--- title: "Vertical drop design, Q 1.8 m3/s (bed 205.05)" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:vertical-drop-q18-bed20505" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/vertical-drop-q18-bed20505 site: BCT Engineering --- # Vertical drop design, Q 1.8 m3/s (bed 205.05) Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2069 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-q18-bed20505 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Design crest, length and thickness of impervious floor of a vertical drop structure for the data given below: Discharge = 1.8 m³/s; Bed level U/S = 205.05 m; Side slope of channel = 1:1; Bed level D/S = 204.35 m; FSL U/S = 205.95 m; Bed width U/S and D/S = 1.5 m. Top width of crest = 0.5 m (for initial assumption); Cd = 0.415; sp.gr. of masonry drop structure = 2.2; Bligh's coeff = 6.0. ## Short answer Crest length equals bed width 1.50 m; crest level is 205.23 m with head 0.75 m. Floor total length 6.65 m, thickness 0.50 m at the wall tapering to 0.30 m. ## Answer #### Given data and assumptions - $Q = 1.8$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 205.95$ m, u/s bed $= 205.05$ m, d/s bed $= 204.35$ m. - Full supply depth u/s $D_u = 0.900$ m, d/s $D_d = 0.900$ m (d/s FSL $= 205.250$ m); drop in water level $H_L = 0.700$ m. - Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.2$. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.900)(0.900) = 2.160\ \text{m}^2 \\ v &= Q/A = 0.833\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.035\ \text{m} \end{aligned} $$ #### 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, $L = B$: $$ H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{1.8}{1.838\times1.5}\right)^{2/3} = 0.753\ \text{m} $$ $$ \text{Crest level} = FSL_u + h_a - H = 205.950 + 0.035 - 0.753 = 205.23\ \text{m} $$ Height of crest above d/s bed $d_c = 205.23 - 204.35 = 0.88$ m. #### 3. Cistern (energy dissipation) Discharge per metre length of crest $q = Q/L = 1.200$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 1.635$ 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 $X = 0.166$ m. Provide **$X = 0.25$ m** (cistern floor level $= 204.10$ m). - With this $X$: $y_1 = 0.209$ m, $F_1 = 4.00$, $y_2 = 1.084$ m. Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.084 - 0.209) = 4.38$ m. #### 4. Crest (drop) wall Top width (assumed) $T = 0.50$ m. Height of wall from cistern floor to crest $H_w = 205.23 - 204.10 = 1.13$ m. Bottom width for a gravity masonry wall: $$ B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.13 + 0.75}{\sqrt{2.2}} = 1.27\ \text{m} $$ Adopt bottom width $B_w = 1.27$ m (not less than the top width), with the d/s face vertical or slightly battered. #### 5. Length of impervious floor (Bligh's theory) Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s: $$ L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{1.200/75}} = 3.95\ \text{m} $$ The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 4.38$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 4.27$ m, so the u/s floor needed is $$ b_u = L - [2(d_1+d_2) + B_w + L_d] = -4.45\ \text{m} $$ Bligh's creep is already satisfied, so a nominal u/s floor of **1.0 m** is provided. Actual creep $= 9.65$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 6.65$ m. #### 6. Thickness of the floor Pressure head at the toe of the wall: creep to that point $= 3.47$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 3.47}{L_{act}} = 0.448$ m. At the end of the cistern the creep is 7.85 m and $h_2 = 0.131$ m. $$ t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.448}{1.20} = 0.50\ \text{m} $$ Adopt floor thickness **0.50 m** under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m). #### Summary | Component | Value | |---|---| | Crest level | 205.23 m | | Crest length / head $H$ | 1.50 m / 0.75 m | | Crest wall top / bottom width | 0.50 m / 1.27 m | | Cistern depth / length | 0.25 m / 4.38 m | | Floor: u/s 1.0 m, d/s 4.38 m, cut-offs 0.6 / 0.9 m | total creep 9.65 m | | Floor thickness | 0.50 m (taper to 0.30 m) | **Answer:** crest level 205.23 m, crest length 1.50 m, crest wall 0.50 m top and 1.27 m bottom width, cistern depth 0.25 m and length 4.38 m, floor length 6.65 m, floor thickness 0.50 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 approach velocity head ha = 0.035 m obtained? A. From v = Q/A with A = (B + mD)D = 2.16 m^2, then ha = v^2/2g B. From ha = HL/2 C. From the crest length divided by 2g D. From the tail water depth squared Answer: A. Area is (1.5 + 0.9)(0.9) = 2.16 m^2, v = 1.8/2.16 = 0.833 m/s, and ha = v^2/2g = 0.035 m. (Key point: Velocity head of approach) 2. What is the crest level, using H = 0.753 m, ha = 0.035 m and FSL_u = 205.95 m? A. 205.20 m B. 205.23 m C. 206.74 m D. 204.35 m Answer: B. Crest level = 205.950 + 0.035 - 0.753 = 205.23 m. (Key point: Crest level) 3. The required cistern depth is 0.166 m but 0.25 m is provided. Why is this acceptable? A. A deeper cistern gives a tail water cushion above the required value, so the jump is safely contained B. Because 0.25 m makes y2 smaller than the d/s depth C. Because the required value must be doubled by law D. Because it makes the crest higher Answer: A. X = 0.166 m is the minimum for the jump; adopting a practical larger value keeps the jump submerged (Dd + X exceeds y2 = 1.084 m). (Key point: Cistern depth adopted) 4. In Bw = (Hw + H)/sqrt(G), what happens if a lighter masonry (smaller G) is used? A. The bottom width decreases B. The bottom width stays the same C. The bottom width increases D. The top width is automatically zero Answer: C. G is in the denominator, so a lighter material needs a larger base to remain stable. 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