Tài liệu Handbook of Mechanical Engineering Calculations P13 doc

P
•
A
•
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T3
ENVIRONMENTAL
CONTROL
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
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ENVIRONMENTAL CONTROL
13.3
SECTION 13
WASTEWATER TREATMENT
AND CONTROL
Kevin D. Wills, M.S.E., P.E.
Consulting Engineer
Stanley Consultants, Inc.
Design of a Complete-Mix Activated
Sludge Reactor
13.3
Design of a Circular Settling Tank
13.10
Thickening of a Waste-Activated Sludge
Using a Gravity-Belt Thickener

13.12
Design of an Aerobic Digester
13.14
Design of an Aerated Grit Chamber
13.18
Design of Solid-Bowl Centrifuge for
Sludge Dewatering
13.20
Sizing of a Traveling-Bridge Filter
13.25
Design of a Rapid-Mix Basin and
Flocculation Basin
13.28
Sizing a Polymer Dilution / Feed System
13.30
Design of a Trickling Filter Using NRC
Equations
13.31
Design of a Plastic Media Trickling Filter
13.35
Sizing a Rotary-Lobe Sludge Pump
13.38
Design of an Anaerobic Digester
13.43
Design of a Chlorination System for
Wastewater Disinfection
13.46
Sanitary Sewer System Design
13.48
Selection of Sewage-Treatment Method

13.51
DESIGN OF A COMPLETE-MIX ACTIVATED
SLUDGE REACTOR
Domestic wastewater with an average daily flow of 4.0 Mgd (15,140 m
3
/d) has a
five day Biochemical Oxygen Demand (BOD
5
) of 240 mg /L after primary settling.
The effluent is to have a BOD
5
of 10 mg /L or less. Design a complete-mix activated
sludge reactor to treat the wastewater including reactor volume, hydraulic retention
time, quantity of sludge wasted, oxygen requirements, food to microorganism ratio,
volumetric loading, and WAS and RAS requirements.
Calculation Procedure:
1. Compute the reactor volume
The volume of the reactor can be determined using the following equation derived
from Monod kinetics:
␪
QY(S
Ϫ
S)
co
V
ϭ
r
X (1
ϩ
k

␪
)
adc
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13.4
ENVIRONMENTAL CONTROL
where V
r
ϭ
Reactor volume (Mgal) (m
3
)
␪
c
ϭ
Mean cell residence time, or the average time that the sludge remains
in the reactor (sludge age). For a complete-mix activated sludge pro-
cess,
␪
c
ranges from 5 to 15 days. The design of the reactor is based
on
␪
c
on the assumption that substantially all the substrate (BOD) con-
version occurs in the reactor, A
␪

c
of 8 days will be assumed.
Q
ϭ
Average daily influent flow rate (Mgd)
ϭ
4.0 Mgd (15,140 m
3
/d)
Y
ϭ
Maximum yield coefficient (mg VSS/mg BOD
5
). For the activated
sludge process for domestic wastewater Y ranges from 0.4 to 0.8. A Y
of 0.6 mg VSS/mg BOD
5
will be assumed. Essentially, Y represents
the maximum mg of cells produced per mg organic matter removed.
S
O
ϭ
Influent substrate (BOD
5
) concentration (mg/L)
ϭ
240 mg/L
S
ϭ
Effluent substrate (BOD

5
) concentration (mg/L)
ϭ
10 mg/L
X
a
ϭ
Concentration of microorganisms in reactor
ϭ
Mixed Liquor Volatile
Suspended Solids (MLVSS) in mg/L. It is generally accepted that the
ratio MLVSS/MLSS
Ϸ
0.8, where MLSS is the Mixed Liquor Sus-
pended Solids concentration in the reactor. MLSS represents the sum
of volatile suspended solids (organics) and fixed suspended solids (in-
organics). For a complete-mix activated sludge process, MLSS ranges
from 1,000 to 6,500 mg /L. An MLSS of 4,500 mg/L will be assumed.
ϭϾ
MLVSS
ϭ
(0.8)(4500 mg / L)
ϭ
3600 mg/L.
k
d
ϭ
Endogenous decay coefficient (d
Ϫ
1

) which is a coefficient representing
the decrease of cell mass in the MLVSS. For the activated sludge pro-
cess for domestic wastewater k
d
ranges from 0.025 to 0.075 d
Ϫ
1
.A
value of 0.06 d
Ϫ
1
will be assumed.
Therefore:
(8 d)(4.0 Mgd)(0.6 mg VSS/mg BOD )(240
Ϫ
10)mg/L
5
V
ϭ
r
Ϫ
1
(3600 mg /L)(1
ϩ
(0.06 d )(8d))
33
ϭ
0.83 Mgal (110,955 ft ) (3140 m )
2. Compute the hydraulic retention time
The hydraulic retention time (

␪
) in the reactor is the reactor volume divided by the
influent flow rate: V
r
/Q. Therefore,
␪
ϭ
(0.83 Mgal)/(4.0 Mgd)
ϭ
0.208 days
ϭ
5.0 hours. For a complete-mix activated sludge process,
␪
is generally 3–5 hours.
Therefore, the hydraulic retention time is acceptable.
3. Compute the quantity of sludge wasted
The observed cell yield, Y
obs
ϭ
Y/1
ϩ
k
d
␪
c
ϭ
0.6/(1
ϩ
(0.06 d
Ϫ

1
)(8 d))
ϭ
0.41
mg/ mg represents the actual cell yield that would be observed. The observed cell
yield is always less than the maximum cell yield (Y).
The increase in MLVSS is computed using the following equation:
P
ϭ
YQ(S
Ϫ
S)(8.34 lb /Mgal/ mg /L)
x obs O
where P
x
is the net waste activated sludge produced each day in (lb VSS/d).
Using values defined above:
mg VSS mg mg lb/ Mgal
P
ϭ
0.41 (4.0 Mgd) 240
Ϫ
10 8.34
ͩͪͩ ͪͩͪ
x
mg BOD L L mg/ L
5
ϭ
3146 lb VSS/ d (1428.3 kg VSS/ d)
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.5
This represents the increase of volatile suspended solids (organics) in the reactor.
Of course the total increase in sludge mass will include fixed suspended solids
(inorganics) as well. Therefore, the increase in the total mass of mixed liquor sus-
pended solids (MLSS)
ϭ
P
x(ss)
ϭ
(3146 lb VSS/d) / (0.8)
ϭ
3933 lb SS / d (1785.6
kg SS/ d). This represents the total mass of sludge that must be wasted from the
system each day.
4. Compute the oxygen requirements based on ultimate carbonaceous oxygen
demand (BOD
L
)
The theoretical oxygen requirements are calculated using the BOD
5
of the waste-
water and the amount of organisms (P
x
) wasted from the system each day. If all
BOD
5

were converted to end products, the total oxygen demand would be computed
by converting BOD
5
to ultimate BOD (BOD
L
), using an appropriate conversion
factor. The ‘‘Quantity of Sludge Wasted’’ calculation illustrated that a portion of
the incoming waste is converted to new cells which are subsequently wasted from
the system. Therefore, if the BOD
L
of the wasted cells is subtracted from the total,
the remaining amount represents the amount of oxygen that must be supplied to
the system. From stoichiometry, it is known that the BOD
L
of one mole of cells is
equal to 1.42 times the concentration of cells. Therefore, the theoretical oxygen
requirements for the removal of the carbonaceous organic matter in wastewater for
an activated-sludge system can be computed using the following equation:
lb O / d
ϭ
(total mass of BOD utilized, lb /d)
2L
Ϫ
1.42 (mass of organisms wasted, lb /d)
Using terms that have been defined previously where f
ϭ
conversion factor for
converting BOD
5
to BOD

