Why is turbulence dissipation $epsilon$ defined this way?

Question

When deriving the turbulence kinetic energy equation $k$ by taking the trace of Reynolds stress transport PDE, this term appears :
$$ 2nu overline{frac{partial u'_i}{partial x_k}frac{partial u'_j}{partial x_k}}$$
Where $nu$ is the kinematic viscosity and $u'$ is velocity fluctuations.
This term is turbulence kinetic energy dissipation rate $epsilon$, which represents the rate at which Kolmogorov eddies energy is converted back into the flow’s internal energy.
I understand the physic of energy cascade and turbulence dissipation, but I don't understand how did we relate the above definition to turbulence dissipation?

Daniel Hatton · Answer

The two factors under the Reynolds-averaging overbar (and inside the Einstein-convention-implied summation over $k$) are:

$partial!u_i/partial!x_k$, the (quotient by dynamic viscosity of the) $i$ component of shear force per area on surfaces of a fluid element whose normal is in the $k$-direction; and
$partial!u_j/partial!x_k$, the difference per $k$-direction size of the fluid element between the $j$ component of velocity on opposite faces of the fluid element with normals in the $pm k$-direction.

Taking the trace is equivalent to combining those component-by-component products of scalars into the scalar product of the force and velocity vectors.  So you end up with the scalar product of force per area with difference per length (perpendicular to the area) in velocity, which gives net rate of work done per volume.
(I should add that the reason for using the difference in velocity, rather than the sum of velocities, is that the shear forces on opposite faces are oppositely-directed.)

Why is turbulence dissipation $epsilon$ defined this way?

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