Sludge treatment and disposal 1

SLUDGE TREATMENT AND
DISPOSAL
ENVIRONMENTAL ENGINEERING II

SLUDGE TREATMENT
GENERAL
There are two end products of the waste water treatment.
The treatment plant effluent
Directly discharged in the natural water sources or into
receiving soils.
The sludge
Contains highly putrescible substances and pathogenic
organisms.
Treated further for safe disposal
2

SLUDGE AND ITS MOISTURE CONTENT
RAW SLUDGE
Deposited in a primary sedimentation tank.
Odorous and contains highly putrescible matter – very
objectionable.
Moisture content 95% .
SECONDARY SLUDGE
Deposited in a secondary clarifier.
contains putrescible matter, but less objectionable
compared to raw sludge.
Moisture content 96% - 98% (trickling filter) and 98% -
99% (activated sludge plant)
3

Problem : A sedimentation tank is treating 4.5 million litres of
sewage per day containing 275 ppm of suspended solids. The tank
removes 50% of suspended solids. Calculated the quantity of
sludge produced per day in bulk and weight, if
(a) moisture content of sludge is 98% ;
(b) moisture content of sludge is 96% .
4

Solution:
Given: volume of sewage treated = 4.5 x 106 L/day
Suspended solids present = 275 ppm
= 275mg/L
= 275 x 10-6 kg/L
ie., In 1 L of sewage 275 x 10-6 kg of suspended solids is
present.
therefore, in 4.5 x 106 L the mass of suspended solids present
= 275 x 10-6 (kg/L) x 4.5 x 106 (L/day)
= 1237.5 kg/day
Also given, % of solids removed in sedimentation tank = 50 %
Thus, the mass of solids removed in sedimentation tank
= 50% of 1237.5 kg/day
= 618.75 kg/day
5

Solution:
Case (a) : moisture content of sludge is 98%
ie., 2kg of solids or dry sludge makes 100 kg of wet sludge.
Thus, mass of wet sludge made by 618.75 kg of solids
= 30937.5 kg ~ 30940 kg
Assume, specific gravity of wet sludge = 1.02
Unit weight of water = 1000kg/m3
So, unit weight of wet sludge
= 1.02 x 1000 kg/m3 = 1020kg/m3
6

Solution:
Case (a) : moisture content of sludge is 96%
Weight of sludge = 15470 kg of wet sludge
Volume of sludge= 15.17 cu.m
Assignment
Problem : There is a sewage sludge with volume containing a
certain moisture content p1%. What will be the volume of this
sludge if its moisture content is reduced to p%?
Solution:
V = vol. of sludge produced at p%
V1 = vol. of sludge produced at p1%
7

The process of stabilization of sludge withdrawn from
the sedimentation basin.
By decomposing the organic matter under controlled
anaerobic conditions.
40% - 60% of organic solids converted into CO2 and
CH4 gas.
Remaining organic matter will be chemically stable
and odourless with 90% - 95% of moisture content.
This process reduces the sludge into three forms
Digested sludge
Supernatant liquor
Gases of decomposition
8
SLUDGE DIGESTION PROCESS

Digested sludge
Stable
Humus like solid mater
Tarry black in colour
Musty earthy odour
With reduced moisture content
Volume= one third times the undigested sludge
Free from pathogenic bacteria
But may contain cysts and eggs of bacteria, worms and
protozoa
Dewatered, dried up and used as fertilizer
9

Supernatant liquor
Liquified finely divided solid matter
BOD about 3000ppm
Contains 1500ppm to3000ppm of suspended solids, so
re-treated with the raw sludge.
Gases of decomposition
65-70% CH4
30% CO2
Traces of inert gases like N, H2S
Collected and used as fuel
10

STAGES IN SLUDGE DIGESTION PROCESS
Acid fermentation/Acid production
Acid regression
Alkaline fermentation
11

Acid fermentation/Acid production
Anaerobic and facultative bacteria (acid formers) acts.
Acid formers stabilize the organic solids through
hydrolysis.
Soluble products fermented to volatile acids and
organic alcohols of low molecular weight.
oPropionic acid, acetic acid etc
Evolution of CH4 , CO2 and H2S gases.
Lowers pH value to less than 6 – highly acidic in
nature.
Evolution of highly putrescible odours.
Continues for 15 days.
12

