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Sedimentation

Lecture No. 6

1. Purpose

Probably the most common waste and wastewater treatment process. Also known as clarification Sedimentation is defined as the separation of a suspension into a clarified fluid and a

more concentrated suspension. The more concentrated suspension is typically knownas sludge.

The sedimentation process is designed to remove a majority of the settleable solids by

gravity. Sedimentation is an efficient process; in addition, downstream processes haveto deal with less load.

Sedimentation is divided into two classifications:- grit chambers, plain sedimentation, Type I, discrete, unhindered settling- sedimentation tanks, clarifiers, Type II, hindered settling

The key to successful settling is proper upstream coagulation and flocculation. Main configurations of the settling tanks:

- horizontal, rectangular basins, , favored- upflow sedimentation tanks- upflow reactor clarifiers

2. Considerations

A. Overall Treatment Process.

If there are big particles in the water, >15um, as might be found in river water, a gritchamber is in order. A conservatively designed sedimentation basin should be used toobtain a settle water turbidity of

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Sedimentation, Page. No.2

B. Nature of Suspended Matter.

Raw water contains 2 basic types of suspended matter: discrete (nonflocculable) suchas sand and silt and colloidal suspensions such as clay, microorganisms and substancesthat cause color.

Discrete particles are relatively easy to remove and removal strategies include: gritchamber, plain sedimentation, cyclone separator.

C. Settling Velocity of the Particles.

The sedimentation process is based on the gravitational settling of the particles. Type I settling can be described by Stokes Law:

v = (s - )d2

In which v=settling velocity, fps or m/s

s = mass density of the particle, kg/m3

= mass density of the fluid, kg/m3

d = diameter of the particle, ft or m

Example:Given: An alum flocFind: 1.)specific gravity 2.)particle size in mm 3.)If the settling rate is .04fpm, convert this valueto gpm/ft2.

From: T3.2.5-1. p.143.1.) specific gravityspecific gravity=1.001

2.)particle size in mm

particle size = 1-4mm

3.)If the settling rate is .04fpm, convert this value to gpm/ft2..04ft/min x 7.48gal/ft3 = .30gpm/ft2

The efficiency of an ideal horizontal flow sedimentation tank is a function of thesettling velocity, v0, the surface area of the tank, A, and the flow, Q, through the tank.v0 is commonly referred to as the surface loading or overflow rate with units ofgpm/ft2. According to Hazen, the efficiency, removal rate, of a tank is independent ofdepth and detention time. In practice, a shallow tank will have better removal rates anda longer detention time favors flocculation.

v0 = Q/A

ExampleGiven: A tank: L=100, W=20, D=12, Q=10MGDFind: v0 ,gpm/ft

2

v0 = Q/A = 10MGD x 694.4gpm/MGD/ (20x12)v0 = 28.93 gpm/ft

2

Ideal settling involves the following elements:- Type I, discrete settling- Even distribution of flow entering the basin

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- 3 zones: inlet, outlet and sludge- Uniform distribution of particles throughout the depth of the entrance zone.- Particles that enter the sludge zone stay in the sludge.

H, inlet

Sludge

OutletH

L

V, horizontal

Vs, settling

vs or v0 is the settling velocity of the smallest particle size that is 100% removed.

Removed means captured in the sludge layer. A smaller particle will be lighter andtherefore will settle at a velocity slower than v0. A smaller, slower particle will have ashallower, less steep slope and be inclined to be removed via the outlet. When such alight particle, vv0, its slope will besteeper than the v0 and 100% of these heavier particles will reach the sludge layer.

Detention time.t = H/Vs = L/VhVh = L/tbutVh = Q/Ax-sectionAx-section = HWAx-section =HWVh = Q/HWSubstitutingL/t = Q/HWt = LHW/Q, LHW=Vt = V/Q

Overflow ratet = H/Vs = LHW/Qtherefore,Vs = Q/LW, LW=Plan area, Ap, also known as the surface area.Vs = Q/ApShows that the overflow rate is equivalent to the settling velocity of smallest particle thatis 100% removed.

