RANS derivation and analysis
Governing equations (usually Navier Stokes Equation) of fluid flow when solved are completely capable of giving solutions for both laminar flow and turbulent flow, the approach of solving the governing equations to simulate turbulence is called direct numerical simulation and even though it gives pretty accurate results it is quite impractical for engineering simulations because to get results in small time steps (inverse of the frequency of fluctuation of velocity) should be used which incurs high computational power. That is why turbulence models are preferred which can be used to solve turbulence for larger time steps without incurring the high computational cost. A turbulence model achieves this by integrating the governing equations over a large time, a process called averaging and applying Reynolds decomposition to convert the governing equation into Reynolds Averaged Navier Stokes equation (RANS) form.
Reynold’s Decomposition:
`u(x,y,t)=u ̅(x,y,t)+u'(x,y,t)`
`v(x,y,t)=v ̅(x,y,t)+v^' (x,y,t)`
Where, `( u) ̅=1/t∫_0^tu(x,y,t)dt` and `v ̅=1/t∫_0^tv(x,y,t)dt`
Now,
`u(x,y,t)=u ̅(x,y,t)+u'(x,y,t)`
Time averaging on both sides, we get
`1/t∫_0^tu(x,y,t)dt =1/t∫_0^tu ̅ (x,y,t)dt+1/t∫_0^tu^' (x,y,t)dt`
We know that,
`1/t ∫_0^tu(x,y,t)dt=u ̅(x,y,t)` and
`1/t∫_0^tu ̅(x,y,t)dt =u ̅(x,y,t)` as averaging the average term gives an average term.
Applying this to the above equation we get,
`u ̅(x,y,t)= u ̅(x,y,t)+ 1/t∫_0^tu'(x,y,t)dt`
`1/t ∫_0^tu'(x,y,t)dt =0` .....Equation 1
Similarly, `1/t ∫_0^tv'(x,y,t)dt =0` ....Equation 2
Now applying RANS decomposition to the continuity equation,
We know that the continuity equation for the 2D is
`∂u//∂x+∂v//∂y=0`
Applying RANS Decomposition we get,
`(∂(u^'+u ̅))//∂x+(∂(v^'+v ̅))//∂y=0`
Using integral time-averaging process,
`1/t∫_0^t(∂(u ̅))//∂x) dt+ 1/t ∫_0^t(∂(u'))//∂x) dt+ 1/t ∫_0^t(∂(v'))//∂y) dt+ 1/t∫_0^t(∂(v ̅))//∂y) dt=0`
Applying equation 1 and equation 2, we get
`∂(u ̅)//∂x+∂(v ̅)//∂y=0`
Now, the momentum equation for 2D flows is,
`ρ(∂u//∂t+u ∂u//∂x+v ∂u//∂y)= -∂P//∂x+ µ((∂^2 u)/(∂y^2 ))`
Now, adding u times continuity here to the LHS as continuity equation is zero, the equation will not be affected.
`(∂u//∂t+u ∂u//∂x+v ∂u//∂y)+u(∂u//∂x+∂v//∂y)= -1/ρ ∂P//∂x+ µ/ρ((∂^2 u)/(∂y^2 ))`
`∂u//∂t+2u ∂u//∂x+v ∂u//∂y+u ∂v//∂y = -1/ρ ∂P//∂x+ µ/ρ((∂^2 u)/(∂y^2 ))`
`∂u//∂t+2u ∂u//∂x+ (∂(uv))//∂y= -1/ρ ∂P//∂x+ µ/ρ((∂^2 u)/(∂y^2 ))`
`∂u//∂t+∂u^2//∂x+ ∂(uv)//∂y= -1/ρ ∂P//∂x+ µ/ρ((∂^2 u)/(∂y^2 ))`
Applying RANS decomposition on momentum equation,
`1/t ∫_0^t(∂(u ̅ )//∂t) dt+ (∂(u^' )//∂t) dt + (∂(u ̅ )^2//∂x) dt + (∂(u^' )^2//∂x) dt + ∂(2u'(u) ̅//∂x) dt + (∂(u ̅v ̅ )//∂y)dt + (∂(u ̅v^' )//∂y)dt + (∂(u^' v ̅ )//∂y)dt + (∂(u^' v^' )//∂y)dt =1/t ∫_0^t-1/ρ (∂P ̅)/∂x dt- 1/ρ (∂P^')/∂x dt + µ/ρ (∂^2 u ̅)/(∂y^2 ) dt + µ/ρ (∂^2 u')/(∂y^2 ) dt)`
Where the terms,
`1/t ∫_0^t(∂(u^' )//∂t)dt,1/t ∫_0^t(∂(2u'(u) ̅)//∂x)dt,1/t∫_0^t(∂(u ̅v^' )//∂x)dt,1/t ∫_0^t(∂(u^' v ̅ )//∂t)dt,1/t∫_0^t(- 1/ρ (∂P^')//∂x)dt,1/t ∫_0^t(1/ρ (∂P^')//∂x)dt`
`1/t ∫_0^t(∂(u^' v^' )//∂y)dt``≠` 0 because for long term observations, fluctuations of the u’ and v’ interact with each other for the transfer of momentum and energy and
`1/t ∫_0^t(∂(u^' )^2//∂x) dt` 0 as squaring a function of time makes all negative values in turn into positive, so, time-averaging the term here cannot be zero.
