses perform deformation work at the expense of kinetic energy of the flow
6. Continuum: Governed by equations of fluid mechanics
7. Flows: Turbulent flows are flows who characterized by properties of flows not fluid

Methods of Analysis:

Let\'s take a look at instantaneous velocity measurements in turbulent flows.

$\"\"$

$⋆$ $***$ Fluctuations can be of small scale and high frequency. They are computationally too expensive to simulate directly in practical engineering applications.
$⋆$ $***$ Instead, the instantaneous (exact) governing equations can be time-averaged, ensemble-averaged, or otherwise manipulated to remove the small scales, resulting in a modiﬁed set of equations that are computationally less expensive to solve. However, the modiﬁed equations contain additional unknown variables, and turbulence models are needed to determine these variables in terms of known quantities.

Turbulence Modeling:

As we know the time scale associated with turbulence is very very less if we have N-S equations with very small time steps and we have large numbers of computational cells then we don\'t need turbulence model. This process is quite impractical to attempt.
That is why we need to use Turbulence modeling where we can capture the effect of turbulence even at coarser grid size and larger time step.

Task of turbulence modeling:

We are trying to ﬁnd approximate simpliﬁed solutions for the Navier - Stokes equations in the manner that either describes turbulence in terms of mean properties or limits the spatial /temporal resolution requirements associated with the full model.

There are three modeling frameworks:

1. Direct Numerical Simulation (DNS)
2. Reynolds-Averaged Navier - Stokes Equations (RANS)
3. Large Eddy Simulation (LES)

Reynolds-Averaged Navier - Stokes Equations (RANS):

Two alternative methods can be employed to transform the Navier-Stokes equations in such a way that the small-scale turbulent ﬂuctuations do not have to be directly simulated: Reynolds averaging and ﬁltering.
Both methods introduce additional terms in the governing equations that need to be modeled in order to achieve “closure”. (Closure implies that there are a suﬃcient number of equations for all the unknowns.)

The rationale for Reynolds averaging is that we are not interested in the part of ﬂow solution that can be described as “turbulent ﬂuctuations” instead, it is the mean (velocity, pressure, lift, drag) that is of interest. Looking at turbulent ﬂow, it may best steady in the mean in spite of turbulent ﬂuctuations.

RANS equations represent transport equations for the mean flow quantities only, with all the scales of the turbulence being modeled. The approach of permitting a solution for the mean ﬂow variables greatly reduces the computational eﬀort.

Reynold\'s Decomposition:

It is standard practice to describe a turbulent velocity field as a superposition of the time-averaged velocity and the fluctuating component as follows:

$\"\"$

$In 3D flow,\$ $\"In 3D flow,\"$

$u (x, y, z, t) = \bar{u} (x, y, z) + u^{'} (x, y, z, t)$ $u(x,y,z,t)=baru (x,y,z)+u^\'(x,y,z,t)$ ___ Equation 1

$v (x, y, z, t) = \bar{v} (x, y, z) + v^{'} (x, y, z, t)$ $v(x,y,z,t)=barv (x,y,z)+v^\'(x,y,z,t)$ ___ Equation 2

$z (x, y, z, t) = \bar{z} (x, y, z) + z^{'} (x, y, z, t)$ $z(x,y,z,t)=barz (x,y,z)+z^\'(x,y,z,t)$ ___ Equation 3

Time- Averaged Quantities:

$\bar{u} (x, y) = \frac{1}{t_{int\}} \cdot \int_{0}^{t_{int\}} u (x, y, z) d t$ $baru(x,y)=1/(t_(\"int\"))*int_0^(t_\"int\")u(x,y,z)dt$

since we are integrating , We get average quantities and this average quantities doesnot change with time. Therefore, $\bar{u}$ $baru$ is f(x,y)

Equation--1 implies

$\frac{1}{t_{int\}} \cdot \int_{0}^{t_{int\}} u (x, y, z, t) d t = \frac{1}{t_{int\}} \cdot \int_{0}^{t_{int\}} \bar{u} (x, y, z) d t + \frac{1}{t_{int\}} \cdot \int_{0}^{t_{int\}} u' (x, y, z, t) d t$ $1/(t_(\"int\"))*int_0^(t_\"int\")u(x,y,z,t)dt=1/(t_(\"int\"))*int_0^(t_\"int\")baru(x,y,z)dt+1/(t_(\"int\"))*int_0^(t_\"int\")u\'(x,y,z,t)dt$

where, $\bar{u} is not a function of time\$ $baru\" is not a function of time\"$

hence, when we integrate fluctuating component it\'s always be ZERO.

