I have the solution of a Navier-Stokes simulation with an incompressible, Newtonian fluid with laminar flow. Now I compute the wall shear stress (vector) as $$\tau_n = \mu (\nabla u) n,$$ where $\mu$ is the dynamic viscosity, $n$ the normal vector to the wall (pointing into the fluid) and $\nabla u$ the Jacobian matrix.

How is the scalar wall shear stress defined? I have seen two definitions:

  • $||\tau_n||_2$
  • $\langle t,\tau_n\rangle$, where $t$ is the direction of the fluid near the wall, i.e., $t=u/||u||_2$ (due to a no-slip condition this cannot be measured at the wall).

The first definition seems more natural to me. However, in my simulation both values are almost identical (minus numerical errors). It seems that $$\frac{(\nabla u)n}{||(\nabla u)n||_2}=\frac{u}{||u||_2}.$$ Is this always the case?

This would also answer my question: Does the direction of $\tau_n$ equal the direction of $u$ (near the wall)? If so, can we prove it?

Pure maths doesn't prove the equality (in case it can be proven). So either it doesn't hold or I need to use physical properties.

Due to the laminar flow, we have $n^T u=0$ (and thus $(\nabla u)^T n=0$) and due to incompressibility $div(u)=0$. I wasn't able to prove or disprove the equality using these properties.

Note that my knowledge about physics is very limited (to school knowledge).


Yikes. You're very confused.

Your first equation, which is incorrect, is trying to calculate the viscous stress vector (not shear stress vector) at the wall. In general, the viscous stress vector will have a component normal to the wall and a component tangent to the wall. The component tangent to the wall is called the shear stress. The correct equation for calculating the viscous stress vector exerted by the wall on the fluid is: $$\vec{\tau}=\mu \left[{\vec{\nabla} \vec{u}+(\vec{\nabla} \vec{u})^T}\right]\cdot \vec{n}$$where $\vec{n}$ is the unit normal vector drawn from the fluid through the wall. (This uses the sign convention that tensile stresses are positive and compressive stresses are negative)

The normal component of the viscous stress vector at the wall is obtained by dotting the stress vector $\vec{\tau}$ with the unit normal $\vec{n}$:$$\tau_n=\vec{\tau}\cdot\vec{n}$$ The shear component of the viscous stress vector at the wall is obtained by dotting the stress vector $\vec{\tau}$ with the unit tangent to the wall $\vec{t}$:$$\tau_t=\vec{\tau}\cdot\vec{t}$$So you can also write that:$$\vec{\tau}=\tau_n\vec{n}+\tau_t\vec{t}$$ The unit tangent vector is oriented parallel to the fluid velocity in close proximity to the wall. More generally, the shear stress vector at the wall, including direction, can be obtained by subtracting the normal component of the viscous stress vector from the total viscous stress vector:

shear stress vector $=\vec{\tau}-\tau_n\vec{n}=\vec{\tau}-(\vec{\tau}\cdot\vec{n})\vec{n} $

Hope this helps.

ADDENDUM For the case of $\vec{u}=u_x\vec{i}_x+u_y\vec{i}_y+u_z\vec{i}_z$ and $\vec{n}=\vec{i}_y$, I get the following at the wall: $$(\vec{\nabla}\vec{u})^T\cdot \vec{n}=\frac{\partial v_y}{\partial y}\vec{i_y}$$ $$(\vec{\nabla}\vec{u})\cdot \vec{n}=\frac{\partial v_x}{\partial y}\vec{i}_x+\frac{\partial v_z}{\partial y}\vec{i}_z+\frac{\partial v_y}{\partial y}\vec{i}_y$$

$$\vec{\tau}=\mu\left(\frac{\partial v_x}{\partial y}\vec{i}_x+\frac{\partial v_z}{\partial y}\vec{i}_z+2\frac{\partial v_y}{\partial y}\vec{i}_y\right)$$

I can see now what you're saying about the shear stress terms in $\vec{\tau}$, but there is also a normal stress component too. But you are correct that the shear stress component is parallel to the velocity vector near the wall.

  • $\begingroup$ Concerning the definition of the viscous stress vector: Later in my post I had the equation $(\nabla u)^T n=0$ (due to laminar flow). This would imply that both definitions are the same. This would also imply that the normal part of the viscous stress vector is zero. This is how I got "my" definition of the wall shear stress vector. Concerning the question if the velocity points in the same direction as the wall shear stress vector: How do you prove that? Let $t_2=t \times n$, then is $\tau\cdot t_2=0$? $\endgroup$ – nx105 Apr 17 '16 at 19:04
  • $\begingroup$ I wrote things out in component form for the case where $\vec{v}=v_x\vec{i}_x+v_y\vec{i}_y$ and the wall is perpendicular to $\vec{i}_y$ (i.e., $\vec{n}=\vec{i}_y$) and didn't get $(\vec{\nabla}\vec{v})^T\cdot \vec{n}=0$ $\endgroup$ – Chet Miller Apr 17 '16 at 20:21
  • $\begingroup$ I'm going to add to my response what I get in component form. Then maybe we can discuss further. $\endgroup$ – Chet Miller Apr 17 '16 at 21:48

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