---
title: "Runoff volume using Horton's equation, 850 km2"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:horton-runoff-volume-850km2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-runoff-volume-850km2
site: BCT Engineering
---

# Runoff volume using Horton's equation, 850 km2

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2072 Chaitra, 10 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/horton-runoff-volume-850km2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The ordinates of a rainfall mass curve of a storm over a basin of area 850 km$^2$ measured in mm at one hour interval are 0, 10, 22, 30, 39, 45.5, 50, 55.5, 60, 64 and 68. If the infiltration during this storm can be represented by Horton's equation with $f_0 = 6.5$ mm/h, $f_c = 1.5$ mm/h and $k = 0.15$ /h, estimate the resulting runoff volume.

## Short answer

Runoff depth = 27.1 mm; runoff volume = 2.30 × 10⁷ m³ (23.0 million m³).

## Answer

**Given:** basin area $= 850$ km² $= 850\times10^6$ m²; $f_0 = 6.5$ mm/h, $f_c = 1.5$ mm/h, $k = 0.15$ h⁻¹; hourly mass curve (mm): 0, 10, 22, 30, 39, 45.5, 50, 55.5, 60, 64, 68.

#### Step 1: Hourly rainfall
10, 12, 8, 9, 6.5, 4.5, 5.5, 4.5, 4, 4 mm (total 68 mm in 10 h).

#### Step 2: Infiltration capacity in each hour
Horton's cumulative capacity: $F(t) = f_c t + \dfrac{f_0 - f_c}{k}(1 - e^{-kt})$ = $1.5t + 33.33(1 - e^{-0.15t})$.

| Hour | Rain (mm) | $F(t)$ at end (mm) | Capacity in hour (mm) | Infiltration = min (mm) | Runoff (mm) |
|---|---|---|---|---|---|
| 1 | 10 | 6.143 | 6.143 | 6.143 | 3.857 |
| 2 | 12 | 11.639 | 5.496 | 5.496 | 6.504 |
| 3 | 8 | 16.579 | 4.940 | 4.940 | 3.060 |
| 4 | 9 | 21.040 | 4.461 | 4.461 | 4.539 |
| 5 | 6.5 | 25.088 | 4.048 | 4.048 | 2.452 |
| 6 | 4.5 | 28.781 | 3.693 | 3.693 | 0.807 |
| 7 | 5.5 | 32.169 | 3.388 | 3.388 | 2.112 |
| 8 | 4.5 | 35.294 | 3.125 | 3.125 | 1.375 |
| 9 | 4 | 38.192 | 2.898 | 2.898 | 1.102 |
| 10 | 4 | 40.896 | 2.704 | 2.704 | 1.296 |

The rainfall exceeds the infiltration capacity in every hour, so infiltration takes place at capacity throughout:

$$
F(10) = 1.5\times10 + 33.33\,(1 - e^{-1.5}) = 15 + 25.90 = 40.90\ \text{mm}
$$

#### Step 3: Runoff
$$
R = 68 - 40.90 = 27.10\ \text{mm} = 0.0271\ \text{m}
$$

$$
V = 0.027104\times850\times10^6 = 2.30\times10^7\ \text{m}^3
$$

**Answer: Runoff depth = 27.1 mm; runoff volume = 2.30 × 10⁷ m³ (23.0 million m³).**

## Check yourself

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

1. Horton's cumulative infiltration capacity is
   A. F(t) = f0 e^(-kt)
   B. F(t) = fc + (f0 - fc)e^(-kt)
   C. F(t) = fc t + ((f0 - fc)/k)(1 - e^(-kt))
   D. F(t) = f0 t
   Answer: C. Integrating fp = fc + (f0 - fc)e^(-kt) gives the cumulative form. (Key point: Cumulative capacity)

2. Why can infiltration be taken as F(10) for the whole storm?
   A. Rainfall is below capacity in every hour
   B. fc is zero
   C. Rainfall exceeds capacity in every hour
   D. Rainfall is constant
   Answer: C. Since rain exceeds capacity every hour, infiltration is at capacity throughout. (Key point: Infiltration at capacity)

3. F(10) with fc = 1.5, f0 = 6.5, k = 0.15 is about
   A. 68 mm
   B. 25.9 mm
   C. 15 mm
   D. 40.9 mm
   Answer: D. 15 + 33.33(1 - e^-1.5) = 15 + 25.90 = 40.9 mm. (Key point: Numbers)

4. Runoff depth 27.1 mm over 850 km2 gives volume of about
   A. 2.30 x 10^7 m3
   B. 2.30 x 10^4 m3
   C. 2.7 x 10^6 m3
   D. 5.8 x 10^7 m3
   Answer: A. 0.0271 m x 850x10^6 m2 = 2.30x10^7 m3. (Key point: Runoff volume)

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