Week 8: Literature review - RANS derivation and analysis
Aim: Literature Review – Reynolds-Averaged Navier-Stokes derivation and analysis.
Objective:
- To apply Reynolds decomposition to the NS equations and come up with the expression for Reynolds stress.
- To explain the understanding of the terms Reynolds stress.
- To explain what is turbulent viscosity and how is it different from molecular viscosity?
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.
Turbulence:
In fluid dynamics, turbulence or turbulent flow is fluid motion characterized by chaotic pressure and flow velocity changes. 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 commonly observed in everyday phenomena such as surf, fast-flowing rivers, billowing storm clouds, or smoke from a chimney, and most fluid flows occurring in nature or created in engineering applications are turbulent. 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. In general terms, in turbulent flow, unsteady vortices appear of many sizes which interact with each other, consequently, drag due to friction effects increases. This increases the energy needed to pump fluid through a pipe. Turbulence can be exploited, for example, by devices such as aerodynamic spoilers on aircraft that "spoil" the laminar flow to increase drag and reduce lift.
Turbulence modeling:
Turbulence modelling is the construction and use of a mathematical model to predict the effects of turbulence. In computational fluid dynamics (CFD), even simple flows are difficult to compute. And the computational complexity grows exponentially when accounting for turbulence: irregular fluid motions that span a wide range of scales in space and time. Selecting the right method for modelling turbulence physics is critical and confusing.
RANS:
The Reynolds Averaged Navier-Stokes equations (or RANS equations) are time-averaged equations of motion for fluid flow. The idea behind the equations is Reynolds decomposition, whereby an instantaneous quantity is decomposed into its time-averaged and fluctuating quantities, an idea first proposed by Osborne Reynolds. 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.
1. To apply Reynolds decomposition to the NS equations and come up with the expression for Reynolds stress.
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) = baru(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 `bar u`, are the deviations from the expectation value (or fluctuations). The fluctuations are defined as the expectation value subtracted from quantity 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.
In this time-step, the time-averaging of the fluctuating component is zero.
`1/t_(Int) int_0^(t_(Int)) u'(x,y,z,t) dt =0`
Proof:
`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) dt = 1/t_(Int) int_0^(t_(Int))bar(u)(x,y,z) dt + 1/t_(Int) int_0^(t_(Int))u'(x,y,z,t) dt`
Here you are time averaging an average quantity `1/t_(Int) int_0^(t_(Int))bar(u)(x,y,z) dt` and if you do it, you'll get the same average quantity.
`bar(u)(x,y,z) = bar(u)(x,y,z) + int_0^(t_(Int))u'(x,y,z,t) dt`
and `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
1. 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 turn out to be zero as shown above.
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`
Upon further simplification, `(del bar(u) )/ (del x) + (del bar(v) )/ (del y) = 0`, which is the RANS Continuity Equation.
2. 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 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))`
By applying time-averaging on both 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`.
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 above equation is similar to the original momentum equation except it is in time-averaged form and two additional terms on the RHS. The additional time integration terms in RHS are not equal to zero & can be proved, however, to simplify the model certain important assumptions are to be applied.
Assuming `- 1/T int_0^T(delu'^2)/(delx) dt = 0`, considering velocities in x-direction to be of negligible gradient and stronger in y-direction (only true in boundary layer), we 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")]`
and `rho/t_(Int) int_0^(t_(Int)) (u'v') dt = tau` is the Reynolds stress term.
2. Reynolds Stress:
The Reynolds stress is 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. The theory of the Reynolds 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 thermal velocities of molecules at a given point in a fluid. Thus, by analogy, Reynold's stress is sometimes thought of as consisting of an isotropic pressure part, termed the turbulent pressure, and an off-diagonal part which may be thought of as an effective turbulent viscosity. Reynolds stress at any given point in a turbulent fluid is somewhat subject to interpretation, depending upon how one defines the average.
3.
a) Turbulent Viscosity: The coefficient that relates the average shear stress and rate of velocity gradient in the turbulent flow is called turbulent viscosity. The turbulent transfer of momentum by eddies gives rise to internal fluid friction on a large scale. Its also known as eddy viscosity. These are enormous in nature and to capture these to extreme details is near impossible, if done computationally it's very costly. Thus, we devise a way to capture the net effective mixing of eddies using viscosity models. Turbulent viscosity increases the shear rate and affects the velocity and heat transfer.
b) Molecular Visocity: It's a fluid property and is the same as viscosity. Its coefficient is the same as dynamic viscosity. Molecular viscosity is the transport of mass motion momentum solely by random motions of individual molecules not moving together in coherent groups. Molecular viscosity is analogous in laminar flow to eddy viscosity in a turbulent flow. It can be construed like some sort of internal friction between two layers of fluid developed while flowing.
Difference between Turbulent viscosity and Molecular viscosity:
The turbulent viscosity is an imaginary concept but molecular viscosity is dynamic that is present in all fluids. The turbulent viscosity is present only in turbulent flow regions but molecular viscosity is present in all types of flows. Turbulent viscosity depends on fluid flow whereas molecular viscosity depends only on fluid properties. The turbulent viscosity is non-linear & extremely hard to estimate whereas molecular viscosity is easy to measure.