Question

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.

Answer 1

Answer:

$\mu$ = 2.6906 × $10^{-3}$ lb-s/in²

Explanation:

given data

velocity V = 64 in/sec

separated by a gap x = 0.41 in

relative motion by shear stress $\tau$  = 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.

$\tau = \mu \times \frac{du}{dy}$      ....................1

so here $\mu$ coefficient of dynamic viscosity and $\frac{du}{dy}$ is velocity gradient

and

$\tau = \mu \times \frac{v1 - v2 }{h2-h0}$  

put here value and we get

0.42 =  $\mu \times \frac{64}{0.41}$

$\mu$ = 2.6906 × $10^{-3}$ lb-s/in²