Week 8: Literature review - RANS derivation and analysis
Aim:
To derive the Reynolds Averaged Navier-Stokes Equations and provide a brief introduction on Reynolds stress terms and turbulent viscosity.
Objectives:
- Brief introduction incompressible Navier-Stokes equations (continuity and momentum equations)
- Explaning the Reynolds Decomposition
- Deriving the RANS equations and explaining the Reynolds Stress term
- Introducing to the turbulence closure problem and solution in the form of eddy viscosity model
Theory:
The Navier-Stokes equations are the representation of the conservation laws of mass, momentum and energy in the form of partial differential equations derived by applying Reynolds transport theorm on infintesimal control volume. For the current derivation, we will not include the energy equations for the sake of brevity and we will assume that the flow is incompressible for simplfying the derivations.
The vector form of the incompressible Navier-Stokes equations are as follows:
Continuity Equation: `grad . vec(u) = 0`
Momentum Equation: `{partial vec u}/{partial t} + vec u. grad vec u = - 1/{rho}grad p + nu grad^2 vec u`
All the terms in the above equations have their standard interpretations which can be found in any fluid dynamics textbook or online resources.
To get the averaged flow property, we use the Reynolds decomposition technique which will seperate the mean flow parameter from its fluctuations. An example of this mathematical technqiue is as follows
Velocity Decomposition: `vec u(mathbf X, t) = bar vec u(mathbf X)+ vec {u'}(mathbf X,t)`
Here,
- `mathbf X` is the vector quantity for representing the three dimensional space.
- `bar vec u(mathbf X)` denotes the ensamble averaged velocity component.
- `vec u' (mathbf X,t)`denotes the deviations form the mean velocity.
Similar, decomposition is also carried out for pressure and is not shown here for the sake of brevity.
Derivation for continuity equation:
Substituting the Reynolds Decompostion in the Continuity Equation we get the following result
`grad . (bar vec u + vec u' ) = 0`
Splitting the vector into its scalar components and representing the above eqaution in the form of parial derivatives, we get the following equation.
`{partial bar U}/{partial x} + {partial bar V}/{partial y} + {partial bar W}/{partial z} + {partial u'}/{partial x} + {partial v'}/{partial y} + {partial w'}/{partial z} = 0`
Taking the time average of the above equation we get,
`grad . bar vec u = 0`
Taking the difference of the Navier Stokes and the Time Averaged eqaution, we get
`grad . vec u'` = 0
Hence, from the Reynolds Decomposition and time averaging, we get two continuity equations. The first equation denotes the continuity of the mean flow velocity while the second equation denotes the continuity of the velocity fluctuations.
Derivation for momentum equation:
For this derivation, we will only look at the derivation of the X momentum equation for the sake of brevity.
The X momentum equation is as follows
`u_t + u u_x + v u_y + wu_z = -1/rhop_x + nu grad^2 u`
In the above and the following equations, we will represent the partial derivatives in the subscript, so that the equations are easier to write and understand. An example of this format is : `u_x = {partial u }/ {partial x}`
Substituting the Reynolds Decomposition in the X momentum equation we get,
`barU_t + u'_t + barU barU_x + bar Uu'_x + u' bar U_x + u' u'_x + bar V bar U_y + bar V u'_y + v' bar U_y + v' u'_y + bar W bar U_z + bar W u'_z + w' bar U_z + w' u'_z`
=
`-1/rho ( bar P_x + p'_x ) + nu (grad^2 bar U + grad^2 u')`
Since, the ensemble averaged velocity is only function of the spatial terms, `bar U_t = 0`. Taking the time average of the above equation and removing the average of the fluctuating terms we get the following equation
`bar U bar U_x + bar V bar U_y + bar W bar U_z + bar {u'u_x} +bar{ v'u_y }+ bar{w'u_z} = -1/rho bar P_x + nu grad^2 bar U`
Using the chain rule on the fluctuating convective terms we get,
`bar{u'u_x} = bar{(u' u')_x} - bar{u' u'_x}`
`bar{v'u_y} = bar{(v' u')_y} - bar{u' v'_y}`
`bar{w'u_z} = bar{(w' u')_z} - bar{u' w'_z}`
Substituting the above equations into the time averaged x momentum equation we get,
`bar U bar U_x + bar V bar U_x + bar W bar U_z + (bar{u'u'})_x + (bar{u'v'})_y + (bar{u'w'})_z - (bar{u'u'_x} + bar{u'v'_y} + bar{u'w_z}) = -1/rho barP_x + nu grad^2 bar U`
From the continuity equation, we know that `bar{u'u'_x} + bar{u'v'_y} + bar{u'w_z} = 0`.
Hence, the final form of the time averaged momentum equation is as follows
`bar U bar U_x + bar V bar U_x + bar W bar U_z = -1/rho barP_x + nu grad^2 bar U - (bar{u'u'})_x - (bar{u'v'})_y - (bar{u'w'})_z `
We can observe from the above equation that the mean flow properties in the momentum equation depend on the fluctuations of the flow properties.
The time average of the products of the fluctating velocity components is non-zero and they are termed as Reynolds stress and they are computationally expensive to resolve
This mathematical problem of having more unknowns than the number of equations is called as the turbulence closure problem.
Eddy viscosity model:
One of the traditional methods to solve this closure problem is to use a model for the Reynolds Stress terms. The core concept is to model the Reynolds stress terms in the form of the mean flow properties and consider its effects similar to that of shear stress generated by the viscous terms. The turbulent viscosity is dominant outside the viscous sublayer while the molecular viscosity dominates inside the viscous sublayer.
The turbulent viscosity model can be represented as follows
`bar{u'_iu'_j} = nu_t({partial bar U_i}/{partial x_j} + {partial bar U_j}/{partial x_i}) - 2/3 k delta_{ij}`
where,
`nu_t` is the kinematic turbulent viscosity
`k` is the turbulent kinetic energy
Conclusions:
We have successfully derived the time averaged continuity and momentum equation using the Reynolds Decomposition technqiue. The relationship between the Reynolds Stress terms and the turbulent viscosity is briefly discussed.