Sedimentation

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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. Th
e 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 loa
d.



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 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, filt
ration etc. In the filter unit, the quality of the
output water is proportional to the quality of the input water. See F.3.2.5
-
1, p.141.



Example

Given: Final filtered water quality of .5NTU

Find: Water quality of settled.

From: F.3.2.5
-
1, p.141.

Anywhere f
rom 1
-
6 units, silica scale



Sedimentation, Page. No.
2

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.



D
iscrete 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 =
g
18


(

s

-


)d
2


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


s

= mass density of the particle, kg/m
3



= mass density of the fluid, kg/m
3

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/ft
2
.


From: T3.2.5
-
1. p.143
.

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/ft
2
.

.04ft/min x 7.48gal/ft
3

=
.30gpm/ft
2



The efficiency of an ideal horizontal flow sedimentation tank is a function of the
settling velocity, v
0
, the surface area of the tank, A, and the flow, Q, throug
h the tank.
v
0

is commonly referred to as the surface loading or overflow rate with units of
gpm/ft
2
. 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 rem
oval rates
and a longer detention time favors flocculation.

v
0

= Q/A



Example

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

Find: v
0

,gpm/ft
2

v
0

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

v
0

= 28.93 gpm/ft
2





Ideal settling involves the following elements:

Sedimentation, Page. No.
3


-

T
ype 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.


H, inlet
Sludge
Outlet
H
L
V, horizontal
Vs, settling




v
s

or v
0

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 v
0
. A smaller, sl
ower particle will have a
shallower, less steep slope and be inclined to be removed via the outlet. When such a
light particle, v<v
0
, starts below the top of the water line at the inlet a portion of such
light particles will be removed. If a particle is he
avier than v
0
, v>v
0
, its slope will be
steeper than the v
0

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:

Sedimentation, Page. No.
4

Given: 2 tanks,

=100’, d=10’, Q=14MGD

Find: t, OR

1.) t

t =
V
/Q = 2tanks x

4

100
2

x 10’ / 14MG/10
6
gal x
7.48 gal
ft
3

t = .08388days = 2.01hours = 120.8min

2.) OR

OR = Q/Ap = 14MGD x 10
6
gal/MG / (2tanks x

4

100)
2


OR = 891.7 gpd/ft
2




Removal r
a
tes

For v>v
0
, 100% removed

For v<v
0
, some will be removed, but how much

fraction of particles removed = v
1
/v
0

= H
1
/H

mathematically,

fraction removed = (1
-
X
s
) +
1
v
0

vdx
Xs
0


(1
-
X
s
) = fraction of the
particles with v>v
0
, all of these particles will be removed.


1
v
0

vdx
Xs
0


= fraction of the particles with v<v
0
, a portion will be removed.


V, settling velocity
fraction of
particles with
less than
the stated
velocity
0
1.0
V
V
0
X
X
s
1-X
s
100% removed
V/V
0




Example:

Given: A settling basin is designed to have a surfac
e overflow rate of 32.6 m/day = .37mm/s
(800gpd/ft
2
).

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.

=1.027,

=0.9997


Sedimentation, Page. No.
5


Sample calculations for the table below:

v, Stokes Law:

v =
g
18


(

s
-


)d
2

=
9800
18 x 1.027

(1.2
-

.997)d
2


v = 107.62d
2

for d=.10mm

v = 107.62(.10)
2

v=1.076 say 1.08

for d=.04

v = 107.62(.04)
2

v=.172

Reynol
ds number, if the Nr < .5, Stokes Law applies.

Nr=

v/


= (.10mm x 1.08mm/s) / 1.011x10
-
6
m
/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.


Particle
size, mm

0.10

0.08

0.07

0
.06

0.04

0.02

0.01

Weight
fraction
greater
than
size,
percent

10

15

40

40% of
the
particles
> .07

70

93

99

100

100% of
the
particles
> .01

Weight
fraction
less than
size,
percent

90

90% of
the
particles
pass the
.10 sieve

85

60

30

7

7% of the
particles
p
ass the
.04 sieve

1

0

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

Sedimentation, Page. No.
6

V, settling velocity
fraction of
particles with
less than
the stated
velocity
0
1.0
V
X
1-X
s
100% removed
V/V
0
Xs=.27
V
0
=.37mm/s(800gal.day,ft
2
)


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 as
sume increments of

x, say 0.04, and pick off the corresponding v, velocity,
from the graph. The resulting product

x(v) is the area for that increment. The increments are then
summed to obtain the total area.


Total

x(v) = .0635


The overall removal is:

fraction removed = (1
-
X
s
) +
1
v
0

vdx
Xs
0


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

fraction remove
d = .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 carr
ying 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 t
wo
-
thirds of the tank length.



The magnitude to the density current can by evaluated via Harleman’s formula:


v = [8g





hs
f
(
1+

)
]
.5

units p.147



or




x

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


x(v)

0.0024

0.0064

0.0088

0.0104

0.0120

0.0136

.0099

Sedimentation, Page. No.
7

v = [2g






h

k
)
]
.5

units p.147







The temperature diffe
rence 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 rec
tangular.



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 a
nd minimum
maintenance cost.



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


Surface loading: .34
-
1 gpm/ft
2
; 490
-
1440 gpd/ft
2

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 inl
et 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.

Sedimentation, Page. No.
8


-

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 co
mpromising 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 reco
mmended
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/


< 20,000

N
F

= v
2
/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 spot
s 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 traveli
ng 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. T
he 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.

Sedimentation, Page. No.
9

3. Operations and Maintenance




Floc settling. The majority of the floc should settle in the first half o
f 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 withdr
awal process.


4. Example Problems


Given: Grit Chamber Design, Q=85MGD, .15mm minimum size to be removed, 10

C. Consult the
g
rit chamber criteria on page 171
. 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 .15m
m sand from T3.2.5
-
1.

v
0

= 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 = 350ft
2

v = Q/
A = 3945 cfm / 350ft
2

v = 11.27fpm

L = K(h/ v
0
)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. 171
:

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

L
/
D = 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

Sedimentation, Page. No.
10


4.) surface loading

OR = Q/A = 3945 cfm / LW = 3945cfm x
7.48 gal
ft
3

/ (56.5ft x 35ft)

OR = 14.92g
pm/ft
2
, 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, Sedimentation

Read Chapter 3 pp. 139
-
194

Problems:


6A.

Given: Final filtered water quality of 1 NTU

Find: Wa
ter 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/ft
2
.


6C.

Given: The Weymouth Filtration Plant has a flow of 300MGD using square tanks t
o
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 ove
rall 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%



1
0/10=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

Sedimentation, Page. No.
11

4.) surface loading



6E.

Given: Design t
he 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 f
or your project.