--- title: "Procedure for designing a Sarda type fall" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-design-procedure" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-design-procedure site: BCT Engineering --- # Procedure for designing a Sarda type fall Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2079 Chaitra, 7 marks; 2063 Asoj (old course), 6 marks (asked 2 times) - 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/sarda-fall-design-procedure - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Explain the procedure of designing a Sarda type fall (fall structure). ## Short answer Design uses Sarda's formula Q = 1.835 L H^{3/2}(H/B)^{1/6} to find head H; crest level = u/s TEL - H. Cistern length 5√(E H_L), depth (1/4)(E H_L)^{2/3}. Floor length from Bligh's L = C H_L, thickness from uplift. Provide cut-offs and block protection. ## Answer A Sarda-type fall is a vertical-drop fall with a raised rectangular or trapezoidal crest wall. The crest is designed with Sarda's formula, the energy is dissipated in a cistern, and the floor is designed by Bligh's or Khosla's theory. The design steps are as follows. #### Part 1. Hydraulic design (7 marks) **Data:** $Q$, u/s and d/s FSL, bed levels, bed width $B_c$, side slopes, soil. 1. **Crest length:** usually equal to the bed width of the canal, $L = B_c$ (it may be made slightly longer to allow for a future increase in discharge). 2. **Head over crest (Sarda's formula):** for a rectangular crest $Q = 1.835\,L\,H^{3/2}\left(\dfrac{H}{B}\right)^{1/6}$, and for a trapezoidal crest $Q = 1.99\,L\,H^{3/2}\left(\dfrac{H}{B}\right)^{1/6}$, where $B$ is the crest top width. Assume $B$, then solve for $H$ by trial. A rectangular crest is used for small discharges (up to about 14 m³/s) and a trapezoidal crest for larger ones. 3. **Crest level** $=$ u/s TEL $- H$ (u/s FSL $+ \dfrac{v^2}{2g} - H$). 4. **Crest dimensions** ($d$ = height of the crest above the d/s bed, $G$ = specific gravity of masonry): - Rectangular crest: top width $B = 0.55\sqrt{d}$, base width $= \dfrac{H + d}{\sqrt{G}}$ for stability. - Trapezoidal crest: top width $B = 0.55\sqrt{H + d}$, with u/s batter about 1:3 and d/s batter about 1:8. 5. **Cistern:** with $E$ = total head over the crest and $H_L$ = drop in water level, length of cistern $L_c = 5\sqrt{E\,H_L}$ and depth of cistern below the d/s bed $X = \dfrac{1}{4}\left(E\,H_L\right)^{2/3}$. 6. **Cistern level** $=$ d/s bed $- X$. #### Part 2. Floor and protection (6 marks) 7. **Bligh's theory:** total creep length $L = C\,H_L$, with $H_L$ = difference of u/s and d/s FSL. Provide cut-offs u/s ($d_1$) and d/s ($d_2$). 8. **Length of floor:** d/s floor length $L_d = 2.21C\sqrt{H_L\sqrt{q/75}}$ (not less than the cistern length); u/s floor from the balance of creep: $b_u = CH_L - 2(d_1 + d_2) - B_w - L_d$. If Khosla's method is used, find the total floor length from the safe exit gradient: $G_E = \dfrac{H}{d}\dfrac{1}{\pi\sqrt{\lambda}}$. 9. **Floor thickness:** at any section $t = \dfrac{4}{3}\dfrac{h}{G-1}$, where $h$ is the residual uplift head obtained from the creep line. 10. **Protection works:** d/s block protection of length about $2D$ to $3D$ with an inverted filter, and u/s block protection; wing walls splayed at 45° upstream and at 1:1 or 2:1 downstream, with the wings ending in the bank. 11. **Check:** stability of the crest wall (no tension, base pressure within the safe bearing capacity) and the exit gradient. 12. Draw the longitudinal section and plan. ```text FSL u/s ~~~~~~~~~| u/s bed __________|crest FSL d/s ~~~~ cut-off d1 | | \___cistern X___ d/s bed u/s floor |wall| Lc | d/s floor | d2 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the crest length of a Sarda fall found? A. L = B - 0.6D B. L = 0.6B C. L = B + 0.6D, B bed width and D full supply depth D. L = 5(y2 - y1) Answer: C. Crest length is the bed width plus an allowance of 0.6 times the u/s depth. (Key point: Crest length) 2. Which formula gives head for a trapezoidal crest? A. Q = 1.835 L H^(3/2) B. Q = 0.55 sqrt(H+d) C. Q = 1.7 L H D. Q = 1.99 L H^(3/2) Answer: D. Rectangular crest uses 1.835 and trapezoidal 1.99. (Key point: Head over crest) 3. How is the cistern length found? A. Lc = 5(y2 - y1), with depth X so conjugate depth y2 = Dd + X B. Lc = 2.21 C sqrt(...) C. Lc = B + 0.6D D. Lc = h/(G-1) Answer: A. The hydraulic jump forms inside the cistern, so its length is 5 times the jump height. (Key point: Cistern) 4. How is floor thickness computed? A. t = 0.55 sqrt(H + d) B. t = (4/3) h/(G - 1) from residual uplift head C. t = 0.06 Q^(1/3) D. t = 5(y2 - y1) Answer: B. Thickness must resist uplift; the 4/3 factor gives a safety margin. 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