ical oxygen demand be reduced 90 percent. The plant will handle only domestic sanitary
sewage. What are the daily oxygen demand and the daily suspended-solids content of the
sewage? If an industrial plant discharges into this system sewage requiring 4500 Ib
(2041.2 kg) of oxygen per day, determine the population equivalent of the industrial
sewage.
Calculation Procedure:
- Compute the daily sewage flow
With an average flow per capita of 200 gal/day (8.8 mL/s), this sewage treatment plant
must handle per capita (200 gal/day)( 100,000 population) = 20,000 gal/day (896.2 L/s). - Compute the sewage oxygen demand
Usual domestic sewage shows a 5-day oxygen demand of 0.12 to 0.17 Ib/day (0.054 to
0.077 kg/day) per person. With an average of 0.15 Ib (0.068 kg) per person per day, the
daily oxygen demand of the sewage is (0.15)( 100,000) = 15,000 Ib/day (78/7 g/s). - Compute the suspended-solids content of the sewage
Usual domestic sewage contains about 0.25 Ib (0.11 kg) of suspended solids per person
per day. Using this average, we see the total quantity of suspended solids that must be
handled is (0.25)( 100,000) = 25,000 Ib/day (0.13 kg/s). - Select the sewage-treatment method
Table 6 shows the efficiency of various sewage-treatment methods. Since the desired re-
duction in suspended matter, biochemical oxygen demand (BOD), and bacteria is known,
this will serve as a guide to the initial choice of the equipment.
Study of Table 6 shows that a number of treatments are available which will reduce
the suspended matter by 80 percent. Hence, any one of these methods might be used. The
same is true for the desired reduction in bacteria and BOD. Thus, the system choice re-
solves to selection of the most economical group of treatment units.
For a city of this size, four steps of sewage treatment would be advisable. The first
step, preliminary treatment, could include screening to remove large suspended solids,
grit removal, and grease removal. The next step, primary treatment, could include sedi-
mentation or chemical precipitation. Secondary treatment, the next step, might be of a bi-
ological type such as the activated-sludge process or the trickling filter. In the final step,
the sewage might be treated by cholorination. Treated sewage can then be disposed of in
fields, streams, or other suitable areas.
Choose the following units for this sewage-treatment plant, using the data in Table 6
as a guide: rocks or screens to remove large suspended solids, grit chambers to remove
grit, skimming tanks for grease removal, plain sedimentation, activated-sludge process,
and chlorination.
Reference to Table 6 shows that screens and plain sedimentation will reduce the sus-
pended solids by the desired amount. Likewise, the activated-sludge process reduces the
BOD by up to 95 percent and the bacteria up to 95+ percent. Hence, the chosen system
satisfies the design requirements. - Compute the population equivalent of the industrial sewage
Use the relation Pe = RID, where Pe = population equivalent of the industrial sewage, per-
sons; R — required oxygen of the sewage, Ib/day; D = daily oxygen demand, Ib per person
per day. So Pe = 4500/0.15 = 30,000 persons.
Related Calculations. Where sewage is combined (i.e., sanitary and storm
sewage mixed), the 5-day per-capita oxygen demand is about 0.25 Ib/day (0.11 kg/day).
Where large quantities of industrial waste are part of combined sewage, the per-capita