---
title: "Runoff and effective rainfall, 4-hr storm, 80 km2"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:runoff-80km2-subbasins-phi"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/runoff-80km2-subbasins-phi
site: BCT Engineering
---

# Runoff and effective rainfall, 4-hr storm, 80 km2

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2070 Asar, 14 marks; 2067 Mangsir (old course), 12 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/runoff-80km2-subbasins-phi
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

A 4-hour storm occurs over a 80 km$^2$ watershed. The details of the catchment are as follows:

| Sub basin (km$^2$) | $\phi$ index (mm/h) | 1st hour (mm) | 2nd hour (mm) | 3rd hour (mm) | 4th hour (mm) |
|---|---|---|---|---|---|
| 15 | 10 | 16 | 48 | 22 | 10 |
| 25 | 15 | 16 | 42 | 20 | 8 |
| 35 | 21 | 12 | 40 | 18 | 6 |
| 5 | 16 | 15 | 42 | 18 | 8 |

Calculate the runoff from the catchment and the hourly distribution of the effective rainfall for the whole catchment.

## Short answer

Excess depths from each sub-basin are area-weighted: effective rain is 1.44, 25.50, 3.94 and 0 mm for hours 1–4. Total runoff depth is 30.875 mm, volume 2.47×10⁶ m³.

## Answer

Effective (excess) rainfall in each hour of a sub-basin is $P_t - \phi$ (if positive; otherwise zero). The sub-basin values are then area-weighted for the whole catchment.

#### Step 1: Effective rainfall (mm) in each sub-basin
| Area (km²) | $\phi$ (mm/h) | 1st h | 2nd h | 3rd h | 4th h | Total (mm) |
|---|---|---|---|---|---|---|
| 15 | 10 | 6 | 38 | 12 | 0 | 56 |
| 25 | 15 | 1 | 27 | 5 | 0 | 33 |
| 35 | 21 | 0 | 19 | 0 | 0 | 19 |
| 5 | 16 | 0 | 26 | 2 | 0 | 28 |

Example (15 km², $\phi = 10$): 16-10 = 6, 48-10 = 38, 22-10 = 12, 10-10 = 0.

#### Step 2: Area-weighted effective rainfall for 80 km²
$$
\bar P_{e,t} = \frac{\sum A_i P_{e,i,t}}{80}
$$

| Hour | $\sum A_iP_{e,i}$ (mm·km²) | Effective rainfall (mm) |
|---|---|---|
| 1 | 115 | 1.4375 |
| 2 | 2040 | 25.5000 |
| 3 | 315 | 3.9375 |
| 4 | 0 | 0.0000 |

Check: $15\times6 + 25\times1 + 0 + 0 = 115$ (1st hour); $15\times38 + 25\times27 + 35\times19 + 5\times26 = 2040$ (2nd hour); $15\times12 + 25\times5 + 5\times2 = 315$ (3rd hour).

#### Step 3: Runoff from the catchment
Total effective rainfall $= 1.4375 + 25.5 + 3.9375 + 0 = 30.875$ mm.

$$
V = 30.875\times10^{-3}\times 80\times10^6 = 2,470,000\ \text{m}^3 = 2.47\ \text{million m}^3
$$

Average runoff rate over the 4 h $= 2.47\times10^6/(4\times3600) = 171.5$ m³/s.

**Answer: Runoff depth = 30.875 mm, volume = 2.47 × 10⁶ m³; hourly effective rainfall = 1.44, 25.50, 3.94 and 0 mm.**

## Check yourself

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

1. Effective rainfall in an hour for a sub-basin is:
   A. P + phi
   B. P x phi
   C. P - phi if positive, otherwise zero
   D. phi - P
   Answer: C. Rain above the phi-index rate is effective; below it all infiltrates. (Key point: Effective rainfall)

2. How is the catchment effective rainfall found?
   A. Simple average of sub-basin values
   B. Area-weighted: sum of A x Pe / 80
   C. Largest sub-basin value
   D. Sum of sub-basin values
   Answer: B. Sub-basins are weighted by area over total 80 km2. (Key point: Weighting)

3. The catchment effective rainfall in the 2nd hour is:
   A. 25.5 mm
   B. 38 mm
   C. 2040 mm
   D. 19 mm
   Answer: A. 2040 mm.km2 / 80 = 25.5 mm. (Key point: 2nd hour)

4. Total effective rainfall 30.875 mm over 80 km2 gives runoff volume:
   A. 2.47 x 10^6 m3
   B. 2.47 x 10^3 m3
   C. 30.875 x 10^6 m3
   D. 2.47 x 10^9 m3
   Answer: A. 0.030875 m x 80 x 10^6 m2 = 2.47 x 10^6 m3. (Key point: Volume)

## Related in Engineering Hydrology

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