SEDIMENTATION
Settling phenomena involved in an urban wastewater treatment plant (UWWTP)
Sabino DE GISI ENEA
Lectures for the course of “Wastewater Treatment”Second Cycle Degree (MSc Level) in Environmental Engineering
University of Padua, ITALY Prof Alessandro SPAGNI
29/11/201204/12/2012
Padua
Framework
• Introduction
• Type of sedimentation processes
• Primary and secondary sedimentation in an urban wastewater treatment plant (UWWTP)
• Goal of lessons
• Solid-Flux Analysis
• Technologyes of sedimentation tanks
• Exercises
• Design of a primary sedimentation unit
• Design of a secondary sedimentation unit in a CAS system (convenctional activated sludge)
Introduction
• Sedimentation is the separation of suspended particles from water by gravitational settling;
• The primary purpose is to produce a clarified effluent;
• It is one of the most widely used unit operations in wastewater treatment. It is used for:
• grit removal, particulate-matter removal in the primary settling basin;
• biological floc-removal in the activated sludge settling basin;
• solids concentration in sludge thickeners.
What is sedimentation?
Type of sedimentation
• According literature (Metcalf & Eddy, 2003), 4 type of settlingcan occur:
• discrete particle;
• flocculant;
• hindered (also calledzone);
• compression.
• For urban wastewater, the attention is focused above all discrete particle sedimentationand hindered/flocculant particlesedimentation.
Settling phenomena involved in wastewater treatment
Time (t)
Depth
(h)
Clear water region
Discrete settling region (type 1)
Discrete settling region (type 2)
Hindered zone
Compression region
Type of sedimentation
Settling phenomena involved in wastewater treatment
(Metcalf & Eddy, 2003)
Type of sedimentation
Settling phenomena involved in wastewater treatment
Water processing flow diagram for a large UWWTP
Type of sedimentation
Settling phenomena involved in wastewater treatment
Sludge processing flow diagram for a large UWWTP
Primary sludge
Parameter Unit Range Most frequent value
Production g/ab/year 30 - 90 50
Suspendid solids kgSS/m3 30 - 120 50
Volatile solids %SS 65 - 90 75 - 80
Calorific power kCal/kgSS 3,800 – 5,600 4,350
N %SS 1,5 - 5 2,5
P as P2O5 %SS 0,5 – 2,8 1,6
Primary and Secondary sedimentation
Metcalf & Eddy (2003)
Primary sludge characterization
Secondary biological sludge characterization
Secondary sludge
Parameter Unit Range Most frequent value
Production g/ab/year 30 - 50 40
Suspendid solids kgSS/m3 5 - 20 15
Volatile solids %SS 55 - 90 80
Calorific power kCal/kgSS 2,700 – 4,500 3,600
N %SS 5 - 10 7-8
P as P2O5 %SS 3 - 11 7
Goal
• The goal of this lesson is the design of the secondary sedimentation tanks and the presentation of the Solids-Flux theory;
• The design of primary sedimentation tanks will be developed in the next lession with the use of exercises.
Solid-Flux Analysis
• Is a method for calculation the area required for hindered settling based on an analysis of the solids (mass) flux;
• Data derived from settling tests must be available when applying this method;
• Work hypthotesis is a settling basin operating at steady state with a constant flux of solids in moving downward;
• The moviment of solids is due to 2 contribution:
• gravity (hindered) settling;
• bulk transport due to the underflow being pumped out and recycled.
