Understanding Viscosity: Solving a Fluid Mechanics Problem

[fayan77]

Homework Statement

the picture has all the information

Homework Equations

I am assuming we will be needing the viscosity equation, tao= mue(dv/dy) where v is velocity

The Attempt at a Solution

I do not know how to approach this exercise.

 

[berkeman]

Homework Statement

the picture has all the information

Homework Equations

I am assuming we will be needing the viscosity equation, tao= mue(dv/dy) where v is velocity

The Attempt at a Solution

I do not know how to approach this exercise.

You have not shown any effort in your post. What have you tried so far? What other approaches would you consider if we suggested they might work?

 

[fayan77]

Well because I do not know how to go about it, I can't try. I know what you are getting at, I must make an effort. But I honestly do not know how to start, maybe if you gave me some insight I would be able to try something.

 

[Chestermiller]

If you had a viscous fluid between two parallel plates, with one plate moving at velocity V and the other plate stationary, would you know what to do?

 

[fayan77]

I think so, would it be the same as sliding a flat surface over a container full of fluid? if so, the fluid would have some viscosity and will retard the flat surface after giving it a push and F= mue(velocity/area).

What I don't understand in my original question is how will I go about the disk being "a" distance from the top and "b" distance from the bottom? I will be using the Couette flow equation if I am not mistaken

 

[fayan77]

I forgot the force will be directly proportional to the Area

 

[Chestermiller]

I think so, would it be the same as sliding a flat surface over a container full of fluid? if so, the fluid would have some viscosity and will retard the flat surface after giving it a push and F= mue(velocity/area).

This equation is not correct (or readable). If $\tau$ is the shear stress, $\mu$ is the viscosity, V is the relative velocity of the plates, and h is the distance between the plates, what is the relationship for calculating $\tau$ from the other parameters?

What I don't understand in my original question is how will I go about the disk being "a" distance from the top and "b" distance from the bottom? I will be using the Couette flow equation if I am not mistaken

Until you are able to answer my question above, you will not be able to answer this question.

 

[fayan77]

To answer your question I would use F = mue (velocity/h) A

for my problem:

Mue= .01
velocity = 1000 rpm = 16.67(.02) (m/s)
h= it should be hi-hf but I do not know where it is measuring from
A= pi(.02)^2

 

[Chestermiller]

You are aware that there is a difference between angular velocity and linear velocity, correct? The relative linear velocity is a function of radial location r. What is the relative velocity of the plates as a function of r?

 

[fayan77]

Yes, velocity =omega(R) = 1000 rpm X 2pi rad/minute X R

 

[Chestermiller]

So what is the shear stress on each side of the disk as a function of r?

 

[fayan77]

Ok, so I would treat each side of the disk independently and add them at the end for example the bottom portion of the system consists of the stationary container and the top would be the moving disk I would add this force to the second system which would be the top portion consisting of the top stationary container and the bottom moving disk.

Fbottom = [.01(pi)(R)^2(2000pi)R]/.002

Ftop = [.01(pi)(R)^2(2000pi)R]/.004

Ftotat= add the above up

 

[Chestermiller]

Are you supposed to get the force or the torque?

 

[fayan77]

Torque, so I will multiply F times R

 

[Chestermiller]

Torque, so I will multiply F times R

No way. The force F is equal to zero. So your calculation of F is incorrect. Do you know why?

The torque, on the other hand, is not zero. Please reconsider how you would calculate the torque (hint: integration is involved)?

 

[fayan77]

Im confused. Torque= FRsin(theta) if F=0 then torque=0 unless I am missing something

 

[Chestermiller]

Im confused. Torque= FRsin(theta) if F=0 then torque=0 unless I am missing something

You are. The force is not uniformly distributed on the disc, and, at each angle, it is pointing in a different direction. All the vectorial contributions to the force on the differential areas of the disk surface cancel out.

To get the torque, you need to integrate the differential contributions of the shear stress over the surface area of the disk: $\tau (2\pi r)rdr$

 

[fayan77]

could you perhaps draw what you are talking about so I could visualize what is going on? The oil is going to oppose the motion of the disk on top and the bottom so they will be in the same direction hence we add... I drew what I am understanding

 

[Chestermiller]

could you perhaps draw what you are talking about so I could visualize what is going on? The oil is going to oppose the motion of the disk on top and the bottom so they will be in the same direction hence we add... I drew what I am understanding

Your diagram is correct. What is the DIRECTION of the NET force?

