Lecture 3 Screening & Sedimentation_1.ppt

Screening and Sedimentation

Some of the material in the lecture slides is adapted from several textbooks and electronic resources

Flow Diagram of a WWTP

Treatment LevelsTreatment Level

Description

Preliminary Removal of wastewater constituents such as rags, sticks, floatables, grit, and grease that may cause maintenance or operational problems with the treatment operations, processes, and ancillary systems

Primary Removal or portion of the suspended solids and organic matter from wastewater

Secondary Removal of biodegradable organic matter (in solution or suspension) and suspended solids. Disinfection is also typically included in the definition of conventional secondary treatment

Tertiary Removal or residual suspended solids (after secondary treatment), usually by granular medium filtration or microscreens. Disinfection is also typically a part of tertiary treatment. Nutrient removal is often included in this definition

Advanced Removal of dissolved and suspended materials remaining after normal biological treatment when required for various water reuse applicationsLevels of Wastewater Treatment

Adapted from MEDAWARE ME8/AIDCO/2001/0515/59341-P033, Task 4: Urban Wastewater Treatment Technologies Part I, December 2004

ScreeningThese are

Physical processesInexpensiveCan be incorporated into WWT operations

Location of screens and bar racks – where??

Located at intakes from rivers, lakes and reserviors for water treatmentLocated at the well into which main trunk sewer discharges for WWTSometimes near pumping stations

What’s their function?Remove coarse solids, sticksWhy??

To avoid damage to pumps or clog downstream sections (pipes and channels)

ScreeningSpacing of screen bars can have

Coarse openings, 50-150 mmMedium, 20-50 mmFine screens, with an opening of 10 mmLess as proposed by few researchers

Critical limits for screensVelocity in the approach channel to the screens should not be <0.6 m/sDepth to width in the approach channel ranges from 1 to 2

Installation of screensWith an inclination?

To facilitate the removal of debris

Typical bar racks or bar screens Bar Racks (or bar screens):

Bar racks consist of bars spaced 10 to 50 mm apart and are usually mechanically cleaned. The screens are placed in rectangular channels.

The raking, done by means of endless chains on sprockets, moves the debri (screenings) upward and drops in a collection bin.

The screenings are odorous and also attract flies.

They are disposed by incineration or land filling or by returning to the wastewater flow after shredding (using comminutors).

Head Loss

Head loss through the screensIs a function of the flow velocity and the openings

∆h and h1 – h2 = [vSc2 - v2]/ (2gCd

2) from Bernoulli’s equation, where

h1 and h2 are upstream and downstream depths of flow,

Cd is the discharge coefficient (typical value ~0.84),

v and vSc are upstream velocity and velocity of flow through the screen

Orifice equation is often used

Water profile through the screen

Head Loss and Cleaning

Open area is often reduced by the space taken by the mesh

Screens to be manually cleanedThe area should be 50% of the open area (the half clogged condition)Head loss is estimated at the maximum flow condition

Cleaning of screensBy hand or automatically

Screenings areNon-putresciblecollected and hauled away to incinerator or landfill disposal site

Screens at WWTCoarse screens (with openings of 50-150 mm)

Used ahead of wastewater pumps

Openings of 25 mm are suitable for most other devices or processes

Installed at the beginning of the treatment plant after the water is pumped from the trunck sewer or influent wet well

Which are protected by coarse bar racks

Medium to large installations, mechanically cleaned screens are used

to reduce labor costsfor better flow conditionsimprove capture

Screening

Coarse screens

6-150mm

Microscreens <50μm

Fine screens <6mm

Manually cleaned

Mechanically cleaned

Static wedgewire

Drum Step

Chain-driven Reciprocating rake

Catenary Continuous belt

FIGURE 8: Definition Sketch for Types of Screens Used In Wastewater Treatment Source: Metcalf & Eddy, 2003

Types of Screens

Screening Volume Variation

This is a design chart based on data collected from several installations (around 133) in the US (showing volume variation with bar opening)

Coarse Screenings Characteristics

This tabular information provides information on screenings collected from separate and combined sewer systems. Combined sewers can produce several times the amount of screenings collected from separate sewer areas during storm flows (peak collection can vary as high as 20:1 on an hourly basis

Sedimentation / Clarification

Removal of particulate matter, chemical floc, and precipitates from suspension through gravity settling

Design variables include detention time, overflow rate, weir loading, and horizontal velocity

Basins Rectangular (more common) or circular Upflow or radial flow

What is it?Physical separation of suspended material from a water by the action of gravity

Sedimentation

Inlet – evenly distributes the flow across x-sectional area

~ 25% of tank in theory

Settling – gravity settling

Outlet – remove effluent

Sedimentation Basin Zones

Sedimentation Basin Zones

Sludge storage – holds settled material

~ 75% of the settable solids may settle in the first 1/5 portion of the tank

Depth for sludge storage should be about 0.3m near the outlet and 2m near the inlet

~ 1 to 10% slope

1% for mechanically cleaned5 to 10% for manually cleaned

Primary Sedimentation Basins

Design principlesBasic design principles evolved for sedimentation in water treatment apply for wastewater treatment also.The solids settling at the bottom of the sedimentation basin (referred to as primary sludge) is removed by mechanical scrapers. Grease and oil which float to the surface are removed by a skimming system.