L
(0.68 is commonly used):
lb/ Mgal
Q(S
Ϫ
S) 8.34
ͩ
O
ͪ
mg/L
lb O / d
ϭϪ
(1.42)(P )
2 x
ƒ
Using the above quantities:
(4.0 Mgd)(240 mg/ L
Ϫ
10 mg /L)(8.34)
lb O / d
ϭϪ
(1.42)(3146 lb /d)
2
0.68
ϭ
6816 lb O /d (3094.5 kg O /d)
22
This represents the theoretical oxygen requirement for removal of the influent
BOD
5

. However, to meet sustained peak organic loadings, it is recommended that
aeration equipment be designed with a safety factor of at least 2. Therefore, in
sizing aeration equipment a value of (2)(6816 lb O
2
/d)
ϭ
13,632 lb O
2
/d (6188.9
kg O
2
/d) is used.
5. Compute the food to microorganism ratio (F:M) and the volumetric loading
(V
L
)
In order to maintain control over the activated sludge process, two commonly used
parameters are (1) the food to microorganism ratio (F:M) and, (2) the mean cell
residence time (
␪
c
). The mean cell residence time was assumed in Part 1 ‘‘Compute
Reactor Volume’’ to be 8 days.
The food to microorganism ratio is defined as:
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WASTEWATER TREATMENT AND CONTROL
13.6
ENVIRONMENTAL CONTROL

F:M
ϭ
S /
␪
X
Oa
where F:M is the food to microorganism ratio in d
Ϫ
1
.
F:M is simply a ratio of the ‘‘food’’ or BOD
5
of the incoming waste, to the
concentration of ‘‘microorganisms’’ in the aeration tank or MLVSS. Therefore, us-
ing values defined previously:
240 mg /L
Ϫ
1
F:M
ϭϭ
0.321 d
(0.208 d)(3600 mg/ L)
Typical values for F:M reported in literature vary from 0.05 d
Ϫ
1
to 1.0 d
Ϫ
1
depending on the type of treatment process used.
A low value of F:M can result in the growth of filamentous organisms and is

the most common operational problem in the activated sludge process. A prolifer-
ation of filamentous organisms in the mixed liquor results in a poorly settling
sludge, commonly referred to as ‘‘bulking sludge.’’
One method of controlling the growth of filamentous organisms is through the
use of a separate compartment as the initial contact zone of a biological reactor
where primary effluent and return activated sludge are combined. This concept
provides a high F:M at controlled oxygen levels which provides selective growth
of floc forming organisms at the initial stage of the biological process. An F:M
ratio of at least 2.27 d
Ϫ
1
in this compartment is suggested in the literature. However,
initial F:M ratios ranging from 20–25 d
Ϫ
1
have also been reported.
The volumetric (organic) loading (V
L
) is defined as:
V
ϭ
SQ/V
ϭ
S /
␪
LOrO
V
L
is a measure of the pounds of BOD
5

applied daily per thousand cubic feet
of aeration tank volume. Using values defined previously:
33 3
V
ϭ
(240 mg /L)/ (0.208 d)
ϭ
1154 mg /L
⅐
d
ϭ
72 lb /10 ft
⅐
d (1.15 kg/ Mm
⅐
d)
L
Volumetric loading can vary from 20 to more than 200 lb /10
3
ft
3
⅐
d (0.32 to 3.2
kg/Mm
3
⅐
d), and may be used as an alternate (although crude) method of sizing
aeration tanks.
6. Compute the waste activated sludge (WAS) and return activated sludge (RAS)
requirements

Control of the activated sludge process is important to maintain high levels of
treatment performance under a wide range of operating conditions. The principle
factors used in process control are (1) maintaining dissolved-oxygen levels in the
aeration tanks, (2) regulating the amount of Return Activated Sludge (RAS), and
(3) controlling the Waste Activated Sludge (WAS). As outlined previously in Part
5 ‘‘Compute the Food to Microorganism Ratio and the Volumetric Loading,’’ the
most commonly used parameters for controlling the activated sludge process are
the F:M ratio and the mean cell residence time (
␪
c
). The Mixed Liquor Volatile
Suspended Solids (MLVSS) concentration may also be used as a control parameter.
Return Activated Sludge (RAS) is important in maintaining the MLVSS concentra-
tion and the Waste Activated Sludge (WAS) is important in controlling the mean
cell residence time (
␪
c
).
The excess waste activated sludge produced each day (see step 3 ‘‘Compute the
Quantity of Sludge Wasted’’) is wasted from the system to maintain a given F:M
or mean cell residence time. Generally, sludge is wasted from the return sludge line
because it is more concentrated than the mixed liquor in the aeration tank,
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.7
FIGURE 2 Aeration tank mass balance.
FIGURE 1 Settling tank mass balance.

hence smaller waste sludge pumps are required. The waste sludge is generally
discharged to sludge thickening and digestion facilities. The alternative method of
sludge wasting is to withdraw mixed liquor directly from the aeration tank where
the concentration of solids is uniform. Both methods of calculating the waste sludge
flow rate are illustrated below.
Use Figs. 1 and 2 when performing mass balances for the determination of
RAS and WAS.
X
ϭ
Mixed Liquor Suspended Solids (MLSS) - see Part 1 ‘‘Compute the
Reactor Volume.’’
Q
r
ϭ
Return activated sludge pumping rate (Mgd)
X
r
ϭ
Concentration of sludge in the return line (mg /L). When lacking site
specific operational data, a value commonly assumed is 8000 mg/L.
Q
e
ϭ
Effluent flow rate (Mgd)
X
e
ϭ
Concentration of solids in effluent (mg / L). When lacking site specific
operational data, this value is commonly assumed to be zero.
Q

w
ϭ
Wasted Activated Sludge (WAS) pumping rate from the reactor (Mgd)
ϭ
Q
w
Ј
Waste Activated Sludge (WAS) pumping rate from the return line (Mgd)
Other variables are as defined previously.
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WASTEWATER TREATMENT AND CONTROL
13.8
ENVIRONMENTAL CONTROL
The actual amount of liquid that must be pumped to achieve process control
depends on the method used and the location from which the wasting is to be
accomplished. Also note that because the solids capture of the sludge processing
facilities (i.e., thickeners, digesters, etc.) is not 100 percent and some solids are
returned, the actual wasting rate will be higher than the theoretically determined
value.
(a) Waste Activated Sludge (WAS) pumping rate from the return line. If the
mean cell residence time is used for process control and the wasting is from the
sludge return line (Fig. 1), the wasting rate is computed using the following:
VX
r
␪
ϭ
c
(QX

ϩ
QX)
w
Ј
ree
Assuming that the concentration of solids in the effluent from the settling tank
(X
e
) is low, then the above equation reduces to:
VX VX
rr
␪
Ϸ ⇒
Q
ϭ
cw
Ј
QX
␪
X
w
Ј
rcr
Using values defined previously:
(0.83 Mgal)(4500 mg/ L)
3
Q
ϭϭ
0.0584 Mgd
ϭ

