Week 8: Literature review - RANS derivation and analysis

Objectives:

• Apply Reynold’s decomposition to the Navier's Stoke equations and come up with the expression for Reynold’s stress.
• Explain Reynold’s stress.
• Turbulent viscosity and it's difference from molecular viscosity.

Turbulence:

In fluid dynamics, turbulence or turbulent flow is fluid motion characterized by chaotic changes in pressure and flow velocity. It is in contrast to a laminar flow, 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.

All the governing equations are based on the laws of conservation. The Navier- Stokes equations are applied to a mathematical model to examine changes in those properties during dynamic or thermal interactions. The equations are adjustable regarding the content of the problem and are expressed based on the principles of conservation of mass, momentum, and energy.

Continuity Equation - Conservation of Mass:

`(∂u)/(∂x)+(∂v)/(∂y)=0`

x-Momentum Equation - Conservation of Momentum ( Newton's Second law):

`ρ[(∂u)/(∂t)+u(∂u)/(∂x)+v(∂u)/(∂y)]=-(∂P)/(∂x)+μ((∂^2u)/(∂x^2)+(∂^2u)/(∂y^2))`

Energy/Transport Equation - Conservation of Energy ( First law of thermodynamics):

`ρCP[(∂ϕ)/(∂t)+u(∂ϕ)/(∂x)+v(∂ϕ)/(∂y)]=K[(∂^2ϕ)/(∂x2)+(∂^2ϕ)/(∂y^2)]+μ((∂u)/(∂y))^2`

 Reynold's decomposition:

u¯(x,y,z)+u′(x,y,z,t)">`u(x,y,z,t) = bar(u)(x,y,z)+u′(x,y,z,t)`

v¯(x,y,z)+v′(x,y,z,t)">`v(x,y,z,t) = bar(v)(x,y,z)+v′(x,y,z,t)`

w¯(x,y,z)+w′(x,y,z,t)">`w(x,y,z,t) = bar(w)(x,y,z)+w′(x,y,z,t)`

where,

       u = instantaneous velocity

       `bar(u)(x,y,z)`= function of space

       u'(x,y,z,t) = function of time

       u' = fluctuating component

on integrating,

          u¯(x,y)=1t∫0tu(x,y,t)dt">u¯(x,y)=1t∫0tu(x,y,t)dt

            v¯(x,y)=1t∫otv(x,y,z)dt">v¯(x,y)=1t∫otv(x,y,z)dt

Note: Integrating a fluctuating component  = 0.

          1t∫0tu′(x,y,z,t)dt=0">1t∫0tu′(x,y,z,t)dt=0

on simplifying,

u(x,y,z,t) = u¯(x,y)+u′(x,y,z,t)">u¯(x,y)+u′(x,y,z,t)

1t∫0tu(x,y,z,t)dt=1t∫0tu¯(x,y)+1t∫0tu′(x,y,z,t)">1t∫0tu(x,y,z,t)dt=1t∫0tu¯(x,y)+1t∫0tu′(x,y,z,t)

                                      ↓">↓

u¯(x,y,z)=u¯(x,y,z)+1t∫0tu′(x,y,z,t)">u¯(x,y,z)=u¯(x,y,z)+1t∫0tu′(x,y,z,t)

 1t∫0tu′(x,y,z,t)dt=0">1t∫0tu′(x,y,z,t)dt=0

Applying the Decomposition :

∂u∂x+∂v∂y=0">∂u∂x+∂v∂y=0

∂∂x(u¯+u′)+∂∂y(v¯+v′)=0">∂∂x(u¯+u′)+∂∂y(v¯+v′)=0

1t∫∂∂x(u¯+u′)dt+∂∂y(v¯+v′)dt=0">1t∫∂∂x(u¯+u′)dt+∂∂y(v¯+v′)dt=0

1t∫∂u∂xdt+1t∫∂u′∂xdt+1t∫∂v¯∂y+1t∫∂v′∂y=0">1t∫∂u∂xdt+1t∫∂u′∂xdt+1t∫∂v¯∂y+1t∫∂v′∂y=0

∂u∂x1t∫dt+∂v¯∂y1t∫dt=0">∂u∂x1t∫dt+∂v¯∂y1t∫dt=0

∂u¯∂x1t∫0tdt+∂v¯∂y1t∫0tdt=0">∂u¯∂x1t∫0tdt+∂v¯∂y1t∫0tdt=0

∂u¯∂xt-0t+∂v¯∂yt-0t">∂u¯∂xt-0t+∂v¯∂yt-0t

∂u¯∂x+∂v¯∂y=0">∂u¯∂x+∂v¯∂y=0- Continuity Equation , thus proved.

