Secondary Treatment ProcessesSecondary Treatment Processes

Two TypesTwo Types

1. Attached growth or Fixed Film1. Attached growth or Fixed Film

2. Suspended Growth2. Suspended Growth

Organisms attached to some inert media like rocks or Organisms attached to some inert media like rocks or plastic. plastic.

Organisms are suspended in the treatment basin fluid. Organisms are suspended in the treatment basin fluid. This fluid is commonly called the “mixed liquor”.This fluid is commonly called the “mixed liquor”.

Attached Growth or Fixed Film ReactorsAttached Growth or Fixed Film Reactors

Trickling FiltersTrickling Filters

Rock MediaRock Media Typically 4 – 8 feet deep.Typically 4 – 8 feet deep.

Trickling FiltersTrickling Filters

With time, the “slime” layer With time, the “slime” layer becomes thicker and thicker until becomes thicker and thicker until oxygen and organic matter can not oxygen and organic matter can not penetrate to the organisms on the penetrate to the organisms on the inside.inside.

The organisms on the inside then The organisms on the inside then die and become detached from the die and become detached from the media, causing a portion of the media, causing a portion of the “slime” layer to “slough off”.“slime” layer to “slough off”.

This means the effluent from a This means the effluent from a trickling filter will have lots of trickling filter will have lots of solids (organisms) in it which must solids (organisms) in it which must be removed by sedimentationbe removed by sedimentation

Trickling FiltersTrickling Filters

Single Stage Trickling FilterSingle Stage Trickling Filter

Two Stage Trickling FilterTwo Stage Trickling Filter

Bio-towersBio-towers

Rotating Biological Contactors (RBC’s)Rotating Biological Contactors (RBC’s)

In trickling filters the moving In trickling filters the moving wastewater passes over the stationary wastewater passes over the stationary rock media. In an RBC, the moving rock media. In an RBC, the moving media passes through the stationary media passes through the stationary wastewater.wastewater.

Commonly used out in series and Commonly used out in series and parallelparallel

Suspended Growth ProcessesSuspended Growth Processes

Activated SludgeActivated Sludge Designed based on loading Designed based on loading (the amount of organic (the amount of organic matter added relative to the matter added relative to the microorganisms available) microorganisms available)

Commonly called the food-Commonly called the food-to-microorganisms ratio, to-microorganisms ratio, F/MF/M

F measured as BOD. M F measured as BOD. M measured as volatile measured as volatile suspended solids suspended solids concentrationconcentration

F/M is the pounds of F/M is the pounds of BOD/day per pound of BOD/day per pound of MLSS in the aeration MLSS in the aeration tanktank

Design of Activated SludgeDesign of Activated Sludge

Influent organic compounds provide the food for the Influent organic compounds provide the food for the microorganisms and is called substrate (S) microorganisms and is called substrate (S)

The substrate is used by the microorganisms for growth, to produce The substrate is used by the microorganisms for growth, to produce energy and new cell material.energy and new cell material.

The rate of new cell production as a result of the use of substrate may The rate of new cell production as a result of the use of substrate may be written mathematically as:be written mathematically as:

dX/dt = - Y dS/dtdX/dt = - Y dS/dt

Y is called the yield and is the mass of cells produced per mass of Y is called the yield and is the mass of cells produced per mass of substrate used (g SS/g BOD)substrate used (g SS/g BOD)

Monod Model for Substrate UtilizationMonod Model for Substrate Utilization

dX/dt = dX/dt = X = ( X = ( S X) /(K S X) /(Kss + S) + S)

= = S /(K S /(Kss + S) + S)

dX/dt = - Y dS/dt dX/dt = - Y dS/dt

So: dS/dt = - dX/dt (1/Y) = (So: dS/dt = - dX/dt (1/Y) = ( S X) / [Y (K S X) / [Y (Kss + S)] + S)]

Mean Cell Residence Time, Mean Cell Residence Time, cc

Mean cell residence time (MCRT, Mean cell residence time (MCRT, cc) is the mass of cells in ) is the mass of cells in

the system divided by the mass of cells wasted per day.the system divided by the mass of cells wasted per day.

