Cippoletti weir slope and discharge formula

Cippoletti weir sides have slope 1:4

A Cippoletti weir is a trapezoidal weir whose side slope makes the extra flow through the triangular ends equal to the loss from the end contractions of a rectangular weir. For crest length $L$ and head $H$:

• Rectangular weir with 2 end contractions (Francis): $Q_1 = \frac{2}{3}C_d\sqrt{2g}\,(L - 0.2H)H^{3/2}$
• Loss due to contractions: $\Delta Q = \frac{2}{3}C_d\sqrt{2g}\,(0.2H)H^{3/2}$
• Two triangular ends (slope $\tan\theta$ from the vertical): $Q_2 = \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$

Equate $Q_2 = \Delta Q$:

$$\frac{8}{15}\tan\theta = \frac{2}{3}(0.2)\;\Rightarrow\;\tan\theta = 0.25 = \frac{1}{4}$$

So the side slope is 1 horizontal : 4 vertical.

Discharge formula for the trapezoidal notch (from first principles)

Base width $b = 0.1$ m, top width $0.5$ m, depth $0.3$ m. The half-width increases by $(0.5 - 0.1)/2 = 0.2$ m over 0.3 m of height, so the side slope is $\tan\theta = 0.2/0.3 = 0.6667$ (horizontal per unit vertical). This is not a Cippoletti slope (0.25).

Take a horizontal strip at depth $h$ below the water surface (upstream head $H$ above the base). Its width is $b + 2\tan\theta\,(H - h)$, its thickness is $dh$, and the velocity through it is $\sqrt{2gh}$:

$$dQ = C_d\left[b + 2\tan\theta\,(H - h)\right]\sqrt{2gh}\;dh$$ $$\begin{aligned} Q &= C_d\sqrt{2g}\int_0^H\left[(b + 2H\tan\theta)h^{1/2} - 2\tan\theta\,h^{3/2}\right]dh\\ &= C_d\sqrt{2g}\left[\frac{2}{3}(b + 2H\tan\theta)H^{3/2} - \frac{4}{5}\tan\theta\,H^{5/2}\right] \end{aligned}$$ $$Q = \frac{2}{3}C_d\sqrt{2g}\,b\,H^{3/2} + \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$$

(a rectangular notch plus a triangular notch).

Discharge at H = 0.228 m

$$\begin{aligned} \frac{2}{3}C_d\sqrt{2g}\,bH^{3/2} &= \frac{2}{3}(0.6)(4.4294)(0.1)(0.228)^{1.5} = 0.01929\ \text{m}^3/\text{s}\\ \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2} &= \frac{8}{15}(0.6)(4.4294)(0.6667)(0.228)^{2.5} = 0.02346\ \text{m}^3/\text{s} \end{aligned}$$ $$Q = 0.01929 + 0.02346 = 0.04274\ \text{m}^3/\text{s}$$

Answer: $Q = \frac{2}{3}C_d\sqrt{2g}\,bH^{3/2} + \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$; for the given notch $Q \approx 0.0427$ m³/s (42.7 L/s).