Relation between Wall shear stress and rate of strain tensor

samiahmed · May 2, 2020, 13:47

In cfd-online, we can find that the wall shear stress is defined by:

$\tau_w = \mu \left(\frac{\partial u}{\partial y} \right)_{y=0}$

But why it depends only on the u and y? AFAIK the rate of strain is a symmetric tensor (has 6 components).

How can one get the expression of the wall shear stress from the rate of strain tensor?

Thank you

FMDenaro · May 2, 2020, 14:02

Quote:

Originally Posted by samiahmed

In cfd-online, we can find that the wall shear stress is defined by:

$\tau_w = \mu \left(\frac{\partial u}{\partial y} \right)_{y=0}$

But why it depends only on the u and y? AFAIK the rate of strain is a symmetric tensor (has 6 components).

How can one get the expression of the wall shear stress from the rate of strain tensor?

Thank you

On a wall the velocity components vanish owing to the no-slip condition, thus write your tensors components on the wall and see ...

samiahmed · May 2, 2020, 14:57

Thank you for your reply, My goal is to get the expression of the wall shear stress from viscous stress tensor (details below), but I am stuck in the final steps. I wonder why the terms: ${\partial v\over \partial x}, {\partial u\over \partial x}$ , and ${\partial v\over \partial y}\$ vanish at the wall?

Here is what I've done so far:

Let's consider a 2D incompressible laminar flow, of a newtonian fluid;

The viscous stress tensor is given by:

$\tau_{ij} = 2\mu D_{ij} \tag{1}$

Where $D_{ij}$ is the rate of strain and is given by:

$D_{ij} = {1\over 2} \left({\partial u_i\over \partial x_j} + {\partial u_j\over \partial x_i}\right) \tag{2}$

The viscous stress vector at a surface with a normal $\mathbf n = (n_1, n_2)^T$ is given by:

$\mathbf T = \mathbf\tau\cdot\mathbf n$

Or:
$T_i = \tau_{ij}n_j$

In other words:

$\begin{pmatrix} T_1 \\ T_2 \end{pmatrix} = \begin{pmatrix} \tau_{11} & \tau_{12} \\ \tau_{21} & \tau_{22} \end{pmatrix} \begin{pmatrix} n_1\\ n_2 \end{pmatrix} = \begin{pmatrix} \tau_{11}n_1 + \tau_{12}n_2\\ \tau_{21}n_1 + \tau_{22}n_2 \end{pmatrix} = \begin{pmatrix} 2\mu{\partial u\over\partial x}n_1 + \mu\left({\partial u\over\partial y} + {\partial v \over\partial x }\right)n_2\\ \mu\left({\partial u\over\partial y} + {\partial v \over\partial x }\right)n_1+ 2\mu{\partial v\over\partial y}n_2 \end{pmatrix}$

Now, I am stuck, How to proceed further to get the wall shear stress expression, could you please help me?

duri · May 2, 2020, 15:45

Quote:

Originally Posted by samiahmed

But why it depends only on the u and y? AFAIK the rate of strain is a symmetric tensor (has 6 components).

Yes it depend on all the strain components and it is quite difficult to derive than to calculate a generic formula. The formula you mentioned is for 2D flow with wall parallel to y-axis. It indicates derivative of resultant tangential velocity normal to the wall.

duri · May 2, 2020, 15:54

Quote:

Originally Posted by samiahmed

Thank you for your reply, My goal is to get the expression of the wall shear stress from viscous stress tensor (details below), but I am stuck in the final steps. I wonder why the terms: ${\partial v\over \partial x}, {\partial u\over \partial x}$ , and ${\partial v\over \partial y}\$ vanish at the wall?

Here is what I've done so far:

Let's consider a 2D incompressible laminar flow, of a newtonian fluid;

The viscous stress tensor is given by:

$\tau_{ij} = 2\mu D_{ij} \tag{1}$

Where $D_{ij}$ is the rate of strain and is given by:

$D_{ij} = {1\over 2} \left({\partial u_i\over \partial x_j} + {\partial u_j\over \partial x_i}\right) \tag{2}$

The viscous stress vector at a surface with a normal $\mathbf n = (n_1, n_2)^T$ is given by:

$\mathbf T = \mathbf\tau\cdot\mathbf n$

Or:
$T_i = \tau_{ij}n_j$

In other words:

$\begin{pmatrix} T_1 \\ T_2 \end{pmatrix} = \begin{pmatrix} \tau_{11} & \tau_{12} \\ \tau_{21} & \tau_{22} \end{pmatrix} \begin{pmatrix} n_1\\ n_2 \end{pmatrix} = \begin{pmatrix} \tau_{11}n_1 + \tau_{12}n_2\\ \tau_{21}n_1 + \tau_{22}n_2 \end{pmatrix} = \begin{pmatrix} 2\mu{\partial u\over\partial x}n_1 + \mu\left({\partial u\over\partial y} + {\partial v \over\partial x }\right)n_2\\ \mu\left({\partial u\over\partial y} + {\partial v \over\partial x }\right)n_1+ 2\mu{\partial v\over\partial y}n_2 \end{pmatrix}$

Now, I am stuck, How to proceed further to get the wall shear stress expression, could you please help me?

