Good afternoon everyone. In defining the shear velocity at the wall $u_\tau=\sqrt{\frac{\tau_w}{\rho}}$ for the incompressible turbulence, $w$ being the wall, I found in some literature that this relation holds: $$\tau_w=\mu \frac{\partial U}{\partial y}|_{y=0}= -\frac{\partial p_w}{\partial x}$$ $p$ being the pressure, $\mu$ the dynamic viscosity and $x$ the direction of the flow velocity $U$. However I don't know neither if the second equality is correct or where it comes from. Could anyone help?


1 Answer 1


I'm trying to show the second equality, what I've done so far:

From Navier-Stokes (incompressible) in the x direction (take z also along the wall and y orthogonal to them, assuming no field forces) and let $\mathbf{u} = u \mathbf{i} + v \mathbf{j} + w \mathbf{k}$:

$$\rho \frac{D u}{Dt} = \mu \Delta u - \frac{\partial p}{\partial x} + \rho g_x$$

notice that the particle is stopped (zero constant velocity and partial derivatives of it along x and z) along the wall ("no slip") and therefore, from the Lagrangian frame, it has zero (material) derivative. It will reduce to:

$$\mu \frac{\partial^2 u}{\partial y^2} = \frac{\partial p}{\partial x}$$

but I don't know what to do from here. Maybe the sign will be reversed since the shear stress in the fluid is a reaction to the shear stress in the wall?

Edit: checking the dimensions, could it happen that something is wrong with the equality? Since the first two terms evaluate to Pa, and the final one to Pa/m.

Using the reasoning above, could it be something like $$\frac{\partial \tau_w}{\partial y} = -\frac{\partial p_w}{\partial x}$$?

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    $\begingroup$ Thanks for the reply. Yes, the minus could come from the fact that there is a pressure drop in the flow due to friction while the stress is positive; however have you thought how to solve the fact of the second derivative? About the dimensions, I think you're true and this formula can't be correct then. $\endgroup$ Dec 31, 2020 at 15:33
  • $\begingroup$ I wrote a small edit few secs ago, maybe it is related to the derivative of shear stress $\endgroup$
    – Petrini
    Dec 31, 2020 at 15:34
  • $\begingroup$ Got it. Yes, it sounds reasonable. Thank you $\endgroup$ Dec 31, 2020 at 15:35

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