Velocity Profile Analysis for Force Balance Equation in Fluid Dynamics

[hotjohn]

Homework Statement

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 ?

Homework Equations

The Attempt at a Solution

 

[Chestermiller]

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.

 

[hotjohn]

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.

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

 

[haruspex]

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

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.

 

[hotjohn]

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.

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

 

[haruspex]

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

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.

 

[Chestermiller]

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

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?

 

[hotjohn]

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.

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 ?

 

[Chestermiller]

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 ?

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 $dA=2\pi r dr$.

 

[hotjohn]

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 $dA=2\pi r dr$.

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

 

[hotjohn]

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?

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 ?

 

[Chestermiller]

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

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

 

[Chestermiller]

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 ?

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

 

[hotjohn]

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

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

 

[Chestermiller]

do u mean the shear stress act on the cylindrical volume , but not on the end surface of free body ?

Cylindrical surface, not volume. And, yes, it doesn't act horizontally on the end surface of the free body.

since 2pi r dx = cylindrical volume ?

This is surface area, not volume.

2 pi dx is the circumference , times dx , we would get the volume

2 pi r is the circumference, times dx, we get the surface area.

 

[haruspex]

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

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.

 

[hotjohn]

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.

why 2 pi r dx is sufrace area ?

 

[hotjohn]

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.

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 ?

 

[haruspex]

when we take thickness multiply by the 2 pi r ,

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 $2\pi r_2 h$.

 

[hotjohn]

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 $2\pi r_2 h$.

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 ?

 

[haruspex]

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 ?

No, all four forces act horizontally right and left.

 

[hotjohn]

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 $2\pi r_2 h$.

$2\pi r_1 h$ 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 ?

 

[haruspex]

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 ?

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

 

[Chestermiller]

 

[hotjohn]

View attachment 94650

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 ?

 

[haruspex]

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 ?

8-11 considers two radii r and r+Δr on the same side of the central axis. Because of the velocity (fastest in the cental axis) the relative velocities create opposite drags either side of the elemenrt
8-12 considers an entire disc of radius r, so the opposite sidea of it are on opposite sides of the central axis. The same velocity profile leads to drag in the same direction on opposite sides.
See Chet's diagram above. The two τ(r+Δr) are in the same direction, but in opposite direction to τ(r).

 

[hotjohn]

8-11 considers two radii r and r+Δr on the same side of the central axis. Because of the velocity (fastest in the cental axis) the relative velocities create opposite drags either side of the elemenrt
8-12 considers an entire disc of radius r, so the opposite sidea of it are on opposite sides of the central axis. The same velocity profile leads to drag in the same direction on opposite sides.
See Chet's diagram above. The two τ(r+Δr) are in the same direction, but in opposite direction to τ(r).

for 8-11, since the velocity profile is drawn from left to right ( same direction) for both r and r+Δr ,m why it will create opposite drags?

 

[hotjohn]

one more question , why the τ is given by - μ / ( du /dr ) ? why y = R -r ? where is r measured from ? where is y measured from ? it's not shown in the diagram

 

[haruspex]

Δ

for 8-11, since the velocity profile is drawn from left to right ( same direction) for both r and r+Δr ,m why it will create opposite drags?

The velocity is greatest at the central axis. The annulus just inside radius r is moving faster than that between r and r+Δr, so drags it in the forward direction. The annulus just beyond r+Δr Is moving more slowly than that between r and r+Δr, so drags it in the backward direction.

 

[hotjohn]

Δ
The velocity is greatest at the central axis. The annulus just inside radius r is moving faster than that between r and r+Δr, so drags it in the forward direction. The annulus just beyond r+Δr Is moving more slowly than that between r and r+Δr, so drags it in the backward direction.

i still don't understand why the resultant drag is in opposite direction , could you explain further?

 

[haruspex]

one more question , why the τ is given by - μ / ( du /dr ) ? why y = R -r ? where is r measured from ? where is y measured from ? it's not shown in the diagram

If τ is the shear force per unit area of interface and μ is the viscosity then that equation comes straight from the definition of viscosity.
The author seems to be defining y as R-r, but I don't know why.

 

[haruspex]

i still don't understand why the resultant drag is in opposite direction , could you explain further?

An object rests on a block on a carpet. You yank the carpet. The box slides on the carpet. The carpet's drag on the box is towards you. The object slides on the block. The box's drag on the object is towards you, so the object's drag on the box is away from you. The two drags on the box are in opposite directions.

 

[hotjohn]

An object rests on a block on a carpet. You yank the carpet. The box slides on the carpet. The carpet's drag on the box is towards you. The object slides on the block. The box's drag on the object is towards you, so the object's drag on the box is away from you. The two drags on the box are in opposite directions.

why there is i assume you mean an object rest on a box on a carpet...
but , i can't visualize the situation although i have tried out myself . Based on my observation , the object will moves with the box ?

 

[haruspex]

why there is i assume you mean an object rest on a box on a carpet...
but , i can't visualize the situation although i have tried out myself . Based on my observation , the object will moves with the box ?

It might move with the box, or it might slip. Either way, the box drags the object towards you, so by action and reaction the drag on the box from the object must be away from you.

 

[hotjohn]

It might move with the box, or it might slip. Either way, the box drags the object towards you, so by action and reaction the drag on the box from the object must be away from you.

ok , back to the topic . by taking the example aforementioned as anology , which is box ? which is object ? which is carpet? r ? r+ del (r) ?

 

[haruspex]

ok , back to the topic . by taking the example aforementioned as anology , which is box ? which is object ? which is carpet? r ? r+ del (r) ?

The box is the annulus between r and r+Δr. The carpet is the next smaller annulus (faster moving) and the object is the next larger (slower moving).

 

[hotjohn]

The box is the annulus between r and r+Δr. The carpet is the next smaller annulus (faster moving) and the object is the next larger (slower moving).

all the box , object and carpet are subjected to the same velocity , right ? this is not same as the situation in the water , which at different point , the partcile would have different velocity ?

 

[haruspex]

all the box , object and carpet are subjected to the same velocity , right ? this is not same as the situation in the water , which at different point , the partcile would have different velocity ?

As I framed it, the object slides on the box and the box slides on the carpet, so the three are at different velocities.

 

[Chestermiller]

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 ?

Drawing it this way really stems from the Cauchy Stress Relationship. This relationship has nothing to do with the type of material that is being deformed. Hot John, unfortunately, until you are taught in your courses the details of how to work with 2nd order tensors like the stress tensor, I'm afraid you're just going to have to accept the fact that you need to draw the specific stress components acting on opposite sides of a body with opposite directions. You had no trouble doing this with the pressure stresses on the ends of the body (which are just normal components of the stress tensor). Mathematically, the normal pressure force is done this way for the same basic reason that the shear stresses are drawn with opposite directions on the other faces of the fluid.