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Assume that water flows along a inlined plane through a circular irrigation waterway. The inclined plane has its bottom length $B=50 m$, and has height $H=10 m$. So the angle of inclination $\theta$ satisfies $\operatorname{sin}\theta=B/(H^2+B^2)^{1/2}=0.1961$ .

Assume that the circular irrigation waterway has its diameter $D = 0.2 m$.

Water has its density $\rho=998.20 kg/m^3$ and has viscosity $\mu=0.0010087 kg/m\cdot s$ (at $20^{\circ}$C) (true?)

Now we want to calculate the velocity profile of the water. I found an associated result, which can be obtained via the Navier-Stokes equation :

enter image description here

I'm trying to substitute the above conditions into the formula $u(y)={\rho g sin\theta \over \mu}(hy-{1 \over 2}y^2)$

If we can set $h$ as the diameter $D=0.2 m$, then for $y=D$ also, we get

$$u(D) = {998.20 \cdot 9.8 \cdot sin {\theta} \over {\mu}} \cdot {D^2 \over 2} = {998.20 \cdot 9.8 \cdot{0.1961} \over 0.0010087} \cdot {D^2 \over 2} =38035 m/s $$

(True?)

And uhm..why we get velocity so high? Think that this result is correct? Or..Is there a mistake which I made? Is there a point that I'm missing? Then, where?

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If I understood the problem and your associated result correctly, they do not quite describe the same situation. In the problem, water is flowing down a cylindrical tube, so Poisefeuille flow is more relevant, whereas the flow of oil is planar. I think this also explains the overestimation of velocity since the open air on top of the planar flow does not slow down the fluid as much as a closed tube would.

Hope this helps and tell me if you find some mistakes.

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  • $\begingroup$ O.K. I will try to find or use a formula of velocity associated the Poisefeuille flow. For the open air on top of the planar flow, is the degree of not being able to slow down that much? Thanks for comment. $\endgroup$
    – Plantation
    Apr 26, 2022 at 1:46

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