PHYSICAL - CHEMICAL TREATMENT OF WATER AND WASTEWATER - CHAPTER 15 docx

Removal of Nitrogen
by Nitrification–
Denitrification

We define a biological reaction as a reaction mediated by organisms. It encompasses
both the organisms and the underlying chemical reactions. To fully apply the knowl-
edge of biological reactions to the treatment of water and wastewater, the chemical
nature of these reactions must be given center stage. In other words, to control the
process of removing nitrogen by nitrification–denitrification, the intrinsic chemical
reactions must be unraveled and fully understood. The organisms only serve as
mediators (i.e., the producer of the enzymes needed for the reaction). Thus, on the
most fundamental level, nitrogen removal is a chemical process (more accurately, a
biochemical process), and the treatment for removal of nitrogen by nitrification–
denitrification as used in this textbook is chemical in nature and the process is a
chemical unit process. In fact, nitrification–denitrification removal of nitrogen can
be effected by purely enzymatic means by providing the needed enzymes externally
without ever using microorganisms.
Similar to phosphorus, nitrogen is a very important element that has attracted
much attention because of its ability to cause eutrophication in bodies of water. As
stated in the chapter on phosphorus removal, the Chesapeake Bay in Maryland and
Virginia is fed by tributaries from farmlands as far away as New York. Because of
the use of nitrogen in fertilizers for these farms, the bay receives an extraordinarily
large amount of nitrogen input that has triggered excessive growths of algae in the
water body. Presently, large portions of the bay are eutrophied.
This chapter discusses removal of nitrogen using the unit process of nitrification
followed by denitrification. Half reactions are utilized in the discussion of the
chemical reactions. Whether or not a particular reaction will occur can be determined
by the free energy change of the reactants and products. Thus, half reactions are
normally tabulated in terms of free energies. To understand the exact meaning of
free energy as it relates to half reactions and thus to nitrogen removal, microbial

thermodynamics is discussed. Carbon requirements, alkalinity dose requirements,
and reaction kinetics as they apply to nitrogen removal are all discussed. A section
on whether or not to remove nitrogen is also included.

15.1 NATURAL OCCURRENCE OF NITROGEN

The element nitrogen is a nonmetal. It belongs to Group VA in the Periodic Table
in the second period. Its electronic configuration is [He]2

s

2

2

p

3

. [He] means that
the helium configuration is filled. The valence configuration represented by the 2,
15

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660

the

L

shell, shows five electrons in the orbitals: 2 electrons in the

s

orbitals and 3
electrons in the

p

orbitals. This means that, like phosphorus, nitrogen can have a
maximum oxidation state of

+

5; its smallest oxidation state is 3

−

. Examples are
nitrous oxide (N

2

O, 1

+

); nitric oxide (NO, 2

+

); dinitrogen trioxide (N

2

O

3

, 3

+

);
nitrogen dioxide (NO

2

, 4

+

); dinitrogen tetroxide (N

2

O

4

, 4

+

); and dinitrogen pentox-
ide (N

2

O

5

, 5

+

). Our interest in nitrogen as it occurs in nature is in the form that
makes it fertilizer to plants. These forms are the

nitrites, nitrates

, and

ammonia

. The
nitrogen in ammonia exists in its smallest oxidation state of 3

−

; in nitrites, it exists
as 3

+,

and in nitrates, it exists as 5

+

. These nitrogen species are utilized by algae
as nutrients for growth. Also, because organic nitrogen hydrolyzes to ammonia, we
will, in general, be concerned with this form of the nitrogen species.

15.2 TO REMOVE OR NOT TO REMOVE NITROGEN

The formula of algae is (CH

2

O)

106

(NH

3

)

16

H

3

PO

3

(Sincero and Sincero, 1996). Gleaning
from this formula, to curtail its production in any water body such as the Chesapeake
Bay, it is necessary to control only any one of the elements of nitrogen, phosphorus,
oxygen, hydrogen, or carbon. It must be stressed that only one needs to be controlled,
because absence of any element needed for the construction of the algal body
prevents the construction of the body. This is analogous to a car. To disable this car,
you only need to remove one wheel and you can never drive the car.
Of course, oxygen, hydrogen, and carbon should never be controlled, because
there are already plenty of them around. From the algae formula, the ratio of N to
P is 16

/

1

=

16 mole for mole or 14(16)

/

31

=

7.2 mass for mass. Table 15.1 shows
various values of nitrogen and phosphorus concentrations in the water column and
the corresponding N/P ratios in some coastal areas of Maryland (Sincero, 1987).
For those ratios greater than 7.2, phosphorus will run out first before nitrogen does.
In these situations, phosphorus should be controlled first and nitrogen should be left
alone in the discharge, until further investigation reveals that the ratio has reversed.

