Reynolds Decomposition
Author: Elisa Lovecchio
The description of the dynamics of a turbulent fluid Wikipedia>Turbulence is extremely complicated because of the complexity, unpredictability and chaotic motion that characterizes such a system, visible on a very wide range of scales.
To say it with Richardson's words:
Big whirls have little whirls that feed on their velocity,
and little whirls have lesser whirls and so on to viscosity. (Wikipedia>Richardson)
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Reynolds decomposition is a useful mathematical representation elaborated by Osborne Reynolds (1842–1912) Wikipedia>Reynolds for analyzing the dynamics of a turbulent fluid through the conceptual and analitical separation of its average regular motion and of the chaotic turbulent flutuations.
The basic concept consists in the decomposition of all the dynamic and thermodynamic instantaneous variables (eg, the velocity component $\textcolor{black}{u}$) that characterize the state of the the fluid in two components:
the average value $\textcolor{black}{\bar{u}}$ and the fluctuations, or deviations, $\textcolor{black}{u'}$ around the average.
This decomposition allows to interpret the complex turbulent dynamics as the superposition of a so called “steady” average component that represent the laminar motion and that can be easily studied in a quantitative fashion, and the contribution of the turbulent eddies and fluctuations which are characterized by a high degree of randomness. The latter can hence be measured as deviations from the mean flow, and studied through their statistical properties such as the moments that characterize their probability distributions.
In absence of turbulence (low Reynolds numbers) we expect then $\textcolor{black}{\vec{u}=\bar{\vec{u}}}$.
Averaging
There are several ways to define the steady component of the flow, each one corresponding to a different definition of average. The most common definitions of average that's useful to know are:
- average in time (so called “mean”):
\begin{equation*}
\textcolor{black}{ ^t\bar{u}=\frac{1}{\Delta t} \int_{0}^{\Delta t} u(t,\vec{x}) dt }
\end{equation*}
- average in space:
\begin{equation*}
\textcolor{black}{ ^s\bar{u} =\frac{1}{ V} \int \int \int u(t,\vec{x}) d^3x}
\end{equation*}
- ensemble average:
\begin{equation*}
\textcolor{black}{ ^a\bar{u} = \frac{1}{N} \sum_{i=1}^{N} u_i}
\end{equation*}
where an ensemble is defined as a set of $\textcolor{black}{N}$ realizations ($\textcolor{black}{N \rightarrow \infty}$) of the event corresponding to the same macroscopic thermodynamic state, in other words a set of repetitions of the same experiment or simulation.
When working with real observations, it's often straightforward to define the average as the mean, where the mean is the average in time of the measured variable (eg, the velocity of the water flow) over a single measurement of length $\textcolor{black}{\Delta t}$.
For an ergodic Wikipedia>Ergodicity process (a process that is assumed to explore the whole phase space in a finite amount of time) the three definitions correspond to each other and we can hence use any of the three to define the average values of the quantities of interest. In the case of turbulent flow, it is reasonable and common to assume the process to be ergodic.
Rules for averaging
The next rules are true for all the previous definitions of average, given their linear Wikipedia>LinearMap character:
\begin{equation*}
\textcolor{black}{ \overline{\lambda a} = \lambda \overline{a}}\qquad \qquad \textcolor{black}{ \overline{a+b} = \overline{a} + \overline{b}}\qquad \qquad \textcolor{black}{ \overline{\overline a} = \overline a } \qquad \qquad \textcolor{black}{\overline{a \cdot b}=\overline{a}\cdot\overline{b}+\overline{a'\cdot b'}} \qquad \qquad \textcolor{black}{\overline{\frac{\partial a}{\partial t}}=\frac{\partial \overline a}{\partial t}}
\end{equation*}
for any scalar $\textcolor{black}{\lambda}$ and any variables $\textcolor{black}{a=\bar a + a'}$ and $\textcolor{black}{b=\bar b + b'}$.
