Cipolletti weir slope and approach velocity

Slope of Cipolletti weir is 1 horizontal : 4 vertical

A Cipolletti weir is a trapezoidal weir whose side slopes are chosen so that the extra discharge through the two triangular ends makes up for the reduction in discharge caused by the end contractions of a rectangular weir.

Let $L$ be the crest length, $H$ the head, and $\theta$ the angle of the side slope with the vertical.

Rectangular weir with two end contractions (Francis):

$$Q_1 = \frac{2}{3}C_d\sqrt{2g}\,(L - 0.1\times 2H)H^{3/2}$$

Rectangular weir without end contractions:

$$Q_0 = \frac{2}{3}C_d\sqrt{2g}\,LH^{3/2}$$

so the loss due to the contractions is

$$Q_0 - Q_1 = \frac{2}{3}C_d\sqrt{2g}\,(0.2H)H^{3/2}$$

Discharge through the two triangular ends (together they form one triangular notch of half-angle $\theta$):

$$Q_2 = \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$$

For a Cipolletti weir, $Q_2 = Q_0 - Q_1$:

$$\frac{8}{15}\tan\theta\,H^{5/2} = \frac{2}{3}(0.2)H^{5/2}\;\Rightarrow\;\tan\theta = \frac{0.1333\times 15}{8} = 0.25$$

So $\tan\theta = \dfrac{1}{4}$: 1 horizontal to 4 vertical ($\theta = 14.04^\circ$ from the vertical).

The discharge is then $Q = \dfrac{2}{3}C_d L\sqrt{2g}\,H^{3/2}$ (with $L$ the base length), with no correction for end contractions.

Velocity of approach

The velocity of approach $V_a = Q/A_{channel}$ adds a velocity head $h_a = V_a^2/2g$ to the head over the weir, so the effective head is $H + h_a$:

$$Q = \frac{2}{3}C_dL\sqrt{2g}\left[(H + h_a)^{3/2} - h_a^{3/2}\right]$$

(For a triangular notch, $Q = \dfrac{8}{15}C_d\tan\dfrac{\theta}{2}\sqrt{2g}\left[(H + h_a)^{5/2} - h_a^{5/2}\right]$.) Since $V_a$ depends on $Q$, solve by trial: first find $Q$ ignoring $h_a$, then find $V_a = Q/A$ and $h_a$, and recompute $Q$; repeat until it converges. If $h_a$ is negligible (a large approach channel), simply use $Q = \tfrac23 C_d L\sqrt{2g}\,H^{3/2}$.