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I have been studying the basics of CFD from a book titled 'An Introduction to Computational Fluid Dynamics' by H.K. Versteeg. In the turbulence modelling section, the author shows how the production of turbulent kinetic energy is maximum in the lowest layers of a turbulent boundary layer which happens to be right next to the wall. The figure shows how the maximum value of u' occurs when y/delta is minimum.

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The question I have is, doesn't this oppose the Kolmogorov Length Scale theory? Since near the wall we should have the smallest eddies, instead of maximum turbulent kinetic energy production, shouldn't the kinetic energy dissipation be maximum?

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Near the wall, the various Kolmogorov hypotheses do not apply. Specifically, Kolmogorov's hypotheses only apply for high Reynolds number, homogeneous flows and right up against the wall, neither of those two things is true -- the Reynolds number locally is quite low because it is moving very slowly, and the flow is highly non-homogeneous due to the friction from the wall. Although the eddies are small there, they aren't really turbulent eddies because there is no range of scales -- all scales are at the smallest scales.

Instead, to understand why turbulent production is largest at the wall, recall the term responsible for production of turbulent kinetic energy in the T.K.E. governing equation:

$$ \text{Prod.} = - \overline{u_i^\prime u_j^\prime} \frac{\partial \overline{u_i}}{\partial x_j} $$

So, the production of turbulent kinetic energy is proportional to the gradient of the mean velocity. And where is the gradient of the mean velocity largest in a boundary layer? Right against the wall!

To put everything together, right against the wall there is little to no turbulence (it depends on how close "right against" really means of course), but there are strong gradients in the mean velocities. So the turbulent kinetic energy is small, but it is being produced quite strongly. There is little dissipation of turbulent kinetic energy, because there is not much turbulent kinetic energy there yet -- however, there is strong dissipation of resolved kinetic energy, because that is where the mean velocity is driven to zero.

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  • $\begingroup$ I congratulate you with an, in my view, excellent answer! $\endgroup$
    – Bernhard
    Nov 27, 2020 at 22:50
  • $\begingroup$ Thank you for this answer! Cleared up quite a messy theory in my head! $\endgroup$ Nov 28, 2020 at 8:04

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