TABLE 15.1
Nitrogen and Phosphorus Ratios, Maryland Coastal Area

Organic N,
mg/L
NH

3

–N,
mg/L
NO

2

–N,

mg/L
NO

3

–N,
mg/L
Total N,
mg/L
Total P,
mg/L N/P Ratio

0.79 0.01 0.002 0.03 0.83 0.59 1.4
0.63 0.04 0.003 0.01 0.68 0.15 4.6
0.47 0.03 0.002 0.02 0.52 0.09 5.8
2.59 0.01 0.002 0.03 2.63 0.29 9.1
1.99 0.01 0.002 0.02 2.02 0.30 6.7
3.19 0.07 0.002 0.03 3.29 0.18 18.3
1.59 0.01 0.002 0.02 1.62 0.13 12.5
0.49 0.01 0.002 0.02 0.52 0.04 13.1
0.59 0.01 0.002 0.02 0.62 0.11 5.7
1.19 0.01 0.002 0.03 1.23 0.27 4.6

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661

For those situations where the ratio is less than 7.2, nitrogen will run out first. In these
cases, should nitrogen be controlled?

Certain forms of algae, the blue-greens, can synthesize nitrogen from the air into
ammonia, which they need for growth (Sincero, 1984). These particular species are
very resistant and can survive anywhere where there is a water body. Thus, if this is
the case in a particular body of water, it may be a waste of money to remove nitrogen,
because the alga could simply get the nitrogen it needs from the air. Phosphorus
should be removed, instead. These situations can become very political, however,
especially with some environmentalists. Some authorities even claim that it is still
advisable to remove both nitrogen and phosphorus (D’Elia, 1977). It is in this situation
that modeling of the effect of the discharge of the nutrients nitrogen and phosphorus
on the eutrophication potential of the water body should be investigated accurately
and in great detail.

15.3 MICROBIAL THERMODYNAMICS

The study of the relationships between heat and other forms of energy is called

thermodynamics

. All living things utilize heat, therefore, the science of thermody-
namics may be used to evaluate life processes. An example of a life process is the
growth of bacteria when wastewater is fed to them to treat the waste. Knowledge
of microbial thermodynamics is therefore important to professionals involved in
cleaning up wastewaters.
Variables involved in the study of the relationship of heat and energy are called

thermodynamic variables

. Examples of these variables are temperature, pressure,
free energy, enthalpy, entropy, and volume. In our short discussion of thermodynam-
ics, we will address enthalpy, entropy, and free energy. As mentioned, whether or

not a particular reaction, such as a biological reaction, is possible can be determined
by the free energy change between products and reactants. Free energy, in turn, is
a function of the enthalpy and entropy of the reactants and products.

15.3.1 E

NTHALPY



AND

E

NTROPY

Let

H

represent the enthalpy,

U

the internal energy,

P

the pressure, and the volume
of a particular system undergoing a process under study. The enthalpy

H

is defined as
(15.1)
Internal energy refers to all the energies that are present in the system such as kinetic
energies of the molecules, ionization energies of the electrons, bond energies, lattice
energies, etc. The system possesses all these energies by virtue of its being and are
all integral (that is, internal) with the system.
Let us derive the relationship between enthalpy and the heat exchange during a
biological reaction, where

biological reaction

is a chemical reaction mediated by
organisms. Biological reactions are carried out at constant pressure; hence, the heat
exchange is a heat exchange at constant pressure. Designate this exchange as

Q

p

.
The first law of thermodynamics states that any heat added to a system minus any
work

W

that the system is doing at the same time manifests itself in the form of an
V

–
HUP
V
–
+=

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662

increase of the internal energy. In differential form,
(15.2)
The only work done in biological reactions is the work of pushing the surroundings
(the atmosphere) in which the reaction is occurring. This is a pressure-volume work;
hence, the term.
Because the biological reaction is at constant pressure, differentiate the enthalpy
equation at constant pressure. This produces
(15.3)
This may be combined with Equation (15.2) to eliminate

dU

producing
(15.4)
This equation concludes that change in enthalpy is a heat exchange at constant
pressure between the system under study and its surroundings.
Before we discuss entropy, define reversible process and reversible cycle. A

reversible process

is a process in which the original state or condition of a system
can be recovered back if the process is done in the opposite direction from that in
which it is currently being done. To perform a reversible process, the steps must be
conducted very, very slowly, in an infinitesimal manner, and without friction. From
the definition of a reversible process, the definition of a reversible cycle follows. A

reversible cycle

is a cycle in which the reversible process is applied in every step
around the cycle.
Heat added to a system causes its constituent particles to absorb the energy resulting
in the system being more chaotic than it was before. If the heat is added reversibly, the
ratio of the infinitesimal heat added to the temperature

T

during the infinitesimal time
that the heat is added defines the

change in entropy

. If this addition is done around a
reversible cycle, the state or condition of the system at the end of the cycle will revert
back to its original state or condition at the beginning of the cycle. This must be so,
since the whole process is being done reversibly in every step along the way around
the cycle. Hence, the change in entropy around a reversible cycle is zero.
Let