58,400 gal /day (221 m /d)
w
Ј
(8 d) (8000 mg /L)
To determine the WAS pumping rate using this method, the solids concentration
in both the aeration tank and the return line must be known.
If the food to microorganism ratio (F:M) method of control is used, the WAS
pumping rate from the return line is determined using the following:
P
ϭ
QX(8.34 lb/Mgal / mg/L)
x(ss) w
Ј
r
Therefore:
3933 lb /d
3
Q
ϭϭ
0.059 Mgd
ϭ
59,000 gal /day (223.3 m /d)
w
Ј
(8000 mg /L)(8.34)
In this case, the concentration of solids in the sludge return line must be known.
Note that regardless of the method used for calculation, if wasting occurs from the
return line, the WAS pumping rate is approximately the same.
(b) Waste Activated Sludge (WAS) pumping rate from the aeration tank. If the
mean cell residence time is used for process control, wasting is from the aeration

tank (Fig. 2), and the solids in the plant effluent (X
e
) are again neglected, then the
WAS pumping rate is estimated using the following:
VV
rr
␪
Ϸ ⇒
Q
Ϸ
cw
Q
␪
w
c
Using values defined previously:
0.83 Mgal
3
Q
ϭϭ
0.104 Mgd
ϭ
104,000 gal /day (393.6 m /d)
w
8d
Note that in case (a) or (b) above, the weight of sludge wasted is the same (3933
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WASTEWATER TREATMENT AND CONTROL

WASTEWATER TREATMENT AND CONTROL
13.9
lb SS /d) (1785.6 kg SS / d), and that either wasting method will achieve a
␪
c
of 8
days. As can be seen, wasting from the aeration tank produces a much higher waste
flow rate. This is because the concentration of solids in the bottom of the settling
tank (and hence the return line) is higher than in the aeration tank. Consequently,
wasting a given mass of solids per day is going to require a larger WAS pumping
rate (and larger WAS pumps) if done from the aeration tank as opposed to the
return line. The Return Activated Sludge (RAS) pumping rate is determined by
performing a mass balance analysis around either the settling tank or the aeration
tank. The appropriate control volume for either mass balance analysis is illustrated
in Fig. 1 and 2 respectively. Assuming that the sludge blanket level in the settling
tank remains constant and that the solids in the effluent from the settling tank (X
e
)
are negligible, a mass balance around the settling tank (Fig. 1) yields the following
equation for RAS pumping rate:
XQ
Ϫ
XQ
rw
Ј
Q
ϭ
r
X
Ϫ

X
r
Using values defined previously, the RAS pumping rate is computed to be:
(4500 mg /L)(4.0 Mgd)
Ϫ
(8000 mg /L)(0.0584 Mgd)
Q
ϭ
r
8000 mg /L
Ϫ
4500 mg /L
3
ϭ
5.0 Mgd (18,925 m / d)
As outlined above, the required RAS pumping rate can also be estimated by
performing a mass balance around the aeration tank (Fig. 2). If new cell growth is
considered negligible, then the solids entering the tank will equal the solids leaving
the tank. Under conditions such as high organic loadings, this assumption may be
incorrect. Solids enter the aeration tank in the return sludge and in the influent flow
to the secondary process. However, because the influent solids are negligible com-
pared to the MLSS in the return sludge, the mass balance around the aeration tank
yields the following equation for RAS pumping rate:
X(Q
Ϫ
Q )
w
Q
ϭ
r

X
Ϫ
X
r
Using values defined previously, the RAS pumping rate is computed to be:
(4500 mg /L)(4.0 Mgd
Ϫ
0.104 Mgd)
3
Q
ϭϭ
5.0 Mgd (18,925 m / d)
r
8000 mg /L
Ϫ
4500 mg /L
The ratio of RAS pumping rate to influent flow rate, or recirculation ratio (
␣
), may
now be calculated:
Q 5.0 Mgd
r
␣
ϭϭ ϭ
1.25
Q 4.0 Mgd
Recirculation ratio can vary from 0.25 to 1.50 depending upon the type of ac-
tivated sludge process used. Common design practice is to size the RAS pumps so
that they are capable of providing a recirculation ratio ranging from 0.50 to 1.50.
It should be noted that if the control volume were placed around the aeration

tank in Fig. 1 and a mass balance performed, or the control volume placed around
the settling tank in Fig. 2 and a mass balance performed, that a slightly higher RAS
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WASTEWATER TREATMENT AND CONTROL
13.10
ENVIRONMENTAL CONTROL
pumping rate would result. However, the difference between these RAS pumping
rates and the ones calculated above is negligible.
DESIGN OF A CIRCULAR SETTLING TANK
Domestic wastewater with an average daily flow of 4.0 Mgd (15,140 m
3
/d) exits
the aeration tank of a standard activated sludge treatment process. Design a circular
settling tank to separate the sludge from the effluent. The settling tank will work
in conjunction with the aeration tank. Assume a peaking factor of 2.5.
Calculation Procedure:
1. Determine the peak flow
Conventional examples of circular settling tank design utilize settling tests to size
the tanks. However, it is more common that settling facilities must be designed
without the benefit settling tests. When this situation develops, published values of
surface loading and solids loading rates are generally used. Because of the large
amount of solids that may be lost in the effluent if design criteria are exceeded,
surface loading rates should be based on peak flow conditions. Using a peaking
factor of 2.5, the daily peak flow (Q
p
) is:
3
Q

ϭ
2.5
ϫ
4.0 Mgd
ϭ
10.0 Mgd (37,850 m / d)
p
2. Find the settling tank surface area using surface loading criteria
The recommended surface loading rates (settling tank effluent flow divided by set-
tling tank area) vary depending upon the type of activated sludge process used.
However, surface loading rates ranging from 200 to 800 gal / day/ft
2
(8.09 to 32.4
L/m
2
⅐
d) for average flow, and a maximum of 1,000 gal /day/ft
2
(40.7 L /m
2
⅐
d)
for peak flow are accepted design values.
The recommended solids loading rate on an activated sludge settling tank also
varies depending upon the type of activated sludge process used and may be com-
puted by dividing the total solids applied by the surface area of the tank. The
preferred units are lb / ft
2
⅐
h (kg/m

2
⅐
h). In effect, the solids loading rate represents
a characteristic value for the suspension under consideration. In a settling tank of
fixed area, the effluent quality will deteriorate if solids loading is increased beyond
the characteristic value for the suspension. Without extensive experimental work
covering all seasons and operating variables, higher rates should not be used for
design. The recommended solids loading rates vary depending upon the type of
activated sludge process selected. However, solids loading rates ranging from
0.8 to 1.2 lb/ ft
2
⅐
h (3.9 to 5.86 kg/m
2
⅐
h) for average flow, and 2.0 lb / ft
2
⅐
h (9.77
kg/m
2
⅐
h) for peak flow are accepted design values.
For a Q
p
of 10.0 Mgd and a design surface loading rate of 1,000 gal/ day /ft
2
at
peak flow, the surface area (A) of a settling tank may be calculated:
6

10
ϫ
10 gal/ day
2
1000 gal /day/ ft
ϭ ⇒
A
6
10
ϫ
10 gal/ day
22
A
ϭϭ
10,000 ft (929 m )
2
1000 gal /day/ ft
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.11
3. Find the settling tank surface area using solids loading criteria
The total solids load on a clarifier consists of contributions from both the influent
and the return activated sludge (RAS). Assume the following (see ‘‘Design of A
Complete-Mix Activated Sludge Reactor’’):
3
Q
ϭ

RAS flow rate
ϭ
1.25(Q)
ϭ
1.25(4.0 Mgd)
ϭ
5.0 Mgd (18,925 m / d)
r
X
ϭ
MLSS in aeration tank
ϭ
4,500 mg /L
Therefore, the maximum solids loading occurs at peak flow and maximum RAS
flow rate. The maximum solids entering the clarifier is calculated using:
Max. Solids (lb/ d)
ϭ
(Q
ϩ
Q )(X)(8.34 lb
⅐
L/mg
⅐
Mgal)
pr
Using values given above;
Max Solids (lb/ d)
ϭ
(10 Mgd
ϩ

