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Water demand calculator

A water system has to meet the busiest hour, not the average day. Enter the population and the per capita use, or solve for either, and the calculator returns the average demand, the maximum day and the peak hour flow that pipes and treatment plants must handle.

Calculator

The ratio of the busiest day to the average day. Canadian guidance often uses 1.5 to 2.5 depending on the community size.

The ratio of the busiest hour to the average day, expressed as a flow. Small systems have higher peaks than large ones.

Average demand (L/day)
2,500,000
Maximum day demand (L/day)
3,750,000
Peak hour flow
72.338
Water per year
912,500
Population served
10,000
Use per person (L/day)
250
What this means
10000 people at 250 L a day use 2500000 L on an average day, 3750000 L on the maximum day and 72.3 L/s at the peak hour. The annual volume is 912500 m3. Design the source, treatment plant and trunk mains for the maximum day and the peak hour, not the average.

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How this calculator works

Water demand is usually built up from population and per capita use. The average day demand is the population multiplied by the litres each person uses, which in Canadian communities typically falls between 225 and 450 litres a day including residential, commercial and industrial use and system losses. Demand is not flat: weekends, hot weather and irrigation push some days higher, so design uses a maximum day factor, commonly 1.5 to 2.5 times the average. Within a day there is also a peak hour, when morning routines coincide, which can be two to twelve times the average day flow for small systems. The calculator applies both factors and converts the peak to litres a second, the unit used to size pipes and pumps. A system sized for the average day will fail on the busiest morning, so the maximum day and peak hour are the design figures.

Formula

average demand (L/day) = population x use per person. maximum day = average x maximum day factor. peak hour flow (L/s) = average x peak hour factor / 86,400. annual volume (m3) = average x 365 / 1,000. Solving for population: average / use per person. Solving for use per person: average / population.

Worked example

A town of 10,000 people using 250 L per person a day has an average demand of 2,500,000 L a day. With a maximum day factor of 1.5 the busiest day is 3,750,000 L. With a peak hour factor of 2.5 the peak flow is 2,500,000 x 2.5 / 86,400 = 72.3 L/s. The annual volume is 2,500,000 x 365 / 1,000 = 912,500 m3.

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Frequently asked questions

Where do the per capita figures come from?

Municipal design guidelines and water use surveys. Canadian residential use is roughly 200 to 250 L per person a day indoors, and total municipal use including businesses and losses is often 300 to 450. Use your own utility data where you have it.

Why is the peak hour factor higher for small systems?

A few simultaneous uses make a large difference when the base flow is small. A town of a thousand people can have a peak hour five or six times the average, while a city of a million smooths out to two or three times.

Does this include fire flow?

No. Fire flow is a separate requirement, often far larger than the peak domestic flow, and it is set by the fire code and the building types served. A water system must meet both the peak hour demand and the fire flow requirement.

What about water conservation?

Lower per capita use reduces every figure in proportion. Fixture replacement, leak repair and outdoor watering restrictions can cut demand by 20 to 40 percent, which can defer an expensive expansion of the source or treatment plant.

Sources

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