--- title: "Sarda fall: Q 12, FSL 501.40" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-q12-fsl5014" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-q12-fsl5014 site: BCT Engineering --- # Sarda fall: Q 12, FSL 501.40 Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2078 Poush, 10 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/sarda-fall-q12-fsl5014 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design the vertical drop of Sarda type from the following data provided. Discharge of canal = 12 m³/s; u/s FSL of canal = 501.40 m; u/s bed level = 500 m; Drop height of canal bed = 1.20 m; d/s FSL of canal = 500.20 m; Bed width of canal = 10 m; Side slope of canal = 1:1; depth of flow 1.40 m for both u/s and d/s side; Bligh's coefficient = 7. ## Short answer Using Bligh's theory (C=7): rectangular crest at 500.65 m (top width 0.75 m, base 1.74 m); cistern 4.84 m long, 0.24 m deep; impervious floor length 9.14 m (u/s 1.00 m, d/s 6.40 m) with u/s cut-off 1.07 m, d/s 1.30 m. ## Answer #### Given data and assumptions - $Q = 12$ m³/s, bed width $B = 10$ m, side slope 1:1, u/s FSL $= 501.40$ m, u/s bed $= 500.00$ m, d/s bed $= 498.80$ m. - Full supply depth u/s = d/s $D = 1.4$ m (d/s FSL $= 500.20$ m); drop in water level $H_L = 1.2$ m. - Bligh's coefficient $C = 7$; specific gravity of the masonry/concrete $G = 2.24$ (assumed). - D/s bed $= 500.20 - 1.40 = 498.80$ m, which agrees with the 1.20 m drop of the bed. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD)D = (10 + 1\times1.4)(1.4) = 15.960\ \text{m}^2 \\ V_a &= Q/A = 0.752\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.029\ \text{m} \end{aligned} $$ #### 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: $$ H = 0.753\ \text{m}, \qquad \text{crest level} = 501.40 - 0.753 = 500.65\ \text{m}, \qquad d = 500.65 - 498.80 = 1.85\ \text{m} $$ Top width $B_c = 0.55\sqrt{1.85} = 0.75$ m. Base width of the crest wall: $$ B_w = \frac{H + d}{\sqrt{G}} = \frac{0.753 + 1.85}{\sqrt{2.24}} = 1.74\ \text{m} $$ #### 3. Cistern Energy head over the crest $E = H + h_a = 0.753 + 0.029 = 0.782$ m. Using the UPIRI formulas: $$ \begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.782\times1.2} = 4.84\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.938)^{2/3} = 0.240\ \text{m} \end{aligned} $$ Cistern floor level $= 498.80 - 0.240 = 498.56$ 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: $H_s = d = 1.85$ m. Required creep length $L = C\,H_s = 7\times1.85 = 12.93$ m. - Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.07$ m; d/s $d_2 = D/2 + 0.6 = 1.30$ m. - Horizontal floor length $b = L - 2(d_1 + d_2) = 12.93 - 4.73 = 8.20$ m. - Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.40$ m, which is more than $L_c$; adopt $L_d = 6.40$ m. - U/s floor $= b - L_d - B_w = 8.20 - 6.40 - 1.74 = 0.06$ m, so a minimum u/s floor of **1.0 m** is provided. Actual creep $= 13.87$ m $\ge 12.93$ m, safe. - Total floor length $= 1.00 + 1.74 + 6.40 = 9.14$ m. #### 5. Floor thickness Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 4.87$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 4.87}{L} = 1.85\times\dfrac{13.87 - 4.87}{13.87} = 1.199$ m (with $L = 13.87$ m, the creep provided). $$ t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.199}{1.24} = 1.29\ \text{m} $$ At the end of the cistern the creep is 9.71 m, $h_2 = 0.554$ m and $t_2 = 0.60$ m. Adopt a d/s floor **1.29 m** thick near the crest wall, reducing to 0.60 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 500.65 m, head $H$ = 0.753 m | | Crest width (top / base) | 0.75 m / 1.74 m | | Cistern (length / depth) | 4.84 m / 0.24 m, floor level 498.56 m | | Floor (u/s / wall base / d/s) | 1.00 / 1.74 / 6.40 m, total 9.14 m | | Cut-offs (u/s / d/s) | 1.07 m / 1.30 m | | D/s floor thickness | 1.29 m, reducing to 0.60 m | **Answer:** crest level 500.65 m (rectangular crest 10 m long, top width 0.75 m, base 1.74 m); cistern 4.84 m long and 0.24 m deep; impervious floor 9.14 m long (u/s 1.00 m, d/s 6.40 m) with cut-offs of 1.07 m (u/s) and 1.30 m (d/s); d/s floor thickness 1.29 m, reducing to 0.60 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. What is the crest level and how is it computed? A. Crest level = FSL_u + H = 502.11 m B. Crest level = FSL_u - H = 501.400 - 0.714 = 500.69 m C. Crest level = u/s bed + H = 500.71 m D. Crest level = d/s bed + drop = 500.00 m Answer: B. The head H = 0.714 m over the crest is subtracted from the u/s FSL, giving 500.69 m. (Key point: Crest level for Sarda fall) 2. How is the cistern length determined? A. 5 x (y2 - y1), five times the difference of conjugate and initial depths of the jump B. Equal to the crest length C. Equal to 2 x drop D. 5 x tail water depth Answer: A. The jump length is taken as 5(y2 - y1) = 5(1.208 - 0.152) = 5.28 m. (Key point: Cistern length) 3. Why is only a nominal 1.0 m u/s floor provided? A. Because the u/s floor is never needed in any fall B. Because the required u/s floor is negative: creep from cut-offs, wall and d/s floor already exceeds C x H_L C. Because the u/s floor must equal the cut-off depth D. Because Bligh's creep is not required for Sarda falls Answer: B. The computed b_u was -2.81 m, meaning creep is already satisfied (12.21 m vs 8.40 m), so only a nominal floor is provided. (Key point: Why u/s floor is nominal) 4. What total creep length is required with C = 7 and H_L = 1.2 m? A. 0.17 m B. 5.8 m C. 8.4 m D. 12.2 m Answer: C. L = C x H_L = 7 x 1.2 = 8.4 m. 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