Sedimentation

PHEH 401 for SAN4

Asst prof Shalasai Huangprasert

Dept. of Env H Sc , Public Health ,MAHIDOL U.

1. Purpose

· Probably the most common waster 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 known as 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 have to 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, Ö, 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 grit chamber is in order. A conservatively designed sedimentation basin should be used to obtain a settle water turbidity of <2NTU. The sedimentation would NOT do it alone, but in combination with chemicals, filtration etc. In the filter unit, the quality of the output water is proportional to the quality of the input water..

· Example

Given: Final filtered water quality of .5NTU

Find: Water quality of settled.

Anywhere from 1-6 units, silica scale

B. Nature of Suspended Matter.

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

· Discrete particles are relatively easy to remove and removal strategies include: grit chamber, 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 = (r_s - r)d²

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

r_s = mass density of the particle, kg/m³

r = mass density of the fluid, kg/m³

d = diameter of the particle, ft or m

· Example:

Given: An alum floc

Find: 1.)specific gravity 2.)particle size in mm 3.)If the settling rate is .04fpm, convert this value to gpm/ft².

1.) specific gravity

specific 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 the settling 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 of gpm/ft². According to Hazen, the efficiency, removal rate, of a tank is independent of depth and detention time. In practice, a shallow tank will have better removal rates and a longer detention time favors flocculation.

v₀ = Q/A

· Example

Given: A tank: L=100’, W=20’, D=12’, Q=10MGD

Find: v₀ ,gpm/ft²

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

v₀ = 28.93 gpm/ft²

· Ideal settling involves the following elements:

- Type I, discrete settling

- Even distribution of flow entering the basin

- 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.

· v_s 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 and therefore will settle at a velocity slower than v₀. A smaller, slower particle will have a shallower, less steep slope and be inclined to be removed via the outlet. When such a light particle, v<v₀, starts below the top of the water line at the inlet a portion of such light particles will be removed. If a particle is heavier than v₀, v>v₀, its slope will be steeper than the v₀ and 100% of these heavier particles will reach the sludge layer.

· Detention time.

t = H/V_s = L/V_h

V_h = L/t

but

V_h = Q/Ax-section

A_x-section = HW

A_x-section =HW

V_h = Q/HW

Substituting

L/t = Q/HW

t = LHW/Q, LHW=V

t = V/Q

· Overflow rate

t = H/V_s = LHW/Q

therefore,

V_s = Q/LW, LW=Plan area, A_p, also known as the surface area.

V_s = Q/A_p

Shows that the overflow rate is equivalent to the settling velocity of smallest particle that is 100% removed.

· Example:

Given: 2 tanks, f=100’, d=10’, Q=14MGD

Find: t, OR

1.) t

t = V/Q = 2tanks x 100² x 10’ / 14MG/10⁶gal x

t = .08388days = 2.01hours = 120.8min

2.) OR

OR = Q/Ap = 14MGD x 10⁶gal/MG / (2tanks x 100)²

OR = 891.7 gpd/ft²

· Removal rates

For v>v₀, 100% removed

For v<v₀, some will be removed, but how much

fraction of particles removed = v₁/v₀ = H₁/H

mathematically,

fraction removed = (1-X_s) +

(1-X_s) = fraction of the particles with v>v₀, all of these particles will be removed.

= fraction of the particles with v<v₀, a portion will be removed.

· Example:

Given: A settling basin is designed to have a surface overflow rate of 32.6 m/day = .37mm/s (800gpd/ft²).

Find: The overall removal obtained for a suspension with the size distribution given below. The specific gravity of the particles is 1.2 and T=20°C. m=1.027, r=0.9997

Particle size, mm

0.10

0.08

0.07

0.06

0.04

0.02

0.01

Weight fraction greater than size, percent

40% of the particles > .07

100

100% of the particles > .01

Weight fraction less than size, percent

90% of the particles pass the .10 sieve

7% of the particles pass the .04 sieve

Sample calculations for the table below:

v, Stokes Law:

v = (rs - r)d² = (1.2 - .997)d²

v = 107.62d²

for d=.10mm

v = 107.62(.10)²

v=1.076 say 1.08

for d=.04

v = 107.62(.04)²

v=.172

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

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

Nr=.10

Weight fraction, %	10.0	15.0	40.0	70.0	93.0	99.0	100
v, mm/s, from above calc.	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 fraction remaining %	90.0	85.0	60.0	30.0	7.0	1.0	0

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

All particles with a settling velocity greater than .37mm/s will be 100% removed. From the graph, the fraction (1-X_s) is equal to 0.73 or 73%; a portion of the remaining 27% will be removed, 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 of Dx, say 0.04, and pick off the corresponding v, velocity, from the graph. The resulting product Dx(v) is the area for that increment. The increments are then summed to obtain the total area.

Dx	0.04	0.04	0.04	0.04	0.04	0.04	.027
v	0.06	0.16	0.22	0.26	0.30	0.34	0.37
Dx(v)	0.0024	0.0064	0.0088	0.0104	0.0120	0.0136	.0099

Total Dx(v) = .0635

The overall removal is:

fraction removed = (1-X_s) +

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 at the outlet carrying much floc with it.

