---
title: "Storm water from a 45 ha catchment"
subject: "Sanitary Engineering"
chapter: 2
key: "semester_6/sanitary_engineering:storm-water-catchment-surfaces"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/storm-water-catchment-surfaces
site: BCT Engineering
---

# Storm water from a 45 ha catchment

Sanitary Engineering · Chapter 2: Quantity of Wastewater

- Asked in: 2071 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/storm-water-catchment-surfaces
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-2

## Question

The catchment area of a city is 45 hectares. Assuming that the surface on which rain falls is classified as follows:

| Type of Surface | % Area | Runoff Coefficient |
|---|---|---|
| 1. Forest and Wooden Area | 10 | 0.15 |
| 2. Open ground + Unpaved street | 10 | 0.20 |
| 3. Parks + Lawns + Gardens | 15 | 0.15 |
| 4. Gravel Road | 20 | 0.25 |
| 5. Asphalt Pavements | 20 | 0.85 |
| 6. Water tight Roof Surfaces | 25 | 0.90 |

Calculate the quantity of storm water if time of entry is 20 minutes and time of flow is 10 minutes.

## Short answer

Weighted runoff coefficient is 0.5025. With tc=30 min, intensity is 19 mm/hr, and A=45 ha, the storm water quantity by Q=CiA/360 is 1.193 m3/s (1193 L/s).

## Answer

Storm water quantity by the Rational method: $Q=\dfrac{CiA}{360}$.

#### Step 1: Weighted runoff coefficient
| Surface | % area | C | % x C / 100 |
|---|---|---|---|
| Forest and wooded | 10 | 0.15 | 0.0150 |
| Open ground + unpaved street | 10 | 0.20 | 0.0200 |
| Parks, lawns, gardens | 15 | 0.15 | 0.0225 |
| Gravel road | 20 | 0.25 | 0.0500 |
| Asphalt pavement | 20 | 0.85 | 0.1700 |
| Watertight roof | 25 | 0.90 | 0.2250 |
| **Total** | 100 | | **0.5025** |

$$
C=\frac{\sum C_jA_j}{\sum A_j}=0.5025
$$

#### Step 2: Time of concentration and intensity
$$
t_c=t_e+t_f=20+10=30\ \text{min}
$$

No intensity formula is given, so the common formula $i=\dfrac{760}{t+10}$ mm/hr is assumed:

$$
i=\frac{760}{30+10}=19.0\ \text{mm/hr}
$$

#### Step 3: Discharge
$$
Q=\frac{0.5025\times19.0\times45}{360}=1.193\ \text{m}^3/\text{s}
$$

**Answer: Storm water quantity = 1.193 m³/s (1193 l/s).**

## Check yourself

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

1. The weighted runoff coefficient for the catchment is
   A. 0.5025
   B. 0.25
   C. 0.90
   D. 0.15
   Answer: A. Sum of (% x C)/100 = 0.5025. (Key point: Weighted C)

2. Time of concentration is
   A. 20 min
   B. 10 min
   C. 30 min
   D. 200 min
   Answer: C. te + tf = 20 + 10 = 30 min. (Key point: tc)

3. Using i = 760/(t+10), the intensity at tc = 30 min is
   A. 38 mm/hr
   B. 25.3 mm/hr
   C. 76 mm/hr
   D. 19.0 mm/hr
   Answer: D. 760 / 40 = 19.0 mm/hr. (Key point: Intensity)

4. The storm water quantity is
   A. 1.193 m3/s
   B. 11.93 m3/s
   C. 0.119 m3/s
   D. 2.39 m3/s
   Answer: A. 0.5025 x 19 x 45 / 360 = 1.193 m3/s. (Key point: Discharge)

## Related in Sanitary Engineering

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