Denitrification and Nitrogen Removal in Cell Recycle Systems

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BIOLOGICAL PROCESSES WITH MULTIPLE ZONES

PROCESS CONSIDERATIONS FOR NITROGEN REMOVAL VIA

NITRIFICATION AND DENITRIFICATION IN CELL RECYCLE

SYSTEM

DENITRIFICATION

ASSIMILATORY (NO3 used for cell synthesis) NO3-  organic N

DISSIMILATORY (NO3 is e- acceptor in respiration) NO3-  N2

typically. Dissimilatory process carried out by facultative heterotrophs

(genera: Pseudomonas, Paracoccus, Hyphomicrobium, Alcaligenes, and

others) is used in wastewater treatment. May also produce NO2-, N2O,

even NH4+ PROCESS DESIGN BASED ON DISSIMILATORY

DENITRIFICATION, OR NITRATE RESPIRATION.

ENERGETICS OF RESPIRATION.

ORDER OF ENERGY PROVIDED BY e- ACCEPTORS:

O2 > NO3- > NO2- > Fe3+ > SO42-

So anoxia is a condition for denitrification by facultative bacteria, which would

use oxygen preferentially for energy production if available.

DENITRIFICATION PROCESS STOICHIOMETRY (MOLAR)

CH2O + 0.8NO3- + 0.8 H+  0.4N2 +1.75H2O +1.25CO2 (1)

COMPARE WITH AEROBIC RESPIRATION

CH2O + O2  H2O + CO2

Compare e- acceptor consumption/gram carbohydrate:

1mole(32g/mole) O2/0.8mole(14g/mole)-NO3-N = 2.86 g- O2/g- NO3-N

each gram on nitrate consumed in respiration saves 2.86 gram oxygen

COD-BASED WITH GROWTH (CELLS AS REFERENCE):

(-1/YH)SS – ((1-YH)/2.86YH)SNO + XBH = 0 (2)

ALKALINITY EFFECTS

NITRIFICATION (MOLAR STOICHIOMETRY, NEGLECT GROWTH):

NH4+ + 2O2  NO3- + 2H+ + H2O (3)

(2eq alkalinity consumed/mole NH4+-N oxidized)*50 g-CaCO3/14 g-N/mole =..

…= 7.1 g-CaCO3 consumed/g-NH4+-N oxidized

and 4.57 g-O2/g-NH4+-N oxidized

DENITRIFICATION

From (1):

1eq alkalinity produced/mole NO3--N reduced)*50 g-CaCO3/14 g-N/mole = …

…= 3.6 g-CaCO3 consumed/g-NO3--N reduced

Denitrification Recovers Alkalinity and Oxygen:

Net alkalinity consumption = 7.1 – 3.6 =…

…= 3.5 g-CaCO3 consumed/g-NH4+-N oxidized and reduced)

Net oxygen consumption = 4.57 – 2.86 = …

… = 1.8 g-O2/g-NH4+-N oxidized and reduced)

SUSPENDED GROWTH REACTORS WITH MULTIPLE ZONES

NITRIFICATION FOLLOWED BY DENITRIFICATION (ALSO "AIR ON-OFF"

OPERATION)

(CH3OH, etc)

MODIFIED LUDZACK-ETTINGER PROCESS (MLE)

THREE-STAGE BARDENPHO PROCESS

(CH3OH, etc)

AEROBIC ANOXIC

ANOXIC AEROBIC

ANOXIC

AEROBIC ANOXIC

CONSTRAINED DESIGN PROEDURE FOR DENITRIFICATION

PROCESSES THAT USE NITRATE RECIRCULATION (MLE AND

BARDENPHO). NOT MASS BALANCE APPROACH AS WAS USED FOR

COD AND AMMONIA OXIDATION (NITRIFICATION)

