A First Course in Turbulence - Chapter 3

The Dynamics of Turbulence
This chapter aims to address two questions:
#How is the kinetic energy of the turbulence maintained?
#Why are vorticity and vortex stretching so important to the study of turbulence?
To answer these questions this chapter will proceed by:
#deriving the mean flow and turbulence kinetic energy equations
#observing that turbulence extacts kinetic energy from the mean flow at large scales, which is balanced by a viscous dissipation at the smallest scales
#recognizing that vorticity and vortex stretching are the mechanisms by which energy is transfered from large to small scales for dissipation purposes.
==Kinetic Energy of the Mean Flow==
For review, the momentum and continuity equations for steady, incompressible mean flow are:
:U_j \frac{\partial U_i}{\partial x_j} = \frac{\partial}{\partial x_j} \left( \frac{T_{ij}}{\rho} \right)
:\frac{\partial U_i}{\partial x_i} = 0
Where the stress tensor is:
:T_{ij} = -P\delta_{ij} + 2\mu S_{ij} - \rho \overline{u_i u_j}
And the mean strain rate is:
:S_{ij} \equiv \frac{1}{2} \left(\frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i}\right)
The mean of the turbulent velocity fluctuations is zero, so in order to investigate the interactions between the mean and turbulent flow equations, we must look look at the mean and turbulent kinetic energy. To get the equation for the kinetic energy of the mean flow, the momentum equation is multiplied by U_i. The following is the derivation of the mean flow kinetic energy equation.
:\rho U_i U_j \frac{\partial U_i}{\partial x_j} = U_i\frac{\partial T_{ij}}{\partial x_j}
On the LHS,
:\rho U_j \frac{\partial}{\partial x_j} \left( \frac{1}{2} U_i U_i \right) = \rho U_i U_j \frac{\partial U_i}{\partial x_j}
And on the RHS,
:\frac{\partial}{\partial x_j} (T_{ij} U_i) = U_i \frac{\partial T_{ij}}{\partial x_j} + T_{ij}\frac{\partial U_i}{\partial x_j} \quad\Rightarrow\quad U_i \frac{\partial T_{ij}}{\partial x_j} = \frac{\partial}{\partial x_j} (T_{ij} U_i) - T_{ij}\frac{\partial U_i}{\partial x_j}
The product T_{ij}\frac{\partial U_i}{\partial x_j} is equal to the product of T_{ij} and S_{ij}, since the stress and strain tensors are symmetric.
The LHS and RHS combine to form the mean flow kinetic energy equation:
:\rho U_j \frac{\partial}{\partial x_j} \left( \frac{1}{2} U_i U_i \right) = \frac{\partial}{\partial x_j} (T_{ij} U_i) - T_{ij}S_{ij}
The first term on the RHS represents the transport of mean flow kinetic energy by stress. This term integrates to zero over the volume if either T_{ij} or U_i vanishes on the surface, which indicates that the energy flux into or out of the volume is zero. This is shown by the divergence theorem:
:\int_V \frac{\partial}{\partial x_j} (T_{ij} U_i) dV = \int_S n_j T_{ij} U_i dS
Where n_j is a unit vector normal to the surface. This means that the transport of mean flow kinetic energy term redistributes kinetic energy within the control volume.
The second term is the deformation work. This represents kinetic energy that was lost or retrieved from the straining of the control volume.
TL states:
The distinction between spatial energy transfer and deformation work is crucial to the understanding of the dynamics of turbulence.
===Pure Shear Flow===
TL uses a pure shear flow (specifically a Couette flow ), where all variables only depend on x2 and U1 is the only component of Ui, to illustrate the kinetic energy equation of the mean flow. Figure 3.1 in TL on page 61 illustrates this example.
