tric,as,rbc-basics - unit 2

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Two Types

1. Attached growth or Fixed Film

2. 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”.

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Attached Growth or Fixed Film Reactors

Trickling Filters

Rock Media

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Trickling 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 sedimentation

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Trickling Filters

Single Stage Trickling Filter

Two Stage Trickling Filter

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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 parallel

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Suspended Growth Processes

Activated 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 Commonly called the food-to-microorganisms ratio, F/ to-microorganisms ratio, F/

M M

F measured as BOD. M F measured as BOD. M

measured as volatile measured as volatile

suspended solids suspended solids

concentration concentration

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 tank

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Design 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/dt dX/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)

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Monod Model for Substrate Utilization

dX/dt =



X = (



S X) /(K

_s_s

+ S)



=



S /(K

_s_s

+ S)

dX/dt = - Y dS/dt

So: dS/dt = - dX/dt (1/Y) = (

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Mean Cell Residence Time,



_c_c

Mean cell residence time (MCRT,



_c_c

) is the mass of cells in

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

Consider the system:



_c_c

= 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:



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Mass Balance on Microorganisms:

V dX/dt = Q X

₀₀

– Q X + Y(dS/dt)V

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

₀₀

= 0

So: dS/dt = X/Y (

So: dS/dt = X/Y (



S /(K

_s_s

+ S)

1/



_c_c

= (



S /(K

_s_s

+ S)

S = K

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Example

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 m

a 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: K

kinetic constants are: K_s_s = 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 = K

_s_s

/(



_c_c

– 1)

Solve for

Solve for _c_c: : _c_c = (K = (K_s_s + S)/(S + S)/(S )) = (500+10)/(10 x 4) = 12.75 days)) = (500+10)/(10 x 4) = 12.75 days



_c_c = V/Q = V/Q

V =

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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 m

reduction in substrate be if the reactor volume was 24 m33_?_?



_c_c = V/Q = 24/3 = 8 days = V/Q = 24/3 = 8 days

S = K

_s_s

/(



_c_c

– 1) = 500/[4(8) – 1] = 16.1 mg/L

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Now consider a CSTR with cell recycle:



_c_c = (X V) / [(Q = (X V) / [(Q_w_wXX_r_r) + (Q – Q) + (Q – Q_w_w)X)X_c_c]]

Since X

Since X_c_c = 0: = 0:



_c_c = (XV)/(Q = (XV)/(Q_w_wXX_r_r))

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

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Mass balance on microorganisms:

V dX/dt = Q X

V dX/dt = Q X₀₀ – Q – Q_w_wXX_r_r – (Q – Q – (Q – Q_w_w)X)X_c_c + +  X V X V

X

X₀₀ = X = X_c_c = 0 = 0



 = (X= (X_r_r Q Q_w_w)/(X V) = 1/ )/(X V) = 1/ _c_c

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 /(K_s_s + S) + S)

By substitution: q =

By substitution: q = (( S) /[Y(K S) /[Y(K_s_s + 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 S

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

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Since q =

Since q =  / Y / Y



_c_c = 1/ (q Y) = 1/ (q Y)

And: x = (S

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Problem

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

S

S₀₀ – S = – S = (( S X t ) /[Y(K S X t ) /[Y(K_s_s + S)] + S)]

An activated sludge system operates at a flow rate of 400 m

An 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: K

determined to be: K_s_s = 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 tank

A. The volume of the aeration tank B. The sludge age (MCRT)

B. The sludge age (MCRT)

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

t = [Y(S

t = [Y(S₀₀ – S) ( K – S) ( K_s_s + S)] / ( + S)] / (  S X) S X)

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



_c_c = 1/ (qY) = 1/ (qY)

Q = (S

Q = (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)



_c_c = 1/ (qY) = 1 / (0.523 x 0.5) = 3.8 days = 1/ (qY) = 1 / (0.523 x 0.5) = 3.8 days

Also

Also _c_c = (X V) / (X = (X V) / (X_r_r Q Q_w_w))

X

X_r_r Q Q_w_w = (X V) / = (X V) / _c_c = [(4000)(51.6)( 10 = [(4000)(51.6)( 1033 L/m L/m33)( 1/10)( 1/1066 kg/mg)] / 3.8 kg/mg)] / 3.8

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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(K_s_s + S)] + S)]

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

X = (S

X = (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



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Solids 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

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Aeration

Diffused Aeration

Coarse Bubble

Fine Bubble

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Modeling Gas Transfer

Henry’s Law

S = K P

S = solubility of the gas, mg gas/L S = solubility of the gas, mg gas/L P = partial pressure of the gas

P = partial pressure of the gas K = solubility constant

K = solubility constant

If a gas is 60% O

If 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 O

pressure 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

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Example

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 P S = 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 P S = K P

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

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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, K The constant of proportionality is called the gas transfer coefficient, K_L_Laa

dC/dt = K

_l_l

a (S – C)

S – C = D, so: dD/dt = K

_l_l

a D

Integrating:

Ln (D/D

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Example

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 20

conducted at 20oo_{Cusing the system shown below, with the following results:}_{Cusing the system shown below, with the following results:}

Time, min Air-max Wonder Diffuser Diffuser

0 2 3.5

1 4 4.8

2 4.8 6

3 5.7 6.7

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Time, min Air-max Wonder Diffuser Diffuser

0 7.2 5.7

1 5.2 4.4

2 4.4 3.2

3 3.5 2.5

D = S - C

K

K_l_la = slope of the linesa = slope of the lines

Air-Max:

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

Wonder:

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Sludge 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 solids mg/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 = 100 SVI = (400 x 1000)/ 4000 = 100