---
title: "Forebay design (14.5 m³/s, two penstocks)"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:forebay-design-14-5cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/forebay-design-14-5cumec
site: BCT Engineering
---

# Forebay design (14.5 m³/s, two penstocks)

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2079 Baishakh, 6 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-14-5cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

Design a forebay structure with turbine discharge of 14.5 m³/s with two penstocks 1.8 m diameter each. Take retention time 3 minutes and limiting velocity 0.22 m/s. Draw neat sketch of plan and section.

## Short answer

Volume 2610 m³, area 65.9 m², depth 5.1 m. Forebay 40 m long × 13 m wide × 5.1 m deep.

## Answer

**Given**: $Q=14.5$ m³/s; 2 penstocks of diameter 1.8 m; retention (detention) time 3 min; limiting velocity in the forebay 0.22 m/s.

#### Step 1: Volume of the forebay
The forebay stores the water for the retention time:
$$
V=Q\,t=14.5\times(3\times60)=2610\ \text{m}^3
$$

#### Step 2: Cross-sectional area from the velocity limit
$$
A=\frac{Q}{v}=\frac{14.5}{0.22}=65.9\ \text{m}^2
$$

#### Step 3: Depth of water (submergence of the penstock intake)
Velocity in each penstock: $V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=2.849$ m/s.
Minimum submergence to avoid vortices (Gordon, with $C=0.7245$ for an asymmetrical approach, which is on the safe side):
$$
S=0.7245\,V_p\sqrt{D}=0.7245\times2.849\times\sqrt{1.8}=2.77\ \text{m}
$$
Water depth = $S+D+$ bottom clearance of 0.5 m $=2.77+1.8+0.5=5.07$ m, adopt $H=5.1$ m.

#### Step 4: Width and length
$$
B=\frac{A}{H}=\frac{65.9}{5.1}=12.92\ \text{m},\qquad L=\frac{V}{A}=\frac{2610}{65.9}=39.6\ \text{m}
$$
Adopt $B=13.0$ m and $L=40$ m (the volume is then $B\,L\,H=2652$ m³ $\ge V$).

```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 | 2610 m³ |
| Water depth | 5.1 m (+0.5 m free board) |
| Width | 13.0 m |
| Length | 40 m |
| Submergence of penstock crown | 2.77 m |

**Answer**: forebay about 40 m long x 13.0 m wide x 5.1 m deep (water), with a required volume of 2610 m³.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For Q = 14.5 m^3/s and a limiting velocity 0.22 m/s, how is the cross-section found and what is it?
   A. A = Q/v = 65.9 m^2
   B. A = Q x v = 3.19 m^2
   C. A = Q/v = 14.5 m^2
   D. A = V/Q = 180 m^2
   Answer: A. A = Q/v = 14.5/0.22 = 65.9 m^2. (Key point: Area from velocity limit)

2. How is the forebay length obtained after A is known?
   A. L = Q x v
   B. L = V/A = 2610/65.9, about 39.6 m
   C. L = penstock diameter x 10
   D. L = A/H
   Answer: B. The stored volume divided by the cross-section gives the length, about 40 m. (Key point: Length)

3. Which sum gives the water depth H here (S = 2.77 m, D = 1.8 m)?
   A. S + D + 0.5 m bottom clearance = 5.07 m, adopt 5.1 m
   B. S + 0.5 = 3.27 m
   C. S x D = 4.99 m
   D. S + D/2 = 3.67 m
   Answer: A. Depth = submergence + pipe diameter + bottom clearance. (Key point: Depth)

## Related in Hydropower Engineering

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- [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
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