Properties of the average and fluctuating fields
From the previous equations it is possible to derive some simple properties:
- the mean of the fluctuations is zero: $\textcolor{black}{u = \bar u + u' \rightarrow \overline{u} = \overline{\bar u + u'} \rightarrow \bar u = \bar u + \bar u' \rightarrow \bar u'= 0}$
- both the average and fluctuations vector fields have zero divergence: $\textcolor{black}{\vec \nabla \vec u = \vec \nabla (\bar{\vec u} + \vec u') = 0 \quad \forall (\vec x, t) \rightarrow \vec \nabla \bar{\vec u} = \vec \nabla \vec u' = 0}$
The fluctuations are hence stochastic variables with mean equal to zero.
Momentum equation for the average flow
To describe the average motion of the fluid flow we start from the general instantaneous Navier-Stokes equations for mass and momentum conservation:
\begin{equation*}
\boxed{
\textcolor{black}{
\begin{cases}
\dfrac{\partial \rho}{\partial t}=\vec \nabla (\rho \vec u)} \qquad \qquad &\text{(continuity eqn: mass conservation)}\\
\dfrac{D \vec u}{D t} = - \dfrac{\vec \nabla p}{\rho} + \vec F_g + \dfrac{1}{\rho} \vec \nabla T \qquad \qquad &\text{(eqn of motion: momentum conservation)}
\end{cases}
}
}
\end{equation*}
where $\textcolor{black}{p}$ is pressure, $\textcolor{black}{\rho}$ represent density, the advective and temporal evolution terms are included in the operator $\textcolor{black}{\frac{D}{Dt} = \frac{\partial}{\partial t} + \vec \nabla \cdot \vec u}$ with $\textcolor{black}{\vec \nabla = (\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})}$ and $\textcolor{black}{\vec F_g}$ is the external body force per unit mass, in our case gravity: $\textcolor{black}{\vec F_g = - \vec g}$.
The term $\textcolor{black}{\vec \nabla T}$ represent the shear stress tensor and depends on the properties of the fluid.
Newtonian fluids, incompressibility and Boussinesq approximation
Water in a Newtonian fluid: this not only means that it is assumed to be isotropic (which means that it has the same mechanical properties in every direction), but it also implies that the viscous stresses generated by its flow are proportional to the local strain, in other words to the local change of the fluid's velocity vector. The proportionality constant is called viscosity $\textcolor{black}{\mu}$ and for newtonian fluids this quantity does not depend on the velocity of the flow.
Non-newtonian fluids such as ketchup YT_video>Ketchup or Oobleck FYFD>Oobleck behave sometimes in a pretty counter-intuitive fashion, showing phenomena such as shear thickening or thinning due to their non constant viscosity Wikipedia>NonNewtonian.
Water is also assumed to be incompressible (a pretty realistic assumption for a fluid) so that $\textcolor{black}{\rho = \rho_0}$ constant in time and space, which implies for the continuity equation:
\begin{equation*}
\textcolor{black}{\frac{\partial \rho}{\partial t} = 0, \quad \textcolor{black}{\vec \nabla \rho = 0 \quad \Rightarrow \quad \boxed{\vec \nabla \cdot \vec u = 0}}
\end{equation*}
that means that there are no sources or wells in the velocity field, or in other words no convergence or divergence of the flow. This approximation is equivalent to the so called Boussinesq approximation Wikipedia>BoussinesqApprox. for gases, and denies the existence of sound waves that propagate through the medium via compression.
For an incompressible newtonian fluid, the shear stress tensor $\textcolor{black}{\vec \nabla T}$ can then be written component by component as:
\begin{equation*}
\textcolor{black}{T_{i,j}=\mu \big( \frac{\partial u_i}{\partial x_j}+\frac{\partial v_j}{\partial x_i} \big)}
\end{equation*}
so that the last term of the momentum conservation equation can be expressed in the form:
\begin{equation*}
\textcolor{black}{\frac{1}{\rho} \vec \nabla T = \frac{\mu}{\rho} \nabla^2 \vec u}
\end{equation*}
Averaged momentum equation
At this point we can apply Reynolds decomposition and average the whole momentum equation.
