1. SEDIMENTATION
Settling phenomena involved in an urban
wastewater treatment plant (UWWTP)
Lectures for the course of “Wastewater Treatment”
Second Cycle Degree (MSc Level) in Environmental Engineering
University of Padua, ITALY
Prof Alessandro SPAGNI
29/11/2012
04/12/2012
Padua
Sabino DE GISI
ENEA
2. 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)
3. Introduction
What is sedimentation?
• 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.
4. Type of sedimentation
Settling phenomena involved in
wastewater treatment
• According literature (Metcalf &
Eddy, 2003), 4 type of settling
can occur: Clear water region
• discrete particle; Discrete settling region (type 1)
• flocculant;
Depth (h)
Discrete settling region (type 2)
• hindered (also called
zone); Hindered zone
• compression.
Compression
• For urban wastewater, the region
attention is focused above all
Time (t)
discrete particle sedimentation
and hindered/flocculant particle
sedimentation.
5. Type of sedimentation
Settling
phenomena involved
in wastewater
treatment
(Metcalf & Eddy, 2003)
6. Type of sedimentation
Settling phenomena involved in
wastewater treatment
Water processing flow diagram for a large UWWTP
7. Type of sedimentation
Settling phenomena involved in
wastewater treatment
Sludge processing flow diagram for a large UWWTP
8. Primary and Secondary sedimentation
Primary sludge characterization
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
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
Metcalf & Eddy (2003)
9. 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.
10. Solid-Flux Analysis
Some information
• 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.
11. Solid-Flux Analysis
Symbols
Influent
Effluent
q = inlet flowrate
qf = return
activated sludge Clarified WW
Borderline
X0 = SS
concentration in the
oxidation basin
Sludge
Xf = SS Section (i) with
concentration of surface A
activated sludge
Z
Activated
Sludge
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
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)
12. Solid-Flux Analysis
x1 < x2 < x3 < xi
Calculation of SF1 (mass sedimentation solid flux)
h(t1) • The determination of SF1 is carried out with a series of
laboratory cylinders in which mixed aeration samples
h(t2)
are introduced with different values of solids
h(t3) concentration (x).
• For each cylinder, the position of the interface
water/sludge is reported, as a function of time.
h(t4)
• 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
h v SF1
SF1(i) = xi · vi
v3
v2 v1
v1
x3 v2
x2
x1
t x1 x2 x x
Fig. 1 Fig. 2 Fig. 3
13. Solid-Flux Analysis
Calculation of SF2 (extraction solid flux) SF2
SF2 = xi · qf/A = xi ·u with u = qf/A = cost
where SF2 is the equation of a straight line passing SF2(i)
through the origin and with the angular coefficient equal
to u
Calculation of SF (solid flux)
x
SF = SF1 + SF2
First of all, the total solid-flux curve has a maximum value.
Then, a minimum value can be observe.
SF This minimum value is called limiting flux (SFL).
SF 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.
SFL
The tank surface useful for a correct function of secondary
sedimentation (A) is equal to:
SF2 (q+qf) · x0
A=
SF1 SFL
xL xf x
14. Solid-Flux Analysis
Determination of Return Activated Sludge flowrate (qf) - RAS
Influent
x0 q + qf
Effluent
Oxidation basin
Secondary
(CAS System) Air sedimentation tank
qf, xf
RAS
Flux
Mass balance on V.C. (steady state condition)
, xf to sludge line
MSS,In +/- Gen = MSS,Out + ∆(t)
Ricircolation
0 0 RAS ratio
x
0 = MSS,In – MSS,out 0 = x0 ⋅ q0 + qf ⋅ xf – (q+qf) ⋅ x qf =
xf - x
·q
0
15. Solid-Flux Analysis
How we can use the Solid-Flux Analysis for the design of a new wastewater
treatment plant?
Influent
x0 q + qf
Effluent
Oxidation basin
A
Air Secondary
(CAS System)
sedimentation tank
Goal qf, xf
Design of surface (A) of the RAS
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)
16. 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) 3. SF1 curve
q = inlet flowrate to the
CAS system (m3/s)
Solid-Flux SF (kgSS/m2/h)
x0 = oxidation basin
concentration
(kgSS/m3) Experimental
Speed vi (m/h)
curve SF1
i.e. 4-6 kgSS/m3.
xf = target = value of
RAS concentration
(kgSS/m3)
i.e. 8-12 kgSS/m3.
