11/6/2023

Water Supply and Wastewater

Management

6506CVQR

Week 9

Sludge Treatment
By
Dr. Reem F. Digna

Lecture Outline

In this lecture, we will discuss the following points:

 Sources of sludge;
 Calculating the sludge quantity;

 Sludge thickening; and

 Sludge thickening conditioning and dewatering.

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Learning Outcomes

The learning outcomes of the current lecture are:

 Define the sources of sludge.

 Estimate the sludge quantity.

 Understand the basics of sludge treatment, including thickening, conditioning and

dewatering.

Sludge Treatment
What is the Sludge?
Sludge is a mixture of water and solid materials removed from water or wastewater. The
sludge is divided into two types, Primary and secondary sludge.
The primary sludge is settleable solids removed during the primary treatment stage, while the
secondary sludge comes from the secondary treatment stage (the bioreactors).
The sludge produced during the treatment process should be removed from the treatment
lines continuously because the capacity of tanks in the water/wastewater lines is not large
enough to store sludge.
What does it mean?

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Sludge Treatment
Sludge handling
Sludge must be removed from the treatment lines, treated, and disposed of.
The final disposal of sludge depends on the chemical composition of the sludge.
Generally, sludge could be utilized as fertilizer, landfilling, or incineration (in-situ or outside the
plant as part of energy production).
However, the sludge must be treated before the disposal.
The treatment includes many stages, such as sludge thickening and stabilization.
Removal of water from sludge (dewatering process) and the final disposal may account for as
much as 40% of the cost of water treatment.

Sludge Treatment
Sludge productivity (quantity)
1- Quantity of Primary sludge
𝒌𝒈
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒍𝒖𝒅𝒈𝒆( ) = 𝑬 × 𝑺𝑺 × 𝑸𝒊𝒏
𝒅𝒂𝒚
SS= suspended solids in the influent (kg/m3 )
E = The efficiency of the primary sedimentation tank (Figure 1)
Qin = influent flow (m3 /day)
To change the mass of sludge from kg/day to m3/day
𝒎𝟑 𝑴𝒂𝒔𝒔 𝒊𝒏 𝒌𝒈/𝒅𝒂𝒚
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒍𝒖𝒅𝒈𝒆( )=
𝒅𝒂𝒚 𝑿𝒑
Xp = concentration of solids in primary sludge, kg/m3

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Sludge Treatment
Sludge productivity (quantity)
1- Quantity of Primary sludge
𝒎𝟑 𝑴𝒂𝒔𝒔 𝒊𝒏 𝒌𝒈/𝒅𝒂𝒚
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒍𝒖𝒅𝒈𝒆( )=
𝒅𝒂𝒚 𝑿𝒑
Xp = =concentration of solids in primary sludge, kg/m3
𝑿𝒑 = 𝝆𝒔𝒍𝒖𝒅𝒈𝒆 × S
𝝆𝒔𝒍𝒖𝒅𝒈𝒆 = sludge density (kg/m3 )
S = solids fraction in sludge expressed as decimal fraction
Note: The overflow rate (OR) in Figure 1 is calculated as follows:
𝑸𝒊𝒏
𝑶𝒗𝒆𝒓𝒇𝒍𝒐𝒘(𝑶𝑹) =
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒂𝒏𝒌

Sludge Treatment
Example
A treatment plant with a radius of 20m receives 1000 m3 /hr of wastewater that contains a
suspended solids concentration of 1000 mg/L. The solids fraction in sludge is 4.5%, and the
sludge density is 1001 kg/m3 . Determine:
1. Quantity of primary sludge in kg/day
2. Quantity of primary sludge in m3/day

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Sludge Treatment
Solution
1- Quantity of primary sludge in kg/day
𝑘𝑔
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑆𝑙𝑢𝑑𝑔𝑒( ) = 𝐸 × 𝑆𝑆 × 𝑄
𝑑𝑎𝑦
The area of the tank : 𝐴 = 𝜋𝑟 = 𝜋20 =1256 𝑚
𝑚 𝑚
𝑄 = 1000 = 24000
𝑑𝑎𝑦 𝑑𝑎𝑦
𝑂𝑅 = = = 19.11 m/day
From Figure-1 : E=71%
SS=1000 mg/L = 1 kg/m3

Mass of Sludge = 0.71 × 1 × 24000 =17040

Sludge Treatment
Solution
1- Quantity of primary sludge in m3/day
𝑘𝑔
𝑋 =𝜌 × 𝑆 = 1001 × 0.045 = 45.045
𝑚
𝑘𝑔
𝑚 𝑀𝑎𝑠𝑠 𝑖𝑛
𝑑𝑎𝑦 17040 𝑚
Mass of Sludge = = = 378.288
𝑑𝑎𝑦 𝑋 45.045 𝑑𝑎𝑦

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Sludge Treatment
Sludge thickening

 Sludge thickening is usually the first step in sludge treatment, and it aims at reducing
the sludge volume by removing the majority of water content.
 Removing water from the sludge leads to the concentration of the solids content of
the sludge.
 The thickening process could be done using different methods, such as gravity,
flotation, centrifuges, gravity belts and rotary drums.

