# Calculation of velocity of water flowing along inclined plane by applying Navier-Stokes theorem

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 :

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?