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(Solved): What is the maximum population that can be served by an 8-in. sanitary sewer laid at minimum grade u ...



What is the maximum population that can be served by an 8-in. sanitary sewer laid at minimum grade using a design flow of 400 gpcd (gallon per capita per day) and a flowing-full velocity of 2.0 ft/sec? (b) Compute the diameter of a storm drain to serve the same population based on population density = 30 persons per acre, coefficient of runoff = 0.40, 10-year rainfall frequency curve, a duration (time of concentration) = 20 min, and a velocity of flow = 5.0 ft/sec.

TABLE 1 Minimum Slopes for Various Sized Sewers at
a Flowing-Full Velocity of 2.0 ft/sec and
Corresponding Discharges
Sewer
D

TABLE 1 Minimum Slopes for Various Sized Sewers at a Flowing-Full Velocity of 2.0 ft/sec and Corresponding Discharges Sewer Diameter (n.) 10 12 15 18 21 24 Minimum Slope (1/100 ft) 0.33 0.25 0.19 0.14 0.11 0.092 0.077 0.066 27 30 36 *Based on Manning's formula with in-0.013. 0.057 0.045 Flowing-Full Discharge (cu ft/sec) 0.7 1.1 1.6 2.4 3.5 4.8 6.3 8.0 9,8 14.1 (ppm) 310 490 700 1080 1570 2160 2820 3570 4410 6330


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Solution : a. Find The amount of maximum population that can be served : We know that ; Q = AV Where Q = Discharge in cfs A = Cross sectional area of sewer d = 8 inch = 8 / 12 = 0.667 sq ft V = 2 ft/s Q = 0.350 X 2 = 0.70 cf
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