---
title: "Rational method peak flow from storm depths"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:rational-storm-20-40-60-80"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/rational-storm-20-40-60-80
site: BCT Engineering
---

# Rational method peak flow from storm depths

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2079 Bhadra, 6 marks
- Video: [Hydrology: Rational Method Peak Runoff Determination and Example Problem](https://www.youtube.com/watch?v=AhYI4SqZWOg) — High Water Studio
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/rational-storm-20-40-60-80
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

A rainfall storm for the duration of 20 mins, 40 mins, 60 and 80 mins are 40 mm, 70 mm, 90 mm and 100 mm respectively. If the slope of the catchment is 0.01, runoff coefficient is 0.7, maximum length of the travel is 1,100 m and catchment area is 3 km$^2$, estimate the peak flow by rational method.

## Short answer

Time of concentration by Kirpich: t_c = 25.2 min. Interpolated storm intensity i = 113.8 mm/h. Peak flow by rational method (C=0.7, A=3 km²): Qp = 0.278×0.7×113.8×3 = 66.4 m³/s.

## Answer

The rational method gives $Q_p = 0.278\,C\,i\,A$ ($Q_p$ in m³/s, $i$ in mm/h, $A$ in km²), where $i$ is the average intensity of a storm whose duration equals the time of concentration $t_c$ of the catchment.

#### Step 1: Time of concentration (Kirpich's formula)

$$
\begin{aligned} t_c &= 0.0195\,L^{0.77}\,S^{-0.385}\ \text{(min; }L\text{ in m)}\\ &= 0.0195\,(1100)^{0.77}\,(0.01)^{-0.385} = 25.23\ \text{min} \end{aligned}
$$

#### Step 2: Intensity for each storm duration

| Duration (min) | Depth (mm) | Intensity i = P/t (mm/h) |
|---|---|---|
| 20 | 40 | 120.0 |
| 40 | 70 | 105.0 |
| 60 | 90 | 90.0 |
| 80 | 100 | 75.0 |

#### Step 3: Intensity at $t_c$

$t_c = 25.23$ min lies between 20 and 40 min. Interpolating the storm depth linearly:

$$
P = 40 + \frac{25.23 - 20}{40 - 20}(70 - 40) = 47.84\ \text{mm}
$$

$$
i = \frac{P}{t_c} = \frac{47.84}{0.4205} = 113.8\ \text{mm/h}
$$

#### Step 4: Peak flow

$$
Q_p = 0.278 \times 0.7 \times 113.8 \times 3 = 66.4\ \text{m}^3/\text{s}
$$

**Answer: Peak flow $\approx$ 66 m³/s** (with $t_c$ = 25.2 min, $i$ = 114 mm/h). Kirpich's formula was used because no other method for $t_c$ was specified.

## Check yourself

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

1. In the rational method, which storm duration is used to find the intensity i?
   A. The longest storm duration given (80 min)
   B. The time of concentration of the catchment
   C. The shortest storm duration given (20 min)
   D. The average of all given durations
   Answer: B. Peak flow occurs when the whole catchment contributes, so i is for a duration equal to t_c. (Key point: Rational formula and duration)

2. Which formula was used for t_c and what is its value?
   A. Kirpich: t_c = 0.0195 L^0.77 S^-0.385, about 25.2 min
   B. Kirpich with S = 1, about 2 min
   C. Time equals L divided by 1 m/s, about 18 min
   D. t_c = 80 min, the longest storm
   Answer: A. Kirpich's formula with L = 1100 m and S = 0.01 gives 25.23 min. (Key point: Time of concentration)

3. t_c = 25.2 min lies between 20 and 40 min. How is i at t_c obtained?
   A. Use the 20 min intensity of 120 mm/h directly
   B. Use the 40 min depth of 70 mm over 25.2 min
   C. Average all four intensities, 97.5 mm/h
   D. Interpolate depth (about 47.8 mm) and divide by t_c, giving about 114 mm/h
   Answer: D. Depth interpolated at t_c is 47.84 mm; i = 47.84/(25.23/60) = 113.8 mm/h. (Key point: Intensity at t_c)

4. With Q = 0.278 C i A, what is the peak flow?
   A. About 0.66 m^3/s
   B. About 95 m^3/s using i = 120 mm/h and C = 1
   C. About 66 m^3/s with A = 3 km^2, i in mm/h
   D. About 233 m^3/s by omitting C
   Answer: C. Q = 0.278 x 0.7 x 113.8 x 3 = 66.4 m^3/s. (Key point: Peak flow)

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