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A Newtonian fluid with a viscosity of 50 cP is placed between two large parallel separated by a distance of 8 mm. Each plate has an area of 2 m^2, The upper plate the positive x-direction with a velocity of 0.4 m/s while the lower plate is kept stationary Calculate the steady force applied to the upper plate. The fluid in part (a) is replaced with another Newtonian fluid of viscosity 5 cP.. If a steady force applied to the upper plate is the same as that of part(a), calculate the velocity of the upper plate.

Respuesta :

Answer:

F= 5 N

V= 4 m/s

Explanation:

Given that fluid is Newtonian fluid .As we know that for Newtonian fluid the shear stress given as

[tex]\tau =\mu \dfrac{du}{dy}[/tex]

We also know that

Force = shear stress x area of plate

Now by putting the values

[tex]\tau =\mu \dfrac{du}{dy}[/tex]

Here velocity of fluid profile is linear.

[tex]\tau =\mu \dfrac{U}{h}[/tex]

[tex]\tau =50\times 10^{-2}\times 0.1\times \dfrac{0.4}{0.008}[/tex]

τ= 2.5 Pa

So force = 2.5 x 2

F= 5 N

Now when fluid is replaced by another fluid but force is constant.The height of plate is also constant.

Lets take velocity of fluid is V in the new condition

So  we can say that

[tex]\mu_1U=\mu_2V[/tex]

Now by putting the value

50 x 0.4 = 5 x V

V= 4 m/s

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