--- title: "Vertical drop crest level and width, Q 30" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-crest-q30-fsl315" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-crest-q30-fsl315 site: BCT Engineering --- # Vertical drop crest level and width, Q 30 Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2079 Asoj, 6 marks - 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-crest-q30-fsl315 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design crest level and crest width of a vertical drop fall of a canal with the following data: u/s FSL = 315 masl, u/s CBL = 312 masl, d/s CBL = 310 masl, canal bed width = 15 m, full supply discharge = 30 m³/s, Canal side slope = 2 (H): 1 (V), Length of crest = canal bed width. ## Short answer Crest level is 313.95 m and crest length equals the canal bed width of 15 m. The rectangular crest wall has a top width of 1.10 m and a bottom width of 3.51 m. ## Answer #### Given data and assumptions - $Q = 30$ m³/s, bed width $B = 15$ m, side slope 2:1, u/s FSL $= 315.00$ m, u/s bed $= 312.00$ m, d/s bed $= 310.00$ m. - Full supply depth u/s $D_u = 3.000$ m; d/s depth taken equal to the u/s depth, $D_d = 3.000$ m (d/s FSL $= 313.000$ m); drop in water level $H_L = 2.000$ m. - Crest length equals the canal bed width, $L = B$. Rectangular crest with $C_d = 0.415$; specific gravity of masonry $G = 2.24$ (assumed). #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD_u)D_u = (15 + 2\times3.000)(3.000) = 63.000\ \text{m}^2 \\ v &= Q/A = 0.476\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.012\ \text{m} \end{aligned} $$ #### 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, $L = B$: $$ H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{30}{1.838\times15}\right)^{2/3} = 1.058\ \text{m} $$ $$ \text{Crest level} = FSL_u + h_a - H = 315.000 + 0.012 - 1.058 = 313.95\ \text{m} $$ Height of crest above d/s bed $d = 313.95 - 310.00 = 3.95$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 2.000$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 5.012$ 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 $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$: - Without any depression the conjugate depth (1.89 m) is less than the tail water depth $D_d = 3.000$ m, so the jump is already drowned and hydraulically no depression is needed. A nominal cistern depth is still provided to hold a water cushion under the jet: **$X = 0.25$ m** (cistern floor level $= 309.75$ m). - With this $X$: $y_1 = 0.201$ m, $F_1 = 7.10$, $y_2 = 1.918$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.918 - 0.201) = 8.59$ m. #### 4. Crest (drop) wall For a rectangular crest, top width $T = 0.55\sqrt{d} = 0.55\sqrt{3.95} = 1.09$ m; adopt **1.10 m**. Height of wall from cistern floor to crest $H_w = 313.95 - 309.75 = 4.20$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{4.20 + 1.06}{\sqrt{2.24}} = 3.51\ \text{m} $$ Adopt bottom width $B_w = 3.51$ m (not less than the top width), with the d/s face vertical or slightly battered. #### Summary | Component | Value | |---|---| | Crest level | 313.95 m | | Crest length / head $H$ | 15.00 m / 1.06 m | | Crest wall top / bottom width | 1.10 m / 3.51 m | | Cistern depth / length | 0.25 m / 8.59 m | **Answer:** crest level 313.95 m, crest length 15.00 m, crest wall 1.10 m top and 3.51 m bottom width, cistern depth 0.25 m and length 8.59 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. How is the head H over the crest found when L = B = 15 m? A. H = (Q/(1.838 L))^(2/3) = 1.058 m B. H = Q/L = 2 m C. H = FSL - bed = 3 m D. H = (Q/L)² = 4 m Answer: A. From Q = 1.838 L H^1.5, H = (30/(1.838 x 15))^(2/3) = 1.058 m. (Key point: Crest head formula) 2. How is the crest level obtained? A. Upstream bed level 312 m B. Crest = FSL_u + h_a - H = 315 + 0.012 - 1.058 = 313.95 m C. Crest = FSL_u + H = 316 m D. Crest = d/s bed + H = 311.06 m Answer: B. The total energy level (FSL plus velocity head) minus the head over the crest gives 313.95 m. (Key point: Crest level) 3. Why is only a nominal cistern depth of 0.25 m provided? A. Because the floor is very long B. Because the crest is very low C. Because the conjugate depth is less than the tail water, so the jump is already drowned D. Because Bligh's coefficient is 6 Answer: C. Without depression the conjugate depth is below the 3 m tail water, so no depression is needed hydraulically; 0.25 m just holds a water cushion. (Key point: Cistern depth) 4. How are the cistern length and wall bottom width computed? A. Lc = H_w; Bw = T B. Lc = 5(y2 - y1) = 8.59 m; Bw = (Hw + H)/sqrt(G) = 3.52 m C. Lc = 2 y2; Bw = G Hw D. Lc = q; Bw = 0.55 sqrt(H) Answer: B. The jump length is 5 times the sequent depth difference, and the gravity wall bottom width is (Hw + H)/sqrt(G) with G = 2.24. 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