Need some intuition for laminar flow in different geometries

[Mangoes]

Hi guys,

I've been working through some notes on Transport Phenomena by Bird and I've basically been just developing velocity and shear stress profiles for various (simple) models by a differential momentum balance and I'm trying to understand why a certain result happens.

When looking at (steady-state, laminar) flow in the z direction in an annular region between two coaxial pipes of radii kR and R respectively (k<1) you get the following results:

(1) a parabolic velocity profile occurs with maximum velocity occurring at a plane r = λR - this isn't surprising when looking at previous models

(2) a shear stress profile occurs with zero shear stress at the plane r = λR. However, the direction of the shear stress changes direction when crossing this plane.

I've gone through the derivation so I get that the math shows the results (1) and (2), but I'm just not understanding any intuition/reason why the shear stress would change directions at the plane of maximum velocity unlike, say, laminar flow in a pipe. I've noticed that this also happens when two immiscible fluids flow along each other in laminar flow... and again, not really able to figure out why this happens. Can anyone provide some insight here?

 

[Chestermiller]

Suppose you have pressure-driven laminar flow between two infinite parallel plates. You encounter the same situation in that case too, correct?

 

[Mangoes]

Well, for the case of parallel infinite plates, I arrived to:

$$\frac {dv_x}{dy} = -\frac{1}{μ} \frac {dP}{dx}$$

This can be integrated twice and the constants of integration can be found by the no-slip condition at the top and bottom plates. I might be making a mistake, but I'm not getting that the shear stress changes sign for this geometry. I'm getting a parabolic velocity distribution and a shear that looks like |x| with x = 0 being the point in the middle of the plates in the fluid (where maximum velocity is) if that makes enough sense.

To clarify, here's a visual representation of what's happening in laminar steady flow for an annular region (from Bird):

https://pasteboard.co/GPfLkBI.png

 

[Chestermiller]

Well, for the case of parallel infinite plates, I arrived to:

$$\frac {dv_x}{dy} = -\frac{1}{μ} \frac {dP}{dx}$$

This equation is not quite correct. It should read:
$$\frac {dv_x}{dy} = -\frac{y}{μ} \frac {dP}{dx}$$

Note that dv/dy does change sign at y = 0.

My intuition is telling me that you are struggling with this (as most of us have also done) because BSL are not precise enough about how they specify the traction stresses on surfaces, and/or, equivalently, how they specify the momentum flux (resulting from pressure and viscous stresses) across surfaces surrounding a control-volume/shell. I can help you with this. To start you on a better way of looking at this, please look over Appendix A.3 of BSL, and get back with me when you have gotten some feel for it. (You don't have to follow every last detail)

 

[Mangoes]

This equation is not quite correct. It should read:
$$\frac {dv_x}{dy} = -\frac{y}{μ} \frac {dP}{dx}$$

Note that dv/dy does change sign at y = 0.

Sorry about that, I originally meant to write:

$$\frac{τ_{yx}}{dy} = -\frac{dp}{dx}$$

I just realized I changed my mind or something when I wrote the equation in my other post and the accompanying paragraph didn't make sense. That must have been really confusing, even I can't really see what I was trying to say in my other post, sorry about that, I think I needed a break. I had a second look and see what you meant now, the direction of the shear stress does indeed change signs through the plane y = 0. My point of confusion is with regards to why the magnitude of the shear stress is symmetric about y = 0, but not the direction.

My intuition is telling me that you are struggling with this (as most of us have also done) because BSL are not precise enough about how they specify the traction stresses on surfaces, and/or, equivalently, how they specify the momentum flux (resulting from pressure and viscous stresses) across surfaces surrounding a control-volume/shell. I can help you with this. To start you on a better way of looking at this, please look over Appendix A.3 of BSL, and get back with me when you have gotten some feel for it. (You don't have to follow every last detail)

Anyways, I went ahead and looked through the section on tensors in the appendix as you said.

 

[Chestermiller]

Sorry about that, I originally meant to write:

$$\frac{τ_{yx}}{dy} = -\frac{dp}{dx}$$

I just realized I changed my mind or something when I wrote the equation in my other post and the accompanying paragraph didn't make sense. That must have been really confusing, even I can't really see what I was trying to say in my other post, sorry about that, I think I needed a break. I had a second look and see what you meant now, the direction of the shear stress does indeed change signs through the plane y = 0. My point of confusion is with regards to why the magnitude of the shear stress is symmetric about y = 0, but not the direction.
Anyways, I went ahead and looked through the section on tensors in the appendix as you said.

Do you still feel like you would like me to cover what I was going to cover, regarding the automatic mathematical approach for determining the momentum flux through, and the traction vector on, a surface? Or would you prefer to wait to see whether you encounter any other issues?