Concentration xi (kgSS/m3) Concentration xi (kgSS/m3)
17. Solid-Flux Analysis
Design of the secondary sedimentation tank with Solid-Flux Analysis
4. SFL calculation 5. SF2 and u calculation
Solid-Flux SF (kgSS/m2/h)
SFL SF2
SFL Q SFL
P
u
xf
Concentration xi (kgSS/m3)
xf xf
SFL
U=
xf
18. Solid-Flux Analysis
Design of the secondary sedimentation tank with Solid-Flux Analysis
6. SF calculation 7. Calculation of qf
Solid-Flux SF (kgSS/m2/h)
SF1
SF
x
qf = ·q
SFL xf - x
SF2 8. Calculation Surface (A)
xf (q+qf) · x0
A=
Concentration xi (kgSS/m3) SFL
The value of A surface of the secondary
sedimentation tank allows to thicken the sludge to
the xf value fixed
32. Technology
Thomson effluent weir
Central pivot
Bridge Scum box
Distribution system
Secondary sedimentation tanks
33. Exercise 1
Design of primary sedimentation units
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
Desing Flowrate
q q PM q PM q PB q PM
Trattamenti biologici
Trattamenti preliminari Trattamenti primari Disinfezione
e terziari
q PM - qPB
q - q PM q PM
q
Design flowrates considered in a wastewater treatment plant Ricettore
qPM = max flowrate inlet in the plant
(q24)C = average flowrate inlet in the plant
34. Exercise 1
Design of primary sedimentation units
1. Some consideration
Characteristics parameters
Range on Range on
Parameter Unit
(q24)C qPM
Primary sedimentation followed by biological secondary treatment
HRT (τ) h 2–3 0,66 – 0,83
3 2
Surface hydraulic load (Cis) m /m /h 1.25 – 2.08 3-5
Depth(h) m 2-5 -
3
Weir load (Cs) m /m/d 125 - 500 -
Characteristics parameters for the design and verification of a primary
sedimentation tank (Metcalf & Eddy, 2003)
35. 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.
36. 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:
q PM 1,354.16(m 3 /h)
S sed, min = = = 270.83 m 2
C is, max ⋅2 5(m 3 /m 2 /h)
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:
4 ⋅ S sed,min 4 ⋅ 270.83
D sed,min = = (m) = 18.57 m → 19 m
π 3.14
π ⋅ D2 3.14 ⋅ 19 2 (m 2 )
S real = real
= = 283.38 m 2
4 4
37. 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:
(q 24 ) C 270.83(m 3 /h)
C is = = = 0.955 m 3 /m 2 /h < 2 m 3 /m 2 /h
S real ⋅ 2 283.38(m 2 )
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:
q PM
V= ⋅ τ max = 1,354.16 (m 3 /h) ⋅ 0.66 (h) = 893.75 m 3
2
V 893.75(m 3 )
h= = 2
= 3.15 m > 2.5 m
S real 283.38(m )
38. Exercise 1
Design of primary sedimentation units
7. Calculation of the volume and depth of the single tank and
choice of the commercial tank
Once determined values of the Table 1. Technical data of radial sedimentation unit
“type PRTP” of Ecoplants Inc.
principal geometrical variables
(volume, diameter and surface), Diameter and depth [m] Engine
Flowrate
Model Surface [m2] 3 power
[m /d] D h
choice of the commercial settler [kW]
PRTP-50 19.6 480 5 3.6 0.12
model should be made. With PRTP-60 28.3 690 6 3.6 0.12
reference to the real diameter of PRTP-70 38.5 940 7 3.6 0.12
PRTP-80 50.3 1,230 8 3.6 0.12
19m and considering table 1, PRTP-90 63.6 1,550 9 3.6 0.18
Ecoplants PRTP-190 tank is PRTP-100 78.5 2,240 10 3.5 0.18
PRTP-110 95.0 2,710 11 3.5 0.18
assumed. PRTP-120 113.1 3,220 12 3.5 0.18
The final characteristic parameters PRTP-130 132.7 3,780 13 3.5 0.25
PRTP-140 153.9 4,390 14 3.5 0.25
of the single sedimentation tanks PRTP-150 176.7 5,040 15 3.5 0.25
are: PRTP-160 201.1 5,730 16 3.5 0.25
PRTP-170 227.0 7,400 17 3.2 0.25
h = 3.2 m; PRTP-180 254.5 8,300 18 3.2 0.37
D = 19 m; PRTP-190 283.5 9,240 19 3.2 0.37
PRTP-200 314.2 10,240 20 3.2 0.37
S = 283.5 m2; PRTP-210 346.4 11,290 21 3.2 0.37
V = 283.5 (m2) x 3.2 (m) = PRTP-220
PRTP-230
380.1
415.5
12,390
13,540
22
23
3.2
3.2
0.37
0.37
907.2 m3; PRTP-240 452.4 14,750 24 3.2 0.55
PRTP-250 490.9 16,000 25 3.2 0.55
Engine power = 0.37 kW. 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
39. 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)
40. Exercise 1
Design of primary sedimentation units
8. Verification of HRT on (q24)C
V 907.2(m3 )
τ min = = = 3.35 h
(q 24 )C /2 270.83(m /h)
3
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).