Surface loading is the controlling parametertypical values ranging from 25 to 50 m3/m2.d

Detention time: 1.5 to 2.5 h

Weir loading: 120 - 200 m3/m.d

Depth: 2 - 5 m (~3.5 m)

Length to width ratio for rectangular basins: 3 to 5:1

Efficiency of primary settling basins: Suspended solids removal: 50 - 60% BOD removal: 30 - 35%

Sedimentation Types What is it?

Physical separation of suspended material from a water by the action of gravity

Settling properties of particles are often categorized into one of three classes:

Type I - particles settle discretely at a constant settling velocity (i.e., no flocculation)

Type II - particles flocculate during sedimentation (since they flocculate their size is constantly changing (i.e., vs is )

Type III - particles settle as a mass (i.e., lime softening)

Hindered settling - Particle concentration is sufficiently high to cause particles to settle as a structured mass

Compression settling - Particle concentration in the lower regions of the basin are sufficiently high to provide structural support for particles above

Type I - Discreet Settling

Particles settle

without significant inter-particle interaction due to low concentration of particles

Analysis is by use of Stokes' equation

Application is typically grit removal

Type II Flocculant settling

Particles agglomerate, thus increasing their size during the sedimentation process

Analysis is by empirical method using laboratory sedimentation data

Flocculant sedimentation is usually the dominant process in

primary clarification, settling following attached growth bioprocesses and water treatment clarification

Type III Hindered and Compression Settling

Hindered settling Particle concentration is sufficiently high to cause particles to settle as a structured mass

Compression settling Particle concentration in the lower regions of the basin are sufficiently high to provide structural support for particles above

Analysis for these types are done usingboth empirical and theoretical methods

Hindered and compression settling usually dominant in activated sludge final clarifiers and sludge thickening tanks

Sedimentation

A discrete particle's velocity can be described by an explicit equation developed by Stokes.

What is the key parameter needed prior to designing a basin?

Settling velocities of particles must be known

Physical properties determine the settling velocities

Forces Acting on a particle

FF

F

What are the three forces acting on a particle settling in water or another fluid?

Type I Analysis

Fd = the drag, acting in the upward direction as the particle settles

Fb = the buoyancy due to the water displaced by the particle, acting in the upward direction

Fg = the gravitational force, acting in the downward direction

FdFb

Fg

Type I Analysis

Fg = mp g [Gravity term]g = gravitational constant, [9.8 m/s2]

mp = particle mass, [Kg]

F + F = F bdg

][ termDragVAC2

1 = F 2

wdd Cd = drag coefficient, dimensionless

A = particle cross-sectional area, [m2]

w = density of water, [Kg/m3]

V = velocity, [m/s]

The buoyant force acting on the particle is: [Buoyancy

term]

mw = mass of water displaced, [Kg]

gm = F wb

gV + vAC2

1 = gV pw

2wdpp

Fg

FbFD

When above three relationships are substituted into the overall force balance equation, one can get

wd

pwp 2

1

AC

gV) - 2( = V

Stokes Equation

wd

pwp 2

1

C

gd) - (

3

4 = V

Solving for the settling velocity, V,

If we assume a spherical particle (the assumption is reasonably accurate), the relationships for particle area and volume can be used, yielding:

Stoke’s Law

Using the previous relationships, the particle settling velocity can be estimated

as a function of the properties of the particle and water, and the particle diameter

This relationship is known as Stokes' Law, and the velocity is known as the Stokes' velocity.

It is the terminal settling velocity for a particle.

18

gd) - ( = V

2wp

p

Grit RemovalTypical grit chambers are designed to retain particles with a diameter greater than 0.21 mm or 0.0083 in.

The vertical velocity of water in a grit chamber or settling basin is often termed the overflow rate.

Although it is a velocity, it is usually expressed as m3/m2-day or Gal/ft2-day.O/F = overflow rate (velocity), [m3/m2-day]

Q = flow rate, [m3/day]A = clarifier area, [m2]

18

gd) - ( = V

A

Q = O/F

2wp

p

Application

To remove a specific size particle

set its Stokes' velocity equal to the overflow rate.

Example

Estimate the settling velocity of sand (density = 2650 Kg/m3) with a mean diameter of 0.21 mm.

Assume the sand is approximately spherical. (Sand is not spherical but the assumption works well.)

Using a scale up factor of 1.4 to account for inlet and outlet losses, estimate the area required for a grit chamber to remove the sand if the flow rate is 0.10 m3/sec.

