---
title: "Balancing reservoir for 0.25 MLD town"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:reservoir-balancing-025mld"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/reservoir-balancing-025mld
site: BCT Engineering
---

# Balancing reservoir for 0.25 MLD town

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2081 Baisakh, 8 marks; 2080 Bhadra, 8 marks; 2070 Chaitra, 4 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/reservoir-balancing-025mld
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

For the water supply of a town with a daily requirement of 0.25 MLD, it is proposed to construct a distribution reservoir. The consumption pattern is as follows:

| Time (hr.) | Consumption |
|---|---|
| 7.00-8.00 | 30% of days supply |
| 8.00-17.00 | 35% of days supply |
| 17.00-18.30 | 30% of days supply |
| 18.30-7.00 | 5% of days supply |

The pumping is to be done at a constant rate of 0.032 ML/hr for 8.00 to 16.00 hours. Determine the required capacity of balancing reservoir using analytical method.

## Short answer

Using mass curve analysis, the maximum cumulative deficit is 82 m³ and surplus 96.22 m³. Required balancing reservoir capacity is 178 m³ (0.178 ML).

## Answer

#### Given data
- Daily requirement = 0.25 MLD = 250 m³/day
- Pumping at 0.032 ML/h = 32 m³/h from 8:00 to 16:00 (8 h), so pumped volume = 256 m³
- Demand: 7–8 h = 30% (75 m³), 8–17 h = 35% (87.5 m³), 17–18.5 h = 30% (75 m³), 18.5–7 h = 5% (12.5 m³)

#### Hourly demand rates
- 18.5–7 h (12.5 h): 12.5/12.5 = 1.0 m³/h
- 7–8 h: 75 m³/h
- 8–17 h (9 h): 87.5/9 = 9.72 m³/h
- 17–18.5 h (1.5 h): 75/1.5 = 50 m³/h

#### Mass-curve table (analytical method)
| Period (h) | Pumped in (m³) | Demand (m³) | Net (m³) | Cumulative (m³) |
|---|---|---|---|---|
| 0–7 | 0 | 7.00 | −7.00 | −7.00 |
| 7–8 | 0 | 75.00 | −75.00 | −82.00 |
| 8–16 | 256.00 | 77.78 | +178.22 | +96.22 |
| 16–17 | 0 | 9.72 | −9.72 | +86.50 |
| 17–18.5 | 0 | 75.00 | −75.00 | +11.50 |
| 18.5–24 | 0 | 5.50 | −5.50 | +6.00 |

The cumulative at 24 h is +6.0 m³, which is the small excess (256 − 250) of pumping over demand.

#### Capacity
- Maximum cumulative deficit = 82.00 m³ (at 8:00)
- Maximum cumulative surplus = 96.22 m³ (at 16:00)
- Balancing capacity = 82.00 + 96.22 = 178.22 m³

**Answer: Required balancing reservoir capacity ≈ 178 m³ (0.178 ML), i.e. about 71% of the daily demand. The tank must hold 82 m³ at midnight so that the 7–8 h peak can be met before pumping starts.**

## Check yourself

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

1. Pumping at 32 m3/h for 8 h (8:00 to 16:00) gives what volume?
   A. 250 m3
   B. 256 m3
   C. 32 m3
   D. 178 m3
   Answer: B. 32 x 8 = 256 m3, slightly more than the 250 m3 daily demand. (Key point: Pumped volume)

2. What is the demand rate between 8:00 and 17:00 (35% of 250 m3)?
   A. 87.5 m3/h
   B. 9.72 m3/h
   C. 75 m3/h
   D. 3.5 m3/h
   Answer: B. 87.5 m3 over 9 h gives 9.72 m3/h. (Key point: Hourly demand rate)

3. How is the balancing capacity found from the cumulative column?
   A. Maximum surplus only
   B. Maximum deficit only
   C. Maximum surplus plus maximum deficit
   D. Average of the cumulative values
   Answer: C. Capacity = 96.22 (max surplus) + 82.00 (max deficit) = 178.22 m3. (Key point: Capacity rule)

4. At what time does the maximum cumulative deficit of 82 m3 occur, and why is it important?
   A. At 8:00; the tank must hold 82 m3 so the 7-8 h peak is met before pumping starts
   B. At 16:00; pumping stops
   C. At 24:00; demand is least
   D. At 17:00; the evening peak starts
   Answer: A. Demand of 75 m3 in 7-8 h plus 7 m3 earlier is drawn before pumping begins at 8:00. (Key point: Critical point)

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