Example:Given: 2 tanks, =100, d=10, Q=14MGDFind: t, OR1.) tt = V/Q = 2tanks x 1002 x 10 / 14MG/106gal x

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t = .08388days = 2.01hours = 120.8min2.) OROR = Q/Ap = 14MGD x 106gal/MG / (2tanks x 100)2OR = 891.7 gpd/ft2

Removal rates

For v>v0, 100% removedFor vv0, all of these particles will be removed.

vdxXs

0 = fraction of the particles with v .07

70 93 99 100100% of

theparticles

> .01

Weightfractionless than

size,percent

9090% of

theparticlespass the .10 sieve

85 60 30 77% of theparticlespass the .04 sieve

1 0

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v = 107.62(.10)2

v=1.076 say 1.08for d=.04v = 107.62(.04)2

v=.172Reynolds number, if the Nr < .5, Stokes Law applies.

Nr=v/ = (.10mm x 1.08mm/s) / 1.011x10-6m/s x (1000mm/m)2

Nr=.10

Plot the above numbers:v vs. weight fraction remaining , e.g. 1.08, 90.0; 0.689,85 etc.

V, settling velocity

fraction ofparticles with

less than

the statedvelocity

0

1.0

V

X

1-Xs 100% removed

V/V0

Xs=.27

V0=.37mm/s(800gal.day,ft2)

All particles with a settling velocity greater than .37mm/s will be 100% removed. From thegraph, the fraction (1-Xs) is equal to 0.73 or 73%; a portion of the remaining 27% will beremoved, graphically this is the area above the settling curve, but below the Xs line. One way to

obtain this desire area is to assume increments ofx, say 0.04, and pick off the corresponding v,velocity, from the graph. The resulting product x(v) is the area for that increment. Theincrements are then summed to obtain the total area.

Total x(v) = .0635

The overall removal is:

fraction removed = (1-Xs) + vdxXs

0

Weightfraction,

%

10.0 15.0 40.0 70.0 93.0 99.0 100

v, mm/s,fromabovecalc.

1.08 0.689 0.527 0.387 0.172 0.043 0.011

Nr, .10 0.05 0.04 0.02 0.01 0.001 0.0001

Weight

fractionremaining

%

90.0 85.0 60.0 30.0 7.0 1.0 0

x 0.04 0.04 0.04 0.04 0.04 0.04 .027v 0.06 0.16 0.22 0.26 0.30 0.34 0.37

x(v) 0.0024 0.0064 0.0088 0.0104 0.0120 0.0136 .0099

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fraction removed = 0.73 + 1/.37(0.0635)fraction removed = .898 = 89.9%

D. Flow Short Circuiting

Three types:- improper design: poor inlet design, short distance between the inlet and the outlet.- when floc is carried over the filter: the influent tends to dive down at the inlet and rise atthe outlet carrying much floc with it.- density flow: severe type of the second, typically caused by switching from one source ofwater supply to another. Can be minimized by installing intermediate diffuser wallperpendicular to the flow direction in the middle or at two-thirds of the tank length.

The magnitude to the density current can by evaluated via Harlemans formula:

v = [8g ].5 units p.147

or

v = [2g ].5 units p.147

The temperature difference involved density flows are 0.2-0.5C and the flow velocityof the density flow is 2.6-6fpm with a design or intended flow velocity of 1.3fpm.

E. Type of Sedimentation Tank

The types include: upflow clarifiers, reactor clarifiers and horizontal rectangular. Upflow clarifiers, reactor clarifiers are susceptible to hydraulic and solids shock

loadings.

Most large water treatment plants use horizontal rectangular clarifiers primarilybecause of the flexible performance, predictable settling efficiency and minimummaintenance cost.

Design criteria include T3.2.5-2, p.150:

Surface loading: .34-1 gpm/ft2; 490-1440 gpd/ft2

Water depth: 3-5 m; 10-16 ftDetention time; 1.5-3 hoursWidth:length ratio > 1:5, minimum 4:1Width:depth ratio: 3:1 with a maximum of 6:1Freeboard: 2ftWeir loading:

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The preferred configuration of the multiple rectangular tanks is common wallconstruction all connected to a common inlet and outlet.