5 Mgd)(4,500 mg/ L)(8.34 lb
⅐
L/mg
⅐
Mgal)
ϭ
562,950 lb /d
ϭ
23,456 lb /h (10,649 kg/ h) of Suspended Solids
Therefore, using a solids loading rate of 2.0 lb/ft
2
⅐
h (9.77 kg/ m
2
⅐
h) at peak flow,
the surface area of a settling tank may be calculated:
23,456 lb /h
22
A
ϭϭ
11,728 ft (1089.5 m )
2
2.0 lb /ft
⅐
h
In this case, the solids loading dominates and dictates the required settling tank
area.
4. Select the number of settling tanks
Generally, more than one settling thank would be constructed for operational flex-

ibility. Two tanks will be sufficient for this example. Therefore, the surface area of
each tank will be 11,728 ft
2
/2
ϭ
5,864 ft
2
(544.8 m
2
). The diameter of each settling
tank is then 86.41 ft. Use 87 ft (26.5 m). The total area of the two settling tanks
is 11,889 ft
2
(110.4.5 m
2
). For an average flow rate of 4.0 Mgd (15,140 m
3
/d) and
a RAS flow rate of 5.0 Mgd (18,925 m
3
), the total solids entering the settling tanks
at average daily flow is:
Solids (lb /d)
ϭ
(4.0 Mgd
ϩ
5.0 Mgd)(4,500 mg/ L) (8.34 lb
⅐
L/mg
⅐

Mgal)
ϭ
337,770 lb /d
ϭ
14,074 lb /h (6389.6 kg/ h)
Therefore, the solids loading on the settling tanks at design flow is:
14,074 lb /h
22
Solids Loading
ϭ Ϸ
1.18 lb /ft
⅐
h (5.77 kg/ m
⅐
h)
2
11,889 ft
which is within the solids loading rate design criteria stated above.
Related Calculations. Liquid depth in a circular settling tank is normally mea-
sured at the sidewall. This is called the sidewater depth. The liquid depth is a factor
in the effectiveness of suspended solids removal and in the concentration of the
return sludge. Current design practice favors a minimum sidewater depth of 12 ft
(3.66 m) for large circular settling tanks. However, depths of up to 20 ft (6.1 m)
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WASTEWATER TREATMENT AND CONTROL
13.12
ENVIRONMENTAL CONTROL
have been used. The advantages of deeper tanks include greater flexibility in op-

eration and a larger margin of safety when changes in the activated sludge system
occur.
THICKENING OF A WASTE-ACTIVATED SLUDGE
USING A GRAVITY BELT THICKENER
A wastewater treatment facility produces 58,000 gal/ day (219.5 m
3
/d) of waste
activated sludge containing 0.8 percent solids (8,000 mg /L). Design a gravity belt
thickener installation to thicken sludge to 5.0 percent solids based on a normal
operation of 6 h/d and 5 d / wk. Use a gravity belt thickener loading rate of 1,000
lb/ h (454 kg /h) per meter of belt width. Calculate the number and size of gravity
belt thickeners required, the volume of thickened sludge cake, and the solids capture
in percent.
Calculation Procedure:
1. Find the dry mass of sludge that must be processed
Gravity belt thickening consists of a gravity belt that moves over rollers driven by
a variable speed drive unit. The waste activated sludge is usually pumped from the
bottom of a secondary settling tank, conditioned with polymer and fed into a
feed/ distribution box at one end. The box is used to distribute the sludge evenly
across the width of the moving belt. The water drains through the belt as the sludge
is carried toward the discharge end of the thickener. The sludge is ridged and
furrowed by a series of plow blades placed along the travel of the belt, allowing
the water released from the sludge to pass through the belt. After the thickened
sludge is removed, the belt travels through a wash cycle.
The 58,000 gal/ day (219.5 m
3
/d) of waste activated sludge contains approxi-
mately 3933 lb/d (1785.6 kg/d) of dry solids: See Design of a Complete-Mix
Activated Sludge Reactor, step 3—‘‘Compute the Quantity of Sludge Wasted,’’ and
step 6—‘‘Compute the WAS and RAS Requirements.’’

Based on an operating schedule of 5 days per week and 6 hours per day, the
dry mass of sludge that must be processed is:
Weekly Rate: (3,933 lb /d)(7 d /wk)
ϭ
27,531 lb /wk (12,499 kg/ wk)
Daily Rate: (27,531 lb /wk)/ (5 d/ wk)
ϭ
5506 lb /d (2499.7 kg/ d)
Hourly Rate: (5506 lb /d)/ (6 h/ d)
ϭ
918 lb /h (416.8 kg/ h)
2. Size the belt thickener
Using the hourly rate of sludge calculated above, and a loading rate of 1,000 lb/h
per meter of belt width, the size of the belt thickener is:
918 lb /h
Belt Width
ϭϭ
0.918 m (3.01 ft)
1000 lb /h
⅐
m
Use one belt thickener with a 1.0 m belt width. Note that one identical belt thickener
should be provided as a spare.
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.13
The thickened sludge flow rate (S) in gal/day (m

3
/d) and the filtrate flow rate
(F) in gal /day (m
3
/d) are computed by developing solids balance and flow balance
equations:
(a) Solids balance equation. Solids in
ϭ
solids out, which implies that: solids
in sludge feed
ϭ
solids in thickened sludge
ϩ
solids in filtrate. Assume the follow-
ing:
•
Sludge feed specific gravity (s.g.)
ϭ
1.01
•
Thickened sludge s.g.
ϭ
1.03
•
Filtrate s.g.
ϭ
1.0
•
Suspended solids in filtrate
ϭ

900 mg/L
ϭ
0.09%
Therefore, the solids balance equation on a daily basis becomes:
5506 lb
ϭ
(S, gal/day)(8.34 lb /gal)(1.03)(0.05)
ϩ
(F, gal/day)(8.34 lb/gal)(1.0)(0.0009) (1)
⇒
5506 lb /d
ϭ
0.4295(S)
ϩ
0.0075(F)
(b) Flow balance equation. Flow in
ϭ
flow out, which implies that: influent
sludge flow rate
ϩ
washwater flow rate
ϭ
thickened sludge flow rate
ϩ
filtrate flow
rate. Daily influent sludge flow rate
ϭ
(58,000 gal/day)(7 / 5)
ϭ
81,200 gal/day

(307.3 m
3
/d).
3. Compute the thickened sludge and filtrate flow rates
Washwater flow rate is assumed to be 16 gal / min (1.0 L/s). Washwater flow rate
varies from 12 gal/min (0.757 L /s) to 30 gal/min (1.89 L/ s) depending on belt
thickener size. Therefore, with an operating schedule of 6 h/d (360 min / d) the
flow balance equation on a daily basis becomes:
81,200 gal /day
ϩ
(16 gal /min)(360 min /d)
ϭ
S
ϩ
F
⇒
86,960 gal /day
ϭ
S
ϩ
F (2)
Putting (EQ1) and (EQ 2) in matrix format, and solving for thickened sludge flow
rate (S) and filtrate flow rate (F):
0.4295 0.0075 S 5506
ϭ
ͫͬͫͬͫͬ
1 1 F 86,960
3
S
ϭ

11,502 gal /day (43.5 m /d) of thickened sludge at 5.0% solids
3
F
ϭ
75,458 gal /day (285.6 m /d) of filtrate
Therefore, the volume of thickened waste activated sludge exiting the gravity belt
thickener is 11,502 gal / day (43.5 m
3
/d) at 5.0 percent solids.
4. Determine the solids capture
The solids capture is determined using the following:
Solids in Feed
Ϫ
Solids in Filtrate
Solids Capture (%)
ϭϫ
(100%)
Solids in Feed
Using values defined previously:
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WASTEWATER TREATMENT AND CONTROL
13.14
ENVIRONMENTAL CONTROL
5506 lb /d
Ϫ
[(75,458 gal /day)(8.34 lb /gal)(1.0)(0.0009)]
Solids Capture (%)
ϭ

