Question

A community with a population of 50,000 has residences built 15 ft apart. The average per capita water consumption is 120 gpd, and the ratio of maximum day consumption to average daily demand is 1.6, and the maximum hourly demand to average daily demand ratio is 1.5. The largest commercial establishment in the community has 8,000 ft^2 of floor space, with class 2 construction with F=1. The facility houses furniture and has an occupancy factor O=1.17. Assume other factors to be zero. Determine average day, maximum day, maximum hourly demands. Also find water system capacity.

Answer 1

50000 × 120 gallon per day = 6,000,000 gallon per day =6 million gallon per day
( This means that on an average day, the water supply works need to have available nearly 6 million gallons of water over a period)
6,000,000 / (60) = 1,000.000 gpm (gallon per minute) of finished water delivered into the water supply distribution system
(i) Maximum day demand- The highest water demand of the year during any 24h period. (it is around 160%-190% 0f the ADD). It is used in the design of ground storage tanks.
MDD=120* 1.8 =216.0 gpd

(II) Peak hour demand. The highest water demand of the year during any 1-h period.
Usually equals to 2.5-3.5Qav, it is used to design the network and high storage tanks.

(III) Minimum day demand = 0.7* 6,000,000 = 4,200,000
(IV) Max daily demand – demand on the day of the year that uses the most water = 1.8 x 6,000,000= 10,800,000

(V) Max hourly demand – the demand during the hour that uses the most water = 3.25 x avg daily demand = 3.25* 6,000,000= 19,500,000