Week 8: Literature review - RANS derivation and analysis

Aim: Reynolds-averaged Navier-Stokes derivation and analysis

Introduction and Theory: The RANS equations are time-averaged equations of motion for fluid flow. The idea behind the equations is Reynolds decomposition, wherein an instantaneous quantity is decomposed into its time-averaged and fluctuating quantities. The RANS equations are primarily used to describe turbulent flows. These equations can be used with approximations based on knowledge of the properties of flow turbulence to give approximate time-averaged solutions to the Navier–Stokes equations. 

Derivation:

Reynolds Decomposition:

Reynolds decomposition is a mathematical technique used to separate the expectation value of a quantity from its fluctuations. For example, for a quantity  `u` the decomposition would be

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

where `bar u` denotes the expectation value of  `u`, (often called the steady component/time, spatial or ensemble average), and `u'`, are the deviations from the expectation value (or fluctuations). The fluctuations are defined as the expectation value subtracted from quantity `u` such that their time average equals zero.

Time - Averaged velocity in x and y directions.

`bar(u)(x,y,z) = 1/t_(Int)int_0^(t_(Int)) u(x,y,z,t) dt`

`bar(v)(x,y,z) = 1/t_(Int)int_0^(t_(Int)) v(x,y,z,t) dt`

The integration time-step is chosen such that the time-averaging of velocity gives us a constant value which is the averaged quantity.

Also, in this time-step, the time-averaging of the fluctuating component is zero.

i.e. `1/t_(Int) int_0^(t_(Int)) u'(x,y,z,t) dt =0`

To prove this,

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

Applying time averaging on both sides,

`1/t_(Int) int_0^(t_(Int))u(x,y,z,t) = 1/t_(Int) int_0^(t_(Int))bar(u)(x,y,z) + 1/t_(Int) int_0^(t_(Int))u'(x,y,z,t)`

which gives,

`bar(u)(x,y,z) = bar(u)(x,y,z) + int_0^(t_(Int))u'(x,y,z,t)`

We got `1/t_(Int) int_0^(t_(Int))bar(u)(x,y,z) = bar(u)(x,y,z)`, because the time-averaging of an averaged quantity is the same averaged quantity itself.

Simplifying, we get,

`1/t_(Int) int_0^(t_(Int)) u'(x,y,z,t) dt =0`

Governing Equations for the boundary layer:

Assumptions: 2D Incompressible.

Using the order of magnitude analysis, some terms have been deleted.

1. Continuity Equation

`(del u )/ (del x) + (del v )/ (del y) = 0`

2. X - Momentum Equation

`rho((del u)/(del t) + u(del u)/(del x) + v(del u)/(del y) ) = -(del P)/(del x) + mu((del ^2 u)/(del y^2))`

3. Energy Equation

`rho_c((del T)/(del t) + u(del T)/(del x) + v(del T)/(del y) ) = K((del^2 T)/(del x^2)) + mu((del u)/(del y))^2`

Substituting Reynolds decomposition into Continuity equation,

 `(del u )/ (del x) + (del v )/ (del y) = 0`

`(del (bar(u) + u') )/ (del x) + (del (bar(v) + v') )/ (del y) = 0`

Applying time averaging ,

`1/t_(Int)int_0^(t_(Int))[(del (bar(u) + u') )/ (del x) + (del (bar(v) + v') )/ (del y)] dt = 0`

`1/t_(Int) int_0^(t_(Int)) (del bar(u) )/ (del x) dt + 1/t_(Int) int_0^(t_(Int)) (del u' )/ (del x) dt + 1/t_(Int) int_0^(t_(Int)) (del bar(v) )/ (del y) dt + 1/t_(Int) int_0^(t_(Int)) (del v' )/ (del y) dt =0`

`1/t_(Int) int_0^(t_(Int)) (del u' )/ (del x) dt` and `1/t_(Int) int_0^(t_(Int)) (del v' )/ (del x) dt` will evaluate to zero as proved before.

Hence we obtain,

`1/t_(Int) int_0^(t_(Int)) (del bar(u) )/ (del x) dt + 1/t_(Int) int_0^(t_(Int)) (del bar(v) )/ (del y) dt =0`

On further simplification, we obtain,

`(del bar(u) )/ (del x) + (del bar(v) )/ (del y) = 0`

which is the RANS continuity equation.