The kinetic energy equation becomes:
:0 = \frac{\partial}{\partial x_2} (T_{12}U_1) - T_{12}\frac{\partial U_1}{\partial x_2}
The LHS is equal to zero because the only non-zero velocity is in the x1 direction, but the only non-zero derivatives are in the x2 direction. In the second term on the RHS, the symmetric stress and strain tensors reduce to T_{12}\frac{\partial U_1}{\partial x_2} because the only non-zero terms in the strain tensor are \frac{\partial U_1}{\partial x_2}. It would appear than this term should be multiplied by a half, because of the previous definition of the strain tensor, but since the tensors are symmetric, there are two components, which eliminates the half.
still have stuff to do here
===The Effects of Viscosity===
Expanding the stress tensor in the kinetic energy equation of the mean flow results in:
:\rho U_j \frac{\partial}{\partial x_j} \left( \frac{1}{2} U_i U_i \right) = \frac{\partial}{\partial x_j} \left( -\frac{P}{\rho}U_j + 2\nu U_i S_{ij} - \overline{u_i u_j} U_i \right) - 2\nu S_{ij}S_{ij} + \overline{u_i u_j}S_{ij}
The first three terms on the RHS are transport terms. The divergence theorem shows that if either T_{ij} or U_i vanishes on the surface, these terms integrate to zero over the volume, as was previously stated. If this is the case, these terms only redistribute energy inside the volume. The terms are called:
*-\frac{P}{\rho}U_j : pressure work
*2\nu U_i S_{ij} : transport of mean flow kinetic energy by viscous stresses
*\overline{u_i u_j} U_i : transport of mean flow kinetic energy by Reynolds stresses
==Kinetic Energy of the Turbulence==
The following is the derivation of the turbulent kinetic energy equation.
Step 1
Starting with the Navier-Stokes equations, as given in Eqn. (2.1.5) in TL on page 28:
:\frac{\partial \tilde u_i}{\partial t} + \tilde u_j \frac{\partial \tilde u_i}{\partial x_j} = -\frac{1}{\rho} \frac{\partial \tilde p}{\partial x_i} + \nu \frac{\partial^2 \tilde u_i }{\partial x_j \partial x_j}
These are multiplied by \tilde u_i, which is the equivalent of taking the dot product of the equations with the vector \tilde u. In this step \partial/\partial x_j(\partial \tilde u_i/\partial x_j) has been replaced by 2 \partial \tilde s_{ij}/\partial x_j, the instantaneous rate of strain tensor.
:\tilde u_i\frac{\partial \tilde u_i}{\partial t} + \tilde u_i\tilde u_j \frac{\partial \tilde u_i}{\partial x_j} = -\frac{1}{\rho}\tilde u_i \frac{\partial \tilde p}{\partial x_i} + 2 \nu \tilde u_i \frac{\partial \tilde s_{ij} }{\partial x_j}
Then we decompose the instantaneous parameters into mean and fluctuating components and take the time average of all of the components:
:\begin{align}&\underbrace{\overline{(U_i + u_i)\frac{\partial (U_i + u_i)}{\partial t}}}_{\mathbf{I}} + \underbrace{\overline{(U_i + u_i)(U_j + u_j) \frac{\partial (U_i + u_i)}{\partial x_j}}}_{\mathbf{II}} \\ &\qquad = \underbrace{-\frac{1}{\rho}\overline{(U_i + u_i) \frac{\partial (P + p)}{\partial x_i}}}_{\mathbf{III}} + \underbrace{2\nu \overline{(U_i + u_i) \frac{\partial (S_{ij} + s_{ij}) }{\partial x_j}}}_{\mathbf{IV}}\end{align}
Let's go thought this term-by-term to simplify it. First, term I:
:\begin{align}
	&\overline{(U_i + u_i)\frac{\partial (U_i + u_i)}{\partial t}} = \cancel{U_i \frac{\partial U_i}{\partial t}} + \cancel{U_i \frac{\partial \overline{u_i}}{\partial t}} + \cancel{\overline{u_i} \frac{\partial U_i}{\partial t}} + \overline{u_i \frac{\partial u_i}{\partial t}}\\
	&\qquad = \frac{\partial}{\partial t} \frac{1}{2}\overline{u_i u_i}
\end{align}
Where:
:\begin{align}
	\frac{\partial u_i u_i}{\partial t} &= 2u_i\frac{\partial u_i}{\partial t} \\
	\therefore \overline{u_i \frac{\partial u_i}{\partial t}} &= \frac{\partial}{\partial t}\frac{1}{2}\overline{u_i u_i}
\end{align}
This term represents the time rate of change of turbulent kinetic energy, but since we are taking the time average of the modified Navier-Stokes equations, this term must be zero.
