Week 8: Literature review - RANS derivation and analysis

Aim:-

To study Reynold's decomposition to the Navier Stokes equations

Objective:-

1. To study the decomposition of the Reynold's Number to the Navier Stokes equation and come up with the expression for Reynold's stress.
2. To understand the terms Reynold's stress
3. To understand turbulent viscosity and its difference with Molecular viscosity.

Theory:-

1. Reynold's Averaged Navier Stokes (RANS):-

1. The Reynold's Averaged Navier Stokes equation are time averaged equation of motion for fluid flow.
2. The idea behind the equation is Reynold's decomposition whereby an instantaneous quality is decomposed into its time averaged and fluctuating quantities.
3. The RANS equation are primarily used to describe turbulent flow
4. These equation can be used with approximation based on knowledge of the properties of flow turbulence to give approximate time averaged solution to the Navier Stokes equation.

Derivation for Reynold's Averaged Navier Stokes equation:-

Consider 2 D flow,

According to Reynold decomposition,

`u(x,t)=bar(u)(x)+u'(x,t)` where u = (x,y,z)........................................(1)

u `rightarrow` actual term

`bar(u) rightarrow`space term

u' `rightarrow` fluctuating term

Navier Stokes contiunity equation:-

`frac{partialu}{partialx}+frac{partialv}{partialy}=0`

`frac{partial(bar(u)+u')}{partialx}+frac{partial(bar(v)+v')}{partialy}=0`...........From equation(1)

Applying Reynold's Averaged Navier Stokes method,

`frac{1}{t_i}int_0^(t_i)frac{partialbar(u)}{partialx}dt +frac{1}{t_i}int_0^(t_i)frac{partialu'}{partialx}dt+frac{1}{t_i}int_0^(t_i)frac{partialbar(v)}{partialy}dt+frac{1}{t_i}int_0^(t_i)frac{partialv'}{partialy}dt=0`

Since u' and v' are fluctuating term, therefore their average is zero.

Hence,

`frac{partialu'}{partialx}=frac{partialv'}{partialy}=0`

`frac{partialbar(u)}{partialx}frac{1}{t_i}int_0^(t_i)dt +frac{partialbar(v)}{partialy}frac{1}{t_i}int_0^(t_i)dt=0`

`frac{partialbar(u)}{partialx}frac{(t_i-0)}{t_i}+frac{partialbarv}{partialy}frac{(t_i-0)}{t_i}=0`

`frac{partialbar(u)}{partialx}+frac{partialbar(v)}{partialy}=0`...............................equation(2)

Hence equation (2) represent RANS contiunity eqaution

Navier Stokes Momentum Equation:-

`rho(frac{partialu}{partialt}+frac{upartialu}{partialx}+frac{vpartialv}{partialy}) =frac{-dP}{dx}+frac{mupartial^2u}{partialy^2}`

`frac{partialu}{partialt}+frac{upartialu}{partialx}+frac{vpartialv}{partialy}=frac{-1}{rho}frac{dP}{dx}+frac{mu}{rho}frac{partial^2u}{partialy^2}`

`frac{partialu}{partialt}+frac{upartialu}{partialx}+frac{vpartialv}{partialy}=frac{-1}{rho}frac{dP}{dx}+nufrac{partial^2u}{partialy^2}`.................................................................(3)

Since kinematic viscosity is equal to ratio of dynamic viscosity to the density.

`nu=mu/rho`

Consider, `u(frac{partialu}{partialx}+frac{partialv}{partialy})` is adding on the Left Hand Side of Equation (3)

Since, `frac{partialu}{partialx}+frac{partialv}{partialy}=0` from Navier Stokes contiunity equation.

Therefore,

`frac{partialu}{partialx}+frac{upartialu}{partialx}+frac{vpartialv}{partialy}+u(frac{partialu}{partialx}+frac{partialv}{partialy})=frac{-1}{rho}frac{dP}{dx}+nufrac{partial^2u}{partialy^2}`

`frac{partialu}{partialx}+2ufrac{partialu}{partialx}+vfrac{partialv}{partialy}+ufrac{partialv}{partialy}=frac{-1}{rho}frac{dP}{dx}+nufrac{partial^2u}{partialy^2}`

`frac{partialu}{partialx}+2ufrac{partialu}{partialx}+frac{partialuv}{partialy}=frac{-1}{rho}frac{dP}{dx}+nufrac{partial^2u}{partialy^2}`

Since, `frac{partialuv}{partialy}=ufrac{partialv}{partialy}+vfrac{partialu}{partialy}`

`frac{partialu}{partialx}+frac{partialu^2}{partialx}+frac{partialuv}{partialy}=frac{-1}{rho}frac{dP}{dx}+nufrac{partial^2u}{partialy^2}`

