How much sewage will a circular drain 3 feet in diameter running half full convey, the fall being 1 in 400?
Here h = (πr2 ∕ 2)/(2πr ∕ 2) = r ∕ 2 = ¾.
- 1 in 400 = x in 5,280 feet (i.e. a mile).
- f = 13·2 in a mile.
- v = 55 × 4·4 = 242 feet per minute.
- = 55√h × 2f.
- S = πr2∕2 = 3·1416 × 9/(4 × 2) = 3·5343.
- v × S = 242 × 3·5343 = 855·8 cubic feet, discharged per minute.
In what way does the size and shape of a sewer affect the velocity of the sewage flowing through it? If a 12-inch pipe sewer, laid at a gradient of 1 in 175, gives a velocity of 3½ feet per second, what would be the velocity if the sewer had a gradient of 1 in 700 (the pipe running half full in each case); and would this latter velocity suffice to keep the sewer clear of deposit?
An elliptical sewer gives greater velocity to flow of small quantities of sewage than a circular one because it exposes a smaller surface for friction.
By formula = v = 55√h × 2f.
- h = ¼ ∴ √h = ½.
- f = 1 in 175 = 30 feet in one mile.
- v = (55 ∕ 2)√60 = 212·85 ft. per min., i.e. slightly over 3½ ft. per sec.
In the second case f = 1 in 700 = 7·56 feet in one mile.
v = (55 ∕ 2)√15·12 = 106·97 feet per minute.
Thus in the first case there is a velocity of 3·55 feet per second, and in the second case of 1·78 feet per second. The latter velocity is quite insufficient to keep the sewer free from deposit, 3 feet per second being the minimum velocity required for that purpose.
Given a sewer 3 feet in diameter, with a fall of 1 in 1,760, what would be the relative discharge if the fall were 1 in 5,280?