Some information
where:SF1 = solid flux due to gravity – mass sedimentation (M L-2 T-1)SF2 = solid flux due to the extraction of the sludge from the bottom of the tank (M L-2 T-1)
Solid-Flux Analysis
Influent
Effluent
Borderline
Sludge
Clarified WW
Activated Sludge
q = inlet flowrate
qf = return activated sludge
X0 = SS concentration in the oxidation basin
Xf = SS concentration of activated sludge
Symbols
Z
Section (i) with surface A
With reference a generic section (i) with a fixed value of the concentration xi, solid-flux (SF)i is defined as the quantity of solids that crosses an horizontal surface unit per unit of time:
SF = SF1 + SF2 = (xi · vi) + (xi · qf/A) = (xi · vi) + (xi ·u) with u = qf/A = cost
Solid-Flux Analysis
Calculation of SF1 (mass sedimentation solid flux)x1 x2 x3 xi< < <
h(t1)
h
h(t2)
h(t3)
h(t4)
t
x1
x2
x3
v3
v1
v2
x
v
x1 x2
v1
v2
x
SF1
SF1(i) = xi · vi
• The determination of SF1 is carried out with a series of laboratory cylinders in which mixed aeration samples are introduced with different values of solids concentration (x).
• For each cylinder, the position of the interface water/sludge is reported, as a function of time.
• As visible in figure 1, mass sedimentation velocity for the initial concentration of the slurry (xi), is calculated as is the angular coefficient of the straight line
Fig. 1 Fig. 2 Fig. 3
Solid-Flux Analysis
Calculation of SF2 (extraction solid flux)
SF2 = xi · qf/A = xi ·u with u = qf/A = cost
where SF2 is the equation of a straight line passing through the origin and with the angular coefficient equal to u
SF2
x
SF2(i)
Calculation of SF (solid flux)
SF = SF1 + SF2
SF
SFL
SF
SF2
SF1
xxL xf
� First of all, the total solid-flux curve has a maximum value. Then, a minimum value can be observe.
� This minimum value is called limiting flux (SFL).
� The sedimentation tank must be fed with a flow value less than the limiting flux (SFL). Otherwise, the solids exit out of the tank within the clarified effluent.
� The tank surface useful for a correct function of secondary sedimentation (A) is equal to:
(q+qf) · x0
SFL
A =
xf - x
Influent
Solid-Flux Analysis
Determination of Return Activated Sludge flowrate (qf) - RAS
x0
Effluent
Secondary sedimentation tank
RAS
Flux
to sludge line
Air
Oxidation basin
Mass balance on V.C. (steady state condition)
MSS,In +/- Gen = MSS,Out + ∆(t)
0 = MSS,In – MSS,out 0 = x0 ⋅ q0 + qf ⋅ xf – (q+qf) ⋅ x
0 0
0
xqf =
qf, xf
q + qf
, xf
· q
Ricircolation RAS ratio
(CAS System)
Solid-Flux Analysis
Influentx0
Effluent
Secondary sedimentation tank
RAS
Air
Oxidation basin
qf, xf
q + qf
(CAS System)
A
Goal
Design of surface (A) of the secondary sedimentation tank in a CAS system.
Hypotesis
Sludge data (speed, concentration) regarding the project wastewater (and the mixed liquor) are taken from literature
SF = SF1 + SF2 = (xi · vi) + (xi · qf/A) = (xi · vi) + (xi ·u)
How we can use the Solid-Flux Analysis for the design of a new wastewater treatment plant?
Solid-Flux Analysis
Design of the secondary sedimentation tank with Solid-Flux Analysis
1. Data input (activated sludge at different value of xi)
x (kgSS/m3) 1 1.5 2 3 4 5 6 8 10
v (m/h) 6.72 6.10 4.80 2.40 1.00 0.55 0.34 0.15 0.07
FS1 (kgSS/m2/h) 6.72 9.15 9.60 7.20 4.00 2.75 2.04 1.20 0.70
2. Interpolation curve (vi ; xi)
Concentration xi (kgSS/m3)
Speed v
i(m
/h)
3. SF1 curve
Concentration xi (kgSS/m3)Solid-F
lux S
F (
kgSS/m
2/h
)
� q = inlet flowrate to the CAS system (m3/s)
� x0 = oxidation basin concentration (kgSS/m3)
i.e. 4-6 kgSS/m3.
� xf = target = value of RAS concentration (kgSS/m3)
i.e. 8-12 kgSS/m3.
