--- title: "Hydraulic jump stilling basin at spillway toe" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:stilling-basin-hydraulic-jump-80cumec-2070" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-hydraulic-jump-80cumec-2070 site: BCT Engineering --- # Hydraulic jump stilling basin at spillway toe Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2070 Chaitra, 9 marks - Video: [Stilling basin | Design steps & Numerical | Hydropower ● USBR Type 1,2,3](https://www.youtube.com/watch?v=YBgwX5fYtt8) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-hydraulic-jump-80cumec-2070 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Design a hydraulic jump stilling basin at the toe of the spillway with the following data: discharge = 80 m³/s; width of the spillway = 8 m; spillway crest level = 96.00 m; river bed level = 65.00 m; tail water level = 71.00 m; coefficient of discharge = 0.7; downstream bed slope (i) = 1:500 and Manning's roughness coefficient = 0.016; ratio of length of stilling basin and sequent depth = 5.1. ## Short answer Head 2.86 m, y1=0.384 m, y2=7.10 m, Fr1=13.42; floor depressed 1.10 m to RL 63.90 m; basin length 37 m. ## Answer **Given**: $Q=80$ m³/s, width $B=8$ m, crest RL 96.00 m, river bed RL 65.00 m, tail water RL 71.00 m (depth $y_{TW}=6$ m), $C_d=0.7$, bed slope 1:500, $n=0.016$ (not needed because the tail water is given), $L/y_2=5.1$. #### Step 1: Unit discharge and head on crest $$ q=\frac{80}{8}=10\ \text{m}^3/\text{s/m},\qquad q=\tfrac23C_d\sqrt{2g}\,H^{3/2}\Rightarrow10=\tfrac23(0.7)(4.429)H^{3/2}\Rightarrow H=2.86\ \text{m} $$ #### Step 2: Conditions at the toe (floor at river-bed level) Fall of crest $=96-65=31$ m, so the specific energy at the toe (losses neglected) $E=31+H=33.86$ m. $$ y_1+\frac{q^2}{2g\,y_1^2}=E\ \Rightarrow\ y_1=0.390\ \text{m},\quad V_1=25.6\ \text{m/s},\quad Fr_1=\frac{V_1}{\sqrt{gy_1}}=13.10 $$ $$ y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=7.04\ \text{m} $$ The tail water depth (6.0 m) is less than $y_2=7.04$ m, so the jump would be swept away. The floor must be depressed. #### Step 3: Depress the floor by $z$ With the floor lowered by $z$: $E=31+z+H$ and tail-water depth over the floor $=6+z$. Condition $6+z=y_2(z)$; by trial: | $z$ (m) | $E$ (m) | $y_1$ (m) | $Fr_1$ | $y_2$ (m) | $6+z$ (m) | |---|---|---|---|---|---| | 0 | 33.86 | 0.390 | 13.10 | 7.04 | 6.00 | | 1.10 | 34.96 | 0.384 | 13.42 | 7.10 | 7.10 | Depression $z\approx1.10$ m, so **floor RL $=65-1.10=63.90$ m** (say RL 63.90 m). #### Step 4: Length of basin $$ L=5.1\,y_2=5.1\times7.10=36.2\ \text{m}\ \Rightarrow\ \text{adopt }37\ \text{m} $$ #### Step 5: Appurtenances - Chute blocks at the toe, height and width $\approx y_1=0.384$ m (say 0.4 m), alternating with equal gaps. - End sill of height $\approx0.2y_2=1.42$ m; $Fr_1=13.42>4.5$ and $V_1=26.0$ m/s $>18$ m/s, so a USBR Type II layout (chute blocks and dentated end sill) is used. - Training walls up to $1.2y_2$ above floor, and riprap protection downstream of the end sill. ```text crest 96.0 ____ \ \ \ \__[chute blocks]__________[sill]__ bed 65.0 ===\ _ _ _ _ _ _ _ _ _ _ _ _ _ _/===== TWL 71.0 floor 63.90 |<------ L = 37 m ------>| ``` **Answer**: $H=2.86$ m, $y_1=0.384$ m, $y_2=7.10$ m, $Fr_1=13.42$; floor depressed by 1.10 m to RL 63.90 m, basin length 37 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the energy at the toe obtained? A. E = fall of crest (31 m) + head H (2.86 m) = 33.86 m, losses neglected B. E = H only C. E = 96 m D. E = 65 m Answer: A. Specific energy at the toe equals the total head above the floor, here 33.86 m. (Key point: Toe energy) 2. Why is the floor depressed? A. Tail water 6.0 m is less than y2 = 7.04 m, so the jump would be swept away B. Tail water is greater than y2 C. Fr1 is less than 1 D. Manning's n is not needed Answer: A. The tail water must be at least the sequent depth; lowering the floor by z adds depth. (Key point: Jump check) 3. What condition determines the depression z? A. 6 + z = y2(z), found by trial, z about 1.10 m B. z = y2 - y1 C. z = H D. z = 6 m Answer: A. Because E and y2 both rise with z, trial gives z = 1.10 m, floor RL 63.90 m. (Key point: Depression condition) 4. How is the length of the basin found? A. L = 5.1 y2 = 5.1 x 7.10 = 36.2 m, adopt 37 m B. L = y2 = 7.1 m C. L = 5.1 y1 D. L = 5.1 H Answer: A. The question gives L/y2 = 5.1, applied with the final y2 of 7.10 m. 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