 

[Mangoes]

Do you still feel like you would like me to cover what I was going to cover, regarding the automatic mathematical approach for determining the momentum flux through, and the traction vector on, a surface? Or would you prefer to wait to see whether you encounter any other issues?

Well, if I'm not misunderstanding you, I don't think my confusion really stems from misunderstanding tensors - I'm still just struggling to see why the tensors point in the directions they do in some situations. I don't know if there's been a misunderstanding, so I just want to reiterate what's confusing me visually.

This is the situation which I find extremely intuitive - the inside of a pipe:

https://pasteboard.co/GPiohQx.png

There's a parabolic velocity field and a symmetrical shear stress field that's always pointing towards +z regardless of the angle θ you're looking at the pipe wall with.

Now, going back to your example, there's still a parabolic velocity field, but now, the direction of the shear stress depends on whether you're looking at the fluid above or below the line y = 0. Something similar happens when you look at fluid flowing through an annular region... unlike fluid flowing in a pipe, the direction of the shear stress *does* depend on what angle θ you're using to look at the pipe wall with (https://pasteboard.co/GPfLkBI.png). I just find that really strange and was hoping that there was some molecular or qualitative explanation for why the shear stresses flip after the plane of maximum velocity in some cases.

 

[Chestermiller]

Well, if I'm not misunderstanding you, I don't think my confusion really stems from misunderstanding tensors - I'm still just struggling to see why the tensors point in the directions they do in some situations. I don't know if there's been a misunderstanding, so I just want to reiterate what's confusing me visually.

This is the situation which I find extremely intuitive - the inside of a pipe:

https://pasteboard.co/GPiohQx.png

There's a parabolic velocity field and a symmetrical shear stress field that's always pointing towards +z regardless of the angle θ you're looking at the pipe wall with.

Now, going back to your example, there's still a parabolic velocity field, but now, the direction of the shear stress depends on whether you're looking at the fluid above or below the line y = 0. Something similar happens when you look at fluid flowing through an annular region... unlike fluid flowing in a pipe, the direction of the shear stress *does* depend on what angle θ you're using to look at the pipe wall with (https://pasteboard.co/GPfLkBI.png). I just find that really strange and was hoping that there was some molecular or qualitative explanation for why the shear stresses flip after the plane of maximum velocity in some cases.

For flow in a pipe, if you convert to cartesian coordinates, on the plane x = 0, the $\tau_{yz}$ shear stress does change sign as one goes from y = -R to y = +R, and, on the plane y = 0, the $\tau_{xz}$ shear stress does change sign as one goes from x = -R to x = +R.

Also, in axial annular flow, the $\tau_{rz}$ shear stress does not depend on what angle $\theta$ you're using to look at the pipe wall; the flow is axisymmetric. But it does depend on r, and changes sign as you go from the inner (zero axial velocity) boundary to the outer (zero axial velocity) boundary (just like pressure driven flow between two infinite parallel plates).

 

[Chestermiller]

OK. Here is my explanation of how you can use the "molecular momentum flux tensor (as BSL call it), $\vec{\Pi}$, in solving viscous fluid flow problems. First of all, let me say that this can also be referred to as the "compressive stress tensor," and is the same entity as the (tensile) stress tensor of solid mechanics, except that it has the opposite sign. Also note that most other fluid mechanics texts use the tensile stress tensor.

Here's how it works: If dA represents a differential element of surface area within a deforming fluid and $\vec{n}$ represents a unit normal passing through dA from one side of dA to the other, the flux of momentum (i.e., per unit area) passing through dA from the side of dA in which the tail of $\vec{n}$ resides to the side of dA in which the nose of $\vec{n}$ resides is determined by the dot product of $\vec{\Pi}$ with $\vec{n}$: $$ \vec{t}=\vec{\Pi}\centerdot \vec{n}$$ An equivalent interpretation of this is that $\vec{t}$ is the (vector) force per unit area exerted by the fluid on the side of dA in which the tail of $\vec{n}$ resides on the fluid on the side of dA in which the nose of $\vec{n}$ resides. This relationship between the stress vector $\vec{t}$ acting on an area, the compressive stress tensor $\vec{\Pi}$, and the unit normal $\vec{n}$ is referred to as the Cauchy Stress Relationship. It is used to automatically and correctly generate the stress vector acting on an element of surface area within a fluid, with no complicated reasoning or thinking involved.

