Energy balance

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Energy balance

The energy balance in absence of external forcing for Navier-Stokes equation follows from Eqs. (1.1,1.2). The total kinetic energy of the fluid is

[画像:\begin{displaymath} E = \int d^3 r {1\over 2} \rho \vert{\mbox{\boldmath$u$}}\vert^2 \end{displaymath}] (1.9)

and its temporal variation is

$\displaystyle \frac {dE}{dt}$ $\textstyle =$ [画像:$\displaystyle \int d^3 r {\mbox{\boldmath$u$}} \cdot \frac {\partial {\mbox{\boldmath$u$}}}{\partial t}$]

$\textstyle =$ [画像:$\displaystyle \int d{\mbox{\boldmath$r$}} \left[ -u_i u_j \partial_j u_i - u_i \partial_i p +\nu u_i \partial_j \partial_j u_i \right]$] (1.10)

Assuming periodic boundary condition on a cubic volume of size $L$

\begin{displaymath} \begin{array}{ll} {\mbox{\boldmath$u$}}(x + n L, y + m L, z ... ... x,y,z \in \mathbb{R},円 \forall n,m,q \in \mathbb{Z}\end{array}\end{displaymath} (1.11)

or null boundary condition on a volume $V$

\begin{displaymath} {\mbox{\boldmath$u$}} {\big\arrowvert}_{\partial V} = 0 \end{displaymath} (1.12)

the first two terms in the integral vanishes and using the identity

[画像:$\displaystyle (\nabla \times {\bf u}) \cdot (\nabla \times {\bf u})$] $\textstyle =$ [画像:$\displaystyle (\epsilon_{ijk} \partial_j u_k) (\epsilon_{ilm} \partial_l u_m)$]

$\textstyle =$ [画像:$\displaystyle \partial_j(u_k \partial_j u_k) - \partial_j \partial_k(u_j u_k) - u_k \partial_j \partial_j u_k$] (1.13)

one gets

[画像:\begin{displaymath} \frac {dE}{dt} = \nu \int d^3 r \rho {\mbox{\boldmath$u$}} \... ... - \nu \int d^3 r \rho \vert{\mbox{\boldmath$\omega$}}\vert^2 \end{displaymath}] (1.14)

where we have have introduced the vorticity of the fluid $ {\mbox{\boldmath$\omega$}} = \nabla \times {\mbox{\boldmath$u$}} $. Defining the total enstrophy as

[画像:\begin{displaymath} Z = \int d^3 r {1\over 2} \rho \vert\nabla \times {\mbox{\boldmath$u$}}\vert^2 \end{displaymath}] (1.15)

the energy balance reads:

[画像:\begin{displaymath} \frac {dE}{dt} = - 2 \nu Z \end{displaymath}] (1.16)

which shows that in absence of external forcing and for $\nu =0$ the kinetic energy is conserved by the dynamics, i.e. it is an inviscid invariant. On the contrary in the limit $\nu \to 0$ the energy dissipation rate does not vanish, but reaches a constant value [7]:

[画像:\begin{displaymath} \lim_{\nu \to 0} 2 \nu Z \equiv \epsilon \end{displaymath}] (1.17)

This phenomenon is known as dissipative anomaly, and implies that in the limit $\nu \to 0$ the total enstrophy must grow as $Z \sim \nu^{-1}$ to compensate the decreasing viscosity. The unbounded growth of enstrophy in three dimensions is the physical origin of the dissipative anomaly, and it is possible because of the vortex stretching, which produces diverging velocity gradient in the limit $Re \to \infty$.

next up previous contents
Next: Energy transfer Up: Navier-Stokes equation Previous: Reynolds number Contents

Stefano Musacchio 2004年01月09日