---
title: "Phi-index and rainfall excess duration from mass curve"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-index-mass-curve-3-8cm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-mass-curve-3-8cm
site: BCT Engineering
---

# Phi-index and rainfall excess duration from mass curve

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2075 Asoj, 6 marks
- Video: [Numerical 3- infiltration indices-phi index](https://www.youtube.com/watch?v=rqyFbrYxKE4) — 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-mass-curve-3-8cm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The mass curve of an isolated storm over a watershed is given below.

| Time from start (hr) | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Cumulative rainfall (cm) | 0 | 0.6 | 1.4 | 1.9 | 2.8 | 3.7 | 5.4 | 6.2 | 7 | 7.8 | 8.2 |

If the storm produced a direct run off of 3.8 cm at the outlet of the watershed, estimate the $\phi$-index of the storm and duration of rainfall excess.

## Short answer

By trial, φ-index is 0.889 cm/h for 0.5-h intervals; rainfall excess lasts 4.5 hours, producing 3.8 cm runoff.

## Answer

**Step 1: Rainfall in each 0.5-h interval (cm)**

| Interval (h) | 0-0.5 | 0.5-1 | 1-1.5 | 1.5-2 | 2-2.5 | 2.5-3 | 3-3.5 | 3.5-4 | 4-4.5 | 4.5-5 |
|---|---|---|---|---|---|---|---|---|---|---|
| Rain (cm) | 0.6 | 0.8 | 0.5 | 0.9 | 0.9 | 1.7 | 0.8 | 0.8 | 0.8 | 0.4 |

Total rainfall $P = 8.2$ cm; direct runoff $R = 3.8$ cm; total loss $= 4.4$ cm over 5 h.

**Step 2: Trial for $\phi$** (let $\phi'$ be the loss per 0.5 h)
Assume only the 0.4 cm interval is below $\phi'$ (the other 9 intervals exceed it):

$$
(8.2 - 0.4) - 9\phi' = 3.8 \Rightarrow \phi' = \frac{4.0}{9} = 0.444\ \text{cm per 0.5 h}
$$

$$
\phi = \frac{0.444}{0.5} = 0.889\ \text{cm/h}
$$

**Step 3: Check**
The first nine intervals (0 to 4.5 h) each have rain of at least 0.5 cm, which is more than 0.444 cm, and the last interval (0.4 cm) is below it. Sum of excess $= 7.8 - 9\times0.444 = 3.8$ cm. Valid.

**Step 4: Duration of rainfall excess**
Excess occurs in 9 intervals of 0.5 h, from 0 to 4.5 h:

$$
t_e = 9\times0.5 = 4.5\ \text{h}
$$

**Answer: $\phi$-index = 0.889 cm/h (about 0.89 cm/h); duration of rainfall excess = 4.5 h.**

## Check yourself

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

1. Interval rainfall comes from:
   A. Cumulative values
   B. Differences of cumulative rainfall
   C. Averages
   D. Totals
   Answer: B. Differences give each interval. (Key point: Interval rain)

2. phi per 0.5 h is:
   A. 0.444 cm
   B. 0.889 cm
   C. 0.4 cm
   D. 0.8 cm
   Answer: A. (7.8 - 3.8)/9 = 0.444 and phi = 0.889 cm/h. (Key point: phi)

3. Duration of excess is:
   A. 5 h
   B. 4.5 h
   C. 2.5 h
   D. 9 h
   Answer: B. Nine intervals of 0.5 h. (Key point: Duration)

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