FACTORS

1. TIME BIOMASS SPENDS IN APPROPRIATE PROCESS ZONE

2. COD SUBSTRATE AVAILABLE FOR DENITRIFICATION

3. NITRATE RECIRCULATION RATE

1. Define a solids retention time for each zone based on fraction of total

solids retention time cells are aerated and anoxic to zone volumes:

fXM,aer = aer/ = i(ViXBi)aer/(VXB)system

fXM,anox = anox/ = i(ViXBi)anox/(VXB)system

where fXM,i = fraction that biomass of concentration XBi spends in a zone, i,

characterized as aerated (aer) or unaerated (anox), with volume, Vi, where i =

i(ViXBi)/QwXw, = (VXB)system QwXw, and i i = .

Assume that biomass concentration is uniform throughout system (cell

recirculation is more significant in determining XBi than growth. Then:

aer/ = i(Vi)aer/(V)system (4)

and

anox/ = i(Vi)anox/(V)system (5)

For example, for an MLE process:

Given: V1 = 3,000 m3, V2 = 7,000 m3, and system = 7 d, then

anox = 2.1 d and aer = 4.9 d

For an MLE PROCESS with one aerobic and one anoxic zone:

Vsystem = Vaer + Vanox

Substituting (6) into (4) and (5) for the following relation:

Vanox/Vaer = anox/( anox) = ( aer)/ aer

And

anox + aer (6)

2. COD REQUIREMENT FOR DENITRIFICATION

O2 consumption rate, RO, for aerobic respiration,

RO (g-O2/day) = Q(SSO – SS)(1 – YHobs) (g-O2/d)

For denitrification, equivalent nitrate consumption rate, RON, expressed

as oxygen equivalents:

RON = 2.86RNO (g- O2/day) = 2.86Q(SNNO-SNO)

1) Anox 2) Aerobic

Where

RON = rate of nitrate consumption expressed as oxygen equivalents

RNO = mass rate of nitrate consumption (kg/d)

SNNO = soluble nitrate nitrogen influent to anoxic tank or zone (mg/L N)

SNO = anoxic tank concentration of soluble nitrate nitrogen (mg/L N)

Q = flow rate (m3/d)

Since nitrate is expressed as oxygen equivalents, now can write

RON = 2.86Q(SNNO-SNO) = Q(SSO – SS)N(1 – YHobs) (g-O2/d) (6)

Since the mass of substrate COD oxidized during denitrification in the

anoxic zone, (SSO – SS)N(1 – YHobs), accounts for the mass of nitrate

reduced, expressed as oxygen.

Where (SSO – SS)N represents the COD consumed during denitrification

COD required per amount of denitrification in terms of N = SS,N/ N

= Q(SSO – SS)N/Q(SNNO – SNO)

Rearranging and substituting SS,N/ N into (6):

SS,N/ N = 2.86/(1 - YHobs) = 2.86(1+b anox)/(1+b anox -YH(1+fDb anox))

(7)

note different effects of on oxygen consumption and on denitrification:

for aerobic processes principal concern is the increase in oxygen required

because as aer increases, decay becomes more important in the biological

process, since soluble COD and ammonia are recycled: as , RO

(more decay, more soluble COD produced, more oxygen consumption,

HIGHER COST)

For denitrification the principal concern is the simultaneous utilization of

nitrate and COD. As anox , the consumption of influent substrate per

unit mass nitrate reduced, COD/ N, since decay creates more soluble

substrate. Two effects: 1) If influent COD is relatively low for the amount

of nitrate to be reduced in denitrification, long values of anox can be

advantageous since the amount of nitrate to be reduced is higher than

influent COD. 2) However, if influent COD is high, then the extra COD

supplied by biomass decay will consume nitrate in the anoxic zone, leading

to higher consumption of oxygen later in the aerobic zone, if all soluble

COD is to be oxidized, and HIGHER COST.

