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EAT 356 WATER & WASTE
WATER ENGINEERINGMAHYUN AB WAHAB
SCHOOL OF ENVIRONMENTALENGINEERING
UNIVERSITI MALAYSIA PERLIS
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TODAYS OBJECTIVE
CO3:
Ability to DEFINE, DESCRIBE,DESIGN and EXPLAIN basicstructure involved in physical
and biological unit processapplied in wastewatertreatment in Malaysia
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30 minutes ASSIGNMENT
One sewerage with 225mm
diameter put at 1:200 slope.
Calculate high and sewage
velocity if its flowrate is
432m3/day. Assume n = 0.013
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Please comment
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Introduction
Why we treat sewage or wastewater?
cityWastewaterTreatment
Plant
river
BOD5= 280mg/l
SS = 360 mg/l
Standard A
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Degree of discharge
Discharge by settlement
EXAMPLE for normal composition of sewage
Organic and inorganic
Organic
solid
(1)
Inorganic
(mineral)
(2)
Total
Organic
(1+2)
BOD5
Settlement
suspended solid
39 15 54 19
Non settlement
Suspended solid
26 10 36 23
TOTAL 90 42
Dissolved solid 80 80 160 12
Table 1 :Example of one sewage sample (all unit in mg/l), Bache, 1989
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What are the conclusion??
54/90 = 60% from total suspended solid can settle
19/42 = 45% from BOD5can discharge through
settlement Settlement (physical operation) ONLYcannot
remove all BOD5 In other word, if we just use sedimentation tankto
treat sewage, its still have 55% BOD5 in our finaleffluent
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Determination of unit operation
In treating sewage, a few different unitoperation involved, either physical,
biological or chemical unit operation. This unit operation also can classified to
primary (physical), secondary (biological) or
tertiary treatment. Every unit has different discharge efficiency
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Example
Primary
treatment
Secondary
treatment
BOD 280SS 360
*BOD 150*SS 144
BOD 20SS 50
o ASSUME IN THIS EXAMPLE % PRIMARY TREATMENT
DISCHARGE FOR BOD IS 40% AND SS IS 60%
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Example
From previous figure, if we have to complystandard A, we need biological unitoperation that can discharge
((150-20)/150) (100) BOD5 =% BOD5 ((144-50)/144) (100) SS = ..% SS
So, we have to choose suitable unitoperation
o ASSUME IN THIS EXAMPLE % PRIMARY TREATMENT
DISCHARGE FOR BOD IS 40% AND SS IS 60%
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TYPE OF TREATMENT
Optional treatmentPhysical Chemical Biology
screensing Precipitation Activated sludge
Mixing Disinfection Rotating Biological
Contractors (RBC)Flocculation Absorption Aerated lagoon
Sedimentation Neutralization Trickling FilterFloatation Oxidation
Filtration Settling
General
A few technique to treat sewerage depend on type ofsewerage.
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Cont
Conventional Normally we just apply physical and biological
treatment for municipal sewerage
When we use chemical treatment?
type of treatment can classified as belowPre-treatment
First stage in treat sewerage
Physical treatment and one of the primary treatment
Objective : protect incoming treatment from course objectsuch as wood, rock, metal and etc.
Example : screens, communitor, grit chamber,sedimentation tank
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Cont
Primary treatment Refer to pre-treatment process and primary
sedimentation tank
Objective : remove suspended solid fromwastewater, so that biological plant will not overloaded.
Main function : remove big portion of suspendedsolid from wastewater and a few of BOD
5 Wastewater volume in primary treatment have tocontrol. Not too slow and too fast. WHY?
Effluent from primary sedimentation tank are calledsludge
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Cont
Secondary treatment
Refer to biological plant
Objective : remove organic biodegradable and suspended
solid that cannot remove in primary treatment
Sludge treatment
Beside primary sedimentation tank, sludge also can find insecondary sedimentation tank (after biological plant)
Its mean, output from biological plant have to settle in othertank, commonly called as secondary sedimentation tank.
Sludge also can perform in other process such as septic tank,Imhoff tank and oxidation pond.
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Cont
Tertiary treatment
Advanced treatment
Done when high standard of wastewaterrequired
Its also done when we want to remove specificparameter such as nitrogen and phosphorus
Normally include absorption, reverse osmosisand etc.
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Cont
Practically, not all phase or type oftreatment are use
Its engineer responsibility to decide whichunit suitable for the wastewater they haveto treat.