- density flow: severe type of the second, typically caused by switching from one source of water supply to another. Can be minimized by installing intermediate diffuser wall perpendicular 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 Harleman’s formula:

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

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

· The temperature difference involved density flows are 0.2-0.5°C and the flow velocity of 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 primarily because of the flexible performance, predictable settling efficiency and minimum maintenance 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 ft

Detention time; 1.5-3 hours

Width:length ratio > 1:5, minimum 4:1

Width:depth ratio: 3:1 with a maximum of 6:1

Freeboard: 2ft

Weir loading: <15gpm/ft; 21,600 gpd/ft

· The preferred configuration of the multiple rectangular tanks is common wall construction 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 and hydraulic instability.

· The most simple and effective method for distributing the water from the flocculation tanks to the sedimentation tanks is a perforated baffle wall whose requirements are as follows:

- 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 floc breakage.

- 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 is perpendicular to the proper flow direction. Perforated baffles are not recommended for 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 the launders. Launders are long troughs which channel the water to an outlet. Long launders have major advantages: the water level of the tank remains substantially constant; wave action is minimized; weirs and modules are easily attached to the launders.

G. Shape of the Tank

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

· Flow characteristics of the sedimentation basin can be estimated by the Reynolds, N_R, and Froude, N_F, numbers:

N_R = vR/n < 20,000

N_F = v²/gR > 10^-5 units p.161

An ordinary basin has N_R >15,000 and N_F <10^-6 both of which indicate an inferior condition.

· One of the least desired shapes has 180° turn at its midlength because they are inefficient due to turbulence and dead spots at the turn.

H. Sludge Collection System

· Choices include: chain and flight; traveling bridge with squeegees; traveling bridge with suction; float supported sludge suction; underwater bogies with squeegee.

· Any thing with moving parts such as the traveling bridge should not be used in very cold, ice prone parts of the country.

· The chain and flight can service a maximum length of 200ft, 60m.

· The traveling bridge can service any length of tank; but it is effective if the length exceeds 260-300ft, 80-90m. The speed is typically 1fpm.

· The underflow rate associated with sludge removal of the horizontal flow and long rectangular basins is typically .1-.2% of the plant flow. The concentration of solids in the sludge is .2-5%.

I. Detailed Design Criteria

See T3.2.5-2, p.150 and p.171 for detailed design criteria for grit chambers, rectangular sedimentation tanks and sedimentation tanks with high-rate settler.

3. Operations and Maintenance

· Floc settling. The majority of the floc should settle in the first half of the tank. Visual observation is important. If the water is clear in the middle and full of floc at the end, a density flow is indicated.

· Abnormal phenomena. sludge floating (bulking); scum; fly larvae; algae; corrosion.

· Optimization of the sludge withdrawal process.

4. Example Problems

Given: Grit Chamber Design, Q=85MGD, .15mm minimum size to be removed, 10°C. Consult the grit chamber criteria on page 160. W=35ft. d=10’

Find:

1.) The number and shape of the tanks

2.) Tank dimensions

3.) t

4.) surface loading

1.) The number and shape of the tanks

Two rectangular tanks. If one goes down for a problem or maintenance, the other is still available to do the job. An alternative for a smaller plant is one tank and a by-pass channel.

2.) Tank dimensions

The settling velocity of the .15mm sand from T3.2.5-1 on page 143

v₀ = 15mm/s = 3.0fpm

From page 160:

water depth = 10-16ft, say 10ft.

Q = 85MGD x 1.547cfs/MGD

Q = 131.5cfs x 60s/min = 7890 cfm

Q/tank = 7890 cfm / 2

Q/tank = 3945 cfm

A = WD = 10’x 35’

A = 350ft²

v = Q/A = 3945 cfm / 350ft²

v = 11.27fpm

L = 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 56.5’

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

check ratios

From p. 160:

WL is from 1:4: to 1:8

DL is a minimum of 1:8

W/L = 56.5’/35’

W/L = 1:1.614, NG

LD = 56.5/10 = 5.65:1, NG

3.) t

t=V/Q = LWD/Q = 56.5ft x10ft x35ft / 3945 cfm

t = 5.01min, p.160 should be between 6-15, NG

4.) surface loading

OR = 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 , Sedimentation

Problems:

6A. Given: Final filtered water quality of 1 NTU

Find: Water quality of settled.

6B. Given: A silt and clay floc, size.06mm

Find: 1.)specific gravity 2.)mesh size 3.)If the settling rate is .75fpm, convert this value to gpm/ft².

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

Find:

1.)The volume and surface area

2.) The number of square tanks such that no dimension exceeds 200' which is an equipment limitation.

6D. Given:Depth=10', overflow rate=.0417fps and the settling data below.

Find: The overall removal percentage assuming Type I, discrete settling

time required portion of particles Vs (fpm)

to settle 10' with velocity less Vs=distance/time

(minutes) than those indicated

3.33 60% 10/3.33=3

5.0 40% 10/5.0=2

10.0 20% 10/10=1

6E. Given: Grit Chamber Design, Q=55MGD, .03mm minimum size to be removed, 10°C. Consult the grit chamber criteria on page 160. W=30ft.

Find:

1.) The number and shape of the tanks

2.) Tank dimensions

3.) t

4.) surface loading

6E. Given: Design the sedimentation tanks for your project. Use example 3, p.163 as a guide.

Find: Include the following items:

1.) The number of tanks

2.) Tank dimensions

3.) The configuration of the tank inlet and diffuser wall.

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