Week 8: Literature review - RANS derivation and analysis

RANS MODELLING:

The RANS approach decomposes flow variables (e.g. velocity) into mean and fluctuating terms.

       $u_i={\overline{u}}_i+u_i\text{'}$

i.e, instantaneous velocity= mean velocity(we are solving for) + fluctuating velocity( we are going for modelling).

 Now, inserting the above equation into the NS equation and averaging the result yields the RANS equation.

Governing equation for the boundary layer:

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

        continuity equation

 $\rho (\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial }y)=-\frac{\partial p}{\partial x}+\mu (\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial x^2})$

        momentum equation

  If we take origina; N.S equation and solve it over small time scale and if we have large number of computational cell, then

we do not need turbulence model.

So, wit the help of turbulence model we can capture the effect of turbulence with coarser grid and larger time step.

Hence,we are taking the governing equation and integrating over a time that is much larger than turbulent time scale ,this is

called averaging.

now , Reynolds decomposition follows:

                     $u(x,y,z,t)=\overline{u}(x,y,z)+u\text{'}(x,y,z,t)$

similarly for V and W velocity components.

 lets apply decomposition to continuity equation:

  $\frac{\partial }{\partial x}(\overline{u}+u\text{'})+\frac{\partial }{\partial y}(\overline{v}+v\text{'})=0$

 $\frac{1}{t_i}{\int }_0^{t_i}(\frac{\partial }{\partial x}(\overline{u}+u\text{'})dt+\frac{1}{t_i}{\int }_0^{t_i}\frac{\partial }{\partial y}(\overline{v}+v\text{'})dt=0$

 $\frac{1}{t_i}\int \frac{\partial }{\partial x}\overline{u}dt+\frac{1}{t_i}\int \frac{\partial }{\partial x}u\text{'}dt+\frac{1}{t_i}\int \frac{\partial }{\partial y}\overline{v}dt+\frac{1}{t_i}\int \frac{\partial }{\partial y}v\text{'}dt=0$

 $\frac{1}{t_i}\int \frac{\partial }{\partial x}\overline{u}dt+\frac{1}{t_i}\int \frac{\partial }{\partial y}\overline{v}dt=0$

$\frac{\partial }{\partial x}\overline{u}\cdot \left(\frac{1}{t_i}\right){\int }_0^{t_i}dt+\frac{\partial }{\partial y}\overline{v}\cdot \left(\frac{1}{t_i}\right){\int }_0^{t_i}dt=0$

$\frac{\partial }{\partial x}\overline{u}\left(\frac{t_i-0}{t_i}\right)+\frac{\partial }{\partial y}\overline{v}\left(\frac{t_i-0}{t_i}\right)=0$

$\frac{\partial }{\partial x}\overline{u}+\frac{\partial }{\partial y}\overline{v}=0$

  time averaging velocity component satisfies the continuity equation.

 Now momentum equation:

$\rho \{\frac{\partial u}{\partial t}+u\cdot \frac{\partial u}{\partial x}+v\cdot \frac{\partial u}{\partial y}\}=-\frac{\partial P}{\partial x}+\mu \cdot \frac{{\partial }^{2}u}{\partial y^2}$

$\frac{\partial u}{\partial t}+u\cdot \frac{\partial u}{\partial x}+v\cdot \frac{\partial u}{\partial y}=\frac{-\partial p}{\partial x}+\left(\frac{\mu }{\rho }\right)\cdot \left(\frac{{\partial }^{2}u}{\partial y^2}\right)$

$\frac{\partial u}{\partial t}+u\cdot \frac{\partial u}{\partial x}+v\cdot \frac{\partial u}{\partial y}+u(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y})=\left(\frac{1}{\rho }\right)\frac{-\partial P}{\partial x}+\nu \frac{{\partial }^{2}u}{\partial y^2}$

                              In this equation x component of velocity is multiplied by continuity equation.

Now rearranging the terms we get,

 $\frac{\partial u}{\partial t}+u\cdot \frac{\partial u}{\partial x}+v\cdot \frac{\partial u}{\partial y}+u\cdot \frac{\partial u}{\partial x}+u\cdot \frac{\partial v}{\partial y}=-\left(\frac{1}{\rho }\right)\frac{\partial P}{\partial x}+\nu \cdot \frac{{\partial }^{2}u}{\partial y^2}$

$\frac{\partial u}{\partial t}+\left(\frac{\partial u^2}{\partial x}\right)+\frac{\partial }{\partial y}(u\cdot v)=-\left(\frac{1}{\rho }\right)\cdot \frac{\partial p}{\partial x}+\nu \frac{{\partial }^{2}u}{\partial y^2}$

 Now Reynolds decomposition follows:

