---
title: "Phi-index from mass curve, 3.5 cm runoff"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-index-3-5cm-runoff"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-3-5cm-runoff
site: BCT Engineering
---

# Phi-index from mass curve, 3.5 cm runoff

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2082 Bhadra, 6 marks
- Video: [#Numericals on Phi-Index# II Runoff Part 4II Hydrology II Prof. Sagar Kolekar](https://www.youtube.com/watch?v=aA3SzG9tIus) — Education Beyond the Wall
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-3-5cm-runoff
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

An isolated storm in a catchment produced a runoff of 3.5 cm. The mass curve of the average rainfall depth over the catchment was as below. Calculate the $\phi$-index for the storm.

| Time from beginning of storm (hr) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Accumulated average rainfall (cm) | 0 | 0.5 | 1.65 | 3.55 | 5.65 | 6.8 | 7.75 |

## Short answer

From the mass curve, hourly rainfall depths are 0.50, 1.15, 1.90, 2.10, 1.15 and 0.95 cm. Total rainfall is 7.75 cm and runoff 3.5 cm; assuming only the 0.50 cm hour is below phi, solve (7.75 - 0.50 - 5 phi) = 3.5, giving phi = 0.75 cm/h.

## Answer

Hourly rainfall from the mass curve (difference of successive values):

| Hour | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Rain (cm) | 0.50 | 1.15 | 1.90 | 2.10 | 1.15 | 0.95 |

Total rainfall $P = 7.75$ cm; runoff $R = 3.5$ cm; total loss $= 7.75 - 3.5 = 4.25$ cm.

Hour 1 has the smallest rain (0.5 cm). Try $\phi > 0.5$ cm/h, so hour 1 gives no excess and the other 5 hours do:
$$
(7.75 - 0.50) - 5\phi = 3.5 \;\Rightarrow\; \phi = \frac{7.25-3.5}{5} = 0.75\ \text{cm/h}
$$

Check: $0.75 > 0.50$ (hour 1 below $\phi$, correct) and $0.75 < 0.95$ (all other hours above $\phi$, correct).

| Hour | Rain | Excess over 0.75 |
|---|---|---|
| 1 | 0.50 | 0 |
| 2 | 1.15 | 0.40 |
| 3 | 1.90 | 1.15 |
| 4 | 2.10 | 1.35 |
| 5 | 1.15 | 0.40 |
| 6 | 0.95 | 0.20 |
| Sum | | 3.50 |

**Answer: $\phi$-index = 0.75 cm/h.**

## Check yourself

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

1. Hourly rain from mass curve is
   A. Sum
   B. Average
   C. Product
   D. Difference of successive values
   Answer: D. Differences give 0.5, 1.15, ... (Key point: Hourly rain)

2. Hour 1 excluded because
   A. Largest
   B. Smallest rain 0.5, below φ
   C. Zero
   D. Equal
   Answer: B. It gives no excess. (Key point: Assumption)

3. (7.25 - 3.5)/5 =
   A. 0.5 cm/h
   B. 1.0 cm/h
   C. 0.75 cm/h
   D. 3.5 cm/h
   Answer: C. 3.75/5 = 0.75. (Key point: φ)

4. Check requires
   A. φ = 7.75
   B. φ > 2
   C. 0.5 < φ < 0.95
   D. φ < 0.5
   Answer: C. Hour 1 below, others above φ. (Key point: Check)

## Related in Engineering Hydrology

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