Settling Design

Wastewater Engineering and Design

By Assist. Prof. Dr. Wipada Sanongraj 1

1

Chapter 2: Physical Treatment Units; Design of Sedimentation Tanks


2

Outline

n Physical Treatment Unit

n Types of Sedimentationn Principle of Sedimentation

n Primary Sedimentation Tank

n Settling Thickener



3Figure 2-1: Physical Unit Treatment (Metcalf and Eddy, 1991)

4

Operation Applicationn Flow metering Process control, process monitoring

and discharge reports

n Screening Removal of coarse and settleablesolids by interception

n Flow equalization Equalization of flow and mass loading of BOD and suspended solids

n Mixing Mixing chemicals and gases with wastewater, and maintaining solids in suspension

n Flocculation Promotes the aggregation of small particles into larger particles to enhance their removal by gravity sedimentation

Physical Treatment Unit



5

Operation Applicationn Sedimentation Removal of settleable solids

and thickening of sludges

n Flotation Removal of finely divided suspended solids and particles biological sludges

n Filtration Removal of fine resudualsuspended solids remaining after biological or chemical treatment

n Microscreening Same as filtration

n Gas transfer Addition and removal of gases

n Volatilization Emission of volatile and semi-volatile and gas stripping

organic compounds from WW

Physical Treatment Unit

6

Types of Sedimentation

Bar Screening



7

Four Types of Settling

Type I: Discrete Settling

1. Particles settle out individually and do not interact with one another.

2. Suspension of low solids concentration.

Types of Sedimentation

8

3. Occur in:a) Grit chambers

and some primary clarifiers used in wastewater treatment plants.

b) Some presedimentationbasins in water treatment.

Type I: Discrete Settling



9

Type II: Flocculants Settling

1. As particles flocculate (due to velocity differences), they increase in mass and velocity:

a) Mixing due to hydraulic gradients in clarifier produces particle collisions:

10

b) Larger particles overtakes smaller ones and flocculate:




11

2. Suspension of low solids concentration3. Occurs in:

a. Most primary and all secondary clarifiers in wastewater treatment.

b. Most sedimentation basins in water treatment.


12




13

Type III: Zone Settling or Hindered Settling

1. Formation of dense mat of particles that settle out as a unit:

Settling of different size particles will eventually form thick settling blanket that settles at a constant velocity.

14

2. Suspension of intermediate concentration.3. Intraparticle forces are sufficient enough to affect adjacent particles.4. Occurs in:

a. Thickeners (sludge disposal)b. Bottom of clarifiers and sometimes

sedimentation basins.

Type III: Zone Settling or Hindered Settling



15

Type IV: Compression Settling

1. Settling occurs by compression where water is forced out of the interstitial voids between particles.

In order to settle, water must pass through particles arranged like porous media. Furthermore, there is a high head loss due to a large surface to volume ratio. Therefore, settling occurs slowly.

16

2. Suspension by very high concentration.3. All particles are influenced by the presence of other particles.4. Occurs in drying beds and some filtration process.

Type IV: Compression Settling



17

Grit Materials

Sand, gravel, egg shells, coffee grounds, fruit rinds, seeds, bones (whos)

18

Principles of Sedimentation

Vp = Volume of particles = 1/6 Dp3 (cm3)Ap = Projected surface area of particle = /4Dp2 (cm2) Fp = Drag force = friction factor * inertial force (g-cm/s2)Fb = Bouyant force (g-cm/s2)Fg = Force due to gravity (g-cm/s2)Vs = Velocity of particle (cm/s)mp = Mass of particle (g)g = Acceleration due to gravity (cm/s2)rl = Density of water (g/cm3)rp = Density of particle (g/cm3)



19

Momentum Balance

( )dbg

sp FFFFdt

Vmd--==

( )2

2s

pldplppsp VACgVgV

dt

Vmdrrr --=

Determination of the Drag Coefficient in water

Cd = f(NR) & Particle Shape

Cd can be determined if the pressure and shear stress characteristics surrounding an object are known. However, Cd can also be determined if the total drag is measured by a force dynamometer.

