Sedimentation is one of the most basic processes of water treatment. Plain sedimentation, such as the use of a pre-sedimentation basin (grit chamber) and sedimentation tank (or basin) following coagulation—flocculation, is most commonly used in water treatment facilities. The grit chamber is generally installed upstream of a raw water pumping station to remove larger particles or objects. It is usually a rectangular horizontal-flow tank with a contracted inlet and the bottom should have at minimum a 1:100 longitudinal slope for basin draining and cleaning purposes. A trash screen (about 2-cm opening) is usually installed at the inlet of the grit chamber.

Sedimentation is a solid–liquid separation by gravitational settling. There are four types of sedimentation: discrete particle settling (type 1); flocculant settling (type 2); hindered settling (type 3); and compression settling (type 4). Sedimentation theories for the four types are discussed in Chapter 1.7 and elsewhere (Gregory and Zabel, 1990).

The terminal settling velocity of a single discrete particle is derived from the forces (gravitational force, buoyant force, and drag force) that act on the particle. The classical discrete particle settlingtheories have been based on spherical particles. The equation is expressed as follow

u = [4gp – ρ)d/3Cd.ρ]1/2     …..  (1)

where

u = settling velocity of particles, m/s or ft/s

g = gravitational acceleration, m/s2 or ft/s2

ρp = density of particles, kg/m3 or lb/ft3

ρ = density of water, kg/m3 or lb/ft3

d = diameter of particles, m or ft

Cd = coefficient of drag

The values of drag coefficient depend on the density of water (ρ), relative velocity (u), particle diameter (d), and viscosity of water (µ), which gives the Reynolds number R as :

R = ρud/µ     …..  (2)

The value of Cd decreases as the Reynolds number increases. For R less than 2 or 1, Cd is related to R by the linear expression as follows :

Cd  = 24/R    …..  (3)

At low values of R, substituting Eq. (2) and (3) into Eq. (1) gives

u = gp – ρ)d2/18µ   …..  (4)

This expression is known as the Stokes equation for laminar flow conditions.

In the region of higher Reynolds numbers (2 < R < 500–1000), Cd becomes (Fair et al. 1968):

Cd = (24/R) + (3/√R) + 0.34    …..  (5)

In the region of turbulent flow (500-1000 < R < 200,000), the Cd remains approximately constant at 0.44. The velocity of settling particles results in Newton’s equation (ASCE and AWWA 1990):

u = 1.74 [(ρp – ρ)gd/ρ]1/2   …..  (6)

When the Reynolds number is greater than 200,000, the drag force decreases substantially and Cd becomes 0.10. No settling occurs at this condition.

### Overflow Rate

For sizing the sedimentation basin, the traditional criteria used are based on the overflow rate, detention time, weir loading rate, and horizontal velocity. The theoretical detention time is computed from the volume of the basin divided by average daily flow (plug flow theory):

t = 24V/Q        ……………     (6.66)

where:

t = detention time,

h 24 = 24 h per day

V = volume of basin, m3 or million gallon (Mgal)

Q = average daily flow, m3/d or Mgal/d (MGD)

The overflow rate is a standard design parameter which can be determined from discrete particle settling analysis. The overflow rate or surface loading rate is calculated by dividing the average daily flow by the total area of the sedimentation basin as follows:

u = Q/A = Q/lw          ……… (6.67)

where:

u = overflow rate, m3/(m2 . d) or gpd/ft2

Q = average daily flow, m3/d or gpd

A = total surface area of basin, m2 or ft2

and w = length and width of basin, respectively, m or ft

For alum coagulation u is usually in the range of 40 to 60 m3/(m2 _ d) (or m/d) (980 to 1470 gpd/ft2) for turbidity and color removal. For lime softening, the overflow rate ranges 50 to 110 m/d (1230 to 2700 gpm/ft2). The overflow rate in wastewater treatment is lower, ranging from 10 to 60 m/d (245 to 1470 gpm/ft2). All particles having a settling velocity greater than the overflow rate will settle and be removed.

It should be noted that rapid particle density changes due to temperature, solid concentration, or salinity can induce density current which can cause severe short-circuiting in horizontal tanks (Hudson, 1972).

EXAMPLE: A water treatment plant has four clarifiers treating 4.0 MGD (0.175 m3/s) of water. Each clarifier is 16 ft (4.88 m) wide, 80 ft (24.4 m) long, and 15 ft (4.57 m) deep. Determine: (a) the detention time, (b) overflow rate, (c) horizontal velocity, and (d) weir loading rate assuming the weir length is 2.5 times the basin width.

