CHAPTER XXVII.

PROBLEMS AS TO FLOW IN SEWERS.

In order to prevent deposit of solid matter, sewers should be constructed with a sufficient gradient, and of a shape which presents the least surface for friction in proportion to the amount of liquid to be conveyed.

All brick sewers should be egg-shaped, with the narrow end downwards. The egg is formed by two circles touching one another, the diameter of the upper circle being twice that of the lower.

This shape possesses the great advantage that when the depth of the stream is diminished the amount of wetted surface of sewer (_wetted perimeter_) is diminished in equal proportion, whereas in every other form of sewer it is relatively increased. Thus the friction, which depends on the extent of the wetted perimeter, is kept down to a minimum. Where, as in outfall sewers, the volume of sewage is large, and does not vary greatly in amount, the circular form may be preferable, as it is cheaper and stronger than the egg-shaped sewer. Below 18 inches internal diameter, sewers should be circular in section, and made of stoneware, not brick.

The _velocity of flow_ depends upon (1) the hydraulic mean depth of the stream, and (2) its inclination or fall.

The _hydraulic mean depth_ means the depth of a rectangular channel whose sectional area (and therefore the volume of whose current) equals that of the curved sewer or pipe, concerning which the calculation is made, and whose width equals the entire wetted perimeter of the sewer or pipe. It is thus equal to sectional area/wetted perimeter.

In the case of circular pipes, if we take the diameter to be 1, and assume the pipe to be running full, the sectional area = πr^2, where π = 3·1416 and r = half diameter.

The wetted perimeter = 2πr, that is, the circumference of the circle formed by the pipe.

Therefore hydraulic mean depth = h = πr^2∕2πr = 1∕4

Similarly when the pipe runs half full—

h = (πr^2∕2)/(2πr) = 1∕4

The solution of problems where a smaller arc of a circle is occupied by fluid requires trigonometrical methods, and is not usually needed in practice.

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.+

_Supposing a sewer to have a gradient of 1 in 300, how much would the velocity and discharge be increased by altering the gradient to 1 in 100?_

1 in 300 = 17·6 in mile. 1 in 100 = 52·8 in mile.

As h is not given, we must assume it = 1∕4, as it does in circular sewers running full or half full.

v = 55 √(h x 2f) = 55 √(35·2∕4) = 163 feet per minute. v^{1} = 55 √(104∕4) = 281 „ „

The increase in discharge may be similarly calculated.

_Describe the relation existing in a sewer between gradient, volume, velocity, and size._

By the formula v = 55 √(_h.f._) Where v = velocity in feet per minute. h = hydraulic mean depth = (area of cross section of stream) ─────────────────────────────── (wetted perimeter). f = fall in feet in two miles.

In circular sewers h = diameter/4.

Thus the velocity varies as the square root of h or f.

The volume discharged varies with the value of the factor v × s where s = sectional area of stream.

If h remains constant, with a varying volume of s, then the volume discharged may remain constant. Thus h and v in a circular sewer are the same, whether the sewer runs full or half full. In a V-shaped channel the velocity remains the same whatever the depth of the stream, as its bed and area preserve the same proportions. An egg-shaped sewer approximates the V shape in form.

Similar volumes of sewage have velocities which vary not only with the amount of fall, but the size of the sewer. The friction, as represented by the wetted perimeter, would be much less with sewage half filling a circular sewer, than with the same amount of sewage forming a broad shallow stream on the invert of a large sewer.