From a table providing the properties of water:

the density of water at 20°C is 998 Kg/m3. the viscosity of water at 20°C is 1.01 x 10-3 N-sec/m2 (Newton = Kg-m/s2).

Solution Cont’d

Use Stokes' settling velocity

Calculate the overflow rate

(3.9 cm/sec)

Estimate the area required for the grit chamber

(3.6 m2) - the area required for the grit chamber to remove 0.21 mm grit from the wastewater.

Grit Chamber Grit

Inert and inorganic material (sand, gravel, road grit, metal pieces, bone chips, glass pieces, etc) in the wastewater

FunctionTo protect equipment from abrasion (for pumps and other mechanical devices), to prevent pipe clogging, and to prevent accumulation in settling basin, digestors, etc.

Design principle: The grit chamber must be designed to remove only grit and not organic matter. Organic matter will be kept in suspension, and any organic matter that might settle must be resuspended by scour.

Grit with a specific gravity of 2.65 has a settling velocity of 30 mm/s.

Organic particle with a specific gravity of 1.10 has a settling velocity of 3 mm/s.

The design of a grit chamber exploits the difference in settling velocities of these particles.

Horizontal-flow, grit chambers

In a horizontal-flow, velocity controlled grit chamber, the velocity of wastewater flow is maintained at 0.30 m/s so that only grit will settle out. This grit chamber will have to be designed to maintain constant velocity of 0.3 m/s.

Maintaining a constant velocity of 0.3 m/s can be achieved in:

a channel of parabolic cross-section, controlled by a downstream standing wave flume, or a channel of rectangular cross-section, controlled by a proportional-flow weir at the outlet

The grit chambers are narrow channels also with a length to depth ratio of 20:1. The detention time is about 1 min.

Aerated grit chambers are also used for grit removal.

The grit, removed by mechanical collectors, is disposed of by burial or used as a fill material. If it still contains organic material, it is sent to sanitary landfills.

Horizontal Flow Grit Chamber

Scraper mechanism moves grit to center for removal

In order for grit to be removed in a horizontal flow grit chamber, it must reach the bottom of the chamber (or tank) before the flow carrying it reaches the exit. Thus removal of a particle is dependent on

its settling velocity,

the flow rate into the tank,

the basin or tank size.

Sedimentation Design

t = V / Qvo = Q / A

vo = Overflow Rate (gal/day)A = Total Surface Area (ft2)

If vs > vo particles will settle!

Sketch for an ideal, horizontal flow, rectangular basin

H - effective depth of the settling zonevf – longitudinal velocity of the waterB – width of the basinv1 and v0 – settling velocities applied to different particles entering at the top of basin

v2 – applies to a particle entering the settling zone at a height h, above the sludge zone


Design volume must be related toThe influent flow rate and particle settling velocityParticle that takes the longest time to remove will be the one that enters at the top of the effective zoneVo = Design settling velocity = Q/A

Implying that design is independent of the depth and depends only on the surface overflow or loading rate

If particle velocity < vo it will eventually exit with the effluent overflowing from the basinIf particle velocity> vo it will be removed after being introduced into the basin regardless of the residence time of water in the basin

Sketch for an ideal, horizontal flow, rectangular basin

Vf

D

L

Settling Velocities & Limits

What minimum settling velocity is required to remove a grit particle entering the chamber at the water surface?

Settling Velocities and Limits

What minimum settling velocity is required to remove a grit particle entering the chamber at the water surface?

Vf

D

L

Vc

Vr

Vf

D

L


What happens when a particle with a settling velocity of Vc enters at some level below the water surface?

D

L

Settling Velocities and LimitsWhat happens when a particle with a settling velocity of Vc enters at some level below the water surface? It will impact before the end of the chamber.

Vc

Vr

Vf

Vf

D

L


What happens when a particle with a settling velocity of less than Vc enters at the water surface?

V<VcVp

Vf

D

L


What happens when a particle with a settling velocity of less than Vc enters at the water surface?It will not be removed.

V<VcVp

D

L


What happens when a particle with a settling velocity of less than Vc enters at a point below water surface?It is dependent on depth and settling velocity.

Vf

V<VcVp

Vf

V<VcVp

Vf

V<VcVp

Vf

V<VcVp

Vf

V<VcVp

{No

Vf

V<VcVp

Vf

V<VcVp

Vf

V<VcVp

Vf

V<VcVp

{Yes


{{

No

Yes

Vf

Vr

Vp

VcD

L

Grit Chamber RemovalAny particle with a critical settling velocity greater than or equal to Vc will always be removed before exiting the chamber. Particles with velocities less than Vc will be removed only partially.

{{

No

Yes

Vf

Vr

Vp

Vc D

L

Grit Chamber RemovalLet us suppose a particle has a settling velocity of Vc as it enters the chamber.

If the entry point is near the bottom of the chamber, it will be removed. If it is near the top of the tank it will not be removed.