5506 lb /d
ϫ
(100%)
ϭ
89.7%
Since only 89.7 percent of the solids entering the gravity belt thickener are
captured, the thickener will actually be required to operate (360 min/ d) /(0.897)
ϭ
400 min/d (6 hours 40 minutes per day) five days per week in order to waste the
3,933 lb/ d (1785.6 kg/d) of dry solids required. This implies that the actual volume
of thickened sludge will be (11,502 gal / day)/(0.897)
ϭ
12,823 gal / day (48.5 m
3
/
d). The actual filtrate flow rate will be (75,458 gal / day)/(0.897)
ϭ
84,123 gal /day
(318.4 m
3
/d).
The thickened sludge is generally pumped immediately to sludge storage tanks
or sludge digestion facilities. If the thickener is operated 6.67 h /d, the thickened
sludge pumps (used to pump the thickened sludge to downstream processes) will
be sized based on the following thickened sludge flow rate:
12,823 gal /day
S
ϭϭ
32 gal /min (2.02 L/ s)
(6.67 h /d)(60 min /h)

Hydraulic (thickened sludge) throughput for a gravity belt thickener ranges from
25 (1.6 L / s) to 100 (6.3 L / s) gal/ min per meter of belt width. The filtrate flow of
84,123 gal / day is generally returned to the head of the wastewater treatment facility
for reprocessing.
DESIGN OF AN AEROBIC DIGESTER
An aerobic digester is to be designed to treat the waste sludge produced by an
activated sludge wastewater treatment facility. The input waste sludge will be
12,823 gal / day (48.5 L /d) (input 5 d /wk only) of thickened waste activated sludge
at 5.0 percent solids—See Thickening of a Waste Activated Sludge Using a Gravity
Belt Thickener. Assume the following apply:
1. The minimum liquid temperature in the winter is 15
Њ
C (59
Њ
F), and the maximum
liquid temperature in the summer is 30
Њ
C (86
Њ
F).
2. The system must achieve a 40 percent Volatile Suspended Solids (VSS) reduc-
tion in the winter.
3. Sludge concentration in the digester is 70 percent of the incoming thickened
sludge concentration.
4. The volatile fraction of digester suspended solids is 0.8.
Calculation Procedure:
1. Find the daily volume of sludge for disposal
Factors that must be considered in designing aerobic digesters include temperature,
solids reduction, tank volume (hydraulic retention time), oxygen requirements and
energy requirements for mixing.

Because the majority of aerobic digesters are open tanks, digester liquid tem-
peratures are dependent upon weather conditions and can fluctuate extensively. As
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.15
FIGURE 3 VSS reduction in aerobic digester vs. liquid temperature
ϫ
sludge age. (Metcalf & Eddy, Wastewater Engineering: Treatment,
Disposal, and Reuse, 3rd Ed., McGraw-Hill.)
with all biological systems, lower temperatures retard the process, whereas higher
temperatures accelerate it. The design of the aerobic digester should provide the
necessary degree of sludge stabilization at the lowest liquid operating temperature
and should supply the maximum oxygen requirements at the maximum liquid op-
erating temperature.
A major objective of aerobic digestion is to reduce the mass of the solids for
disposal. This reduction is assumed to take place only with the biodegradable con-
tent (VSS) of the sludge, although there may be some destruction of the inorganics
as well. Typical reduction in VSS ranges from 40 to 50 percent. Solids destruction
is primarily a direct function of both basin liquid temperature and sludge age, as
indicated in Fig. 3. The plot relates VSS reduction to degree-days (temperature
ϫ
sludge age).
To ensure proper operation, the contents of the aerobic digester should be well
mixed. In general, because of the large amount of air that must be supplied to meet
the oxygen requirement, adequate mixing is usually achieved. However, mixing
power requirements should always be checked.
The aerobic digester will operate 7 days per week, unlike the thickening fa-

cilities which operate intermittently due to larger operator attention requirements.
The thickened sludge is input to the digester at 12,823 gal /day (48.5 L/d), 5 days
per week. However, the volume of the sludge to be disposed of daily by the digester
will be lower due to its operation 7 days per week (the ‘‘bugs’’ do not take the
weekends off). Therefore the volume of sludge to be disposed of daily (Q
i
) is:
33
Q
ϭ
(12,823 gal /day)(5/ 7)
ϭ
9,159 gal /day
ϭ
1,224 ft / d (34.6 m /d)
i
2. Determine the required VSS reduction
The sludge age required for winter conditions is obtained from Fig. 3 using the
minimum winter temperature and required VSS reduction.
To achieve a 40 percent VSS reduction in the winter, the degree-days required
from Fig. 3 is 475
Њ
C
⅐
d. Therefore, the required sludge age is 475
Њ
C
⅐
d/15
Њ

C
ϭ
31.7 days. During the summer, when the liquid temperature is 30
Њ
C, the degree-
days required is (30
Њ
C)(31.7 d)
ϭ
951
Њ
C
⅐
d. From Fig. 3, the VSS reduction will
be 46 percent.
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WASTEWATER TREATMENT AND CONTROL
13.16
ENVIRONMENTAL CONTROL
The total mass of solids processed by the digester will be 3,933 lb/ d (1785.6
kg/ d) which is the total mass of solids wasted from the treatment facility—See
Design of a Complete-Mix Activated Sludge Reactor, step 3. The total mass of VSS
input to the digester is:
(0.8)(3,933 lb /d)
ϭ
3146 lb /d (1428.3 kg/ d).
Therefore, during the winter:
•

VSS reduction
ϭ
(3,146 lb/ d)(0.40)
ϭ
1258 lb VSS reduced/ d (571.1 kg/d)
•
Digested (stabilized) sludge leaving the digester
ϭ
3933 lb/d
Ϫ
1258 lb/d
ϭ
2675 lb/d (1214.5 kg / d).
3. Compute the volume of digested sludge
The volume of digested sludge is:
W
s
V
ϭ
(
␳
)(s.g.)(% solids)
where V
ϭ
Sludge volume (ft
3
)(m
3
)
W

s
ϭ
Weight of sludge (lb) (kg)
␳
ϭ
density of water (62.4 lb / ft
3
) (994.6 kg / m
3
)
s.g.
ϭ
specific gravity of digested sludge (assume s.g.
ϭ
1.03)
% solids
ϭ
percent solids expressed as a decimal (incoming sludge: 5.0%)
Therefore, the volume of the digested sludge is:
2675 lb /d
3
V
ϭϭ
832 ft / d
ϭ
6223 gal /day (23.6 L/ d)
3
(62.4 lb /ft )(1.03)(0.05)
During the summer:
•

VSS reduction
ϭ
(3,146 lb/ d)(0.46)
ϭ
1447 lb VSS reduced/ d (656.9 kg/d)
•
Digested (stabilized) sludge leaving the digester
ϭ
3933 lb/d
Ϫ
1447 lb/d
ϭ
2486 lb/d (1128.6 kg / d).
•
Volume of digested sludge:
2486 lb /d
3
V
ϭϭ
774 ft / d
ϭ
5790 gal /day (21.9 L/ d)
3
(62.4 lb /ft )(1.03)(0.05)
4. Find the oxygen and air requirements
The oxygen required to destroy the VSS is approximately 2.3 lb O
2
/lb VSS (kg /
kg) destroyed. Therefore, the oxygen requirements for winter conditions are:
(1258 lb VSS/ d)(2.3 lb O / lb VSS)

