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Fluid Mechanics: Cone-Plate viscometer

  1. Jan 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Untitled.png

    2. Relevant equations
    Fluid Mechanics

    τ =μ (du/dy)
    3. 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.
     
  2. jcsd
  3. Jan 22, 2015 #2
    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.
     
  4. Jan 22, 2015 #3

    Nathanael

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    Homework Helper

    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: Jan 22, 2015
  5. Jan 22, 2015 #4
    Does it really require all that? Isn't using similarity to relate x and R, considering a string element of radius x, just enough?
     
  6. Jan 22, 2015 #5

    Nathanael

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    Homework Helper

    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"
     
  7. Jan 22, 2015 #6
    Oh ok.
     
  8. Jan 22, 2015 #7
    Excellent explanation. Thank you very much, Sir.
     
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