Rural area demand 2025 AD, incremental increase

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).

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).