 

[fayan77]

Clockwise,

 

[Chestermiller]

The diagram for the torque is not correct. You are aware that the differential force contributions acting on the surface of the disk are pointing in the local tangential (theta) direction at each location on the surface, correct? And the local tangential direction changes with position around the circumference of a circle. And the cross product of a unit vector in the circumferential direction with a radial vector to a given location on the surface is oriented in the axial direction? So the force contributions cancel, while the torque contributions reinforce.

 

[fayan77]

I am not following. what are you talking about when you say cross product between unit vector and radial vector? and how are the force contributions pointing in the tangential direction?

 

[Chestermiller]

I am not following. what are you talking about when you say cross product between unit vector and radial vector? and how are the force contributions pointing in the tangential direction?

The velocity of the disk surface at each location is pointing in the circumferential direction. Do you not see that?

 

[fayan77]

Yes I see that, that is what I drew in my first diagram. Therefore the fluid will oppose that motion in the same manner but opposite direction. But I don't understand when you say the force will cancel out.

 

[Chestermiller]

Yes I see that, that is what I drew in my first diagram. Therefore the fluid will oppose that motion in the same manner but opposite direction. But I don't understand when you say the force will cancel out.

The shear force per unit area of disk surface exerted by the fluid (at radial location r) is given by:
$$\mathbf{f}=\mu \left(\frac{\Omega r}{a}+\frac{\Omega r}{b}\right)\mathbf{i}_{\theta}$$where $\mathbf{i}_{\theta}$ is the unit vector in the circumferential direction. The differential area of surface that this shear force per unit acts upon is $dA=rdrd\theta$. So, the total force exerted fy the fluid on the disk is: $$\mathbf{F}=\int_0^{2\pi}\int_0^R{\mu \left(\frac{\Omega r}{a}+\frac{\Omega r}{b}\right)\mathbf{i}_{\theta}rdrd\theta}$$

Are we in agreement on this so far?

 

[fayan77]

I understand everything except the i sub theta "unit vector" I'm trying to find my multivariable calc notes to see if there is an example to help me understand the need for the unit vector

 

[Chestermiller]

I understand everything except the i sub theta "unit vector" I'm trying to find my multivariable calc notes to see if there is an example to help me understand the need for the unit vector

If you have a force, it has to have a direction. The unit vector establishes that direction. In the case of cylindrical polar coordinates, the unit vector in the theta direction is pointing along the circumference of each circle. This is the same direction as the velocity vector.

 

[fayan77]

I'm afraid I am not understanding. Can you give me an example where we integrate using unit vectors? other than this new topic for me.

 

[Chestermiller]

I'm afraid I am not understanding. Can you give me an example where we integrate using unit vectors? other than this new topic for me.

Maybe you would be more comfortable working in Cartesian coordinates. If the disk is rotating clockwise and we focus on a differential element of area dA=dxdy on the surface of the disk, then the differential force acting on this differential area has components in the x- and y directions, given by:
$$dF_x=-\mu\left(\frac{\Omega y}{a}+\frac{\Omega y}{b}\right)dxdy\tag{1}$$
$$dF_y=+\mu\left(\frac{\Omega x}{a}+\frac{\Omega x}{b}\right)dxdy\tag{2}$$

Okay so far?

 

[fayan77]

Why is force in x negative? And why is Omega y
Same for force in y?

 

[Chestermiller]

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.

 

[fayan77]

What is the area?

 

[Chestermiller]

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.

 

[fayan77]

So what is... On the picture

 

[Chestermiller]

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.

 

[fayan77]

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?

 

[Chestermiller]

Maybe the above figure can help explain.

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

 

[fayan77]

We are on the same page now!

 

[fayan77]

I understand this. Can I ignore the i unit vector?

 

[Chestermiller]

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?

 

[fayan77]

No, but isn't it the same to the polar one?

 

[Chestermiller]

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.

 

[Chestermiller]

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):

$$\mathbf{i}_{\theta}=\cos{\theta}\mathbf{i}_y-\sin{\theta}\mathbf{i}_x$$

 

[fayan77]

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)?

 

[fayan77]

Nevermind got it!

 

[fayan77]

Didn't know limits of integration for Cartesian form so I used polar, hopefully this is correct

 

[Chestermiller]

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).

 

[fayan77]

Ok I got zero after I simplified, how is it possible to have no force but have a torque!?

 

[Chestermiller]

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?

 

[fayan77]

Yes I understand the mathematical concept torque= from but when I find force it is 0 so how is this physically possible?