ϭ
2893 lb O /d (1313.4 kg/ d)
22
The volume of air required at standard conditions (14.7 lb/ in
2
and 68
Њ
F) (96.5
kPa and 20
Њ
C) assuming air contains 23.2 percent oxygen by weight and the density
of air is 0.075 lb/ft
3
is:
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.17
2893 lb O /d
2
33
Volume of Air
ϭϭ
166,264 ft / d (4705.3 m /d)
3
(0.075 lb /ft )(0.232)
For summer conditions:
•

Oxygen required
ϭ
(1447 lb/ d)(2.3 lb O
2
/d)
ϭ
3328 lb O
2
/d (1510.9 kg / d)
•
Volume of Air
ϭ
3328 lb O
2
/d/(0.075 lb/ ft
3
)(0.232)
ϭ
191,264 ft
3
/d (5412.8
m
3
/d)
Note that the oxygen transfer efficiency of the digester system must be taken
into account to get the actual volume of air required. Assuming diffused aeration
with an oxygen transfer efficiency of 10 percent, the actual air requirements at
standard conditions are:
•
Winter: volume of air

ϭ
166,264 ft
3
/d/(0.1)(1,400 min/d)
ϭ
1155 ft
3
/min (32.7
m
3
/min)
•
Summer: volume of air
ϭ
191,264 ft
3
/d/(0.1)(1,440 min /d)
ϭ
1328 ft
3
/min (37.6
m
3
/min)
To summarize winter and summer conditions:
Parameter Winter Summer
Total Solids In, lb / d (kg/ d) 3933 (1785.6) 3933 (1785.6)
VSS In, lb / d (kg/ d) 3146 (1428.3) 3146 (1428.3)
VSS Reduction, (%) 40 46
VSS Reduction, lb /d (kg / d) 1258 (571.1) 1447 (656.9)

Digested Sludge Out, gal/ day (L / d) 6223 (23.6) 5790 (21.9)
Digested Sludge Out, lb/ d (kg / d) 2675 (1214.5) 2486 (1128.6)
Air Requirements @ S.C., ft
3
/min (m
3
/min) 1155 (32.7) 1328 (37.6)
5. Determine the aerobic digester volume
From the above analysis it is clear that the aerobic digester volume will be calcu-
lated using values obtained under the winter conditions analysis, while the aeration
equipment will be sized using the 1328 ft
3
/min (37.6 m
3
/min) air requirement
obtained under the summer conditions analysis.
The volume of the aerobic digester is computed using the following equation,
assuming the digester is loaded with waste activated sludge only:
QX
ii
V
ϭ
X(KP
ϩ
1/
␪
)
d
v
c

where V
ϭ
Volume of aerobic digester, ft
3
(m
3
)
Q
i
ϭ
Influent average flow rate to the digester, ft
3
/d (m
3
/d)
X
i
ϭ
Influent suspended solids, mg /L (50,000 mg/ L for 5.0% solids)
X
ϭ
Digester total suspended solids, mg/L
K
d
ϭ
Reaction rate constant, d
Ϫ
1
. May range from 0.05 d
Ϫ

1
at 15
Њ
C (59
Њ
F)
to 0.14 d
Ϫ
1
at 25
Њ
C (77
Њ
F) (assume 0.06 d
Ϫ
1
at 15
Њ
C)
P
v
ϭ
Volatile fraction of digester suspended solids (expressed as a decimal)
ϭ
0.8 (80%) as stated in the initial assumptions.
␪
c
ϭ
Solids retention time (sludge age), d
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WASTEWATER TREATMENT AND CONTROL
13.18
ENVIRONMENTAL CONTROL
Using values obtained above with winter conditions governing, the aerobic digester
volume is:
3
(1,224 ft / d)(50,000 mg/ L)
33
V
ϭϭ
21,982 ft (622.2 m )
Ϫ
1
(50,000 mg /L)(0.7)[(0.06 d )(0.8)
ϩ
1/ 31.7 d]
The air requirement per 1,000 ft
3
(2.8 m
3
) of digester volume with summer con-
ditions governing is:
3
1328 ft / min
3333 3
Volume of Air
ϭϭ
60.41 ft / min /10 ft (0.97 m / min/Mm )

33
21.982 10 ft
The mixing requirements for diffused aeration range from 20 to 40 ft
3
/min/10
3
ft
3
(0.32 to 0.64 m
3
/min/Mm
3
). Therefore, adequate mixing will prevail.
DESIGN OF AN AERATED GRIT CHAMBER
Domestic wastewater enters a wastewater treatment facility with an average daily
flow rate of 4.0 Mgd (15,140 L/d). Assuming a peaking factor of 2.5, size an
aerated grit chamber for this facility including chamber volume, chamber dimen-
sions, air requirement, and grit quantity.
Calculation Procedure:
1. Determine the aerated grit chamber volume
Grit removal in a wastewater treatment facility prevents unnecessary abrasion and
wear of mechanical equipment such as pumps and scrappers, and grit deposition in
pipelines and channels. Grit chambers are designed to remove grit (generally char-
acterized as nonputrescible solids) consisting of sand, gravel, or other heavy solid
materials that have settling velocities greater than those of the organic putrescible
solids in the wastewater.
In aerated grit chamber systems, air introduced along one side near the bottom
causes a spiral roll velocity pattern perpendicular to the flow through the tank. The
heavier particles with their correspondingly higher settling velocities drop to the
bottom, while the rolling action suspends the lighter organic particles, which are

carried out of the tank. The rolling action induced by the air diffusers is independent
of the flow through the tank. Then non flow dependent rolling action allows the
aerated grit chamber to operate effectively over a wide range of flows. The heavier
particles that settle on the bottom of the tank are moved by the spiral flow of the
water across the tank bottom and into a grit hopper. Screw augers or air lift pumps
are generally utilized to remove the grit from the hopper.
The velocity of roll governs the size of the particles of a given specific gravity
that will be removed. If the velocity is too great, grit will be carried out of the
chamber. If the velocity is too small, organic material will be removed with the
grit. The quantity of air is easily adjusted by throttling the air discharge or using
adjustable speed drives on the blowers. With proper adjustment, almost 100 percent
grit removal will be obtained, and the grit will be well washed. Grit that is not well
washed will contain organic matter and become a nuisance through odor emission
and the attraction of insects.
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WASTEWATER TREATMENT AND CONTROL
WASTEWATER TREATMENT AND CONTROL
13.19
FIGURE 4 Aerated grit chamber. (Metcalf & Eddy, Wastewater Engineering: Treat-
ment, Disposal, and Reuse, 3rd Ed., McGraw-Hill.)
Wastewater will move through the aerated grit chamber in a spiral path as
illustrated in Fig. 4. The rolling action will make two to three passes across the
bottom of the tank at maximum flow and more at lesser flows. Wastewater is in-
troduced in the direction of the roll.
At peak flow rate, the detention time in the aerated grit chamber should range
from 2 to 5 minutes. A detention time of 3 minutes will be used for this example.
Because it is necessary to drain the chamber periodically for routine maintenance,
two redundant chambers will be required. Therefore, the volume of each chamber

is:
(peak flow rate, gal /day)(detention time, min)
3
V (ft )
ϭ
3
(7.48 gal /ft )(24 h/ d)(60 min/ h)
Using values from above, the chamber volume is:
6
(2.5)(4
ϫ
10 gal/ day)(3 min)
333
V (ft )
ϭϭ
2785 ft (78.8 m )
3
(7.48 gal /ft )(24 h/ d)(60min /h)
2. Determine the dimensions of the grit chamber
Width-depth ratio for aerated grit chambers range from 1:1 to 5:1. Depths range
from 7 to 16 feet (2.1 to 4.87 m). Using a width-depth ratio of 1.2:1 and a depth
of 8 feet (2.43 m), the dimensions of the aerated grit chamber are:
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WASTEWATER TREATMENT AND CONTROL
13.20
ENVIRONMENTAL CONTROL
Width
ϭ

