Secondary Treatment Processes

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Secondary Treatment Processes
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Two Types
1. Attached growth or Fixed Film
Organisms attached to some inert media like rocks
or plastic.
2. Suspended Growth
Organisms are suspended in the treatment basin
fluid. This fluid is commonly called the mixed
liquor.
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Attached Growth or Fixed Film Reactors
Trickling Filters
Rock Media
Typically 4 8 feet deep.
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Trickling Filters
With time, the slime layer becomes thicker and
thicker until oxygen and organic matter can not
penetrate to the organisms on the inside.
The organisms on the inside then die and become
detached from the media, causing a portion of the
slime layer to slough off.
This means the effluent from a trickling filter
will have lots of solids (organisms) in it which
must be removed by sedimentation
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Trickling Filters
Single Stage Trickling Filter
Two Stage Trickling Filter
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Bio-towers
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Rotating Biological Contactors (RBCs)
In trickling filters the moving wastewater passes
over the stationary rock media. In an RBC, the
moving media passes through the stationary
wastewater.
Commonly used out in series and parallel
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Suspended Growth Processes
Activated Sludge
Designed based on loading (the amount of organic
matter added relative to the microorganisms
available)
Commonly called the food-to-microorganisms ratio,
F/M
F measured as BOD. M measured as volatile
suspended solids concentration
F/M is the pounds of BOD/day per pound of MLSS in
the aeration tank
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Design of Activated Sludge
Influent organic compounds provide the food for
the microorganisms and is called substrate (S)
The substrate is used by the microorganisms for
growth, to produce energy and new cell material.
The rate of new cell production as a result of
the use of substrate may be written
mathematically as
dX/dt - Y dS/dt
Y is called the yield and is the mass of cells
produced per mass of substrate used (g SS/g BOD)
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Monod Model for Substrate Utilization
dX/dt - Y dS/dt
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Mean Cell Residence Time, 2c
Mean cell residence time (MCRT, 2c) is the mass
of cells in the system divided by the mass of
cells wasted per day.
Consider the system
2c VX/QX V/Q
At SS the amount of solids wasted per day must
equal the amount produced per day
2c XV/Y(dS/dt)V X/Y(dS/dt)
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Mass Balance on Microorganisms
V dX/dt Q X0 Q X Y(dS/dt)V
S.S. (dX/dt) V 0, and QX0 0
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Example
Solve for 2c 2c (Ks S)/(S ))
(50010)/(10 x 4) 12.75 days
2c V/Q
V 2c Q 12.75 (3) 38.25 m3
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Given the conditions in the previous example,
What would the percent reduction in substrate be
if the reactor volume was 24 m3?
2c V/Q 24/3 8 days
Reduction (600 16.1)/600 x 100 97.3
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Now consider a CSTR with cell recycle
2c (X V) / (QwXr) (Q Qw)Xc
Since Xc 0
2c (XV)/(QwXr)
Removal of substrate often expressed in terms of
substrate removal velocity, q
q (mass of substrate removed/time)/(mass of
microorganisms under aeration)
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Mass balance on microorganisms
V dX/dt Q X0 QwXr (Q Qw)Xc X V
X0 Xc 0
(Xr Qw)/(X V) 1/ 2c
The substrate removal velocity, q, can also be
expressed as q /Y
If we equate these two equations for q and solve
for S0 S
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Since q / Y
2c 1/ (q Y)
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Problem
The hydraulic retention time may be found from
the following equation
0.5(300 30)(200 30) / 2 (30) (400)
0.129 days 3.1 hr
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2c 1/ (qY)
Q (S0 S) / (X t ) (300 30) /
(4000)(0.129)
0.523 (kg BOD removed/day) / (kg SS in the
reactor)
2c 1/ (qY) 1 / (0.523 x 0.5) 3.8 days
Also 2c (X V) / (Xr Qw)
Xr Qw (X V) / 2c (4000)(51.6)( 103 L/m3)(
1/106 kg/mg) / 3.8
54.3 kg/day
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Using the same data what MLSS is necessary to
produce an effluent concentration of 15 mg BOD/L?
2(15) / 0.5(200 15) 0.28 day-1
(300 15) / 0.129(0.28) 7890 mg/L
2c 1 / (q Y) 1 / 0.28(0.5) 7.2 days
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Solids Separation
The success of the activated sludge process
depends on the efficiency of the secondary
clarifier, which depends on the settling
characteristics of the sludge (biosolids).
Some system conditions result in sludge that is
very difficult to settle. In this case the
return activated sludge becomes thin (low MLSS)
and the concentration of organisms in the
aeration tank goes down. This produces a higher
F/M ratio (same food input, but fewer organisms)
and a reduced BOD removal efficiency.
One condition that commonly causes this problem
is called bulking sludge. Bulking sludge occurs
when a type of bacteria called filamentous
bacteria grow in large numbers in the system.
This produces a very billowy floc structure with
poor settling characteristics.
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Aeration
Diffused Aeration
Coarse Bubble
Fine Bubble
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Mechanical Aeration
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Modeling Gas Transfer
Henrys Law
S K P
S solubility of the gas, mg gas/L P partial
pressure of the gas K solubility constant
If a gas is 60 O2 and 40 N2 and the total
pressure of the gas is 1 atm (101 KPa), the
partial pressure of O2 0.6 x 101 60.6 Kpa.
The total pressure is equal to the sum of the
partial pressures (Daltons Law)
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Example
At one atmosphere, the solubility of pure oxygen
is 46 mg/L in water with no suspended solids.
What would be the solubility if the gas were
replaced by air?
With Pure oxygen
S K P
46 K x 1
K 46 mg/(L-atm)
With air
S K P
Since air is 20 oxygen, P 1 x 0.2 0.2 atm
S 46 x 0.2 9.2 mg/L
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Oxygen Transfer
The rate of oxygen transfer is proportional to
the difference in the oxygen concentration that
exists in the system and the saturation
concentration
dC/dt (S C)
The constant of proportionality is called the gas
transfer coefficient, KLa
dC/dt Kla (S C)
S C D, so dD/dt Kla D
Integrating
Ln (D/D0) -Kla t
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Example
Two diffusers are to be tested for their oxygen
transfer capability. Tests were conducted at
20oCusing the system shown below, with the
following results
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Kla slope of the lines
Air-Max Kla 2.37 min-1 Wonder Kla 2.69
min-1
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Sludge Volume Index, SVI
A mixed liquor has 4000 mg/L suspended solids.
After 30 minutes of settling in a 1 L cylinder,
the sludge occupied 400 ml.
SVI (400 x 1000)/ 4000 100
Good settling if SVI lt 100, if SVI gt 200 .
problems