--- title: "Crest width, cistern length of vertical drop, Q 1.55" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q155-drop075" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q155-drop075 site: BCT Engineering --- # Crest width, cistern length of vertical drop, Q 1.55 Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2073 Bhadra, 8 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-q155-drop075 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design a crest width, cistern length and its level of a vertical drop structure for the data given below. Full supply discharge u/s and d/s = 1.55 cumecs; Drop height = 0.75 m; FSL u/s and d/s = 105.997 and 105.247; Full supply depth u/s and d/s = 0.929 m; Bed levels u/s and d/s = 105.068 and 104.318; Bed width u/s and d/s = 1.1 m. Top width of crest = 0.5 m for initial assumption, Cd = 0.415 for rectangular crest. The drop structure is of masonry with specific gravity 2.0. Side slope of the canal is 1:1. The Bligh's coefficient is 7.0 for sandy loam soil at foundation. ## Short answer Crest top width 0.50 m, bottom width 1.39 m. Cistern depth 0.25 m (floor level 104.07 m) and length 4.67 m. ## Answer #### Given data and assumptions - $Q = 1.55$ m³/s, bed width $B = 1.1$ m, side slope 1:1, u/s FSL $= 106.00$ m, u/s bed $= 105.07$ m, d/s bed $= 104.32$ m. - Full supply depth u/s $D_u = 0.929$ m, d/s $D_d = 0.929$ m (d/s FSL $= 105.247$ m); drop in water level $H_L = 0.750$ m. - Bligh's coefficient $C = 7$; specific gravity of masonry/concrete $G = 2$. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD_u)D_u = (1.1 + 1\times0.929)(0.929) = 1.885\ \text{m}^2 \\ v &= Q/A = 0.822\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.034\ \text{m} \end{aligned} $$ #### 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, $L = B$: $$ H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{1.55}{1.838\times1.1}\right)^{2/3} = 0.838\ \text{m} $$ $$ \text{Crest level} = FSL_u + h_a - H = 105.997 + 0.034 - 0.838 = 105.19\ \text{m} $$ Height of crest above d/s bed $d_c = 105.19 - 104.32 = 0.88$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 1.409$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.713$ 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 $X = 0.247$ m. Provide **$X = 0.25$ m** (cistern floor level $= 104.07$ m). - With this $X$: $y_1 = 0.242$ m, $F_1 = 3.77$, $y_2 = 1.176$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.176 - 0.242) = 4.67$ m. #### 4. Crest (drop) wall Top width (assumed) $T = 0.50$ m. Height of wall from cistern floor to crest $H_w = 105.19 - 104.07 = 1.13$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.13 + 0.84}{\sqrt{2}} = 1.39\ \text{m} $$ Adopt bottom width $B_w = 1.39$ m (not less than the top width), with the d/s face vertical or slightly battered. #### Summary | Component | Value | |---|---| | Crest level | 105.19 m | | Crest length / head $H$ | 1.10 m / 0.84 m | | Crest wall top / bottom width | 0.50 m / 1.39 m | | Cistern depth / length | 0.25 m / 4.67 m | **Answer:** crest level 105.19 m, crest length 1.10 m, crest wall 0.50 m top and 1.39 m bottom width, cistern depth 0.25 m and length 4.67 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 d/s bed 104.32 m and cistern depth X = 0.25 m, what is the cistern floor level? A. 104.57 m B. 104.07 m C. 105.19 m D. 103.82 m Answer: B. The cistern is a depression below the d/s bed: 104.32 - 0.25 = 104.07 m. (Key point: Cistern floor level) 2. How is the crest level found here (H = 0.838 m, ha = 0.034 m)? A. 105.997 - 0.838 = 105.16 m B. 105.997 + 0.034 + 0.838 = 106.87 m C. 105.997 + 0.034 - 0.838 = 105.19 m D. Equal to the d/s bed level Answer: C. The crest lies H below the total energy level FSL + ha, giving 105.19 m. (Key point: Crest level) 3. With y1 = 0.242 m and y2 = 1.176 m, the cistern length 5(y2 - y1) equals A. 4.67 m B. 1.17 m C. 7.09 m D. 5.88 m Answer: A. 5 x (1.176 - 0.242) = 5 x 0.934 = 4.67 m, the length of the jump. 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