Solved: Flow over Spillway: Calculating Q with Bernoulli & Torricelli

[kubaanglin]

Homework Statement

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When the level of water in a reservoir is too high, the water spills out over a spillway,
as illustrated in the figure below

Neglecting viscosity, show that the water flow $Q$ over the spillway is given by
$$Q=\frac{2}{3}w\sqrt{2gy^3}$$

Homework Equations

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Bernoulli's Equation: $P_1+\rho gh_1+\frac{1}{2}\rho v_1^2=P_2+\rho gh_2+\frac{1}{2}\rho v_2^2$
Torricelli's Equation: $v=\sqrt{2gh}$

The Attempt at a Solution

$Q=Av=(wy)(\sqrt{2gy})=w\sqrt{2gy^3}$

I am not sure how to get $\frac{2}{3}$ in this equation as indicated by the answer. I feel that I am oversimplifying this by assuming the height in Torricelli's equation is just $y$ when it should account for flow over an area as opposed to a point. Would this mean the height is $\frac{y}{2}$ to get the average velocity of the fluid?

 

[haruspex]

it should account for flow over an area as opposed to a point.

Yes. The consequence for the total flow varies as the height above the spillway increases from0 to y.

Would this mean the height is $\frac{y}{2}$ to get the average velocity of the fluid?

No, that's still too simplistic. Consider a thin horizontal slice from x to x+dx above the spillway. What is the flow due to that?

 

[kubaanglin]

Oh, that makes sense.

$Q=Av=(wy)\int_0^y (\sqrt{2gy})=(wy\sqrt{2g})\int_0^y \sqrt{y}=(wy\sqrt{2g})(\frac{\sqrt{y^3}}{\frac{3}{2}})=\frac{2}{3}w\sqrt{2gy^3}$

Thank you!

 

[haruspex]

Oh, that makes sense.

$Q=Av=(wy)\int_0^y (\sqrt{2gy})=(wy\sqrt{2g})\int_0^y \sqrt{y}=(wy\sqrt{2g})(\frac{\sqrt{y^3}}{\frac{3}{2}})=\frac{2}{3}w\sqrt{2gy^3}$

Thank you!

Good job.