--- title: "Stilling basin invert level and USBR type" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:stilling-basin-usbr-300cumec-12m" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-usbr-300cumec-12m site: BCT Engineering --- # Stilling basin invert level and USBR type Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2078 Bhadra, 10 marks - Video: [Determination of Stilling Basin Invert Elevation and Its Effect on Controlling Hydraulic Jumps](https://www.youtube.com/watch?v=ILsYPavq3oM) — BP International - 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-usbr-300cumec-12m - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question The discharge of water over a spillway 12 m wide is 300 m³/s into a stilling basin of the same width. The lake level behind the spillway has an elevation of 50 m and the river water surface elevation downstream of the stilling basin is 25 m. Assume a 10% energy loss in flow down the spillway, find the invert level elevation of the floor of the stilling basin so that a hydraulic jump forms in the basin. Select an appropriate USBR stilling basin and list all the dimensions. ## Short answer Floor invert RL 14.27 m. y1=1.011 m, y2=10.73 m, Fr1=7.85. USBR Type II basin length 47 m, chute blocks 1.0 m, dentated end sill 2.15 m. ## Answer **Given**: $Q=300$ m³/s, width $B=12$ m, so $q=\dfrac{300}{12}=25$ m³/s/m; lake level RL 50 m; tail water level RL 25 m; 10% of the energy is lost on the spillway. Let $Z$ be the RL of the basin floor. Then the energy at the toe above the floor is $$ E_1=0.9\,(50-Z),\qquad y_1+\frac{q^2}{2g\,y_1^2}=E_1 $$ and the jump forms at the toe if the tail water depth equals the sequent depth: $$ 25-Z=y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right),\qquad Fr_1=\frac{q}{y_1\sqrt{gy_1}} $$ #### Solution by trial | Floor RL $Z$ (m) | $E_1$ (m) | $y_1$ (m) | $Fr_1$ | $y_2$ (m) | $Z+y_2$ (m) | |---|---|---|---|---|---| | 12.0 | 34.20 | 0.979 | 8.24 | 10.93 | 22.93 | | 16.0 | 30.60 | 1.038 | 7.55 | 10.57 | 26.57 | | 14.27 | 32.16 | 1.011 | 7.85 | 10.73 | 25.00 | The condition $Z+y_2=25$ is met for $$ \boxed{Z=14.27\ \text{m}}\ \text{(invert level of the basin floor)},\quad y_1=1.011\ \text{m},\ V_1=24.7\ \text{m/s},\ Fr_1=7.85,\ y_2=10.73\ \text{m} $$ #### Selection of the USBR basin $Fr_1=7.85>4.5$, $V_1=24.7$ m/s $>18$ m/s and $q=25>18.5$ m³/s/m, so a **Type II** basin is suitable (Type III is limited to $V_1<18$ m/s; Type IV is for $2.5 TWL 25 |<-------- L = 47 m -------->| ``` **Answer**: floor invert RL **14.27 m**; USBR Type II basin, length about 47 m, chute blocks 1.0 m, dentated end sill 2.15 m, with $y_1=1.011$ m and $y_2=10.73$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is energy at the toe above the floor (RL Z) expressed? A. E1 = 0.9 (50 - Z), because 10% is lost on the spillway B. E1 = 50 - Z C. E1 = 0.1 (50 - Z) D. E1 = 25 - Z Answer: A. With a 10% loss, 90% of the head between lake level and the floor remains at the toe. (Key point: Unit discharge and energy) 2. What condition fixes the floor level Z? A. Tail water depth over the floor (25 - Z) equals y2 B. 25 - Z equals y1 C. Z equals y2 D. Fr1 equals 1 Answer: A. The jump forms in the basin when downstream depth equals the sequent depth; trial gives Z = 14.27 m. (Key point: Jump condition) 3. Why is a Type II basin chosen? A. Fr1 = 7.85 > 4.5 and V1 = 24.7 m/s > 18 m/s B. Fr1 is between 2.5 and 4.5 C. V1 is below 18 m/s D. q is very small Answer: A. Type IV is for 2.5 < Fr1 < 4.5 and Type III is limited to V1 < 18 m/s. (Key point: Basin selection) 4. Which dimensions are correct for the Type II basin? A. L about 4.3 y2 = 47 m, dentated end sill 0.2 y2 = 2.15 m B. L = y2, sill 0.2 y1 C. L = 10 m, sill 5 m D. L = 4.3 y1, sill y2 Answer: A. Length is about 4.3 y2 and sill height 0.2 y2 with y2 = 10.73 m. 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