---
title: "Reservoir level change, water balance, 1600 ha"
subject: "Engineering Hydrology"
chapter: 1
key: "semester_5/engineering_hydrology:reservoir-level-change-1600ha"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/reservoir-level-change-1600ha
site: BCT Engineering
---

# Reservoir level change, water balance, 1600 ha

Engineering Hydrology · Chapter 1: Introduction

- Asked in: 2075 Asoj, 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-1600ha
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-1

## Question

The catchment area of a reservoir is 1600 ha. A uniform precipitation of 8 mm/hr for 2 hour was observed on a particular day. 55% run off reached the reservoir. A canal carrying a flow of 1 m$^3$/s is taken from the reservoir. The rate of evaporation was 0.8 mm/h/m$^2$. Assuming seepage loss is 40% of evaporation loss, find the change in the reservoir level for 6 hours, if the water spread of the reservoir was 45 ha.

## Short answer

Runoff volume = 55% of 16 mm × 1600 ha = 140,800 m³; net storage change after 6 h of canal, evaporation, and seepage is 116,176 m³; level rise = 116,176 / (45 ha) = 0.258 m.

## Answer

**Given:** catchment area $A = 1600$ ha $= 16 \times 10^6$ m²; rainfall 8 mm/h for 2 h; runoff coefficient 0.55; canal flow 1 m³/s; evaporation 0.8 mm/h/m² of water spread; seepage = 40% of evaporation; water spread $= 45$ ha $= 4.5 \times 10^5$ m²; period $T = 6$ h.

Water balance: $\Delta S = \text{Inflow} - \text{Canal outflow} - \text{Evaporation} - \text{Seepage}$

#### Step 1: Inflow
Rainfall depth $= 8 \times 2 = 16$ mm $= 0.016$ m

$$
V_{in} = 0.55 \times 0.016 \times 16\times10^6 = 140\,800\ \text{m}^3
$$

#### Step 2: Outflows over 6 h
- Canal: $1 \times 6 \times 3600 = 21\,600$ m³
- Evaporation: $0.0008 \times 4.5\times10^5 \times 6 = 2\,160$ m³
- Seepage: $0.4 \times 2160 = 864$ m³

#### Step 3: Change in storage
$$
\Delta S = 140\,800 - 21\,600 - 2\,160 - 864 = 116\,176\ \text{m}^3
$$

#### Step 4: Change in level
$$
\Delta h = \frac{116\,176}{4.5\times10^5} = 0.258\ \text{m}
$$

**Answer: The reservoir level rises by about 0.258 m (25.8 cm) in 6 hours.**

## Check yourself

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

1. How is inflow volume obtained?
   A. Rainfall depth x water spread
   B. Rainfall depth x canal flow
   C. Runoff coefficient x rainfall depth x catchment area
   D. Catchment area x evaporation rate
   Answer: C. V = 0.55 x 0.016 x 16x10^6 = 140800 m^3. (Key point: Inflow)

2. Evaporation loss is computed over:
   A. The catchment area, 1600 ha
   B. The canal cross-section
   C. Only the rainfall area
   D. The reservoir water spread, 45 ha
   Answer: D. Evaporation occurs from the water surface. (Key point: Evaporation area)

3. How is seepage found?
   A. 40% of evaporation volume
   B. 40% of inflow
   C. 40% of canal volume
   D. 40% of rainfall depth
   Answer: A. Seepage = 0.4 x 2160 = 864 m^3. (Key point: Seepage)

4. What is the level change?
   A. 0.258 m fall
   B. 0.258 m rise = ΔS / water spread
   C. 2.58 m rise
   D. ΔS / catchment area
   Answer: B. 116176/450000 = 0.258 m. (Key point: Level change)

## Close variants set in other exams

- [Reservoir level change, water balance, 15.5 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-level-change-15-5km2) — Engineering Hydrology, chapter 1, asked 2×

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