## Literature review RANS derivation

OBJECTIVE:

Reynold's decomposition to the NS equations and come up with the expression for Reynold's stress.

• Explain your understanding of the terms Reynold's stress
• What is turbulent viscosity? How is it different from molecular viscosity?

The Reynolds-averaged Navier–Stokes equations (or RANS equations) are time-averaged equations of motion for fluid flow. The idea behind the equations is Reynolds decomposition, whereby an instantaneous quantity is decomposed into its time-averaged and fluctuating quantities. The RANS equations are primarily used to describe turbulent flows. These equations can be used with approximations based on knowledge of the properties of flow turbulence to give approximate time-averaged solutions to the Navier stokes equation.

For a stationary flow of an incompressible Newtonian fluid, these equations can be written in Einstein notation  as:

The left-hand side of this equation represents the change in mean momentum of fluid element owing to the unsteadiness in the mean flow and the convection by the mean flow. This change is balanced by the mean body force, the isotropic stress owing to the mean pressure field, the viscous stresses, and apparent stress owing to the fluctuating velocity field, generally referred to as the Reynolds stress. This nonlinear Reynolds stress term requires additional modelling to close the RANS equation for solving and has led to the creation of much different turbulence models. The time-average operator is a Reynolds operator.

The Reynolds operators are used here for the derivation

(∂u_i)/(∂x_i )=0

(∂u_i)/(∂x_i )+u_j (∂u_i)/(∂x_i )=f_i-1/ρ δρ/(δx_i )+v (δ^(2 ) u_i)/(δx_(i ) x_j )

u=velocity

u ̅=time average component of velocity

u ́=fluctuating speed component.

fi=vector representing external forces.

Resulting equation :

(∂u_i)/(∂x_i )=0 (∂u ̅)/∂t+u ̅_j (∂u ̅_i)/(∂u_j )+u ́_j (∂u ́_j)/(∂x_j )=f_i-1/ρ (δρ ̅)/(δx_i )+ v (δ^(2 ) u ̅_i)/(δx_(i ) x_j )

Momentum equation ;

(∂u ̅)/∂t +u ̅_j (∂u ̅_i)/(∂u_i )=f_i-1/ρ (δρ ̅)/(δx_i )+ v (δ^(2 ) u ̅_i)/(δx_(i ) x_j )+(∂u ̅_i ∂u ̅_j)/(∂x_i ∂x_j )

After further operating on equation

ρ (∂u ̅)/∂t+ρu ̅_j (∂u ̅_i)/(∂u_j )=ρf_i+(∂(-ρ δ_(ij )+2μS ̅_ij-ρ(u ́_i u ́_j)) ̅)/(∂x_j )

The time dependent components gets cancelled and equation reduces to :

ρu ̅_j (∂u ̅_i)/(∂u_j )=ρf_i+(∂(-p δ_(ij )+2μS ⃗_ij-ρ(u ́_i u ́_j)) ̅)/(∂x_j )

turbulent viscosity? How is it different from molecular viscosity?

The turbulent transfer of momentum by eddies giving rise to internal fluid friction, in a manner analogous to the action of molecular viscosity in laminar flow, but taking place on a much larger scale.

The value of the coefficient of eddy viscosity (an exchange coefficient) is of the order of 1 m2 s-1, or one hundred thousand times the molecular kinematic viscosity. Eddy viscosity is often represented by the symbol K, and the turbulence parameterization that uses eddy viscosity is called K-theory. In this theory, the eddy flux in kinematic units is related to the mean vertical gradient, such as in this example for the vertical flux of horizontal momentum:

where w is vertical velocity, U is the horizontal wind in the x-direction, the overbar represents an average, and the prime denotes the deviation or perturbation from an average. Eddy viscosity is a function of the flow, not of the fluid. It is greater for flows with more turbulence. The eddy viscosity or K-theory approach is a parameterization for the eddy momentum flux (Reynolds stress) that works reasonably well when only small eddies are present in the flow, but that behaves poorly when large-eddy coherent structures, such as thermals in the convective mixed layer, are present.

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