For example,
influent screens PrimarySedimentation
tank
Aerated lagoonActivated
sludge
SecondarySedimentation
tank
effluent
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Table : Sewage Treatment Method
Sewage
inflow
Preliminary
Treatment
Primary
Treatment
Secondary
Treatment Tertiary Treatment
effluent
discharge
screensing sedimentationActivatedsludge filtration
grit removal floatation biofiltration disinfection
grease tank sedimentation tertiary ponds
pre-aeration
flowmeasurem
ent
flow balancing
removal ofrags,
rubbish,grit, oil,grease
removal ofsettleable and
Floatablematerials
Biologicaltreatment to
Removeorganic andSuspended
solids
biological andchemical treatmentto remove nutrients
and pathogens
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Primary treatment
Screens
Objective : remove course object to avoid
problem in the next unit treatment First unit in wastewater treatment unit series
2 type of screens
Mechanical screensManual screens
2 type of screens
Course screens
Fine screens
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Cont
Mechanical screens
Manual screens
Manual screens
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Cont
Parameter Design criteria
Manually raked Mechanical raked
Flowrate, Q Qpeak= ? Qpeak= ?
Maximum clear spacing 25mm 25mmSlope to vertical 0-45 0-45
Max. approach velocity 1.0m/s 1.0m/s
Max. flow through velocity 1.0m/s 1.0m/s
Min. freeboard 150mm 150mm
Estimated volume of screens pervolume of sewage
30m3/106 m3 See figure
Storage period of screens 7 days 7 days
Screens thickness 25mm N/A
Washing and dewatering of screens No Yes
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Cont
After we know a few design criteria, so, width ofscreens chambercan determine by this formula
Where,
W = chamber width (m)
B = screensing thickness (mm)
S = maximum clear spacing (mm)
F = Qpeak (m3/s)
V = maximum flow through velocity (m/s)
D = depth (m)
( )B S FW XS VD
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Cont
Screens output quantity based on sewageage and clear spacing.
Sewage volume is between 1.3x10-6 to3.67x10-5per m3 flowrate with average valueis 1.5x10-5m3/m3flowrate (Mc Ghee, 1991)
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SKETCH ON SCREENS DESIGN
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Example
One wastewater treatment plan are planned tocater 50,000 people. Determine width of screenschamber and appropriate number of screens
thickness, based on these design criteria: People, P = 50000
Maximum clear spacing = 25mm
Screens thickness = 10mm
q = standard water consumption for Malaysia
Qpeak= 4.7 p-0.11DWF
Volume = 0.9m/s
Depth = 0.85m
FROM QUESTION WE KNOW THAT
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FROM QUESTION WE KNOW THAT,
Screens thickness B = 10 mm 0.01 m
Maximum clear spacing S = 25 mm 0.025 m
Volume V = 0.9 m/s
Depth D = 0.85 m
QPEAK
= 4.7 p-0.11 DWF
DWF = qP
11250 m3/day
Q peak = 34384.577 m3/day
0.39797 m3/s
W = (B+S) x F
S VD
W = 1.4 x 0.5202
W = 0.73 m
W is taken as 0.8 m
Number of screensing thickness
No. of B = 22.8571
No. of B is taken as 23 pieces
so, new width for this chamber is
W = (B+S)*(no. of B)
0.805 m
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Comminutor
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Cont
Another primary unit operation like screens
However, its not functionas a screens but as a
grinder. Grind course thing to smaller size anddischarge
Comminutorwill not effect the rest of operation
with their output But, its still have their disadvantage.Comminutor will increase load for nextoperation(especially biological operation)
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Grit Removal
Grit is a inorganic material such as.and..and..
etcGrit will damage pumps by abrasion andcause serious operation difficulties insedimentation tanks and sludge
digesters by accumulation around andplugging of outlets and pump suctions
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Normally, grit remover is designed toremove inorganic material especially sandwith 0.2mm diameter or bigger (Hammer
and Hammer, 1996) In this research finding also shows 0.2mm
grit with specific gravity 2.65, will have1.2m/min velocity, it is higher thansuspended velocity of organic solid inwastewater (Tebbutt, 1991)
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Cont
Grit remover is one of the most important part incombined sewerage system.
Can u think what are the principle using in gritchamber?
A lots of different types of grit chamber, such as :
Grit chamber
Gravity Channel
Aerated Grit Channels
Vortex Grit Traps
Detritor
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Cont
Every grit removal have different designapproach
It is common practice to remove thismaterial by grit chambers .
So that, we will only learn how to design grit
chamber.
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Design Grit Chamber
Grit chambers are basin to remove theinorganic particles to prevent damage to the
pumps, and to prevent their accumulation insludge digestors.
Grit chamber is a shallow tank, commonlyrectangle, and designed to settle grit
Grit chambers are usually located after the barscreens and before the primary sedimentationtanks.
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Grit chambers are generally designed as longchannels
Grit chambers are designed to be cleanedmanually or by mechanically operateddevices
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Cont
Parameter Design characteristics
Flowrate, Q Qpeak = ?
Minimum retention time, t 1 minutes
Surface Load Rate, SLR (m3/m2.day)
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Design principle
Flowrate, Q
Using Qpeak= .
