---
title: "Total rainfall and phi-index, 100 km2 basin"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:total-rainfall-phi-100km2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/total-rainfall-phi-100km2
site: BCT Engineering
---

# Total rainfall and phi-index, 100 km2 basin

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2070 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/total-rainfall-phi-100km2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Precipitation falls on a 100 km$^2$ drainage basin according to the following schedule:

| Time (minute) | 30 | 60 | 90 | 120 |
|---|---|---|---|---|
| Rainfall intensity (cm/hr) | 4 | 2 | 6 | 5 |

Determine the total storm rainfall. Also, find out $\phi$-index for the basin if the net storm runoff is 3 cm.

## Short answer

Total storm rainfall is 8.5 cm. The φ-index is found by trial, excluding intervals with intensity below it; φ = 3.0 cm/h.

## Answer

Each intensity acts for 30 min ($\Delta t = 0.5$ h).

#### Total storm rainfall
$$
P = (4 + 2 + 6 + 5)\times0.5 = 8.5\ \text{cm}
$$

#### $\phi$-index
Net storm runoff $R = 3$ cm, so the total loss $= 8.5 - 3 = 5.5$ cm.

Trial 1: assume all four intervals exceed $\phi$: $8.5 - 4(0.5\phi) = 3 \Rightarrow \phi = 2.75$ cm/h. But $2 < 2.75$, so the 2 cm/h interval gives no runoff.

Trial 2: only the intervals of 4, 6 and 5 cm/h contribute:

$$
\left[(4 - \phi) + (6 - \phi) + (5 - \phi)\right]\times0.5 = 3 \Rightarrow 15 - 3\phi = 6 \Rightarrow \phi = 3.0\ \text{cm/h}
$$

Check: $4, 6, 5 > 3.0$ and $2 < 3.0$. Excess depth $= (1 + 3 + 2)\times0.5 = 3.0$ cm. Valid.

**Answer: Total storm rainfall = 8.5 cm; $\phi$-index = 3.0 cm/h.**

## Check yourself

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

1. Intensities 4, 2, 6, 5 cm/h for 0.5 h each give
   A. 8.5 cm
   B. 4.25 cm
   C. 3 cm
   D. 17 cm
   Answer: A. (4+2+6+5) x 0.5 = 8.5 cm. (Key point: Total rainfall)

2. Trial 1 with all intervals contributing gives φ = 2.75, which fails because
   A. The 2 cm/h interval is below φ
   B. Runoff is negative
   C. 6 is above φ
   D. Rain is zero
   Answer: A. An interval below φ gives no excess, so it must be excluded. (Key point: First trial)

3. Excluding the 2 cm/h interval, φ is
   A. 4 cm/h
   B. 1.8 cm/h
   C. 3.0 cm/h
   D. 2.75 cm/h
   Answer: C. (15 - 3φ) x 0.5 = 3 gives φ = 3.0. (Key point: Final φ)

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