Understanding Viscosity: Solving a Fluid Mechanics Problem

In summary: This should be simple to understand once you see it.In summary, the student does not know how to approach this assignment and has not attempted to do anything yet.
  • #1
fayan77
84
0

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.
 

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  • #2
fayan77 said:

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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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
 
  • #6
I forgot the force will be directly proportional to the Area
 
  • #7
fayan77 said:
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.
 
  • #8
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
 
  • #9
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?
 
  • #10
Yes, velocity =omega(R) = 1000 rpm X 2pi rad/minute X R
 
  • #11
So what is the shear stress on each side of the disk as a function of r?
 
  • #12
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
 
  • #13
Are you supposed to get the force or the torque?
 
  • #14
Torque, so I will multiply F times R
 
  • #15
fayan77 said:
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)?
 
  • #16
Im confused. Torque= FRsin(theta) if F=0 then torque=0 unless I am missing something
 
  • #17
fayan77 said:
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##
 
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  • #18
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
 

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  • #19
fayan77 said:
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?
 
  • #20
Clockwise,
 

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  • #21
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.
 
  • #22
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?
 
  • #23
fayan77 said:
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?
 
  • #24
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.
 
  • #25
fayan77 said:
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?
 
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  • #26
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
 
  • #27
fayan77 said:
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.
 
  • #28
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.
 
  • #29
fayan77 said:
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?
 
  • #30
Why is force in x negative? And why is Omega y
Same for force in y?
 
  • #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.
 
  • #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.
 
<h2>1. What is viscosity?</h2><p>Viscosity is a measure of a fluid's resistance to flow. It is a property of fluids that describes how easily they can be deformed or moved. High viscosity fluids are thick and resist flow, while low viscosity fluids are thin and flow easily.</p><h2>2. How is viscosity measured?</h2><p>Viscosity is typically measured using a viscometer, which is a device that measures the time it takes for a fluid to flow through a small opening. The longer it takes, the higher the viscosity. Viscosity can also be calculated using the fluid's density, velocity, and other properties.</p><h2>3. What factors affect viscosity?</h2><p>Viscosity can be affected by several factors including temperature, pressure, and the type of fluid. Generally, higher temperatures decrease viscosity, while higher pressures increase viscosity. Additionally, different fluids have different inherent viscosities.</p><h2>4. How is viscosity important in fluid mechanics?</h2><p>Viscosity is an important concept in fluid mechanics because it affects the behavior of fluids in motion. It determines how easily a fluid can be moved or deformed, and can impact the speed and efficiency of fluid flow. Viscosity is also important in understanding the behavior of fluids in industrial and environmental processes.</p><h2>5. How can I solve a fluid mechanics problem involving viscosity?</h2><p>To solve a fluid mechanics problem involving viscosity, you will need to use equations and principles from fluid mechanics, such as Bernoulli's equation and the continuity equation. You will also need to know the properties of the fluid, such as its density and viscosity, and the conditions of the system, such as pressure and velocity. It is important to carefully set up the problem and use the appropriate equations to find a solution.</p>

1. What is viscosity?

Viscosity is a measure of a fluid's resistance to flow. It is a property of fluids that describes how easily they can be deformed or moved. High viscosity fluids are thick and resist flow, while low viscosity fluids are thin and flow easily.

2. How is viscosity measured?

Viscosity is typically measured using a viscometer, which is a device that measures the time it takes for a fluid to flow through a small opening. The longer it takes, the higher the viscosity. Viscosity can also be calculated using the fluid's density, velocity, and other properties.

3. What factors affect viscosity?

Viscosity can be affected by several factors including temperature, pressure, and the type of fluid. Generally, higher temperatures decrease viscosity, while higher pressures increase viscosity. Additionally, different fluids have different inherent viscosities.

4. How is viscosity important in fluid mechanics?

Viscosity is an important concept in fluid mechanics because it affects the behavior of fluids in motion. It determines how easily a fluid can be moved or deformed, and can impact the speed and efficiency of fluid flow. Viscosity is also important in understanding the behavior of fluids in industrial and environmental processes.

5. How can I solve a fluid mechanics problem involving viscosity?

To solve a fluid mechanics problem involving viscosity, you will need to use equations and principles from fluid mechanics, such as Bernoulli's equation and the continuity equation. You will also need to know the properties of the fluid, such as its density and viscosity, and the conditions of the system, such as pressure and velocity. It is important to carefully set up the problem and use the appropriate equations to find a solution.

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