Significance of the viscous stress tensor

[dRic2]

Hi,

when working with NS equations the stress tensor can be written as $\nabla \tau = - \nabla P + \nabla \tau_{v}$, where $\tau_{v}$ is
\begin{pmatrix}
\tau_{xx} & \tau_{xy} & \tau_{xz} \\
\tau_{xy} & \tau_{yy} & \tau_{yz} \\
\tau_{zx} & \tau_{zy} & \tau_{zz}
\end{pmatrix}

This question was stuck in my mind since last years, but I used to forget to ask: what is the physical interpretation of $\tau_{xx}$, $\tau_{yy}$, $\tau_{zz}$? I know where they come from "mathematically", but I don't get the difference with pressure... I thought viscous stresses could only be like $\tau_{ij}$ with $i≠j$.

PS: Out of the blue: In a very turbulent flow is it ok to consider $\tau_{ij} = 0$ with $i≠j$?

Thanks
Ric

 

[Chestermiller]

Hi,

when working with NS equations the stress tensor can be written as $\nabla \tau = - \nabla P + \nabla \tau_{v}$, where $\tau_{v}$ is
\begin{pmatrix}
\tau_{xx} & \tau_{xy} & \tau_{xz} \\
\tau_{xy} & \tau_{yy} & \tau_{yz} \\
\tau_{zx} & \tau_{zy} & \tau_{zz}
\end{pmatrix}

This question was stuck in my mind since last years, but I used to forget to ask: what is the physical interpretation of $\tau_{xx}$, $\tau_{yy}$, $\tau_{zz}$? I know where they come from "mathematically", but I don't get the difference with pressure... I thought viscous stresses could only be like $\tau_{ij}$ with $i≠j$.

Why do you wonder about the difference with pressure? Regarding the i,j thing, why did you think this?

PS: Out of the blue: In a very turbulent flow is it ok to consider $\tau_{ij} = 0$ with $i≠j$?

Thanks
Ric

In turbulent flow, the time averaged values of the viscous stresses are definitely not equal to zero.

 

[dRic2]

Why do you wonder about the difference with pressure? Regarding the i,j thing, why did you think this?

$\tau_xx$ (for example) is a normal vector respect to $zy$ surface. I thought viscous stress can only be tangent vectors (like $\tau_{xy}$) because of Newton's law $\tau = -\mu \frac {\partial v_x} {\partial y}$.

 

[dRic2]

In turbulent flow, the time averaged values of the viscous stresses are definitely not equal to zero.

Sorry, I'm so dumb...

I don't know how I came up this...

What I really meant is: "In a turbulent flow is it ok to consider the fluctuations of $\tau$ to be zero? I mean if I average NS equations I get $\tau = \tau_{v} + \tau_{t}$ where $\tau_{t}$ is:

\begin{pmatrix}
\bar v'^2_{xx} & \bar {v'_xv'_y} & \bar {v'_xv'_z} \\
\bar v'^2_{yx} & \bar {v'^2_y} & \bar {v'_yv'_z} \\
\bar v'^2_{zx} & \bar {v'_zv'_y} & \bar {v'^2_z}
\end{pmatrix}

Is it ok to do this:

\begin{pmatrix}
\bar v'^2_{xx} & 0& 0\\
0 & \bar {v'^2_y} &0 \\
0& 0 & \bar {v'^2_z}
\end{pmatrix}

?

 

[Chestermiller]

$\tau_xx$ (for example) is a normal vector respect to $zy$ surface. I thought viscous stress can only be tangent vectors (like $\tau_{xy}$) because of Newton's law $\tau = -\mu \frac {\partial v_x} {\partial y}$.

That's only the simplified version for a specific type of flow situation. The 3D version of the law is what is required so that law is independent of the coordinate system used by the observer. You are aware that, even for the state of stress you have indicated, if the coordinate axes were rotated, the components of the stress tensor in the new coordinate system will contain normal stresses, correct?

 

[Chestermiller]

Sorry, I'm so dumb...

I don't know how I came up this...

