Understanding Viscosity: Solving a Fluid Mechanics Problem

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SUMMARY

This discussion focuses on solving a fluid mechanics problem involving viscosity, specifically using the viscosity equation τ = μ(dv/dy), where τ is shear stress, μ is viscosity, and dv/dy is the velocity gradient. Participants explore the application of the Couette flow equation to a scenario with a rotating disk and stationary fluid, emphasizing the importance of understanding shear forces and torque calculations. Key insights include the necessity of integrating differential contributions of shear stress to determine torque, as well as the distinction between linear and angular velocities in the context of fluid dynamics.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly viscosity and shear stress.
  • Familiarity with the Couette flow equation and its applications.
  • Knowledge of integration techniques in calculus, especially in polar coordinates.
  • Basic concepts of torque and its calculation in rotational systems.
NEXT STEPS
  • Study the derivation and applications of the Couette flow equation in fluid dynamics.
  • Learn how to calculate torque using integration of shear stress over a surface area.
  • Explore the differences between linear and angular velocities in rotating systems.
  • Investigate the use of unit vectors in vector calculus, particularly in polar coordinates.
USEFUL FOR

Students and professionals in mechanical engineering, fluid mechanics, and physics, particularly those dealing with problems involving viscosity and rotational dynamics.

  • #31
fayan77 said:
Why is force in x negative?
If the disk is rotating clockwise, the force per unit area is in the counterclockwise direction. So, in the first and second quadrant, the component of the force per unit area is in the negative x direction.
And why is Omega y
Same for force in y?
The magnitude of the force per unit area is ##\mu\left(\frac{\Omega\sqrt{x^2+y^2}}{a}+\frac{\Omega\sqrt{x^2+y^2}}{b}\right)##. So the magnitude of the component of force in the x direction is geometrically $$\mu\left(\frac{\Omega\sqrt{x^2+y^2}}{a}+\frac{\Omega\sqrt{x^2+y^2}}{b}\right)\frac{y}{\sqrt{x^2+y^2}}=\mu\left(\frac{\Omega y}{a}+\frac{\Omega y}{b}\right)$$Similarly for the force in the y direction.
 
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  • #32
What is the area?
 

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  • #33
fayan77 said:
What is the area?
As I said in post #29, the differential area over which these components of force per unit area apply is dA=dxdy.
 
  • #34
So what is... On the picture
 

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  • #35
fayan77 said:
So what is... On the picture
That's the cosine of the angle that the velocity vector makes with the x direction. We are trying to resolve the velocity vector into components in the x and y directions, so we must multiply the magnitude of the velocity vector by the appropriate trigonometric functions.
 
  • #36
I think I understand now. is it analogous to when finding the electric field? Because electric field is a vector we have to decompose to components then integrate?
 
  • #37
Disk Torque.png


Maybe the above figure can help explain.

Regarding electric field applications: I don't know much about the application to problems like that.
 

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  • #38
We are on the same page now!
 
  • #39
I understand this. Can I ignore the i unit vector?
 

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  • #40
fayan77 said:
We are on the same page now!
OK. Can you integrate Eqns. 1 and 2 of my post #29 to get the net horizontal and vertical force components on the disk? Do you know how to work with Eqns. 1 and 2 to get the torque on the disk?
 
  • #41
No, but isn't it the same to the polar one?
 
  • #42
fayan77 said:
I understand this. Can I ignore the i unit vector?
No way. This does not take into account the fact that the direction of the force per unit area (as described by the unit vector) changes with angular position on the disk. That's why I thought it would be better if you switched to Cartesian coordinates. L'd like to see what you get for the x and y components of the force Fx and Fy when you integrate over the area of the disk.
 
  • #43
fayan77 said:
No, but isn't it the same to the polar one?
No. Your formulation of the polar one is incorrect because it does not take into account the change in direction of the differential force over the area of the disk. Force is a vector, and if, in your system, its direction varies with position, it is imperative that you account in summing its contribution to the net force. If you want to use polar coordinates, you can express the unit vector in the theta direction as follows in terms of the (constant) unit vectors in the x and y directions (and then integrate using polar coordinates):

$$\mathbf{i}_{\theta}=\cos{\theta}\mathbf{i}_y-\sin{\theta}\mathbf{i}_x$$
 
  • #44
Ok got it. Going back to the picture of the circle and differential area can you explain how force x has direction y/√(x²+y²)? Shouldn't it be x/√(x²+y²)?
 
  • #45
Nevermind got it!
 
  • #46
Didn't know limits of integration for Cartesian form so I used polar, hopefully this is correct
 
  • #47
fayan77 said:
Didn't know limits of integration for Cartesian form so I used polar, hopefully this is correct
After you show that the net force is zero, let’s talk about how to get the torque (which is not zero).
 
  • #48
Ok I got zero after I simplified, how is it possible to have no force but have a torque!?
 
  • #49
fayan77 said:
Ok I got zero after I simplified, how is it possible to have no force but have a torque!?
You'll see. At each location on the disk, the differential force dF on each differential area is oriented perpendicular to the moment arm drawn from the origin r. Therefore, its differential contribution to the torque is $$dT=\left[\mu \left(\frac{\Omega r}{a}+\frac{\Omega r}{b}\right)rdrd\theta\right] r=\mu \left(\frac{1 }{a}+\frac{1}{b}\right)\Omega r^3drd\theta$$What does this give you when you integrate with respect to r and theta?
 
  • #50
Yes I understand the mathematical concept torque= from but when I find force it is 0 so how is this physically possible?
 
  • #51
fayan77 said:
Yes I understand the mathematical concept torque= from but when I find force it is 0 so how is this physically possible?
Suppose you have two parallel forces acting on a rigid body. The forces are equal in magnitude, but opposite in direction. What is the net force acting on the body?

Even through the two parallel forces are equal in magnitude and opposite in direction, they do not have the same line of action. If the magnitudes of each of the two forces is F and the perpendicular distance between their lines of action is d, what is the net torque on the body?
 
  • #52
A couple system?
 
  • #53
fayan77 said:
A couple system?
Yes. Please answer my questions.
 
  • #54
Net force is 0 and net torque is Fd
 
  • #55
fayan77 said:
Net force is 0 and net torque is Fd
Correct. Now, in our problem, for each differential element of area on the disk, there is a corresponding differential area element located 180 degrees away in which the force per unit area is parallel and opposite in direction. Do you see that?
 
  • #56
Yes, so instead of integrating df (force) I should integrate dam (moment).
 
  • #57
fayan77 said:
Yes, so instead of integrating df (force) I should integrate dam (moment).
Well, you already showed by integrating that the net force is equal to zero. Of course, we expected that from the symmetry of the system.

To get the moment, integrating is required.
 
  • #58
Oh ok! now I don't care about direction since all the vectors are all tangential so we just integrate radius so I can can pull out 2pi out of integral?
 
  • #59
fayan77 said:
Oh ok! now I don't care about direction since all the vectors are all tangential so we just integrate radius so I can can pull out 2pi out of integral?
Yes. You just integrate the equation in post 49, pulling out the 2pi.
 
  • #60
Thank you so much! I can honestly say I fully understand this problem!
 

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