{{

No

Yes

Vf

Vr

Vp

VcD

L

Example: Grit Chamber Design

Design a grit chamber to remove sand particles (p = 2650 kg/m3) with a mean diameter of 0.21 mm. The wastewater flow is 10,000 m3/d. A velocity of 0.3 m/s will be automatically maintained, and the depth must be 1.5 times the width at maximum flow.

Assumptions:

The sand is spherical and the temperature of the wastewater is 20oC.


Calculate settling velocity0.039 m/s

Calculate the cross-sectional area0.39 m2

Calculate the width and depth0.51 m and 0.76 m

Determine the detention time required for a particle to fall the entire tank depth

19.4 s

Determine the length to achieve this detention time

5.8 mThe tank must have dimensions

W = 0.51 m, D = 0.76 m, L = 5.8 m


Calculate the cross-sectional area

v

QAx

s

min

min

d

m

s

d

m3

6014403.0000,10xA

2m 39.0xA


Calculate the width and depth

25.15.1 WWWAx

m 51.05.1

39.0

5.1

5.05.0

xA

W

m 76.05.151.05.1 WD


Determine the detention time required for a particle to fall the entire tank depth

Determine the length to achieve this detention time

Thus, the tank must have dimensions

W = 0.51 mD = 0.76 mL = 5.8 m

s m/s

m 4.19

039.0

76.0

sd v

Dt

m m/s s 85304.19 ..vtL d

Grit Chamber RemovalWe can calculate the fraction removed for any given particle size (and thus settling velocity). The velocity vectors depicting this can be shown graphically. Removal rate calculations will now be developed.

The vertical velocity of water in a sedimentation basin, V, is a function of the volumetric flow rate, Q, and the area of the basin, A, or, Vc = Q/A

However, the volumetric flow rate is also equal to the basin volume, V, divided by the detention time , or, Q =V/θ

Vc = V/Aθ

Since V/A is the basin height (or depth), Vc =h/θ

meaning that the critical settling velocity in a sedimentation basin is a function of the basin depth, h, and the hydraulic detention time, .

Grit Characteristics

Grit is composed of a variety of particle sizes

Particle settling velocity varies with particle density and size

Determine the size distribution and settling velocities

Screening Grit

The grit is placed in a screening device with successively smaller screen sizes

As the screens and grit are vibrated, grit passes through successive screens until it is retained on a screen with openings smaller than the grit

Each fraction can then be weighed to determine its contribution

Settling Velocities

The mean settling velocity of each fraction can then be determined by experiment or Stokes' Law

Grit RemovalTo calculate the fraction of grit removed we note that all grit particles with a settling velocity equal to or greater than Vc are removed, or

t

cpcp m

VVm1VVR

Where R is the removal contribution for particles with a settling velocity greater than or equal to the critical settling velocity, Vc

m is the mass fraction with a settling velocity greater than or equal to the critical settling velocity, vc

A fraction, but not all, of the grit particles with a settling velocity less than Vc are also removed

Grit RemovalFor each size fraction, the removal is equal to the ratio of the average particle velocity divided by the critical removal velocity, or (VP/VC)

The fractional removal of particles with a settling velocity less than Vc is then:

Since both Vc and mt are constants, they can be removed from the summation, yielding:

The total removal is then the sum of the two fractions, or:

The application of this is demonstrated in the example to follow.

t

in

1i c

i,pcp m

m

V

VVVR

n

1iii,p

tccp mV

mV

1VVR

n

1iii,p

tct

cpmV

mV

1

m

VVmR

Sedimentation Designh/H < v/vo Fractional removal h/H = v/vo

Proposed by Bharagave and Rajgopal (1989),J. Environ Eng Div., ASCE, pp. 1191-1198.

U is the ratio of sizes through which 60 and 10% by wt of the particles pass

(the size through which 10% by wt of the particles pass)


U is the ratio of sizes through which 60 and 10% by wt of the particles pass

P10 is the size through which 10% by wt of the particles pass

Design Continued

Weir loading – divide the average daily quantity of flow by the total effluent weir length (gal/d.ft)

________________________________ Type of Flow Weir Loading

______________ _________m3/d•m ____Light alum floc – low TU 143 – 179Heavy alum floc – high TU 179 – 268Heavy floc from lime softening 268 – 322_____________________________________

Sedimentation Rules of Thumb

Detention time 4 hours

Horizontal velocity 0.5 ft/min

Max weir loading 20,000 gal/day•ft

vo between 500 - 800 gal/day •ft2

vs > vo

Sedimentation Concepts

Two critical variables in the design of sedimentation basins are vs and vo

If vs > vo, particles will settle

For design, find vs and let vo = 80% vs

P = 100 vs / vo

P = % of particles removed

It is best to determine vs based on sedimentation lab data

ExampleEstimate the amount of grit removed from a grit chamber with an inflow of 12,500 m3/day and effective area of 5 m by 5 m (or 25 m2). A sieve analysis with approximate settling velocities is shown in the following table.