Acid regression
Volatile organic acids and nitrogenous compounds of the
first stage acted upon by bacteria.
Forms acid carbonates and ammonia compounds.
Evolution of H2S and CO2 gases in small amount.
Offensive odour
pH value rises to 6.8
Entraps of gases of decomposition and forms formy scum
layer.
Continues for 3 months.
BOD remains high.
13

Alkaline fermentation
Proteins and organic acids attacked and broken up by
anaerobic bacteria (methane formers).
Forms ammonia, organic acids and gases.
Liquid separates out from solids and the digestive sludge (or
ripened sludge) is formed.
Digestive sludge collected at the bottom of digestion tank.
Alkaline in nature – pH value rises little above 7.
Large amount of CH4 and small amount of CO2 and N gases
are evolved.
Continues for 1 month.
BOD rapidly falls down.
14

FACTORS AFFECTING SLUDGE DIGESTION PROCESS
Temperature
pH value
Seeding with digested sludge
Mixing and stirring of raw sludge with digested sludge
15

FACTORS AFFECTING SLUDGE DIGESTION PROCESS
Temperature
Rate of digestion increases at higher temperature.
Fig 9.27
16

FACTORS AFFECTING SLUDGE DIGESTION
PROCESS
Temperature
i. Zone of thermophilic digestion
 High temperature zone – 400C to 600C.
 Acted upon by heat loving thermophilic organisms.
 Optimum temperature is about 540C – digestion within
10 – 15 days.
ii. Zone of mesophilic digestion
 Moderate temperature zone – 250C to 400C.
 Acted upon by mesophilic organisms.
 Optimum temperature is about 290C – digestion in 30
days.
17

PROCESS
pH value
Should not lower below 6.5
Optimum value – 7.2 to 7.4 (in the final stage)
Acidity increases due to
oOverdosing of raw sludge
oOver withdrawal of digested sludge
oSudden admission of industrial wastes
Remedy: Add hydrated lime – 2.3 to 4.5 kg per 1000
persons to the raw sludge.
Raw sludge of about 3- 5% of the weight of the digested
sludge should only be added daily.
18

PROCESS
Seeding with digested sludge
Seed with digested sludge from another tank
Speeds up the digestion process
Mixing and stirring of raw sludge with digested sludge
Proper agitation methods mixes raw and digested sludge to
form homogeneous mass of sludge.
Bacterial enzymes present in digested sludge gets
thoroughly mixed up – helps in better decomposition.
19

SLUDGE DIGESTION TANK OR DIGESTORS.
20

Design considerations
Cylindrical in shape, circular in plan – dia 3 to 12m
Slope of bottom hopper floor – 1:1 to 1:3
Depth of digestion tank – 6m
Except in large plants not more than 2 units are
provided.
The capacity provided ranges from 21 to 61 lpcd.
21

If the progress of sludge digestion is assumed to be
linear, then capacity of digestor (in cu.m) is
Where,
V1 = raw sludge added per day, cu.m/d
V2 = equivalent digested sludge produced
per day, on completion of digestion V2 = (V1/3)
t = digestion periods in days
22

When the daily digested sludge could not be removed,
even though digestion gets completed , then consider
separate capacity . (Monsoon Storage)
Thus total capacity ,
23

Realistic case: When the change during digestion is
assumed to be parabolic then the average volume of
digesting sludge
Then, total capacity without monsoon storage
And total capacity with monsoon storage
24

Design problem
Design a digestion tank for the primary sludge with the
help of following data:
i. Average flow = 20 Mld
ii. Total suspended solids in raw sewage = 300mg/L
iii. Moisture content of digested sludge = 85%
Assume any other suitable data required.
25

Solution
Assumptions made
% of solids removed in primary settling tank=65%
Moisture content of fresh sludge =95%
Sp.gravity of wet sludge =1.02
Digestion period, t =30days
Depth of digestor =6m
26