Solution:

Step 1. Compute detention time t for each clarifier

Q = 4 MGD/4 = 1,000,000 gal/day x 1 ft3/7.48 gal x 1 day/24 h

= 5570 ft3/h

=  92.83 ft3/min

a). t = V/Q = (16 ft x 80 ft x 15 ft)/5570

=3.447 h

Step 2. Compute overflow rate u, using Eq. (6.67)

b). u = Q/lw = 1,000,000 gpd

= 16 ft x 80 ft x 781 gpd/ft2

=  31.83 m3/m2 . d

Step 3. Compute horizontal velocity v

c). v = Q/wd  = 92.83 ft3/min / 16 ft x 15 ft

= 0.387 ft/min = 11.8 cm/min

d). uw = Q/2.5w = 1,000,000 gpd / 2.5 x 16 ft

= 25,000 gpd/ft =  310 m2/m . d

### Grit Chamber (pre sedimentation)

Grit originates from domestic wastes, stormwater runoff, industrial wastes, pumpage from excavations, and groundwater seepage. It consists of inert inorganic material such as sand, cinders, rocks, gravel, cigarette filter tips, metal fragments, etc. In addition grit includes bone chips, eggshells, coffee grounds, seeds, and large food wastes (organic particles). These substances can promote excessive wear of mechanical equipment and sludge pumps, and even clog pipes by deposition.

Composition of grit varies widely, with moisture content ranging from 13 to 63 percent, and volatile content ranging from 1 to 56 percent. The specific gravity of clean grit particles may be as high as 2.7 with inert material, and as low as 1.3 when substantial organic matter is agglomerated with inert. The bulk density of grit is about 1600 kg/m3 or 100 lb/ft3 (Metcalf and Eddy, Inc. 1991). Grit chambers should be provided for all wastewater treatment plants, and are used on systems required for plants receiving sewage from combined sewers or from sewer systems receiving a substantial amount of ground garbage or grit. Grit chambers are usually installed ahead of pumps and comminuting devices.

Grit chambers for plants treating wastewater from combined sewers usually have at least two hand cleaned units, or a mechanically cleaned unit with bypass. There are three types of grit settling chamber: hand cleaned, mechanically cleaned, and aerated or vortex-type degritting units. The chambers can be square, rectangular, or circular. A velocity of 0.3 m/s (1 ft/s) is commonly used to separate grit from the organic material. Typically, 0.0005 to 0.00236 m3/s (1 to 5 ft3/min) of air per foot of chamber length is required for a proper aerated grit chamber; or 4.6 to 7.7 L/s per meter of length. The transverse velocity at the surface should be 0.6 to 0.8 m/s or 2 to 2.5 ft/s (WEF 1996a). Grit chambers are commonly constructed as fairly shallow longitudinal channels to catch high specific gravity grit (1.65). The units are designed to maintain a velocity close to 0.3 m/s (1.0 ft/s) and to provide sufficient time for the grit particle to settle to the bottom of the chamber.

EXAMPLE: The designed hourly average flow of a municipal wastewater plant is 0.438 m3/s (10 Mgal/d). Design an aerated grit chamber where the detention time of the peak flow rate is 4.0 min (generally 3 to 5 min).

Solution:

Step 1. Determine the peak hourly flow Q

Using a peaking factor of 3.0

Q = 0.438 m3/s x 3

= 1.314 m3/s

= 30 Mgal/d

Step 2. Calculate the volume of the grit chamber

Two chambers will be used; thus, for each unit

Volume = 1.314 m3/s x 4 min x 60 s/min : 2

= 157.7 m3

= 5570 ft3

Step 3. Determine the size of a rectangular chamber

(Assume): Select the width of 3 m (10 ft), and use a depth-to-width ratio of 1.5 : 1 (typically 1.5 : 1 to 2.0 : 1)

Depth = 3 m x 1.5 = 4.5 m

= 15 ft

Length = volume/(depth x width) = 157.7 m3/(4.5 m x 3 m)

= 11.7 m

= 36 ft

Note: Each of the two chambers has a size of 3 m x 4.5 m x 11.7 m or 10 ft x 15 ft x 36 ft.

Step 4. Compute the air supply needed

Use assume: 5 std ft3/min (scfm) or (0.00236 m3/s/ft length.

Air needed = 0.00236 m3/(s . ft) x 36 ft

= 0.085 m3/s or = 5 ft3/min x ft x 36 ft

= 180 ft3/min

Step 5. Estimate the average volume of grit produced

Assume 52.4 mL/m3 (7 ft3/Mgal) of grit produced

Volume of grit = 52.4 mL/m3 x 0.438 m3/s x 86,400 s/d

= 1,980,000 mL/d

= 1.98 m3/d

or = 7 ft3/Mgal x 10 Mgal/d

= 70 ft3/d 