What I really meant is: "In a turbulent flow is it ok to consider the fluctuations of $\tau$ to be zero? I mean if I average NS equations I get $\tau = \tau_{v} + \tau_{t}$ where $\tau_{t}$ is:

\begin{pmatrix}
\bar v'^2_{xx} & \bar {v'_xv'_y} & \bar {v'_xv'_z} \\
\bar v'^2_{yx} & \bar {v'^2_y} & \bar {v'_yv'_z} \\
\bar v'^2_{zx} & \bar {v'_zv'_y} & \bar {v'^2_z}
\end{pmatrix}

Is it ok to do this:

\begin{pmatrix}
\bar v'^2_{xx} & 0& 0\\
0 & \bar {v'^2_y} &0 \\
0& 0 & \bar {v'^2_z}
\end{pmatrix}

?

Sure. These are called the turbulent stresses.

 

[dRic2]

Thanks for the quick replies, I'm in kind of a hurry now. I'll be back later/tomorrow! :)

 

[dRic2]

You are aware that, even for the state of stress you have indicated, if the coordinate axes were rotated, the components of the stress tensor in the new coordinate system will contain normal stresses, correct?

Do you mean something like this?

I think it is clear. But it is different: here you "create" the normal components by rotating the axis. Consider a chunk of fluid flowing in rectangular pipe (so we won't bother changing coordinates). Let's set the origin of the axis in one of the corner of the pipe (no rotation). What is the meaning of $\tau_{ii}$?

Ps: I remember the Cauchy stress tensor for a rigid (static) body has to be symmetric because of conservation of momentum. Does this apply also to viscous stress tensor?

 

[dRic2]

Sure. These are called the turbulent stresses.

Why $\bar {v'_iv'_j} = 0$ is reasonable ?

 

[Chestermiller]

Why $\bar {v'_iv'_j} = 0$ is reasonable ?

It's not, and the average of the product of these fluctuations isn't zero.

 

[Chestermiller]

Do you mean something like this?

I think it is clear. But it is different: here you "create" the normal components by rotating the axis. Consider a chunk of fluid flowing in rectangular pipe (so we won't bother changing coordinates). Let's set the origin of the axis in one of the corner of the pipe (no rotation). What is the meaning of $\tau_{ii}$?

For this flow, it is zero. If you want to see the full derivation showing why the normal components can be non-zero, see chapters 1 and 2 in Transport Phenomena by Bird, Stewart, and Lightfoot

Ps: I remember the Cauchy stress tensor for a rigid (static) body has to be symmetric because of conservation of momentum. Does this apply also to viscous stress tensor?

Yes.

 

[dRic2]

For this flow, it is zero. If you want to see the full derivation showing why the normal components can be non-zero, see chapters 1 and 2 in Transport Phenomena by Bird, Stewart, and Lightfoot

Things are coming together in my mind, thank you!

 

[dRic2]

It's not, and the average of the product of these fluctuations isn't zero.

I might have misunderstood your previous answer:

Sure. These are called the turbulent stresses.

I thought it was referred to this:

Sorry, I'm so dumb...

I don't know how I came up this...

What I really meant is: "In a turbulent flow is it ok to consider the fluctuations of $\tau$ to be zero? I mean if I average NS equations I get $\tau = \tau_{v} + \tau_{t}$ where $\tau_{t}$ is:

\begin{pmatrix}
\bar v'^2_{xx} & \bar {v'_xv'_y} & \bar {v'_xv'_z} \\
\bar v'^2_{yx} & \bar {v'^2_y} & \bar {v'_yv'_z} \\
\bar v'^2_{zx} & \bar {v'_zv'_y} & \bar {v'^2_z}
\end{pmatrix}

Is it ok to do this:

\begin{pmatrix}
\bar v'^2_{xx} & 0& 0\\
0 & \bar {v'^2_y} &0 \\
0& 0 & \bar {v'^2_z}
\end{pmatrix}

?

 

[Chestermiller]

I might have misunderstood your previous answer:
I thought it was referred to this:

What I meant was that all the elements of the turbulent stress tensor can be non-zero, including the shear components.

 

[dRic2]

Sorry, I'm so dumb...

Is it ok to do this:

\begin{pmatrix}
\bar v'^2_{xx} & 0& 0\\
0 & \bar {v'^2_y} &0 \\
0& 0 & \bar {v'^2_z}
\end{pmatrix}

?

So this is generally false, right?

 

[Chestermiller]

So this is generally false, right?

Yes.