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Sludge Treatment
Sludge thickening
1- Gravity thickening
 It is a simple and commonly used method for sludge thickening.
 The gravity thickening is achieved by leaving the wet sludge in concrete tanks (usually
circular concrete tanks) for a suitable period, which allows the water to flow out of the
sludge.
 The gravity thickeners are similar in design to the sedimentation tanks, but the bed slop in
the gravity thickeners is more than that in the sedimentation tanks.
 The gravity thickeners are relatively slow and not suitable for light weight sludge.

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Sludge Treatment
Sludge thickening
1.1 Design Gravity thickening
 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑟 =

 Volume of applied sludge =

×
×
 Volume of thickened sludge =
×

 Solid retention time =

 Gravity thickeners usually have:

1. A diameter in the range of 10-24 m.
2. Depths in the range of 3-4 m.
3. Bed slops in the range of 1:4 to 1:6

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Sludge Treatment
Example
A treatment plant produces 450 kg per day of sludge with a density of 1001 kg/m3 . The solids
percentage in the produced sludge is 4.5%, and the solids loading is 45 kg/m2.day. The
thickness of the sludge blanket in the tank is 1.0 m, and the underflow and the solids capturing
rates are 9% and 96%, respectively. Determine:
1. The volumes of the applied and thickened sludge.
2. The solids retention time.

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Sludge Treatment

Solution

.
 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑟 = = = 10m2
.

 Volume of applied sludge = = = 9.99 m3/day

× . × /

× × .
 Volume of thickened sludge = = = 4.795 m3/day
× . × /

×
 Solid retention time = = 2.086 day = 50.05 hrs
. /

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Sludge Treatment
Sludge thickening
2- Centrifugation based thickening
 In this method; the sludge is subjected to a
centrifuges force many times the force of
gravity, which results in the separation of the
solids from the solution.
Source: https://www.membranechemicals.com/water-treatment/sludge-thickening/
 There are basically three types of centrifuges:
 Disc nozzle
 Basket
Animation | Alfa Laval wastewater treatment decanter centrifuge for sludge
 Solid bowl thickening and dewatering - YouTube
Decanter Centrifuge Working Principle - YouTube

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Sludge Treatment
Sludge thickening

2- Centrifugation based thickening

 Generally, the centrifuges are small in size, simple, and have different capacities. But the
centrifuges are expensive, have high power consumption, and are not efficient in the
thickening of lightweight sludge.
 This method is usually used for thickening the sludge of the activated sludge. This
method should not be used for the thickening of primary sludge because it may contain
abrasive material.

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Sludge Treatment
Sludge thickening

3- Dissolved Air Flotation

 The gravity thickening process does not apply to low-density sludge. Thus, Dissolved Air
Flotation is more efficient with this type of sludge.
 In this method, the air is injected into the sludge under high pressure (several
atmospheres). The pressurized flow (sludge and air) is discharged into the flotation tank that
works under a pressure of 1.0 atmosphere.
 The pressurized air floats the sludge to the surface (separating it from the solution), which
will be collected by a skimming mechanism.

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Sludge Treatment
Sludge thickening

3- Dissolved Air Flotation

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Sludge Treatment
Sludge thickening

3- Dissolved Air Flotation

 The Dissolved Air Flotation process usually
includes chemical additives to aid the
flotation process., such as organic polymers.
Also, variation in pressurizing process is
expected (all or a small portion of the
sludge).
HUBER Dissolved Air Flotation Plant HDF - Animation - YouTube

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Sludge Treatment
Sludge thickening
3- Dissolved Air Flotation
 One of the key factors in the dissolved air flotation method is the ratio between the amount of
injected air and the solids (A/S ratio). The following equation could be used to calculate the A/S ratio:
𝑨 𝟏.𝟑×𝑨𝒔 ×(𝒇×𝑷 𝟏)×𝑹 𝑹
= , 𝑨=
𝑺 𝑺𝒂 ×𝑸 𝑺𝒖𝒓𝒇𝒂𝒄𝒆 𝒍𝒐𝒂𝒅𝒊𝒏𝒈 𝒓𝒂𝒕𝒆

