Water Demand Calculator (Population, Per-Capita Use & Design Flows)

How it works

Future population P = P₀ × (1 + r)ⁿ · average daily demand Qavg = P × per-capita demand ÷ 86,400 (L/s) · maximum day Qmd = K₁ × Qavg · peak hour Qmh = K₂ × Qavg · regulating storage ≈ 25% of the average daily volume · wastewater: Qww = return factor × Qavg, Harmon peaking factor PF = 1 + 14/(4 + √(P/1000)).

Water systems are sized for the population at the end of the design period, usually 20–30 years, projected here with geometric growth. The average daily demand comes from the per-capita consumption (dotación), which already includes commercial use and losses in most codes. Treatment plants and supply mains are designed for the maximum day, distribution networks for the peak hour, and tanks hold about a quarter of a day’s use to balance the hourly swings. The sewer side receives 70–85% of the water supplied, and Harmon’s formula from Ten States Standards gives its peak factor, which falls as the population grows. Fire flow, when required, is added to the maximum day.

Worked example

A town of 10,000 people growing 2% a year reaches 10,000 × 1.02²⁰ = 14,859 people in 20 years. At 200 L/person/day the average demand is 14,859 × 200/86,400 = 34.40 L/s, the maximum day is 1.3 × 34.396 = 44.71 L/s and the peak hour 2 × 34.396 = 68.79 L/s. The town uses 2,971.8 m³ a day, so regulating storage of about 743.0 m³ is needed. The sewer receives 27.52 L/s on average, and with a Harmon peaking factor of 2.78 the peak wastewater flow is 76.56 L/s.

Sources: Recommended Standards for Water Works and Wastewater Facilities (Ten States Standards) — design flows and Harmon peaking factor · AWWA M32 — Computer Modeling of Water Distribution Systems: demand peaking factors (via EPA)

Frequently asked questions