---
title: "Service reservoir capacity, 150,000 population"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:res-c75"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/res-c75
site: BCT Engineering
---

# Service reservoir capacity, 150,000 population

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2075 Chaitra, 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-c75
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Water is to be supplied to a municipality in Nepal with forecasted population of 150,000 with 110 litres per capita per day. The variation in water demand is mentioned below. Calculate the capacity of service reservoir considering pumping at 6-9am and 6-9pm respectively. Neglect the fire demand and use analytical method.

| Time | 6am - 9am | 9am - 12 noon | 12 noon - 3pm | 3pm - 6pm | 6pm - 9pm | 9pm - 6am |
|---|---|---|---|---|---|---|
| % Demand | 30 | 10 | 10 | 20 | 25 | 5 |

## Short answer

Daily demand is 16,500 m³ and pumping rate is 2,750 m³/h for 6 h. Mass-curve gives maximum surplus 3,300 m³ and maximum deficit 3,300 m³, so service reservoir capacity = 6,600 m³ (40% of daily demand).

## Answer

#### Given data
- Population 150,000 × 110 lpcd = 16,500 m³/day
- Pumping 6–9 h and 18–21 h (6 h), at a rate of 16,500/6 = 2,750 m³/h
- Fire demand neglected

#### Mass-curve table (analytical method)
Inflow: 2,750 m³/h during 6–9 h and 18–21 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 |
|---|---|---|---|---|
| 06:00–09:00 | 4950.00 | 8250.00 | +3300.00 | +3300.00 |
| 09:00–12:00 | 1650.00 | 0.00 | -1650.00 | +1650.00 |
| 12:00–15:00 | 1650.00 | 0.00 | -1650.00 | 0.00 |
| 15:00–18:00 | 3300.00 | 0.00 | -3300.00 | -3300.00 |
| 18:00–21:00 | 4125.00 | 8250.00 | +4125.00 | +825.00 |
| 21:00–06:00 | 825.00 | 0.00 | -825.00 | 0.00 |

Total demand = 16500.00 m³; total inflow = 16500.00 m³.

#### Capacity
- Maximum cumulative surplus = 3300.00 m³
- Maximum cumulative deficit = 3300.00 m³
- Balancing capacity = max. surplus + max. deficit = 3300.00 + 3300.00 = **6600.00 m³**

**Answer: Service reservoir capacity = 6,600 m³ (40% of the daily demand).**

## Check yourself

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

1. Daily demand is 16,500 m3 and pumping is for 6 h in total. What is the uniform pumping rate?
   A. 687.5 m3/h
   B. 1,650 m3/h
   C. 2,750 m3/h
   D. 5,500 m3/h
   Answer: C. Total daily demand must be pumped in the 6 pumping hours: 16,500/6 = 2,750 m3/h. (Key point: Pumping rate)

2. In the 6-9 am period the demand is 30% of 16,500 m3 and pumping is on. What is the inflow minus demand?
   A. +3,300 m3
   B. -1,650 m3
   C. +8,250 m3
   D. +4,950 m3
   Answer: A. Inflow = 3 h x 2,750 = 8,250 m3; demand = 4,950 m3; difference = +3,300 m3. (Key point: Inflow in a period)

3. The cumulative column peaks at +3,300 m3 and falls to -3,300 m3. What is the reservoir capacity?
   A. 3,300 m3
   B. 6,600 m3
   C. 0 m3
   D. 16,500 m3
   Answer: B. Maximum surplus 3,300 + maximum deficit 3,300 = 6,600 m3. (Key point: Capacity formula)

## Related in Water Supply Engineering

- [Service reservoir capacity, 10,000 m3/day city](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-s73) — Water Supply Engineering, chapter 7
- [Service reservoir of 30 m diameter, 1,00,000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-bk79) — Water Supply Engineering, chapter 7
- [Demand of new municipality 2016, 45,000 pop](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-a76) — Water Supply Engineering, chapter 3, asked 2×
- [Demand of rural municipality 2018, 15,000 pop](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-c74) — Water Supply Engineering, chapter 3
- [Slow sand filter, 15,000 population, 150 L/h/m2](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Assf-a75) — Water Supply Engineering, chapter 6
- [Impounded reservoir, 400 m3/day demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aimpounded-reservoir-b66) — Water Supply Engineering, chapter 2

## Related in other subjects

- [Conventional ASP design, population 1,00,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-100000-150lpcd) — Sanitary Engineering, chapter 7
- [Minimum reservoir capacity and wasted water](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Areservoir-minimum-capacity-wasted-water-1000hp) — Hydropower Engineering, chapter 3
- [Conventional ASP design, population 50,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-50000-80lpcd) — Sanitary Engineering, chapter 7, asked 2×