---
title: "Total rainfall, net runoff, W-index of 120-min storm"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-w-index-120min-storm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-w-index-120min-storm
site: BCT Engineering
---

# Total rainfall, net runoff, W-index of 120-min storm

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2081 Baisakh, 1+2+5 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-120min-storm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

In a 120 minute storm the rates of rainfall observed in successive 20 minute intervals are 5, 8, 15, 12, 9 and 2 mm/hr. Assuming the $\phi$-index value as 2.5 mm/hr and an initial loss of 0.5 mm, determine the total rainfall, net runoff and W-index of the storm.

## Short answer

Total rainfall = 17.0 mm, net runoff = 11.67 mm, W-index = 2.9 mm/h. Rain depths per 20 min were computed; excess over φ = 2.5 mm/h summed, initial loss 0.5 mm subtracted, and W-index found using time when rain exceeded φ.

## Answer

Each interval is 20 min = 1/3 h, so rain depth = rate $/3$.

| Interval | Rate (mm/h) | Rain (mm) | Excess over $\phi = 2.5$ (mm/h) | Excess depth (mm) |
|---|---|---|---|---|
| 1 | 5 | 1.667 | 2.5 | 0.833 |
| 2 | 8 | 2.667 | 5.5 | 1.833 |
| 3 | 15 | 5.000 | 12.5 | 4.167 |
| 4 | 12 | 4.000 | 9.5 | 3.167 |
| 5 | 9 | 3.000 | 6.5 | 2.167 |
| 6 | 2 | 0.667 | 0 (below $\phi$) | 0 |

#### Total rainfall
$$
P = \frac{5+8+15+12+9+2}{3} = \frac{51}{3} = 17.0\ \text{mm}
$$

#### Net runoff
Excess rainfall above $\phi$ $= 0.833+1.833+4.167+3.167+2.167 = 12.167$ mm.
The initial loss (0.5 mm) is taken from the start of the storm, so:
$$
R = 12.167 - 0.5 = 11.67\ \text{mm}
$$

#### W-index
The W-index is the average infiltration rate during the time $t$ when the rain intensity exceeds the infiltration capacity. The rain rate is above $\phi = 2.5$ mm/h only in the first five intervals, so
$$
t = 5\times20 = 100\ \text{min} = 1.667\ \text{h}
$$
$$
W = \frac{P - R - I_a}{t} = \frac{17.0 - 11.67 - 0.5}{1.667} = 2.90\ \text{mm/h}
$$

The last interval (rain 2 mm/h, less than $\phi$) is not counted in $t$.

**Answer: Total rainfall = 17.0 mm; net runoff = 11.67 mm; W-index = 2.9 mm/h.**

## Check yourself

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

1. How is rainfall depth found for each 20-min interval of a rate in mm/h?
   A. Rate x 20
   B. Rate / 20
   C. Rate / 3 (rate x 1/3 h)
   D. Rate x 3
   Answer: C. 20 min is 1/3 h, so depth = rate/3; total P = 51/3 = 17.0 mm. (Key point: Rainfall depth)

2. Why is the excess zero in interval 6 (2 mm/h)?
   A. Initial loss absorbed it
   B. It is the last interval
   C. Rain ended
   D. Rainfall intensity is below phi (2.5 mm/h), so no runoff
   Answer: D. Excess occurs only when intensity exceeds phi. (Key point: Interval below phi)

3. How is net runoff obtained?
   A. Total rainfall minus 0.5
   B. Sum of excess (12.167 mm) minus initial loss 0.5 mm = 11.67 mm
   C. Total rainfall minus phi x 2
   D. Sum of excess plus 0.5 mm
   Answer: B. Excess over phi totals 12.167 mm; the initial loss is deducted to give 11.67 mm. (Key point: Net runoff)

4. What is the W-index here?
   A. 11.67/2 = 5.8 mm/h
   B. (17.0 - 11.67)/2 = 2.67 mm/h
   C. (17.0 - 0.5)/2 = 8.25 mm/h
   D. (17.0 - 11.67 - 0.5)/2 = 2.42 mm/h
   Answer: D. W = (P - R - I_a)/t with t = 2 h, giving 2.42 mm/h. (Key point: W-index)

## Close variants set in other exams

- [Total rainfall, net runoff, W-index of 140-min storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-w-index-140min-storm) — Engineering Hydrology, chapter 3, asked 2×

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