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Newtons Law of Viscosity

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Metals: 5x10-4 Pa-s to 1x10-2 Pa-s above melt temperature ... n 1, shear thickening. n 1, shear thinning. n = 1, Newtonian. Flow Between Parallel Plates ... – PowerPoint PPT presentation

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Title: Newtons Law of Viscosity


1
Newtons Law of Viscosity
  • ?xy -????dv(x)/dy
  • Shear Stress Tensor
  • Velocity Vector
  • Dependent on Geometry of Problem
  • Simplest cases of fluid flow 1D Flow

2
Shear Stress Tensor
Txx Txy Txz Tyx Tyy
Tyz Tzx Tzy Tzz
?xy
Tii Are Normal Stresses in the Shear Stress Tensor
3
Typical Viscosities
  • Metals 5x10-4 Pa-s to 1x10-2 Pa-s above melt
    temperature
  • Ceramics Molten Viscosities, In the range of
    1x10-3 Pa-s to 10Pa-s again in relation to the
    melt temperature
  • Polymers Molecular Weight Dependent, but 1 to
    1x105Pa-s above Tm or Tg
  • Log ???? 3.4 log Mw N

4
Viscosity Relations
  • ??Newtonian ?xy -??dvx/dy
  • Bingham ?xy ?0 ??dvx/dy
  • Power Law Fluids
  • ?xy A??dvx/dy)n
  • ngt1, shear thickening
  • nlt1, shear thinning
  • n 1, Newtonian

5
Flow Between Parallel Plates
y
2?
x
vx
x0
xL
Assumptions Incompressible Fluid Fully
Developed Flow Profile...Steady State Develop a
momentum balance within a small block of fluid
6
Momentum Balances
  • Pressure momentum flux per square area
  • Shear Stress Also a momentum flux per unit area
  • Goal Do an equivalent sum of momentum flux as
    one would do a sum or forces around a free body
    diagram

7
Flow Between Parallel Plates
  • Momentum Flux Across Surface y
  • LW?yx evaluated at y y
  • Momentum Flux Across Surface y ?y
  • LW ?yx evaluated at y y ?y
  • Momentum Flux Across Surface at x 0
  • W ?yvx(?vx), evaluated at x 0
  • Momentum Flux Across Surface at x L
  • W ?yvx(?vx), evaluated at x L

8
Flow Between Parallel Plates
  • Pressure Force on liquid at x 0
  • W?yPx0 W?yP0
  • Pressure Force on liquid at x L
  • W?yPxL W?yPL

9
Momentum Flux Sum, Integration
  • convective momentum (that carried through the
    cube) is the same given that the fluid velocity
    profile is the same...fully developed
  • LW?yx at y y ?y - ?yx at y y W ?yPL -
    P0 0
  • ?yx at y y ?y - ?yx at y y / ?y P0 -
    PL/L
  • d ?yx/dy P0 - PL/L
  • ?yx P0 - PLy/L C1
  • B. C. 1 Shear Stress 0 at y 0, midpoint in
    fluid
  • ?yx P0 - PLy/L

10
Applying Newtons Law
  • ?xy -??dvx/dy P0 - PLy/L
  • vx -P0 - PL/L ??y2/2 C2
  • B. C. 2 No slip at wall... vx 0 at y ?
  • C2 P0 - PL/L ?????2/2
  • Plug back into equation and exactly solve for vx

11
Summary of Results
  • Flow through // plates can be exactly solved.
  • Assumes viscosity is uniform and known, other
    assumptions include no slip at the wall, an
    incompressible fluid and a fully developed flow
    profile.
  • Other concerns could include friction which could
    alter m, and plate roughness.

12
How about a different geometry
  • Flow down an incline plane
  • Assumptions
  • Fully Developed Flow
  • Newtonian Fluid
  • Shear is a function of normal distance from plane
  • Gravity acting to aid flow
  • Pressurenot a contributing since flow isnt
    pressurized

13
Flow Down the Incline Plane
  • Momentum Balance
  • Neglect convection
  • Sforces 0
  • tyx _at_yy - tyx_at_yyDy DZDx rgcos(q) DxDyDz
    0
  • d tyx/dy rgcos(q)
  • Boundary Conditions no slip at wall
  • no shear stress at free surface,
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