Substituting Reynolds decomposition into Momentum equation,

 `rho((del u)/(del t) + u(del u)/(del x) + v(del u)/(del y) ) = -(del P)/(del x) + mu((del ^2 u)/(del y^2))`

Dividing by `rho` on both the sides, we get,

`((del u)/(del t) + u(del u)/(del x) + v(del u)/(del y) ) = -(1/rho)(del P)/(del x) + nu((del ^2 u)/(del y^2))`

`(del u)/(del t) + u(del u)/(del x) + v(del u)/(del y) + u((del u)/(del x) + (del v)/(del y)) = -(1/rho)(del P)/(del x) + nu((del ^2 u)/(del y^2))`

`(del u)/(del t) + 2u(del u)/(del x) + v(del u)/(del y) + u(del v)/(del y) = -(1/rho)(del P)/(del x) + nu((del ^2 u)/(del y^2))`

`(del u)/(del t) + (del u^2)/(del x) + (del uv)/(del y) = -(1/rho)(del P)/(del x) + nu((del ^2 u)/(del y^2))`

Applying Time-averaging on both the sides,

`1/t_(Int)int_0^(t_(Int))[(del (bar(u) + u'))/(del t) + (del (bar(u) + u')^2)/(del x) + (del [(bar(u) + u')(bar(v) + v')])/(del y)] dt = 1/t_(Int)int_0^(t_(Int))[-(1/rho)(del (P + P'))/(del x) + nu(del ^2 (bar(u) + u'))/(del y^2)]dt`

`1/t_(Int)int_0^(t_(Int))(del bar(u))/(del t) + 1/t_(Int)int_0^(t_(Int))(del u')/(del t) + 1/t_(Int)int_0^(t_(Int))(del (bar(u))^2)/(del x) + 1/t_(Int)int_0^(t_(Int))(del (u')^2)/(del x) + 1/t_(Int)int_0^(t_(Int))(del (2bar(u)u'))/(del y) + 1/t_(Int)int_0^(t_(Int))(del (bar(u)bar(v)))/(del y) + 1/t_(Int)int_0^(t_(Int))(del bar(u)v')/(del y) + 1/t_(Int)int_0^(t_(Int))(del u'bar(v))/(del y) + 1/t_(Int)int_0^(t_(Int))(del u'v')/(del y)`

`= 1/t_(Int)int_0^(t_(Int))(-1/rho)(del bar(P))/(del x) + 1/t_(Int)int_0^(t_(Int))(-1/rho)(del P')/(del x) + 1/t_(Int)int_0^(t_(Int))(nu)(del^2 bar(u))/(del y^2) + 1/t_(Int)int_0^(t_(Int))(nu)(del^2 u')/(del y^2)`

The following terms will evaluate to zero because of the time averaging of fluctuating components:

`1/t_(Int)int_0^(t_(Int))(del u')/(del t) dt`, `1/t_(Int)int_0^(t_(Int))(del (2bar(u)u'))/(del y) dt`, `1/t_(Int)int_0^(t_(Int))(del (bar(u)v'))/(del y) dt`, `1/t_(Int)int_0^(t_(Int))(del (bar(v)u'))/(del y) dt`, `1/t_(Int)int_0^(t_(Int))(-1/rho)(del P')/(del x) dt`, `1/t_(Int)int_0^(t_(Int))(nu)(del^2 u')/(del y^2) dt`

Therefore on further simplification we get,

`(del bar(u))/ (del t) + (del bar(u)^2)/ (del x) + (del bar(u)bar(v))/ (del y) = (-1/rho)(del bar(P))/(del x) + nu(del^2 bar(u))/(del y^2) - 1/t_(Int)int_0^(t_(Int)) (del (u')^2)/(del x) dt - 1/t_(Int)int_0^(t_(Int)) (del u'v')/(del y) dt`

The following term will be ignored:

`1/t_(Int)int_0^(t_(Int)) (del (u')^2)/(del x) dt`

Because, the gradient of u in the x direction will be small compared to the y direction due to the boundary layer.

Hence , we finally get,

`ubrace((del bar(u))/ (del t) + (del bar(u)^2)/ (del x) + (del bar(u)bar(v))/ (del y))_("Inertial terms") = ubrace((-1/rho)(del bar(P))/(del x))_("Pressure term") + (1/rho)del / (del y)[ubrace(mu(del bar(u))/(del y))_("Molecular viscosity")+ ubrace(rho/t_(Int) int_0^(t_(Int)) (u'v') dt)_("Turbulent viscosity")]`

Where,

`rho/t_(Int) int_0^(t_(Int)) (u'v') dt = rho bar(u'v') = tau`; which is the Reynolds stress term.

Reynolds Stress:

Reynolds stress results from the turbulent transport of momentum. It can be defined as the component of the total stress tensor in a fluid obtained from the averaging operation over the Navier-Stokes equations to account for turbulent fluctuations in fluid momentum.

Turbulence and Molecular viscosity:

Turbulent viscosity is also known as eddy viscosity. It is due to the velocity fluctuations in a turbulent flow at high Reynolds number. It depends on the flow, it is not a fluid property. However, molecular viscosity is a fluid property. Molecular viscosity is the same as dynamic viscosity. Molecular viscosity depends on the type of fluid and the ambient pressure and temperature conditions.