The 3D Instantaneous Navier-Stokes Equation for the motion of a newtonian incompressible turbulent fluid is:
\begin{equation*}
\textcolor{black}{
\boxed{
\frac{\partial \vec u}{\partial t} + \bigg( u \frac{\partial}{\partial x} + v \frac{\partial}{\partial y} + w \frac{\partial}{\partial z} \bigg) \vec u = - \frac{\vec \nabla p}{\rho_0} + \vec g + \nu \nabla^2 \vec u
}
}
\end{equation*}
where $\textcolor{black}{\nu=\frac{\mu}{\rho_0}}$ is called kinematic viscosity and $\textcolor{black}{\vec g = (0,0,-g)}$.
Let's consider the 1D equation, for example for the x-component, and apply Reynolds decomposition to the variables $\textcolor{black}{u = \bar{\vec u} + \vec u'}$ and $\textcolor{black}{p = \bar p + p'}$.
The whole equation will then be averaged according to the rules:
\begin{equation*}
\textcolor{black}{
\overline{ \frac{\partial}{\partial t}(\bar u + u') + \bigg( (\bar u + u') \frac{\partial}{\partial x} + (\bar v + v') \frac{\partial}{\partial y} + (\bar w + w') \frac{\partial}{\partial z}\bigg)(\bar u + u') } = \overline{ - \frac{1}{\rho_0} \frac{\partial (\bar p + p')}{\partial x} + \nu \nabla^2 (\bar u + u') }
}
\end{equation*}
where $\textcolor{black}{\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}}$.
Given the properties of the two vector fields we have:
\begin{equation*}
\textcolor{black}{
\overline{ \frac{\partial}{\partial t}(\bar u + u')} = \frac{\partial \bar u}{\partial t} \qquad \qquad \overline{\frac{1}{\rho_0} \frac{\partial (\bar p + p')}{\partial x} } = \frac{1}{\rho_0}\frac{\partial \bar p}{\partial x} \qquad \qquad \overline{ \nu \nabla^2 (\bar u + u') } = \nu \nabla^2 \bar u
}
\end{equation*}
while only the non-linear advective term produces some cross-terms which depend on the turbulent fluctuations
\begin{equation*}
\textcolor{black}{
\overline{ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z} } & = \bigg( \bar u \frac{\partial \bar u}{\partial x} + \bar v \frac{\partial \bar u}{\partial y} + \bar w \frac{\partial \bar u}{\partial z} \bigg) + \bigg( \overline{u' \frac{\partial u'}{\partial x}} + \overline{v' \frac{\partial u'}{\partial y}} + \overline{w' \frac{\partial u'}{\partial z}} \bigg)}
}
\end{equation*}
\begin{equation*}
\textcolor{black}{
= \bigg( \bar u \frac{\partial \bar u}{\partial x} + \bar v \frac{\partial \bar u}{\partial y} + \bar w \frac{\partial \bar u}{\partial z} \bigg) + \Bigg[ \bigg( \overline{\frac{\partial u'^2}{\partial x}} + \overline{\frac{\partial (u'v')}{\partial y}} + \overline{\frac{\partial (u'w')}{\partial z}} \bigg) - \bigg( \overline{u' \frac{\partial u'}{\partial x}} + \overline{u' \frac{\partial v'}{\partial y}} + \overline{u' \frac{\partial w'}{\partial z}} \bigg) \Bigg]
}
\end{equation*}
where the last term is zero, being $\textcolor{black}{\vec \nabla \vec u' = 0}$.
