--- title: "Canal fall design, Q 12, drop 1.5 m" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:canal-fall-design-q12-drop15" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/canal-fall-design-q12-drop15 site: BCT Engineering --- # Canal fall design, Q 12, drop 1.5 m Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2065 Kartik (old course) - 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/canal-fall-design-q12-drop15 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design a fall for a canal having a discharge of 12 cumecs for the data given below. Full supply level u/s and d/s = 204 and 202.5; Bed level u/s and d/s = 202.5 and 201; Bed width u/s and d/s = 8 m; Drop height = 1.5 m; Side Slope of canal = 1:1; Bligh's creep coefficient = 8. ## Short answer Rectangular crest at 203.13 m, 8 m long, top width 0.80 m, base 2.00 m. Cistern 5.82 m long, 0.31 m deep, floor level 200.69 m. Total floor 12.16 m; d/s floor thickness 1.29 m reducing to 0.51 m. ## Answer #### Given data and assumptions - $Q = 12$ m³/s, bed width $B = 8$ m, side slope 1:1, u/s FSL $= 204.00$ m, u/s bed $= 202.50$ m, d/s bed $= 201.00$ m. - Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 202.50$ m); drop in water level $H_L = 1.5$ m. - Bligh's coefficient $C = 8$; specific gravity of the masonry/concrete $G = 2.24$ (assumed). - The 1.5 m fall is designed as a Sarda-type vertical drop. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD)D = (8 + 1\times1.5)(1.5) = 14.250\ \text{m}^2 \\ V_a &= Q/A = 0.842\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.036\ \text{m} \end{aligned} $$ #### 2. Crest Since $Q < 14$ m³/s, a **rectangular crest** is used, with crest length equal to the bed width, $L = 8$ m, and $$ Q = 1.835\,L\,H^{3/2}\left(\frac{H}{B_c}\right)^{1/6}, \qquad B_c = 0.55\sqrt{d} $$ where $H$ is the head over the crest (measured from the u/s FSL), $B_c$ is the top width of the crest and $d$ is the height of the crest above the d/s bed. Solving by trial: $$ H = 0.867\ \text{m}, \qquad \text{crest level} = 204.00 - 0.867 = 203.13\ \text{m}, \qquad d = 203.13 - 201.00 = 2.13\ \text{m} $$ Top width $B_c = 0.55\sqrt{2.13} = 0.80$ m. Base width of the crest wall: $$ B_w = \frac{H + d}{\sqrt{G}} = \frac{0.867 + 2.13}{\sqrt{2.24}} = 2.00\ \text{m} $$ #### 3. Cistern Energy head over the crest $E = H + h_a = 0.867 + 0.036 = 0.903$ m. Using the UPIRI formulas: $$ \begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.903\times1.5} = 5.82\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(1.355)^{2/3} = 0.306\ \text{m} \end{aligned} $$ Cistern floor level $= 201.00 - 0.306 = 200.69$ m. #### 4. Impervious floor (Bligh's theory) Maximum seepage head occurs with no flow, water up to the crest on the u/s side and the d/s side dry: $H_s = d = 2.13$ m. Required creep length $L = C\,H_s = 8\times2.13 = 17.06$ m. - Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.10$ m; d/s $d_2 = D/2 + 0.6 = 1.35$ m. - Horizontal floor length $b = L - 2(d_1 + d_2) = 17.06 - 4.90 = 12.16$ m. - Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.90$ m, which is more than $L_c$; adopt $L_d = 6.90$ m. - U/s floor $= b - L_d - B_w = 12.16 - 6.90 - 2.00 = $ **3.26 m**. - Total floor length $= 3.26 + 2.00 + 6.90 = 12.16$ m. #### 5. Floor thickness Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 7.46$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 7.46}{L} = 2.13\times\dfrac{17.06 - 7.46}{17.06} = 1.200$ m (with $L = 17.06$ m, the creep provided). $$ t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.200}{1.24} = 1.29\ \text{m} $$ At the end of the cistern the creep is 13.28 m, $h_2 = 0.473$ m and $t_2 = 0.51$ m. Adopt a d/s floor **1.29 m** thick near the crest wall, reducing to 0.51 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m. #### Summary | Component | Value | |---|---| | Crest | rectangular, length 8 m, level 203.13 m, head $H$ = 0.867 m | | Crest width (top / base) | 0.80 m / 2.00 m | | Cistern (length / depth) | 5.82 m / 0.31 m, floor level 200.69 m | | Floor (u/s / wall base / d/s) | 3.26 / 2.00 / 6.90 m, total 12.16 m | | Cut-offs (u/s / d/s) | 1.10 m / 1.35 m | | D/s floor thickness | 1.29 m, reducing to 0.51 m | **Answer:** crest level 203.13 m (rectangular crest 8 m long, top width 0.80 m, base 2.00 m); cistern 5.82 m long and 0.31 m deep; impervious floor 12.16 m long (u/s 3.26 m, d/s 6.90 m) with cut-offs of 1.10 m (u/s) and 1.35 m (d/s); d/s floor thickness 1.29 m, reducing to 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. How is the crest level of the 1.5 m Sarda-type fall found? A. Crest level = u/s FSL + H B. Crest level = u/s FSL - H, where H is the head over the crest from Q = 1.835 L H^(3/2) C. Crest level = d/s bed + 1.5 m D. Crest level = u/s bed - H Answer: B. The water surface u/s sits H above the crest, so crest level = FSL(u/s) - H = 204.000 - 0.814 = 203.19 m. (Key point: Crest level) 2. What crest length is adopted for the Sarda fall and why does it exceed the bed width? A. L = B + 0.6 D_u = 8.90 m, a little wider than the bed to reduce the head over the crest B. L = B = 8 m, equal to the bed width C. L = B - 0.6 D_u, to flume the canal D. L = 2B, to double the discharge capacity Answer: A. The Sarda crest length is B + 0.6 D_u, giving 8.90 m, which lowers the head H needed to pass 12 m3/s. (Key point: Crest length) 3. In the cistern design, why is only a nominal depth of 0.25 m provided? A. Because the cistern length is already large B. Because the conjugate depth without depression is less than the tail water depth, so the jump is already drowned C. Because Bligh's creep length is satisfied D. Because the floor thickness is already 1.10 m Answer: B. When the required conjugate depth is below the available tail water, the jump is submerged and no real depression is needed, only a nominal water cushion. (Key point: Cistern and jump) 4. The calculated u/s floor length b_u is negative (-1.90 m). What does this mean and what is done? A. The design fails and the fall must be redesigned B. Cut-offs, wall and d/s floor already provide more than C x H_L creep, so a nominal 1.0 m u/s floor is adopted C. The d/s floor must be shortened D. The u/s floor must be 1.90 m long Answer: B. A negative b_u means the required creep L = C H_L = 12 m is already exceeded; a nominal floor is provided and actual creep (14.90 m) is checked to be at least 12 m. 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