Consider the system:Consider the system:

c c = VX/QX = V/Q= VX/QX = V/Q

At SS the amount of solids wasted per At SS the amount of solids wasted per day must equal the amount produced per day must equal the amount produced per day:day:

cc = XV/[Y(dS/dt)V] = X/[Y(dS/dt) = XV/[Y(dS/dt)V] = X/[Y(dS/dt)

Mass Balance on Microorganisms:Mass Balance on Microorganisms:

V dX/dt = Q XV dX/dt = Q X00 – Q X + Y(dS/dt)V – Q X + Y(dS/dt)V

S.S.: (dX/dt) V = 0, and QXS.S.: (dX/dt) V = 0, and QX00 = 0 = 0

So: dS/dt = X/Y (So: dS/dt = X/Y ( S /(K S /(Kss + S) + S)

1/1/cc = ( = ( S /(K S /(Kss + S) + S)

S = KS = Kss/(/(cc – 1) – 1)

ExampleExample

A CSTR without cell recycle receives an influent with 600 mg/L BOD at A CSTR without cell recycle receives an influent with 600 mg/L BOD at a rate of 3 ma rate of 3 m33/day. The BOD in the effluent must be 10 mg/L. The /day. The BOD in the effluent must be 10 mg/L. The kinetic constants are: Kkinetic constants are: Kss = 500 mg/L and = 500 mg/L and = 4 days = 4 days-1-1. How large should . How large should

the reactor be?the reactor be?

S = KS = Kss/(/(cc – 1) – 1)

Solve for Solve for cc: : cc = (K = (Kss + S)/(S + S)/(S )) = (500+10)/(10 x 4) = 12.75 days)) = (500+10)/(10 x 4) = 12.75 days

cc = V/Q = V/Q

V = V = cc Q = 12.75 (3) = 38.25 m Q = 12.75 (3) = 38.25 m33

Given the conditions in the previous example, What would the percent Given the conditions in the previous example, What would the percent reduction in substrate be if the reactor volume was 24 mreduction in substrate be if the reactor volume was 24 m33??

cc = V/Q = 24/3 = 8 days = V/Q = 24/3 = 8 days

S = KS = Kss/(/(cc – 1) = 500/[4(8) – 1] = 16.1 mg/L – 1) = 500/[4(8) – 1] = 16.1 mg/L

Reduction = [(600 – 16.1)/600] x 100 = 97.3%Reduction = [(600 – 16.1)/600] x 100 = 97.3%

Now consider a CSTR with cell recycle:Now consider a CSTR with cell recycle:

cc = (X V) / [(Q = (X V) / [(QwwXXrr) + (Q – Q) + (Q – Qww)X)Xcc]]

Since XSince Xcc = 0: = 0:

cc = (XV)/(Q = (XV)/(QwwXXrr))

Removal of substrate often expressed in terms of Removal of substrate often expressed in terms of substrate removal velocitysubstrate removal velocity, q:, q:

q = (mass of substrate removed/time)/(mass of microorganisms under aeration)q = (mass of substrate removed/time)/(mass of microorganisms under aeration)

= [(S= [(S00 – S)/ t ] V /(XV) – S)/ t ] V /(XV)

= (S= (S00 – S)/(X t ) – S)/(X t )

Mass balance on microorganisms:Mass balance on microorganisms:

V dX/dt = Q XV dX/dt = Q X00 – Q – QwwXXrr – (Q – Q – (Q – Qww)X)Xcc + + X V X V

XX00 = X = Xcc = 0 = 0

= (X= (Xrr Q Qww)/(X V) = 1/ )/(X V) = 1/ cc

The substrate removal velocity, q, can also be expressed as: q = The substrate removal velocity, q, can also be expressed as: q = /Y/Y

Since Since = = S /(K S /(Kss + S) + S)

By substitution: q = By substitution: q = (( S) /[Y(K S) /[Y(Kss + S)] + S)]

But q is also equal to: q But q is also equal to: q = (S= (S00 – S)/(X t ) – S)/(X t )

If we equate these two equations for q and solve for SIf we equate these two equations for q and solve for S00 – S: – S:

SS00 – S = – S = (( S X t ) /[Y(K S X t ) /[Y(Kss + S)] + S)]

Since q = Since q = / Y / Y

cc = 1/ (q Y) = 1/ (q Y)

