--- title: "USBR stilling basin design (80 m³/s)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:stilling-basin-usbr-80cumec-2082" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/stilling-basin-usbr-80cumec-2082 site: BCT Engineering --- # USBR stilling basin design (80 m³/s) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2082 Baishakh, 7 marks - 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-80cumec-2082 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Design an appropriate USBR-type stilling basin using the following data: spillway discharge = 80 m³/s, width of spillway = 8 m, spillway crest level = 96 m, riverbed level = 65 m, river bed slope = 1:500 and Manning's n = 0.016. ## Short answer Using Manning's equation for tailwater and energy equation at toe, the required floor depression is 4.66 m, giving basin floor RL 60.34 m. For Fr1=14.45, select USBR Type II basin with length 32 m, chute blocks 0.4 m, dentated end sill 1.46 m. ## Answer **Assumptions**: spillway discharge coefficient $C_d=0.7$ in $q=\tfrac23C_d\sqrt{2g}H^{3/2}$; energy losses on the spillway face are neglected; the basin has the same width as the spillway (8 m); the tail-water depth is the normal depth in the river channel (rectangular, 8 m wide). #### Step 1: Unit discharge and head over crest $$ q=\frac{80}{8}=10\ \text{m}^3/\text{s/m},\qquad 10=\tfrac23(0.7)(4.429)H^{3/2}\Rightarrow H=2.86\ \text{m} $$ #### Step 2: Tail-water depth (Manning) River section 8 m wide, $S=1/500$, $n=0.016$. Solving $Q=\frac1nAR^{2/3}S^{1/2}$ for $Q=80$ m³/s gives normal depth $$ y_n=2.63\ \text{m} $$ #### Step 3: Conditions at the toe (without depression) Total head above the river bed $E=(96-65)+H=33.86$ m. $$ y_1+\frac{q^2}{2g\,y_1^2}=E\ \Rightarrow\ y_1=0.390\ \text{m},\quad V_1=25.6\ \text{m/s},\quad Fr_1=13.10 $$ $$ y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=7.04\ \text{m} $$ Since $y_n=2.63\ \text{m}4.5$ and $V_1=27.4$ m/s $>18$ m/s, so the **USBR Type II** basin (with chute blocks, dentated end sill and no baffle blocks) is chosen; a Type III basin is not suitable above about 18 m/s. | Component | Size | |---|---| | Sequent depth $y_2$ | 7.29 m | | Basin length $L\approx4.3\,y_2$ (USBR chart) | 31.3 m, say 32 m | | Chute blocks | height, width and spacing $\approx y_1=0.365$ m (say 0.4 m) | | Dentated end sill | height $0.2y_2=1.46$ m; width and spacing $0.15y_2=1.09$ m | | Floor RL | 60.34 m (to be lined with high-strength concrete against abrasion and cavitation) | | Tail water | $y_n+z=7.29\ \text{m}\ge y_2$, so the jump is held in the basin | ```text crest 96 ____ \ \ spillway face \ \_ chute blocks dentated sill 65 river ===\_|#|#|__________|^|^|^|=== river floor 60.34 <----- L = 32 m ------> ``` **Answer**: $y_1=0.365$ m, $y_2=7.29$ m, $Fr_1=14.45$; USBR Type II basin, floor depressed by 4.66 m to RL 60.34 m, length 32 m, chute blocks 0.4 m, dentated end sill 1.46 m high. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the available tail-water depth found? A. Normal depth from Manning's equation in the 8 m wide river for 80 m3/s, yn = 2.63 m B. From the crest level C. From q/sqrt(g) D. Equal to the head over the crest Answer: A. No tail water is given, so the river's normal depth is used. (Key point: Tail water depth) 2. Why must the basin floor be depressed? A. yn = 2.63 m is less than the sequent depth y2 = 7.04 m, so the jump would be swept away B. The tail water exceeds y2 C. Froude number is below 1 D. The slope is too flat Answer: A. Lowering the floor increases the available tail-water depth (yn + z) until it equals y2. (Key point: Why depress the floor) 3. What is the final floor level? A. 60.34 m, from z = 4.66 m depression B. 65 m C. 69.66 m D. 64.66 m Answer: A. Floor RL = 65 - 4.66 = 60.34 m, where yn + z = y2 = 7.29 m. (Key point: Floor RL) 4. Why is a USBR Type II basin chosen? A. Fr1 = 14.45 > 4.5 and V1 = 27.4 m/s > 18 m/s, so a Type III with baffle blocks is unsuitable B. Fr1 is less than 2.5 C. V1 is below 10 m/s D. The discharge is tiny Answer: A. Type III is limited to velocities below about 18 m/s; Type II uses chute blocks and a dentated sill. 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