---
title: "Rural area demand 2025 AD, incremental increase"
subject: "Water Supply Engineering"
chapter: 3
key: "semester_5/water_supply_engineering:demand-bk68"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/demand-bk68
site: BCT Engineering
---

# Rural area demand 2025 AD, incremental increase

Water Supply Engineering · Chapter 3: Quantity of Water

- Asked in: 2068 Baisakh (old course), 8 marks
- Video: [population forecasting, Incremental increase method with example, water supply engineering numerical](https://www.youtube.com/watch?v=oAKAJPr_-Ug) — Civil Engineering Exam
- 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-bk68
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3

## Question

Estimate the total water requirement for a rural area for the year 2025 AD by forecasting the population by incremental increase method with the following data.

| Year | 1950 | 1960 | 1970 | 1980 | 1990 | 2000 |
|---|---|---|---|---|---|---|
| Population | 7150 | 7680 | 8425 | 9265 | 11780 | 14339 |

There are 4 schools with 125 day scholar students and staffs in each school, livestock (3520 chicken/ducks and 170 big animals), 2 health posts with 5 beds capacity and other offices with 345 staffs altogether.

## Short answer

Population in 2025 by incremental increase method: 20,153. Total water requirement: 1,074 m3/day (12.4 l/s) including 15% losses.

## Answer

The last census is 2000, so for 2025, $n = 2.5$ decades.

#### Population by incremental increase method
**Incremental increase method**

Increase per decade: 530, 745, 840, 2,515, 2,559; incremental increase: 215, 95, 1,675, 44.
$\bar x = 1{,}437.80$, $\bar y = 507.25$

$$
P_{2025} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 14{,}339 + 2.5(1{,}437.80) + \frac{2.5(3.5)}{2}(507.25) = 20{,}153
$$


#### 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. Health post beds are taken at 100 l/bed/d; each of the 2 health posts is taken to have 5 beds. School staff are counted with the 125 day scholars of each school (4 schools, 500 persons at 5 l/d).

| Demand | Quantity | Rate | Demand (l/day) |
|---|---|---|---|
| Domestic (45 lpcd) | 20,153 | 45 l/c/d | 906,872 |
| Day scholars and staff (4 x 125) | 500 | 5 l/c/d | 2,500 |
| Big animals | 170 | 45 l/d | 7,650 |
| Chickens and ducks | 3,520 | 0.2 l/d | 704 |
| Health posts (2 x 5 beds) | 10 | 100 l/bed/d | 1,000 |
| Office staff | 345 | 45 l/staff | 15,525 |
| **Net demand** | | | **934,251** |

Losses and wastage $= 15\%$ of net demand $= 140{,}138$ l/day

$$
Q_{avg} = 934{,}251 + 140{,}138 = 1{,}074{,}389\ \text{l/day} = 1{,}074.39\ \mathrm{m^3/day} = 12.44\ \text{l/s}
$$

**Answer: Population in 2025 = 20,153; total water requirement = 1,074 m$^3$/day (12.4 l/s).**

## Check yourself

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

1. Which formula is used for 2025 with n = 2.5?
   A. P = Pn + n x
   B. P = Pn + n x + n(n+1)/2 y
   C. P = Pn(1+r)^n
   D. P = Pn + n x - n y
   Answer: B. x is average increase and y is average incremental increase. (Key point: Incremental increase method)

2. What is y (incremental increase) in this method?
   A. The average decade increase (1,437.8)
   B. The last decade increase
   C. The average of the changes between successive decade increases (507.25)
   D. The total population increase
   Answer: C. Increases per decade 530, 745, 840, 2515, 2559; y averages 215, 95, 1675, 44. (Key point: Average values)

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