at n = 0.013 and Cf= 78.5, and project to the exact or next higher value of Cf. Table 5
shows that C 7 Is 64.70 for 15-in (381-mm) pipe and 105.1 for 18-in (457-mm) pipe. Since
the actual value of Cf is 78.5, a 15-in (381-mm) pipe would be too small. Hence, an 18-in
(457-mm) pipe would be used. This size agrees with that found in procedure a.
- Compute the size of the lateral sewer
The lateral sewer serves one-third of the total area. Since the total sanitary flow from the
entire area is 4.65 ft^3 /s (0.13 m^3 /s), the flow from one-third of the area, given an even dis-
tribution of population and the same pipe slope, is 4.65/3 = 1.55 ft
3
/s (0.044 m
3
/s). Using
either procedure in step 3, we find the required pipe size = 12 in (305 mm). Hence, three
12-in (302-mm) laterals will discharge into the main sewer, assuming that each lateral
serves an equal area and has the same slope.
5. Check the suitability of the main sewer size
Compute the value of d
2- 5
for each of the lateral sewer pipes discharging into the main
sewer pipe. Thus, for one 12-in (305-mm) lateral line, where d = smaller pipe diameter,
in, d
25
= 12
25
= 496. For three pipes of equal diameter, 3d
2- 5
= 1488 = D
25
, where D =
larger pipe diameter, in. Solving gives D^25 = 1488 and D = 17.5 in (445 mm). Hence, the
18-in (4570 mm) sewer main has sufficient capacity to handle the discharge of three 12-in
(305-mm) sewers. Note that Fig. 4, Section 14, shows that the flow velocity in both the
lateral and main sewers exceeds the minimum required velocity of 2 ft/s (0.6 m/s).
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- Compute the sewer size with infiltration
Infiltration is the groundwater that enters a sewer. The quantity and rate of infiltration de-
pend on the character of the soil in which the sewer is laid, the relative position of the
groundwater level and the sewer, the diameter and length of the sewer, and the material
and care with which the sewer is constructed. With tile and other jointed sewers, infiltra-
tion depends largely on the type of joint used in the pipes. In large concrete or brick sew-
ers, the infiltration depends on the type of waterproofing applied.
Infiltration is usually expressed in gallons per day per mile of sewer. With very careful
construction, infiltration can be kept down to 5000 gal/(daymi) [0.14 L/(km-s)] of pipe
even when the groundwater level is above the pipe. With poor construction, porous soil,
and high groundwater level, infiltration may amount to 100,000 gal/(daymi) [2.7
L/(knvs)] or more. Sewers laid in dense soil where the groundwater level is below the
sewer do not experience infiltration except during and immediately after a rainfall. Even
then, the infiltration will be in small amounts.
Assuming an infiltration rate of 20,000 gal/(daymi) [0.54 L/(knrs)] of sewer and a
sewer length of 1.2 mi (1.9 km) for this city, we see the daily infiltration is 1.2 (20,000) =
24,000 gal (90,850 L).
Checking the pipe size by either method in step 3 shows that both the 12-in (305-mm)
laterals and the 18-in (457-mm) main are of sufficient size to handle both the sanitary and
infiltration flow.
Related Calculations. Where a sewer must also handle the runoff from fire-
fighting apparatus, compute the quantity of fire-fighting water for cities of less than
200,000 population from Q = 1020(P)^0 5 [1 - 0.01(P)^05 ], where Q = fire demand, gal/min;
P = city population in thousands. Add the fire demand to the sanitary sewage and infiltra-
tion flows to determine the maximum quantity of liquid the sewer must handle. For cities
having a population of more than 200,000 persons, consult the fire department headquar-
ters to determine the water flow quantities anticipated.
Some sanitary engineers apply a demand factor to the average daily water require-
ments per capita before computing the flow rate into the sewer. Thus, the maximum
monthly water consumption is generally about 125 percent of the average annual demand
but may range up to 200 percent of the average annual demand. Maximum daily demands
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