---
title: "Runoff and infiltration volume, Horton f=15+50e^-0.9t"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:horton-15-50e-0-9t"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-15-50e-0-9t
site: BCT Engineering
---

# Runoff and infiltration volume, Horton f=15+50e^-0.9t

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2069 Chaitra, 2+2+2+2 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-15-50e-0-9t
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The infiltration of a catchment can be represented by the equation $f = 15 + 50e^{-0.9t}$. If the rainfall intensity of 45 mm/hr occurs continuously for 10 hour from a catchment of area 12 km$^2$, calculate

i) Total runoff volume generated from that catchment
ii) Total infiltration volume at the period
iii) Time from the start of rainfall from which runoff started
iv) Show all (above three) results in infiltration curves

## Short answer

Runoff starts when infiltration capacity equals rainfall intensity, at tp = 0.57 h (≈34 min). Total infiltration volume over 10 h is 2.40×10⁶ m³, and total runoff volume is 3.00×10⁶ m³.

## Answer

**Given:** $f_p = 15 + 50e^{-0.9t}$ mm/h; rainfall intensity $i = 45$ mm/h for 10 h; area $= 12$ km² $= 12\times10^6$ m².

#### (iii) Time when runoff starts
Initially $f_p(0) = 65 > 45$, so all rain infiltrates. Runoff starts when $f_p$ falls to $i$:

$$
15 + 50e^{-0.9t_p} = 45 \Rightarrow e^{-0.9t_p} = 0.6 \Rightarrow t_p = \frac{\ln(1/0.6)}{0.9} = 0.568\ \text{h} \;(\approx 34\ \text{min})
$$

#### (ii) Total infiltration in 10 h
Up to $t_p$ infiltration equals rainfall; after $t_p$ it occurs at capacity:

$$
F = i\,t_p + \int_{t_p}^{10}(15 + 50e^{-0.9t})dt = 45\times0.568 + 15(10 - 0.568) + \frac{50}{0.9}\left(e^{-0.9t_p} - e^{-9}\right)
$$

$$
F = 25.54 + 141.49 + 33.33 = 200.35\ \text{mm}
$$

$$
V_{inf} = 200.35\times10^{-3}\times12\times10^6 = 2,404,248\ \text{m}^3 \approx 2.40\times10^6\ \text{m}^3
$$

#### (i) Total runoff volume
Rainfall $= 45\times10 = 450$ mm.

$$
R = 450 - 200.35 = 249.65\ \text{mm}, \qquad V_R = 2,995,752\ \text{m}^3 \approx 3.00\times10^6\ \text{m}^3
$$

#### (iv) Infiltration curve
```text
 mm/h
 65 |*
    | *
 45 |--+--------------------- rainfall i = 45
    |   *  <- runoff starts, t = 0.57 h
    |     *
 15 |         *_______________ fc = 15
    +-----------------------------> t (h)
      0.57                    10
   (area under fp curve after tp = infiltration;
    area above fp, below i line = runoff)
```

**Answer: (i) Runoff volume = 3.00 × 10⁶ m³ (249.6 mm); (ii) infiltration volume = 2.40 × 10⁶ m³ (200.4 mm); (iii) runoff starts 0.57 h (about 34 min) after the start of rain.**

## Check yourself

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

1. Runoff begins when
   A. fp falls to the rainfall intensity i
   B. fp = 0
   C. fp = fc
   D. t = 10 h
   Answer: A. Before that all rain infiltrates. (Key point: Runoff start)

2. 15 + 50e^(-0.9t) = 45 gives tp of
   A. 0.06 h
   B. 0.568 h
   C. 1.2 h
   D. 0.9 h
   Answer: B. e^(-0.9t) = 0.6 so t = ln(1/0.6)/0.9 = 0.568 h. (Key point: tp)

3. Total infiltration in 10 h is
   A. 200.4 mm
   B. 450 mm
   C. 65 mm
   D. 150 mm
   Answer: A. Rain to tp (25.5) plus capacity integral after tp = 200.35 mm. (Key point: Infiltration)

4. Runoff depth is
   A. 249.6 mm
   B. 200 mm
   C. 450 mm
   D. 65 mm
   Answer: A. Rain 450 mm minus infiltration 200.4 mm. (Key point: Runoff)

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