Here is an example: Suppose we have two infinite parallel plates separated by a distance h, with the lower plate stationary, and the upper plate moving in the x direction with velocity V. This is called "simple shear flow," and the fluid velocity profile within the gap between the plates is given by:
$$v_x=V\frac{y}{h}$$
The compressive stress tensor in this system is given by: $$\vec{\Pi}=P(\vec{i_x}\vec{i_x}+\vec{i_y}\vec{i_y}+\vec{i_z}\vec{i_z})+\tau_{xy}(\vec{i_x}\vec{i_y}+\vec{i_y}\vec{i_x})$$where P is the fluid pressure, $\tau_{xy}$ is the xy shear stress on a plane of constant y in the x direction (and also a plane of constant x in the y direction), and the i's are the ordinary unit vectors. For this flow, $$\tau_{xy}=-\mu\frac{dV}{dy}=-\mu\frac{V}{h}$$
Now, suppose we focus on the slab of fluid situated between the horizontal planes y and $y+\Delta y$. Suppose we wish to determine the stress vector exerted by the fluid above the slab on the fluid inside the slab. What we do to get this is dot the compressive stress tensor with a unit normal directed from above the slab, into the slab. This is just $(-\vec{i_y})$ (i.e., a unit vector in the negative y direction). So we have: $$\vec{t}=\vec{\Pi}\centerdot (-\vec{i_y})=P(\vec{i_x}\vec{i_x}+\vec{i_y}\vec{i_y}+\vec{i_z}\vec{i_z})\centerdot (-\vec{i_y})+ \tau_{xy}(\vec{i_x}\vec{i_y}+\vec{i_y}\vec{i_x})\centerdot (-\vec{i_y})=-P\vec{i_y}-\tau_{xy}\vec{i_x}$$So, we find that $$\vec{t}=-P\vec{i_y}+\mu\frac{V}{h}\vec{i_x}$$

This is the force per unit area exerted by the fluid immediately above the plane $y+\Delta y$ on the material immediately below the plane. Since the fluid above is moving faster than that below, the viscous drag is in the positive x direction. Another equivalent interpretation is that this is the momentum flux flowing downward from the fluid above, through the plane $y+\Delta y$, into the fluid within our slab.

Now it's your turn. Please calculate the stress vector exerted by the fluid immediately below the plane at y on the fluid immediately above this plane? (This is the same as the flux of momentum flowing through the plane from the fluid below the plane at y to the fluid in our slab immediately above the plane).

 

[Mangoes]

Now it's your turn. Please calculate the stress vector exerted by the fluid immediately below the plane at y on the fluid immediately above this plane? (This is the same as the flux of momentum flowing through the plane from the fluid below the plane at y to the fluid in our slab immediately above the plane).

Hey, thank you for taking the time to write that.

So, for the stress vector exerted by the fluid immediately below the plane at y on the fluid immediately above the plane, the normal vector is now $\vec{i_y}$, so:

$$\vec{t} = \vec{\Pi}\centerdot \vec{i_y} = P\vec{i_y} + \tau_{xy}\vec{i_x} = P\vec{i_y}-\mu\frac{V}{h}\vec{i_x}$$

Letting Δy→0, seems like a rehash of Newton's third law. That aside though, if I'm interpreting correctly, I'm seeing that the direction of the shear stress on a plane is dependent on which side of the plane is going faster, which is mathematically described by the velocity gradient. If this is correct, this is more or less the physical intuition I was looking for. Seems pretty obvious in hindsight, but I might be oversimplifying it or looking at it from the wrong perspective.

For flow in a pipe, if you convert to cartesian coordinates, on the plane x = 0, the $\tau_{yz}$ shear stress does change sign as one goes from y = -R to y = +R, and, on the plane y = 0, the $\tau_{xz}$ shear stress does change sign as one goes from x = -R to x = +R.

This was especially clarifying for my confusion between the two systems, thank you.

 

[Chestermiller]

Hey, thank you for taking the time to write that.

So, for the stress vector exerted by the fluid immediately below the plane at y on the fluid immediately above the plane, the normal vector is now $\vec{i_y}$, so:

$$\vec{t} = \vec{\Pi}\centerdot \vec{i_y} = P\vec{i_y} + \tau_{xy}\vec{i_x} = P\vec{i_y}-\mu\frac{V}{h}\vec{i_x}$$

Letting Δy→0, seems like a rehash of Newton's third law. That aside though, if I'm interpreting correctly, I'm seeing that the direction of the shear stress on a plane is dependent on which side of the plane is going faster, which is mathematically described by the velocity gradient. If this is correct, this is more or less the physical intuition I was looking for. Seems pretty obvious in hindsight, but I might be oversimplifying it or looking at it from the wrong perspective.
This was especially clarifying for my confusion between the two systems, thank you.

I'm thrilled that I was able to help you. The material we covered here will benefit you greatly in your continued work in fluid mechanics.