Solution
Average flow =20 Mld = 20 x 106 L/day
TSS in raw sewage =300mg/L = 300 x 10-6 kg/L
∴ Mass of TSS in 20 Mld of sewage per day
=20 x 106 L/day x 300 x 10-6 kg/L
=6000 kg/day
Mass of solids removed in primary settling tank
= mass of solids forming the raw sludge
= 65% x 6000 kg/day = 3900 kg/L
∵ Moisture content of fresh sludge = 95%
Mass of wet sludge made by 3900kg/L of dry solids
= 78000 kg
27

Solution
Density of wet sludge = sp.gr x density of water
= 1.02 x 1000kg/m3
= 1020 kg/m3
Volume of raw sludge, V1 = (mass / density)
=(78000kg/d) / (1020kg/m3)
=76.47 m3/d
Volume of digested sludge,V2
= 25.49m3/d
28

Solution
Capacity of digestor ,
= 1274.5 cu.m ~ 1275 cu.m
∵ Depth of tank = 6m
Cross sectional area of the tank= (1275 cu.m / 6m)
= 212.5 sq.m
Dia of tank
=16.5m
29

Solution
Provide a cylindrical digestion tank of 6m deep and
16.5m diameter, with an additional hoppered bottom of
1:1 slope for collection of digested sludge.
Sketch the section with necessary dimensions.
Design problem
Design a sludge digestion tank for 40,000 people. The
sludge content per capita per day is 0.068kg. The
moisture of the sludge is 94%. The sp. gravity of the
wet sludge is 1.02 and 3.5% of the digestor volume is
daily filled with fresh sludge, which is mixed with the
digested sludge.
30

The digested sludge from the digestor contains lots of water
which should be removed by dewatering and then disposed off.
Sludge drying beds
Mechanical methods
Dewatering, drying and disposal of sludge by sludge drying
beds.
Drying of the digested sludge on open beds of land – sludge
drying beds.
Suitable for hot countries – India.
31
DEWATERING OF SLUDGE

Sludge drying beds
→15m x 30m in plan
→45 to 60 cm deep
→Area – 0.05 to 0.2sq.m per
capita.
→Bottom layer –thick graded
layers of gravel or crushed
stone – size varying from
15cm at bottom to 1.25cm at
top.
→Top layer – 10 to 15cm thick
coarse sand layer.
→Open jointed under drained
pipe – 15 cm dia – 5 to 7 cm
spacing – below gravel layer
in valleys
32

Sludge drying beds
→Surrounded by brick wall
rising about 1m above the sand
surface.
→Distribution troughs -
Opening:15 cm x 20 cm –
spacing: 2m
→Sewage sludge from digestion
tank is spread over the top to a
depth of 20 to 30cm.
→2 weeks - 2 months to dry the
sludge.
33

Mechanical methods of dewatering sludge
By vacuum filtration or by high speed centrifuges.
50% moisture removed
High speed centrifuges requires only a small area.
Before vacuum filtration the sludge is washed by the process of
elutriation to remove the organic and fatty acids.
34

Disposal by dumping into sea.
Disposal by burial into trenches.
Disposal by incineration.
Multiple hearth furnace
Fluid bed furnace
Flash type furnace
Infra red (Electric or radiant heat) furnace.
The process of reducing the moisture content in the sludge.
Using sludge thickener or concentrator unit.
Three types of thickening units.
Gravity thickeners
Flotation thickeners
Centrifugal thickeners
35
SLUDGE DISPOSAL
SLUDGE THICKENING

Completely covered, horizontal continuous flow type primary
sedimentation tank.
Detention period =12 to 36 hrs
Has an extra provision for digestion of the settled sludge.
Works under the principle of anaerobic decomposition.
Removes 60-70% of the dissolved matter in it.
36
SEPTIC TANK

38
SEPTIC TANK
DESIGN CONSIDERATIONS
A septic tank should be capable
of storing the sewage flow
during the detention period and
an additional volume of sludge
for 6 months to 3 years,
depending upon periodicity of
cleaning.

