---
title: "Infiltration index from 8-hr rainfall, Bagmati Basin"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-index-bagmati-11-6cm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-bagmati-11-6cm
site: BCT Engineering
---

# Infiltration index from 8-hr rainfall, Bagmati Basin

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2080 Baisakh, 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-bagmati-11-6cm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The following data represents the temporal distribution of rainfall in Bagmati Basin for the duration of 8 hrs. If surface runoff generated by the rainfall is 11.6 cm calculate the infiltration index. Assume no initial losses.

| Duration (hr) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Rainfall per hr (cm) | 0.8 | 1.8 | 3 | 4.6 | 3.6 | 3.2 | 2 | 1 |

## Short answer

Total rainfall is 20 cm, runoff 11.6 cm. Excluding the hours with rainfall 0.8 and 1.0 cm below the index, the remaining 6 hours sum to 18.2 cm. Using 18.2 - 6φ = 11.6 gives φ = 1.10 cm/h.

## Answer

Total rainfall:
$$
P = 0.8+1.8+3.0+4.6+3.6+3.2+2.0+1.0 = 20.0\ \text{cm}
$$
Runoff $R = 11.6$ cm, so total loss $= 20.0 - 11.6 = 8.4$ cm. With no initial loss, all of this is infiltration, and the time step is 1 h.

#### Trial
Assume the hours with rain below $\phi$ are the 0.8 cm and 1.0 cm hours, so these give no excess. Excess is from 6 hours (1.8, 3.0, 4.6, 3.6, 3.2, 2.0), total rain $= 18.2$ cm.

$$
18.2 - 6\phi = 11.6 \;\Rightarrow\; \phi = \frac{6.6}{6} = 1.10\ \text{cm/h}
$$

Check: $0.8 < 1.1$ and $1.0 < 1.1$ (no excess in those hours, correct); every other hour has rain $> 1.1$.

| Hour | Rain (cm) | Excess over 1.1 (cm) |
|---|---|---|
| 1 | 0.8 | 0 |
| 2 | 1.8 | 0.7 |
| 3 | 3.0 | 1.9 |
| 4 | 4.6 | 3.5 |
| 5 | 3.6 | 2.5 |
| 6 | 3.2 | 2.1 |
| 7 | 2.0 | 0.9 |
| 8 | 1.0 | 0 |
| Total | | 11.6 |

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

## Check yourself

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

1. With no initial loss, what is the total infiltration loss?
   A. 31.6 cm
   B. 11.6 cm
   C. 8.4 cm
   D. 20.0 cm
   Answer: C. Loss = P - R = 20.0 - 11.6 = 8.4 cm. (Key point: Total loss)

2. Which hours are assumed to give no excess?
   A. Hours 2 and 3
   B. Hours 1 and 8 (0.8 and 1.0 cm), with rain below phi
   C. Hours 4 and 5
   D. None
   Answer: B. The lowest-rain hours are assumed below phi and excluded from the count n = 6. (Key point: Trial assumption)

3. Which equation gives phi?
   A. 11.6/8
   B. 18.2 - 6 phi = 11.6
   C. 18.2 - 8 phi = 11.6
   D. 20 - 8 phi = 11.6
   Answer: B. Rain of the six excess hours (18.2) minus 6 phi equals runoff, so phi = 1.10 cm/h. (Key point: Equation)

4. How is the answer verified?
   A. By comparing with the average
   B. By Horton's equation
   C. 1.10 exceeds 0.8 and 1.0, and every other hour exceeds 1.10; excess totals 11.6
   D. No check is needed
   Answer: C. Consistency of the assumed excess hours confirms phi, and the excess sums to the runoff. (Key point: Verification)

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