--- title: "Vertical drop design, Q 7.5 m3/s" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q75" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q75 site: BCT Engineering --- # Vertical drop design, Q 7.5 m3/s Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2079 Jestha, 8 marks - Video: [Design of Sarda Fall water Resources engineering](https://www.youtube.com/watch?v=Rncmf2oZ16o) — Study with Civil buddy - 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-q75 - 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 = 7.5 m³/sec; Side slope of the channel = 1:1; FSL U/S = 156.10 m; Bed level U/S = 155.15 m; Bed slope [level] D/S = 154.45 m; Bed width U/S and D/S = 1.5 m; Top width of crest = 0.55 m (for initial assumption); Cd = 0.415; Sp. Gr. of masonry drop structure = 2.25; Bligh's Coefficient = 6.0. ## Short answer The crest is set at the upstream bed level of 155.15 m with a length of 2.27 m. The total impervious floor length is 11.13 m and its thickness near the drop wall is 0.51 m, tapering to 0.30 m at the end. ## Answer #### Given data and assumptions - $Q = 7.5$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 156.10$ m, u/s bed $= 155.15$ m, d/s bed $= 154.45$ m. - Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 155.400$ 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 $$ \begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.327\ \text{m}^2 \\ v &= Q/A = 3.222\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.529\ \text{m} \end{aligned} $$ #### 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 $h_a$). With $L = B = 1.5$ m the head needed would put the crest below the u/s bed. The crest is therefore fixed at the u/s bed level and its length increased (splayed wing walls): $$ \begin{aligned} H &= (FSL_u + h_a) - \text{crest} = 156.629 - 155.150 = 1.479\ \text{m} \\ L &= \frac{Q}{1.838\,H^{3/2}} = 2.27\ \text{m} \end{aligned} $$ Height of crest above d/s bed $d_c = 155.15 - 154.45 = 0.70$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 3.307$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 2.179$ 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 $X = 1.076$ m. Provide **$X = 1.10$ m** (cistern floor level $= 153.35$ m). - With this $X$: $y_1 = 0.443$ m, $F_1 = 3.58$, $y_2 = 2.032$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(2.032 - 0.443) = 7.94$ m. #### 4. Crest (drop) wall Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 155.15 - 153.35 = 1.80$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.80 + 1.48}{\sqrt{2.25}} = 2.19\ \text{m} $$ Adopt bottom width $B_w = 2.19$ 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: $$ L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{3.307/75}} = 5.08\ \text{m} $$ The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 7.94$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 5.19$ m, so the u/s floor needed is $$ b_u = L - [2(d_1+d_2) + B_w + L_d] = -8.93\ \text{m} $$ Bligh's creep is already satisfied, so a nominal u/s floor of **1.0 m** is provided. Actual creep $= 14.13$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 11.13$ m. #### 6. Thickness of the floor Pressure head at the toe of the wall: creep to that point $= 4.39$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 4.39}{L_{act}} = 0.483$ m. At the end of the cistern the creep is 12.33 m and $h_2 = 0.089$ m. $$ t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.483}{1.25} = 0.51\ \text{m} $$ Adopt floor thickness **0.51 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 | 155.15 m | | Crest length / head $H$ | 2.27 m / 1.48 m | | Crest wall top / bottom width | 0.55 m / 2.19 m | | Cistern depth / length | 1.10 m / 7.94 m | | Floor: u/s 1.0 m, d/s 7.94 m, cut-offs 0.6 / 0.9 m | total creep 14.13 m | | Floor thickness | 0.51 m (taper to 0.30 m) | **Answer:** crest level 155.15 m, crest length 2.27 m, crest wall 0.55 m top and 2.19 m bottom width, cistern depth 1.10 m and length 7.94 m, floor length 11.13 m, floor thickness 0.51 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. Why is the crest fixed at the u/s bed level and the crest length increased? A. Because the cistern is too deep B. Because Bligh's C is high C. Because the floor is thick D. With L = B = 1.5 m the required head would put the crest below the u/s bed, so crest is kept at bed level and splayed to 2.27 m Answer: D. A 1.5 m crest would need a head putting the crest below the bed, so crest level 155.15 m is adopted and L = Q/(1.838 H^1.5) = 2.27 m. (Key point: Crest fixed at u/s bed) 2. How is the cistern depth X decided? A. So the conjugate depth y2 equals the tail water depth over the cistern, D_d + X (required X = 1.076 m, 1.10 m provided) B. As half the drop C. As equal to the crest length D. As 0.30 m minimum Answer: A. The jump forms in the cistern when y2 = D_d + X, solved with the energy equation, giving X = 1.10 m. (Key point: Cistern design) 3. How does Bligh's creep check turn out? A. Required creep 4.20 m is not met B. Required 6 x 0.70 = 4.20 m is exceeded easily; actual creep 14.13 m is safe, so a nominal 1.0 m u/s floor is provided C. Required creep is 14.13 m D. Creep is not checked Answer: B. L = C H_L = 4.2 m; cut-offs, wall and the d/s floor needed for the cistern already exceed this. (Key point: Creep length) 4. Why is the d/s floor length adopted as 7.94 m rather than the Bligh value 5.08 m? A. Because Bligh's formula gives 7.94 m B. Because the u/s floor is negative C. The d/s floor must contain the cistern (hydraulic jump length) D. To thicken the floor Answer: C. The floor from the wall toe has to contain the jump length of 7.94 m, which exceeds the Bligh requirement. (Key point: Floor length driver) 5. How is floor thickness near the wall found? A. t = h1/(G + 1) B. t = 4/3 x h1/(G - 1) = 0.51 m, using residual uplift head at that point C. t = H_L D. t = 0.30 m everywhere Answer: B. Thickness resists uplift: t = (4/3) h/(G - 1) with h1 = 0.483 m and G = 2.25, giving 0.51 m, tapering to a minimum of 0.30 m. 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