$\"\"$

Therefore, when we integrate fluctuating terms for longer time period this term would be ZERO.

CONTINUITY EQUATION:

$\frac{d u}{d x} = 0$ $(du)/dx=0$

$Substitute, \ u = \bar{u} + u'$ $\"Substitute, \" u= baru+u\'$

$\frac{d (\bar{u} + u')}{d x} = 0$ $(d( baru+u\'))/(dx)=0$

Take time average of whole equation,

$\frac{1}{t_{int\}} \int_{0}^{t_{int\}} [\frac{d}{d x} (\bar{u} + u') d t] = 0$ $1/t_(\"int\")int_0^(t_(\"int\"))[d/dx(baru+u\')dt]=0$

$\frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d}{d x} (\bar{u}) d t + \frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d}{d x} (u') d t = 0$ $1/t_(\"int\")int_0^(t_(\"int\"))d/dx(baru)dt+1/t_(\"int\")int_0^(t_(\"int\"))d/dx(u\')dt=0$

Since, time integrating of fluctuating component is Zero and also, integral limits is not a function of X.

$\frac{d (\bar{u})}{d x} \frac{1}{t_{int\}} \int_{0}^{t_{int\}} d t = 0$ $(d(baru))/dx1/t_(\"int\")int_0^(t_(\"int\"))dt=0$

$\frac{d (\bar{u})}{d x} (\frac{t_{int\} - 0}{t_{int\}}) = 0$ $(d(baru))/dx([t_(\"int\")-0]/t_(\"int\"))=0$

$\frac{d (\bar{u})}{d x} = 0$ $(d(baru))/dx=0$

$\frac{d (u')}{d x} = 0$ $(d(u\'))/dx=0$

Time averaged flow filed and unsteady velocity fluctuations field obeys continuity.

CONSERVATION OF MOMENTUM:

Conservation form of N-S equation is,

$ρ \cdot (\frac{d u}{d t} + u \frac{d u}{d x} + v \frac{d u}{d y}) = - \frac{d P}{d x} + μ \cdot \frac{d^{2} u}{{d y}^{2}}$ $rho*((du)/dt+u(du)/dx+v(du)/dy)=-(dP)/dx+mu*(d^2u)/dy^2$

$where\, ν = \frac{μ}{ρ}$ $\"where\", nu=mu/rho$

Equation will be rewrite as,

$(\frac{d u}{d t} + u \frac{d u}{d x} + v \frac{d u}{d y}) = - (\frac{1}{ρ}) \frac{d P}{d x} + ν \cdot \frac{d^{2} u}{{d y}^{2}}$ $((du)/dt+u(du)/dx+v(du)/dy)=-(1/rho)(dP)/dx+nu*(d^2u)/dy^2$

$From Product rule, \ \frac{d}{d y} (u \cdot v) = u \frac{d u}{d y} + v \frac{d u}{d y}$ $\"From Product rule, \" d/dy(u*v)=u(du)/dy+v(du)/dy$

$\frac{d u}{d t} + 2 u \cdot \frac{d u}{d x} + \frac{d}{d y} (u \cdot v) = - (\frac{1}{ρ}) \frac{d P}{d x} + ν \cdot \frac{d^{2} u}{{d y}^{2}}$ $(du)/dt+2u*(du)/dx+d/dy(u*v)=-(1/rho)(dP)/dx+nu*(d^2u)/dy^2$

$\frac{d u}{d t} + \frac{d u^{2}}{d x} + \frac{d}{d y} (u \cdot v) = - (\frac{1}{ρ}) \frac{d P}{d x} + ν \cdot \frac{d^{2} u}{{d y}^{2}}$ $(du)/dt+(du^2)/dx+d/dy(u*v)=-(1/rho)(dP)/dx+nu*(d^2u)/dy^2$

$Put\, u = \bar{u} + u' & \ v = \bar{v} + v'$ $\"Put\", u=baru+u\'\" & \"v=barv+v\'$