For the x-component of the averaged Navier-Stokes equation we obtain:
\begin{equation*}
\textcolor{black}{
\boxed{
\frac{D \bar u}{D t} = \frac{\partial \bar u}{\partial t} + \bigg( \bar u \frac{\partial \bar u}{\partial x} + \bar v \frac{\partial \bar u}{\partial y} + \bar w \frac{\partial \bar u}{\partial z} \bigg) = - \frac{1}{\rho_0} \frac{\partial \bar p}{\partial x}+ \nu \nabla^2 \bar u - \bigg( \frac{\partial \overline{u'^2}}{\partial x} + \frac{\partial \overline{u'v'}}{\partial y} + \frac{\partial \overline{u'w'}}{\partial z} \bigg)
}
}
\end{equation*}
The same procedure is valid for the other two spatial components of the velocity field.
The whole Reynolds Averaged Navier-Stokes Equation (RANS) can be expressed as follow:
\begin{equation*}
\textcolor{black}{
\boxed{
\frac{D \bar{\vec u}}{D t} = - \frac{\vec \nabla \bar p}{\rho_0} + \nu \nabla^2 \bar u - \frac{1}{\rho_0} \vec \nabla \hat \tau
}
}
\end{equation*}
where $\textcolor{black}{\hat \tau}$ is the Reynolds Stress Tensor, that represents the influence of the turbulent fluctuations on the mean flow.
Reynolds stress tensor
The Reynolds stress tensor $\textcolor{black}{\hat \tau}$ is defined as follows:
\begin{equation*}
\textcolor{black}{
\hat \tau = \rho_0 \cdot
\begin{pmatrix}
\overline{u'^2} & \overline{u'v'} & \overline{u'w'} \\ \overline{u'v'} & \overline{v'^2} & \overline{v'w'} \\ \overline{u'w'} & \overline{v'w'} & \overline{w'^2}
\end{pmatrix}
}
\end{equation*}
where the terms $\textcolor{black}{\overline{u_i' u_j'}}$ can be interpreted as the correlation between the two velocity components $\textcolor{black}{\overline{u_i'}}$ and $\textcolor{black}{\overline{u_j'}}$.
The closure problem and K-Theory
To predict the average quantities associated to the turbulent flow the main problem we have to face is the fact that the number of equations that describe the problem is lower then the number of unknown variables.
Reynolds stresses $\textcolor{black}{\overline{u_i' u_j'}}$ are in fact supplementary unknown quantities, not easy to measure and determine.
Even if we try to resolve the problem multiplying the RANS equations by the fluctuations $\textcolor{black}{u_i'}$ and mediating again, what we obtain is nothing more than a new equation in the third moments $\textcolor{black}{\overline{u_i' u_j' u_k'}}$ of the distribution of the fluctuations: trying to eliminate the second moments creates again a system with more unknowns than equations.
For this reason the system for the coupled evolution of the statistical moments of the fluctuations is non-closed, and to solve it it's necessary to make some hypothesis that link the unknown variables to measurable quantities. This hypothesis can be based for example on empirical observations.
The simplest way to close the system of equations is linking the second moments to the mean values, the so-called K-Theory model.
The assumption is based on the concept of eddy viscosity introduced by Boussinesq in 1887 Wikipedia>EddyViscosity. A new quantity $\textcolor{black}{k}$ called turbulent viscosity is defined in analogy to the molecular viscosity and the turbulent stresses are supposed to be proportional to the gradient of the average velocity, as it happens in the laminar flow. The turbulent viscosity is strongly dependent on the local degree of turbulence of the flow.
The Reynolds stress tensor takes this form
\begin{equation*}
\textcolor{black}{
\overline{u_i' u_j'} = k \bigg(\frac{\partial \bar u_i}{ \partial x_j} + \frac{\partial u_j}{ \partial x_i} \bigg) \Rightarrow \frac{1}{\rho_0} \vec \nabla \hat \tau = \frac{k}{\rho_0} \nabla^2 \bar{\vec u}
}
\end{equation*}
so that with this assumption the momentum equation for the mean flow has exactly the same structure of that for the instantaneous flow.