So, the momentum equation becomes,
`∂(u ̅ )//∂t + ∂(u ̅ )^2//∂x + ∂(u ̅v ̅ )//∂y = -1/ρ (∂P ̅)//∂x+ µ/ρ (∂^2 u ̅)/(∂y^2 )-1/t∫_0^t(∂(u^' )^2//∂x)dt- 1/t∫_0^t(∂(u^' v^' )//∂y)dt`
Now,
`1/t∫_0^t(∂(u^' )^2//∂x) dt`though not zero, the value of this term is negligible as the variation of u velocity along x-direction in the boundary layer is small.
So, the momentum equation becomes,
`∂(u ̅ )//∂t + ∂(u ̅ )^2//∂x + ∂(u ̅v ̅ )//∂y = -1/ρ (∂P ̅)//∂x+ µ/ρ (∂^2 u ̅)/(∂y^2 )-1/t∫_0^t(∂(u^' v^' )//∂y) dt`
`∂(u ̅ )//∂t + ∂(u ̅ )^2//∂x + ∂(u ̅v ̅ )//∂y = -1/ρ (∂P ̅)//∂x +1/ρ ∂//∂x(µ∂(u ̅ )//∂y-ρ/t ∫_0^tu'v'dt)`
Where the terms,
`∂(u ̅ )//∂t + ∂(u ̅ )^2//∂x + ∂(u ̅v ̅ )//∂y` are the inertial terms and `-1/ρ (∂P ̅)//∂x`is the pressure term.
`µ∂(u ̅ )//∂y`is viscous stress.
`ρ/t∫_0^tu'v'dt`is reynold’s stress.
Now, when the equation
`∂(u ̅ )//∂t + ∂(u ̅ )^2//∂x + ∂(u ̅v ̅ )//∂y = -1/ρ ∂P ̅//∂x +1/ρ ∂//∂y(µ∂u ̅ //∂y-ρ/t ∫_0^tu'v'dt)`
Is applied inside the viscous sub-region of the boundary layer with negligible pressure gradient it becomes,
`1/ρ ∂//∂y(µ∂(u ̅ )//∂y-ρ/t ∫_0^tu'v'dt)``≠`0
As inertial terms are so small in the viscous sublayer that they are approximately zero,
`∂//∂y(µ∂(u ̅ )//∂y-ρ/t ∫_0^tu^' v^' dt)`= 0
`µ∂u ̅ //∂y-ρ/t ∫_0^tu^' v^' dt=constant`
Now when the above equation is applied to a region outside the viscous sublayer region, the turbulent viscosity dominates so the equation becomes,
`-ρ/t∫_0^tu^' v^' dt= τ`
`τ = ρ((u'v') ̅)`
Reynolds stress is the stress that is induced into the fluid by convective momentum transfer due to the random motions caused by turbulent fluctuations of the velocities.
Now, when the Reynolds stress is written in the form of viscous stress, we get,
`τ= ρϵ (∂u ̅//∂y)` where is the turbulent or eddy viscosity.
The total stress experienced by the fluid is,
`τ= τs+ τv`
`τ= µ∂(u ̅ )//∂y+ µ∂(u ̅ )//∂y`
`τ=ρ(µ/ρ ∂(u ̅ )//∂y+ ϵ ∂(u ̅ )//∂y)`
`τ=ρ(γ ∂(u ̅ )//∂y+ ϵ ∂(u ̅ )//∂y)`
`τ=ρ(γ+ ϵ)(∂u ̅ //∂y)`
Here, `γ` is the molecular viscosity.
Molecular viscosity is the property of the fluid and it is not influenced by the flow conditions, eddy viscosity is smaller than the molecular viscosity when the viscous sublayer is observed.
Turbulent viscosity arises solely due to the random flow fluctuations that cannot be easily predicted, turbulent viscosity is larger than in the regions outside of the viscous sublayer of the flow.