---
title: "Total rainfall, net runoff, W-index of 140-min storm"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-w-index-140min-storm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-w-index-140min-storm
site: BCT Engineering
---

# Total rainfall, net runoff, W-index of 140-min storm

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2067 Shrawan (old course), 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-w-index-140min-storm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

In a 140-minute storm, the following rates of rainfall were observed in successive 20-minute intervals: 3.0, 3.0, 9.0, 6.6, 1.2, 1.2 and 6.0 mm/hr. Assume the $\phi$ index value as 3.0 mm/hr and an initial loss of 0.8 mm. Determine the total rainfall, net runoff and W-index for the storm.

## Short answer

Convert each 20-minute rate to depth by dividing by 3; total rainfall is 10.0 mm. Rainfall excess over phi = 3 mm/h gives net runoff 4.2 mm, and W-index = (10.0 - 4.2 - 0.8)/(140/60) = 2.14 mm/h.

## Answer

Each interval is 20 min = 1/3 h, so depth in an interval = rate $\times \frac{1}{3}$.

| Interval | Rate (mm/h) | Rain (mm) | $\phi$ loss (mm) | Excess (mm) |
|---|---|---|---|---|
| 1 | 3.0 | 1.000 | 1.000 | 0 |
| 2 | 3.0 | 1.000 | 1.000 | 0 |
| 3 | 9.0 | 3.000 | 1.000 | 2.000 |
| 4 | 6.6 | 2.200 | 1.000 | 1.200 |
| 5 | 1.2 | 0.400 | 0.400 | 0 |
| 6 | 1.2 | 0.400 | 0.400 | 0 |
| 7 | 6.0 | 2.000 | 1.000 | 1.000 |

The initial loss (0.8 mm) is absorbed in the first interval, where the rain is only 1.0 mm and is already at the $\phi$ rate, so it does not change the excess rainfall.

#### Total rainfall
$$
P = \frac{3.0+3.0+9.0+6.6+1.2+1.2+6.0}{3} = \frac{30.0}{3} = 10.0\ \text{mm}
$$

#### Net runoff
$$
R = 2.0 + 1.2 + 1.0 = 4.2\ \text{mm}
$$

#### W-index
$$
W = \frac{P - R - I_a}{t} = \frac{10.0 - 4.2 - 0.8}{140/60} = \frac{5.0}{2.333} = 2.14\ \text{mm/h}
$$

**Answer: Total rainfall = 10.0 mm; net runoff = 4.2 mm; W-index = 2.14 mm/h.**

## Check yourself

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

1. 20-min interval depth equals rate x
   A. 2/3 h
   B. 1/3 h
   C. 1 h
   D. 20 h
   Answer: B. 20 min = 1/3 h. (Key point: Interval depth)

2. Net runoff is
   A. 10 mm
   B. 4.2 mm
   C. 2.1 mm
   D. 5 mm
   Answer: B. Excess 2.0 + 1.2 + 1.0 = 4.2 mm. (Key point: Runoff)

3. Initial loss 0.8 mm does not change excess because
   A. First interval rain 1.0 mm is already lost at φ rate
   B. It is ignored
   C. It is bigger than rain
   D. It is negative
   Answer: A. Rain there is wholly lost anyway. (Key point: Initial loss)

4. W = (P - R - Ia)/t gives
   A. 3.0 mm/h
   B. 2.14 mm/h
   C. 1.2 mm/h
   D. 4.2 mm/h
   Answer: B. (10 - 4.2 - 0.8)/2.333 = 2.14. (Key point: W-index)

## Close variants set in other exams

- [Total rainfall, net runoff, W-index of 120-min storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-w-index-120min-storm) — Engineering Hydrology, chapter 3, asked 2×

## Related in Engineering Hydrology

- [Surface runoff with initial loss and phi-index, 100-min storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Asurface-runoff-initial-loss-0-6) — Engineering Hydrology, chapter 3
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- [Total rainfall and phi-index, 100 km2 basin](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Atotal-rainfall-phi-100km2) — Engineering Hydrology, chapter 3