---
title: "Storage change from inflow equation I = 2t"
subject: "Engineering Hydrology"
chapter: 1
key: "semester_5/engineering_hydrology:storage-change-i2t"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/storage-change-i2t
site: BCT Engineering
---

# Storage change from inflow equation I = 2t

Engineering Hydrology · Chapter 1: Introduction

- Asked in: 2076 Asoj, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/storage-change-i2t
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-1

## Question

In a certain catchment, inflow rate into the catchment due to rainfall is given by equation $I = 2t$ m$^3$/s. If loss in the catchment is neglected, determine the change in storage in catchment within 3 hr duration.

## Short answer

Integrate inflow I=2t m³/s over 3 h (10800 s) giving ΔS = ∫₀¹⁰⁸⁰⁰ 2t dt = (10800)² = 1.1664×10⁸ m³ increase.

## Answer

The water balance gives $I - O = \Delta S$. Losses are neglected, so outflow $O = 0$ and the change in storage equals the total inflow volume.

**Given:** $I = 2t$ m³/s, with $t$ in seconds; duration $T = 3$ h $= 3 \times 3600 = 10800$ s.

$$
\Delta S = \int_0^{T} I\,dt = \int_0^{10800} 2t\,dt = \left[t^2\right]_0^{10800}
$$

$$
\Delta S = (10800)^2 = 1.1664 \times 10^{8}\ \text{m}^3
$$

**Answer: Change in storage = 1.1664 × 10⁸ m³ (116.64 million m³) increase in 3 hours.**

(If $t$ were taken in hours, the answer would be different; $t$ in seconds is consistent with the unit m³/s.)

## Check yourself

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

1. With losses neglected, how is ΔS found?
   A. Multiply I by 3 only
   B. Differentiate I with respect to t
   C. Divide I by duration
   D. Integrate inflow over time since O = 0
   Answer: D. ΔS = ∫I dt when outflow is zero. (Key point: Method)

2. Why must t be in seconds?
   A. Because I is in m^3/s
   B. Because storage is in hours
   C. Because rainfall is in mm
   D. Because 3 is odd
   Answer: A. Consistency with m^3/s needs t in seconds. (Key point: Units)

3. What upper limit is used for 3 hr?
   A. 3 s
   B. 10800 s
   C. 180 s
   D. 3600 s
   Answer: B. 3 x 3600 = 10800 s. (Key point: Duration)

4. What is ΔS?
   A. 2.16 x 10^4 m^3
   B. 1.08 x 10^4 m^3
   C. 1.1664 x 10^8 m^3
   D. 5.83 x 10^7 m^3
   Answer: C. ∫2t dt = t^2 = 10800^2 = 1.1664 x 10^8 m^3. (Key point: Result)

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