---
title: "Phi-index of an 8-hour storm"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-index-15cm-8-7"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-15cm-8-7
site: BCT Engineering
---

# Phi-index of an 8-hour storm

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2076 Asoj, 5 marks
- Video: [Numerical 1. Infiltration indices- Phi index.](https://www.youtube.com/watch?v=Kz6UFKXvYIc) — learn with aashi
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-15cm-8-7
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

A storm with a 15.0 cm precipitation produced a direct runoff of 8.7 cm. The time distribution of storm is as follows.

| Time from start (hr) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Incremental rainfall in each hr (cm) | 0.6 | 1.35 | 2.25 | 3.45 | 2.7 | 2.4 | 1.5 | 0.75 |

Estimate the $\Phi$-index of the storm.

## Short answer

By trial, φ-index equals 0.825 cm/h; only six hours with rainfall exceeding φ contribute to 8.7 cm runoff.

## Answer

#### Method
The $\phi$-index is the constant rate of loss such that the rainfall in excess of it equals the direct runoff. Losses = $15.0 - 8.7 = 6.3$ cm over 8 h.

#### Trial
Assume all hours exceed $\phi$: $15 - 8\phi = 8.7 \Rightarrow \phi = 0.7875$ cm/h. But the 1st hour (0.6) is below $\phi$, so exclude it (loss in that hour is 0.6 cm). Try 7 hours: $(15 - 0.6) - 7\phi = 8.7$, so $\phi = 0.8143$ cm/h. The 8th hour (0.75) is also below $\phi$, so exclude it too. Try 6 hours:

$$
(15 - 0.6 - 0.75) - 6\phi = 8.7 \Rightarrow 13.65 - 6\phi = 8.7 \Rightarrow \phi = 0.825\ \text{cm/h}
$$

#### Check
| Hour | Rain (cm) | Excess = rain - 0.825 (cm) |
|---|---|---|
| 1 | 0.6 | 0 |
| 2 | 1.35 | 0.525 |
| 3 | 2.25 | 1.425 |
| 4 | 3.45 | 2.625 |
| 5 | 2.7 | 1.875 |
| 6 | 2.4 | 1.575 |
| 7 | 1.5 | 0.675 |
| 8 | 0.75 | 0 |
| **Total** | 15.0 | **8.70** |

The excess equals the 8.7 cm runoff, and the 6 hours with excess have rainfall > 0.825 cm/h, so the trial is valid.

**Answer: $\phi$-index = 0.825 cm/h.**

## Check yourself

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

1. phi-index is:
   A. Average rainfall
   B. Constant rate of loss so that rain in excess equals direct runoff
   C. Peak intensity
   D. Total runoff
   Answer: B. Rain above phi gives the runoff. (Key point: Definition)

2. If an hour's rainfall is below the trial phi:
   A. It is kept in the sum
   B. It is excluded; its rainfall is a loss and phi is recalculated
   C. phi is doubled
   D. Runoff is changed
   Answer: B. Excluded hours lose all rainfall. (Key point: Procedure)

3. phi-index is:
   A. 0.7875 cm/h
   B. 0.8143 cm/h
   C. 0.825 cm/h
   D. 0.6 cm/h
   Answer: C. (15 - 0.6 - 0.75 - 6 phi) = 8.7, so phi = 0.825. (Key point: Result)

4. Which hours are below phi?
   A. 1st and 8th
   B. 1st only
   C. 4th and 5th
   D. None
   Answer: A. Rain 0.6 and 0.75 are below 0.825. (Key point: Hours excluded)

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