Fluid Mechanics: velocity u for circular pipe

  • Thread starter BobJonesX
  • Start date
  • #1
8
0

Homework Statement


For low-speed(laminar) , steady flow through a circular pipe, the velocity u varies with radius and takes what form? Please see this link for picture of the pipe: http://www.sfu.ca/~ptaherib/teaching/ENSC_283_2013/Suggested%20Problems/Suggested%20Problems%20Chapter%201/P1.12.pdf [Broken]
(it's the picture on the very first page of this pdf).
In the posted pdf, the equation for u(r) is given; however, the question that I am given is asking me to somehow come up with the equation for u(r) by myself. How do I do this ? I've looked all over my textbook and can't even seem to find anything remotely related to this; in the very least, what specific topic of fluid mechanics does this fall under ?

Thanks.

Homework Equations




The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
20,871
4,546
Have you covered the Navier Stokes equations in your course yet? Otherwise, see first couple of chapters of Transport Phenomena by Bird, Stewart, and Lightfoot.

Chet
 
  • #3
8
0
Have you covered the Navier Stokes equations in your course yet? Otherwise, see first couple of chapters of Transport Phenomena by Bird, Stewart, and Lightfoot.

Chet
Ok, Thanks. So, do the Navier Stokes equations include this specific equation?
Or am I supposed to derive the u(r) equation using the Stokes equations ?
By the way, my textbook is Fluid Mechanics by Frank M. White, so if possible, could somebody tell me which chapter this type of problem is covered in ?
 
  • #4
20,871
4,546
Ok, Thanks. So, do the Navier Stokes equations include this specific equation?
Or am I supposed to derive the u(r) equation using the Stokes equations ?
You can derive it from them. There is also a more intuitive method discussed in the early chapters of Transport Phenomena (using shell momentum balances).

Chet
 
  • #5
8
0
You can derive it from them. There is also a more intuitive method discussed in the early chapters of Transport Phenomena (using shell momentum balances).

Chet
Unfortunately, I only have access to Fluid Mechanics (seventh edition) by Frank White.
I looked up the Navier Stokes equations there and all I'm getting are a bunch of general-case partial
derivative equations (e.g. du/dx + dv/dy = 0 ).
I must not be getting it; can someone please either show me where in the Fluid Mechanics book where exactly
(which page) I should be looking at, or can someone please show me which equations I need to use to start off with and
how exactly I"m supposed to derive the expression for u(r) ? (If possible, could you provide a link to a picture instead of
writing out all the symbols here ?) Or, can someone please provide a link to a webpage which directly addresses
this problem? I'm just lost...Or, I'm wondering if my teacher intended for me to just memorize this equation? Because I think I might
have seen it before, but now derivation was provided....

Thanks.
 
  • #6
20,871
4,546
I can help you work through the shell force balance derivation on your own, if you're willing to give it a try. First, some questions:

Are you familiar with the concept of shear stress?
Do you know how the viscosity comes in in determining the shear stress on a surface in a shear flow?
What do you think the axial velocity u is at the wall of the pipe?
Do you think that the axial velocity u gets (a) higher or (b) lower with increasing distance from the pipe axis?

Do you feel up to continuing?

Chet
 
  • #7
8
0
I
I can help you work through the shell force balance derivation on your own, if you're willing to give it a try. First, some questions:

Are you familiar with the concept of shear stress?
Do you know how the viscosity comes in in determining the shear stress on a surface in a shear flow?
What do you think the axial velocity u is at the wall of the pipe?
Do you think that the axial velocity u gets (a) higher or (b) lower with increasing distance from the pipe axis?

Do you feel up to continuing?

Chet
I'm familiar with shear stress.
I've just been introduced to viscocity.
At the wall of the pipe, I'd imagine that the velocity would be almost zero.
I would imagine that u(r) increases as it moves closer to the center of the pipe.
 
  • #8
20,871
4,546
I
I'm familiar with shear stress.
I've just been introduced to viscocity.
At the wall of the pipe, I'd imagine that the velocity would be almost zero.
I would imagine that u(r) increases as it moves closer to the center of the pipe.
Good answers. We are going to make use of viscosity in solving this problem. Do you know the relationship between shear stress, velocity gradient, and viscosity? If so, please write it down for me.

We are going to focus on the fluid in the pipe contained between the cross section at axial location z, and z + Δz. Consider the annular shell of fluid between radial locations r and r + Δr. We are going to treat this shell of fluid as a free body, and do a force balance on it. There are forces acting on this free body on all four of its surfaces: at z, z + Δz, r, and r + Δr. Are we together so far? Start thinking about what the forces are that are acting on the four surfaces of the free body in the axial direction.

