--- title: "Cipolletti weir slope and approach velocity" subject: "Fluid Mechanics" chapter: 6 key: "semester_3/fluid_mechanics:cipolletti-slope-velocity-of-approach" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cipolletti-slope-velocity-of-approach site: BCT Engineering --- # Cipolletti weir slope and approach velocity Fluid Mechanics · Chapter 6: Flow measurement - Asked in: 2078 Kartik, 3+1 marks - Video: [VELOCITY APPROACH OF WEIR OR NOTCH](https://www.youtube.com/watch?v=fq6ak8TBVWA) — FM_for All - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cipolletti-slope-velocity-of-approach - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6 ## Question Show that the slope of Cipolletti weir is 1:4. How can you account for velocity of approach while computing the discharge over weirs? ## Short answer The sides lopes of a Cipolletti weir compensate for end contractions; equating the triangular discharge to the loss gives tanθ = 1/4, i.e., 1 horizontal to 4 vertical. Velocity of approach is accounted by adding the velocity head ha = Va²/(2g) to the weir head, using effective head H+ha, and solving iteratively. ## Answer #### 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 $Q = \tfrac23 C_d L\sqrt{2g}\,H^{3/2}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the sides of a Cipolletti weir sloped? A. So extra flow through the triangular ends compensates the loss from end contractions B. To increase the velocity of approach C. To eliminate the need for a discharge coefficient D. To make the crest length equal to the channel width Answer: A. The side slope is chosen so the extra discharge through the triangular ends equals the reduction caused by end contractions on a rectangular weir. (Key point: Purpose of Cipolletti slope) 2. Which equality gives tan(theta) = 1/4? A. Q2 = Q1 B. Q2 = Q0 + Q1 C. Q2 (triangular ends) = Q0 - Q1 (loss from contractions) D. Q1 = Q0 Answer: C. Setting (8/15) tan(theta) H^(5/2) = (2/3)(0.2)H^(5/2) gives tan(theta) = 0.25, i.e. 1 horizontal to 4 vertical. (Key point: Equating flows) 3. How is velocity of approach accounted for in weir discharge? A. Use effective head H + ha, with ha = Va^2/2g B. Subtract the velocity head from H C. Multiply Q by Va D. Ignore it and increase Cd by 10% Answer: A. Velocity of approach adds a velocity head ha to the head, so Q uses (H + ha)^(3/2) - ha^(3/2). (Key point: Velocity of approach effect) 4. Why is the velocity-of-approach correction solved by trial? A. Va depends on Q, which is itself being calculated B. The head cannot be measured C. Cd is unknown D. The channel area changes with time Answer: A. Va = Q/A depends on Q, so find Q ignoring ha, compute Va and ha, recompute Q, and repeat until it converges. 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