(1.2)(8 ft)
ϭ
9.6 ft (2.92 m)
3
2785 ft
Length
ϭ
(volume)/ [(width)(depth)]
ϭϭ
36.3 ft (11.1 m)
(8 ft)(9.6 ft)
Length-width ratios range from 3:1 to 5:1. As a check, length to width ratio for
the aerated grit chamber sized above is: 36.3 ft/9.6 ft
ϭ
3.78:1 which is acceptable.
3. Determine the air supply required
The air supply requirement for an aerated grit chamber ranges from 2.0 to 5.0 ft
3
/
min/ ft of chamber length (0.185 to 0.46 m
3
/min
⅐
m). Using 5.0 ft
3
/min / ft (0.46
m
3
/min
⅐

m) for design, the amount of air required is:
33 33
Air required (ft /min)
ϭ
(5.0 ft / min /ft)(36.3 ft)
ϭ
182 ft / min (5.2 m /min)
4. Estimate the quantity of grit expected
Grit quantities must be estimated to allow sizing of grit handling equipment such
as grit conveyors and grit dewatering equipment. Grit quantities from an aerated
grit chamber vary from 0.5 to 27 ft
3
/Mgal (3.74 to 201.9 m
3
/L) of flow. Assume
a value of 20 ft
3
/Mgal (149.5 m
3
/L). Therefore, the average quantity of grit ex-
pected is:
33 33
Volume of grit (ft / d)
ϭ
(20 ft / Mgal)(4.0 Mgd)
ϭ
80 ft / d (2.26 m /d)
Some advantages and disadvantages of the aerated grit chamber are listed below:
Advantages Disadvantages
The same efficiency of grit removal is

possible over a wide flow range.
Power consumption is higher than other
grit removal processes.
Head loss through the grit chamber is
minimal.
Additional labor is required for
maintenance and control of the aeration
system.
By controlling the rate of aeration, a grit
of relatively low putrescible organic
content can be removed.
Significant quantities of potentially
harmful volatile organics and odors may
be released from wastewaters containing
these constituents.
Preaeration may alleviate septic conditions
in the incoming wastewater to improve
performance of downstream treatment
units.
Foaming problems may be created if
influent wastewater has surfactants
present.
Aerated grit chambers can also be used
for chemical addition, mixing,
preaeration, and flocculation ahead of
primary treatment.
DESIGN OF A SOLID-BOWL CENTRIFUGE FOR
SLUDGE DEWATERING
A 4.0 Mgd (15,140 m
3

/d) municipal wastewater treatment facility produces 6,230
gal/ day (23.6 m
3
/d) of aerobically digested sludge at 5.0 percent solids. Determine
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WASTEWATER TREATMENT AND CONTROL
13.21
TABLE 1
Facility Capacity & Number of Centrifuges
Facility size, Mgd (m
3
/d)
Dewatering
operation, h / d
Centrifuges operating
ϩ
spare @ gal / min (L/ s)
2 (7,570) 7 1
ϩ
1 @ 25 (1.58)
5 (18,930) 7.5 1
ϩ
1 @ 50 (3.16)
20 (75,700) 15 2
ϩ
1 @ 50 (3.16)
50 (189,250) 22 2

ϩ
1 @ 75 (4.73)
100 (378,500) 22 3
ϩ
2 @ 100 (6.31)
250 (946,250) 22 4
ϩ
2 @ 200 (12.62)
(Design Manual for Dewatering Municipal Wastewater Sludges, U.S. EPA)
design parameters for the specification of a solid bowl centrifuge for dewatering
the sludge including: number of centrifuges, solids feed rate, percent solids recov-
ery, dewatered sludge (cake) discharge rate, centrifugal force, polymer dosage, and
polymer feed rate. Assume the following apply:
•
Feed sludge is aerobically digested at 5.0 percent solids.
•
Dewatered sludge (cake) is to be 25 percent solids.
•
Centrate assumed to be 0.3 percent solids.
•
Polymer solution concentration is 25 percent.
Calculation Procedure:
1. Select the number of centrifuges
The separation of a liquid-solid sludge during centrifugal thickening is analogous
to the separation process in a gravity thickener. In a centrifuge, however, the applied
force is centrifugal rather than gravitational and usually exerts 1,500 to 3,500 times
the force of gravity. Separation results from the centrifugal force-driven migration
of the suspended solids through the suspending liquid, away from the axis of ro-
tation. The increased settling velocity imparted by the centrifugal force as well as
the short settling distance of the particles accounts for the comparatively high ca-

pacity of centrifugal equipment.
Centrifuges are commonly used for thickening or dewatering Waste Activated
Sludge (WAS) and other biological sludges from secondary wastewater treatment.
In the process, centrifuges reduce the volume of stabilized (digested) sludges to
minimize the cost of ultimate disposal. Because centrifuge equipment is costly and
sophisticated, centrifuges are most commonly found in medium to large wastewater
treatment facilities.
The capacity of sludge dewatering to be installed at a given facility is a func-
tion of the size of a facility, capability to repair machinery on-site, and the avail-
ability of an alternative disposal means. Some general guidelines relating the min-
imal capacity requirements are listed in Table 1. This table is based on the
assumption that there is no alternative mode of sludge disposal and that the capacity
to store solids is limited.
Using Table 1, the number of centrifuges recommended for a 4.0 Mgd (15,410
m
3
/d) wastewater treatment facility is one operational
ϩ
one spare for a total of
two centrifuges.
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13.22
ENVIRONMENTAL CONTROL
2. Find the sludge feed rate required
If the dewatering facility is operated 4 h/d, 7 d/wk, then the sludge feed rate is:
Sludge Feed Rate
ϭ

(6,230 gal /day)/ [(4 h/ d)(60 /min/h)]
ϭ
26 gal /min (1.64 L/ s)
Although a 4 h/d operation is below that recommended in Table 1, the sludge
feed rate of 26 gal/min (1.64 L/s) is adequate for the size of centrifuge usually
found at a treatment facility of this capacity. A longer operational day would be
necessary if the dewatering facilities were operated only 5 days per week, or during
extended period of peak flow and solids loading.
Assume a feed sludge specific gravity of 1.03. The sludge feed in lb/h is
calculated using the following equation:
(V)(
␳
)(s.g.)(% solids)(60 min/ h)
W
ϭ
s
3
7.48 gal /ft
W
s
ϭ
Weight flow rate of sludge feed, lb/ h (kg / h)
V
ϭ
Volume flow rate of sludge feed, gal /min (L/ s)
s.g.
ϭ
specific gravity of sludge
% solids
ϭ

percent solids expressed as a decimal
␳
ϭ
density of water, 62.4 lb/ ft
3
(994.6 kg/ m
3
)
Using values obtained above, the sludge feed in lb / h is:
3
(26 gal /min)(62.4 lb /ft )(1.03)(0.05)(60 min/h)
W
ϭ
s
3
7.48 gal /ft
ϭ
670 lb /h of dry solids (304.2 kg /h)
3. Compute the solids capture
Since the solids exiting the centrifuge are split between the centrate and the cake,
it is necessary to use a recovery formula to determine solids capture. Recovery is
the mass of solids in the cake divided by the mass of solids in the feed. If the
solids content of the feed, centrate and cake are measured, it is possible to calculate
percent recovery without determining total mass of any of the streams. The equation
for percent solids recovery is:
CF
Ϫ
C
SC
R