Selected anox should result in maximum consumption of influent COD

and generated nitrate in anoxic zone and minimum oxygen required in

aerobic zone. Graph below shows second (high influent COD) effect: long

anox  less influent COD consumed per mass nitrate reduced  greater

oxygen requirement. Typically 5 < SS,N/ N < 9 g-COD/g-NO3-N

3. NITRATE RECIRCULATION, or MIXED LIQUOR

RECIRCULATION (MLR)

Process constraint, providing sufficient nitrate for maximizing COD

removal and saving oxygen.

Invert (7):

N/ SS,N = (1+b anox-YH(1+fDb anox))/2.86(1+b anox) (g-N/g-COD) (8)

N = Q SS,N(1+b anox-YH(1+fDb anox))/2.86(1+b anox) (g-N/d) (8a)

Available nitrate for denitrification (SNO) is calculated assuming that

influent nitrogen is either stored in heterotrophic cells (neglect autotroph

cell growth), oxidized to nitrate or in effluent:

NO = Q(SNHO+ SNSO + XNSO –NR(SSO-SS) - SNH - SNS) (g-N/d) (9)

Formula assumes influent ammonia, soluble and particulate organic

nitrogen (SNHO+ SNSO + XNSO) minus cell synthesis and effluent soluble

ammonia and organic nitrogen is available nitrate

Define the fraction of available nitrate that can be denitrified, NO, in the

anoxic zone, fNO,D, which is a function of available substrate COD (from 8)

fNO,D = N/NO (10)

For the MLE system shown below, introduce the term fraction of nitrate

produced in the aerobic tank that is recirculated (assuming negligible

nitrate in the influent to the aerobic tank):

fNO,R = (NO3-N flow into anoxic zone)/(NO3-N flow out of aerobic zone)

fNO,R = [(MLR + QRAS)SNO,aer]/[(Q + MLR + QRAS)SNO,aer]

where SNO,aer = nitrate concentration in aerobic tank and in both flows,

MLR = mixed liquor recirculation flow rate from aerobic to anoxic tank,

Q = system influent flow rate, QRAS = recycled biomass flow rate from

secondary clarifier.

divide numerator and denominator by Q and define:

= QRAS/Q

and

= MLR/Q

fNO,R = ( )

Rearranging to solve for the recirculation/recycle flow rate design values:

( fNO,R( ) = fNO,R fNO,R( )

fNO,R fNO,R)

Aerobic Tank

SNO,aer

Anoxic Tank

Q + QRAS

QRAS

MLR

Q+QRAS+MLR

Q + QRAS

( fNO,R fNO,R) (11)

Objective: set and such that fNO,R = fNO,D from (10), that is, design the

recirculation/recycle flows such that the recirculated fraction equals the

fraction of available nitrate that can be denitrified, given a limited supply

of COD. After substituting fNO,D for fNO,R from (10)

= fNO,D/(1-fNO,D) (12)

SUMMARY

Process factors control denitrification in a recirculation process:

I. Ratio of available COD to nitrate reduction (g-COD/g-NO3-N

reduced) which is a function of anox Equation 8)

II. Recirculation of nitrified water, determines available

nitrate nitrogen to be reduced. Equation 11)

III. Recycled biomass flow rate, is also a source of nitrate as in

MLR (which has a lesser effect and is usually fixed by

considerations that have nothing to do with denitrification,

e.g., secondary clarifier performance and solids wasting rate)

Design procedure for MLE process:

1. Select anox (typically between 1 and 3 d), determine system volume

using CSTR-with-solids-recycle formula for VXT then select XT,max

(constrained by rO,max for min or clarifier capacity). Use total volume

and i ratios to determine aerated/anoxic zone volumes.

2. Calculate N/ SS,N using (8)

3. Calculate attainable N (g-N/day) assuming complete consumption of

soluble influent COD from (8a)

4. Calculate NO from (9)

5. calculate fNO,D from (10)

6. Calculate ( from (12). Select based on secondary clarifier

requirements and wasting rate, usually 0.25 < < 1.

7. Effluent nitrate nitrogen = {(1-fNO,D)*NO}/Q