Retention Time, t t = volume / flowrate
Surface Load Rate (SLR)
SLR = Q / As = flowrate / area of tank
= m3/m2.day
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Cont
Horizontal velocity, Vh
Vhis a horizontal velocity enter
the tank You must know how to
differentiate between vertical
velocityand horizontal velocity Vertical velocity occur because
of gravity Vh
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Cont
We know that,
(3.1)
(3.2)
V
t Q
s
QSLR
A
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Cont
Its 2 option when we want to designgrit chamber
Starting from retention time, t
Starting from Surface Load Rate,
(SLR)
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Starting from retention time, t
Starting from Equation 3.1
We can determine retention time (t) according to
value given in previous table. With knowing t andQ value, we will get V.
With knowing depth of tank, we will get As (how?)
We have Q and As, so that we will get SLR
Compare value of SLR we obtained from thiscalculation and reference value in table given. Ifstill in range, its acceptable.
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Starting from SLR
Starting from equation 3.2
Determine SLR value from table
From Q and SLR value, we can obtain As With knowing depth of tank, we will get V
(how?)
From V, we will get retention time, t
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Example
One grit chamber is designed for a domesticwastewater treatment plant. This plant
receives waste from 8000 people. If Qpeakisused in this design, calculate length, width
and depth of this chamber. Given, SLR is1500 m3/m2 and horizontal velocity is 25
cm/s.
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SOLUTION
Qpeak= 4.7 p-011 DWF
SLR = 1,500 m3 /m2 .day
Vh = 25 cm/s Width (W) _ ?
Length (L) _ ?
Depth (d) _ ?
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DWF = q*P
= (0.225 m3/cap.day) (8000)
= .m3/day
Qpeak = 4.7 (8)-0.11 DWF, p = 8
= 3.739 (1800)
= 6730 m3/day
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Given, SLR = Q/As
= 1,500 m3/m2.hari
= 6730/As = 1,500 m3/m. hari As = 4.487m
2
Assume depth, d = 1 m; So,
Volume, V = As(d)
= 4.487 (1)
= 4.487 m3
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Retention Time, t
= V/Q
= (4.487 m3)/(6,730 m3/day) = ..day
= (6.67 x 10-4day) (24 x 3600 second)
= .. second,
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Given Vh = 25 sm/s
Vh = L/t
(Horizontal velocity = Length/time taken) 25 = L/57.6
L = (25 sm/s) * (57.6 s)
= 1,440 sm
= 14.4 m
= .m
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As = BL
4.186 m2 = B (14.5)
B = 0.288 m Take B as = m
We have assume depth, d =1.0m
Conclusion
L = .m; B = . m ; d =.m
L = .. m
B = .. m
d =.. m
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Grease tank
Grease tank is one of the pre-treatment unitdesigned to remove greasy material and lighterthan water.
Some of wastewater consist high composition offat, grease and oil. Example, wastewater from.
So that, we have to remove this greasy materialbefore its enter next unit operation.
But, grease tank is very rare in domesticwastewater treatment
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Equalization tank
Wastewater is held in the equalization tank toallow solids to begin settling.
Activated sludge from the leveling ponds is mixedwith the wastewater in the equalization tanks tobegin biological digestion of organic contaminants.
The equalization tank also helps to maintain a more
constant flow rate through the treatment plant.
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Sedimentation tank
Sedimentation tank function to settle suspendedsolid in wastewater or in easy word, sedimentationtank separate solids from the liquid stream
Sedimentation tank also known as clarifier
Theoretically, purpose of sedimentation tank is todivide two component, which is :
Sludge (settled suspended solid) Effluent
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Cont
The purpose of the scraper mechanism mountedinside the tank, is to collect the settled solids forremoval from the tank by pumping
In circular sedimentation tanks the clarifiermechanism has sludge scrapers attached to arotating arm scraping the sludge towards a centralhopper.
In rectangular clarifiers scrapers are carried alongthe tank bottom collecting the sludge into a troughor hopper at the influent end of the tank.
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Separation in sedimentation tank
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Design primary sedimentation tank
Primary sedimentation tank is one of the physicalunit operation
It just after grit chamber (if needed)
Function of this tank is ..
What will happen if this solid did not remove well?
Primary sedimentation tank usually design toremove 25-40% BOD and 50-70% SS
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Cont
A few types of primary sedimentation tank
Rectangular Tank
Circular Tank Upward flow tank
All this three types has their own advantages
and disadvantages
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PARAMETER UNIT DESIGN CRITERIA
Fl t 3/d Q k ?
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Flowrate m3/day Q peak = ?
Minimum retention time at Qpeak
Hour 1.5-2.0
SLR at Q peakCircular (
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Cont
Using Retention time
t = V/Q
Example
Volume = 1000 m3
Flowrate = 50000m3/ day
Retention time = ??