Applying Reynold's Averaged Navier Stokes Method,

`frac{1}{t_i}int_0^(t_i)frac{partial(bar(u)+u')}{partialt}dt+frac{1}{t_i}int_0^(t_i)frac{partial(bar(u)+u')^2}{partialx}dt+frac{1}{t_i}int_0^(t_i)frac{partial(bar(u)+u')(bar(v)+v')}{partialy}dt=frac{1}{t_i}frac{1}{rho}int_0^(t_i)frac{d(bar(rho)+rho')}{dx}dt+frac{1}{t_i}nuint_0^(t_i)frac{partial^2(bar(u)+u')}{partialy^2}``frac{1}{t_i}int_0^(t_i)frac{partial(bar(u))}{partialt}dt+frac{1}{t_i}int_0^(t_i)frac{partial(u')}{partialt}dt+frac{1}{t_i}int_0^(t_i)frac{partialbar(u^2)}{partialx}dt+frac{1}{t_i}int_0^(t_i)frac{partialu'^2}{partialx}dt+frac{1}{t_i}int_0^(t_i)frac{partialbar(u)bar(v)}{partialy}dt+frac{1}{t_i}int_0^(t_i)frac{partialbar(u)v'}{partialy}dt+frac{1}{t_i}int_0^(t_i)frac{partialu'bar(v)}{partialy}frac{1}{t_i}int_0^(t_i)frac{partialu'v'}{partialy}dt=frac{1}{t_i}int_0^(t_i)frac{dbar(P)}{dx}dt+frac{1}{t_i}int_0^(t_i)frac{dP'}{x}dt+nu[frac{1}{t_i}int_0^(t_i)frac{partial^2bar(u)}{partialy^2}dt+frac{1}{t_i}int_0^(t_i)frac{partial^2u'}{partialy^2}dt]`.....................................................(4)

Since, u', v' and P' are fluctuating term their average is zero, therefore u', v' and P' = 0

And multiplication of fluctuating term their average are not zero, therefore `(bar(u^2)) and u'v' ne0`

Equation 4 becomes,

`frac{partialbar(u)}{partialt}+frac{partialbar(u^2)}{partialx}+frac{partial(bar(u)bar(v))}{partialy}=frac{-1}{rho}frac{dbar(P)}{dx}+nufrac{partial^2bar(u)}{partialy^2}-frac{1}{t_i}int_0^(t_i)frac{partialu'^2}{partialx}dt-frac{1}{t_i}int_0^(t_i)frac{partial(u'v')}{partialy}dt`..........................................(5)

Since u' = 0

Therefore, `frac{1}{t_i}int_0^(t_i)frac{partialu'^2}{partialx}dt=0`

Hence equation 5 becomes,

`frac{partialbar(u)}{partialt}+frac{partialbar(u^2)}{partialx}+frac{partial(bar(u)bar(v))}{partialy}=frac{-1}{rho}frac{dbar(P)}{dx}+nufrac{partial^2bar(u)}{partialy^2}-frac{1}{t_i}int_0^(t_i)frac{partial(u'v')}{partialy}dt`

`frac{partialbar(u)}{partialt}+frac{partialbar(u^2)}{partialx}+frac{partial(bar(u)bar(v))}{partialy}=frac{-1}{rho}frac{dbar(P)}{dx}+frac{1}{rho}[mufrac{partialbaru}{partialy}-frac{1}{t_i}int_0^(t_i)frac{partial(u'v')}{partialy}dt]`..............................................................(6)

Here equation 6 represent RANS Momentum equation

From equation 6,

`frac{partialbar(u)}{partialt}+frac{partialbar(u^2)}{partialx}+frac{partialbar(u)bar(v)}{partialy} rightarrow` Inertial term

`frac{-1}{rho}frac{dP}{dx}rightarrow`Pressure term

`mufrac{partialbar(u)}{partialy}rightarrow`Molecular viscosity

`frac{1}{t_i}int_0^(t_i)frac{partialu'v'}{partialy}dt rightarrow` Momentum diffusivity of turbulence `rightarrow`Reynold's Stress

Reynold's Stress:-

1. The theory of the Reynold's stress is quite analogous to the kinetic theory of gases and indeed the stress tensor in a fluid at a point may be seen to be the ensemble average of the stress due to the therma velocities of molecule at a given point a fluid.
2. Thus, by analogy the Reynold's stress is somethimes thought of as consisting of an isentropic pressure part, termed the turbulent pressure and an diagonal part which may be though of an effective turbulent viscosity.
3. Infact, which must effort has been expended in developing good model for the Reynold's stress as practical matter, when solving the fluid equation using computational fluid dynamics, often the simplest turbulence model prove the most effective.
4. One class of model, closely related to the concept of turbulent viscosity are the k-epsilion turbulence model, based upon coupled transport equation for the turbulent energy density I.C. engine and the turbulent dissipation rate.
5. Reynold's Stress is important for Turbulence Model.

Turbulent Viscosity:-

1. In a turbulent fluid, a linear interface between two different fluid break apart to form small scale structure known as eddies.
2. This effective viscosity is known as turbulent viscosity.
3. The product of this turbulent viscosity and the shear, give the net difference flux of fluid across the gird base.

Turbulent Viscosity differ from molecular viscosity:-

From equation 6,

`tau_(xy)=mufrac{partialbar(u)}{partialy}-rhou'v'`

1. From above equation each shear stress component is a sum of two term.
2. The first term is the shear stress created a result of molecular viscosity.
3. This is property of the fluid and is defined as the fluid internal resistance to the flow.
4. The second term is the shear stress generated as a result of adding formed due to the turbulent transfer of momentum.
5. This gives rises to internal fluid friction called eddy viscosity.
6. The important distinction is that eddy viscosity is the property of flow, unlike molecular viscosity.
7. The turbulent transfer of momentum by eddus created internal fluid friction.
8. It is the fundamental idea of how we define viscosity in turbulent flow, that is internal fluid resistance.
9. Molecular viscosity is the same as viscosity, the coefficient of molecular viscosity is the same value as dynamics viscosity.
10. Molecular viscosity is analogy in laminar flow to eddy viscosity in turbulent flow.

Conclusion:-

Hence, we have derived the Reynold's Averaged Navier Stokes equation