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A particular fluid has a shear stress of 0.005 N m-at a shear rate of 1 s- and a shear stress of 2 N m-2 at a shear rate of 50 - Present an argument as to whether this fluid is Newtonian or non-Newtonian.

Respuesta :

Answer:

The given fluid is Non-Newtonian

Explanation:

Given:

τ₁ = 0.005 N/m²

[tex](\frac{\textup{du}}{\textup{dy}})_1[/tex]  = 1 s⁻¹

τ₂ = 2 N/m²

[tex](\frac{\textup{du}}{\textup{dy}})_2[/tex]  = 50 s⁻¹

Now,

From Newton's law of viscosity

τ = [tex]\mu\times\frac{\textup{du}}{\textup{dy}}[/tex]

Here, μ is the dynamic viscosity of the fluid

thus,

for the first case

0.005 = μ₁ × 1

or

μ₁ = 0.005 Pa.s

similarly

For the second case

2 = μ₂ × 50

or

μ₂ = 0.04 Pa.s

Since,

μ₁ ≠ μ₂

The given fluid is Non-Newtonian

as for the Newtonian Fluid the value of dynamic viscosity remains constant

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