---
title: "Forebay dimension for 3 minutes storage"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:forebay-dimension-3min-storage-2-24m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/forebay-dimension-3min-storage-2-24m
site: BCT Engineering
---

# Forebay dimension for 3 minutes storage

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2075 Ashwin, 3 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-dimension-3min-storage-2-24m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

Find out the dimension of a forebay which accommodates a storage for 3 minutes of operation for a hydropower plant having the following data: design discharge = 20 m³/s; length of penstock = 300 m; diameter of penstock = 2.20 m.

## Short answer

Volume = 3600 m³, area = 100 m², depth 8.4 m. Dimensions: 36 m long × 12 m wide × 8.4 m deep.

## Answer

**Given**: $Q=20.0$ m³/s; 1 penstock (assumed) of diameter 2.2 m and length 300 m; storage (retention) time 3 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:
$$
V=Q\,t=20.0\times(3\times60)=3600\ \text{m}^3
$$

#### 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. For Q = 20 m^3/s, 3 min storage and an assumed 0.2 m/s, what are V and A?
   A. V = 3600 m^3, A = 4 m^2
   B. V = 3600 m^3, A = 100 m^2
   C. V = 1200 m^3, A = 100 m^2
   D. V = 360 m^3, A = 10 m^2
   Answer: B. V = Qt = 3600 m^3 and A = Q/v = 100 m^2. (Key point: Volume and area)

2. Which relation gives the length of the forebay?
   A. L = A/V
   B. L = V/A = 36 m
   C. L = Q/D
   D. L = V x A
   Answer: B. 3600/100 = 36 m. (Key point: Length)

3. Which items make up the 8.4 m depth?
   A. Width x length
   B. Penstock length/36
   C. Gordon submergence 5.65 + diameter 2.2 + clearance 0.5
   D. Free board only
   Answer: C. 5.65 + 2.2 + 0.5 = 8.35, adopt 8.4 m; the width is then 100/8.4, adopted 12 m. (Key point: Depth)

## Related in Hydropower Engineering

- [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 (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 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 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
- [Economic diameter; penstock thickness (1.25 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-economic-diameter-thickness-1-25m) — Hydropower Engineering, chapter 6

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