--- title: "Vertical drop design, Q 5.2 m3/s" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q52" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q52 site: BCT Engineering --- # Vertical drop design, Q 5.2 m3/s Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2077 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-q52 - 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 = 5.2 m³/sec; Side slope of the channel = 1:1; FSL U/S = 206.10 m; Bed level U/S = 205.15 m; Bed slope [level] D/S = 204.45 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 Crest level is 205.15 m (upstream bed) and crest length is 2.14 m. The impervious floor extends 9.09 m in total and its thickness at the drop wall is 0.50 m, reducing to 0.30 m at the downstream end. ## Answer #### Given data and assumptions - $Q = 5.2$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 206.10$ m, u/s bed $= 205.15$ m, d/s bed $= 204.45$ m. - Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 205.400$ 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 $$ \begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.327\ \text{m}^2 \\ v &= Q/A = 2.234\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.254\ \text{m} \end{aligned} $$ #### 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): $$ \begin{aligned} H &= (FSL_u + h_a) - \text{crest} = 206.354 - 205.150 = 1.204\ \text{m} \\ L &= \frac{Q}{1.838\,H^{3/2}} = 2.14\ \text{m} \end{aligned} $$ Height of crest above d/s bed $d_c = 205.15 - 204.45 = 0.70$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 2.430$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.904$ 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 $X = 0.683$ m. Provide **$X = 0.70$ m** (cistern floor level $= 203.75$ m). - With this $X$: $y_1 = 0.367$ m, $F_1 = 3.49$, $y_2 = 1.638$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.638 - 0.367) = 6.35$ m. #### 4. Crest (drop) wall Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 205.15 - 203.75 = 1.40$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.40 + 1.20}{\sqrt{2.25}} = 1.74\ \text{m} $$ Adopt bottom width $B_w = 1.74$ 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: $$ L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{2.430/75}} = 4.71\ \text{m} $$ The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 6.35$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.74$ m, so the u/s floor needed is $$ b_u = L - [2(d_1+d_2) + B_w + L_d] = -6.89\ \text{m} $$ Bligh's creep is already satisfied, so a nominal u/s floor of **1.0 m** is provided. Actual creep $= 12.09$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 9.09$ m. #### 6. Thickness of the floor Pressure head at the toe of the wall: creep to that point $= 3.94$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.94}{L_{act}} = 0.472$ m. At the end of the cistern the creep is 10.29 m and $h_2 = 0.104$ m. $$ t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.472}{1.25} = 0.50\ \text{m} $$ 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 | 205.15 m | | Crest length / head $H$ | 2.14 m / 1.20 m | | Crest wall top / bottom width | 0.55 m / 1.74 m | | Cistern depth / length | 0.70 m / 6.35 m | | Floor: u/s 1.0 m, d/s 6.35 m, cut-offs 0.6 / 0.9 m | total creep 12.09 m | | Floor thickness | 0.50 m (taper to 0.30 m) | **Answer:** crest level 205.15 m, crest length 2.14 m, crest wall 0.55 m top and 1.74 m bottom width, cistern depth 0.70 m and length 6.35 m, floor length 9.09 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. Why is the crest level fixed at the u/s bed level and the crest length increased here? A. Because the bed width is larger than needed B. Because with L = B = 1.5 m the required head would put the crest below the u/s bed, which is not acceptable C. Because the discharge is too small for a rectangular crest D. Because the cistern needs a higher crest Answer: B. A crest as long as the narrow bed would need a large head and a crest below the u/s bed level, so the crest is kept at bed level and made longer using splayed wing walls. (Key point: Why the crest is fixed at the u/s bed) 2. With the crest at the u/s bed level, how is the crest length found? A. L = Q / (1.838 H^(3/2)), with H = (FSL + ha) - crest level B. L = bed width plus twice the side slope times depth C. L = 5(y2 - y1) D. L = C * HL Answer: A. H is now known (206.354 - 205.15 = 1.204 m), so rearranging the weir equation gives L = 2.14 m, longer than the 1.5 m bed width. (Key point: Crest length from head) 3. In Bw = (Hw + H)/sqrt(G), how is Hw measured? A. From the d/s bed to the crest B. From the u/s FSL to the d/s FSL C. From the cistern floor to the crest D. From the u/s bed to the d/s bed Answer: C. The wall stands on the cistern floor, so Hw = crest level - cistern floor level = 205.15 - 203.75 = 1.40 m. (Key point: Crest wall height and width) 4. The floor thickness t = (4/3) h1/(G - 1) is based on h1, which is A. the drop in water level HL B. the residual (uplift) pressure head at the toe of the wall found from the Bligh creep C. the depth of the d/s cut-off D. the conjugate depth y2 Answer: B. Uplift pressure at the toe is HL x (Lact - creep to that point)/Lact = 0.472 m, and the floor weight (G - 1) must balance it. 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