--- title: "Sarda fall: Q 12, U/S bed 103 m" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-q12-bed103-76b" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-q12-bed103-76b site: BCT Engineering --- # Sarda fall: Q 12, U/S bed 103 m Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2076 Baisakh, 12 marks - Video: [DESIGN OF SARDA FALL NUMERICAL WITH DESIGN STEPS (PART 1) IRRIGATION/ Water resources engineering](https://www.youtube.com/watch?v=I3A3mTWh6xk) — 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/sarda-fall-q12-bed103-76b - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design a suitable Sarada Type fall for a canal carrying a discharge of 12 m³/s with the following data. U/S bed level = 103 m; D/S bed level = 101.50 m; Side Slope of canal = 1½:1 (H:V); Full Supply level U/S = 104.50; Bed width of canal = 10 m; Bligh's coefficient = 6. ## Short answer Rectangular crest at 103.74 m, 10 m long, top width 0.82 m, base 2.00 m. Cistern 5.41 m long, 0.28 m deep, floor level 101.22 m. Total floor 9.90 m; d/s floor thickness 1.56 m reducing to 0.68 m. ## Answer #### Given data and assumptions - $Q = 12$ m³/s, bed width $B = 10$ m, side slope 1.5:1, u/s FSL $= 104.50$ m, u/s bed $= 103.00$ m, d/s bed $= 101.50$ m. - Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 103.00$ m); drop in water level $H_L = 1.5$ m. - Bligh's coefficient $C = 6$; specific gravity of the masonry/concrete $G = 2.24$ (assumed). - D/s depth = u/s depth = 1.5 m, so d/s FSL $= 103.0$ m and the drop is 1.5 m. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD)D = (10 + 1.5\times1.5)(1.5) = 18.375\ \text{m}^2 \\ V_a &= Q/A = 0.653\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.022\ \text{m} \end{aligned} $$ #### 2. Crest Since $Q < 14$ m³/s, a **rectangular crest** is used, with crest length equal to the bed width, $L = 10$ 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: $$ H = 0.760\ \text{m}, \qquad \text{crest level} = 104.50 - 0.760 = 103.74\ \text{m}, \qquad d = 103.74 - 101.50 = 2.24\ \text{m} $$ Top width $B_c = 0.55\sqrt{2.24} = 0.82$ m. Base width of the crest wall: $$ B_w = \frac{H + d}{\sqrt{G}} = \frac{0.760 + 2.24}{\sqrt{2.24}} = 2.00\ \text{m} $$ #### 3. Cistern Energy head over the crest $E = H + h_a = 0.760 + 0.022 = 0.782$ m. Using the UPIRI formulas: $$ \begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.782\times1.5} = 5.41\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(1.173)^{2/3} = 0.278\ \text{m} \end{aligned} $$ Cistern floor level $= 101.50 - 0.278 = 101.22$ 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: $H_s = d = 2.24$ m. Required creep length $L = C\,H_s = 6\times2.24 = 13.44$ 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 $b = L - 2(d_1 + d_2) = 13.44 - 4.90 = 8.54$ 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 $= b - L_d - B_w = 8.54 - 6.90 - 2.00 = -0.37$ m, so a minimum u/s floor of **1.0 m** is provided. Actual creep $= 14.80$ m $\ge 13.44$ m, safe. - Total floor length $= 1.00 + 2.00 + 6.90 = 9.90$ m. #### 5. Floor thickness Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 5.20$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 5.20}{L} = 2.24\times\dfrac{14.80 - 5.20}{14.80} = 1.452$ m (with $L = 14.80$ m, the creep provided). $$ t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.452}{1.24} = 1.56\ \text{m} $$ At the end of the cistern the creep is 10.62 m, $h_2 = 0.633$ m and $t_2 = 0.68$ m. Adopt a d/s floor **1.56 m** thick near the crest wall, reducing to 0.68 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 10 m, level 103.74 m, head $H$ = 0.760 m | | Crest width (top / base) | 0.82 m / 2.00 m | | Cistern (length / depth) | 5.41 m / 0.28 m, floor level 101.22 m | | Floor (u/s / wall base / d/s) | 1.00 / 2.00 / 6.90 m, total 9.90 m | | Cut-offs (u/s / d/s) | 1.10 m / 1.35 m | | D/s floor thickness | 1.56 m, reducing to 0.68 m | **Answer:** crest level 103.74 m (rectangular crest 10 m long, top width 0.82 m, base 2.00 m); cistern 5.41 m long and 0.28 m deep; impervious floor 9.90 m long (u/s 1.00 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 1.56 m, reducing to 0.68 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 side slope 1.5:1, how is the u/s flow area found? A. A = B x D = 15 m^2 B. A = (B + mD)D = (10 + 1.5x1.5)x1.5 = 18.375 m^2 C. A = (B + D)D D. A = B x (D + m) Answer: B. For a trapezoid with side slope m, A = (B + mD)D, giving 18.375 m^2 and v = 0.653 m/s. (Key point: Side slope effect on approach flow) 2. What crest length does the Sarda design use, independent of the side slope? A. L = B + 0.6D = 10.90 m B. L = B = 10 m C. L = B + 1.5D = 12.25 m D. L = B + D/0.6 Answer: A. Sarda crest length is B + 0.6 D_u, which is 10.9 m here. (Key point: Sarda crest length) 3. How does the solution show the floor is safe by Bligh's theory? A. Actual creep 12.62 m is at least the required C x H_L = 9.00 m B. Actual creep 12.62 m is less than 9.00 m but accepted C. The floor thickness exceeds creep length D. Creep length is not checked Answer: A. Safe design needs actual creep >= C x H_L; 12.62 m >= 9.0 m. (Key point: Bligh creep adequacy) 4. Which expression gives the cistern length? A. L_c = y2 + y1 B. L_c = 5(y2 - y1) = 5.57 m C. L_c = 2 y2 D. L_c = q x H Answer: B. The jump length is five times the rise in depth: 5(1.255 - 0.141) = 5.57 m. (Key point: Cistern length) ## Related in Irrigation and Drainage Engineering - [Sarda fall: Q 12, bed level 103 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-q12-bed103) — Irrigation and Drainage Engineering, chapter 7 - [Canal fall design, Q 12, drop 1.5 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acanal-fall-design-q12-drop15) — Irrigation and Drainage Engineering, chapter 7 - [Sarda fall: Q 12, FSL 501.40](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-q12-fsl5014) — Irrigation and Drainage Engineering, chapter 7 - [Sarda fall: Q 8, FSL 218.30](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-q8-bed10) — Irrigation and Drainage Engineering, chapter 7 - [Sarda fall for Bagmati branch canal, Q 25](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-bagmati-q25) — Irrigation and Drainage Engineering, chapter 7 - [Sarda fall: Q 9, drop 1.2 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-q9-drop12) — Irrigation and Drainage Engineering, chapter 7 ## Related in other subjects - [Slope-area discharge, 10 km reach, 104.77 m](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-104-77) — Engineering Hydrology, chapter 4