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Fluid Mechanics: Cone-Plate viscometer

  • #1
599
6

Homework Statement


Untitled.png


Homework Equations


Fluid Mechanics

τ =μ (du/dy)

The Attempt at a Solution



I got far enough to write down

dM = μ (Ωr/tanθ) dA

from just substitutions, easy enough.

I get confused when I'm solving for the differential surface area. I somehow need the dA for a cone.
For a circle it's easy enough. A=πr^2 so then dA = 2πr dr.

But how do I go about getting this for a cone? the book lists it as (2πr/cosθ) dr, without any explanation, of course.
 

Answers and Replies

  • #2
374
7
Take a strip element 2πxdr, where x is the radius of circle formed by that strip element. Try relating x to r using similarity of triangles.
 
  • #3
Nathanael
Homework Helper
1,650
239
Well you should first understand how to find the surface area of a "truncated cone" (ignoring the faces)
It is the average circumference of the cone multiplied by the side length (not the height)

So the surface area of a truncated cone (ignoring the faces) is [itex]2\pi r_{avg}L[/itex] or you could say [itex]2\pi L\frac{r_1+r_2}{2}[/itex]
(Side note, this also applies to normal cones; just treat the tip as r2=0)

So with this understanding of truncated cones, look at the following picture I made:
coneintegral.png

The differential surface area would be [itex]2\pi r_{avg}L[/itex] but as you can see from the picture, [itex]L=\frac{dr}{\cos\theta}[/itex] therefore the differential surface area is [itex]\frac{2\pi R_{avg}}{\cos\theta}dr[/itex]

The two radii are r and (r+dr) but since dr is obviously infinitesimal, it suffices to say [itex]r_{avg}=r[/itex]
 
Last edited:
  • #4
374
7
Does it really require all that? Isn't using similarity to relate x and R, considering a string element of radius x, just enough?
 
  • #5
Nathanael
Homework Helper
1,650
239
Does it really require all that? Isn't using similarity to relate x and R, considering a string element of radius x, just enough?
No that is incorrect. The differential area element is [itex]\frac{2\pi x}{\cos\theta}dx[/itex]

Me and the OP were using r not to represent the radius of the base of the cone, but to represent what you called "x"
 
  • #6
374
7
No that is incorrect. The differential area element is [itex]\frac{2\pi x}{\cos\theta}dx[/itex]

Me and the OP were using r not to represent the radius of the base of the cone, but to represent what you called "x"
Oh ok.
 
  • #7
599
6
Well you should first understand how to find the surface area of a "truncated cone" (ignoring the faces)
It is the average circumference of the cone multiplied by the side length (not the height)

So the surface area of a truncated cone (ignoring the faces) is [itex]2\pi r_{avg}L[/itex] or you could say [itex]2\pi L\frac{r_1+r_2}{2}[/itex]
(Side note, this also applies to normal cones; just treat the tip as r2=0)

So with this understanding of truncated cones, look at the following picture I made:
View attachment 78118
The differential surface area would be [itex]2\pi r_{avg}L[/itex] but as you can see from the picture, [itex]L=\frac{dr}{\cos\theta}[/itex] therefore the differential surface area is [itex]\frac{2\pi R_{avg}}{\cos\theta}dr[/itex]

The two radii are r and (r+dr) but since dr is obviously infinitesimal, it is suffices to say [itex]r_{avg}=r[/itex]
Excellent explanation. Thank you very much, Sir.
 

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