S

be the entropy and

Q

rev

be the reversible heat added. In a given differential
step, the heat added is

dQ

rev

. The differential change in entropy in every differential
step is therefore

dS



=



dQ

rev

/

I

. Around the cycle, the change in entropy is the integral,
thus
(15.5)
The symbol

͛

means that the integrand is to be integrated around the cycle and
subscripts

e

and

b

refer to the end and the beginning of the cycle, respectively. If
the process is not around a cycle, the previous subscripts simply mean the end and
dU dQ
p
dW– dQ
p
Pd
V
–==
Pd
V

dH dU Pd
V
–
+=
dH dQ
p
=
dS
∫
°
S
e
S
b
– ∆S
dQ
rev
T

∫
°
0=== =

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663

beginning of the process. In this case, the integral will not be zero and the equation
is written as the integral

(15.6)
(15.7)

Interpretations of enthalpy and entropy.

The heat absorbed by the system
causes more agitation of its constituent elements. This increased agitation and chaos
is the entropy increase and is calculated by Eqs. (15.6) and (15.7). The entropy
increase is an increase in disorder of the constituents of the system. The energy state
of the system is increased, but because the energy supporting this state is nothing
more than supporting chaos, this energy is a wasted energy. The equations therefore
calculate the loss in energy of the system as a result of increased chaos or disorder.
Consider a fuel such as coal, and burn it in a furnace. The burning of the coal
occurs under constant atmospheric pressure. As the coal burns, heat is released; this
heat is energy

Q

p

, which may be used to produce electricity by using a boiler and
a turbine generator. From Equation (15.4),

Q

p

is equal to the enthalpy change

∆

H.

We therefore conclude that before the coal was burned, it possessed an enthalpy

H

which, by virtue of Equation (15.4), is its energy content. By the entropy change
during the process of burning, however, all this energy is not utilized as useful energy
but is subtracted by the change in entropy. The electrical energy that is ultimately
delivered to the consumer is less by an amount equal to the overall entropy change
in the transformation of coal to electricity.
In biological reactions, the fuel is the food. In biological nitrogen removal,
nitrogen in its appropriate form is fed to microorganisms to be utilized as food. This
food possesses enthalpy as does coal; and, similar to coal, its energy content cannot
all be utilized as useful energy by the microorganism as a result of the inevitable
entropy inefficiency that occurs in the process of consuming food.

15.3.2 F

REE

E

NERGY

Will a certain food provide energy when utilized by microorganisms? If the answer
is yes, then the food will be eaten; and if it is in a wastewater, the wastewater will
be cleaned up. The answer to this question can now be quantified by the combination
of the concept of enthalpy and entropy. This combination is summed up in a term

called

free energy

.

Free energy



G

is defined as
(15.8)
Because

H

is an energy content and

S

is a wasted energy,

G

represents the useful
energy (or, alternatively, the maximum energy) the fuel can provide after the wasteful
Sd
b

e
∫
S
e
S
b
– ∆S
Qd
rev
T

b
e
∫
===
Q
rev
T

= T constatnt=
GHTS–=
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energy (represented by S) has been subtracted from the energy content. Thus, the
term free energy.
Biological processes are carried out at a given constant temperature as well as
constant pressure. Thus, differentiating the free-energy equation at constant temper-
ature,
(15.9)
Note: In order for G to be a maximum (i.e., to be a free energy), Q must be the

Q
rev
as depicted in the equation.
15.4 OXIDATION–REDUCTION REACTIONS
OF NITROGEN FOODS
Life processes involve electron transport. Specifically, the mitochondrion and the
chloroplast are the sites of this electron movement in the eucaryotes. In the procary-
otes, this function is embedded in the sites of the cytoplasmic membrane. As far as
electron movement is concerned, life processes have similarity to a battery cell. In
this cell, electrons move because of electrical pressure, the voltage difference. By the
same token, electrons move in an organism because of the same electrical pressure,
the voltage difference. In a battery cell, one electrode is oxidized while the other is
reduced; that is, oxidation–reduction occurs in a battery cell. Exactly the same process
occurs in an organism.
In an oxidation–reduction reaction, a mole of electrons involved is called the
faraday, which is equal to 96,494 coulombs. A mole of electrons is equal to one
equivalent of any substance. Therefore, a faraday is equal to one equivalent.
Let n be the number of faradays of charge or mole of a substance participating
in a reaction, and let the general reaction be represented by the half-cell reaction of
Zn as follows:
(15.10)
The couple Zn/Zn
2+
has an electric pressure between them. Now, take another couple
such as Mg/Mg
2+
whose half-cell reaction would be similar to that of Equation (15.10).
The couples Zn/Zn
2+
and Mg/Mg