 DESIGN CONSIDERATIONS
 Water closets only, then sewage
flow = 40-70lpcd
 8 to 10 persons: 1400L
(min.cap. of septic tank)
 Water closets and sullage, then
sewage flow = 90-150lpcd
 8 t0 10 persons: 2250L
(min.cap. of septic tank)
 Rate of accumulation of sludge
= 30 L/person/year
 Freeboard : 0.3 to 0.5m
39
SEPTIC TANK

Inlet and outlet baffles
 Extended upto 20-22cm above
top sewage line
 Atleast 7.5cm below top
covering slab
 Inlet penetrating level: 30cm
below the top sewage line
 Outlet penetrating level: 40%
of the depth of sewage.
40
SEPTIC TANK

Detention period:
 commonly adopted is 24 hrs.
Length to width ratio
 L= 2B to 3B
 B not less than 90cm
 Depth = 1.2 to 1.8m
41
SEPTIC TANK

DESIGN PROBLEM
Design the dimensions of a septic tank for a small colony of 150
persons provided with an assured water supply from the municipal
head-works at a rate of 120lpcd. Assume any other data needed.
Assumptions made
% of water supplied that becomes sewage = 80%
Detention time = 24hrs
Rate of deposited sludge = 30L/capita/day
Period of cleaning = 1 yr
Depth = 1.2 to 1.8m
 L = 2B to 3B
42
SEPTIC TANK

DESIGN PROBLEM
Solution
The quantity of water supplied = per capita rate x population
= 120 x 150L/day = 18000L/day
The quantity of sewage produced = 80% x 18000 = 14400L/day
The quantity of sewage produced during the detention period
= the capacity of the septic tank
= (14400 L/day) x detention time
= (14400 L/24hrs) x 24hrs
= 14400L
43
SEPTIC TANK

DESIGN PROBLEM
Solution
The volume of sludge deposited
=rate of sludge deposition x no.of persons x period of cleaning
= 30L/capita/year x 150capita x 1yr
= 4500 L
∴ Total required capacity of the tank
= volume of sewage + volume of sludge
= 14400 + 4500 = 18900L = 18.9 cu.m
44
SEPTIC TANK

DESIGN PROBLEM
Solution
Take depth of tank = 1.5m
The surface area of tank = (18.9/1.5)sq.m = 12.6 sq.m
Take L:B = 3:1
L x B = 12.6 sq.m
→ B = 2.05m ~ 2.1m
→L = (12.6sq.m/2.1m) ~ 6m
Provide free board = 0.3m
→Depth = 1.5 + 0.3 =1.8m
∴ Dimension of septic tank = 6m x 2.1m x 1.8m
45
SEPTIC TANK

DISPOSAL OF EFFLUENT FROM THE SEPTIC TANK.
Effluent contains 200 to 250mg/l of putrescible organic matter.
BOD is high – 100 to 200mg/L
Three methods of disposal
1. Soil absorption system
2. Biological filters
3. Upflow anaerobic filters.
46
SEPTIC TANK

Soil absorption system
Disposal of effluent on land.
Only adopted when sufficient land is available.
The soil must be sufficiently porous.
Percolation rate not more than 60minutes per cm.
Types - seepage pit or soakpit
- dispersion trench
47
SEPTIC TANK

 Seepage pit or soak pit
Circular covered pit.
Effluent allowed to be soaked or absorbed into the surrounding soil
Either empty or filled with stone or brick aggregates.
Effluent pipe at a depth of 0.9m from top – anti mosquito measure.
Percolation rate not more than 30min per cm
48
SEPTIC TANK

 Seepage pit or soak pit: empty and lined
49
SEPTIC TANK

 Seepage pit or soak pit: unlined and filled with stone or brick
50
SEPTIC TANK

DESIGN PROBLEM
a.Design a septic tank for the following data:
No. of people = 100
Sewage/capita/day = 120L
De-sludging period = 1 year
Length:width = 4:1
b. What would be the size of soak well if the effluent from this septic
tank is to be discharged in it. Assume percolation rate through the
soak well to be 1250 l/cu.m/day.
51
SEPTIC TANK