SF1
Experimental curve
4. SFL calculation
u
Solid-F
lux S
F (
kgSS/m
2/h
)
Design of the secondary sedimentation tank with Solid-Flux Analysis
xf
P
SFL
SFL
xf
SFL
xf
Q
Solid-Flux Analysis
Concentration xi (kgSS/m3)
U =
SF2
xf
SFL
5. SF2 and u calculation
6. SF calculation
Solid-F
lux S
F (
kgSS/m
2/h
)
Design of the secondary sedimentation tank with Solid-Flux Analysis
SFL
xf
Solid-Flux Analysis
SF
7. Calculation of qf
SF1
SF2
Concentration xi (kgSS/m3)
xf - xx
qf = · q
8. Calculation Surface (A)
(q+qf) · x0
SFL
A =
The value of A surface of the secondary sedimentation tank allows to thicken the sludge to the xf value fixed
Technology
Large urban wastewater
treatment plant
Primary treatment
Secondary treatment
Technology
Primary sedimentation tanks
WW distribution well
Oxidation basins
Technology
Primary sedimentation tanks
Technology
Thomson effluent weir
150 50 150 50
75
150
90°
(values in cm)
Secondary sedimentation tanks
Technology
Technology
Secondary sedimentation tanks
Technology
Secondary sedimentation tanks
Sludge scrapers
Surface skimmer
Bridge
Drive unit
Inlet pipe
Technology
Secondary sedimentation tanks
Scum baffle
Effluent weir
Support
Technology
Secondary sedimentation tanks
Scum box
Inlet scum
Scum pipe
Versus Scum pit
Technology
Secondary sedimentation tanks
Scum box
Route of Scum
Rotation of bridge
Technology
Secondary sedimentation tanks
Effluent launder
Scum baffle
Surface skimmer
Effluent weir
Clarified wastewater’s route
Technology
Secondary sedimentation tanks
Cockpit divider
Technology
Secondary sedimentation tanks
Cockpit divider
Technology
Secondary sedimentation tanks
Thomson effluent weirCentral pivot
Scum boxBridge
Distribution system
Design of primary sedimentation units
Exercise 1
� Design the primary sedimentation units of a large wastewater treatment plant serving 65,000 ab. In order to ensure continuity of operation, two equal size units should be realized. In particular, calculate:
� the geometry of the single sedimentation tank.
1. Some consideration
Trattamenti preliminari Trattamenti primariTrattamenti biologici
e terziariDisinfezione
Ricettore
qPM
q
qPM
q -
qPB
qPM
qPM
- qPB
qPM
qPM
q
Design flowrates considered in a wastewater treatment plant
� qPM = max flowrate inlet in the plant
� (q24)C = average flowrate inlet in the plant
Desing Flowrate
Exercise 1
1. Some consideration
Characteristics parameters for the design and verification of a primary sedimentation tank (Metcalf & Eddy, 2003)
Characteristics parameters
Parameter Unit Range on (q24)C
Range on qPM
Primary sedimentation followed by biological secondary treatment HRT (τ) h 2 – 3 0,66 – 0,83 Surface hydraulic load (Cis) m3/m2/h 1.25 – 2.08 3 - 5 Depth(h) m 2 - 5 - Weir load (Cs) m3/m/d 125 - 500 -
Design of primary sedimentation units
Exercise 1
Design of primary sedimentation units
2. Data input and design parameters
The following parameters are considered:� equivalent population = 65,000 PE;� max flowrate inlet in the plant (qPM) = 65,000 m3/d;� average flowrate (q24)C = 13,000 m3/d;� number of tanks (N) = 2;� shape of single tank: radial;� Cis,max (on qPM) = 5 m3/m2/h;� τmax (on qPM) = 2 h;� hmin = 2.5 m;� CS = 125-500 m3/m/d.