Momentum Equation:

ρ∂u∂t+u∂u∂x+v∂u∂y=-dρdx+μ∂2u∂y2">ρ∂u∂t+u∂u∂x+v∂u∂y=-dρdx+μ∂2u∂y2

∂u∂t+μ∂u∂x+v∂u∂y=1ρ-∂ρ∂x+μρ∂2u∂y2">∂u∂t+μ∂u∂x+v∂u∂y=1ρ-∂ρ∂x+μρ∂2u∂y2

Replace μρ">μρwith δ">δ(kinematic viscosity)

∂u∂t+u∂u∂xv∂u∂y=-∂ρρ∂x+δ∂2u∂y2">∂u∂t+u∂u∂xv∂u∂y=-∂ρρ∂x+δ∂2u∂y2

Multiply u∂u∂x">u∂u∂xwith∂u∂x">∂u∂x

where,

        u(∂u∂x+∂v∂y)=0">u(∂u∂x+∂v∂y)=0

∂u∂t+∂u∂u∂x+v∂u∂y+v∂v∂y=-1ρ∂ρ∂x+δ∂2u(∂y)2">∂u∂t+∂u∂u∂x+v∂u∂y+v∂v∂y=-1ρ∂ρ∂x+δ∂2u(∂y)2

By Product Rule:

          ∂∂y(uv)=u∂v∂y+v∂u∂y">∂∂y(uv)=u∂v∂y+v∂u∂y      

          ∂u∂t+∂u∂u∂x+∂∂y(uv)=-1ρ∂ρ∂x+ν∂2u∂y2">∂u∂t+∂u∂u∂x+∂∂y(uv)=-1ρ∂ρ∂x+ν∂2u∂y2

          ∂u∂t+∂u2∂x+∂∂y(uv)=-1ρ∂ρ∂x+ν∂2u∂y2">∂u∂t+∂u2∂x+∂∂y(uv)=-1ρ∂ρ∂x+ν∂2u∂y2

          1t∫ot∂∂t(u¯+u′)dt+∂∂x(u¯+u′)2dt+∂∂y[(u¯+u′)(v¯+v′)]dt=1t∫0t-1ρ[(∂(ρ¯+ρ′)∂x)]dt+ν∂2∂y2(u¯+u′)dt">1t∫ot∂∂t(u¯+u′)dt+∂∂x(u¯+u′)2dt+∂∂y[(u¯+u′)(v¯+v′)]dt=1t∫0t-1ρ[(∂(ρ¯+ρ′)∂x)]dt+ν∂2∂y2(u¯+u′)dt

          1t∫0t∂u¯∂tdt+∂u′∂tdt+∂u¯2∂xdt+∂u′2∂xdt+(∂∂x(∂u¯u′)dt)+∂∂y(u¯v¯)dt)+∂∂y(u¯v′)dt+d∂y        (u′v¯)dt)+∂∂y(u′v′)dt">1t∫0t∂u¯∂tdt+∂u′∂tdt+∂u¯2∂xdt+∂u′2∂xdt+(∂∂x(∂u¯u′)dt)+∂∂y(u¯v¯)dt)+∂∂y(u¯v′)dt+d∂y        (u′v¯)dt)+∂∂y(u′v′)dt

           1t∫0t-1ρ∂ρ¯∂xdt-1ρ∂ρ′∂xdx+ν∂2u¯∂y2dt+ν∂2u∂y2dt">1t∫0t-1ρ∂ρ¯∂xdt-1ρ∂ρ′∂xdx+ν∂2u¯∂y2dt+ν∂2u∂y2dt

