2. INTRODUCTION
Sewage Treatment
Involves various physical unit operations sequentially
with biological unit processes.
Physical unit operations include screens, grit
chambers, skimming tanks and grease traps and
primary clarifiers.
Biological processes(both aerobic and anaerobic)
may be –
attached growth processes with units like trickling filters
or
suspended growth processes with units like activated
sludge process, aeration tanks, aerated lagoons.
3. PRIMARY & SECONDARY SEWAGE TREATMENTS
Primary Treatment – removal of heavy suspended
matter like faecal solids, kitchen refuse, cloth, waste
paper, etc.
It’s advantage is that it prevents damage or clog of
the pumps and skimmers of primary treatment
clarifiers (trash, tree limbs, leaves, etc.).
Primary treatment involves sequential unit processes
like screening, removal of grit &other floating matter.
The remainder sewage flows to primary
sedimentation tank or clarifier(PST).
4. Secondary Treatment – biological process for
treatment of very fine suspended matter, colloids &
dissolved solids in sewage from PST.
The unit processes are biological oxidation and
activated sludge process (converting sewage to
heavier & bulkier form and then it gets settled in
Secondary Settling Tank)
This process stabilizes & makes the sewage
completely harmless.
The separated sewer-sludge is decomposed
anaerobically in Sludge Digestion Tanks & the
digested sludge is disposed of separately in Sludge
Drying Beds.
5. RETURNED ACTIVATED
SLUDGE
TREATE
RAW GRIT AERATION D
SCREEN SEWAG
WATER CHAMBER TANK E
P.S.T. S.S.T
PRIMAR SECONDA
Y RY
SLUDGE SLUDGE
SLUDGE
SLUDGE
DIGESTION
DRYING BEDS
TANK
6. METHOD FOR NORMAL WATER TREATMENT
P.S.T
COAGULLATION
FLOCULATION
SETTLING
FILTRATION
DININFECTION
STORAGE
DISTRIBUTION
8. GRIT CHAMBERS: -
It is an enlarged channel or long basin in which the
cross-section is increased -------to reduce the
velocity of the flowing sewage sufficiently ------- to
cause heavy inorganic matter such as grit, sand
and gravel of size 0.2mm and larger to settle, with
the lighter organic matter remaining in suspension.
Types of Grit Chambers
Mechanically cleaned and Manually cleaned.
9. DESIGN CRITERIA :
In case of PST both Surface Overflow Rate and
Detention Period are important design
considerations.
This is because of the flocculent nature of the
settling particles.
In case of Secondary Settling Tanks, Solid
Loading Rate and Overflow Rates are important
for the design
10. 1) Detention period: 2 to 3 hours for PST
and 1.5 to 2 hours for SST. Longer periods
result in higher efficiency but too long a
period induces septic condition and should
be avoided.
2) Velocity of Flow: About
30cm./mt.
3) Depth of Flow: 2.5 to 3.0 mts. for PST and
3 to 4 mts. For SST. Depth influences
sludge thickening in SST.
11. Surface Loading or Overflow Rate: It is the
hydraulic loading per unit of surface area of
tank per day and is expressed as litres/m2/day.
Solid Loading is due to sludge solids
contained in the mixed liquor emanating from
aeration tanks/trickling filters and is referred as
Mixed Liquor Suspended Solids(MLSS).
Weir Loading should be such as to ensure
uniform withdrawal over the entire perophery of
the tank to avoid short-circuiting.
12. LOADING RATES FOR SETTLING TANKS
Design Primary Settling Tank Secondary Settling Tank
Parameter Primary Primary Secondary Secondary
Settling Only Settling settling after settling after
followed by trickling filter activated
secondary sludge process
treatment
Overflow Rate 25,000 – 35,000 – 15,000 – 50,000 15,000 – 50,000
50,000 80,000
(litres/m2/day)
Solid Loading - - 70 - 180 70 - 210
(kg/m2/day)
Weir Loading Not to Exceed 1,50,000 Not to exceed 1,85,000
(litres/m/day)
13. SEDIMENTATION TANK DESIGN
Problem: Design a rectangular sedimentation tank to treat 2.4
million litres of raw water per day. The detention period may be
assumed to be 3 hours.
Solution: Raw water flow per day is 2.4 x 106 l. Detention period
is 3h.
1. Volume of tank = Flow x Detention period = 2.4 x 103 x 3/24 =
300 m3
2. depth of tank = 3.0 m.
3. Surface area = 300/3 = 100 m2
4. L/B = 3 (assumed). L = 3B.
3B2 = 100 m2 i.e. B = 5.8 m
L = 3B = 5.8 X 3 = 17.4 m
Hence surface loading (Overflow rate) = 2.4 x 106 = 24,000 l/d/m2 <
40,000 l/d/m2 (OK)
