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Literature review on RANS derivation & its Analysis

Title: Literature review on RANS derivation & its Analysis Objective: 1. Derive the RANS derivation 2. Importance of RANS Theory: The Reynolds-averaged Navier–Stokes equations (or RANS equations) are time-averaged equations…

Dipakv Virkarwe
updated on 21 Aug 2020

Title: Literature review on RANS derivation & its Analysis

Objective: 1. Derive the RANS derivation

2. Importance of RANS

Theory:

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

The basic tool required for the derivation of the RANS equations from the instantaneous Navier-Stroke equation is the Reyold decomposition. Reynolds decomposition refers to separation of the flow variable like velocity( $u$ ) into the mean (time-averaged) component ( $overline{u}$ ) and the fluctuating component ( $u^{prime}$ ). Because the mean operator is aReynold operator, it has a set of properties. One of these properties is that the mean of the fluctuating quantity is equal to zero

$({bar {u^{prime }}}=0)$

$Lets consider momentum equation$

$ρ (\partial u \partial t + u \partial u \partial x + v \partial u \partial y) = - \partial p \partial x + μ \partial 2 u \partial y 2 ρ (\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}) = - \frac{\partial p}{\partial x} + μ \frac{\partial 2 u}{\partial y 2} rho((\partialu)/(\partialt)+u(\partialu)/(\partialx)+v(\partialu)/(\partialy))=-(\partialp)/(\partialx)+mu(\partial2u)/(\partialy2)$

$k i e m a t i c v i s cos i t y (ν) = μ ρ k i e m a t i c v i s cos i t y (ν) = \frac{μ}{ρ} kiematic viscosity(nu)=(mu)/rho$

$(\partial u \partial t + u \partial u \partial x + v \partial u \partial y) = - (1 ρ) \partial p \partial x + ν \partial 2 u \partial y 2 (\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}) = - (\frac{1}{ρ}) \frac{\partial p}{\partial x} + ν \frac{\partial 2 u}{\partial y 2} ((\partialu)/(\partialt)+u(\partialu)/(\partialx)+v(\partialu)/(\partialy))=-(1/rho)(\partialp)/(\partialx)+nu(\partial2u)/(\partialy2)$

$for the simplification u & continuity equation uses$

$(\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}) + u (\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}) = - (\frac{1}{ρ}) \frac{\partial p}{\partial x} + ν \frac{\partial 2 u}{\partial y 2}$ $((∂u)/(∂t)+u(∂u)/(∂x)+v(∂u)/(∂y))+u((∂u)/(∂x)+(∂v)/(∂y))=-(1/rho)(∂p)/(∂x)+nu(∂2u)/(∂y2)$

$(\frac{\partial u}{\partial t} + 2 u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + u \frac{\partial v}{\partial y}) = - (\frac{1}{ρ}) \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial y^{2}}$ $((∂u)/(∂t)+2u(∂u)/(∂x)+v(∂u)/(∂y)+u(∂v)/(∂y))=-(1/rho)(∂p)/(∂x)+nu(∂^2u)/(∂y^2)$

from product rule

$(\frac{\partial u}{\partial t} + \frac{\partial}{\partial x} (u^{2}) + \frac{\partial}{\partial y} (u v)) = - (\frac{1}{ρ}) \frac{\partial p}{\partial x} + ν \frac{\partial^{2} u}{\partial y^{2}}$ $((∂u)/(∂t)+(∂)/(∂x)(u^2)+(∂)/(∂y)(uv))=-(1/rho)(∂p)/(∂x)+nu(∂^2u)/(∂y^2)$

Now apply decomposition rule

$1/(t)int((∂(baru+u'))/(∂t))dt+((∂(baru+u')^2)/(∂x))dt+((∂(baru+u').(barv+v'))/(∂y))dt)=-(1/t)(1/rho)int(∂(barp+p'))/(∂x)+nu(∂^2(baru+u'))/(∂y^2)dt$

Fluctuate velocity will be zero so, wherever u' tem is equll to zero

Assuming that u' velocity in x-direction & v' for y-direction. as there is very less momentum occurs in x-direction . so, u' is consider zero.

$(∂baru)/(∂t)+(∂baru^2)/(∂x)+(∂baru.barv)/(∂y)=-(1/rho).(∂barp)/(∂x)+nu.(∂^2baru)/(∂y^2)-(1/t).int.(∂((u'.v'))/(∂y)).dt$

$(∂baru)/(∂t)+(∂baru^2)/(∂x)+(∂baru.barv)/(∂y)=-(1/rho).(∂barp)/(∂x)+(1/rho).(∂)/(∂y)[nu.∂baru/(∂y)-rho/t int(u'v')dt)]$

To find out Reynold stress correctly called as Turbulence Modeling

Reynpld stress = $-(rho/t int(u'v')dt)$

What is Turbulent Viscosity

Turbulent viscosity is the Aveg. shear stress to the rate of Velocity Gradient in the turbulent flow.

Turbulent Viscosity VS Molecular Viscosity

Rate of momentum transfer causes by Turbulent Eddies is modelled by Turbulet Viscosity. rate of momentum transfer causes by molecular friction is modelled by Molecular viscosity

Conclusion

1. Understand Importance of RANS euation in Turbulent modeling

2. Reynold stress importance in RANS equation

3. Turbulence viscosity & Molecular viscosity importance Understand