And: x = (SAnd: x = (S00 – S) / t q – S) / t q

ProblemProblem

The hydraulic retention time may be found from the following equation:The hydraulic retention time may be found from the following equation:

SS00 – S = – S = (( S X t ) /[Y(K S X t ) /[Y(Kss + S)] + S)]

An activated sludge system operates at a flow rate of 400 mAn activated sludge system operates at a flow rate of 400 m33/day and has an /day and has an influent BOD of 300 mg/L. The kinetic constants for the system have been influent BOD of 300 mg/L. The kinetic constants for the system have been determined to be: Kdetermined to be: Kss = 200 mg/L, Y = 0.5 kg SS/kg BOD, = 200 mg/L, Y = 0.5 kg SS/kg BOD, = 2 day = 2 day-1-1. The . The

mixed liquor suspended solids concentration will be 4000 mg/L. IF the mixed liquor suspended solids concentration will be 4000 mg/L. IF the system must produce an effluent with 30 mg/L BOD, determine:system must produce an effluent with 30 mg/L BOD, determine:

A. The volume of the aeration tankA. The volume of the aeration tank

B. The sludge age (MCRT)B. The sludge age (MCRT)

C. The quantity of sludge wasted per dayC. The quantity of sludge wasted per day

t = [Y(St = [Y(S00 – S) ( K – S) ( Kss + S)] / ( + S)] / ( S X) S X)

= [0.5(300 – 30)(200 + 30)] / [2 (30) (400)] = 0.129 days = 3.1 hr= [0.5(300 – 30)(200 + 30)] / [2 (30) (400)] = 0.129 days = 3.1 hr

V = t Q = 400 (0.129) = 51.6 mV = t Q = 400 (0.129) = 51.6 m33

cc = 1/ (qY) = 1/ (qY)

Q = (SQ = (S00 – S) / (X t ) = (300– 30) / [(4000)(0.129)] – S) / (X t ) = (300– 30) / [(4000)(0.129)]

= 0.523 (kg BOD removed/day) / (kg SS in the reactor)= 0.523 (kg BOD removed/day) / (kg SS in the reactor)

cc = 1/ (qY) = 1 / (0.523 x 0.5) = 3.8 days = 1/ (qY) = 1 / (0.523 x 0.5) = 3.8 days

Also Also cc = (X V) / (X = (X V) / (Xrr Q Qww))

XXrr Q Qww = (X V) / = (X V) / cc = [(4000)(51.6)( 10 = [(4000)(51.6)( 1033 L/m L/m33)( 1/10)( 1/1066 kg/mg)] / 3.8 kg/mg)] / 3.8

= 54.3 kg/day= 54.3 kg/day

Using the same data what MLSS is necessary to produce an effluent Using the same data what MLSS is necessary to produce an effluent concentration of 15 mg BOD/L?concentration of 15 mg BOD/L?

q = q = (( S) /[Y(K S) /[Y(Kss + S)] + S)]

= [2(15)] / [0.5(200 + 15)] = 0.28 day= [2(15)] / [0.5(200 + 15)] = 0.28 day-1-1

X = (SX = (S00 – S) / ( t q ) – S) / ( t q )

= (300 – 15) / [0.129(0.28)] = 7890 mg/L= (300 – 15) / [0.129(0.28)] = 7890 mg/L

cc = 1 / (q Y) = 1 / [0.28(0.5)] = 7.2 days = 1 / (q Y) = 1 / [0.28(0.5)] = 7.2 days

Solids SeparationSolids Separation

The success of the activated sludge process depends on the efficiency of the The success of the activated sludge process depends on the efficiency of the secondary clarifier, which depends on the settling characteristics of the secondary clarifier, which depends on the settling characteristics of the sludge (biosolids).sludge (biosolids).

Some system conditions result in sludge that is very difficult to settle. In 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 this case the return activated sludge becomes thin (low MLSS) and the concentration of organisms in the aeration tank goes down. This concentration of organisms in the aeration tank goes down. This produces a higher F/M ratio (same food input, but fewer organisms) and produces a higher F/M ratio (same food input, but fewer organisms) and a reduced BOD removal efficiency.a reduced BOD removal efficiency.

