Water demand 2030, fully plumbed town

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. $\bar x = 6{,}750.00$, $\bar y = 666.67$

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

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).