Next, term II:
:\begin{align}
	&\overline{(U_i + u_i)(U_j + u_j) \frac{\partial (U_i + u_i)}{\partial x_j}} = \overline{(U_i U_j + U_i u_j + u_i U_j + u_i u_j)\frac{\partial (U_i + u_i)}{\partial x_j}} \\
	&\qquad = U_i U_j \frac{\partial U_i}{\partial x_j} + \cancel{U_i \overline{u_j} \frac{\partial U_i}{\partial x_j}} + \cancel{\overline{u_i} U_j \frac{\partial U_i}{\partial x_j}} + \overline{u_i u_j} \frac{\partial U_i}{\partial x_j} \\
	&\qquad \qquad + \cancel{U_i U_j \frac{\partial \overline{u_i}}{\partial x_j}} + U_i \overline{u_j \frac{\partial u_i}{\partial x_j}} + U_j \overline{u_i \frac{\partial u_i}{\partial x_j}} + \overline{u_i u_j \frac{\partial u_i}{\partial x_j}}
\end{align}
Three terms require further simplification. We will utilize the incompressible continuity equation, namely \partial u_i /\partial x_i = 0.
:\begin{align}
	\frac{\partial u_i u_j}{\partial x_j} &= u_j\frac{\partial u_i}{\partial x_j} + \cancel{u_i \frac{\partial u_j}{\partial x_j}} \\
	\therefore U_i \overline{u_j \frac{\partial u_i}{\partial x_j}} &= U_i \frac{\partial}{\partial x_j}\overline{u_i u_j}\\
	\\
	\frac{\partial u_i u_i}{\partial x_j} &= 2u_i\frac{\partial u_i}{\partial x_j} \\
	\therefore U_j \overline{u_i \frac{\partial u_i}{\partial x_j}} &= U_j \frac{\partial}{\partial x_j}\frac{1}{2}\overline{u_i u_i}\\
	\\
	\frac{\partial u_i u_i u_j}{\partial x_j} &= 2u_i u_j\frac{\partial u_i}{\partial x_j} + \cancel{u_i u_i\frac{\partial u_j}{\partial x_j}} \\
	\therefore \overline{u_i u_j \frac{\partial u_i}{\partial x_j}} &= \frac{1}{2}\frac{\partial}{\partial x_j}\overline{u_i u_i u_j}
\end{align}
Combining these terms, term II becomes:
:
	U_i U_j \frac{\partial U_i}{\partial x_j} + \overline{u_i u_j} \frac{\partial U_i}{\partial x_j} + U_i \frac{\partial}{\partial x_j}\overline{u_i u_j} + U_j \frac{\partial}{\partial x_j}\frac{1}{2}\overline{u_i u_i} + \frac{1}{2}\frac{\partial}{\partial x_j}\overline{u_i u_i u_j}
Next, term III:
:\begin{align}
	&-\frac{1}{\rho}\overline{(U_i + u_i) \frac{\partial (P + p)}{\partial x_i}} = -\frac{1}{\rho}\left( U_i \frac{\partial P}{\partial x_i} + \cancel{U_i \frac{\partial \overline{p}}{\partial x_i}} + \cancel{\overline{u_i} \frac{\partial P}{\partial x_i}} + \overline{u_i \frac{\partial p}{\partial x_i}} \right) \\
	&\qquad =-\frac{1}{\rho}\left( U_i \frac{\partial P}{\partial x_i} + \frac{\partial}{\partial x_i} \overline{pu_i} \right)
\end{align}
Finally, term IV:
:\begin{align}
&2\nu \overline{(U_i + u_i) \frac{\partial (S_{ij} + s_{ij}) }{\partial x_j}} = 2\nu \left( U_i \frac{\partial S_{ij}}{\partial x_j} + \cancel{U_i \frac{\partial \overline{s_{ij}}}{\partial x_j}} + \cancel{\overline{u_i} \frac{\partial S_{ij}}{\partial x_j}} + \overline{u_i \frac{\partial s_{ij}}{\partial x_j}} \right)\\
&\qquad = 2\nu\left( U_i \frac{\partial S_{ij}}{\partial x_j} + \frac{\partial}{\partial x_j} \overline{u_i s_{ij}} - \overline{s_{ij} s_{ij}}\right)