As= Air solubility (mL/L); it depends on the temperature of water (See Table-1)

Sa= influent suspended solids (mg/L), F= fraction of air dissolved (usually =0.5)

R= pressurized recycle (m3/d), Q= mixed liquor flow

P= The pressure (atm)

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Sludge Treatment
Sludge thickening
3- Dissolved Air Flotation Table 1: Values of As in different water temperatures.
 Use the following equation to
change the pressure from kPa to
atm
𝑝 + 101.35
𝑃=
101.35

P= pressure (atm)
p= pressure(kPa)

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Sludge Treatment
Example
What is the pressurized recycle of air in a dissolved air flotation unit to thicken the solids in activated
sludge mixed liquor from 0.4% to about 4%? The water temperature is 20 oC, the A/S ratio is 0.009
mL/mg, and the Sa is 3100 mg/L. Assume:
 Applied recycle system pressure=275 kpa 𝑨 𝟏.𝟑×𝑨𝒔 ×(𝒇×𝑷 𝟏)×𝑹 𝑹
= , 𝑨 = 𝑺𝒖𝒓𝒇𝒂𝒄𝒆 𝒍𝒐𝒂𝒅𝒊𝒏𝒈 𝒓𝒂𝒕𝒆
 Fraction of saturation =0.45 𝑺 𝑺 𝒂 ×𝑸

 Sludge flow rate=500 m3 /d

 Surface loading rate=10 L/m2 .min Req:
Given:
Given: R= pressurized recycle
F= fraction of air dissolved =0.45
As= Air solubility (mL/L); it depends on the (m3/d) =????
temperature of water (See Table-1) = 18.7 mL/L Q= mixed liquor flow = 500 m3 /d
P= The pressure (atm) =(p+101.35)/101.35
Sa= influent suspended solids
p=275kPa
(mg/L)=3100mg/L 𝑨
= 0.009 mL/mg
𝑺
R= pressurized recycle (m3/d) =??????? Surface Loading rate = 10 L/m2.min

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Sludge Treatment
Solution
The air solubility is 18.7 mg/L (Table-1) Table 1: Values of As in different water temperatures.

𝑃 = 𝑝+101.35 = 275+101.35 = 3.713 𝑎𝑡𝑚

101.35 101.35

𝐴
= 1.3×𝐴𝑠×(𝑓×𝑃−1)×𝑅
𝑆 𝑆𝑎×𝑄
𝑚
1.3×18.7 𝐿 ×(0.45×3.713−1)×𝑅
0.009 𝑚 = 𝑚 3
𝑚𝑔 3100 𝐿 ×(500 𝑚 ×1000 𝐿3)
𝑑𝑎𝑦 𝑚
3
𝐿 𝑚
𝑅 = 855389 = 855.389
𝑑𝑎 y 𝑑𝑎 𝑦

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Sludge Treatment
4- Gravity Belt Thickener

Gravity belt thickening separates water from the sludge under

gravity through a moving permeable medium (a moving belt)
on which the sludge sits. This method does not require high
Sludge treatment − gravity belt thickening | Sludge Processing
power, and it has excellent solids capture efficiency (90−98%).
The efficiency of this method depends on the hydraulic loading
rate (HLR) per unit belt width and the belt speed. The HLR is
usually 7 to 35 m3/h per m width of the belt, at belt speed
https://youtu.be/sAqKflyW_Uw
300-1200 m/h. Gravity belt thickener for thickening sludge and slurries - YouTube

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Sludge Treatment

4- Gravity Belt Thickener

However, the need for cationic polymers to neutralise the sludge and the blocking of the

permeable belt are the main drawbacks of this method.

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Sludge Treatment

5- Rotary Drum Thickener:

Rotary Drum Thickener is similar to a gravity belt

thickener and achieves the separation of free water
through a rotating porous media. The porous media
can be a drum with wires, stainless steel fabric,
polyester fabric, or a combination of stainless steel Source:
Sludge treatment − rotary drum thickening | Sludge
and polyester fabric. Processing

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Sludge Treatment

Sludge Stabilization

 After the dewatering process, the sludge still contains microorganisms, some of which may
be pathogens, and are to be odourless.
 The need for the stabilisation process is determined by the final sludge disposal route.

 If the sludge is to be burned (incineration), stabilisation is not necessary.

