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(1)

Unit Process

Unit Process in

Biological Treatment

(2)

Modeling Suspended Growth

Treatment Processes

Description of treatment process:

All biological treatment reactor designs are based on using

mass balances across a defined volume for each constituent of

interest (i.e., biomass, substrate, etc.)

Biomass mass balance:

Accumulation = inflow – outflow + net growth

Q

Q

X

Q

X

r

V

eq

(7

-

32)

QX

V

dt

dX

g r

w e

w

o

(3)
(4)

Modeling Suspended Growth

Treatment Processes

Assuming stead-state and

X

o

= 0, equation 7-32 can be

simplified:

Q

Q

w

X

e

Q

w

X

r

r

g

V

Eq

(7

-

33)

21)

-(7

Eq

X

k

Yr

r

g

su

d

Eq

(7

-

34)

d su

r w e

w

k

X

r

Y

VX

X

Q

X

Q

Q

day

each

system

the

form

removed

organisms

of

mass

reactor

the

in

organisms

of

mass

(5)

Modeling Suspended Growth

Treatment Processes

Equation 7-34 can be written as:

The term 1/SRT is related to

µ

, the specific biomass growth rate:

Q

Q

w

X

e

Q

w

X

r

Eq

(7

-

35)

VX

SRT

36)

-(7

Eq

d su

k

X

r

Y

SRT

1

37)

-(7

Eq

(6)

Modeling Suspended Growth

Treatment Processes

In Eq. (7-36) the term

(-r

su

/

X

) is known as the

specific substrate

utilization rate U

and can be calculated as the following:

Substituting Eq. (7-12) into Eq. (7-36) yields:

Solving Eq. (7-39) for

S

yields:

38)

-(7

Eq

X

S

S

VX

S

S

Q

X

r

U

su o o

(

)

39)

-(7

Eq

d s

k

S

K

YkS

SRT

1

S

K

kXS

r

s su

 

(7)

Modeling Suspended Growth

Treatment Processes

Substrate mass balance:

Accumulation = inflow – outflow + generation

Substituting for

r

su

and assuming steady-state, Eq. (7-41) can be

written as:

41)

-(7

eq

V

r

S

Q

QS

QS

V

dt

dS

su w

o

42)

-(7

eq

S

K

kXS

Q

V

S

S

s o

 

eq

(7

-

43)

(8)
(9)

Modeling Suspended Growth

Treatment Processes

Mixed liquor solids concentration and solids production:

The solids production from a biological reactor represents the

mass of material that must be removed each day to maintain

the process:

Eq. (7-45) can be used to calculate the amount of solids

wasted for any of the mixed liquor components. For the

amount of biomass wasted (

P

X

), the biomass concentration

X

can be used in place of

X

in Eq. (7-45).

3

VSS/m

g

tank,

areation

in

MLVSS

total

VSS/d

g

daily,

wasted

solids

total

where;

45)

-(7

eq

T VSS X

T VSS

X

X

P

SRT

V

X

P

T T

(10)

Modeling Suspended Growth

Treatment Processes

Mixed liquor solids concentration:

The total MLVSS equals the biomass concentration

X

plus the

nbVSS concentration

X

i

:

A mass balance is needed to determine the nbVSS conc.:

Accumulation = inflow – outflow + generation

46)

-(7

eq

i

T

X

X

X

(11)

Modeling Suspended Growth

Treatment Processes

Mixed liquor solids concentration:

At steady-state and substituting Eq. (7-25)

for in Eq. (7-47) yields:

Combining Eq. (7-43) and Eq. (7-49) for X and Xi produces the

following equation that can be used to determine XT :

49)

-(7

eq

)

(

)

)(

(

/

)

(

,

SRT

f

k

X

SRT

X

X

i

o i

d d

dX

i

/

dt

0

X

k

f

r

Xd

d

(

d

)

r

X,i

 

   

(C) (B) (A) influent in VSS adable Nonbiodegr debris Cell biomass ic Hetrotroph

50)

-(7

eq

SRT

X

SRT

X

k

f

SRT

k

S

S

Y

SRT

X

d d o i

d o T

,

1

(12)

Modeling Suspended Growth

Treatment Processes

Solids production:

The amount of VSS produced and wasted daily is as follows:

Eq. (7-43) is substituted for biomass concentration (X) in Eq. (7-51) to show VSS production rate in terms of the substrate removal,

influent VSS, and kinetic coefficients as follows:

 

k

SRT

f

   

k

X

V

QX

eq

(7

-

51)

S

S

QY

P

d d o i

d o VSS

X, ,

1

 

  

 

(C) (B) (A) influent in VSS adable Nonbiodegr debris Cell biomass ic Hetrotroph

52)

-(7

eq

QX

SRT

k

SRT

S

S

QY

k

f

SRT

k

S

S

QY

P

o i

d o d d d o VSS

X, ,

1

1

(13)

Modeling Suspended Growth

Treatment Processes

Solids production:

The effect of SRT on the performance of an activated sludge system for soluble substrate removal is shown in figure 7-13

The total suspended solids (TSS) production can be calculated by modifying Eq. (7-52) assuming that a typical biomass VSS/TSS ratio of 0.85 as follows:

solids inorganic

influent

53)

-(7

eq

VSS

TSS

Q

C

B

A

P

X TSS

(

o o

)

85

.

0

85

.