DESIGN PROBLEM
Assumptions made
For septic tank
Detention time = 24hrs
Rate of deposited sludge = 30L/capita/day
Period of cleaning = 1 yr
Depth = 1.2 to 1.8m
52
SEPTIC TANK

Solution
Design of septic tank : 6m x 1.5m x 1.8m
Design of soak pit
Sewage outflow =12000L/d
Percolation rate =1250L/cu.m/day
Volume of filtering media for the soak well
= outflow/ Percolation rate = 9.6cu.m
Take depth = 2m
Area of soakwell = 4.8 sqm dia = 2.5m
53
SEPTIC TANK

Dispersion trench
Effluent from septic tank to a masonry chamber - distribution box.
From distribution box effluent is uniformly distributed through an
underground network of open jointed pipes into absorption trenches
– dispersion trenches.
Dispersion trenches – filled with gravel and well graded aggregate.
Not used where plants with fibres roots are grown – causes
blockage.
Percolation not more than 60minutes per cm.
54
SEPTIC TANK

55
SEPTIC TANK

56
SEPTIC TANK

Minimum absorption area pits or trenches can be calculated on the
basis of maximum allowable rate of effluent application (q in
l/m2/day) given as q = 130√t
Where, t = standard percolation rate in minutes.
According to IS:2470 (Part II) 1985 q = 204/√t
Shall not be installed closer than 18m from any source of drinking
water.
Not closer than 6m from any habitable building.
57
SEPTIC TANK

Design problem
Estimate the size of a septic tank (length to width ratio = 2.25,
liquid depth = 2m with 300mm freeboard), desludging intervals in
years and the total trench area (sq.m) of the percolation field, for a
small colony of 300 people. Assume water supply of 100 lpcd,
waste water flow at 80% of water consumption, sludge production
of 0.04 cu.m per capita per year, and the retention time of 3 days at
start up. Desludging is done when the tank is one-third full of
sludge. A percolation test indicated an allowable hydraulic loading
of 100L per sq.m per day.
58
SEPTIC TANK

Given:
L/B = 2.25
Dw = 2m
Free-board = 0.3m
Population = 300
Water supply = 100 lpcd
Waste flow = 80% of water supplied
Sludge production = 0.04m3/c/year
Hydraulic loading = 100L/m2/day
59
SEPTIC TANK

Solution
Water supply to the colony =100 lpcd x 300persons =30000 L/d
Sewage produced per day = 80% x 30000 L/d =24000L/d
Sewage produced in 3 days of retention period
=3d x 24000L/d = 72000L=72m3
Desludging is done when the tank is filled upto ⅓ of its capacity (C).
Hence, sludge volume collected is C/3.
∴ Capacity (C)
= max. sewage volume retained + sludge volume retained
⇒C = 72m3+ C/3
⇒C = 108m3
∵C = L x B x Dw
108m3 = (2.25B) x B x 2m
60
SEPTIC TANK

Solution
⇒ B = 4.9m
⇒ L = 2.25 x 4.9m = 11.1m
⇒ D = Dw + freeboard = 2.3m
∴ Tank size = 11.1m x 4.9m x 2.3m
Sludge volume removed while desludging
= C/3= (108m3
/3) = 36m3
Sludge producd per year =0.04m3/capita/year x 300persons
=12m3/year
∴ 36m3 of sludge will be produced in
= (36m3)/(12m3/year) = 3years
∴ Desludging period = 3years
61
SEPTIC TANK

Solution
∴ Trench area required = 240m2.
Design problem
Design the absorption field system for the disposal of septic tank
effluent for a population of 100 persons with sewage flow rate of
135lpcd. The percolation rate for the percolation test carried out at
the site of the absorption field may be taken as 3 minutes.
62
SEPTIC TANK

2. Biological filters
 Used where soil has high percolation rate (> 60min/cm)
 Used in water logged areas
 Septic tank effluent treated further by coating with organic
medium.
 Much of the polluting matter gets oxidized.
 Requires ample ventilation and an efficient system of under
drainage system.
3. Upflow anaerobic filters
 Used where soil has high percolation rate (> 60min/cm)
 Used in water logged areas
63
SEPTIC TANK

