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}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cmu%20%5Ccfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D)
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 ![1.125 \, m/s](https://tex.z-dn.net/?f=1.125%20%5C%2C%20m%2Fs)
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\\](https://tex.z-dn.net/?f=v%28y%29%3D%20A%2B%20By%20%2BCy%5E2%20%5C%5Cv%280%29%20%3D%200%5C%5Cv%2875%20%5C%2C%20mm%29%20%3D%201.125%20%5C%2C%20m%2Fs%5C%5Cv%27%2875%20%5C%2C%20mm%29%20%3D%200%5C%5C)
From the no-slip condition, we can deduce that
and so we are left with just two terms:
![v(y) = By + C y ^2 \\](https://tex.z-dn.net/?f=v%28y%29%20%3D%20By%20%2B%20C%20y%20%5E2%20%5C%5C)
We know that the vertex is at
and so we can rewrite the last equation as:
![v(y) = k(y-75 \, mm) ^2+h](https://tex.z-dn.net/?f=v%28y%29%20%3D%20k%28y-75%20%5C%2C%20mm%29%20%5E2%2Bh)
where k and h are constants to be determined. First we check that
:
![v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \, m/s](https://tex.z-dn.net/?f=v%28%2075%20%5C%2C%20mm%29%20%3D%20k%2875%20%5C%2C%20mm%20-75%20%5C%2C%20mm%29%20%5E2%2Bh%20%3D%20h%20%3D%201.125%20%5C%2C%20m%2Fs%5C%5C%5C%5Ch%3D%20v_%7Bmax%7D%20%3D%201.125%20%5C%2C%20%20m%2Fs)
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}](https://tex.z-dn.net/?f=v%28%200%29%20%3D%20k%28%20-75%20%5C%2C%20mm%29%20%5E2%2B%20%201.125%20%5C%2C%20%20m%2Fs%3D%200%20%5Cquad%20%28no%20%5C%2C%20%5Ctextendash%20slip%29%20%20%5C%5C%5C%5Ck%3D%20-%20%5Ccfrac%7B%201.125%20%5C%2C%20m%2Fs%20%7D%7B%2875%20%5C%2C%20mm%20%29%20%5E2%7D%20%3D%20-%20%5Ccfrac%7B%201125%20%5C%2C%20mm%2Fs%20%7D%7B%2875%20%5C%2C%20mm%20%29%20%5E2%7D%5C%5C%5C%5Ck%3D%20-%20%20%5Ccfrac%7B0.2%7D%7Bmm%20%5Ctimes%20s%7D)
And thus we find that the final expression for the fluid's velocity is:
![v( y) = 1125- 0.2 ( y -75 ) ^2](https://tex.z-dn.net/?f=v%28%20y%29%20%3D%201125-%20%200.2%20%28%20y%20-75%20%29%20%5E2)
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](https://tex.z-dn.net/?f=v%28%20y%29%20%3D%201.125-%20%20200%20%28%20y%20-0.075%20%29%20%5E2)
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}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cmu%20%5Ccfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D)
![\tau =0.048\, \cdot (-400) ( y-0.075 )](https://tex.z-dn.net/?f=%5Ctau%20%3D0.048%5C%2C%20%20%20%5Ccdot%20%20%28-400%29%20%28%20y-0.075%20%20%20%29)
for
we have:
![\tau =0.048\, \cdot (-400) ( 0.050 -0.075 ) = 0.48\, Pa](https://tex.z-dn.net/?f=%5Ctau%20%3D0.048%5C%2C%20%20%20%5Ccdot%20%20%28-400%29%20%28%200.050%20-0.075%20%20%20%29%20%3D%200.48%5C%2C%20Pa)
Which is our final result