Once you got force components take only the tangential force which acts on the wall. Derivation of this is difficult for generic wall orientation in 3D. For 2D it is Fx*ny - Fy*nx. Also apply u.n=0 (wall normal velocity is zero).

FMDenaro · May 2, 2020, 16:03

For incompressible 2d flows you have the divergence-free constraint that, written on the wall reduces to dv/dy=0

samiahmed · May 2, 2020, 16:25

Quote:

Originally Posted by FMDenaro

For incompressible 2d flows you have the divergence-free constraint that, written on the wall reduces to dv/dy=0

Why is that true? how can I get that result? because I guess you are assuming ${\partial u\over\partial x}\vert_{\text{wall}} = 0$ but why?

FMDenaro · May 2, 2020, 16:31

Quote:

Originally Posted by samiahmed

Why is that true? how can I get that result? because I guess you are assuming ${\partial u\over\partial x}\vert_{\text{wall}} = 0$ but why?

because you assume that x is the direction of the wall and the tangential velocity component is zero everywhere along it

samiahmed · May 2, 2020, 16:46

Quote:

Originally Posted by FMDenaro

because you assume that x is the direction of the wall and the tangential velocity component is zero everywhere along it

But that's not necessarily true for a general curved wall (e.g airfoil).

FMDenaro · May 2, 2020, 16:48

Quote:

Originally Posted by samiahmed

But that's not necessarily true for a general curved wall (e.g airfoil).

it is still true if you write the expression in curvilinear coordinates, i.e., normal and tangential directions

samiahmed · May 2, 2020, 17:19

Quote:

Originally Posted by FMDenaro

it is still true if you write the expression in curvilinear coordinates, i.e., normal and tangential directions

Thank you. Now, I think it makes sense:
since we chose to write the equations in curvilinear coordinates, we have:

$\mathbf n \equiv \mathbf j \implies (n_1, n_2)^T \equiv (0, 1)^T$
hence:

$T_w = \begin{pmatrix} \mu\left({\partial u\over\partial y}\right)_w \\ 0 \end{pmatrix}$

And since the wall shear stress is the tangential component of the viscous stress vector at the wall, we get:

$\tau_w = \mu\left({\partial u\over\partial y}\right)_w$

May 2, 2020, 13:47	Relation between Wall shear stress and rate of strain tensor	#1
samiahmed New Member Join Date: Jun 2019 Posts: 15 Rep Power: 6	In cfd-online, we can find that the wall shear stress is defined by: $\tau_w = \mu \left(\frac{\partial u}{\partial y} \right)_{y=0}$ But why it depends only on the u and y? AFAIK the rate of strain is a symmetric tensor (has 6 components). How can one get the expression of the wall shear stress from the rate of strain tensor? Thank you

May 2, 2020, 16:03		#6
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,771 Rep Power: 71	For incompressible 2d flows you have the divergence-free constraint that, written on the wall reduces to dv/dy=0 samiahmed likes this.

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May 2, 2020, 14:57		#3
samiahmed New Member Join Date: Jun 2019 Posts: 15 Rep Power: 6	Thank you for your reply, My goal is to get the expression of the wall shear stress from viscous stress tensor (details below), but I am stuck in the final steps. I wonder why the terms: ${\partial v\over \partial x}, {\partial u\over \partial x}$ , and ${\partial v\over \partial y}\$ vanish at the wall? Here is what I've done so far: Let's consider a 2D incompressible laminar flow, of a newtonian fluid; The viscous stress tensor is given by: $\tau_{ij} = 2\mu D_{ij} \tag{1}$ Where $D_{ij}$ is the rate of strain and is given by: $D_{ij} = {1\over 2} \left({\partial u_i\over \partial x_j} + {\partial u_j\over \partial x_i}\right) \tag{2}$ The viscous stress vector at a surface with a normal $\mathbf n = (n_1, n_2)^T$ is given by: $\mathbf T = \mathbf\tau\cdot\mathbf n$ Or: $T_i = \tau_{ij}n_j$ In other words: $\begin{pmatrix} T_1 \\ T_2 \end{pmatrix} = \begin{pmatrix} \tau_{11} & \tau_{12} \\ \tau_{21} & \tau_{22} \end{pmatrix} \begin{pmatrix} n_1\\ n_2 \end{pmatrix} = \begin{pmatrix} \tau_{11}n_1 + \tau_{12}n_2\\ \tau_{21}n_1 + \tau_{22}n_2 \end{pmatrix} = \begin{pmatrix} 2\mu{\partial u\over\partial x}n_1 + \mu\left({\partial u\over\partial y} + {\partial v \over\partial x }\right)n_2\\ \mu\left({\partial u\over\partial y} + {\partial v \over\partial x }\right)n_1+ 2\mu{\partial v\over\partial y}n_2 \end{pmatrix}$ Now, I am stuck, How to proceed further to get the wall shear stress expression, could you please help me?