64
SEPTIC TANK

ADVANTAGES & DISADVANTAGES OF SEPTIC TANK.
Advantages
Easily constructed
No skilled supervision required
No moving parts so no maintenance problems
Reasonable cost compared to the sanitation provided.
Considerably reduce the SS & BOD
Sludge volume to be disposed off is quite less compared to normal
sedimentation tank. Volume – about 60%, Weight – about 30%.
Effluent can be disposed off without much trouble.
Best for isolated rural areas, isolated hospitals, isolated buildings
etc.
65
SEPTIC TANK

ADVANTAGES & DISADVANTAGES OF SEPTIC TANK.
Disadvantages
If not functioning properly, effluents will become very foul.
Size increases in case of serving many people.
Leakage from top cover may cause bad smell and environmental
pollution.
Periodical cleaning, removal and disposal of sludge remains a
tedious problem.
Working of a septic tank is unpredictable and non uniform.
66
SEPTIC TANK

Designed by Mr. Karl Imhoff
Improvement over septic tank.
Incoming sewage and sludge produced not allowed to get mixed up.
Effluent not allowed to carry with it a large amount of organic load.
Also known as two storey digestion tanks.
Sludge removed is dried up and then disposed off.
67
IMHOFF TANK

CONSTRUCTION DETAILS &WORKING
Double chamber rectangular tank.
Upper chamber: sedimentation chamber or flowing through
chamber.
› Sewage flow at a very low velocity – solids settle down to the
bottom of sedimentation chamber.
› Bottom side slope: 1.25V:1H
› Entrance slot at the lowest point of this chamber.
Lower chamber: digestion chamber.
› Sludge gets digested due to anaerobic decomposition.
› Divided into a number (3 to 4) of inter connected
compartments
69
IMHOFF TANK

Lower chamber: digestion chamber.
› Hoppered bottom: 1:1 side slopes – sludge concentrates here.
› Digested sludge removed, with flow under hydrostatic
pressure of 1.2 to 1.8m, periodically from the hoppered
bottom through the cast-iron desludging pipes.
› Only completely digested sludge removed, left out sludge
used for seeding.
Gas vent or scum chamber:
› Provided above the digestion chamber and along side the
sedimentation chamber.
70
IMHOFF TANK

Neutral zone:
› Provided to prevent the particles of sludge or scum from
entering into the sedimentation chamber from the digestion
chamber.
› Scum and sludge maintained atleast 45cm above and below
the slots, respectively.
71
IMHOFF TANK

SEDIMENTATION CHAMBER
Rectangular shape
Detention period = 2 to 4 hrs, usually 2hr
Flow through velocity ≯ 0.3m/min
Surface loading ≯ 30000L/m2 of plan area/day.
Length of tank ≯ 30m
Length to width ratio = 3 to 5
Depth = 3 to 3.5m
Freeboard = 0.45m
73
IMHOFF TANK

DIGESTION CHAMBER
Capacity = 57 litres per capita. But, in warmer climate
capacity = 30 to 40 litres per capita.
 GAS VENT OR SCUM CHAMBER
Surface area = 25 to 30% of the area of horizontal projection
of the top of the digestion chamber.
Width of a vent ≥ 60cm
74
IMHOFF TANK

ADVANTAGES AND DISADVANTAGES OF IMHOFF
TANK
Advantages
 Combines the advantages of septic tank and sedimentation
tank.
Do not require skilled supervision while operating.
60 to 65% removal of solids.
30 to 40% removal of BOD.
75
IMHOFF TANK

ADVANTAGES AND DISADVANTAGES OF IMHOFF
TANK
Disadvantages
 depth of tank is more – makes it uneconomical.
May give out offensive odours, when improperly operated.
Unsuitable where sewage is highly acidic.
Have a tendency to foam or boil – causes scum to go up –
force the sludge particle to enter the sedimentation tank.
No control over the operation – unsuitable for large
treatment plants – useful only for small cities and
institutions.
Mostly they have become obsolete these days.
76
IMHOFF TANK

Sludge treatment and disposal 1

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