--- title: "Forebay design with spillway length (14 m³/s)" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:forebay-design-14cumec-spillway-length" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/forebay-design-14cumec-spillway-length site: BCT Engineering --- # Forebay design with spillway length (14 m³/s) Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2081 Bhadra, 8 marks - 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-14cumec-spillway-length - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question Design a forebay for a power plant with a design discharge of 14 m³/s. The system uses two penstock pipes, each with a length of 500 m and diameter of 2 m, to convey water. Assume a detention time of 3 minutes and a flow velocity of 0.2 m/s in the forebay. Additionally, design an appropriate spillway length for the forebay. ## Short answer Volume = 2520 m³, area 70 m², depth 4.8 m; forebay 36 m×15 m×4.8 m. Spillway length L = Q/(C h^1.5) = 14/(1.7×0.5^1.5) ≈24 m for 0.5 m head. ## Answer **Given**: $Q=14.0$ m³/s; 2 penstocks of diameter 2.0 m and length 500 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s. #### Step 1: Volume of the forebay The forebay stores the water for the retention time: $$ V=Q\,t=14.0\times(3\times60)=2520\ \text{m}^3 $$ #### Step 2: Cross-sectional area from the velocity limit $$ A=\frac{Q}{v}=\frac{14.0}{0.20}=70.0\ \text{m}^2 $$ #### Step 3: Depth of water (submergence of the penstock intake) Velocity in each penstock: $V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=2.228$ m/s. Minimum submergence to avoid vortices (Gordon, with $C=0.7245$ for an asymmetrical approach, which is on the safe side): $$ S=0.7245\,V_p\sqrt{D}=0.7245\times2.228\times\sqrt{2.0}=2.28\ \text{m} $$ Water depth = $S+D+$ bottom clearance of 0.5 m $=2.28+2.0+0.5=4.78$ m, adopt $H=4.8$ m. #### Step 4: Width and length $$ B=\frac{A}{H}=\frac{70.0}{4.8}=14.58\ \text{m},\qquad L=\frac{V}{A}=\frac{2520}{70.0}=36.0\ \text{m} $$ Adopt $B=15.0$ m and $L=36$ m (the volume is then $B\,L\,H=2592$ 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. #### Step 5: Spillway of the forebay If the turbines trip, the whole discharge must spill over the forebay spillway (side-channel type) while the water level rises by $h=0.5$ m: $$ Q=C\,L_s\,h^{3/2}\ \Rightarrow\ L_s=\frac{Q}{C\,h^{3/2}}=\frac{14.0}{1.7\times0.5^{3/2}}=23.3\ \text{m} $$ ($C=1.7$ for a broad-crested weir.) Adopt $L_s=24$ m, placed along the side wall (or folded), with a spill channel to the river. ```text PLAN headrace forebay penstocks tunnel/canal +------------------+ ====> \ | |==> penstock 1 \ trans| trash rack | ====> / | |==> penstock 2 +--+-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 | 2520 m³ | | Water depth | 4.8 m (+0.5 m free board) | | Width | 15.0 m | | Length | 36 m | | Submergence of penstock crown | 2.28 m | | Spillway length | 24 m (head 0.5 m) | **Answer**: forebay about 36 m long x 15.0 m wide x 4.8 m deep (water), with a volume of 2520 m³, and a spillway 24 m long. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Q = 14 m^3/s and detention 3 min. What storage volume is needed? A. 2520 m^3 B. 4200 m^3 C. 1260 m^3 D. 840 m^3 Answer: A. V = 14 x 3 x 60 = 2520 m^3. (Key point: Storage volume) 2. How is the water depth of the forebay obtained? A. Depth = penstock length / 100 B. Depth = area / velocity C. Depth = Gordon submergence + penstock diameter + bottom clearance 0.5 m D. Depth = volume / penstock area Answer: C. H = S + D + 0.5 = 2.28 + 2.0 + 0.5 = 4.78, adopt 4.8 m. (Key point: Depth from submergence) 3. What role does the 500 m penstock length play in sizing this forebay? A. It sets the forebay length B. It sets the spillway length C. It sets the detention time D. It is not needed for sizing; it matters for head loss and water hammer Answer: D. The method uses Q, retention time, velocity and diameter; penstock length affects only head loss and water hammer. (Key point: Penstock length) 4. The spillway must pass the whole 14 m^3/s with h = 0.5 m using Q = C Ls h^(3/2), C = 1.7. What length results? A. About 11.7 m B. About 23.3 m C. About 8.2 m D. About 46.6 m Answer: B. Ls = 14/(1.7 x 0.5^1.5) = 14/(1.7 x 0.3536) = 23.3 m, adopt 24 m. 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