Applying Reynold\'s Decomposition

$\frac{1}{t_{int\}} \int_{0}^{t_{int\}} [\frac{d (\bar{u} + u')}{d t} + \frac{d {(\bar{u} + u')}^{2}}{d x} + \frac{d}{d y} ((\bar{u} + u') \cdot (\bar{v} + v'))] \cdot d t = \frac{1}{t_{int\}} \int_{0}^{t_{int\}} [- (\frac{1}{ρ}) \frac{d (\bar{P} + P')}{d x} + ν \cdot \frac{d^{2} (\bar{u} + u')}{{d y}^{2}}] \cdot d t$ $1/t_(\"int\")int_0^(t_(\"int\"))[(d(baru+u\'))/dt+(d(baru+u\')^2)/dx+d/dy((baru+u\')*(barv+v\'))]*dt=1/t_(\"int\")int_0^(t_(\"int\"))[-(1/rho)(d(barP+P\'))/dx+nu*(d^2(baru+u\'))/dy^2]*dt$

Since, time averaged of fluctuation trms is zero. Equation reduced to__

$\frac{d u}{d t} + \frac{d {(\bar{u})}^{2}}{d x} + \frac{d {(u')}^{2}}{d x} + \frac{d}{d y} (\bar{u} \cdot \bar{v}) + \frac{d}{d y} (u' \cdot v') = - (\frac{1}{ρ}) \frac{d \bar{P}}{d x} + ν \cdot \frac{d^{2} \bar{u}}{{d y}^{2}}$ $(du)/dt+(d(baru)^2)/dx+(d(u\')^2)/dx+d/dy(baru*barv)+d/dy(u\'*v\')=-(1/rho)(dbarP)/dx+nu*(d^2baru)/dy^2$

$\frac{d u}{d t} + \frac{d {(\bar{u})}^{2}}{d x} + \frac{d}{d y} (\bar{u} \cdot \bar{v}) = - (\frac{1}{ρ}) \frac{d \bar{P}}{d x} + ν \cdot \frac{d^{2} \bar{u}}{{d y}^{2}} - \frac{d {(u')}^{2}}{d x} - \frac{d}{d y} (u' \cdot v')$ $(du)/dt+(d(baru)^2)/dx+d/dy(baru*barv)=-(1/rho)(dbarP)/dx+nu*(d^2baru)/dy^2-(d(u\')^2)/dx-d/dy(u\'*v\')$

We can write above equation as,

$\frac{d u}{d t} + \frac{d {(\bar{u})}^{2}}{d x} + \frac{d}{d y} (\bar{u} \cdot \bar{v}) = - (\frac{1}{ρ}) \frac{d \bar{P}}{d x} + ν \cdot \frac{d^{2} \bar{u}}{{d y}^{2}} - \frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d {(u')}^{2}}{d x} - \frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d}{d y} (u' \cdot v') \cdot d t$ $(du)/dt+(d(baru)^2)/dx+d/dy(baru*barv)=-(1/rho)(dbarP)/dx+nu*(d^2baru)/dy^2-1/t_(\"int\")int_0^(t_(\"int\"))(d(u\')^2)/dx-1/t_(\"int\")int_0^(t_(\"int\"))d/dy(u\'*v\')*dt$

Assume changes along X- direction is negligible compare to changes along Y-diredtion inside Boundary Layer, Hence

$\frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d {(u')}^{2}}{d x} \cdot d t = 0$ $1/t_(\"int\")int_0^(t_(\"int\"))(d(u\')^2)/dx*dt=0$

Equation becomes:

$\frac{d u}{d t} + \frac{d {(\bar{u})}^{2}}{d x} + \frac{d}{d y} (\bar{u} \cdot \bar{v}) = - (\frac{1}{ρ}) \frac{d \bar{P}}{d x} + ν \cdot \frac{d^{2} \bar{u}}{{d y}^{2}} - \frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d}{d y} (u' \cdot v') \cdot d t$ $(du)/dt+(d(baru)^2)/dx+d/dy(baru*barv)=-(1/rho)(dbarP)/dx+nu*(d^2baru)/dy^2-1/t_(\"int\")int_0^(t_(\"int\"))d/dy(u\'*v\')*dt$

