--- title: "Cippoletti weir slope and discharge formula" subject: "Fluid Mechanics" chapter: 6 key: "semester_3/fluid_mechanics:cipolletti-weir-trapezoidal-notch-discharge" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cipolletti-weir-trapezoidal-notch-discharge site: BCT Engineering --- # Cippoletti weir slope and discharge formula Fluid Mechanics · Chapter 6: Flow measurement - Asked in: 2076 Chaitra, 2+6+2 marks - Video: [Cippolletti Notch/Weir || Fluid Mechanics || Derivation of discharge equation over the weir](https://www.youtube.com/watch?v=gBJj714RAhI) — Competitive Engineers - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cipolletti-weir-trapezoidal-notch-discharge - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6 ## Question Prove that in Cippoletti weir the sides have a slope of 1:4. A sharp-edged notch is in the form of a symmetrical trapezium. The horizontal base is 100 mm wide, the top is 500 mm wide and the depth is 300 mm. Develop from first principles a formula relating the discharge to the upstream water level, and estimate the discharge when the upstream water surface is 228 mm above the level of the base of the notch. Assume that Cd = 0.6 and that the velocity of approach is negligible. ## Short answer A Cippoletti weir has side slope 1:4 because the extra triangular end discharge equals the contraction loss. For the given notch, integrating horizontal strips gives Q = (2/3)Cd sqrt(2g) b H^(3/2) + (8/15)Cd sqrt(2g) tanθ H^(5/2). With b=0.1 m, tanθ=0.6667, H=0.228 m, Cd=0.6, Q ≈ 0.0427 m^3/s. ## Answer #### 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition gives the 1:4 side slope of a Cipolletti weir? A. Flow through two triangular ends equals the loss from end contractions B. The notch angle is 90 degrees C. The velocity of approach is zero D. Triangular and rectangular parts have equal width Answer: A. Equating (8/15)tan(theta) to (2/3)(0.2) gives tan(theta) = 0.25. (Key point: Cipolletti slope condition) 2. For the notch with base 100 mm, top 500 mm and depth 300 mm, what is tan(theta) and is it Cipolletti? A. 1.333, not Cipolletti B. 0.6667, Cipolletti C. 0.25, Cipolletti D. 0.6667, not Cipolletti Answer: D. Half-width increase 0.2 m over 0.3 m gives tan(theta) = 0.6667, which differs from the Cipolletti value 0.25. (Key point: Given notch slope) 3. How is the trapezoidal notch discharge derived from first principles? A. Multiply the top width by sqrt(2gH) B. Use the mean width times the full-head velocity C. Differentiate Q with respect to H D. Integrate dQ = Cd*(strip width)*sqrt(2gh)*dh from 0 to H, with width b + 2 tan(theta)(H - h) Answer: D. A strip at depth h has velocity sqrt(2gh) and width b + 2 tan(theta)(H - h). Integrating gives the result. (Key point: Strip integration) 4. What does the resulting formula represent? A. Two rectangular notches B. A triangular notch only C. A rectangular notch plus a triangular notch D. A rectangular notch minus a triangular notch Answer: C. Q = (2/3)Cd sqrt(2g) b H^(3/2) + (8/15)Cd sqrt(2g) tan(theta) H^(5/2). 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