---
title: "Impounded reservoir capacity, 4x10^6 m3 demand"
subject: "Water Supply Engineering"
chapter: 2
key: "semester_5/water_supply_engineering:impounded-reservoir-c72"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/impounded-reservoir-c72
site: BCT Engineering
---

# Impounded reservoir capacity, 4x10^6 m3 demand

Water Supply Engineering · Chapter 2: Sources of Water

- Asked in: 2072 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/impounded-reservoir-c72
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-2

## Question

Determine the storage capacity of the impounded reservoir for a city with a water demand of $4\times10^6\ \mathrm{m^3}$ per month. The run-off discharge in the river is given in the table:

| Month | April | May | June | July | Aug. | Sep. | Oct. | Nov. | Dec. | Jan. | Feb. | March |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Inflow ($10^6\ \mathrm{m^3}$) | 5.1 | 5.9 | 8.3 | 8.9 | 7.5 | 5.0 | 3.5 | 3.0 | 2.0 | 1.6 | 2 | 2.4 |

## Short answer

Using the mass curve method, the maximum cumulative deficit of inflow below the constant monthly demand of 4×10^6 m³ is 9.5×10^6 m³ at the end of March. The required storage capacity is 9.5 million m³.

## Answer

The storage of an impounded reservoir must supply the demand during the period when inflow is less than the demand. It is found by the analytical (mass curve) method: take demand $D = 4\times10^6\ \mathrm{m^3}$/month and find the largest cumulative deficit. All volumes are in million m$^3$.

Check: total inflow $= 55.2$ and total demand $= 12\times4 = 48$, so the river can meet the demand over the year.

The reservoir is assumed full at the end of September, when the deficit period starts.

| Month | Inflow (Mm3) | Demand (Mm3) | Surplus/deficit | Cumulative deficit |
|---|---|---|---|---|
| Oct | 3.50 | 4.00 | -0.50 | -0.50 |
| Nov | 3.00 | 4.00 | -1.00 | -1.50 |
| Dec | 2.00 | 4.00 | -2.00 | -3.50 |
| Jan | 1.60 | 4.00 | -2.40 | -5.90 |
| Feb | 2.00 | 4.00 | -2.00 | -7.90 |
| Mar | 2.40 | 4.00 | -1.60 | -9.50 |

The inflow is below the demand from October to March, and the cumulative deficit reaches its maximum at the end of March.

**Answer: Storage capacity of the impounded reservoir = 9.5 million m$^3$ ($= 9.5\times10^6\ \mathrm{m^3}$).**

## Check yourself

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

1. What does the storage capacity of an impounded reservoir equal in the analytical method?
   A. Total annual demand
   B. Total annual inflow
   C. Largest cumulative deficit of inflow against demand
   D. Largest monthly inflow
   Answer: C. Storage must supply the demand during the period when inflow is less than demand, i.e. the maximum cumulative deficit. (Key point: Principle of storage)

2. For demand 4 Mm3/month, in which months is inflow less than demand?
   A. April to September
   B. October to March
   C. June to August
   D. November to January only
   Answer: B. Inflow is 3.5, 3.0, 2.0, 1.6, 2.0, 2.4 from Oct to Mar, all below 4. (Key point: Deficit period)

3. What is the storage capacity found?
   A. 9.5 Mm3
   B. 7.9 Mm3
   C. 5.9 Mm3
   D. 12.0 Mm3
   Answer: A. Deficits 0.5+1.0+2.0+2.4+2.0+1.6 = 9.5 Mm3, reached at the end of March. (Key point: Result)

4. Why compare total inflow (55.2) with total demand (48)?
   A. To find the reservoir volume directly
   B. To confirm the river can meet the demand over the year
   C. To find evaporation loss
   D. To compute the peak factor
   Answer: B. Since annual inflow exceeds annual demand, a reservoir can carry surplus over to the dry months. (Key point: Feasibility check)

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