ϭ
100
ͩͪͫ ͬ
FC
Ϫ
C
SC
where R
ϭ
Recovery, percent solids
C
s
ϭ
Cake solids, percent solids (25%)
F
ϭ
Feed solids, percent solids (5%)
C
c
ϭ
Centrate solids, percent solids (0.3%)
Therefore, using values defined previously:
(5
Ϫ
0.3)
R
ϭ
100(25/ 5)
ϭ
95.14%

ͫͬ
25
Ϫ
0.3
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WASTEWATER TREATMENT AND CONTROL
13.23
4. Determine the Dewatered Sludge Cake Discharge Rate
The dewatered sludge (cake) discharge rate is calculated using the following:
Cake discharge rate (lb /h) dry solids
ϭ
(sludge feed rate, lb /h)(solids recovery)
ϭ
(670 lb /h)(0.9514)
ϭ
637.5 lb /h (289.4 kg/ h) dry cake
The wet cake discharge in lb/h is calculated using the following:
Dry cake rate, lb /h
Wet cake discharge (lb /h)
ϭ
Cake % solids
637.5 lb /h
⇒
Wet cake discharge (lb /h)
ϭ
0.25
ϭ

2550 lb /h wet cake (1157.7 kg/ h)
The volume of wet cake, assuming a cake density of 60 lb/ft
3
is calculated as
follows:
Wet cake rate, lb /h 2550 lb /h
3
Volume of wet cake (ft /h)
ϭϭ
33
Cake density, lb/ ft 60 lb /ft
33
ϭ
42.5 ft / h (1.2 m /h) wet cake
For a dewatering facility operation of 4 h /d, the volume of dewatered sludge cake
to be disposed of per day is:
33
(42.5 ft / h)(4 h/ d)
ϭ
170 ft / d
ϭ
1272 gal /day (4.81 L/ d)
5. Find the percent reduction in sludge volume
The percent reduction in sludge volume is then calculated using the following:
Sludge volume in
Ϫ
Sludge volume out
% Volume Reduction
ϭ
Sludge volume in

ϫ
100%
6,230 gal /day
Ϫ
1,272 gal /day
ϭ
6,230 gal /day
ϫ
100%
ϭ
79.6%
Centrifuges operate at speed ranges which develop centrifugal forces from
1,500 to 3,500 times the force of gravity. In practice, it has been found that higher
rotational speeds usually provide significant improvements in terms of performance,
particularly on wastewater sludges.
In most cases, a compromise is made between the process requirement and
O&M considerations. Operating at higher speeds helps achieve optimum perform-
ance which is weighed against somewhat greater operating and maintenance costs.
Increasing bowl speed usually increases solids recovery and cake dryness. Today
most centrifuges used in wastewater applications can provide good clarity and solids
concentration at G levels between 1,800 and 2,500 times the force of gravity.
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13.24
ENVIRONMENTAL CONTROL
6. Compute the centrifugal force in the centrifuge
The centrifugal acceleration force (G), defined as multiples of gravity, is a function
of the rotational speed of the bowl and the distance of the particle from the axis

of rotation. In the centrifuge, the centrifugal force, G, is calculated as follows:
2
(2
␲
N) R
G
ϭ
2
32.2 ft /s
where N
ϭ
Rotational speed of centrifuge (rev/ s)
R
ϭ
Bowl radius, ft (cm)
The rotational speed and bowl diameter of the centrifuge will vary depending upon
the manufacturer. However, a rotational speed of 2,450 r/min and a bowl diameter
of 30 inches (72.6 cm) are common for this type of sludge dewatering operation.
Therefore, the centrifugal force is:
2
((2
␲
)(2450 r /min/ 60 s/ min)) (30 in /12 in / ft)(0.5)
G
ϭϭ
2,555 Gs
2
32.2 ft /s
7. Find the polymer feed rate for the centrifuge
The major difficulty encountered in the operation of centrifuges is the disposal of

the centrate, which is relatively high in suspended, non-settling solids. The return
of these solids to the influent of the wastewater treatment facility can result in the
passage of fine solids through the treatment system, reducing effluent quality. Two
methods are used to control the fine solids discharge and increase the capture. These
are: (1) increased residence time in the centrifuge, and (2) polymer addition. Longer
residence time of the liquid is accomplished by reducing the feed rate or by using
a centrifuge with a larger bowl volume. Better clarification of the centrate is
achieved by coagulating the sludge prior to centrifugation through polymer addition.
Solids capture may be increased from a range of 50 to 80 percent to a range of 80
to 95 percent by longer residence time and chemical conditioning through polymer
addition.
In order to obtain a cake solids concentration of 20 to 28 percent for an
aerobically digested sludge, 5 to 20 pounds of dry polymer per ton of dry sludge
feed (2.27 to 9.08 kg/ ton) is required. 15 lb / ton (6.81 kg /ton) will be used for this
example. Usually this value is determined through pilot testing or plant operator
trial and error.
The polymer feed rate in lb /h of dry polymer is calculated using the following:
(polymer dosage, lb/ ton)(dry sludge feed, lb/ h)
Polymer feed rate (lb /h)
ϭ
2000 lb /ton
Using values defined previously, the polymer feed rate is:
(15 lb /ton)(670 lb /h)
Polymer feed rate (lb /h)
ϭ
2000 lb /ton
ϭ
5.0 lb /h of dry polymer (2.27 kg /h)
Polymer feed rate in gal / h is calculated using the following:
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WASTEWATER TREATMENT AND CONTROL
13.25
polymer feed rate lb /h)
Polymer feed rate (gal /h)
ϭ
(8.34 lb /gal)(s.g.)(% polymer concentration)
where s.g.
ϭ
specific gravity of the polymer solution
% polymer concentration expressed as a decimal
Using values defined previously:
5.0 lb /h
Polymer feed rate (gal /h)
ϭ
(8.34 lb /gal)(1.0)(0.25)
ϭ
2.4 gal /h of 25% polymer solution (0.009 L/ h)
The polymer feed rate is used to size the polymer dilution/ feed equipment required
for the sludge dewatering operation.
Related Calculations. Selection of units for dewatering facility design is de-
pendent upon manufacturer’s rating and performance data. Several manufacturers
have portable pilot plant units, which can be used for field testing if sludge is
available. Wastewater sludges from supposedly similar treatment processes but dif-
ferent localities can differ markedly from each other. For this reason, pilot plant
tests should be run, whenever possible, before final design decisions regarding cen-
trifuge selection are made.
SIZING OF A TRAVELING-BRIDGE FILTER

Secondary effluent from a municipal wastewater treatment facility is to receive
tertiary treatment, including filtration, through the use of traveling bridge filters.
The average daily flow rate is 4.0 Mgd (2778 gal/ min) (15,140 m
3
/d) and the
peaking factor is 2.5. Determine the size and number of traveling bridge filters
required.
Calculation Procedure:
1. Determine the peak flow rate for the filter system
The traveling bridge filter is a proprietary form of a rapid sand filter. This type of
filter is used mainly for filtration of effluent from secondary and advanced waste-
water treatment facilities. In the traveling bridge filter, the incoming wastewater
floods the filter bed, flows through the filter medium (usually sand and/or anthra-
cite), and exits to an effluent channel via an underdrain and effluent ports located
under each filtration cell. During the backwash cycle, the carriage and the attached
hood (see Fig. 5) move slowly over the filter bed, consecutively isolating and back-
washing each cell. The washwater pump, located in the effluent channel, draws
filtered wastewater from the effluent chamber and pumps it through the effluent
port of each cell, forcing water to flow up through the cell thereby backwashing
the filter medium of the cell. The backwash pump located above the hood draws
water with suspended matter collected under the hood and transfers it to the back-
wash water trough. During the backwash cycle, wastewater is filtered continuously
through the cells not being backwashed.
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