---
title: "Reservoir level change, water balance, 15.5 km2"
subject: "Engineering Hydrology"
chapter: 1
key: "semester_5/engineering_hydrology:reservoir-level-change-15-5km2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/reservoir-level-change-15-5km2
site: BCT Engineering
---

# Reservoir level change, water balance, 15.5 km2

Engineering Hydrology · Chapter 1: Introduction

- Asked in: 2076 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/reservoir-level-change-15-5km2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-1

## Question

The catchment area of a reservoir is 15.5 km$^2$. A uniform precipitation of 0.5 cm/h for 2 hr was observed on a particular day. 65% of precipitation reached into the reservoir. A canal carrying a flow of 1.1 m$^3$/s is taken from the reservoir. The rate of evaporation was 0.6 mm/hr/m$^2$. Assuming seepage loss to be 45% of evaporation loss, find the change in reservoir level for 10 hrs, if the water spread of the reservoir was 0.6 km$^2$.

## Short answer

Inflow is 65% of 0.01 m rainfall over 15.5 km² = 100,750 m³. Outflows: canal 39,600 m³, evaporation 3,600 m³, seepage 1,620 m³. Net increase 55,930 m³ over 0.6 km² water spread gives a level rise of 0.093 m (9.3 cm).

## Answer

**Given:** catchment area $A = 15.5$ km² $= 15.5\times10^6$ m²; rainfall 0.5 cm/h for 2 h $= 1$ cm $= 0.01$ m; runoff reaching reservoir 65%; canal 1.1 m³/s; evaporation 0.6 mm/h per m²; seepage 45% of evaporation; water spread $= 0.6$ km² $= 6\times10^5$ m²; period $T = 10$ h.

$$
\Delta S = V_{in} - V_{canal} - V_{evap} - V_{seep}
$$

#### Step 1: Inflow
$$
V_{in} = 0.65 \times 0.01 \times 15.5\times10^6 = 100\,750\ \text{m}^3
$$

#### Step 2: Outflows in 10 h
- Canal: $1.1 \times 10 \times 3600 = 39\,600$ m³
- Evaporation: $0.0006 \times 6\times10^5 \times 10 = 3\,600$ m³
- Seepage: $0.45 \times 3600 = 1\,620$ m³

#### Step 3: Storage change
$$
\Delta S = 100\,750 - 39\,600 - 3\,600 - 1\,620 = 55\,930\ \text{m}^3
$$

#### Step 4: Level change
$$
\Delta h = \frac{55\,930}{6\times10^5} = 0.0932\ \text{m}
$$

**Answer: The reservoir level rises by about 0.093 m (9.3 cm) in 10 hours.**

## Check yourself

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

1. Rainfall depth used is:
   A. 0.5 cm/h x 2 h = 1 cm = 0.01 m
   B. 0.5 cm
   C. 10 cm
   D. 0.1 m
   Answer: A. Uniform rain for 2 h. (Key point: Rainfall depth)

2. Inflow volume is:
   A. 0.01 x 0.6 x 10^6
   B. 0.65 x 0.01 x 15.5 x 10^6 = 100750 m^3
   C. 1.1 x 3600
   D. 15.5 x 10^6 m^3
   Answer: B. Runoff fraction x depth x catchment area. (Key point: Inflow)

3. Canal loss in 10 h is:
   A. 1.1 x 10 = 11 m^3
   B. 1.1 x 3600
   C. 1.1 x 10 x 3600 = 39600 m^3
   D. 39.6 m^3
   Answer: C. Discharge x time in seconds. (Key point: Canal)

4. Level change is:
   A. 0.093 m fall
   B. 55930/15.5x10^6
   C. 0.93 m rise
   D. 55930 / 6x10^5 = 0.093 m rise
   Answer: D. Divide by water spread. (Key point: Level)

## Close variants set in other exams

- [Reservoir level change, water balance, 1600 ha](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-level-change-1600ha) — Engineering Hydrology, chapter 1, asked 2×

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