---
title: "Rational method peak flow, Melamchi storm"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:rational-melamchi"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/rational-melamchi
site: BCT Engineering
---

# Rational method peak flow, Melamchi storm

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2079 Baisakh, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/rational-melamchi
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Consider a rainfall storm in Melamchi during a recent flood event spanned for a period of 160 minutes. Observed records showed that during the first 30 minutes, 50 minutes, 84 minutes and 100 minutes the rainfall depths were 25 mm, 40 mm, 70 mm and 80 mm respectively. If the catchment area of the watershed is 100 km$^2$, length of the longest flow path is 6 km, general slope of the catchment is 0.013 and runoff coefficient is 0.6, estimate the peak flow using the Rational method.

## Short answer

Time of concentration by Kirpich formula is 84.2 min. Rainfall intensity for this duration is 50.0 mm/h from the storm hyetograph. Peak flow by rational method Qp = 0.278 × 0.6 × 50.0 × 100 = 834 m³/s.

## Answer

**Rational formula:** $Q_p = 0.278\,C\,i\,A$ (Q in m³/s, i in mm/h, A in km²), where $i$ is the intensity for a duration equal to the time of concentration.

#### Step 1: Time of concentration (Kirpich)
$$
\begin{aligned} t_c &= 0.0195\,L^{0.77}S^{-0.385} \\ &= 0.0195\,(6000)^{0.77}(0.013)^{-0.385} = 84.2 \text{ min} \end{aligned}
$$

#### Step 2: Rainfall intensity for duration $t_c$
Cumulative rainfall from the start of the storm:

| Time (min) | 0 | 30 | 50 | 84 | 100 |
|---|---|---|---|---|---|
| Depth (mm) | 0 | 25 | 40 | 70 | 80 |

The depth in the first 84.2 min, by linear interpolation between 50 and 84 min (this is also the most intense 84.2-min window of the storm):

$$
P = 40 + \frac{84.2-50}{84-50}(70-40) = 70.2 \text{ mm}
$$

$$
i = \frac{70.2}{84.2}\times 60 = 50.0 \text{ mm/h}
$$

#### Step 3: Peak flow
$$
\begin{aligned} Q_p &= 0.278\times 0.6\times 50.0 \times 100 \\ &= 834 \text{ m}^3/\text{s} \end{aligned}
$$

**Answer: peak flow ≈ 834 m³/s** (note: a 100 km² catchment is large for the rational method, so this is a rough estimate).

## Check yourself

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

1. Which formula gives tc here?
   A. Kirpich: tc = 0.0195 L^0.77 S^-0.385 with L in m
   B. tc = L/A
   C. tc = A/L
   D. tc = 0.0195 L S
   Answer: A. Kirpich gives tc about 84.2 min. (Key point: Time of concentration)

2. The design intensity is for a duration equal to:
   A. The full 160 min storm
   B. tc, taken from the storm record by interpolation
   C. The first 30 min
   D. Any duration, since i is constant
   Answer: B. The depth in 84.2 min is about 70.1 mm, so i = 50 mm/h. (Key point: Intensity)

3. Qp = 0.278 C i A with C = 0.6, i = 50 mm/h, A = 100 km2 is:
   A. About 83 m3/s
   B. About 2780 m3/s
   C. About 833 m3/s
   D. About 500 m3/s
   Answer: C. 0.278 x 0.6 x 50 x 100 = 833 m3/s. (Key point: Peak flow)

## Related in Engineering Hydrology

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