Afluid has a dynamic viscosity of 0.048 pa. s and a specific gravity of 0.913. for the flow of such a fluid over a flat solid surface, the velocity at a point 75 mm away from the surface is 1.125 m/s. calculate the shear stress at a point 50 mm away from the solid surface if the velocity distribution of the fluid can be represented by a parabolic velocity profile with the vertex of the parabola at a point 75 mm away from the surface where the velocity is 1.125 m/s? the general equation of parabola can be assumed to be: v = a + by + cy2 , where a, b, and c are constants.

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at and the velocity at that point is

We can put that in mathematical terms as:

From the no-slip condition, we can deduce that and so we are left with just two terms:

We know that the vertex is at and so we can rewrite the last equation as:

where k and h are constants to be determined. First we check that :

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

And thus we find that the final expression for the fluid's velocity is:

where v is in mm/s and y is in mm.

In SI units it would be:

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

for we have:

Which is our final result

answer:

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explanation:

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