Considering RHS of above equation:

$RHS\ = - (\frac{1}{ρ}) \frac{d \bar{P}}{d x} + ν \cdot \frac{d^{2} \bar{u}}{{d y}^{2}} - \frac{1}{t_{int\}} \int_{0}^{t_{int\}} \frac{d}{d y} (u' \cdot v') \cdot d t$ $\"RHS\"=-(1/rho)(dbarP)/dx+nu*(d^2baru)/dy^2-1/t_(\"int\")int_0^(t_(\"int\"))d/dy(u\'*v\')*dt$

$RHS\ = - (\frac{1}{ρ}) \frac{d \bar{P}}{d x} \cdot \frac{1}{ρ} \cdot \frac{d}{d y} [μ \cdot \frac{d \bar{u}}{d y} - \frac{ρ}{t_{int\}} \int_{0}^{t_{int\}} (u' \cdot v') \cdot d t]$ $\"RHS\"=-(1/rho)(dbarP)/dx*1/rho*(d)/dy[mu*(dbaru)/dy-rho/t_(\"int\")int_0^(t_(\"int\"))(u\'*v\')*dt]$

$\frac{1}{ρ} \frac{d \bar{P}}{d x} \to Pressure Term\$ $1/rho(dbarP)/dx->\"Pressure Term\"$

$μ \cdot \frac{d \bar{u}}{d y} \to Molecular Viscocity\$ $mu*(dbaru)/dy->\"Molecular Viscocity\"$

$\frac{ρ}{t_{int\}} \int_{0}^{t_{int\}} (u' \cdot v') \cdot d t \to Momentun Diffusivity\$ $rho/t_(\"int\")int_0^(t_(\"int\"))(u\'*v\')*dt->\"Momentun Diffusivity\"$

Reynold\'s Stress:

It is results from turbulent transport of momentum. The momentum flux is equivalently a stress as it causes defromation of fluid parcels. It is not beacause of existence of real tangential force (as in viscous stress). It is property of flow.

We consider a homogeneous fluid, whose density ρ is taken to be a constant.
For such a fluid, the components τ\'_ij of the Reynolds stress tensor are defined as:

$τ_{i, j} = - ρ \cdot \bar{u' \cdot v'}$ $tau_(i,j)=-rho*bar(u\'*v\')$

where, $ρ \cdot \bar{u' \cdot v'} is Average advection of u\' & v\'\$ $rho*bar(u\'*v\') \" is Average advection of u\' & v\'\"$

$⋆$ $***$ Advection of momentum or velocity fluctuation results in a spread of momentum by turbulence. This term describes the diffusive nature of turbulence.
The negative divergence of the Reynolds stresses represents convective momentum transfer due to the random motion of macroscopic fluid packets (eddies)

Reynold\'s Stress Tensor:
To describe Reynold\'s stress

$τ_{i, j} = - ρ \cdot [\begin{matrix} \bar{u'^{2}} & \bar{u' v}' & \bar{u' w'} \\ \bar{v' u'} & \bar{v'^{2}} & \bar{v' w'} \\ \bar{w' u'} & \bar{w' v'} & \bar{w'^{2}} \end{matrix}]$ $tau_(i,j)=-rho*[[bar(u\'^2),bar(u\'v)\',bar(u\'w\')],[bar(v\'u\'),bar(v\'^2),bar(v\'w\')],[bar(w\'u\'),bar(w\'v\'),bar(w\'^2)]]$

here, $τ_{i, j}$ $tau_(i,j)$ - independent of viscosity
- depends on the turbulent flow field
- results in the diffusion of momentum.

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

Turbulent viscosity:

The coefficient that relates the average shear stress and rate of velocity gradient in the turbulent flow is called Turbulent Viscocity. The viscosity in turbulence is non-homogeneous and it varies in space.

The relation between Turbulent stress $\"\"$ and turbulent viscosity $\"\"$ is

$\"\"$

$⋆$ $***$ 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.

In general, the shearing stress of the turbulent flow may be written as in the following form:

$\"\"$

The first term of Equation is the result of the viscous effect and the second term is the
result of the turbulence effect. If the flow is laminar, u′ and v′ velocity fluctuations will be zero.