20




21

Larminar region: viscous forces control the drag forceTurbulent region: inertial forces of displaced fluid are controlling the drag force

For laminar flow: (NR < 1)

l

splR

VDN

mr

=

Rd N

C24

=

splspl

spl

lspldd VD

VD

VDV

ACF pmpr

rmr 3

2*424

2

222

=

==


22

Remembering the general equation:

( )splplpp

sp VDgVgVdt

Vdm pmrr 3--=

( )p

spl

p

pl

p

pps

m

VD

m

gV

m

gV

dt

Vd pmrr 3--=

Substituting for mp and Vp:

( )

pp

spl

pp

pl

pp

pps

D

VD

V

gV

V

gV

dt

Vd

rppm

rr

rr

6

33

--=

( )pp

sl

p

lps

D

Vg

dt

Vd

rm

rrr

2

18-

-=




23

Let l

ppD

mr

t18

2

=

( )tr

r sp

ls Vgdt

Vd-

-= 1

Using the integration factor method and the initial conditions: Vs = 0 @ t = 0

( )t /lsp

V g 1 1 e- t r

= t - - r Where the relative measure of the fractional approach to SS is:

( )t /1 e- t-


24

l

ppDtimerelaxationmr

t18

2

==

When t then VsVtl

tp

V g 1 r

= t - r

( )t /s tV V 1 e- t= -ornow (1-e-3) = 0.95 (it takes 3t to attain 0.95Vt)




25

Example 2-1: Typical grit particle (sand)

Dp = 100 mm = 0.01 cmrp = 2.65 g/cm3

Since the typical retention time for a particle may be on the order of hours or more, we are not concerned about non-steady state.

26

Therefore, we can assume steady-state:

( ) p ls l s2

p p p

d V 18 V0 g

dt D

r -r m= = - r r

Solve for Vs: (Stokes law for settling)2

p l ps R

l

g( )DV for N 1

18

r -r=

m

Example 2-1: Typical grit particle (sand)



27

Example 2-2: Bio floc particle

ml = 0.01185 g/(cm-s) rp = 1.05 g/cm3 rl = 1.0 g/cm3Dp = 100 mm T = 25 oC

Check NR

Since NR < 1, laminar flow exists and Stokes Law is Valid

28

What about a large particle size (sand)?

rp = 2.65 g/cm3 Dp = 200 mm T = 25 oC

Since the Reynolds number is greater than 1.0, Stokes law is not valid because the drag coefficient of the particle in water does not vary as 24/NR, but as:




29

Rearrange the original force balance:

( ) 2p s sp p l p d l p

d m V VV g V g C A 0

dt 2= r -r - r =

Solving for Vs yield the following equation for Newtons law for settling particles:


30

Using NR = 5.11 from above:

d

24 3C 0.34 6.36

5.11 5.11= + + =

recalculate NR NR = 4.38


Since we cannot solve explicitly for Vs as in Stokess Law expression, we must now use a trial and error solution as demonstrated below:



31

The procedure for calculating the settling velocity of a particle when NR > 1 is:

1. Use Stokes Law to obtain the initial guess for VS.

2. Use this Vs to obtain NR, if NR > 1, then go to next step.

3. Use the following iteration process:

a) Use NR to get Cd

dR R

24 3C 0.34

N N= + +

32

c) Use Vs to find new NRd) Repeat until convergence is met.

b) Use Cd to get new Vs



33

Brownian Motion

In water and WW, colloidal particles are usually present and are hard to remove by settling. A colloidal particle may be thought as a giant molecule and its Brownian motion is really a diffusion process, where diffusion results from the random thermal motion (due to thermal gradients within the fluid).

34

The above particle motion was first observed for water molecules by Dr. Robert Brown. In 1905, Einstein used Stokess relationship and developed the following equation to describe the motion.

Brownian Motion



35

Bp

1 2kTV

X 3 D

= D pm

Where: k = Boltzman constant = 1.38*10-16 (g-cm2)/(s2-K)

T = temperature (K) DX = net x component distance of travel (cm)m= kinematic fluid viscosity (g/cm-s)Dp = mean particle diameter (cm)VB = particle velocity due to Brownian Motion (cm/s)

When VB > Vs (stokes), particles will not settle out of solution because the motion of the particle is governed by collision withthe water molecules.