Weight Fraction

Settling Vel., m/min

0.02 0.05 0.05 0.15 0.05 0.2 0.1 0.3 0.1 0.4 0.2 1 0.2 2

0.28 4

Solution

First, we need to calculate the critical settling velocity:

Vc = 0.35 m/min

Note that the weight fractions with settling velocities greater than 0.35 m/min will be removed completely.

Next, calculate the fractional removal:

Grit removal of 92%.

n

1iii,p

tct

cpmV

mV

1

m

VVmR

daym

m500

m25

day

m500,12

A

QV

2

3

2

3

c

min

m35.0Vc

Solution

First, we need to calculate the critical settling velocity:

Settling Velocities

Weight Fraction


Comment

0.02 0.05 V<Vc; Fract. removed 0.05 0.15 V<Vc; Fract. removed 0.05 0.2 V<Vc; Fract. removed 0.1 0.3 V<Vc; Fract. removed 0.1 0.4 V>Vc; All removed 0.2 1 V>Vc; All removed 0.2 2 V>Vc; All removed 0.28 4 V>Vc; All removed

Settling Velocities

Weight Fraction


Fraction Removed

0.02 0.05 0.00288 0.05 0.15 0.0216 0.05 0.2 0.0288 0.1 0.3 0.0864 0.1 0.4 0.1 0.2 1 0.2 0.2 2 0.2 0.28 4 0.28

Sum = 0.92

n

1iii,p

tct

cpmV

mV

1

m

VVmR

Solution

Note that the weight fractions with settling velocities greater than 0.35 m/min will be removed completely.

Next, calculate the fractional removal:

n

1iii,p

tct

cpmV

mV

1

m

VVmR

]3.01.02.005.0

15.005.005.002.0[0.135.0

10.1

28.02.02.01.0R

92.014.078.0R

Solution

Thus, a grit removal of 92%

Weight Fraction


Fraction Removed

0.02 0.05 0.00288 0.05 0.15 0.0216 0.05 0.2 0.0288 0.1 0.3 0.0864 0.1 0.4 0.1 0.2 1 0.2 0.2 2 0.2 0.28 4 0.28

Sum = 0.92

n

1iii,p

tct

cpmV

mV

1

m

VVmR


Typical loading parameters are:

Detention time of 45 to 90 s at peak flow

Horizontal velocity of 0.8 to 1.3 ft/s

Head loss 30 to 40% of channel depth

Aerated Grit Chamber

Another common method of removing grit is the aerated grit chamber.

The water enters a concrete basin with aerators along one side at the bottom.

The critical design parameters for an aerated grit chamber are:

a detention time of 2 to 5 minutes at peak flow and

an air supply of 2 to 5 ft3/min-ft of chamber length

Air in


Air induces a spiral flow through the chamber

Grit falls out beneath the aerator in a collection channel


Aerated grit chambers have the following typical physical size restrictions:

a length of 25 to 65 ft.depth of 7 to 15 ft.width of 8 to 25 ft.length to width ratio of 3:1 to 5:1width to depth ratio of 1:1 to 5:1

A plant with a peak flow of less than 3 MGD will be over designed.

Example: Aerated Grit Chamber

Estimate the volume of an aerated grit chamber for Carbondale’s Southeast WWTP. Assume a detention time of 3 minutes at a peak flow of 12 MGD.

To complete the design, the length, width, depth and air requirements must also be calculated.

QθV

min60hr

hr24d

min3d

Gal10x12V 6

3ft3340Gal000,25V

Flocculant SettlingFlocculant or Type II settling occurs when particle concentrations are sufficiently high that particles collide and agglomerate as they settle.

As a result of this agglomeration process, the particle size distribution, as well as effective density, does not remain constant but instead, changes with depth.

Thus, Stoke’s Law is difficult to apply.

Flocculant Settling Cont’dNo straightforward analytical solution to this type of settling has been developed.

Flocculant settling is usually analyzed by empirical means using laboratory data obtained from column settling experiments.

Such experiments yield the suspended solids concentrations at various depths over the anticipated settling time.

Settling column description

at least 8 in. in diameter approximately the same depth as the anticipated sedimentation basin

h 0

h 1

h 2

h 3

h 4

h 5

h 6

h 7

h 0

h 1

h 2

h 3

h 4

h 5

h 6

h 8

h 7

Scale Up Factors

When column settling data is used to size a sedimentation basin (primary, secondary, or thickening)

the final calculated area or detention time is usually multiplied by a scale-up factor of 1.25 to 1.5

Why?? to allow for the less than ideal settling which occurs in a “real” basin.

[See Wastewater Engineering, M&E]

Sampling ProcedureA sample of the wastewater to be treated is placed in the column and initially mixed.

An initial set of data is taken from the sampling ports and analyzed to determine the starting solids concentration as well as the initial uniformity of the solids.

At time intervals thereafter, the ports are sampled to determine the suspended solids concentration. This process is continued for approximately two to three hours.

