---
title: "Balancing reservoir: town 1.2 million, 45 L/head"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:res-bk68"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/res-bk68
site: BCT Engineering
---

# Balancing reservoir: town 1.2 million, 45 L/head

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2068 Baisakh (old course), 8 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/res-bk68
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

A newly established town with a population of 1.2 million is to be supplied with water daily at 45 liters per capita. Water have to be stored also for fire demand keeping at least 1% of total demand. The variation in demand is as follows:

| Time | Consumption % |
|---|---|
| 05.00 - 07.00 | 25 |
| 07.00 - 12.00 | 35 |
| 12.00 - 17.00 | 20 |
| 17.00 - 19.00 | 20 |
| 19.00 - 05.00 | 0 |

Determine analytically the balancing reservoir capacity assuming pumping to be done at an uniform rate and the period of pumping is 5.00 A.M. to 10.00 A.M. and 5.00 P.M. to 8.00 P.M. in two shifts.

## Short answer

Balancing capacity is 18,360 m³ (surplus 8,910 m³ plus deficit 9,450 m³ from two-shift pumping at 6,750 m³/h). Adding 1% fire storage (540 m³) gives total reservoir capacity of 18,900 m³.

## Answer

#### Given data
- Population 1,200,000 × 45 lpcd = 54,000 m³/day
- Pumping in two shifts, 5–10 h and 17–20 h (8 h in all), at a rate 54,000/8 = 6,750 m³/h
- Fire storage = 1% of the total daily demand

#### Mass-curve table (analytical method)
Inflow: 6,750 m³/h during 5–10 h and 17–20 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

| Time | Demand (m³) | Inflow (m³) | Inflow − demand | Cumulative |
|---|---|---|---|---|
| 05:00–07:00 | 13500.00 | 13500.00 | +0.00 | +0.00 |
| 07:00–10:00 | 11340.00 | 20250.00 | +8910.00 | +8910.00 |
| 10:00–12:00 | 7560.00 | 0.00 | -7560.00 | +1350.00 |
| 12:00–17:00 | 10800.00 | 0.00 | -10800.00 | -9450.00 |
| 17:00–19:00 | 10800.00 | 13500.00 | +2700.00 | -6750.00 |
| 19:00–20:00 | 0.00 | 6750.00 | +6750.00 | +0.00 |
| 20:00–05:00 | 0.00 | 0.00 | +0.00 | +0.00 |

Total demand = 54000.00 m³; total inflow = 54000.00 m³.

#### Capacity
- Maximum cumulative surplus = 8910.00 m³
- Maximum cumulative deficit = 9450.00 m³
- Balancing capacity = max. surplus + max. deficit = 8910.00 + 9450.00 = **18360.00 m³**
- Fire storage = 1% of daily demand = 0.01 × 54000 = 540.00 m³
- Total capacity = 18360.00 + 540.00 = **18900.00 m³**

**Answer: Balancing capacity 18,360 m³ + fire storage 540 m³ = 18,900 m³ total.**

## Check yourself

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

1. Demand 54,000 m3/day, pumped 5-10 h and 17-20 h. What is the rate?
   A. 2,250 m3/h
   B. 6,750 m3/h
   C. 10,800 m3/h
   D. 54,000 m3/h
   Answer: B. Pumping hours total 5 + 3 = 8 h; 54,000/8 = 6,750 m3/h. (Key point: Pumping rate)

2. How is the fire storage added here?
   A. It is the balancing capacity
   B. 1% of the total daily demand = 540 m3, added to the balancing capacity
   C. It is subtracted from the balancing capacity
   D. It is 1% of the balancing capacity
   Answer: B. Fire storage is a separate volume (0.01 x 54,000 = 540 m3) added on top. (Key point: Fire storage)

3. Balancing = 8,910 + 9,450 = 18,360 m3. What is the total capacity?
   A. 18,360 m3
   B. 18,900 m3
   C. 17,820 m3
   D. 9,450 m3
   Answer: B. 18,360 + 540 = 18,900 m3. (Key point: Total capacity)

## Close variants set in other exams

- [Balancing reservoir, town of 1.3 million](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-c70o) — Water Supply Engineering, chapter 7, asked 2×

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