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A flat square plate of side 20cm moves over other similar plate with a thin layer of 0.4cm of a liquid between them with force 1 Kgw moves of the plates uniformly with a velocity 1m/s. Calculate co-efficient of viscosity of the liquid.

Respuesta :

Answer:

[tex]\mu=0.98\ Pa.s[/tex]

Explanation:

Given:

  • dimension of square plate, [tex]l=0.2\ m[/tex]
  • thickness of fluid layer, [tex]dy=0.004\ m[/tex]
  • force on the fluid due to plate, [tex]F=1\ kgw=1\times 9.8=9.8\ N[/tex]
  • velocity of plate, [tex]du=1\ m.s^{-1}[/tex]

Using Newton's law of viscosity:

[tex]\tau=\mu.\frac{du}{dy}[/tex] ..........................................(1)

where:

[tex]\tau=[/tex] shear force on the surface on the fluid

[tex]\mu=[/tex] coefficient of (dynamic) viscosity

Now, shear force:

[tex]\tau=\frac{shear\ force}{area}[/tex]

[tex]\tau=\frac{9.8}{0.2\times0.2} \ Pa[/tex]

Putting respective values in eq.(1)

[tex]\frac{9.8}{0.2\times0.2}=\mu\times\frac{1}{0.004}[/tex]

[tex]\mu=0.98\ Pa.s[/tex]

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