The data collected then provides a history of the suspended solids concentration versus time for different heights in the column.

This information, in turn, can then be used to estimate the basin suspended solids removal versus time.

h 0

h 1

h 2

h 3

h 4

h 5

h 6

h 7

h 0

h 1

h 2

h 3

h 4

h 5

h 6

h 8

h 7

Flocculant Analysis

If we look at one sampling port on the column, h1, the initial suspended solids concentration is TSSh1-0.

If, in fact, the entire column has a uniform initial TSS concentration, then it is TSS0 at all heights.

At some later time, t, the concentration at sampling point h1 is then TSSh1-t.

h 0

h 1

h 2

h 3

h 4

h 5

h 6

h 7

h 0

h 1

h 2

h 3

h 4

h 5

h 6

h 8

h 7

%100x

TSS

TSSTSSR%

0

t,1h0t,1

%100x

TSS

TSSTSSR%

0

t,2h0t,2

Flocculant Analysis

The removal at that point during time t is then:

The removal at height h2 is

The average removal from h1 to h2 is then:

2

R%R%R% t,2t,1

t,1h

2

R%R%x

h

h

R%xh

hR%

t,2t,1

T

1

t,1hT

1t,1h

Flocculant AnalysisThe contribution this element has to the overall removal at time t is the removal multiplied by the fraction of the total height it represents, or, where hT is the total basin height

Flocculant SettlingThe total removal at any time t is the sum of the removals at each height segment, or,

n

i

titi

T

it

RR

h

hR

0

,1,

2

%%%

Typical Removal Curves

Primary Sed. Design Information

Min Max Typ.

Detention time at average flow, hr

1.5 2.5 2.0

Overflow rate at average flow, Gal/ft2-day

800 1200 1000

Overflow rate at peak flow, Gal/ft2-day

1000 3000 2500

Weir loading, Gal/ft-day

10,000 40,000 25,000

Typical Circular Primary Basin Sizes

Range Typical

Depth, ft 10 to 15 12 to 13

Diameter, ft. 10 to 300+ 40 to 100

Bottom slope, in/ft.

.75 to 2 1

Scraper tip speed, ft/min

6 to 12

Example: Analysis of Flocculant Settling Data

Design a primary settling system for a wastewater plant with an average flow of 3.2 MGD and a peak flow of 9 MGD.

Use two identical circular primary clarifiers, each to receive one-half of the flow. The basins should remove at least 65% of the solids at average flow.

Specify the depth and diameter. Check the weir loading. The data below was obtained from a settling column similar to the one shown previously.

After the basins are sized, recalculate the actual removal at both average and peak flows.

Settling Column Data

Suspended Solids Data (mg/L) at IndicatedTime (min) and Height

Station Ht.,m

0, min 30, min 60, min 90, min 120, min

h0 4 178 N/A N/A N/A N/Ah1 3.5 198 95 62 40 33h2 3 190 120 85 60 42h3 2.5 192 125 102 71 54h4 2 192 128 110 85 66h5 1.5 188 137 120 95 76h6 1 194 144 128 102 79h7 0.5 201 149 130 109 87h8 0 196 154 132 110 91

Process OverviewStep1 : Estimate the removal at different times and heights.

Step 2: Next construct the removal vs. time graph.

Step 3: Estimate the detention time required.

Step 4: Use the standard design criteria to determine the area and depth.

Solution

The first step is to calculate an average or initial TSS concentration, TSS0.

The next step is to use this TSS0 value to determine the TSS removal as a percentage.

These calculations are ideally suited to a spreadsheet analysis, although they can be done equally well by hand given the time.

L/TSSmg192n

TSS

TSS

n

1ii

0

Sample Calculations

The average initial TSS concentration is:

Removal at 4 m,30 min

Since any particles in the water will have a finite settling velocity, in theory, they should settle below the top of the basin at any finite time. Thus, the removal at the top of the basin at any finite time is 100 percent.Suspended Solids Data (mg/L) at Indicated

Time (min) and HeightStation Ht.,

m0, min 30, min 60, min 90, min 120, min

h0 4 178 N/A N/A N/A N/Ah1 3.5 198 95 62 40 33h2 3 190 120 85 60 42h3 2.5 192 125 102 71 54h4 2 192 128 110 85 66h5 1.5 188 137 120 95 76h6 1 194 144 128 102 79h7 0.5 201 149 130 109 87h8 0 196 154 132 110 91

