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Two parallel surfaces move in opposite directions relative to each other at a velocity of 64 in/sec and are separated by a gap of 0.41 in. The gap is filled by a fluid of unknown viscosity. The relative motion is resisted by a shear stress of 0.42 lb/in2 due to the viscosity of the fluid. If the velocity gradient in the space between the surfaces is constant, determine the viscosity of the fluid in lb·s/in2.

Respuesta :

Answer:

[tex]\mu[/tex] = 2.6906 × [tex]10^{-3}[/tex] lb-s/in²

Explanation:

given data

velocity V = 64 in/sec

separated by a gap x = 0.41 in

relative motion by shear stress [tex]\tau[/tex]  = 0.42 lb/in²

solution

we know that shear stress is directly proportional to rate of change of velocity  as per newton's law of viscosity.

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

so here [tex]\mu[/tex] coefficient of dynamic viscosity and [tex]\frac{du}{dy}[/tex] is velocity gradient

and

[tex]\tau = \mu \times \frac{v1 - v2 }{h2-h0}[/tex]  

put here value and we get

0.42 =  [tex]\mu \times \frac{64}{0.41}[/tex]

[tex]\mu[/tex] = 2.6906 × [tex]10^{-3}[/tex] lb-s/in²

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