3. Calculation of the minimum sedimentation tank surface(Ssed,min)
The following calculation are devepoled with refer to a single unit and considering these flowrate values:
� qPM/2 = 32,500 m3/d = 1,354.16 m3/h;� (q24)C/2 = 6,500 m3/d = 270.83 m3/h.
Exercise 1
Design of primary sedimentation units
4. Calculation of the minimum sedimentation tank surface(Ssed,min)
The minimum sedimentation tank surface is equal to:
2
23
3
maxis,
PMminsed, m 270.83
/h)/m5(m
/h)1,354.16(m
2C
qS ==
⋅=
5. Calculation of the real diameter (Dreal) and the real surface (Sreal) of the single sedimentation tank
The real diameter (Dreal) of the single tank is calculated from the minimum diameter (Dsed,min). Its value is equal to:
m 19m 18.57(m)3.14
270.834
π
S4D minsed,
minsed, →=⋅=⋅
=
2222
realreal m 283.38
4
)(m193.14
4
DπS =⋅=
⋅=
Exercise 1
Design of primary sedimentation units
6. Verification of Surface Hydraulic Load (Cis) with reference (q24)C
With reference (q24)C/2, the Surface Hydraulic Load (Cis) for the single tank is less than the maximum allowed value:
7. Calculation of the volume and depth of the single tank and choice of the commercial tank
For volume calculation, a hydraulic detention time (HRT) of 40 min (0.66 h) with reference qPM/2 flowrate is considered. The single tank volume is equal to:
/h/mm 2/h/mm 0.955)283.38(m
/h)270.83(m
2S
)(qC 2323
2
3
real
C24is <==
⋅=
33max
PM m 893.75(h) 0.66/h)(m 1,354.16τ2
qV =⋅=⋅=
m 2.5m 3.15)283.38(m
)893.75(m
S
Vh
2
3
real
>===
Exercise 1
Design of primary sedimentation units
Once determined values of the principal geometrical variables (volume, diameter and surface), choice of the commercial settler model should be made. With reference to the real diameter of 19m and considering table 1, Ecoplants PRTP-190 tank is assumed. The final characteristic parameters of the single sedimentation tanks are:
� h = 3.2 m;� D = 19 m;� S = 283.5 m2;� V = 283.5 (m2) x 3.2 (m) =
907.2 m3;� Engine power = 0.37 kW.
7. Calculation of the volume and depth of the single tank and choice of the commercial tank
Model Surface [m2] Flowrate [m3/d]
Diameter and depth [m] Engine power [kW] D h
PRTP-50 19.6 480 5 3.6 0.12 PRTP-60 28.3 690 6 3.6 0.12 PRTP-70 38.5 940 7 3.6 0.12 PRTP-80 50.3 1,230 8 3.6 0.12 PRTP-90 63.6 1,550 9 3.6 0.18 PRTP-100 78.5 2,240 10 3.5 0.18 PRTP-110 95.0 2,710 11 3.5 0.18 PRTP-120 113.1 3,220 12 3.5 0.18 PRTP-130 132.7 3,780 13 3.5 0.25 PRTP-140 153.9 4,390 14 3.5 0.25 PRTP-150 176.7 5,040 15 3.5 0.25 PRTP-160 201.1 5,730 16 3.5 0.25 PRTP-170 227.0 7,400 17 3.2 0.25 PRTP-180 254.5 8,300 18 3.2 0.37 PRTP-190 283.5 9,240 19 3.2 0.37 PRTP-200 314.2 10,240 20 3.2 0.37 PRTP-210 346.4 11,290 21 3.2 0.37 PRTP-220 380.1 12,390 22 3.2 0.37 PRTP-230 415.5 13,540 23 3.2 0.37 PRTP-240 452.4 14,750 24 3.2 0.55 PRTP-250 490.9 16,000 25 3.2 0.55 PRTP-260 530.9 17,310 26 3.2 0.55 PRTP-270 572.6 18,660 27 3.2 0.55 PRTP-280 615.8 20,070 28 3.2 0.55
Table 1. Technical data of radial sedimentation unit “type PRTP” of Ecoplants Inc.
