--- title: "Forebay design (20 m³/s, 300 m penstock)" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:forebay-design-20cumec-300m" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/forebay-design-20cumec-300m site: BCT Engineering --- # Forebay design (20 m³/s, 300 m penstock) Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2080 Bhadra, 6 marks - Video: [Design of Forebay || Hydropower Engineering Numerical](https://www.youtube.com/watch?v=wBR0XxeCHEw) — 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-20cumec-300m - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question Design a forebay with the following data: design discharge = 20 m³/s, penstock length = 300 m, detention time = 3 minutes, diameter of penstock = 2.2 m. ## Short answer Volume 3600 m³, area 100 m², depth 8.4 m. Forebay 36 m × 12 m × 8.4 m. ## Answer **Given**: $Q=20.0$ m³/s; 1 penstock (only one diameter is given) of diameter 2.2 m and length 300 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s (assumed, as it is not given). #### Step 1: Volume of the forebay The forebay stores the water for the retention time: $$ V=Q\,t=20.0\times(3\times60)=3600\ \text{m}^3 $$ #### Step 2: Cross-sectional area from the velocity limit $$ A=\frac{Q}{v}=\frac{20.0}{0.20}=100.0\ \text{m}^2 $$ #### Step 3: Depth of water (submergence of the penstock intake) Velocity in the penstock: $V_p=\dfrac{Q}{\frac{\pi}{4}D^2}=5.261$ m/s. Minimum submergence to avoid vortices (Gordon, with $C=0.7245$ for an asymmetrical approach, which is on the safe side): $$ S=0.7245\,V_p\sqrt{D}=0.7245\times5.261\times\sqrt{2.2}=5.65\ \text{m} $$ Water depth = $S+D+$ bottom clearance of 0.5 m $=5.65+2.2+0.5=8.35$ m, adopt $H=8.4$ m. #### Step 4: Width and length $$ B=\frac{A}{H}=\frac{100.0}{8.4}=11.90\ \text{m},\qquad L=\frac{V}{A}=\frac{3600}{100.0}=36.0\ \text{m} $$ Adopt $B=12.0$ m and $L=36$ m (the volume is then $B\,L\,H=3629$ m³ $\ge V$). The penstock length (300 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 | 8.4 m (+0.5 m free board) | | Width | 12.0 m | | Length | 36 m | | Submergence of penstock crown | 5.65 m | **Answer**: forebay about 36 m long x 12.0 m wide x 8.4 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 = 20 m^3/s with 3 min detention gives what volume? A. 6000 m^3 B. 1200 m^3 C. 3600 m^3 D. 360 m^3 Answer: C. V = 20 x 180 = 3600 m^3. (Key point: Storage volume) 2. With one penstock of D = 2.2 m carrying 20 m^3/s, how is the penstock velocity found? A. V = Q/(2 x pi/4 x D^2) = 2.63 m/s B. V = Q/(pi/4 x D^2) = 5.26 m/s C. V = Q/D = 9.1 m/s D. V = Q x D Answer: B. One pipe takes the full flow: 20/3.801 = 5.26 m/s; this is used in Gordon's formula. (Key point: Single penstock velocity) 3. The question gives no forebay velocity. What is assumed to find A? A. 0.2 m/s limiting velocity, so A = 100 m^2 B. The penstock velocity C. 1 m/s, so A = 20 m^2 D. 0.5 m/s, so A = 40 m^2 Answer: A. The usual limit 0.2 m/s gives A = 20/0.2 = 100 m^2. (Key point: Assumed velocity) 4. Why is the depth so large (8.4 m) compared with the earlier 2-penstock cases? A. Penstock length is 300 m B. Higher penstock velocity gives larger Gordon submergence (5.65 m) C. The width is smaller D. Retention time is longer Answer: B. S = 0.7245 V sqrt(D) grows with velocity, and depth = S + D + 0.5. (Key point: Depth effect) ## Related in Hydropower Engineering - [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 - [Forebay design (15 m³/s, 4 minutes)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aforebay-design-15cumec-4min) — Hydropower Engineering, chapter 6 - [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 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 - [Penstock thickness with gate closure time](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-thickness-water-hammer-600m) — Hydropower Engineering, chapter 6 ## Related in other subjects - [Design discharge from given data](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adesign-discharge-sewer-75000) — Sanitary Engineering, chapter 2 - [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 - [Penstock thickness (1.3 m, 20 kg/cm²)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-thickness-1-3m) — Hydro Power, chapter 5 - [Penstock thickness (1.20 m, 17.5 kg/cm²)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-thickness-1-2m) — Hydro Power, chapter 5