Brownian Motion

36

The following characteristics are typical for water and wastewater colloids:

Table 2-1: Characteristics of water and wastewater colloids (Metcalf & Eddy, 1991)



37

Example 2-3:

Calculate the smallest settleable sand particle in water at 25 oC.

Assume DX = 1.0 cm

Solution

38

Setting VB = Vs

Coagulation and flocculation can be used to bring particles together to form larger ones that will settle out of solution by gravitational forces.

Example 2-3:



39

Characteristics of Colloidal Particles in Water

1) Size-Very Small

1-1000 nm10-10,000 Angstroms10-4 10-7 cm

2) Surface Area Very Large

Diameter of Spheres Surface Area1 cm 0.0487 in2

10-4 cm 33.8 ft2

10-8 cm 3.8 yd2

10-6 cm 0.7 acres10-7 cm 7.0 acres

3) Charge Colloidal particles are usually NEGATIVELY charge causing repulsion of similar charge and colloidal stability.

40

Design of Type I Sedimentation Basins

Assumptions:1. Plug flow exists in settling zone where fluid

moves across the settling zone with a constant velocity.

2. Designed to remove discrete particle (settling zone only considered)

3. Particles uniformly distributed at inlet.



41

Key Process Variables

1. Hydraulic retention time = t = V/Q = Vol. tank/Flowrate2. Overflow rate = Vo =Q/Atop

Vs = Flowrate/Top cross sectional area = h0/t

Vf = horizontal fluid velocity ho = height of settling zoneL = Length of settling zone

If VsVo, all particle of that diameter are removed.If Vs



43

Say Dp = 200 mm, Vs = 0.0235 m/s

Will all the particles be removed?

Example 2-4:

44

Say Dp = 100 mm, Vs = 0.0076 m/s

l

l

Example 2-4:



45

Removal efficiency is independent of depth (ho) and retention time (t)

Vs = f(g,ml,rl, rp,Dp) Vo = f(Q, Atop)How can we use this information to determine the R.E. of a sedimentation tank with many sizes of particles?

(Metcalf & Eddy, 1991)

46

Break the problem into two parts.

1. For particles which have Vs Vo all particles are removed or (1-Xo) is removed.

2. For Vs Vo a fraction of the particles with such velocities are removed.



47

Break down the curve for all particles with settling velocities Vs



49

How do we obtain the necessary information to make this calculation?

Two ways:1. Use Stokes equation or Newtons equation

depending on NR2. Use a column settling test (most common)

1. Put a suspension in column.

2. Mix well and sample to determine Co.

3. Stop mixing and begin timing.

4. Take samples from same depth.

50

Example 2-5:For the following data and an overflow rate of 2.0 m/hr, what is the R.E.?

Solution1. Plot x vs. Vs2. Determine Xo from overflow rate (Vo) 3. Determine R.E. from



51

ox

o so o

1R .E. (1 X ) V dx

V= - +

Example 2-5:

52

Now suppose youre given R.E. and you need to calculate Vo.

Solution: 1. Plot x vs. Vs2. Assume an Xo3. Go to curve and get Vo4. Calculate:

Example 2-5:



53

5. Calculate R.E.

6. Plot R.E. vs. Vo7. Use graph to get overflow for the R.E. of interest.

Example 2-5:

54

Example 2-6:

For the following data, calculate the overflow rate for a R.E. of 75%.

ox

o so o

1R.E. (1 X ) V dx

V= - +



55

Solution1. Assume Xo = 0.207

2. Assume Xo = 0. 417

Example 2-6:

56

3. Assume Xo = 0. 65

Example 2-6:



57

58Figure 2-1: Two approaches to determining minimum particle size removed in a nonturbulent chamber (Mihelcic, 1999)



59

Type I Sedimentation for a Circular Clarifier

60

But Vf = Q/SA = Q/2prH = area of a cylinder with height H

All particles with Vs > Vo will be 100% removed. The analysis is the same for both circular and rectangular tanks employing type I sedimentation.



