---
title: "Service reservoir of 30 m diameter, 1,00,000 population"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:res-bk79"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/res-bk79
site: BCT Engineering
---

# Service reservoir of 30 m diameter, 1,00,000 population

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2079 Baisakh, 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-bk79
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Determine the capacity of a service reservoir of diameter 30 m for a town having 1,00,000 populations with peak per capita water demand of 110 liters/day. The water demand is to be fulfilled by a river flowing under gravity by a continuous system. The variations of the demand are as follows:

| Time (hours) | 05-07 | 07-12 | 12-17 | 17-19 | 19-05 |
|---|---|---|---|---|---|
| Demand (%) | 25 | 30 | 15 | 20 | 10 |

Also determine the water level in the tank at 12 hrs.

## Short answer

Daily demand is 11,000 m³ with uniform inflow 458.33 m³/h. Mass-curve gives maximum deficit 3,483.33 m³, so capacity ≈ 3,483 m³; storage at 12 h is 641.67 m³, giving water level 0.91 m in the 30 m diameter tank.

## Answer

#### Given data
- Population 100,000 × 110 lpcd = 11,000 m³/day
- River water flows in by gravity at a uniform rate: 11,000/24 = 458.33 m³/h
- Tank diameter = 30 m, area = π × 30²/4 = 706.86 m²
- Demand: 5–7 h 25%, 7–12 h 30%, 12–17 h 15%, 17–19 h 20%, 19–5 h 10%

#### Mass-curve table (analytical method)
Inflow: 458.33 m³/h for 24 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 | 2750.00 | 916.67 | -1833.33 | -1833.33 |
| 07:00–12:00 | 3300.00 | 2291.67 | -1008.33 | -2841.67 |
| 12:00–17:00 | 1650.00 | 2291.67 | +641.67 | -2200.00 |
| 17:00–19:00 | 2200.00 | 916.67 | -1283.33 | -3483.33 |
| 19:00–05:00 | 1100.00 | 4583.33 | +3483.33 | 0.00 |

Total demand = 11000.00 m³; total inflow = 11000.00 m³.

#### Capacity
- Maximum cumulative surplus = 0.00 m³
- Maximum cumulative deficit = 3483.33 m³
- Balancing capacity = max. surplus + max. deficit = 0.00 + 3483.33 = **3483.33 m³**

#### Depth and water level
- Water depth for full capacity $= 3483.33/706.86 = 4.93$ m (adopt about 5 m plus 0.5 m free board)
- The reservoir is empty at 19:00 and full at 05:00. At 12:00 the cumulative value is −2,841.67 m³, so storage = −2,841.67 + 3,483.33 = 641.67 m³
- Water level at 12 h $= 641.67/706.86 = 0.91$ m above the floor

**Answer: Capacity ≈ 3,483 m³; water level in the tank at 12 h ≈ 0.91 m.**

## Check yourself

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

1. For 100,000 people at 110 lpcd with continuous gravity supply, what is the hourly inflow?
   A. 458.33 m3/h
   B. 11,000 m3/h
   C. 110 m3/h
   D. 916.67 m3/h
   Answer: A. Demand = 11,000 m3/day, divided uniformly over 24 h = 458.33 m3/h. (Key point: Daily demand and inflow)

2. Capacity is 3,483.33 m3 and the tank diameter is 30 m. How is the water depth for full capacity found?
   A. Capacity x area
   B. Capacity / (pi x 30^2/4) = about 4.93 m
   C. Capacity / (pi x 30) 
   D. Capacity / 30
   Answer: B. Depth = volume / plan area, with area = pi D^2/4 = 706.86 m2. (Key point: Reservoir depth)

3. At 12:00 the cumulative value is -2,841.67 m3 and the tank is full at 05:00 with empty at 19:00. What is the storage at 12 h?
   A. 2,841.67 m3
   B. 641.67 m3
   C. 3,483.33 m3
   D. 0 m3
   Answer: B. Storage = -2,841.67 + 3,483.33 = 641.67 m3, i.e. 641.67/706.86 = 0.91 m. (Key point: Level at 12 h)

## Related in Water Supply Engineering

- [Service reservoir capacity, 150,000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-c75) — Water Supply Engineering, chapter 7
- [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
- [Balancing reservoir: town 1.2 million, 45 L/head](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-bk68) — Water Supply Engineering, chapter 7, asked 2×
- [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×
- [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
- [Balancing reservoir: village 600 population, 65 lpcd](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-as70) — Water Supply Engineering, chapter 7

## 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
- [Degree of treatment, town with dairy and industries](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-city-industries-71) — Sanitary Engineering, chapter 6
- [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×