---
title: "Demand of rural municipality, base period 3 years"
subject: "Water Supply Engineering"
chapter: 3
key: "semester_5/water_supply_engineering:demand-bk79"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/demand-bk79
site: BCT Engineering
---

# Demand of rural municipality, base period 3 years

Water Supply Engineering · Chapter 3: Quantity of Water

- 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/demand-bk79
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3

## Question

Baseline survey of a newly developing rural municipality in 2021 AD has the following information:

Population = 6000 with annual growth rate of 2%; No. of cows = 3050; No. of buffaloes = 2000; No. of goats = 2500; No. of ducks = 1000; No. of chickens = 6000; No. of rural municipality = 1 and no. of students in a school = 150 boarders and 1500 day scholars.

Neglecting the fire demand, calculate the water demand of the area considering a base period of 3 and design period of 20 years.

## Short answer

With 2% annual growth, design population in 2044 is 9,461. Summing domestic (45 lpcd), livestock, school, and office demands, then adding 15% losses, average daily demand = 784 m³/day (9.1 L/s) and maximum day demand = 13.6 L/s.

## Answer

Survey year 2021; base period 3 years and design period 20 years, so the design year is $2021+3+20 = 2044$, i.e. 23 years after the survey.

#### Population forecast (geometrical, r = 2%)
$$
P_{2044} = P_0(1+r)^{n} = 6000\times(1.02)^{23} = 9{,}461
$$

#### Water demand
**Design values assumed** (Nepal rural water supply practice): domestic 45 lpcd (public taps); day scholar 5 l/d; boarder 45 l/d; office staff 45 l/d with 10 staff per office; big animal (cow, buffalo) 45 l/d; small animal (goat, pig) 5 l/d; poultry 0.2 l/d; health post 1000 l/d; tea shop 500 l/d; losses and wastage 15% of net demand; maximum day factor 1.5 and peak hour factor 3.

Livestock and school numbers are taken as constant. Fire demand is neglected.

| Demand | Quantity | Rate | Demand (l/day) |
|---|---|---|---|
| Domestic (45 lpcd) | 9,461 | 45 l/c/d | 425,763 |
| Boarders | 150 | 45 l/c/d | 6,750 |
| Day scholars | 1,500 | 5 l/c/d | 7,500 |
| Cows + buffaloes | 5,050 | 45 l/d | 227,250 |
| Goats | 2,500 | 5 l/d | 12,500 |
| Ducks + chickens | 7,000 | 0.2 l/d | 1,400 |
| Rural municipality office (10 staff) | 10 | 45 l/staff | 450 |
| **Net demand** | | | **681,613** |

Losses and wastage $= 15\%$ of net demand $= 102{,}242$ l/day

$$
Q_{avg} = 681{,}613 + 102{,}242 = 783{,}855\ \text{l/day} = 783.85\ \mathrm{m^3/day} = 9.07\ \text{l/s}
$$

Peak (maximum day) demand $= 1.5\times783.9 = 1{,}175.8\ \mathrm{m^3/day}$.

**Answer: Design population = 9,461; average daily water demand = 784 m$^3$/day (9.1 l/s); maximum day demand = 13.6 l/s.**

## Check yourself

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

1. Survey 2021, base period 3 and design period 20 years. How many years of growth?
   A. 20
   B. 3
   C. 23
   D. 26
   Answer: C. Design year = 2021+3+20 = 2044, so n = 23 years. (Key point: Design year)

2. What is the design population at 2% growth?
   A. 9,461
   B. 8,000
   C. 12,000
   D. 6,000
   Answer: A. 6000 x 1.02^23 = 9,461. (Key point: Population)

3. What rates are used for animals?
   A. Cows and buffaloes 5 l/d, goats 45 l/d
   B. All animals 45 l/d
   C. Cows and buffaloes 45 l/d, goats 5 l/d, poultry 0.2 l/d
   D. All animals 5 l/d
   Answer: C. Big animals 45, small animals 5, poultry 0.2 l/d. (Key point: Livestock rates)

4. How are losses and max day demand applied?
   A. Losses 15% of net demand; maximum day = 1.5 x average
   B. Losses 15% of domestic only; max day = 3 x average
   C. Losses ignored; peak = 1.5
   D. Losses 50%
   Answer: A. Net demand 681,613 + 15% = 783,855 l/day, and max day = 1.5 x average. (Key point: Losses and design flow)

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