The _quantity of fluid discharged_ in a given time is represented by the product of the sectional area of the stream into its velocity. The greater the hydraulic mean depth the greater is the velocity, if the inclination remains the same. The _velocity of flow_ is determined by =Eytelwein’s formula=, which states that the mean velocity per second of a stream of water similar in form to those now under consideration is nine-tenths of a mean proportional between the hydraulic mean depth and the fall in two English miles, if the channel were prolonged so far. Thus if f = the fall (in feet) in two miles, h = hydraulic mean depth in feet, V = mean velocity per second, Then V = ·9√(_hf_), or if v = velocity per minute, then v = 55√(_hf_). It is more convenient to let f = fall in one mile. Then the formula becomes v = 55√(h × 2f). _How much sewage will a circular drain 3 feet in diameter running half full convey, the fall being 1 in 400?_ Here h = (πr^2∕2)/(2πr/2) = r/2 = 3∕4. 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 = πr^2∕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 = 1∕4 ∴ √h = 1∕2. 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?_ In the first case, 1 in 1,760 = 3 in mile. 1 in 5,280 = 1 in mile. h = r/2 = 3∕4. v = 55√(h × 2f) = 55√(3∕4 × 6) = 165/√ = 118. In second case v = 55√(3∕4 × 2) = 55√(3∕2) = 67·9. Thus the velocity of the two streams would be as +118: 67·9.+