0

(14)
(15)

Modeling Suspended Growth

Treatment Processes

The observed yield:

The observed yield for VSS can be calculated by dividing Eq. (7-52) by the substrate removal rate Q(So-S):

Oxygen requirements:

Oxygen used = bCOD removed – COD of waste sludge

56)

-(7

eq

S

S

X

SRT

k

SRT

Y

k

f

SRT

k

Y

Y

o i o d d d d obs

,

)

(

1

)

)(

)(

(

)

(

1

cells g / O g tissues, cell of COD 1.42 kg/d day, per wasted VSS as biomass kg/d required, Oxygen where 59) -(7 eq 2

bio X o bio X o o

P

R

P

S

S

Q

R

, ,

42

.

1

)

(

(16)

Modeling Suspended Growth

Treatment Processes

Design and operating parameters:

Following are the design and operating parameters that are fundamentals to treatment and performance of the process:

-

SRT

-

Food to microorganisms (F/M) ratio

The SRT can be related to F/M by the following equation:

60)

-(7

eq

biomass microbial

total

rate substrate applied

total

X

S

VX

QS

M

F

o o

/

66)

-(7

eq

k

E

M

F

Y

(17)

Modeling Suspended Growth

Treatment Processes

Design and operating parameters:

-

Organic volumetric loading rate.

Defined as the amount of BOD or COD applied to the

aeration tank volume per day:

(18)

Modeling Suspended Growth

Treatment Processes

Modeling plug-flow reactors:

Developing a kinetic model for the plug-flow reactor is

mathematically difficult (X vary along the reactor). Two assumptions are made to simplify the modeling:

-

The concentration of microorganisms is uniform along the reactor

This assumption applies only when SRT/

5.

-

The rate of substrate utilization is given by:

 

X

72)

-(7

eq

S

K

X

kS

r

s su

(19)
(20)

Modeling Suspended Growth

Treatment Processes

Modeling plug-flow reactors:

Integrating Eq. (7-72) over the retention time in the tank gives:

 

X

(21)

Biological Nitrification

Nitrification is the conversion (by oxidation) of Ammonia (NH

4

-N) to nitrite (NO

2

-N) and then to nitrate (NO

3

-N).

The need for nitrification arises from water quality concerns:

-

Effect of ammonia on receiving water; DO demand, toxicity.

-

Need to provide nitrogen removal for eutrophication control.

-

Need to provide nitrogen removal for reuse applications.

The current drinking water MCL for nitrate is 45 mg/l as

nitrate or 10 mg/l as nitrogen.

(22)

Biological Nitrification

Process description:

-

Nitrification is commonly achieved with BOD removal in the same

single-sludge process.

-

In case of the presence of toxic substances in the wastewater, a

two-sludge system is considered.

(23)

Biological Nitrification

Process description:

-

The oxygen required for complete oxidation of ammonia is 4.57 g O2/ g N oxidized.

-

The alkalinity (alk) requirement is 7.14 g alk as CaCO3 for each g of ammonia nitrogen (as N).

O

H

CO

NO

O

HCO

NH

4

2

3

2

2

2

3

2

2

3

2
(24)
(25)

Biological Denitrification

Process description:

-

Denitrification is the biological reduction of nitrate (NO3) to nitric oxide (NO), nitrous oxide (N2O), and nitrogen (N).

-

The purpose is to remove Nitrogen from wastewater.

-

Compared to alternatives of ammonia stripping, breakpoint

chlorination, and ion exchange, biological nitrogen removal is more cost-effective and used more often.

-

Concerns over eutrophication and protection of groundwater against
(26)
(27)

Biological Denitrification

Stoichiometry:

-

In denitrification, nitrate is used as the electron acceptor instead of

oxygen and the COD or BOD as the carbon source (electron donor).

-

The carbon source can be the influent wastewater COD or external

source (Methanol).

-

One equivalent of alkalinity is produced per equivalent of nitrate

reduced. (3.57 g alk per g nitrate)

(28)

Biological Phosphorus Removal

Process description:

-

Phosphorous removal is done to control eutrophication.

-

Chemical treatment using alum or iron salts is the most commonly

used technology for phosphorous removal.

-

The principle advantages of biological phosphorous removal are

reduced chemical costs and less sludge production.

-

In the biological removal of phosphorous, the phosphorous in the

influent is incorporated into cell biomass which is removed by sludge wasting.

-

Phosphorous accumulating organisms (PAOs) are encouraged to
(29)
(30)
(31)

Anaerobic Fermentation and Oxidation

Process description:

-

Used primarily for the treatment of waste sludge and high strength

organic waste.

-

Advantages include low biomass yield and recovery of energy in the

form of methane.

-

Conversion of organic matter occurs in three steps:

Step1 (Hydrolysis): involves the hydrolysis of higher-molecular-mass compounds into compounds suitable for use as a source of energy and carbon.

Step2 (Acidogenesis): conversion of compounds from step1 into lower-molecular-mass intermediate compounds.

(nonmethanogenic bacteria)

(32)

Anaerobic Fermentation and Oxidation

Process description:

-

For efficient anaerobic treatment, the reactor content should be:

– void of O2

– free of inhibiting conc. of heavy metals and sulfides – pH ~ 6.6 – 7.6

– sufficient alkalinity to ensure pH is not <6.2 (methane bacteria will not function below 6.2).

-

Methanogenic bacteria has slow growth rate, therefore:

– require long detention time for waste stabilization

– yield is low: less sludge production and most organic matter is converted to CH4 gas.

(33)

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