Population 2030 AD by three methods, with fire demand

The last census is 2000, so for 2030, $n = 3$ decades. Three methods are used.

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{62{,}000-40{,}000}{3} = 7{,}333.3$

$$P_{2030} = P_n + n\bar x = 62{,}000 + 3\times 7{,}333.3 = 84{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{62{,}000}{40{,}000}\right)^{1/3}-1 = 15.729\%$

$$P_{2030} = P_n(1+r)^n = 62{,}000\times(1+0.15729)^{3} = 96{,}100$$

Incremental increase method

Increase per decade: 5,000, 10,000, 7,000; incremental increase: 5,000, -3,000. $\bar x = 7{,}333.33$, $\bar y = 1{,}000.00$

$$P_{2030} = P_n + n\bar x + \frac{n(n+1)}{2}\bar y = 62{,}000 + 3(7{,}333.33) + \frac{3(4)}{2}(1{,}000.00) = 90{,}000$$

The mean of the three is adopted as the design population: $P_{2030} = 90{,}033$.

Design quantity of water

Assumptions: domestic and public demand 135 lpcd (IS 1172), losses and wastage 15% of the domestic demand, fire demand by the National Board of Fire Underwriters formula for a duration of 2 hours.

$$\begin{aligned} \text{Domestic} &= 90{,}033\times135 = 12{,}154{,}455\ \text{l/day}\\ \text{Losses} &= 0.15\times12{,}154{,}455 = 1{,}823{,}168\ \text{l/day}\\ Q_f &= 4637\sqrt{P}\,(1-0.01\sqrt{P}),\ P = 90.03\ \text{(thousand)}\\ &= 39{,}823.7\ \text{l/min}\\ \text{Fire volume (2 h)} &= 39{,}823.7\times120 = 4{,}778{,}841\ \text{l} \end{aligned}$$ $$Q_{total} = 12{,}154{,}455+1{,}823{,}168+4{,}778{,}841 = 18{,}756{,}464\ \text{l/day}$$

Answer: Design population (2030) = 90,033; design quantity of water = 18,756,464 l/day (18.76 million l/day), including losses and a 2-hour fire demand.