---
title: "Infiltration depth from Horton's equation, 60-min storm"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:horton-fp-3-0-e-2t"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-fp-3-0-e-2t
site: BCT Engineering
---

# Infiltration depth from Horton's equation, 60-min storm

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2071 Shrawan, 3+3 marks
- Video: [Infiltration Capacity | Horton's Infiltration Equation @TutorJackPh](https://www.youtube.com/watch?v=RSuif0LlIqM) — Engr. Clidez
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-fp-3-0-e-2t
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The infiltration capacity in a basin is represented by Horton's equation as $f_p = 3.0 + e^{-2t}$, where $f_p$ is in cm/hr and 't' in hours. Assuming the infiltration to take place at capacity rates in a 60 minutes storm, estimate the depth of infiltration in (i) the first 30 minutes and (ii) the second 30 minutes of the storm.

## Short answer

Depth is obtained by integrating the infiltration capacity curve fp = 3 + e^{-2t} over time. First 30 minutes: 1.82 cm; second 30 minutes: 1.62 cm.

## Answer

Infiltration occurs at capacity, so the depth of infiltration in any period is the integral of $f_p$.

$$
F = \int_{t_1}^{t_2} (3.0 + e^{-2t})\,dt = 3.0\,(t_2 - t_1) + \frac{1}{2}\left(e^{-2t_1} - e^{-2t_2}\right)
$$

#### (i) First 30 minutes ($t = 0$ to $0.5$ h)
$$
F_1 = 3.0\times0.5 + 0.5\,(1 - e^{-1}) = 1.5 + 0.5\times0.6321 = 1.816\ \text{cm}
$$

#### (ii) Second 30 minutes ($t = 0.5$ to $1.0$ h)
$$
F_2 = 3.0\times0.5 + 0.5\,(e^{-1} - e^{-2}) = 1.5 + 0.5\times(0.3679 - 0.1353) = 1.616\ \text{cm}
$$

**Answer: Infiltration in the first 30 min = 1.816 cm; in the second 30 min = 1.616 cm (total in 60 min = 3.432 cm).**

## Check yourself

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

1. Infiltration depth at capacity rates is found by
   A. Averaging i
   B. Multiplying fp(0) by time
   C. Integrating fp over the period
   D. Differentiating fp
   Answer: C. Depth is the integral of the capacity rate. (Key point: Method)

2. F1 = 3.0(0.5) + 0.5(1 - e^-1) is
   A. 1.5 cm
   B. 0.316 cm
   C. 1.816 cm
   D. 3.43 cm
   Answer: C. 1.5 + 0.5 x 0.6321 = 1.816. (Key point: First 30 min)

3. Why is F2 smaller than F1?
   A. Rain stops
   B. Soil dries
   C. Capacity decreases with time
   D. fc decreases
   Answer: C. The exponential term decays, so capacity falls; F2 = 1.616 cm. (Key point: Second 30 min)

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