2. SLUDGE DIGESTION
Raw sludge contains a lot putrescible organic matter, if disposed
without any treatment the organic matter may decompose producing
foul gases and health hazards.
To avoid these problems the sludge is stabilized by decomposing the
organic matter by controlled anaerobic conditions; this process of
stabilization is called sludge digestion and done in sludge digestion
tanks.
3. Stages in sludge digestion process
a) Acid Fermentation Stage
• Acid formers, convert organic solids to volatile acids and gases like methane, carbon dioxide and
hydrogen sulphide.
• Lowers the pH value to less than 6
b) Acid Regression Stage
• Volatile organic acids from first stage is converted to acid carbonates and ammonia compounds and
hydrogen sulphide and carbondioxide gas is evolved.
• The pH is raised to 6.8
c) Alkaline Fermentation Stage
• Methane formers, convert proteins and organic acids to simple compounds like ammonia and gases
(mainly methane and small amount of other gases like CO2, H2S etc..)
• The pH is about 7.5
4. Factors affecting sludge digestion
a) Temperature
• Higher the temperature, higher the rate of digestion.
• THERMOPHILIC DIGESTION - 40 to 60 degree C; optimum is 50degree C; 7
to 10 days; not advisable because extra cost for increasing temperature.
• MESOPHILIC DIGESTION – 25 to 40 degree C, optimum is 29degree C,
digestion in 10days.
b) pH
• Proper care to be taken to maintain pH.
• Should not be less than 6.5;and optimum is 7.2 to 7.4
5. c) Seeding with digested sludge
• For a new sludge digestion tank, seeding with digested sludge from another plant is
highly beneficial, without seeding it may take a few months for a new plant to
operate properly.
d) Mixing raw sludge with digested sludge
• Incoming raw sewage thoroughly mixed with partly digested sludge to make a
homogeneous mixture by slow moving mechanical devices, so that bacterial mass
from digested sludge decomposes the raw sludge.
7. Design of sludge digestion tank
Design a sludge digestion tank for 40,000 people. The sludge content per capita
per day is 0.068kg. The moisture content of sludge is 94%. The sp. gravity of
wet sludge is 1.02 and 3.5 per cent of the digestor volume is daily filled with
fresh sludge, which is mixed with the digested sludge.
Solution:
Dry sludge produced by 40,000 persons
= 0.068 x 40,000kg = 2720kg/day
8. 94% moisture content means that 6kg of wet sludge will produce 100kg of wet
sludge.
=> 6kg of dry sludge has wet sludge = 100kg
therefore, 2720kg of dry sludge has wet sludge = (100 / 6) * 2720
= 45333kg = 45.3 t/day
Volume of wet sludge produced;
= ( mass of sludge) / ( density of sludge)
= ( 45333.33) / ( 1.02)
= 44.4 m3/ day
Given, 3.5% of digester volume is daily filled with fresh sludge;
=> 3.5% of digestor volume = 44.4 m3/day (fresh sludge)
Sp.Gravity = (density of
sludge) / ( density of
water)
(1.02) = (density) / (1000)
Density = 1020kg/m3
9. Therefore, volume of digestor = (100 / 3.5) x 44.4 m3/day = 1268.9 m3
Providing, 30% additional capacity for fluctuations,
=> volume of digestor = ( 1268.9 ) + ( 30% of 12668.9 ) = 1650 m3
Provide depth of cylindrical digestion tank = 6m
=> Area = ( volume ) / ( depth )
= ( 1650 ) / ( 6 ) = 275m2
=> dia = 18.7m
Therefore, provide a cylindrical sludge digestion tank with 6m deep and 18.7m
in diameter, with an additional hoppered bottom of 1:1 slope for collecting the
digested sludge.