61Figure 2-2: Typical rectangular primary sedimentation tank (Metcalf & Eddy, 1991)

62

Horizontal Flow Grit Chambers

Flow control weirs typically require free discharge and hence a relatively high head loss (typically 30 40% of flow depth)

Proportional weirs may cause higher velocities at the bottom leading to bottom scour

Where effective flow control is not achieved, channels will remove significant quantities of organic material requiring grit washing and classifying

With effective flow control, removal of grit not requiring further classification is possible

Excessive wear on submerged chain and flight equipment and bearings

No unusual construction is required

Difficulty in maintaining a 0.3 m/s velocity over a wide range of flows

Flexibility to alter performance is possible by adjusting outlet control device

DisadvantagesAdvantages



63

Aerated Grit Chambers

Flexibility to remove grit can adapt to varying field conditions

Significant quantities of potentially harmful volatile organic and odors may be released from wastewaters containing these constituents

Pre-aeration may alleviate septic conditions in the incoming wastewater to improve performance of downstream treatment units

Aerated grit chambers ca also be used for chemical addition, mixing, pre-aeration, and flocculation ahead of primary treatment

Some confusion exits about design criteria necessary to achieve a good spiral roll pattern removal system

By controlling the rate of aeration, a grit of relatively low putrescible organic content can be removed

Additional labor is required for maintenance and control of the aeration system

Head loss through the grit chamber is minimal

Power consumption is higher than other grit removal processes

The same efficiency of grit removal is possible over wide flow range

DisadvantagesAdvantages

64

Optimum velocity is 3 m/s0.15-0.4Horizontal velocity, m/s

Function of velocity and channel length

15-90Detention time at peak flow, s

Based on theoretical length20-50Allowance for inlet and outlet turbulence, %

Function of channel depth and grit velocity

3-25Length, m

Depend on channel area and flow rate

0.6-1.5Water depth, m

Dimension

CommentRangeItem

Table 2-2: Horizontal Flow Grit Chambers Typical Design Criteria (Metcalf& Eddy, 1999)



65

Table 2-3: Aerated Grit Chambers Typical Design Criteria (Metcalf & Eddy, 1999)

Provide valves and flow meters to allow proper adjustment

0.6 0.75Transverse roll velocity, m/s

0.45 typical0.27 0.74

Medium to coarse bubble

Air supply m3/min-m

Type of diffuser

3 typical2 5 Minimum detention time (at peak flow), min

Varies widely

4: 1 typical

1.5:1 typical

0.2-5

3:1 5:1

1.1 5.1

Dimensions

Depth, m

L: W ratio

W: D ratio

CommentRangeItem

66

Type II: Primary Sedimentation Tanks

Primary sedimentation tanks are designed to:1. Reduce solids loading to minimize operational

problems in downstream treatment processes.2. Lower the oxygen demand.3. Decrease the rate of energy consumption for

oxidation of particulate matter.

These effects enhance soluble substrate removal during aeration and reduce the volume of waste activated sludge that is generated. Also removes floating material, thereby minimizing operational problems in downstream treatment processes (scum build-up in secondary treatment processes. If efficiently designed, removal rates of 50 to 70% of suspended solids and 25 to 40% of BOD5 can be realized.



67

Flocculent Settling

Depends Upon:1. Overflow Rate2. Depth of Basin3. Velocity Gradients in the Basin4. Concentration of Particles5. Range of Particle Sizes6. Settling Column Analysis of

Flocculating Particles.

68

Example 2-7:

A column analysis of flocculating suspension is performed in the settling column shown below. The initial solids concentration is 250 mg/L. The data is also shown below. What will be the overall removal efficiency of a settling basin that is 3 meters in depth with a retention time of 1 hour and 45 minutes



69

Solution:1) Determine the removal rate at each depth and time.

70

2) Plot iso-concentration lines as shown is the accompanying figure.3) Construct vertical line at to = 105 min.4) Calculate the percentage removal using the following equation:



71

72

This procedure is very cumbersome and time consuming. Another easier method is to calculate the removal efficiency from the average concentration in the column. For a given retention time, t, the average concentration in the column can be calculated using the following equation:



73

74

Now add 1 to both side of the last equation in the previous slide.



75

76



77

Type III Settling - Thickening

Definition: a process for removing water from sludge which produces a more concentrated slurry (i.e. end product is a liquid).

This process is not dewatering which produces end product with the properties of solid.

