Shear Stress and Fluid Mechanics

[bliz]

Homework Statement

It's a relatively simple problem I'm sure, but I'm a bit confused on how exactly to go about it:

There are three parallel plates with water (viscosity of 0.8807 cp @ 30°C) between plates 1 and 2 (plate order of 1 on bottom, 2 in middle, 3 on top), and toluene (viscosity of 0.5179 cp @ 30°C) between plates 2 and 3. The distance between each plate pair is 10 cm and plate 3 (top plate) moves at 3 m/s while plate 1 (bottom plate) is at rest.

a) I need to find the velocity of plate 2 at steady-state
b) And the F/A (shear stress) on plate 3 that is needed to maintain the 3 m/s velocity

Homework Equations

Please excuse any formatting issues you may come across. I am new to forums and still learning how to properly format questions and comments.

Shear stress = $\tau$_yx = -$\mu$*v_x/y where $\mu$ = viscosity and v_x = velocity in the positive x direction (the same direction plate 3 is moving in)

$\tau$_yx = F/A

The Attempt at a Solution

I was a bit unsure initially of how to go about the problem but my attempt was to solve for the shear stress between plates 1 & 2 and plates 2 & 3, and then equate them to get the velocity of the middle plate.

So after multiplying dy by both sides and integrating (y from 0 to 10, and v_x from 0 to v₂ for plates 1 & 2 and from v₂ to 3 for plates 2 & 3), I got the equations:
Plates 1 & 2: $\tau$_yx = -0.08007*v₂
Plates 2 & 3: $\tau$_yx = -0.15937 + 0.5179*v₂
Equating these gave me the velocity of plate 2 to be 1.178 m/s

For the 2^nd part of the question (solving for F/A), I think you only need to solve for shear stress as that equals F/A. So using v₂, I plugged it into one of the equations to get $\tau$_yx = F/A = -0.094322 N.


The problem I have with my methods is that I'm not too sure about equating the two shear stresses as I see no real reason that they should be equal in the first place. Also, the 2^nd part of the question specifically asks about the F/A on plate 3, which really makes me think that the shear stress must be different for each plate. While my first answer seems to make sense, I don't think a negative force value for the second one fits, so I would assume that I'm doing something wrong there. Thanks in advance for any help, the hardest part I have with this stuff is that it's really confusing to me what actually needs to be done and which equations to use to get there.

 

[Chestermiller]

Homework Statement

It's a relatively simple problem I'm sure, but I'm a bit confused on how exactly to go about it:

There are three parallel plates with water (viscosity of 0.8807 cp @ 30°C) between plates 1 and 2 (plate order of 1 on bottom, 2 in middle, 3 on top), and toluene (viscosity of 0.5179 cp @ 30°C) between plates 2 and 3. The distance between each plate pair is 10 cm and plate 3 (top plate) moves at 3 m/s while plate 1 (bottom plate) is at rest.

a) I need to find the velocity of plate 2 at steady-state
b) And the F/A (shear stress) on plate 3 that is needed to maintain the 3 m/s velocity

Homework Equations

Please excuse any formatting issues you may come across. I am new to forums and still learning how to properly format questions and comments.

Shear stress = $\tau$_yx = -$\mu$*v_x/y where $\mu$ = viscosity and v_x = velocity in the positive x direction (the same direction plate 3 is moving in)

$\tau$_yx = F/A

The Attempt at a Solution

I was a bit unsure initially of how to go about the problem but my attempt was to solve for the shear stress between plates 1 & 2 and plates 2 & 3, and then equate them to get the velocity of the middle plate.

So after multiplying dy by both sides and integrating (y from 0 to 10, and v_x from 0 to v₂ for plates 1 & 2 and from v₂ to 3 for plates 2 & 3), I got the equations:
Plates 1 & 2: $\tau$_yx = -0.08007*v₂
Plates 2 & 3: $\tau$_yx = -0.15937 + 0.5179*v₂
Equating these gave me the velocity of plate 2 to be 1.178 m/s

For the 2^nd part of the question (solving for F/A), I think you only need to solve for shear stress as that equals F/A. So using v₂, I plugged it into one of the equations to get $\tau$_yx = F/A = -0.094322 N.


The problem I have with my methods is that I'm not too sure about equating the two shear stresses as I see no real reason that they should be equal in the first place. Also, the 2^nd part of the question specifically asks about the F/A on plate 3, which really makes me think that the shear stress must be different for each plate. While my first answer seems to make sense, I don't think a negative force value for the second one fits, so I would assume that I'm doing something wrong there. Thanks in advance for any help, the hardest part I have with this stuff is that it's really confusing to me what actually needs to be done and which equations to use to get there.