2+
do not possess the same voltage potential. If the
two couples are connected together, they form a cell. Their voltage potentials are
not the same, so a voltage difference would be developed between their electrodes
and electric current would flow.
Let the voltage between the electrodes of the above cell be measured by a poten-
tiometer. Designate this voltage difference by ∆E. In potentiometric measurements,
no electrons are allowed to flow, but only the voltage tendency of the electrons to
flow is measured. No electrons are allowed to flow, therefore, no energy is dissipated
due to friction of electrons “rubbing along the wire.” Thus, any energy associated with
this no-electron-to-flow process represents the maximum energy available. Because
voltage is energy in joules per coulomb of charge, the energy associated can be
dG dH TdS– dH Q
rev
–==
Zn Zn
2+
2e
−
+→
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calculated from the voltage difference. This associated energy corresponds to a no-
friction loss process; it is therefore a maximum energy—the change in free energy—
after the entropic loss has been deducted.
Let n, the number of faradays involved in a reaction, be multiplied by F, the
number of coulombs per faraday. The result, nF, is the number of coulombs involved
in the reaction. If nF is multiplied by ∆E, the total associated energy obtained from
the previous potentiometric experiment results. Because the voltage measurement
was done with no energy loss, by definition, this associated energy represents the

free energy change of the cell (i.e., the maximum energy change in the cell). In
symbols,
(15.11)
The sign has been used. A convention used in chemistry is that if the sign is
negative, the process is spontaneous and if the sign is positive, the process is not
spontaneous.
As mentioned, the battery cell process is analogous to the living cell process of
the mitochondrion, the chloroplast, and the electron-transport system in the cyto-
plasmic membrane of the procaryotes. Thus, Equation (15.11) can represent the basic
thermodynamics of a microbial system.
In the living cell, organic materials are utilized for both energy (oxidation) and
synthesis (reduction). Microorganisms that utilize organic materials for energy are
called heterotrophs. Those that utilize inorganics for energy are called autotrophs.
Autotrophs utilize CO
2
and for the carbon needed for cell synthesis; the het-
erotrophs utilize organic materials for their carbon source. Autotrophs that use inorganic
chemicals for energy are called chemotrophs; those that use sunlight are phototrophs.
The bacteria that consumes the nitrogen species in the biological removal of nitrogen
are chemotrophic autotrophs. Algae are phototrophic autotrophs.
Somehow, in life processes, the production of energy from release of electrons
does not occur automatically but through a series of steps that produce a high energy-
containing compound. This high energy-containing compound is ATP (adenosine
triphosphate). Although ATP is not the only high energy-containing compound, it is
by far the major one that fuels synthesis in the cell. ATP is the energy currency that
the cell relies upon for energy supply.
The energy function of ATP is explained as follows: ATP contains two high-
energy bonds. To form these bonds, energy must be obtained from an energy source
through electron transfer. The energy released is captured and stored in these bonds.
On demand, hydrolysis of the bond releases the stored energy which the cell can

then use for synthesis and cell maintenance.
ATP is produced from ADP (adenosine diphosphate) by coupling the release of
electrons to the reaction of organic phosphates and ADP producing ATP. ATP has
two modes of production: substrate-level phosphorylation and oxidative phospho-
rylation. In the former, the electrons released by the energy source are absorbed by
an intermediate product within the system. The electron absorption is accompanied
by an energy release and ATP is formed. The electron-transport system is simple.
∆GnF ∆E±=
±
HCO
3
−
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Fermentation is an example of a substrate-level phosphorylation process that uses
intermediate absorbers such as formaldehyde. Substrate-level phosphorylation is
inefficient and produces only a few molecules of ATP.
In the oxidative phosphorylation mode, the electron moves from one electron
carrier to another in a series of complex reduction and oxidation steps. The difference
between substrate-level and oxidative phosphorylation is that in the former, the
transport is simple, while in the latter, it is complex. For a hydrogen-containing
energy source, the series starts with the initial removal of the hydrogen atom from
the molecule of the source. The hydrogen carries with it the electron it shared with
the original source molecule, moving this electron through a series of intermediate
carriers such as NAD (nicotinamide adenine dinucleotide). The intermediate that
receives the electron-carrying hydrogen becomes reduced. The reduction of NAD,
for example, produces NADH
2
. The series continues on with further reduction and
oxidation steps. The whole line of reduction and oxidation constitutes the electron-

