Transcript Document

Secondary Treatment Processes
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”.
Attached Growth or Fixed Film Reactors
Trickling Filters
Rock Media
Typically 4 – 8 feet deep.
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
Trickling Filters
Single Stage Trickling Filter
Two Stage Trickling Filter
Bio-towers
Rotating Biological Contactors (RBC’s)
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
Suspended Growth Processes
Activated Sludge
Designed based on loading
(the amount of organic
matter added relative to the
microorganisms available)
Commonly called the foodto-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
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)
Monod Model for Substrate Utilization
: = : S /(Ks + S)
dX/dt = : X = (: S X) /(Ks + S)
dX/dt = - Y dS/dt
So: dS/dt = - dX/dt (1/Y) = (: S X) / [Y (Ks + S)]
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)
Mass Balance on Microorganisms:
V dX/dt = Q X0 – Q X + Y(dS/dt)V
S.S.: (dX/dt) V = 0, and QX0 = 0
So:
dS/dt = X/Y (: S /(Ks + S)
1/2c = (: S /(Ks + S)
S = Ks/(:2c – 1)
Example
A CSTR without cell recycle receives an influent with 600 mg/L BOD at
a rate of 3 m3/day. The BOD in the effluent must be 10 mg/L. The
kinetic constants are: Ks = 500 mg/L and : = 4 days-1. How large
should the reactor be?
S = Ks/(:2c – 1)
Solve for 2c:
2c = (Ks + S)/(S :)) = (500+10)/(10 x 4) = 12.75 days
2c = V/Q
V = 2c Q = 12.75 (3) = 38.25 m3
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
S = Ks/(:2c – 1) = 500/[4(8) – 1] = 16.1 mg/L
Reduction = [(600 – 16.1)/600] x 100 = 97.3%
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)
= [(S0 – S)/ t ] V /(XV)
= (S0 – S)/(X t )
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
Since
: = : S /(Ks + S)
By substitution: q =
(: S) /[Y(Ks + S)]
But q is also equal to: q = (S0 – S)/(X t )
If we equate these two equations for q and solve for S0 – S:
S0 – S =
(: S X t ) /[Y(Ks + S)]
Since q = : / Y
2c = 1/ (q Y)
And: x = (S0 – S) / t q
Problem
An activated sludge system operates at a flow rate of 400 m3/day and has an
influent BOD of 300 mg/L. The kinetic constants for the system have been
determined to be: Ks = 200 mg/L, Y = 0.5 kg SS/kg BOD, : = 2 day-1. The
mixed liquor suspended solids concentration will be 4000 mg/L. IF the system
must produce an effluent with 30 mg/L BOD, determine:
A. The volume of the aeration tank
B. The sludge age (MCRT)
C. The quantity of sludge wasted per day
The hydraulic retention time may be found from the following equation:
S0 – S = (: S X t ) /[Y(Ks + S)]
t = [Y(S0 – S) ( Ks + S)] / ( : S X)
= [0.5(300 – 30)(200 + 30)] / [2 (30) (400)] = 0.129 days = 3.1 hr
V = t Q = 400 (0.129) = 51.6 m3
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
Using the same data what MLSS is necessary to produce an effluent concentration
of 15 mg BOD/L?
q=
(: S) /[Y(Ks + S)]
= [2(15)] / [0.5(200 + 15)] = 0.28 day-1
X = (S0 – S) / ( t q )
= (300 – 15) / [0.129(0.28)] = 7890 mg/L
2c = 1 / (q Y) = 1 / [0.28(0.5)] = 7.2 days
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.
Aeration
Diffused Aeration
Coarse Bubble
Fine Bubble
Mechanical Aeration
Modeling Gas Transfer
Henry’s Law
S=KP
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
(Dalton’s Law)
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=KP
46 = K x 1
K = 46 mg/(L-atm)
With air:
S=KP
Since air is 20% oxygen, P = 1 x 0.2 = 0.2 atm
S = 46 x 0.2 = 9.2 mg/L
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
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:
Time, min
0
1
2
3
Dissolved Oxygen, mg/L
Air-max
Wonder
Diffuser
Diffuser
2
3.5
4
4.8
4.8
6
5.7
6.7
Time, min Air-max
Diffuser
0
1
2
3
D=S-C
Wonder
Diffuser
7.2
5.2
4.4
3.5
5.7
4.4
3.2
2.5
Kla = slope of the lines
Air-Max:
Wonder:
Kla = 2.37 min-1
Kla = 2.69 min-1
Sludge Volume Index, SVI
(volume of sludge after 30 min. settling, ml) x 1000
SVI =
mg/L suspended solids
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 < 100, if SVI > 200 …. problems