\end{align}
Where:
:
\frac{\partial u_i s_{ij}}{\partial x_j} = u_i \frac{\partial s_{ij}}{\partial x_j} + s_{ij} \frac{\partial u_i}{\partial x_j} = u_i \frac{\partial s_{ij}}{\partial x_j} + s_{ij}s_{ij}
Combining terms I, II, III, and IV:
:
\begin{align}
&U_i U_j \frac{\partial U_i}{\partial x_j} + \overline{u_i u_j} \frac{\partial U_i}{\partial x_j} + U_i \frac{\partial}{\partial x_j}\overline{u_i u_j} + U_j \frac{\partial}{\partial x_j}\frac{1}{2}\overline{u_i u_i} + \frac{1}{2}\frac{\partial}{\partial x_j}\overline{u_i u_i u_j} \\
&\qquad = -\frac{1}{\rho}\left( U_i \frac{\partial P}{\partial x_i} + \frac{\partial}{\partial x_i} \overline{pu_i} \right) + 2\nu\left( U_i \frac{\partial S_{ij}}{\partial x_j} + \frac{\partial}{\partial x_j} \overline{u_i s_{ij}} - \overline{s_{ij} s_{ij}}\right)
\end{align}
Step 2
Next we multiply the time-averaged Navier-Stokes equations by U_i, which is the equivalent of taking the dot product of the equations with the vector U. This results in:
:U_i U_j \frac{\partial U_i}{\partial x_j} = -\frac{1}{\rho}U_i \frac{\partial P}{\partial x_i} + 2\nu U_i \frac{\partial S_{ij}}{\partial x_j} - U_i \frac{\partial \overline{u_i u_j}}{\partial x_j}
Step 3
Now we subtract the equation in step 2 from equation in step 1, resulting in:
:\begin{align}
	&\overline{u_i u_j} \frac{\partial U_i}{\partial x_j} + U_j \frac{\partial}{\partial x_j}\frac{1}{2}\overline{u_i u_i} + \frac{1}{2}\frac{\partial}{\partial x_j}\overline{u_i u_i u_j}\\
	&\qquad = -\frac{1}{\rho}\frac{\partial}{\partial x_i} \overline{pu_i} + 2\nu\left( \frac{\partial}{\partial x_j} \overline{u_i s_{ij}} - \overline{s_{ij} s_{ij}}\right)
\end{align}
The desired equation, the turbulent kinetic energy equation, results from some simple rearrangement of the terms:
:U_j \frac{\partial}{\partial x_j}\frac{1}{2}\overline{u_i u_i} = -\frac{\partial}{\partial x_j} \left( \frac{1}{\rho} \overline{u_j p} + \frac{1}{2}\overline{u_i u_i u_j} - 2\nu \overline{u_i s_{ij}} \right) - \overline{u_i u_j}S_{ij} -2\nu\overline{s_{ij} s_{ij}}
===Production Equals Dissipation===
===Taylor Microscale===
===Scale Relations===
===Spectral Energy Transfer===
===Further Estimates===
===Wind-Tunnel Turbulence===
===Pure Shear Flow===
==Vorticity Dynamics==
===Vorticity Vector and Rotation Tensor===
===Vortex Terms in the Equations of Motion===
===Reynolds Stress and Vorticity===
===The Vorticity Equation===
===Vorticity in Turbulent Flow===
===Two-Dimensional Mean Flow===
===The Dynamics of ΩiΩi===
===The Equation for ωiωi===
===Turbulence is Rotational===
===An Approximate Vorticity Budget===
===Multiple Length Scales===
===Stretching of Magnetic Field Lines===
==The Dynamics of Temperature Fluctuations==
===Microscales in the Temperature Field===
===Buoyant Convection===
===Richardson Numbers===
===Buoyancy Time Scale===
===Monin-Oboukhov Length===
===Convection in the Atmospheric Boundary Layer===