transport system. At strategic points of the transport system, ATP is produced from
ADP and inorganic phosphates.
The other version of oxidative phosphorylation used by autotrophs involves the
release of electrons from an inorganic energy source. An example of this is the
release of electrons from , oxidizing to , and the release of electrons
from , oxidizing to .
The transported electrons emerge from the system to reduce a final external
electron acceptor. The type of the final acceptor depends upon the environment on
which the electron transport is transpiring and may be one of the following: for
aerobic environments, the acceptor is O
2
; for anaerobic environments, the possible
acceptors are , , and CO
2
. When the acceptor is , the system is said
to be anoxic.
The values of free energy changes are normally reported at standard conditions.
In biochemistry, in addition to the requirement of unit activity for the concentrations
of reactants and products, pressure of one atmosphere, and temperature of 25°C, the
hydrogen ion concentration is arbitrarily set at pH 7.0. Following this convention,
Equation (15.11) may be written as
(15.12)
The primes emphasize the fact that the standard condition now requires the {H
+
} to
be 10
−7
moles per liter.The subscript o signifies conditions at standard state.
In environmental engineering, it is customary to call the substance oxidized as
the electron donor and the substance reduced as the electron acceptor. The electron

donor is normally considered as food. In the context of nitrogen removal, the foods
are the nitrites, nitrates, and ammonia. Equation (15.10) is an example of an electron
donor reaction. Zn is the donor of the electrons portrayed on the right-hand side of
the half-cell reaction. On the other hand, the reverse of the equation is an example
of an electron acceptor reaction. Zn
+2
would be the electron acceptor. McCarty (1975)
derived values for free energy changes of half-reactions for various electron donors
and acceptors utilized in a bacterial systems. The ones specific for the nitrogen
species removal are shown in Table 15.2.
NH
4
−
NH
4
+
NO
2
−
NO
2
−
NO
2
−
NO
3
−
NO
3

−
SO
4
−2
NO
3
−
∆G′
o
nF∆E
o
′±=
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15.4.1 CRITERION FOR SPONTANEOUS PROCESS
It is a law of nature that things always go in the direction of creating greater chaos—
this is the second law of thermodynamics. Any system, except those at temperature
equals absolute zero, is always disordered.* The energy required to maintain this
disorder, we have found, is called entropy. As mentioned, any system possesses free
energy at any instant, this energy being the net energy remaining after the entropy
required to maintain the current disorder has been subtracted from the enthalpy (energy
content).
When the system goes from state 1 (current state) to state 2, its free energy at
the latter state may or may not be the same as the former. If the free energy at state 2
TABLE 15.2
Half-Cell Reactions for Bacterial Systems in Nitrogen Removal
kcal/electron–mol
Reactions for cell synthesis:
Ammonia as nitrogen source:
—

Nitrate as nitrogen source:
—
Reactions for electron acceptors:
Oxygen as acceptior:
−18.68
Nitrate as acceptor:
−17.13
Nitrite as acceptor:
—
Reactions for electron donors:
Domestic wastewater as donor (heterotrophic reaction):
−7.6
Nitrite as donor:
+9.43
Ammonia as donor:
+7.85
* At absolute zero, all particles practically cease to move and are therefore structured and orderly.
∆∆
∆∆
G
o
′′
′′
1
5

CO
2
1
20

HCO
3
−
1
20

NH
4
+
H
+
e
−
++++
1
20

C
5
H
7
NO
2
9
20

H
2
O+→

1
28

NO
3
−
5
28

CO
2
29
28

H
+
e
−
+++
1
28

C
5
H
7
NO
2
11
28

H
2
O+→
1
4

O
2
H
+
e
−
++
1
2

H
2
O→
1
5

NO
3
−
6
5

H

+
e
−
++
1
10

N
2
3
5

H
2
O+→
1
3

NO
2
−
4
3

H
+
e
−
++
1

6

N
2
2
3

H
2
O+→
1
50

C
10
H
19
NO
3
9
25

H
2
O+
9
50

→ CO
2

1
50

NH
4
+
1
50

HCO
3
−
H
+
e
−
++ ++
1
2

NO
2
−
1
2

H
2
O+
1

2

NO
3
−
H
+
e
−
++→
1
6

NH
4
+
1
3

H
2
O+
1
6

NO
2
−
4
3

H
+
e
−
++→
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© 2003 by A. P. Sincero and G. A. Sincero
is the same as that in state 1, the system must be in equilibrium. If the free energy
at state 2 is greater than that at state 1, then some outside free energy must have been
added to the system. In Table 15.2, this is the case of nitrite as a donor and ammonia
as a donor. External free energies of 9.43 kcal/electron-mol and 7.85 kcal/electron-
mol, respectively, has been added to the system; these values are indicated by the
plus signs. External sources of energy are being required, so these half-cell reactions
cannot occur spontaneously; they are said to be endothermic (i.e., requiring external
energies to effect the reaction).
On the one hand, when the free energy at state 2 is less than that at state 1, some
energy must have been released by the system to the surroundings, thus manifesting
in the decrease of free energy. A decrease in free energy is indicated in the table
with a negative sign. This energy has been released “voluntarily” by the system
without some form of “coercion” from the surroundings. The release is spontaneous,
and therefore the reaction is spontaneous.
Note: Thus, this is the criterion for a spontaneous process: When the free energy
change is negative, the process is spontaneous.
Judging from Table 15.2, when the electrons that travel through the electron
transport system are finally accepted by oxygen, a large amount of energy equal to
18.68 kcal/electron-mol is liberated. This liberated energy is then captured in the
bonds of ATP. The same statement holds for the others whose free energy changes
have negative signs. Thus, any material in wastewater, edible by organisms, will
release energy, resulting in their destruction. The more energy that can be released,