9. Verification of weir load on (q24)C
The last verification regarding the weir load (CS):
(q 24 )C /2 6,500(m3 /d)
Cs = = = 108.9 m 3 /m/d < 500 m 3 /m/d
Ls 59.7(m)
where LS is the length of the tank circumference:
LS = D ⋅ p = 19 (m) ⋅ 3.14 = 59.7 m
41. 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; %TSS = 65%
Inlet TSS
Percentage removal [%]
concentration
(TSS,in) = 350
gTSS/m3;
Percentage TSS
removal (ηTSS) =
65%;
Primary sludge solids
(S) = 4%.
τ = 3,35 h
Time [h]
(q24)C
42. Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
(q 24)C
Influent TSSin
M TSS, in (q 24)C
(q24)C = 541,66 m3/h
TSSin = 350 g/m3 TSS out
Effluent
PTSS,in = ? M TSS, out
(q24)C = 541,66 m3/h
TSSout = ? g/m3
PTSS,out = ?
M TSS, rimossi (secco)
MFanghi (secco + umido)
Primary sludge
PTSS,removed = ? (only dry matter)
Psludge = ? (dry matter + water)
43. Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
(q 24)C
Influent TSS in
M TSS, in (q 24)C
(q24)C = 541,66 m3/h
TSS out Effluent
TSSin = 350 g/m3 M TSS, out
(q24)C = 541,66 m3/h
PTSS,in = ?
TSSout = ? g/m3
PTSS,out = ?
M TSS, rimossi (secco)
MFanghi (secco + umido)
Primary sludge
1) PTSS, in = (q24)C ⋅ TSSin = 541,66 (m3/h) ⋅ 350 (g/m3) = 189,6 kg/h
2) PTSS,out = MTSS, in ⋅ (1- ηTSS) = 189,6 (kg/h) ⋅ (1- 0,65) = 66,3 kg/h
3) PTSS,removed = MTSS, in - MTSS, out = (189,6 – 66,3) kg/h = 123,2 kg/h
44. Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
(q 24)C
Influent TSS in
M TSS, in (q 24)C
(q24)C = 541,66 m3/h
TSS out Effluent
TSSin = 350 g/m3 M TSS, out
(q24)C = 541,66 m3/h
PTSS,in = ?
TSSout = ? g/m3
PTSS,out = ?
This is the solid
part of primary
M TSS, rimossi (secco) sludge
MFanghi (secco + umido)
Primary sludge
1) PTSS, in = (q24)C ⋅ TSSin = 541.66 (m3/h) ⋅ 350 (g/m3) = 189.6 kg/h
2) PTSS,out = MTSS, in ⋅ (1- ηTSS) = 189.6 (kg/h) ⋅ (1- 0.65) = 66.3 kg/h
Water?
3) PTSS,removed = MTSS, in - MTSS, out = (189.6 – 66.3) kg/h = 123.2 kg/h + Solids ?
45. Exercise 1
Design of primary sedimentation units
10. Calculation of primary sludge production (Psludge)
Water?
PTSS PTSS
S= =
Psludge PTSS + Pwater
Solids ?
Psludge = PTSS,removed/S = 123.2 (kg/h)/0.04 = 73,937.5 kgsludge/d
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
46. Exercise 2
Design of secondary sedimentation units
with the 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) 3. SF1 curve
q = 125 (l/s)
x0 = oxidation basin
Solid-Flux SF (kgSS/m2/h)
concentration = 3.4
kgSS/m3
Experimental
Speed vi (m/h)
xf = RAS concentration curve SF1
= 12 kgSS/m3
Concentration xi (kgSS/m3) Concentration xi (kgSS/m3)
47. Exercise 2
Design of secondary sedimentation units
with the Solid-Flux Analysis
4. SFL calculation 5. SF2 and u calculation
Solid-Flux SF (kgSS/m2/h)
SFL = 3.6
SF2
kgSS/m2/h
Q SFL
P
u
xf
Concentration xi (kgSS/m3)
SFL 3.6 kgSS/m2/h
xf = 12 kgSS/m3 xf U= = = 0.300 m/h
12 kgSS/m3
48. Exercise 2
Design of secondary sedimentation units
with the Solid-Flux Analysis
6. SF calculation 7. Calculation of qf
Solid-Flux SF (kgSS/m2/h)
SF1
SF
SFL
x 3.4 kgSS/m3
qf = ·q = · 125 (l/s) = 49 l/s
SF2 xf - x (12-4) kgSS/m3
xf
Concentration xi (kgSS/m3)
8. Calculation Surface (A)
(q+qf) · x0 2,130
A= = = 592 m2
SFL 3.6
49. 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)
50. 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 Division
Via Martiri di Monte Sole 4, 40129 Bologna, ITALY
sabino.degisi@enea.it
