---
title: "Average rainfall by isohyetal method, Bagmati basin"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:isohyetal-bagmati-442"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/isohyetal-bagmati-442
site: BCT Engineering
---

# Average rainfall by isohyetal method, Bagmati basin

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2082 Baisakh, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/isohyetal-bagmati-442
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

An isohyet drawn in the catchment of Bagmati river basin with catchment area of 442 sq. km has the following data. Calculate the average rainfall for the basin.

| SN | Range of isohyet (cm) | Area enclosed by isohyet (sq. km) |
|---|---|---|
| 1 | 17-13 | 125 |
| 2 | 13-9 | 117 |
| 3 | 9-5 | 108 |
| 4 | 5-1 | 92 |

## Short answer

Mean depths of intervals: 15, 11, 7, 3 cm. Weighted by areas (125, 117, 108, 92 km²), Σ(A×depth) = 4194. Divide by 442 km² gives average 9.49 cm (about 95 mm).

## Answer

The average depth between two isohyets is the mean of their values. Weighted by the area, the catchment average is

$$
\bar P = \frac{\sum A_i \bar P_i}{\sum A_i}
$$

| Isohyet range (cm) | Mean depth (cm) | Area (km²) | Area × depth |
|---|---|---|---|
| 17-13 | 15 | 125 | 1875 |
| 13-9 | 11 | 117 | 1287 |
| 9-5 | 7 | 108 | 756 |
| 5-1 | 3 | 92 | 276 |
| **Total** |  | **442** | **4194** |

$$
\bar P = \frac{4194}{442} = 9.49\ \text{cm}
$$

**Answer: Average rainfall of the Bagmati basin = 9.49 cm (about 95 mm).**

## Check yourself

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

1. Which depth is used for the area between two isohyets?
   A. Higher value
   B. Mean of the two isohyet values
   C. Lower value
   D. Product
   Answer: B. E.g. 17-13 gives 15. (Key point: Mean depth between isohyets)

2. Which formula is used?
   A. Sum of P_i / n
   B. Sum A_i
   C. Sum of A_i x mean P_i / sum of A_i
   D. Max P
   Answer: C. Area-weighted mean. (Key point: Formula)

3. What is the average rainfall?
   A. 10 cm
   B. 15 cm
   C. 4194 cm
   D. 9.49 cm
   Answer: D. 4194/442 = 9.49 cm. (Key point: Result)

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