the easier it is to be treated using microorganisms.
15.5 MODES OF NITROGEN REMOVAL
The physical removal of nitrogen using the unit operation of stripping was discus-
sed in a previous chapter. The present chapter concerns only the removal of nitrogen
by biochemical means, as mediated by microorganisms. The technique of the unit
process is to release the nitrogen in the form of the gas N
2
to the atmosphere. This
will first entail nitrifying the nitrogens using the species of bacteria Nitrosomonas
and Nitrobacter. Nitrosomonas oxidizes the ammonium ion into nitrites, deriving
from this oxidation the energy it needs. Nitrobacter then oxidizes the nitrites into
nitrates, also deriving from this oxidation the energy that it needs. These oxidations
into nitrites and nitrates is called nitrification. Nitrification is an aerobic process.
After the nitrogen has been nitrified, the second unit process of denitrification
is then applied. The denitrifying bacteria, which are actually heterotrophs, convert
the nitrates into nitrogen gas, thus ridding the wastewater of nitrogen. Denitrification
is an anaerobic process.
15.6 CHEMICAL REACTIONS IN NITROGEN REMOVAL
In biochemical nitrogen removal, BNR, two steps are required: oxidation of nitrogen
to nitrate and subsequent reduction of the nitrate to gaseous nitrogen, N
2
. The
oxidation steps are mediated by Nitrosomonas and by Nitrobacter, as mentioned
TX249_Frame_C15.fm Page 668 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
previously. The reduction step is mediated by the normal heterotrophic bacteria. We
will now discuss the chemical reactions involved in these oxidations and reduction.
15.6.1 NITRIFICATION: NITROSOMONAS STAGE
From Table 15.2, the generalized donor reaction, acceptor reaction, and synthesis
reaction mediated by Nitrosomonas are, respectively, shown by the following half-

cell reactions:
donor reaction (15.13)
acceptor reaction (15.14)
synthesis reaction (15.15)
The table shows three possibilities for a donor reaction: domestic wastewater, nitrite,
and ammonia.
Note: Ammonia and ammonium are interchangeable, since one converts to the
other.
We are nitrifying ammonia, so the ammonium is the donor. The table also shows other
possibilities for the electron acceptor. Because nitrification is aerobic, however,
oxygen is the acceptor. Lastly, the synthesis reaction has two possibilities: one using
ammonia and the other using nitrate. Of these two possible species, organisms prefer
the ammonium ion to the nitrate ion. Only when the ammonium ions are consumed
will the nitrates be used in synthesis.
If the previous equations are simply added without modifications, electrons will
remain free to roam in solution; this is not possible. In actual reactions, the previous
equations must be modified to make the electrons given up by Equation (15.13) be
balanced by the electrons accepted by Equations (15.14) and (15.15). Starting with
one gram-equivalent of NH
4
–N, based on Equation (15.15), assume m equivalents
are incorporated into the cell of Nitrosomonas, C
5
H
7
NO
2
. Thus, the NH
4
–N equi-

valent remaining for the donor reaction of Equation (15.13) is 1 − m, equals (1 − m)
[((1/20)N)/1] (1/N) = (1 − m) moles. [The 1/20 came from Equation (15.15) used
to compute the equivalent mass of NH
4
–N.] Thus, (1 − m) moles of NH
4
–N is
available for the donor reaction to donate electrons. By Equation (15.13), the mod-
ified donor reaction is then
(15.16)
1
6

NH
4
+
1
3

H
2
O
1
6

NO
2
−
4
3

H
+
e
−
++→+
1
4

O
2
H
+
e
−
1
2

H
2
O→++
1
5

CO
2
1
20

HCO

3
−
1
20

NH
4
+
H
+
e
−
++++
→
1
20

C
5
H
7
NO
2
9
20

H
2
O+
1

20

1
20

1/6
1/6

1
20



1 m–()NH
4
+
1/3
1/6

1
20



1 m–()H
2
O+
→
1/6
1/6

1
20



1 m–()NO
2
−
4/3
1/6

1
20



1 m–()H
+
1
1/6

1
20



1 m–()e
−
++

TX249_Frame_C15.fm Page 669 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
From Equation (15.15), the m equivalents of NH
4
–N is m[((1/20)N)/1] (1/N) =
moles. Thus, the synthesis reaction becomes
(15.17)
From Eqs. (15.16) and (15.17), the electron-moles left for the acceptor reaction is
Hence, the acceptor reaction, Equation (15.14), modifies to
(15.18)
Adding Eqs. (15.16), (15.17), and (15.18) produces the overall reaction for the
Nitrosomonas reaction shown next. Note that, after addition, the electrons e
−
are
gone. The overall reaction should indicate no electrons in the equation, since elec-
trons cannot just roam around in the solution. They must be taken up by some atom.
The overall reaction is
(15.19)
15.6.2 NITRIFICATION: NITROBACTER STAGE
Nitrobacter utilizes the nitrites produced by Nitrosomonas for energy. The unmod-
ified donor, acceptor, and synthesis reactions are written below, as taken from
Table 15.2:
donor reaction (15.20)
acceptor reaction (15.21)
synthesis reaction (15.22)
m
20

