---
title: "Surface runoff with initial loss and phi-index, 100-min storm"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:surface-runoff-initial-loss-0-6"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/surface-runoff-initial-loss-0-6
site: BCT Engineering
---

# Surface runoff with initial loss and phi-index, 100-min storm

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2081 Bhadra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/surface-runoff-initial-loss-0-6
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The mass curve of the rainfall of 100 min duration is given below. If the catchment had an initial loss of 0.6 cm and $\phi$-index of 0.6 cm/hr, calculate the total surface runoff from the catchment.

| Time from start of rainfall (min) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Cumulative rainfall (cm) | 0 | 0.5 | 1.2 | 2.6 | 3.3 | 3.5 |

## Short answer

After subtracting initial loss of 0.6 cm from early rain, φ-index losses of 0.2 cm per 20-min interval applied to remaining rainfall yielded excess depths summing to 2.1 cm, so total surface runoff = 2.1 cm.

## Answer

Rainfall in each 20-minute interval (difference of the mass curve):

| Interval (min) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| Rain (cm) | 0.5 | 0.7 | 1.4 | 0.7 | 0.2 |

$\phi$ loss per 20 min $= 0.6\times\frac{20}{60} = 0.2$ cm.

The initial loss of 0.6 cm is satisfied first: all 0.5 cm of interval 1 and 0.1 cm of interval 2. After that the $\phi$ loss is applied.

| Interval | Rain after initial loss (cm) | $\phi$ loss (cm) | Excess (cm) |
|---|---|---|---|
| 1 | 0 | 0 | 0 |
| 2 | 0.6 | 0.2 | 0.4 |
| 3 | 1.4 | 0.2 | 1.2 |
| 4 | 0.7 | 0.2 | 0.5 |
| 5 | 0.2 | 0.2 | 0 |
| Total | | | 2.1 |

Check: total rain 3.5 - initial loss 0.6 - infiltration (0.2 + 0.2 + 0.2 + 0.2 = 0.8) = 2.1 cm.

**Answer: Total surface runoff = 2.1 cm (depth over the catchment).**

## Check yourself

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

1. How are the rainfall depths in each 20-min interval obtained from the mass curve?
   A. By differences between successive cumulative values
   B. By taking the maximum value
   C. By adding successive cumulative values
   D. By dividing cumulative rainfall by 100
   Answer: A. Incremental rain is the difference between consecutive mass-curve readings (0.5, 0.7, 1.4, 0.7, 0.2 cm). (Key point: Incremental rainfall)

2. What is the phi loss in each 20-min interval for phi = 0.6 cm/hr?
   A. 1.8 cm
   B. 0.6 cm
   C. 0.12 cm
   D. 0.2 cm
   Answer: D. 0.6 x 20/60 = 0.2 cm per interval. (Key point: phi loss per interval)

3. How is the 0.6 cm initial loss handled?
   A. Added to the phi loss in every interval
   B. Subtracted from the final runoff only
   C. Taken first: all 0.5 cm of interval 1 and 0.1 cm of interval 2, then phi loss is applied
   D. Ignored
   Answer: C. Initial loss is satisfied from the start of the rain, after which the phi loss is applied. (Key point: Initial loss first)

4. What is the total surface runoff?
   A. 3.5 cm
   B. 2.1 cm
   C. 2.7 cm
   D. 1.3 cm
   Answer: B. 3.5 - 0.6 - 0.8 = 2.1 cm (interval excess 0.4 + 1.2 + 0.5 + 0). (Key point: Total runoff)

## Related in Engineering Hydrology

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- [Phi-index from mass curve, 3.5 cm runoff](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-index-3-5cm-runoff) — Engineering Hydrology, chapter 3
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