Exercise 1
Design of primary sedimentation units
7. Calculation of the volume and depth of the single tank and choice of the commercial tank
View of “type PRTP-190” sedimentation tank (Ecoplants Inc., Italy)
Exercise 1
Design of primary sedimentation units
8. Verification of HRT on (q24)C
9. Verification of weir load on (q24)C
The last verification regarding the weir load (CS):
( ) h 3.35/h)270.83(m
)907.2(m
/2q
Vτ
3
3
C24min ===
The obtained value is greater than the maximun value generally considered on (q24)C = 3h. With this solution a more safety margin is guaranteed above all during the peak period (qPM).
( )/m/dm 500/m/dm 108.9
59.7(m)
/d)6,500(m
L
/2qC 33
3
s
C24s <===
where LS is the length of the tank circumference:
LS = D ⋅ p = 19 (m) ⋅ 3.14 = 59.7 m
Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
The following parameters are considered:
� average flowrate (q24)C = 13,000 m3/d;
� Inlet TSS concentration (TSS,in) = 350 gTSS/m3;
� Percentage TSS removal (ηTSS) = 65%;
� Primary sludge solids (S) = 4%.
Time [h]
Perc
enta
ge r
em
oval [%
]
τ = 3,35 h
(q24)C
%TSS = 65%
Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)(q
24)
TSSin
TSS, inM
TSS, rimossiM
(q24
)
TSSout
TSS, outM
FanghiM(secco)
(secco + umido)
C
C
Influent
(q24)C = 541,66 m3/h
TSSin = 350 g/m3
PTSS,in = ? Effluent
(q24)C = 541,66 m3/h
TSSout = ? g/m3
PTSS,out = ?
Primary sludge
PTSS,removed = ? (only dry matter)
Psludge = ? (dry matter + water)
Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
(q24
)
TSSin
TSS, inM
TSS, rimossiM
(q24
)
TSSout
TSS, outM
FanghiM(secco)
(secco + umido)
C
C
Influent
(q24)C = 541,66 m3/h
TSSin = 350 g/m3
PTSS,in = ?
Effluent
(q24)C = 541,66 m3/h
TSSout = ? g/m3
PTSS,out = ?
PTSS, in = (q24)C ⋅ TSSin = 541,66 (m3/h) ⋅ 350 (g/m3) = 189,6 kg/h
PTSS,out = MTSS, in ⋅ (1- ηTSS) = 189,6 (kg/h) ⋅ (1- 0,65) = 66,3 kg/h
PTSS,removed = MTSS, in - MTSS, out = (189,6 – 66,3) kg/h = 123,2 kg/h
1)
2)
3)
Primary sludge
Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
(q24
)
TSSin
TSS, inM
TSS, rimossiM
(q24
)
TSSout
TSS, outM
FanghiM(secco)
(secco + umido)
C
C
Influent
(q24)C = 541,66 m3/h
TSSin = 350 g/m3
PTSS,in = ?
Effluent
(q24)C = 541,66 m3/h
TSSout = ? g/m3
PTSS,out = ?
PTSS, in = (q24)C ⋅ TSSin = 541.66 (m3/h) ⋅ 350 (g/m3) = 189.6 kg/h
PTSS,out = MTSS, in ⋅ (1- ηTSS) = 189.6 (kg/h) ⋅ (1- 0.65) = 66.3 kg/h
PTSS,removed = MTSS, in - MTSS, out = (189.6 – 66.3) kg/h = 123.2 kg/h
1)
2)
3)
Primary sludge
This is the solid part of primary
sludge
Water?
+ Solids ?
S = =
Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
Water?PTSS
Psludge PTSS + Pwater
PTSS
Psludge = PTSS,removed/S = 123.2 (kg/h)/0.04 = 73,937.5 kgsludge/d
Solids ?