1/5
1/20

m
20



CO
2
1/20
1/20

m
20



HCO
3
−
1/20
1/20

m
20



NH
4
+

1
1/20

m
20



H
+
+
1
1/20

m
20



e
−
+++
1/20
1/20

m
20




→ C
5
H
7
O
2
N
9/20
1/20

m
20



H
2
O+
1
1/6

1
20



1 m–()
1
1/20

m
20



–
313m–
10

=
1
4

313m–
10



O
2
313m–
10

H
+
313m–
10

e
−

++
1
2

313m–
10



H
2
O→
1
20

NH
4
+
313m–()
40

O
2
m
20

HCO
3
−
m

5

CO
2
+++
m
20

C
5
H
7
NO
2
+
1
20

1 m–()NO
2
−
12m–
20

H
2
O+→
1 m–
10

H
+
+
1
2

NO
2
−
1
2

H
2
O+
1
2

NO
3
−
H
+
e
−
++→
1
4

O

2
H
+
e
−
++
1
2

H
2
O→
1
5

CO
2
1
20

HCO
3
−
1
20

NH
4
+
H

+
e
−
++++
→
1
20

C
5
H
7
NO
2
9
20

H
2
O+
TX249_Frame_C15.fm Page 670 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
In the BNR process, the Nitrobacter reaction follows the Nitrosomonas reaction.
From Equation (15.19), 1/20(1 − m) moles of NO
2
–N have been produced from the
original one equivalent of ammonia nitrogen based on Equation (15.15). These nitrites
serve as the elector donor for Nitrobacter. Thus, the donor reaction of Equation (15.20)
becomes
(15.23)

Let n, based on Equation (15.22), be the equivalents of Nitrobacter cells pro-
duced. This quantity, n equivalents, is equal to
Modifying the synthesis reaction,
(15.24)
Subtracting the electrons used in Equation (15.24), 1/(1/20)(n/20), from the
electrons donated in Equation (15.23), 1/(1/2)[1/20(1 − m)], produces the amount
of electrons available for the acceptor (energy) reaction, (1 − m − 10n)/10. Modifying
the acceptor reaction,
(15.25)
Adding Equations (15.23), (15.24), and (15.25) produces the overall reaction
for Nitrobacter,
(15.26)
1/2
1/2

1
20

1 m–()NO
2
−
1/2
1/2

1
20

1 m–()H
2
O +

1/2
1/2

1
20

1 m–()NO
3
−
1
1/2

+
1
20

1 m–()H
+
1
1/2

1
20

1 m–()e
−
+→
n
C
5

H
7
O
2
N
20

1




C
5
H
7
O
2
N n/20= moles.
1/5
1/20

n
20



CO
2
1/20

1/20

n
20



HCO
3
−
1/20
1/20

n
20



NH
4
+
1
1/20

n
20



H

+
1
1/20

n
20



e
−
++++
→
1/20
1/20

n
20



C
5
H
7
O
2
N
9/20
1/20

n
20



H
2
O+
1
4

1 m–10n–
10

O
2
1 m–10n–
10

H
+
1 m–10n–
10

e
−
1
2

1 m–10n–
10

H
2
O→++
1 m–()
20

NO
2
n
20

NH
4
+
n
5

CO
2
n
20

HCO
3
−
n
20

H
2
O
1 m–10n–
40

O
2
+++ ++
→
n
20

C
5
H
7
O
2
N
1 m–
20

NO
3
−
+
TX249_Frame_C15.fm Page 671 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero

