--- title: "Forebay design (15 m³/s, 4 minutes)" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:forebay-design-15cumec-4min" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/forebay-design-15cumec-4min site: BCT Engineering --- # Forebay design (15 m³/s, 4 minutes) Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2071 Chaitra, 4 marks - Video: [Design of Forebay || Hydropower Engineering Numerical](https://www.youtube.com/watch?v=E9tU-GG_tuM) — OPengineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/forebay-design-15cumec-4min - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question Design a forebay using the following data: Q = 15 m³/s; storage requirement = 4 minutes; length of penstock = 500 m; diameter of penstock = 2 m. ## Short answer With 4 min retention, volume = 15×240 = 3600 m³, area 75 m², depth 7.4 m. Forebay 48 m long × 10.5 m wide × 7.4 m deep. ## Answer **Given**: $Q=15.0$ m³/s; 1 penstock (assumed) of diameter 2.0 m and length 500 m; storage (retention) time 4 min; limiting velocity in the forebay 0.20 m/s (assumed, the usual limit, as it is not given). #### Step 1: Volume of the forebay The forebay stores the water for the retention time: $$ V=Q\,t=15.0\times(4\times60)=3600\ \text{m}^3 $$ #### Step 2: Cross-sectional area from the velocity limit $$ A=\frac{Q}{v}=\frac{15.0}{0.20}=75.0\ \text{m}^2 $$ #### Step 3: Depth of water (submergence of the penstock intake) Velocity in the penstock: $V_p=\dfrac{Q}{\frac{\pi}{4}D^2}=4.775$ m/s. Minimum submergence to avoid vortices (Gordon, with $C=0.7245$ for an asymmetrical approach, which is on the safe side): $$ S=0.7245\,V_p\sqrt{D}=0.7245\times4.775\times\sqrt{2.0}=4.89\ \text{m} $$ Water depth = $S+D+$ bottom clearance of 0.5 m $=4.89+2.0+0.5=7.39$ m, adopt $H=7.4$ m. #### Step 4: Width and length $$ B=\frac{A}{H}=\frac{75.0}{7.4}=10.14\ \text{m},\qquad L=\frac{V}{A}=\frac{3600}{75.0}=48.0\ \text{m} $$ Adopt $B=10.5$ m and $L=48$ m (the volume is then $B\,L\,H=3730$ m³ $\ge V$). The penstock length (500 m) is not needed to size the forebay by this method; it matters for the penstock head loss and the water hammer. ```text PLAN headrace forebay penstock tunnel/canal +------------------+ ====> \ | | \ trans| trash rack |==> penstock ====> / | | +--+-spillway-+-flush+ | | v v to river (spillway/flushing) SECTION canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~ | S (submergence) | rack ____ crown | #### \ penstock flushing ______ |___####____\_D__ bottom clearance 0.5 m ``` | Item | Value | |---|---| | Storage volume | 3600 m³ | | Water depth | 7.4 m (+0.5 m free board) | | Width | 10.5 m | | Length | 48 m | | Submergence of penstock crown | 4.89 m | **Answer**: forebay about 48 m long x 10.5 m wide x 7.4 m deep (water), with a volume of 3600 m³. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Q = 15 m^3/s with 4 min storage requires what volume? A. 2700 m^3 B. 900 m^3 C. 3600 m^3 D. 60 m^3 Answer: C. V = 15 x 240 = 3600 m^3. (Key point: Storage volume) 2. For one penstock D = 2 m carrying 15 m^3/s, the velocity is about? A. 4.77 m/s B. 7.5 m/s C. 1.5 m/s D. 2.39 m/s Answer: A. 15/(pi/4 x 4) = 4.775 m/s. (Key point: Penstock velocity) 3. Which formula gives the minimum submergence to avoid vortices? A. S = 0.7245 V sqrt(D) B. S = 0.5 D C. S = V^2/2g D. S = Q/A Answer: A. Gordon's relation for symmetrical approach: S = 0.7245 V sqrt(D) = 4.89 m here. (Key point: Gordon submergence) 4. Which set matches the adopted forebay? A. L = 40 m, B = 13 m, H = 5.1 m B. L = 48 m, B = 15 m, H = 4.8 m C. L = 36 m, B = 12 m, H = 8.4 m D. L = 48 m, B = 10.5 m, H = 7.4 m Answer: D. L = 3600/75 = 48 m, B = 75/7.4 about 10.5 m, H = 4.89 + 2 + 0.5 = 7.4 m. (Key point: Dimensions) ## Related in Hydropower Engineering - [Forebay dimension for 3 minutes storage](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aforebay-dimension-3min-storage-2-24m) — Hydropower Engineering, chapter 6 - [Forebay design (20 m³/s, 300 m penstock)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aforebay-design-20cumec-300m) — Hydropower Engineering, chapter 6 - [Forebay design with spillway length (14 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aforebay-design-14cumec-spillway-length) — Hydropower Engineering, chapter 6 - [Most economic penstock diameter (cost numerical)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-economic-diameter-cost-numerical) — Hydropower Engineering, chapter 6 - [Forebay design considerations; design for 12 m³/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aforebay-design-considerations-12cumec) — Hydropower Engineering, chapter 6 - [Forebay design (14.5 m³/s, two penstocks)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aforebay-design-14-5cumec) — Hydropower Engineering, chapter 6 ## Related in other subjects - [Economic penstock diameter by empirical formulas](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-economic-diameter-formulas) — Hydro Power, chapter 5 - [Economic diameter of penstock](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-economic-diameter) — Hydro Power, chapter 5, asked 5× - [Surge tank objectives and economic penstock diameter factors](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asurge-tank-objectives-economic-diameter) — Hydro Power, chapter 5 - [Penstock thickness and economic diameter](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-thickness-economic-diameter) — Hydro Power, chapter 5