---
title: "Water demand 2030, fully plumbed town"
subject: "Water Supply Engineering"
chapter: 3
key: "semester_5/water_supply_engineering:demand-a74"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/demand-a74
site: BCT Engineering
---

# Water demand 2030, fully plumbed town

Water Supply Engineering · Chapter 3: Quantity of Water

- Asked in: 2074 Asoj, 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-a74
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-3

## Question

Population of a town in Nepal as obtained from the census report is as follows:

| Year A.D. | 1971 | 1981 | 1991 | 2001 | 2011 |
|---|---|---|---|---|---|
| Population | 15000 | 21000 | 27000 | 34000 | 42000 |

Determine the water demand in the year 2030 if the town has fully plumbed house. Take industrial demand as 20% total demand and water losses wastage as 15% of the total demand. Neglect other demands.

## Short answer

Forecast population for 2030 by mean of arithmetical, geometrical, and incremental methods is 59,993. With domestic 135 lpcd, industrial 20% and losses 15% of total, total water demand is 12,460 m3/day, about 144.2 l/s.

## Answer

Design year 2030; the last census is 2011, so $n = 1.9$ decades.

#### Population forecast
**Arithmetical increase method**

Average increase per decade: $\bar x = \dfrac{42{,}000-15{,}000}{4} = 6{,}750.0$

$$
P_{2030} = P_n + n\bar x = 42{,}000 + 1.9\times 6{,}750.0 = 54{,}825
$$

**Geometrical increase method**

Average growth rate per decade (geometric mean): $r = \left(\dfrac{42{,}000}{15{,}000}\right)^{1/4}-1 = 29.357\%$

$$
P_{2030} = P_n(1+r)^n = 42{,}000\times(1+0.29357)^{1.9} = 68{,}493
$$

**Incremental increase method**

Increase per decade: 6,000, 6,000, 7,000, 8,000; incremental increase: 0, 1,000, 1,000.
$\bar x = 6{,}750.00$, $\bar y = 666.67$

$$
P_{2030} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 42{,}000 + 1.9(6{,}750.00) + \frac{1.9(2.9)}{2}(666.67) = 56{,}662
$$


| Method | Population (2030) |
|---|---|
| Arithmetical | 54,825 |
| Geometrical | 68,493 |
| Incremental | 56,662 |

The mean of the three methods is adopted: $P_{2030} = 59{,}993$.

#### Water demand
For a fully plumbed town the domestic demand is taken as 135 lpcd (IS 1172).

$$
Q_{dom} = 59{,}993\times135 = 8{,}099{,}055\ \text{l/day} = 8{,}099\ \mathrm{m^3/day}
$$

Industrial demand (20%) and losses (15%) are fractions of the total demand $Q$, so domestic demand is 65% of $Q$.

$$
\begin{aligned}
Q &= \frac{8{,}099}{0.65} = 12{,}460\ \mathrm{m^3/day}\\
\text{Industrial} &= 0.20Q = 2{,}492\ \mathrm{m^3/day},\quad \text{Losses} = 0.15Q = 1{,}869\ \mathrm{m^3/day}
\end{aligned}
$$

**Answer: Population in 2030 = 59,993; total water demand = 12,460 m$^3$/day (144.2 l/s).**

## Check yourself

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

1. Census last in 2011 and design year 2030: what n is used in the forecast?
   A. 19 decades
   B. 2.0 decades
   C. 0.19 decades
   D. 1.9 decades
   Answer: D. (2030-2011)/10 = 1.9 decades. (Key point: Value of n)

2. How is the design population 59,993 chosen from the three methods?
   A. As the geometrical value, the largest
   B. As the mean of the arithmetical, geometrical and incremental results
   C. As the arithmetical value only
   D. As the lowest of the three
   Answer: B. (54,825+68,493+56,662)/3 = 59,993 balances the extremes. (Key point: Adopted population)

3. Industrial demand is 20% and losses 15% of total demand. How is total demand Q found from domestic demand?
   A. Q = domestic/0.65, since domestic is 65% of Q
   B. Q = domestic x 1.35
   C. Q = domestic/0.35
   D. Q = domestic x 1.20 x 1.15
   Answer: A. Because both are fractions of the total Q, domestic = (1-0.20-0.15)Q = 0.65Q. (Key point: Industrial and losses fractions)

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