For the calculation of qsludge (volumetric flowrate of primary sludge), the sludge density (ρ)must be considered. Generally, its value is about 1,000 kg/m3. This hyphotesis is true if the wastewater solids value (S) is less than 10%.
qsludge = Psludge/ρsludge = 73,937.5 (kg/d)/ 1,000 (kg/m3) ≅ 74 m3/d
Design of secondary sedimentation units with the Solid-Flux Analysis
Exercise 2
1. Data input (activated sludge at different value of xi)
x (kgSS/m3) 1 1.5 2 3 4 5 6 8 10
v (m/h) 6.72 6.10 4.80 2.40 1.00 0.55 0.34 0.15 0.07
FS1 (kgSS/m2/h) 6.72 9.15 9.60 7.20 4.00 2.75 2.04 1.20 0.70
2. Interpolation curve (vi ; xi)
Concentration xi (kgSS/m3)
Speed v
i(m
/h)
3. SF1 curve
Concentration xi (kgSS/m3)Solid-F
lux S
F (
kgSS/m
2/h
)
� q = 125 (l/s)
� x0 = oxidation basin concentration = 3.4 kgSS/m3
� xf = RAS concentration = 12 kgSS/m3
SF1
Experimental curve
Design of secondary sedimentation units with the Solid-Flux Analysis
Exercise 2
4. SFL calculation
u
Solid-F
lux S
F (
kgSS/m
2/h
)
xf = 12 kgSS/m3
P
SFL = 3.6 kgSS/m2/h
xf
SFL
xf
Q
Concentration xi (kgSS/m3)
U =
SF2
SFL
5. SF2 and u calculation
= 12 kgSS/m3
3.6 kgSS/m2/h= 0.300 m/h
Design of secondary sedimentation units with the Solid-Flux Analysis
Exercise 2
6. SF calculation
Solid-F
lux S
F (
kgSS/m
2/h
)
SFL
xf
SF
7. Calculation of qf
SF1
SF2
Concentration xi (kgSS/m3)
xf - x
xqf = · q
8. Calculation Surface (A)
(q+qf) · x0
SFL
A =
(12-4) kgSS/m3
3.4 kgSS/m3
· 125 (l/s) = 49 l/s=
2,130
3.6= = 592 m2
References
• Bonomo L., Trattamenti delle acque reflue, McGraw-Hill Companies, Srl, Publishing Group Italia, Milano, ISBN: 978-88-386-6518-9, 2008 (in Italian).
• Masotti L., Depurazione delle acque, tecniche ed impianti per il trattamento delle acque di rifiuto, 2nd Edizione, Edizioni Calderini, Bologna, Italia, ISBN: 88-7019-292-X, 1993 (in Italian).
• Metcalf & Eddy, Wastewater Engineering. Treatment and Reuse, 4th ed., McGraw Hill, New York (USA), 2003.
• IRSA-CNR, Istituto di Ricerca Sulle Acque - Consiglio Nazionale delle Ricerche, La protezione delle acqua dagli inquinamenti - Quaderno 2 -Aspetti biochimici e microbiologici dei processi depurativi naturali ed artificiali delle acque di rifiuto, Roma, 1974 (in Italian).
• Passino R., La conduzione degli impianti di depurazione delle acque di scarico, Ed Scient. A Cremonese, Roma, 1980 (in Italian).
• De Feo G., De Gisi S., Galasso M. (2012), Acque reflue, Progettazione e gestione di impianti per il trattamento e lo smaltimento, Dario Flaccovio Editore Srl, ISBN 9788857901183, 1244 pagine (in Italian). (http://www.darioflaccovio.it/libro.php/acque-reflue-df0118_C762)
Sabino DE GISI, Ph.D.
ENEA (Italian National Agency for the New Technology, Energy and Sustainable Economic Development), Technical Unit on Models, Methods and Technologies for the Environmental Assessment (UTVALAMB), Water Resource Management DivisionVia Martiri di Monte Sole 4, 40129 Bologna, [email protected]