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Sludge Treatment

Purposes of the stabilization process

 To break down the organic solids biochemically to make them more stable
and less odorous.
 To reduce the mass of sludge.
 To remove/minimise the pathogens.

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Sludge Treatment

Stabilization methods
 The selection of the sludge stabilization method depends on the disposal method.

 Two basic stabilisation processes are in use, which are:

 Aerobic Digestion: The air is injected into the sludge to accomplish the digestion process.

 Anaerobic Digestion: it is carried out in closed tanks devoid of oxygen.

 Other sludge stabilization processes use chemical and physical means, such as chemical
stabilisation and thermal conditioning.

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Sludge Treatment

Stabilization methods
A. Aerobic Digestion:
 It is accomplished by aerating the organic sludge in an open tank.
 Commonly used at small plants.
 It has an excellent ability to remove odours and has a low capital cost.
 It must be followed by a settling tank unless the sludge is to be disposed of on land in
liquid form.
 Slow (10-20 days).

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Sludge Treatment

Stabilization methods
B. Anaerobic Digestion:
 This process is carried out in airtight tanks (no oxygen), where the anaerobic
microorganisms stabilize the organic matter producing CH4 and CO2 .
 The digested sludge is stable, has a low pathogens count and is suitable for soil
conditioning.
 Although it metabolizes 50-60% of the organic matter, less than 10% of the organic
matter is converted to biomass. In addition, it has a high capital cost.

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Sludge Treatment
Stabilization methods
B. Anaerobic Digestion:

Standard-rate digestion High-rate digestion

• Suitable for treatment plants with a • Suitable for larger plants.

capacity <4000 m3/d. • The unit is heated to increase the
• Sludge is fed into the digester on an
metabolic rate of microorganisms.
intermittent basis.
• The supernatant is withdrawn and • the volume of sludge is unchanged &

returned to the secondary treatment solids content is reduced

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Sludge Treatment
Stabilization methods
B. Anaerobic Digestion:
Design of the anaerobic digestion

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Sludge Treatment
Stabilization methods
B. Anaerobic Digestion:
Design of the anaerobic digestion
The volume of the standard-rate digester (VSD) is determined by the loading rate (𝑉1),
digestion period (𝑡1), sludge storage period (𝑡2) and sludge accumulation rate (𝑉2).
𝑉 +𝑉
𝑉𝑆𝐷 = ×𝑡 +𝑉 ×𝑡
2
Total remaining mass
Digested sludge accumulation rate (𝑉 ) ( )=
1000×Solids content

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Sludge Treatment
Example
A standard-rate digester receives sludge consisting of 75% organic matter and 25% inorganic matter.
The raw sludge loading rate is 77.2 m3/d, and the density of the sludge is 40 kg/m3.
This digester converts 60% of the organic matter into liquid and gas after 25 days. The digested sludge
has a solids content of 4.0%, and it must be stored for 85 days. Calculate the volume of this standard-
rate digester.
𝑉 +𝑉 Total remaining mass
𝑉𝑆𝐷 = ×𝑡 +𝑉 ×𝑡 Digested sludge accumulation rate (𝑉 ) ( )=
2 1000×Solids content
Given: Given: Req:
the loading rate (𝑉1) = 77.2 m3/d F digestion period (𝑡1) =25 days the volume of this standard-rate
Sludge storage period (𝑡2) = 85 days digester (VSD)
Organic matter Fraction= 75%
Solid content = 4%
Inorganic matter Fraction= 25%
Sludge Density = 40 kg/m3

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Sludge Treatment
Solution
 𝑇ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑢𝑑𝑔e = 77.2 × 40 = 3088

 Organic fraction = 3088 × 0.75 = 2316

 Inorganic fraction remaining = 3088 × 0.25 = 772

 Organic fraction remaining = 2316 × (1 - 0.6) = 926.4
 Total mass remaining = 926.4 + 772 = 1698.4 kg/d
Total remaining mass 1698.4
 Digested sludge accumulation rate (𝑉 ) ( )= =
1000×Solids content 1000×0.04
=42.46

. .
 the volume of this standard-rate digester (VSD) 𝑉𝑆𝐷 = ×𝑡 +𝑉 ×𝑡 = × 25 + 42.46 × 85
𝑉𝑆𝐷 = 59.83 × 25 + 3609.1 = 1495.75 + 3609.1 = 𝟓𝟏𝟎𝟒. 𝟖𝟓𝑚

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Next Lecture:

Solutions for Water Supply

Thank you very much

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