--- title: "Vertical drop design, Q 2.1 m3/s" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q21" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q21 site: BCT Engineering --- # Vertical drop design, Q 2.1 m3/s Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2076 Bhadra, 8 marks - Video: [Design of sharda fall | Vertical drop structure | Irrigation Numerical | Most important problem](https://www.youtube.com/watch?v=r_VaWVvL3YY) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q21 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design crest, length and thickness of impervious floor of a vertical drop structure for the following data: Discharge = 2.1 m³/sec; Side slope of the channel = 1:1; FSL U/S = 276.10 m; Bed level U/S = 275.15 m; Bed level D/S = 274.45 m; Bed width U/S and D/S = 1.5 m; Top width of crest = 0.55 m (for initial assumption); For rectangular crest Cd = 0.415; Sp. Gr. of masonry drop structure = 2.25; Bligh's coefficient = 6.0. ## Short answer Crest length equals bed width 1.50 m; crest level is 275.31 m with head 0.83 m. Impervious floor total length 6.92 m, thickness 0.48 m under cistern near wall tapering to 0.30 m. ## Answer #### Given data and assumptions - $Q = 2.1$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 276.10$ m, u/s bed $= 275.15$ m, d/s bed $= 274.45$ m. - Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 275.400$ m); drop in water level $H_L = 0.700$ m. - Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.25$. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.328\ \text{m}^2 \\ v &= Q/A = 0.902\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.041\ \text{m} \end{aligned} $$ #### 2. Crest length, head and crest level Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). Take crest length equal to the bed width, $L = B$: $$ H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{2.1}{1.838\times1.5}\right)^{2/3} = 0.834\ \text{m} $$ $$ \text{Crest level} = FSL_u + h_a - H = 276.100 + 0.041 - 0.834 = 275.31\ \text{m} $$ Height of crest above d/s bed $d_c = 275.31 - 274.45 = 0.86$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 1.400$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.691$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$: - Required $X = 0.208$ m. Provide **$X = 0.25$ m** (cistern floor level $= 274.20$ m). - With this $X$: $y_1 = 0.242$ m, $F_1 = 3.74$, $y_2 = 1.168$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.168 - 0.242) = 4.63$ m. #### 4. Crest (drop) wall Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 275.31 - 274.20 = 1.11$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.11 + 0.83}{\sqrt{2.25}} = 1.29\ \text{m} $$ Adopt bottom width $B_w = 1.29$ m (not less than the top width), with the d/s face vertical or slightly battered. #### 5. Length of impervious floor (Bligh's theory) Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s: $$ L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{1.400/75}} = 4.10\ \text{m} $$ The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 4.63$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.29$ m, so the u/s floor needed is $$ b_u = L - [2(d_1+d_2) + B_w + L_d] = -4.72\ \text{m} $$ Bligh's creep is already satisfied, so a nominal u/s floor of **1.0 m** is provided. Actual creep $= 9.92$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 6.92$ m. #### 6. Thickness of the floor Pressure head at the toe of the wall: creep to that point $= 3.49$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.49}{L_{act}} = 0.453$ m. At the end of the cistern the creep is 8.12 m and $h_2 = 0.127$ m. $$ t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.453}{1.25} = 0.48\ \text{m} $$ Adopt floor thickness **0.48 m** under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m). #### Summary | Component | Value | |---|---| | Crest level | 275.31 m | | Crest length / head $H$ | 1.50 m / 0.83 m | | Crest wall top / bottom width | 0.55 m / 1.29 m | | Cistern depth / length | 0.25 m / 4.63 m | | Floor: u/s 1.0 m, d/s 4.63 m, cut-offs 0.6 / 0.9 m | total creep 9.92 m | | Floor thickness | 0.48 m (taper to 0.30 m) | **Answer:** crest level 275.31 m, crest length 1.50 m, crest wall 0.55 m top and 1.29 m bottom width, cistern depth 0.25 m and length 4.63 m, floor length 6.92 m, floor thickness 0.48 m. Sketch: ```text FSL u/s ~~~~~~~~~~~| | crest u/s bed ___________|__ FSL d/s ~~~~ | u/s floor | \ drop ~~~~~ d1 | | \_____cistern________ d/s bed | |<-Lc->| d/s floor | d2 |____ floor thickness __| ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With L = B, how is the crest level obtained? A. FSL_u - H B. FSL_u + ha + H C. u/s bed + H D. FSL_u + ha - H Answer: D. H is total head including ha, so the crest is H below the total energy level: 276.100 + 0.041 - 0.834 = 275.31 m. (Key point: Crest level) 2. The initial depth y1 at the start of the jump is found from A. the continuity equation alone with y1 = D_d B. Bligh's creep formula C. energy conservation: E1 + X = y1 + q^2/(2g y1^2), where E1 is measured above the d/s bed D. the crest wall width Answer: C. Energy at the u/s level plus the cistern depth X equals depth plus velocity head at the contracted section y1. (Key point: Finding y1 on the cistern floor) 3. What is checked to show the floor is safe against piping? A. Actual creep length is at least C*HL (9.92 m >= 4.20 m) B. Actual creep length must be less than C*HL C. Cistern length is greater than the floor length D. Crest height is greater than the floor thickness Answer: A. Bligh's criterion requires total creep length >= C x HL; here 9.92 m exceeds 6 x 0.70 = 4.20 m. (Key point: Bligh creep check) 4. Why is the floor 0.48 m near the wall but tapers to 0.30 m at its end? A. The d/s end carries a heavier load B. Uplift head is largest near the wall and falls with increasing creep length, so less thickness is needed downstream C. The cistern is deeper at the d/s end D. Because the gravity wall is lighter at the end Answer: B. Residual uplift head drops from 0.453 m near the wall to 0.127 m at the end of the cistern, so the floor can be thinner, with 0.30 m as minimum. (Key point: Tapering floor thickness) ## Close variants set in other exams - [Vertical drop design, Q 7.5 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q75) — Irrigation and Drainage Engineering, chapter 7, asked 4× - [Vertical drop design, Q 5.2 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q52) — Irrigation and Drainage Engineering, chapter 7, asked 4× - [Vertical drop design, Q 1.8 m3/s (bed 205.05)](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q18-bed20505) — Irrigation and Drainage Engineering, chapter 7, asked 4× ## Related in Irrigation and Drainage Engineering - [Vertical drop crest level and width, Q 30](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-crest-q30-fsl315) — Irrigation and Drainage Engineering, chapter 7 - [Crest width, cistern length of vertical drop, Q 1.55](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q155-drop075) — Irrigation and Drainage Engineering, chapter 7, asked 2× - [Vertical drop design, Q 5.1 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q51) — Irrigation and Drainage Engineering, chapter 7 - [Vertical drop design, Q 1.8 m3/s (drop 0.75 m)](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q18-drop075) — Irrigation and Drainage Engineering, chapter 7, asked 2× - [Vertical drop crest and cistern, Q 4.5](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q45-fsl1065) — Irrigation and Drainage Engineering, chapter 7 - [Minimum impervious floor length, 10000 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abarrage-min-floor-length-q10000) — Irrigation and Drainage Engineering, chapter 5, asked 2× ## Related in other subjects - [Channel earthwork with drop at 0+090](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Achannel-e-drop-0-090) — Estimating and Costing, chapter 6 - [Design of side intake with trashrack](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aintake-side-design-3315masl) — Hydropower Engineering, chapter 5 - [Economical trapezoidal channel side length](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomical-trapezoidal-side-length) — Hydro Power, chapter 1