---
title: "Phi-index from runoff volume 92,800 m3"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-index-0-8km2-92800"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-0-8km2-92800
site: BCT Engineering
---

# Phi-index from runoff volume 92,800 m3

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2079 Bhadra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-0-8km2-92800
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Determine $\phi$-index for a watershed with catchment area of 0.8 km$^2$ if temporal distribution of rainfall at different time duration are as the following table. Take surface runoff as 92,800 m$^3$.

| Duration (hr) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Incremental rainfall (cm) | 0.8 | 1.8 | 3.0 | 4.6 | 3.6 | 3.2 | 2.0 | 1.0 |

## Short answer

Runoff depth = 92800 m³ / (0.8×10⁶ m²) = 0.116 m = 11.6 cm. Total rainfall is 20 cm. Excluding the 0.8 and 1.0 cm hours, the remaining 6 hours give 18.2 - 6φ = 11.6, yielding φ = 1.10 cm/h.

## Answer

#### Runoff depth
$$
R = \frac{V}{A} = \frac{92{,}800\ \text{m}^3}{0.8\times10^{6}\ \text{m}^2} = 0.116\ \text{m} = 11.6\ \text{cm}
$$

#### Rainfall
$P = 0.8+1.8+3.0+4.6+3.6+3.2+2.0+1.0 = 20.0$ cm over 8 h (1-h steps). Total loss $= 20.0 - 11.6 = 8.4$ cm.

#### Trial for $\phi$
Assume the 0.8 cm and 1.0 cm hours are below $\phi$ (no excess). The other six hours contribute (1.8 + 3.0 + 4.6 + 3.6 + 3.2 + 2.0) = 18.2 cm.

$$
18.2 - 6\phi = 11.6 \;\Rightarrow\; \phi = 1.10\ \text{cm/h}
$$

Check: $1.10 > 0.8$ and $1.10 > 1.0$; all other hours exceed 1.10 cm. The assumption holds.

| Hour | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Rain (cm) | 0.8 | 1.8 | 3.0 | 4.6 | 3.6 | 3.2 | 2.0 | 1.0 |
| Excess (cm) | 0 | 0.7 | 1.9 | 3.5 | 2.5 | 2.1 | 0.9 | 0 |

The excess sums to 11.6 cm, which agrees with the runoff.

**Answer: $\phi$-index = 1.10 cm/h.**

## Check yourself

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

1. What is the runoff depth from 92,800 m3 over 0.8 km2?
   A. 11.6 cm
   B. 116 cm
   C. 1.16 cm
   D. 9.28 cm
   Answer: A. 92,800/(0.8x10^6) = 0.116 m = 11.6 cm. (Key point: Runoff depth)

2. What is the total loss?
   A. 11.6 cm
   B. 8.4 cm
   C. 20 cm
   D. 18.2 cm
   Answer: B. Rain 20.0 cm minus runoff 11.6 cm. (Key point: Total loss)

3. Which equation gives phi?
   A. 18.2 - 8 phi = 11.6, phi = 0.83
   B. 11.6 - 6 phi = 18.2
   C. 20 - 8 phi = 11.6, phi = 1.05
   D. 18.2 - 6 phi = 11.6, phi = 1.10 cm/h
   Answer: D. Six hours give excess, so 6 phi = 6.6 and phi = 1.10 cm/h. (Key point: Phi value)

4. Why is the trial assumption valid?
   A. Because total excess is 20 cm
   B. 1.10 > 0.8 and 1.0 while all other hours exceed 1.10
   C. Because phi equals the average rain
   D. Because it is the smallest value
   Answer: B. Excluded hours are below phi and included hours above it, and excess sums to 11.6 cm. (Key point: Check)

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