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antoniya [11.8K]
4 years ago
10

A fluid 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 su

rface, 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.
Engineering
1 answer:
sattari [20]4 years ago
5 0

Answer:

Explanation:

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

\tau = \mu \cfrac{\partial v}{\partial y}

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 y=75 \, mm and the velocity at that point is 1.125 \, m/s

We can put that in mathematical terms as:

v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\

From the no-slip condition, we can deduce that A=0 and so we are left with just two terms:

v(y) = By + C y ^2 \\

We know that the vertex is at y= 75 \, mm and so we can rewrite the last equation as:

v(y) = k(y-75 \, mm) ^2+h

where k and h are constants to be determined. First we check that v( 75 \, mm) = 1.125 \,  m/s :

v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \,  m/s

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.

v( 0) = k( -75 \, mm) ^2+  1.125 \,  m/s= 0 \quad (no \, \textendash slip)  \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= -  \cfrac{0.2}{mm \times s}

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

v( y) = 1125-  0.2 ( y -75 ) ^2

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

In SI units it would be:

v( y) = 1.125-  200 ( y -0.075 ) ^2

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

\tau = \mu \cfrac{\partial v}{\partial y}

\tau =0.048\,   \cdot  (-400) ( y-0.075   )

for y= 0.050 \, m we have:

\tau =0.048\,   \cdot  (-400) ( 0.050 -0.075   ) = 0.48\, Pa

Which is our final result

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