%5.37

%100xL/mg192

L/mg120L/mg192

%100xTSS

TSSTSSR%

0

min30,m30min30,m3

Removal at 3 m, 30 min

%5.50

%100xL/mg192

L/mg95L/mg192

%100xTSS

TSSTSSR%

0

min90,m5.10min90,m5.1

Removal at 1.5 m, 90 min

Removal Summary

% Removal at Indicated Time (min) and Height

Station ColumnHt., m

0 30 60 90 120

h0 4 7.5% 100.0% 100.0% 100.0% 100.0%

h1 3.5 -2.9% 50.4% 67.5% 79.2% 82.9%

h2 3 0.8% 37.5% 55.8% 68.8% 77.9%

h3 2.5 0.0% 35.0% 46.7% 62.9% 72.1%

h4 2 0.0% 33.3% 42.5% 55.8% 65.8%

h5 1.5 2.1% 28.8% 37.5% 50.5% 60.4%

h6 1 -0.8% 25.0% 33.3% 46.7% 58.8%

h7 0.5 -4.6% 22.5% 32.5% 43.3% 54.6%

h8 0 -2.1% 19.6% 31.3% 42.5% 52.5%

Average Removal = 0.0% 39.1% 49.7% 61.1% 69.4%

Removal Curve

Estimating Detention Time

The previous graph indicates a detention time of approximately 103 minutes.

Using a scale up factor of 1.25 the actual detention time is:

hr15.2min129

min10325.1

SF

actual

actual

columnactual

Basin Volume

The total basin volume required is then:

3

36

ft300,38V

Gal48.7ft

hr24day

hr15.2dayGal

102.3V

QV

Since there will be two basins, the volume of each is 19,200 ft3.

Question

We now know the volume. How do we determine the diameter (or surface area) and the depth?

We use the standard overflow rates to estimate an area at both average flow and peak flow. We choose the larger area. We can then calculate the diameter and depth.

Basin Area

Initially assuming the maximum overflow rate at both average and peak flows:

Q = AV = A x OFR

2avg

2

6

avg

ft2670A

dayft

Gal1200

dayGal

102.3

OFRQ

A

Min Max Typ.

Detention time ataverage flow, hr

1.5 2.5 2.0

Overflow rate ataverage flow,Gal/ft2-day

800 1200 1000

Overflow rate atpeak flow, Gal/ft2-day

1000 3000 2500

Weir loading,Gal/ft-day

10,000 40,000 25,000

Basin Area

And, for the peak flow:

2avg

2

6

peak

ft3000A

dayft

Gal3000

dayGal

109

OFRQ

A

Note, if the peaking factor is 2.5, the two areas will be equal.

Min Max Typ.


1.5 2.5 2.0


800 1200 1000


1000 3000 2500


10,000 40,000 25,000

Basin Diameter

The larger area is 3000 ft2, so we must use it. Each basin will be one-half the total area, or 1500 ft2. This corresponds to a diameter of:

ft7.43D

ft9.21ft1500A

r

rA

2

2

One can round this to a standard diameter of 45 ft.

2

2

2

ft1590A

ft5.22A

rA

Basin Diameter

A diameter of 45 ft. results in an actual area of:

Basin Depth

We will now use the area and required volume to calculate the minimum depth:

ft07.12d

ft1590

ft200,19AV

d

dAV

2

3

One can round this to 13 ft. to maintain the required detention time.

Recheck of Parameters

Actual detention time at average flow is:

hrs3.2

hr24day

Gal48.7ft

dayGal

106.1

.ft13.ft5.22

Qdr

QV

36

2

2


Actual detention time at peak flow is:

min49hrs083.0

hr24day

Gal48.7ft

dayGal

105.4

.ft13.ft5.22

Qdr

QV

36

2

2

No established standard, but some consulting firms will not design below one hour at peak.


Actual overflow rate at average flow:

ft1006OFR

ft1590

dayGal

106.1

AQ

OFR2

6

Min Max Typ.


1.5 2.5 2.0


800 1200 1000


1000 3000 2500


10,000 40,000 25,000

Average flow of 3.2 MGD and a peak flow of 9 MGD


Actual overflow rate at peak flow:

ft2830OFR

ft1590

dayGal

105.4

AQ

OFR2

6

Min Max Typ.


1.5 2.5 2.0


800 1200 1000


1000 3000 2500


10,000 40,000 25,000

Average flow of 3.2 MGD and a peak flow of 9 MGD


Weir loading rate:

dayft/Gal800,31WLF

.ft45d

Gal105.4

WLF

D

Q

L

QWLF

6

pp

Estimated RemovalTo estimate the removal, first recall we multiplied the required area/detention time, by a scale-up factor of 1.25. To re-estimate the removal at average and peak flows, we must “remove” the safety factor.

min11025.1min138

min4025.1min49

25.1

avg,corrected

peak,corrected

designcorrected

Removal Curve

Estimated Removal

From the removal graph,

Flow Removal

Average flow 67%

Peak flow 44%

Summary

Parameter Values

Number of units

Diameter

Depth

2

45 ft

13 ft

Hindered and Compression Settling

Hindered settling occurs

when the number of particles is sufficiently high that they settle as a structured mass.

Compression settling occurs

where particle numbers are so high that particles above can only settle if lower particles are further compressed—the lower particles actually support upper particles.


