---
title: "Demand of rural municipality 2018, 15,000 pop"
subject: "Water Supply Engineering"
chapter: 3
key: "semester_5/water_supply_engineering:demand-c74"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/demand-c74
site: BCT Engineering
---

# Demand of rural municipality 2018, 15,000 pop

Water Supply Engineering · Chapter 3: Quantity of Water

- Asked in: 2074 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/demand-c74
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3

## Question

Data obtained from a baseline survey of a newly formed rural municipality in year 2018 A.D. are as follows. Population = 15,000, No. of day students in school = 1500, No. of big animals = 6500, No. of small animals = 8000. There are altogether 10 offices, one hospital with total 25 beds, No. of tea shops = 12, No. of health post = 2 and number of police check post = 2. Neglect fire fighting demand. Calculate the water demand of that rural municipality in design year with base period of 2 years and design period of 20 years. Assume the population growth rate of that community is 1.8 % per annum.

## Short answer

Design population in 2040 is 22,210. Applying rural unit rates and 15% losses gives average demand 1,566 m3/day or 18.1 l/s; maximum day demand is about 27.2 l/s.

## Answer

Survey year 2018; base period 2 years and design period 20 years, so the design year is 2040, $n = 22$ years after the survey.

#### Design population
$$
P_{2040} = 15000\times(1.018)^{22} = 22{,}210
$$

#### 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. Hospital 340 l/bed/d (IS 1172). Fire demand is neglected.

| Demand | Quantity | Rate | Demand (l/day) |
|---|---|---|---|
| Domestic (45 lpcd) | 22,210 | 45 l/c/d | 999,436 |
| Day students | 1,500 | 5 l/c/d | 7,500 |
| Big animals | 6,500 | 45 l/d | 292,500 |
| Small animals | 8,000 | 5 l/d | 40,000 |
| Offices (10 staff each) | 100 | 45 l/staff | 4,500 |
| Hospital (25 beds) | 25 | 340 l/bed/d | 8,500 |
| Tea shops | 12 | 500 l/d | 6,000 |
| Health posts | 2 | 1000 l/d | 2,000 |
| Police check posts (10 staff each) | 20 | 45 l/staff | 900 |
| **Net demand** | | | **1,361,336** |

Losses and wastage $= 15\%$ of net demand $= 204{,}200$ l/day

$$
Q_{avg} = 1{,}361{,}336 + 204{,}200 = 1{,}565{,}536\ \text{l/day} = 1{,}565.54\ \mathrm{m^3/day} = 18.12\ \text{l/s}
$$

**Answer: Design population = 22,210; average daily water demand = 1,566 m$^3$/day (18.1 l/s); maximum day demand = 27.2 l/s.**

## Check yourself

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

1. Which domestic rate is used for this rural municipality?
   A. 110 lpcd
   B. 135 lpcd
   C. 45 lpcd (public taps)
   D. 340 lpcd
   Answer: C. Rural water supply with public taps uses 45 lpcd; 110 and 135 lpcd are for urban, house-connected supply. (Key point: Rural domestic rate)

2. How is the demand of 2 police check posts calculated?
   A. 10 staff each, so 20 staff x 45 l/staff = 900 l/day
   B. 2 x 1000 l/day
   C. 2 x 500 l/day
   D. 20 x 5 l/day
   Answer: A. Staff of offices and check posts are taken as 10 each at 45 l/staff/day. (Key point: Office and check-post demand)

3. Given Q_avg = 18.12 l/s, how is the maximum day demand found?
   A. 3 x 18.12 = 54.4 l/s
   B. 1.5 x 18.12 = 27.2 l/s
   C. 18.12/1.5 = 12.1 l/s
   D. 18.12 + 15% = 20.8 l/s
   Answer: B. Maximum day factor of 1.5 is applied to the average demand. (Key point: Maximum day demand)

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