I came across some literature containing plots of k+ versus distance-to-the-wall
(for example, http://web.stanford.edu/class/me469b/handouts/turbulence.pdf
But I was not able to find a mathematical expression for k+ (normalized turbulent kinetic energy, I guess).
Could anyone have a clear idea about it?
You can find the mathematical expression in the attached file
Thanks for the reply. But I doubt in the attached doc file, if k^+ = k / friction_velocity is correct. k and friction_velocity have different physical units.
You are right , k^+ = k / (friction_velocity)^2 ,and you can find more details in the attached file
The kinetic energy KE of a turbulent flow is a conserved quantity in the inviscid force-free limit. This observation is at the heart of the description of turbulence through an energy cascade from large to small scales even if in real stationary turbulent flows, kinetic energy KE(t) is only conserved in average. This quantity therefore appears as a natural global observable to study turbulence. However, without proper normalization, E does depend on the Reynolds number or on the forcing conditions(viz. vessel shape, injection device,etc.) In numerical simulations, the kinetic energy is therefore classically normalized using the typical velocity which appears in the Reynolds number definition. So, we rather adopt a global normalization, through the kinetic energy of the time-averaged flow, to define our normalized global kinetic energy as KE/u2
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