Importance of Thickening:1. Produce a greater volume of product water.2. Produce a smaller volume of sludge.3. Reduces the size of a sludge treatment facility,

(e.g. digester).

78

Purposes of Sedimentation Basins:

1. Clarification

2. Thickening

With respect to design of sedimentation basins (or clarifiers) it is essential to consider both of these functions.



79(Metcalf & Eddy, 1991)




81

Batch Settling Test


82

Design Considerations for Thickening of Sludge



83

In which,Q = influent flow rate to the clarifier/thickener, (L3/t)Co = influent suspended solids concentration, (M/L3)Ce = effluent suspended solids concentration, (M/L3)Ac = area available for clarification, (L2)AT = area available for thickening, (L2)Qu = flow rate leaving bottom of the

clarifier/thickener, (L3/t)Cu = suspended solids concentration leaving the

bottom of the clarifier/thickener, (M/L3)Vi = settling velocity of the suspended solids, (L/t)Ci = suspended solids concentration of the blanket,

(M/L3)Ub = velocity of the solids in the underflow due to

pumping (L/t)

84

Assume: Ce



85

In which,GT = total solids flux in the clarifier (M/L2-t)Gg = settling flux in the clarifier due to gravity (M/L2-t)Gb = bulk flux due to under flow pumping (M/L2-t)

2) Sizing a clarifier for thickening based on solids flux method.

( ).T

masssolidsG Totalflux

area time=

In which, area is equal to the cross sectional area of the sludge blanket

GT = Gg + Gb

86

At the bottom of the clarifier, Ci = Cu so the expression for Gb can be written as:

u ub b u

T

Q CG U C

A= =

Since the product of QuCu is usually unknown, it can be replaced by QCo from the above mass balance, QCo = QuCu.



87

In addition, at the bottom of the clarifier, Gg = 0, therefore, the area for thickening can be rewritten as:

u u o oT

L L L L b

Q C QC QCA

G G C (V U )= = =

+




89


90




91


92

Alternative definition sketch for the analysis of settling data using the solids-flux method of analysis (Metcalf & Eddy, 1991).



93

Example 2-8:

Given the treatment scheme, estimate the maximum concentration of the aerator mixed-liquor biological suspended solids concentration that can be maintained if the sedimentation tank application rate is fixed at 600 gal/ft2-d and the sludge recycle rate, Qr, is 40% of Q. The following settling data was obtained from operation of a pilot plant.

94



95

Solution1. Develop the gravity solids-flux curve from the given data.

96

2. Determine the underflow bulk velocity and plot the curve on the same graph as the gravity solids flux curve.

b 2

3

0.4Q gal 1 ftU 600 0.95

gal hQ 0.4Q ft d h7.48 24ft d

= = + -

Ub is the slope of the curve for the underflow solids flux. The underflow flux curve can be plotted using the following equation:



97

b i bG kC U=

In which,

98

3. Develop the total flux curve for the system by summing the gravity flux curve with the underflow flux curve, and determine the value of the limiting flux and the maximum underflow concentration.



99

21,800 /R uC C mg L= =

100



101

Example 2-9:

102

Given the settling data in the following table derived from an activated sludge pilot plant, determine the limiting solids flux values when the concentration of the recycled solids concentration is 10,500 and 15,000 mg/L, respectively. Determine the recycle rate to the sedimentation tank required for thickening in conjunction with the aeration tank, if the MLSS in the aeration tank is to be maintained at 5,000 mg/L and the underflow concentration from the sedimentation tank is 12,000 mg/L. Neglect the effect of biological growth in the sedimentation tank and assume the flow is equal to 1.0 MGD.



103

104



105

106



107

10,500

108For an underflow concentration of 12,000 mg/L the limiting solidflux, SFL, = 1.8 lb/ft2-h



109

110

22

(1 0.71)(1.0 )(5,000 )8.34 1,650

(1.8 / )(24 / )

lbmgMGD

MGLA ftmglb ft hr hr dL

+= =

-

The corresponding surface hydraulic loading rate is 606 gal/ft2-d.



111

Table 2-4: Typical Design Information for Secondary Clarifiers(Metcalf & Eddy, 1999)

Documents

Settling Design