2.4 Theory of turbulence
2.4.2 Turbulent kinetic energy
During a wave breaking process, the wave energy is converted into two parts : the mean current energy and the turbulent kinetic energy. The turbulent kinetic energy is damped through turbulence viscous dissipation (Huang et al.[115]). In this section the effect of turbulence on kinetic energy is discussed as it is useful to understand how the mean flow feeds kinetic energy to the turbulence.
For an incompressible fluid with constant transport coefficients, the transport equation for the kinetic energy is :
∂
∂t uiui
2
+ ∂
∂xj
ujuiui
2
=−1 ρ
∂ujp
∂xj
+∂ui2νSij
∂xj −2νSij∂ui
∂xj
(2.67) The terms on the RHS of Eqn. (2.67) represent work done pressure forces, work done by viscous stresses and dissipation, respectively.
The above equation is sometimes written as :
∂
∂t uiui
2
=−∂uj
∂xj
p ρ+uiui
2
+∂(ui2νSij)
∂xj −2νSij
∂ui
∂xj
(2.68)
with the first term on the RHS now representing the work done by the total dynamic pressure.
If we now apply the Reynolds decomposition to Eqn. (2.67), the decomposition of the kinetic energy gives :
uiui=uiui+ 2uiu′i+u′iu′i (2.69) Substituting Eqn. (2.69) and Eqn. (2.56)-(2.61) in Eqn. (2.67) and averaging over the entire equation gives :
∂ 12uiui
∂t +∂
1 2u′iu′i
∂t =−∂ui
∂xj
p ρ+1
2ujuj
+ν∂uj2Sij
∂xi −ν2Sij
∂uj
∂xi
− ∂
∂xiu′i p′
ρ +1 2u′ju′j
−∂uj
∂xi
u′iu′j
−1 2
∂uiu′ju′j
∂xi
+ν∂u′j2sij
∂xi −ν2sij
∂u′j
∂xi
(2.70)
The above equation contains both the mean and the turbulent kinetic energy. To obtain an equation for the mean turbulent kinetic energy alone, we need to multiply Eqn. (2.64) byui. The resulting equation
is :
∂
∂t 1
2u′iu′i
+ ∂
∂xi
ui
p ρ+1
2ujuj
=−u′iu′j∂ui
∂xj
+ ∂
∂xj
−u′iu′jui
+ν ∂
∂xi
uj2Sij−ν2Sij
∂uj
∂xj
(2.71) As was the case in the equation for the mean momentum (Eqn. (2.66)), Eqn. (2.71) has additional terms due to turbulence that affect the mean kinetic energy. Becauseu′iu′j is normally negative, the action of this term tends to decrease or take away kinetic energy from the mean motion. The first term on the RHS of Eqn. (2.71) represents the deformation work by the turbulent stresses. The second term on the RHS is the work done by the turbulent stresses. Viscosity does not have a significant impact on the mean kinetic energy transport. This shows that the large structures in turbulent flows are relatively independent of viscosity.
More information about turbulence and its interaction with mean flow can be obtained from the transport equation for the turbulent kinetic energy, by subtracting Eqn. (2.71) from Eqn. (2.70) to give :
∂
1 2u′iu′i
∂t +uj ∂
∂xj
1 2uiui
=− ∂
∂xi
u′i p′
ρ +1 2u′ju′j
−u′iu′j∂uj
∂xi
+ν ∂
∂xi
u′jsij−ν2sij
∂u′j
∂xi
(2.72) The mechanisms that affect the turbulent kinetic energy on the RHS of the above equation are the turbulent diffusion of mechanical energy, deformation work on the mean flow by the turbulent stresses, the viscous work by the the turbulent shear stresses, and the viscous dissipation of the turbulent kinetic energy. As opposed to the case of the mean flow turbulent kinetic energy equation, the viscous terms are important to the turbulent kinetic energy balance. The last term in Eqn. (2.72) has a particular significance. It is always positive and therefore indicates a drain of energy. This is the viscous dissipation term.
For a statistically steady flow, the time derivatives in the equations of the mean and turbulent kinetic energy is zero. Also
sijsij =sij
∂u′j
∂xi
. (2.73)
Applying these two conditions, the equation to themean kinetic energy can be written as :
uj
∂
∂xj
1 2uiui
= ∂
∂xj
−p
ρuj+ 2νuiSij−u′iu′juj
+ 2νSijSij+u′iu′jSij (2.74) and that for theturbulent kinetic energy can be written as :
uj
∂
1 2u′iu′i
∂xj
=− ∂
∂xj
1
ρu′jp′+1
2u′iu′iu′j−2νu′isij
−u′iu′jSij−2νsijsij (2.75) These are the same two equations as (3.1.11) and (3.2.1) given by Tennekes and Lumley [34].
Equation (2.75) can be conveniently written as :
∂k
∂t +∂ujk
∂xj =− ∂
∂xj
1
ρu′jp′+u′jk′−2νu′is′ij
−2νs′ijsij+u′iu′js′ij (2.76)
where theensemble-averaged kinetic energy is : k= 1
2(u′iu′i) (2.77)
and theinstantaneous turbulent kinetic energy is : k′= 1
2(u′iu′i) (2.78)
The first three terms on the right hand side of Eqn. (2.76) are due to the work by the pressure gradient, transport by turbulent velocity fluctuations and transport by viscous stresses, respectively. The last two terms are due to the work associated with deformation. The termu′iu′jSij is the deformation work term which appears in both equation for the mean and turbulent flows with opposite sign. This term serves to exchange energy between the mean flow and turbulence. The last term is the work due to the deformation by viscous stresses and is always a drain of energy. From the above equations, an estimate of the turbulent energy production,℘and dissipation,ǫ, rates can be obtained from the following equations:
℘=−u′iu′js′ij (2.79)
ǫ= 2νs′ijs′ij (2.80)
Both ℘and ǫ can be in local equilibrium, however in most cases the energy is produced at one point, transported and dissipated at another point in space or time as described by Eqn. (2.76). It should be noted that the dissipation work term−u′iu′jSijor equivalently−u′iu′j∂u∂xji appear in both equations for the mean (Eqn. (2.74)) and turbulent kinetic energy (Eqn. (2.75)) but with an opposite sign. We also call this term the production term as the effect of the deformation work is to exchange energy between the mean flow and the turbulence. Usually this term is negative (Reynolds stress is usually negative) so that there is a net flow of kinetic energy from the mean flow to the turbulence. In a purely laminar flow, viscous stresses dissipate the kinetic energy directly into heat. In turbulent flow, however, the deformation work caused by turbulent stresses first converts mean flow kinetic energy into turbulent kinetic energy. It then cascades through the wave number spectrum before it is finally dissipated into heat. The terms in the parentheses on the RHS of the turbulent kinetic energy Eqn. (2.75) are in the form of a divergence of a vector quantity. These are conservative terms that vanish when integrated over an infinite flow field volume. These inertial terms do not create or destroy energy, but simply redistribute the turbulent kinetic energy from place to place in the flow. In a statistically steady homogeneous flow, a further simplification of Eqn. (2.75) can be made. In this case, all spatial gradients of the turbulence transport quantities will be zero. Under these conditions, the only remaining terms give :
−u′iu′jSij = 2νsijsij (2.81)
In other words, Eqn. (2.81) implies that the rate of production of turbulent kinetic energy is balanced by the rate of viscous dissipation.