Chet
 
  • #9
8
0
Good answers. We are going to make use of viscosity in solving this problem. Do you know the relationship between shear stress, velocity gradient, and viscosity? If so, please write it down for me.

We are going to focus on the fluid in the pipe contained between the cross section at axial location z, and z + Δz. Consider the annular shell of fluid between radial locations r and r + Δr. We are going to treat this shell of fluid as a free body, and do a force balance on it. There are forces acting on this free body on all four of its surfaces: at z, z + Δz, r, and r + Δr. Are we together so far? Start thinking about what the forces are that are acting on the four surfaces of the free body in the axial direction.

Chet
Is the relationship the following :

7cae16584200c2bafe7487af29db84f4.png

?

As for forces, is there a normal force ?
 
  • #10
20,871
4,546
Is the relationship the following :

7cae16584200c2bafe7487af29db84f4.png

?

As for forces, is there a normal force ?
Yes. Well done. There are normal forces acting on our free body and there are also shear forces acting on our free body. The normal forces per unit area are the result of the pressures at z and at z + Δz, p(z) and p(z+Δz), respectively, acting on the end faces of the free body. We are going to be focusing on the component of the force balance (on our free body) in the axial direction. What are the areas of the end faces of our free body? Consider the pressure force exerted by the fluid behind our free body on our free body. Is it pointing in the positive z direction or in the negative z direction? Same question for the pressure force exerted by the fluid ahead of our free body on our free body. What is the magnitude of the pressure force acting on our free body at the face at axial location z? What is the magnitude of the pressure force acting on our free body at the face at axial location z + Δz?

Chet
 
  • #11
8
0
Yes. Well done. There are normal forces acting on our free body and there are also shear forces acting on our free body. The normal forces per unit area are the result of the pressures at z and at z + Δz, p(z) and p(z+Δz), respectively, acting on the end faces of the free body. We are going to be focusing on the component of the force balance (on our free body) in the axial direction. What are the areas of the end faces of our free body? Consider the pressure force exerted by the fluid behind our free body on our free body. Is it pointing in the positive z direction or in the negative z direction? Same question for the pressure force exerted by the fluid ahead of our free body on our free body. What is the magnitude of the pressure force acting on our free body at the face at axial location z? What is the magnitude of the pressure force acting on our free body at the face at axial location z + Δz?

Chet
I was just wondering ( before I try to answer the rest of the questions), what causes the shear forces? (i.e. why are they there?)
 
  • #12
20,871
4,546
Sir Isaac Newton discovered experimentally that, when you shear a fluid between parallel plates, the force needed to move the top plate at a constant speed is proportional to the speed difference between the plates. He called the constant of proportionality the viscosity. The molecular cause of this is that the different adjacent layers of fluid are always exchanging molecules with one another, and the molecules from the faster layer transfer momentum to the slower layer to try to speed it up, while those from the faster layer transfer momentum to the slower layer to try to slow it down. The details of this are discussed in Transport Phenomena by Bird, Stewart, and Lightfoot.

Chet
 
  • #13
8
0
Ok. So considering the free body diagram,
how come we don't consider all six faces?
(i.e. are we considering a rectangular prism
of water or is it more of a cylinder with shells? )
 
  • #14
20,871
4,546
Ok. So considering the free body diagram,
how come we don't consider all six faces?
(i.e. are we considering a rectangular prism
of water or is it more of a cylinder with shells? )
It's a cylinder with shells.
 
  • #15
8
0
So then, for the free-body diagram, how come we don't have to
consider six planes ? the positive and negative z planes, the positive and negative
y planes, and the positive and negative x planes ?
 
  • #16
20,871
4,546
So then, for the free-body diagram, how come we don't have to
consider six planes ? the positive and negative z planes, the positive and negative
y planes, and the positive and negative x planes ?
Because the mathematics is very simple when you solve the problem in cylindrical coordinates, and the mathematics is desperately complicated when you solve the problem in cartesian coordinates. Why do you think that they gave you the velocity profile as u(r)?

Chet
 

Related Threads on Fluid Mechanics: velocity u for circular pipe

  • Last Post
Replies
2
Views
4K
Replies
3
Views
928
  • Last Post
Replies
0
Views
3K
Replies
11
Views
818
  • Last Post
Replies
1
Views
901
  • Last Post
Replies
4
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
4K
Top