Hindered and compression settling occurs only in the lower regions of sedimentation basins. It is rate determining step

in most secondary sedimentation basins receiving activated sludge

in gravity thickening used to concentrate waste primary sludge and sometimes waste activated sludge.


Hindered and compression settling are normally analyzed together. The two most common methods of analyzing hindered and compression settling are

the solids flux method and

the Talmadge and Fitch method

Column Settling

00.5

11.5

22.5

0 30 60Time, min

Set

tlin

g D

ista

nce

, m

Flocculent

Hindered Compression

Solids Flux Method

Flux is a mass crossing a boundary or area.

The solids flux method used to analyze sedimentation processes can be developed

by studying the mass of solids passing an arbitrary plane within a sedimentation basin.

Solids FluxThe mass of solids passing the arbitrary plane increases with increasing inlet solids concentration to some limiting value and then decreases rapidly.

This decrease past some maximum inlet solids concentration is caused by the structure of the solids at the bottom of the tank preventing the rapid settling of the solids above.

Solids Flux

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

0 5 10 15Concentration, g/L

So

lid

s F

lux

, K

g/m

2-h

r

@ low concentration – low flux due to less solids@ high concentration – low flux due to extreme hindered settling

Solids Flux

This maximum point is termed the limiting solids flux. Let us develop the equations to describe this process, solve the system, and determine that limiting or maximum solids flux.

Ai

Vp Ci

Solids FluxAn arbitrary area, Ai, with solids of concentration Ci, and velocity Vi crossing the plane.

Solids FluxThe solids flux, Gi, is the concentration times the velocity, or

ipi CVG

Gi = solids flux, mass/area-time

Vp = particle velocity, length/time

Ci = particle concentration, mass/volume or mass/length3

Ai

Qu

Qi+Qu

Qi

Solids Flux

Ai

Qu

Qi+Qu

Qi

QuestionWhat is the fluid velocity at the plane Ai?

i

uf

fiu

A

QV

VAQ

uit GGG

uiiit VCVCG

Solids Flux

Note from fluid mechanics that

A

QV

i

ui

i

iit A

QC

A

QCG

Then,

L

iuii G

CQQA

Solids FluxIf we use the maximum or limiting solids flux, GL, with the inlet solids concentration, we obtain the following

Which is the minimum area required for settling

Solids Flux

The solids have two velocity components:

A gravity component due to the movement of solids as they settle in the tank

A fluid component due to fluid being pumped out the bottom of the tank (bulk fluid movement)

Solids Flux vs Concentration

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

0 5 10 15

So

lid

s F

lux

, Kg

/m2

-hr

Concentration, g/L

Gravity flux – solids transport by gravity

Pumped flux-Underflow withdrawalIncreases the downward movement of solids

Total flux

Limiting flux

Solids transmitting capacity of each layer varies with the concentration of the layerOverall or total flux has a minimum & maximum Minimum point determines the design area of the clarifierMinimum solids handling capacity of the suspension is the limiting flux

Solids Flux

Recall that a scale up factor of 25% to 50% is applied to the area based on column settling data. Or, the area is multiplied by 1.25 to 1.50. See page 228 of M&E, 3rd. Edition.

Example: Solids FluxAn activated sludge wastewater treatment plant currently has an average flow of 2.5 MGD. The aeration basin reactor solids concentration is 2500 mg/L and the desired underflow concentration is 13,000 mg/L. You may assume that the recycle flow will never exceed 50% of the plant flow.

Determine the total secondary settling basin area required and the diameter if two identical basins are utilized.

Example: Solids Flux

Let’s convert the flow to metric;

hr

m394Q

hr24

d

L10

m

Gal

L78.3

d

Gal105.2Q

3

i

3

36

i

The recycle flow is half this, or;

hr

m197Q

3

u

Example: Solids FluxFrom the solids flux graph in the spreadsheet the limiting solids flux is 4.9 Kg/m2-hr. The required area is then:

hrm

Kg9.4

g10

Kg

m

g2500

hr

m197

hr

m394

A

G

CQQA

2

33

33

i

L

iuii


The minimum required area is then;

22i ft3240m301A

A scale up factor of 1.25 to 1.50 (25% to 50%) is usually used for the area determined by column settling studies. Thus,

22i ft4375ft324035.1A


This yields a surface overflow rate of;

dayft

Gal571RateF/OV

ft4375d

Gal105.2

A

QV

2

2

6

i

i

Which is within the acceptable range of 400 to 800 Gal/ft2-day. Note that overflow rates are based on plant flow excluding recycle.


To calculate the area:

ft8.52D

ft4.26ft2190

r

rft21902

ft4375A

2

222

b

Question

What size units would be selected?

Round up to the next available size.

Some manufacturers only provide specific sizes, typically in 5 ft. increments.

Others, may provide any requested size in 1 ft. increments.

Assuming 5 ft. increments, the actual diameter would then be 55 ft. So, two units, 55 ft. in diameter are required.

Lecture 3 Screening & Sedimentation_1.ppt

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