$u=\overline{u}+u\text{'}$

 $\frac{1}{t_i}\cdot {\int }_0^{t_i}\left(\frac{\partial }{\partial t}(\overline{u}+u\text{'})dt\right)+\frac{\partial }{\partial y}(\overline{u}+u\text{'})(\overline{v}+v\text{'})dt+\frac{\partial }{\partial x}{(\overline{u}+u\text{'})}^{2}dt=\frac{1}{t_i}{\int }_0^{t_i}\{-\frac{\partial }{\rho \partial x}(\overline{P}+P\text{'})dt\}+\nu \frac{{\partial }^2}{\partial y^2}(\overline{u}+u\text{'})dt$

 $\frac{1}{t_i}({\int }_0^{t_i}\frac{\partial \overline{u}}{\partial t}dt+\frac{\partial u\text{'}}{\partial t}dt+\frac{\partial {\overline{u}}^2}{\partial x}dt+\frac{\partial u{\text{'}}^2}{\partial x}dt+(\partial \left(\frac{2\cdot \overline{u}\cdot u\text{'}}{\partial x}\right)dt+\frac{\partial }{\partial y}\left(\overline{u}\overline{v}\right)dt+\frac{\partial }{\partial y}(\overline{u}v\text{'})dt+\frac{\partial }{\partial y}(u\text{'}\overline{v})dt+\frac{\partial }{\partial y}(u\text{'}v\text{'})dt=\frac{1}{t_i}{\int }_0^{t_i}(-\frac{1}{\rho }\cdot \frac{\partial \overline{P}}{\partial x}dt-\left(\frac{1}{\rho }\right)\cdot \frac{\partial P\text{'}}{\partial x}dt$

+$\nu \frac{{\partial }^2\overline{u}}{\partial y^2}dt+\nu \frac{{\partial }^{2}u\text{'}}{\partial y^{2}dt}$

$\frac{\partial \overline{u}}{\partial t}+\frac{\partial {\overline{u}}^2}{\partial x}+\frac{\partial }{\partial y}(\overline{u}\cdot \overline{v})=-\left(\frac{1}{\rho }\right)\cdot \frac{\partial \overline{P}}{\partial x}+\nu \cdot \frac{{\partial }^2\overline{u}}{\partial y^2}-\left(\frac{1}{t_i}\right){\int }_0^{t_i}\frac{\partial }{\partial x}\left(u{\text{'}}^{2}dt\right)-\left(\frac{1}{t_i}\right){\int }_0^{t_i}\frac{\partial }{\partial y}(u\text{'}v\text{'})dt$

since the velocity gradient in'x' direction is lower as compared to 'y' direction so, $\frac{1}{t_i}{\int }_0^{t_i}\frac{\partial }{\partial x}{(u\text{'})}^{2}dt$

tends to zero.

Therfore, $\frac{\partial \overline{u}}{\partial t}+\frac{\partial {\overline{u}}^2}{\partial x}+\frac{\partial \overline{u}\overline{v}}{\partial y}=-\frac{1}{\rho }\frac{\partial \overline{P}}{\partial x}+\nu \cdot \frac{{\partial }^2\overline{u}}{\partial y^2}-\left(\frac{1}{t_i}\right){\int }_0^{t_i}\frac{\partial (u\text{'}v\text{'})}{\partial y}dt$

  = $-\left(\frac{1}{\rho }\right)\cdot \frac{\partial \overline{P}}{\partial x}+\left(\frac{1}{\rho }\right)\frac{\partial }{\partial y}[\nu \cdot \frac{\partial \overline{u}}{\partial y}-\frac{\rho }{t_i}{\int }_0^{t-i}(u\text{'}v\text{'})dt]$

  The left hand side denotes the inertial term calculated based on average quantity.

The right hand side denotes momentum diffusion i.e, combination of molecular viscosity(first term) and turbulent

viscosity(2nd term).

Turbulance modelling is all about capturing the time integral of (u'v')dt i.e, Reynolds stress.

In the viscous sub-layer the viscous term is much much greater than the inertial term and assuming insignificant pressure

drop, dp/dx->0

we have the equation : $-\frac{1}{\rho }\cdot \frac{\partial \overline{P}}{\partial x}+\left(\frac{1}{\rho }\right)\frac{\partial }{\partial y}[\nu \cdot \frac{\partial \overline{u}}{\partial y}-\frac{\rho }{t_i}{\int }_0^{t_i}(u\text{'}v\text{'})dt]$

                                            $\mu \cdot \frac{\partial \overline{u}}{\partial y}-\frac{\rho }{t_i}{\int }_0^{t_i}(u\text{'}v\text{'})dt=const.$

inside viscous sub-layer molecular viscosity dominates When we are in outside the viscous sub-layer, generally turbulent viscosity dominates. so Reynolds stress is ${\tau }_s$equal to

 ${\tau }_s=-\frac{\rho }{t_i}{\int }_0^{t_i}(u\text{'}v\text{'})dt$

approx

 Generally the turbulent transfer of momentum by eddies giving rise to an internal fluid friction but taking place inmuch

larger scale.

 Wheras the molecular viscosity is the transfer of momentum (motion) by random motions of individual molecules not moving

together.