           ∂u¯∂t+∂(u¯2)∂x+∂(u¯v¯)∂y=-1ρ∂ρ¯∂x+ν∂2u¯∂y2-1t∫0t∂(u′)2∂xdt-1t∫0t∂(u′v′)∂ydt">∂u¯∂t+∂(u¯2)∂x+∂(u¯v¯)∂y=-1ρ∂ρ¯∂x+ν∂2u¯∂y2-1t∫0t∂(u′)2∂xdt-1t∫0t∂(u′v′)∂ydt

            ∂u¯∂t+∂(uu¯)2∂x+∂(u¯v¯)∂y=-1ρ∂ρ∂x+ν∂2u¯∂y2-1t∫0t∂∂y(u′v′)dt">∂u¯∂t+∂(uu¯)2∂x+∂(u¯v¯)∂y=-1ρ∂ρ∂x+ν∂2u¯∂y2-1t∫0t∂∂y(u′v′)dt

=(-1ρ∂ρ¯∂x)⏟Pressure term1ρ∂∂y[μ∂u¯∂y+ρt∫0t(u′v′)dt]">(-1ρ∂ρ¯∂x)⏟Pressure term1ρ∂∂y[μ∂u¯∂y+ρt∫0t(u′v′)dt]

where,

             μ∂u¯∂y=ViscousStress">μ∂u¯∂y=ViscousStress

             ρt∫0t(u′v′)dt">ρt∫0t(u′v′)dt= Reynold's Stress

 reduced equation.

∂u∂t+∂u2∂x+∂(uv)∂y=-1ρ∂P∂x+ν∂2u∂x2+ν∂2u∂y2">∂u∂t+∂u2∂x+∂(uv)∂y=-1ρ∂P∂x+ν∂2u∂x2+ν∂2u∂y2

substituting the averaged quantities

∂(u¯+u′)∂t+∂(u¯+u′)2∂x+∂((u¯+u′)(v¯+v′))∂y=-1ρ∂(P¯+P′)∂x+ν∂2(u¯+u′)∂x2+ν∂2(u¯+u′)∂y2">∂(u¯+u′)∂t+∂(u¯+u′)2∂x+∂((u¯+u′)(v¯+v′))∂y=-1ρ∂(P¯+P′)∂x+ν∂2(u¯+u′)∂x2+ν∂2(u¯+u′)∂y2

expanding the terms

∂u¯∂t+∂u′∂t+∂u¯2∂x+∂(u′)2∂x+∂(2u¯u′)∂x+∂(u¯v¯)∂y+∂(u¯v′)∂y+∂(u′v¯)∂y+∂(u′v′)∂y=-1ρ∂P¯∂x-1ρ∂P′∂x+ν∂2u¯∂x2+ν∂2u′∂x2+ν∂2u¯∂y2+ν∂2u′∂y2">∂u¯∂t+∂u′∂t+∂u¯2∂x+∂(u′)2∂x+∂(2u¯u′)∂x+∂(u¯v¯)∂y+∂(u¯v′)∂y+∂(u′v¯)∂y+∂(u′v′)∂y=-1ρ∂P¯∂x-1ρ∂P′∂x+ν∂2u¯∂x2+ν∂2u′∂x2+ν∂2u¯∂y2+ν∂2u′∂y2

averaging over time step 'T'

1T∫0T∂u¯∂tdt+1T∫0T∂u′∂tdt+1T∫0T∂u¯2∂xdt+1T∫0T∂(u′)2∂xdt+1T∫0T∂(2u¯u′)∂xdt+1T∫0T∂(u¯v¯)∂ydt+1T∫0T∂(u¯v′)∂ydt+1T∫0T∂(u′v¯)∂ydt+1T∫0T∂(u′v′)∂ydt=-1T∫0T1ρ∂P¯∂xdt-1T∫0T1ρ∂P′∂xdt+1T∫0Tν∂2u¯∂x2dt+1T∫0Tν∂2u′∂x2dt+1T∫0Tν∂2u¯∂y2dt+1T∫0Tν∂2u′∂y2dt">1T∫0T∂u¯∂tdt+1T∫0T∂u′∂tdt+1T∫0T∂u¯2∂xdt+1T∫0T∂(u′)2∂xdt+1T∫0T∂(2u¯u′)∂xdt+1T∫0T∂(u¯v¯)∂ydt+1T∫0T∂(u¯v′)∂ydt+1T∫0T∂(u′v¯)∂ydt+1T∫0T∂(u′v′)∂ydt=-1T∫0T1ρ∂P¯∂xdt-1T∫0T1ρ∂P′∂xdt+1T∫0Tν∂2u¯∂x2dt+1T∫0Tν∂2u′∂x2dt+1T∫0Tν∂2u¯∂y2dt+1T∫0Tν∂2u′∂y2dt