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Cont
Using SLR = Q / As
SLR = flowrate / area tank surface
= flowrate / b.l Unit ???
Q
l
b
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Horizontal velocity, Vh
Horizontal velocity is a even velocity enter the tank
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Cont
Differentiate between Vhand Vo(Settlementvelocity)
Settlement velocity, Vois a vertical velocity cause
by gravity Settlement theory stated design sedimentation
tank (Rectangular or Circular Tank) based onreference particle moving from top of the tank to
base of the tank (from point A to point B) Design have to be done with Vhand Voin almost
same magnitude, so that the particle can settleideally.
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Cont
REMEMBER : depth of tank NOTinfluencing particle settlement
Vh= Q / Ah Where,
Ah
= area of cross section
= b x d
d
b
Ah
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Cont
3 settlement cases
If settlement velocity > horizontal velocity
= particle will settle in front of tank If horizontal velocity > settlement velocity
= particle will settle at the end of tank
If horizontal velocity ~ settlement velocity
= ideal settlement will occur
Why we have to avoid case 1 and 2 ?
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Weir Overflow Rate, WOR
Settlement tanks must therefore be designed deepenough to allow all particles to settle, and also tohave flow such that settled solids are not disturbed
and carried over the weir at the outlet of thesettlement tank. In designing sedimentation tank,
Another important parameter in settlement tanks
is the rate at which water flows over the weir,known as the Weir Overflow Rate (WOR)
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Cont
WOR also have its own important function.
WOR control settled sludge.
WOR is a barrier at tank perimeter controllingdischarge of effluent from this tank to the nextunit operation
Length of WOR have to design properly so that,
settled sludge will not suspend again and flowwith effluent
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Cont
Weir overflow rate
Production of sludge
Settled sludge in primary sedimentation tank can be
estimate using this equation
Dry weight =
(kg.day)
(1DWF)suspended
solid
% removal ofsuspended
solid
x x
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Cont
Sludge is produced by settled suspended solid
Sludge production can be estimated from outputper capita
Rough estimation by Fuaad, 1990 is
0.0014m3/people.day or 50gram/capita.day
In designing sedimentation tank, volume of sludge
storage also needed. 2-5% of sedimentation tankvolume is using as a rough estimation.
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Cont
In calculating sludge storage volume,dislodging frequency has to determine.
If dislodging is done every day, so thatstorage volume needed is just one day.
Design approach of sedimentation tank is
similar with design of grit chamber. Its canbe started from SLR or from retention time.
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Design step by step
1. List out all the data given
2. What are the equation we have?
1. t =
2. SLR = ..3. Find As
4. If we got the depth of tank (From reference table) , we will getvolume of tank, V
5. From V, we will know retention time, t6. Check your design
1. SLR
2. Horizontal velocity (for rectangular sedimentation tank)
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EXAMPLE
One primary sedimentation tank are receiving influencefrom one housing area with 100,000 people. Waterconsumption is 200 liter/capita.day. If SLR is 30m3/m2.day,
design: One rectangular primary sedimentation tank
One circular primary tank with slope 7.5. Determinesludge produced.
Using this design data: Use Qpeak as design flowrate
SS in influence is 400mg/l
Efficiency of SS removal is 70%
Assume no slope
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EXAMPLE
One small city consist 15000 person. Designone upward flow tank to cater this city using
given design criteria: q = standard
Retention time at maximum flowrate is 2 hours
SLR = 35m3/m2.day
Sludge production = 0.0016m3/person.day
Dislodging frequency = every day
Slope to horizontal 60
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Secondary sedimentation tank, SST
Secondary sedimentation tank is a tank AFTER biologicaltank
Its purposely to settle sludge that produce in biological
tank. This sludge is pumped back into the inlet end of theprimary sedimentation tanks and settles with the rawsludge
We know in biological tank, a lots of sludge produced from
synthesis and microorganism oxidation process This tank has to design properly to make sure effluent
discharged comply to standard
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Cont
Compare to primary sedimentation tank that has 3option of tank, in SST, normally we designcirculartank
This SST is very important to specific unitoperation, especially ASP and aerated lagoon.
Without this tank, that two process cannot operate
because sludge produce cannot settle. As a result,effluent will consist high number of SS
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Cont
The sludge that collects in the SST is calledaerated sludge or activated sludge because
it is fully aerated.
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Design parameter for secondary sedimentation tank
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Parameter Value for PE 5000
Retentiontime
< 2 hours
SLR 30 m3/m2.day
WOR 150-180 m3/m.day
Minimum
depth
3 m
Solid LoadRate @ Qpeak
< 150 kg/m2.day
Flowrate Qpeak
B i A ti t d Sl d P
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Basic Activated Sludge Process
Primarysedimentation
tank
Aerated tank Secondarysedimentation
tank
Return sludge
effluent
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