You did a very nice job of solving this. Setting the shear stresses in the two regions equal to one another was the correct thing to do. There is no net force acting on plate 2, so the shear stress on the top of the plate must match the shear stress on the bottom of the plate. The problem you are having is basically one of sign conventions. All of us who have worked in fluid mechanics have struggled with this at the beginning. As you progress, you will learn about the stress tensor and how it can be used to find the stress vector (aka the traction vector) on any arbitrarily oriented surface within a fluid. This is accomplished by using the so-called Cauchy stress relationship. Also, you are using a sign convention in which compressive stresses are regarded as positive, and tensile stresses are regarded as negative. More often, in both solid mechanics and fluid mechanics, tensile stresses are regarded as positive, and compressive stresses are regarded as negative. In that case, there would be no negative sign in your equations for the shear stresses. This has been a source of confusion for many students. In their book Transport Phenomena, Bird, Stewart, and Lightfoot used the more unconventional symbology that you are using, so that the form of the stress equation would be similar to that for conductive heat transfer and for mass transfer.

Your result for plate 3 is correct. But this is the shear force exerted by the fluid on the plate above. This is minus the shear force you would have to exert on the plate from above to keep it moving. So the shear force you have to exert on the plate to keep it moving is positive. Until you learn about the Cauchy stress relationship, you will have to depend on your intuition to get the signs correct.

Chet

 

[bliz]

You did a very nice job of solving this. Setting the shear stresses in the two regions equal to one another was the correct thing to do. There is no net force acting on plate 2, so the shear stress on the top of the plate must match the shear stress on the bottom of the plate. The problem you are having is basically one of sign conventions. All of us who have worked in fluid mechanics have struggled with this at the beginning. As you progress, you will learn about the stress tensor and how it can be used to find the stress vector (aka the traction vector) on any arbitrarily oriented surface within a fluid. This is accomplished by using the so-called Cauchy stress relationship. Also, you are using a sign convention in which compressive stresses are regarded as positive, and tensile stresses are regarded as negative. More often, in both solid mechanics and fluid mechanics, tensile stresses are regarded as positive, and compressive stresses are regarded as negative. In that case, there would be no negative sign in your equations for the shear stresses. This has been a source of confusion for many students. In their book Transport Phenomena, Bird, Stewart, and Lightfoot used the more unconventional symbology that you are using, so that the form of the stress equation would be similar to that for conductive heat transfer and for mass transfer.

Your result for plate 3 is correct. But this is the shear force exerted by the fluid on the plate above. This is minus the shear force you would have to exert on the plate from above to keep it moving. So the shear force you have to exert on the plate to keep it moving is positive. Until you learn about the Cauchy stress relationship, you will have to depend on your intuition to get the signs correct.

Chet

Thanks for the reply, I appreciate the help. And I do have BSL as my textbook so it seems that my improper use of the negative sign was understandable.

 

[Chestermiller]

Thanks for the reply, I appreciate the help. And I do have BSL as my textbook so it seems that my improper use of the negative sign was understandable.

Hi Bliz. Welcome to Physics Forums.
I wouldn't call the use of the negative sign as "improper." You just need to bear in mind when using the notation in BSL precisely what it is saying mathematically.

 

[rezeew]

Dear student,

Thank you for your question. I understand that you are struggling with understanding how to approach this problem and which equations to use. I will try my best to explain the concepts and steps involved in solving this problem.

Firstly, let's start with some background information on shear stress and fluid mechanics. Shear stress is a force that acts parallel to the surface of an object, causing it to deform or move. In fluid mechanics, shear stress is caused by the movement of fluid particles past each other, and it is directly related to the viscosity of the fluid. Viscosity is a measure of a fluid's resistance to flow, with higher viscosity fluids being more resistant to flow.

Now, let's apply these concepts to the problem at hand. We have three parallel plates with two different fluids between them, water and toluene. The top plate (plate 3) is moving at a constant velocity of 3 m/s, while the bottom plate (plate 1) is at rest. We are asked to find the velocity of the middle plate (plate 2) and the shear stress on plate 3.

To solve for the velocity of plate 2, we can use the equation for shear stress, \tauyx = -\mu*vx/y. This equation tells us that the shear stress is directly proportional to the velocity of the fluid particles (vx) and inversely proportional to the distance between the plates (y). We can also see that the shear stress is dependent on the viscosity of the fluid (\mu).

Using this equation, we can solve for the shear stress between plates 1 and 2 and between plates 2 and 3. We know the distance between each plate pair is 10 cm, so y = 0.1 m. We also know the viscosity of water and toluene at 30°C. Plugging these values into the equation, we get:

Shear stress between plates 1 and 2: \tauyx = - (0.8807 cp * v2) / 0.1 m
Shear stress between plates 2 and 3: \tauyx = - (0.5179 cp * (3 m/s - v2)) / 0.1 m

Note that we use v2 for the velocity of plate 2 in the second equation because the velocity of plate 2 is unknown and we want to solve for it. Now, we