--- title: "Vertical drop design, Q 5.1 m3/s" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q51" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q51 site: BCT Engineering --- # Vertical drop design, Q 5.1 m3/s Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2078 Chaitra, 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-q51 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design a vertical type drop structure for the following data: Discharge = 5.1 m³/sec; Bed level U/S = 255.15 m, D/S = 254.45 m; Normal flow depth U/S, D/S = 0.95 m; Bed width of canal U/S, D/S = 2.5 m; Side slope of canal = 1:1. Take the top width of crest wall as 0.5 m for initial assumption. Assume other suitable data if necessary. ## Short answer Crest level is 255.15 m with a length of 2.5 m. The drop wall has a top width of 0.5 m and bottom width of 1.52 m. The impervious floor total length is 8.10 m with a thickness of 0.50 m near the wall. ## Answer #### Given data and assumptions - $Q = 5.1$ m³/s, bed width $B = 2.5$ m, side slope 1:1, u/s bed $= 255.15$ m, d/s bed $= 254.45$ m. - U/s FSL $=$ bed $+$ normal depth $= 255.15 + 0.95 = 256.10$ m; d/s FSL $= 254.45 + 0.95 = 255.40$ m. Full supply depth $D_u = D_d = 0.950$ m; drop in water level $H_L = 0.700$ m. - Assumed (not given): Bligh's coefficient $C = 6$ (sandy loam soil), specific gravity of masonry $G = 2.24$, and $C_d = 0.415$ for a rectangular crest. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD_u)D_u = (2.5 + 1\times0.950)(0.950) = 3.278\ \text{m}^2 \\ v &= Q/A = 1.556\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.123\ \text{m} \end{aligned} $$ #### 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, $L = B$: $$ H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{5.1}{1.838\times2.5}\right)^{2/3} = 1.072\ \text{m} $$ $$ \text{Crest level} = FSL_u + h_a - H = 256.100 + 0.123 - 1.072 = 255.15\ \text{m} $$ Height of crest above d/s bed $d_c = 255.15 - 254.45 = 0.70$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 2.040$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.773$ 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 $X = 0.494$ m. Provide **$X = 0.50$ m** (cistern floor level $= 253.95$ m). - With this $X$: $y_1 = 0.330$ m, $F_1 = 3.43$, $y_2 = 1.446$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.446 - 0.330) = 5.58$ m. #### 4. Crest (drop) wall Top width (assumed) $T = 0.50$ m; check $0.55\sqrt{d_c} = 0.55\sqrt{0.70} = 0.46$ m $< 0.50$ m, so it is adequate. Height of wall from cistern floor to crest $H_w = 255.15 - 253.95 = 1.20$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.20 + 1.07}{\sqrt{2.24}} = 1.52\ \text{m} $$ Adopt bottom width $B_w = 1.52$ 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: $$ L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{2.040/75}} = 4.51\ \text{m} $$ The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 5.58$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.52$ m, so the u/s floor needed is $$ b_u = L - [2(d_1+d_2) + B_w + L_d] = -5.90\ \text{m} $$ Bligh's creep is already satisfied, so a nominal u/s floor of **1.0 m** is provided. Actual creep $= 11.10$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 8.10$ m. #### 6. Thickness of the floor Pressure head at the toe of the wall: creep to that point $= 3.72$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.72}{L_{act}} = 0.465$ m. At the end of the cistern the creep is 9.30 m and $h_2 = 0.114$ m. $$ t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.465}{1.24} = 0.50\ \text{m} $$ Adopt floor thickness **0.50 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 | 255.15 m | | Crest length / head $H$ | 2.50 m / 1.07 m | | Crest wall top / bottom width | 0.50 m / 1.52 m | | Cistern depth / length | 0.50 m / 5.58 m | | Floor: u/s 1.0 m, d/s 5.58 m, cut-offs 0.6 / 0.9 m | total creep 11.10 m | | Floor thickness | 0.50 m (taper to 0.30 m) | **Answer:** crest level 255.15 m, crest length 2.50 m, crest wall 0.50 m top and 1.52 m bottom width, cistern depth 0.50 m and length 5.58 m, floor length 8.10 m, floor thickness 0.50 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. In Q = Cd*sqrt(2g)*L*H^(3/2) used for the crest, what does H represent? A. Only the depth of water u/s, 0.95 m B. The total head over the crest, including the velocity head of approach C. The drop in water level between u/s and d/s D. The height of the crest above the d/s bed Answer: B. H is the total head over the crest, so the approach velocity head ha is added to the u/s FSL when fixing the crest level: crest level = FSL + ha - H. (Key point: Head over the crest) 2. How is the cistern depth X decided? A. X is made equal to the drop in water level B. X is made equal to the height of the crest above the d/s bed C. X is chosen so that the conjugate depth y2 of the jump equals the tail water depth over the cistern, Dd + X D. X is chosen so that the initial depth y1 equals the tail water depth Answer: C. The jump forms properly when its sequent depth y2 matches the tail water over the cistern floor, Dd + X, which is solved together with the energy equation for y1. (Key point: Cistern depth from the hydraulic jump) 3. The cistern length is taken as 5(y2 - y1). With y1 = 0.330 m and y2 = 1.446 m, what is it? A. 1.12 m B. 5.58 m C. 7.23 m D. 1.65 m Answer: B. The cistern must contain the hydraulic jump, whose length is about 5 times (y2 - y1) = 5 x 1.116 = 5.58 m. (Key point: Length of cistern) 4. Why is only a nominal 1.0 m u/s floor provided, even though Bligh's creep length is required? A. Because the u/s floor does not matter for seepage at all B. Because the cut-offs, crest wall and the long d/s floor (needed for the cistern) already give creep 11.10 m, more than the required C*HL = 4.20 m C. Because the u/s floor is replaced by the cistern D. Because Bligh's theory does not apply to drops Answer: B. The required creep is only 6 x 0.70 = 4.20 m and the elements already needed provide far more, so b_u comes out negative and a nominal length is used. 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