since the time integral of fluctuation component is zero and the rules mentioned above. 

1T∫0T∂u¯∂tdt+1T∫0T∂u′∂tdt+1T∫0T∂u¯2∂xdt+1T∫0T∂(u′)2∂xdt+1T∫0T∂(2u¯u′)∂xdt+1T∫0T∂(u¯v¯)∂ydt+1T∫0T∂(u¯v′)∂ydt+1T∫0T∂(u′v¯)∂ydt+1T∫0T∂(u′v′)∂ydt=-1T∫0T1ρ∂P¯∂xdt-1T∫0T1ρ∂P′∂xdt+1T∫0Tν∂2u¯∂x2dt+1T∫0Tν∂2u′∂x2dt+1T∫0Tν∂2u¯∂y2dt+1T∫0Tν∂2u′∂y2dt">1T∫0T∂u¯∂tdt+1T∫0T∂u′∂tdt+1T∫0T∂u¯2∂xdt+1T∫0T∂(u′)2∂xdt+1T∫0T∂(2u¯u′)∂xdt+1T∫0T∂(u¯v¯)∂ydt+1T∫0T∂(u¯v′)∂ydt+1T∫0T∂(u′v¯)∂ydt+1T∫0T∂(u′v′)∂ydt=-1T∫0T1ρ∂P¯∂xdt-1T∫0T1ρ∂P′∂xdt+1T∫0Tν∂2u¯∂x2dt+1T∫0Tν∂2u′∂x2dt+1T∫0Tν∂2u¯∂y2dt+1T∫0Tν∂2u′∂y2dt

The above equation reduces to

1T∫0T∂u¯∂tdt+1T∫0T∂u¯2∂xdt+1T∫0T∂(u′)2∂xdt+1T∫0T∂(u¯v¯)∂ydt+1T∫0T∂(u′v′)∂ydt=-1T∫0T1ρ∂P¯∂xdt+1T∫0Tν∂2u¯∂x2dt+1T∫0Tν∂2u¯∂y2dt">1T∫0T∂u¯∂tdt+1T∫0T∂u¯2∂xdt+1T∫0T∂(u′)2∂xdt+1T∫0T∂(u¯v¯)∂ydt+1T∫0T∂(u′v′)∂ydt=-1T∫0T1ρ∂P¯∂xdt+1T∫0Tν∂2u¯∂x2dt+1T∫0Tν∂2u¯∂y2dt

Upon integrating, the equation reduces to

∂u¯∂t+∂u¯2∂x+1T∫0T∂(u′)2∂xdt+∂(u¯v¯)∂y+1T∫0T∂(u′v′)∂ydt=-1ρ∂P¯∂x+ν∂2u¯∂x2+ν∂2u¯∂y2">∂u¯∂t+∂u¯2∂x+1T∫0T∂(u′)2∂xdt+∂(u¯v¯)∂y+1T∫0T∂(u′v′)∂ydt=-1ρ∂P¯∂x+ν∂2u¯∂x2+ν∂2u¯∂y2

Rearranging the terms

∂u¯∂t+∂u¯2∂x+∂(u¯v¯)∂y=-1ρ∂P¯∂x+ν∂2u¯∂x2+ν∂2u¯∂y2-1T∫0T∂(u′v′)∂ydt-1T∫0T∂(u′)2∂xdt">∂u¯∂t+∂u¯2∂x+∂(u¯v¯)∂y=-1ρ∂P¯∂x+ν∂2u¯∂x2+ν∂2u¯∂y2-1T∫0T∂(u′v′)∂ydt-1T∫0T∂(u′)2∂xdt