F. Inlet and Outlet of the Basin

Flow imbalance at the inlet will lead to flow short-circuiting, jetting, turbulences andhydraulic instability.

The most simple and effective method for distributing the water from the flocculationtanks to the sedimentation tanks is a perforated baffle wall whose requirements are asfollows:

- The wall should cover the entire cross section of the basin- The wall should be uniformly perforated- A maximum of ports should be provided to minimize jets and dead zones- The headloss through an individual port should be .12-.35

- The headloss through an individual port should be less than .4 to prevent flocbreakage.

- The size of the ports should be uniform in diameter, 3-8 to avoid clogging- The ports should be placed no more than approximately 10-20 on center to

avoid compromising the structural strength of the wall.- The flow should be directed at the basin outlet.

The water exiting the basin should be uniformly collected across an area that isperpendicular to the proper flow direction. Perforated baffles are not recommendedfor the outlet because they are not effective in dealing with density currents.

V-notched weir plates are used for the outlets and are generally attached to thelaunders. Launders are long troughs which channel the water to an outlet. Long

launders have major advantages: the water level of the tank remains substantiallyconstant; wave action is minimized; weirs and modules are easily attached to thelaunders.

G. Shape of the Tank

Rectangular basins that are both wide and deep tend to hydraulically unstable andfoster density flow patterns. Basins that are narrow, shallow and long have flowstability and minimize short circuiting.

Flow characteristics of the sedimentation basin can be estimated by the Reynolds, NR,and Froude, NF, numbers:

NR = vR/ < 20,000NF = v2/gR > 10-5 units p.161

An ordinary basin has NR >15,000 and NF

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Q/tank = 3945 cfmA = WD = 10x 35A = 350ft2

v = Q/A = 3945 cfm / 350ft2

v = 11.27fpmL = K(h/ v0)v, where K=1.5, equation from rear end of author.L = 1.5(10/3.0)11.27

L = 56.35ft say 56.5L = 56.5ft, D=10ft, W=35ft

check ratiosFrom p. 171:WL is from 1:4: to 1:8DL is a minimum of 1:8W/L = 56.5/35W/L = 1:1.614, NGL/D = 56.5/10 = 5.65:1, NG

3.) tt=V/Q = LWD/Q = 56.5ft x10ft x35ft / 3945 cfmt = 5.01min, p.160 should be between 6-15, NG

4.) surface loadingOR = Q/A = 3945 cfm / LW = 3945cfm x / (56.5ft x 35ft)OR = 14.92gpm/ft2, p.160 should be 4-10, therefore NG.

Go over example problems in book, p.171,especially part iii, baffle wall design

HOMEWORK No. 6, SedimentationRead Chapter 3 pp. 139-194Problems:

6A. Given: Final filtered water quality of 1 NTUFind: Water quality of settled.

6B. Given: A silt and clay floc, size.06mmFind: 1.)specific gravity 2.)mesh size 3.)If the settling rate is .75fpm, convert this value togpm/ft2.

6C. Given: The Weymouth Filtration Plant has a flow of 300MGD using square tanks to adepth of 12'. DT=2hours.

Find:1.)The volume and surface area2.) The number of square tanks such that no dimension exceeds 200' which is anequipment limitation.

6D. Given:Depth=10', overflow rate=.0417fps and the settling data below.Find: The overall removal percentage assuming Type I, discrete settlingtime required portion of particles Vs (fpm)

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to settle 10' with velocity less Vs=distance/time(minutes) than those indicated3.33 60% 10/3.33=35.0 40% 10/5.0=210.0 20% 10/10=1

6E. Given: Grit Chamber Design, Q=55MGD, .03mm minimum size to be removed, 10C.Consult the grit chamber criteria on page 160. W=30ft.Find:1.) The number and shape of the tanks2.) Tank dimensions3.) t4.) surface loading

6E. Given: Design the sedimentation tanks for your project. Use example 3, p.163 as aguide.Find: Include the following items:1.) The number of tanks2.) Tank dimensions3.) The configuration of the tank inlet and diffuser wall.

6F. Given: Design the sedimentation tanks for your project.

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