Wastewater sedimentation

50
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 29/11/2012 04/12/2012 Padua

Transcript of Wastewater sedimentation

Page 1: Wastewater sedimentation

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

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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)

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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?

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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

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Type of sedimentation

Settling phenomena involved in wastewater treatment

(Metcalf & Eddy, 2003)

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Type of sedimentation

Settling phenomena involved in wastewater treatment

Water processing flow diagram for a large UWWTP

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Type of sedimentation

Settling phenomena involved in wastewater treatment

Sludge processing flow diagram for a large UWWTP

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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

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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.

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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

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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

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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

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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 =

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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)

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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?

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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

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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

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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

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Technology

Large urban wastewater

treatment plant

Primary treatment

Secondary treatment

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Technology

Primary sedimentation tanks

WW distribution well

Oxidation basins

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Technology

Primary sedimentation tanks

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Technology

Thomson effluent weir

150 50 150 50

75

150

90°

(values in cm)

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Secondary sedimentation tanks

Technology

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Technology

Secondary sedimentation tanks

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Technology

Secondary sedimentation tanks

Sludge scrapers

Surface skimmer

Bridge

Drive unit

Inlet pipe

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Technology

Secondary sedimentation tanks

Scum baffle

Effluent weir

Support

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Technology

Secondary sedimentation tanks

Scum box

Inlet scum

Scum pipe

Versus Scum pit

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Technology

Secondary sedimentation tanks

Scum box

Route of Scum

Rotation of bridge

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Technology

Secondary sedimentation tanks

Effluent launder

Scum baffle

Surface skimmer

Effluent weir

Clarified wastewater’s route

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Technology

Secondary sedimentation tanks

Cockpit divider

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Technology

Secondary sedimentation tanks

Cockpit divider

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Technology

Secondary sedimentation tanks

Thomson effluent weirCentral pivot

Scum boxBridge

Distribution system

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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

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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

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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.

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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 =⋅=

⋅=

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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

>===

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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.

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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)

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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

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

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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%

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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)

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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

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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 ?

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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

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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

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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

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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

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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)

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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]