One condition that commonly causes this problem is called One condition that commonly causes this problem is called bulking bulking sludge. sludge. Bulking sludge occurs when a type of bacteria called Bulking sludge occurs when a type of bacteria called filamentous bacteria grow in large numbers in the system. This filamentous bacteria grow in large numbers in the system. This produces a very billowy floc structure with poor settling characteristics.produces a very billowy floc structure with poor settling characteristics.

AerationAeration

Diffused AerationDiffused Aeration

Coarse BubbleCoarse Bubble

Fine BubbleFine Bubble

Mechanical AerationMechanical Aeration

Modeling Gas TransferModeling Gas Transfer

Henry’s LawHenry’s Law

S = K PS = K P

S = solubility of the gas, mg gas/LS = solubility of the gas, mg gas/LP = partial pressure of the gasP = partial pressure of the gasK = solubility constantK = solubility constant

If a gas is 60% OIf a gas is 60% O22 and 40% N and 40% N22 and the total and the total

pressure of the gas is 1 atm (101 KPa), the partial pressure of the gas is 1 atm (101 KPa), the partial pressure of Opressure of O22 = 0.6 x 101 = 60.6 Kpa. The total = 0.6 x 101 = 60.6 Kpa. The total

pressure is equal to the sum of the partial pressures pressure is equal to the sum of the partial pressures (Dalton’s Law)(Dalton’s Law)

ExampleExample

At one atmosphere, the solubility of pure oxygen is 46 mg/L in water with no 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 suspended solids. What would be the solubility if the gas were replaced by air?air?

S = K PS = K P

46 = K x 1 46 = K x 1

K = 46 mg/(L-atm)K = 46 mg/(L-atm)

With Pure oxygen:With Pure oxygen:

With air:With air:

S = K PS = K P

Since air is 20% oxygen, P = 1 x 0.2 = 0.2 atmSince air is 20% oxygen, P = 1 x 0.2 = 0.2 atm

S = 46 x 0.2 = 9.2 mg/LS = 46 x 0.2 = 9.2 mg/L

Oxygen TransferOxygen Transfer

The rate of oxygen transfer is proportional to the difference in The rate of oxygen transfer is proportional to the difference in the oxygen concentration that exists in the system and the the oxygen concentration that exists in the system and the saturation concentration:saturation concentration:

dC/dt dC/dt (S – C) (S – C)

The constant of proportionality is called the gas transfer coefficient, KThe constant of proportionality is called the gas transfer coefficient, KLLaa

dC/dt = KdC/dt = Klla (S – C)a (S – C)

S – C = D, so: dD/dt = KS – C = D, so: dD/dt = Klla Da D

Integrating:Integrating:

Ln (D/DLn (D/D00) = -K) = -Klla ta t

ExampleExample

Two diffusers are to be tested for their oxygen transfer capability. Tests were Two diffusers are to be tested for their oxygen transfer capability. Tests were conducted at 20conducted at 20ooCusing the system shown below, with the following results:Cusing the system shown below, with the following results:

Time, min Air-max WonderDiffuser Diffuser

0 2 3.51 4 4.82 4.8 63 5.7 6.7

Dissolved Oxygen, mg/L

Time, min Air-max WonderDiffuser Diffuser

0 7.2 5.71 5.2 4.42 4.4 3.23 3.5 2.5

D = S - C

KKlla = slope of the linesa = slope of the lines

Air-Max:Air-Max: KKlla = 2.37 mina = 2.37 min-1-1

Wonder:Wonder: KKlla = 2.69 mina = 2.69 min-1-1

Sludge Volume Index, SVISludge Volume Index, SVI

SVI = SVI = (volume of sludge after 30 min. settling, ml) x 1000(volume of sludge after 30 min. settling, ml) x 1000

mg/L suspended solidsmg/L suspended solids

A mixed liquor has 4000 mg/L suspended solids. After 30 minutes of settling A mixed liquor has 4000 mg/L suspended solids. After 30 minutes of settling in a 1 L cylinder, the sludge occupied 400 ml.in a 1 L cylinder, the sludge occupied 400 ml.

SVI = (400 x 1000)/ 4000 = 100SVI = (400 x 1000)/ 4000 = 100

Good settling if SVI < 100, if SVI > 200 …. problemsGood settling if SVI < 100, if SVI > 200 …. problems