# Homework Help: Velocity profile

1. Jan 20, 2016

### hotjohn

1. The problem statement, all variables and given/known data
if we divide the force balance equation by 2pi r dx , we would get

(P_(x+ Δx) ) - (P_x) + ( τ_(x + Δx) ) - τ_(x) = 0 , am i right ? why the notes give different eqaution ?

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

File size:
81.4 KB
Views:
84
• ###### DSC_0391.JPG
File size:
82 KB
Views:
82
2. Jan 20, 2016

### Staff: Mentor

The analysis in the book looks correct to me. Why do you have the τ's evaluated at x and x + Δx? They are shear stresses on the shell element being analyzed at r and r + Δr.

3. Jan 20, 2016

### hotjohn

Can you explain how do you get the same formula derivation with the book? I can't understand it....

4. Jan 20, 2016

### haruspex

There are four horizontal forces acting on the element: two opposing but slightly different pressures acting on the sides, and two opposing but slightly different shear forces acting above and below.
The shear force varies with radius, not as a function of x.

5. Jan 21, 2016

### hotjohn

can you expalin why the area of τ is 2 pi r dx ? why not 2 pi rdr?

6. Jan 21, 2016

### haruspex

The element is a ring, radius r, rectangular cross section dxdr. The shear forces act on the surfaces parallel to the pipe. These are bands of width dx. The radius of one is r, the other r+dr. So the surface areas are 2 pi r dx and 2 pi (r+dr)dx.

7. Jan 21, 2016

### Staff: Mentor

The surfaces that the $\tau$'s are acting upon are cylindrical. In terms of R and L, what is the curved surface area of a cylinder?

8. Jan 21, 2016

### hotjohn

ok , I understand that the shear stress act on the area = 2 pi r dx , can you explain why the pressure act on the area = 2 pi r dr ?

9. Jan 21, 2016

### Staff: Mentor

It acts on the end of the shell, with cross sectional area $\Delta A = \pi (r+\Delta r)^2-\pi r^2=\pi[(r+\Delta r)^2-r^2)=\pi \Delta r(2r+\Delta r)$. In the limit of small $\Delta r$ compared to r, this becomes $dA=2\pi r dr$.

10. Jan 21, 2016

### hotjohn

in the pipe , why not the r = constant, which is r ? why there is delta r ?

11. Jan 21, 2016

### hotjohn

why the shear stress not act on a round cross sectional area ? if it act on the round surface area , the area would be 2 pi r dr , am I right ? the dx is for the rectangular cross sectional area , right ?

12. Jan 21, 2016

### Staff: Mentor

r is not the pipe radius. r is the radial coordinate measured from the centerline.

13. Jan 21, 2016

### Staff: Mentor

The shear stress is acting on surfaces of constant r and is oriented in the x direction. At radial coordinate r, the area of the free body surface that the shear stress acts upon is $2\pi r dx$. There is no shear stress acting on the end surfaces of the free body. (Only the pressure acts on the end surfaces).

Chet

14. Jan 21, 2016

### hotjohn

do u mean the shear stress act on the cylindrical volume , but not on the end surface of free body ? since 2pi r dx = cylindrical volume ? 2 pi dx is the circumference , times dx , we would get the volume

15. Jan 21, 2016

### Staff: Mentor

Cylindrical surface, not volume. And, yes, it doesn't act horizontally on the end surface of the free body.
This is surface area, not volume.
2 pi r is the circumference, times dx, we get the surface area.

16. Jan 21, 2016

### haruspex

Shear force is like friction.
Imagine pulling on a long tight sock. The radius of your leg is r, the thickness of the sock dr, its length x. The normal pressure is P.
The surface area of the sock contacting your leg is $2\pi r x$, so the frictional force is $2\pi r x P \mu$. No dr in there anywhere.

17. Jan 21, 2016

### hotjohn

why 2 pi r dx is sufrace area ?

18. Jan 21, 2016

### hotjohn

when we take thickness multiply by the 2 pi r , we eou;ld get vplume , right ? can you explain further ? or can you attach a diagram ?

19. Jan 22, 2016

### haruspex

That's only multiplying two distances, so it can give area but not a volume.
Look at http://keisan.casio.com/exec/system/1340330749.
For the shear force, we are interested in the cylindrical surfaces. In the image, these have areas $2\pi r_1 h$ and $2\pi r_2 h$.

20. Jan 22, 2016

### hotjohn

do u mean the shear stress act on the pipe inward and outward of tha paper? while the pressure act on the surface area of pipe from the right adn the left ?

21. Jan 22, 2016

### haruspex

No, all four forces act horizontally right and left.

22. Jan 22, 2016

### hotjohn

$2\pi r_1 h$ is the lateral area right ? shear force must act perpendicular to the lateral area , right ? so it the shear force should be inward and outward of the book , right ?

23. Jan 22, 2016

### haruspex

No, shear force acts along the surface. That's why it is called shear force, not normal force.

24. Jan 22, 2016

### Staff: Mentor

25. Feb 3, 2016

### hotjohn

can someone try to explain why the shear stress force at down and above in diagram 8-11 is opposite in direction ? why in diagram 8-12 , they are in the same direction ?