15.6.3 OVERALL NITRIFICATION
The Nitrosomonas and the Nitrobacter reactions may now be added to produce the
overall nitrification reaction as shown next.
(15.27)
The literature reports cell yields (productions) for Nitrosomonas of 0.04 to 0.29
milligrams of the bacteria per milligram of NH
4
–N destroyed and cell yields for
Nitrobacter of 0.02 to 0.084 milligrams of the bacteria per milligrams of NO
2
–N
destroyed (Mandt and Bell, 1982). Yield or specific yield simply means the amount
of organisms produced per unit amount of substrate consumed. Also, for nitrification
to be the dominant reaction, the BOD
5
/TKN ratios should be less than 3. At these
ratios, the nitrifier population is about 10% and higher. For BOD
5
/TKN ratios of
greater than 5.0, the process may be considered combined carbonaceous-nitrification
reaction. At these ratios, the nitrifier population is less than 4%. In addition, to ensure
complete nitrification, the dissolved oxygen concentration should be, at least, about
2.0 mg/L.
15.6.4 DENITRIFICATION: HETEROTROPHIC SIDE REACTION STAGE
Aside from the normal anoxic reaction, two side reactions must be considered in
denitrification: the continued oxidation using the leftover dissolved oxygen from the
nitrification reaction stage, and nitrite reduction. Immediately after nitrification is
stopped, a large concentration of dissolved oxygen still exists in the reactor—this
would be around 2.0 mg/L. In nitrification, the nitrifiers are mixed with the heterotrophic
bacteria. The heterotrophic bacteria are fast growers compared to Nitrosomonas and

Nitrobacter, so they overwhelm the reaction and the overall chemical process is the
normal heterotrophic carbonaceous reaction in the last stage of oxidation. By control
of the process, the growth rates of the nitrifiers and the heterotrophs are balanced
during nitrification and the two types of bacteria grow together. As soon as the oxygen
supply is cut off, however, the nitrifiers cannot compete against the heterotrophs for
the ever decreasing concentration of dissolved oxygen. Thus, the activities of the
nitrifiers “fade away,” and the heterotrophs predominate in the last stage of oxidation
after aeration is cut off.
The other side reaction is the reduction of nitrite to the nitrogen gas. Although
the process is aimed at oxidizing nitrogen to the nitrate stage, some nitrite can still
be found. In the absence of oxygen, after the heterotrophs have consumed all the
remaining oxygen, no other reaction can occur except for the reduction of nitrite.
This is discussed further, after the discussion on the regular denitrification reaction.
Now, derive the chemical reaction for the heterotrophic stage. Let r be the equiv-
alents of O
2
(based on the oxygen acceptor reaction) used at this stage. The removal
1 n+
20

NH
4
+
mn+
20

HCO
3
−
mn+

5

CO
2
27m–5n–
20

O
2
+++
mn+
20

C
5
H
7
NO
2
1 m–
10

H
+
1 m–
20

NO
3
−

12m– n–
20

H
2
O++ +→
TX249_Frame_C15.fm Page 672 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
of nitrogen is done in conjunction with the treatment of sewage. Thus, sewage
(C
10
H
19
NO
3
) must be the electron donor. Using sewage, the ammonium ion is
produced, see Table 15.2. As mentioned before, given and in solution,
organisms prefer to use first, before for synthesis. Thus, of the two
competing synthesis reactions, the one using the ammonium is then the one favored
by the bacteria. Letting q be the equivalents of cells, based on the synthesis reaction,
produced during the last stage of the aerobic reaction, the synthesis reaction may
be modified as follows:
(15.28)
From the r equivalents of O
2
(based on the oxygen acceptor reaction), the
acceptor reaction is
(15.29)
From these equations, the total electron moles needed from the donor is r + q.
Thus, the donor reaction is modified as follows:

(15.30)
Adding Eqs. (15.28), (15.29), and (15.30), the overall aerobic reaction is produced:
(15.31)
From Equation (15.31), let Y
c
be the cell yield in moles per unit mole of sewage.
Then, Y
c
= ((q/20)/(r + q/50)) = 5q/2(r + q). And,
(15.32)
The reaction involving sewage is called a carbonaceous reaction, because it is the
reaction where organisms utilize the carbon of sewage for synthesis of the cells.
Thus, Y
c
is also called a carbonaceous cell yield.
NO
3
−
NH
4
+
NH
4
+
NO
3
−
q
5

CO
2
q
20

HCO
3
−
q
20

NH
4
+
qH
+
qe
−
q
20

C
5
H
7
NO
2
9q
20

H
2
O+→++++
r
4

O
2
rH
+
re
−
++
r
2

H
2
O→
rq+
50

C
10
H
19
NO
3
9 rq+()
25

H
2
O+
9 rq+()
50

CO
2
rq+
50

NH
4
+
+→
rq+
50

HCO
3
−
rq+()H
+
rq+()e
−
+++
rq+
50

C
10
H
19
NO
3
r
4

O
2
q
20

C
5
H
7
NO
2
9rq–
50

CO
2
+→+
2r 3q–
100

+ NH

4
+
2r 3q–
100

HCO
3
−
14r 9q+
100

H
2
O++
q
2rY
c
52Y
c
–

=
TX249_Frame_C15.fm Page 673 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
15.6.5 DENITRIFICATION: NORMAL ANOXIC STAGE
Let s, based on the synthesis reaction, be the equivalents of cells produced for the
regular anoxic denitrification reaction. The organisms in denitrification are het-
erotrophic, which must use sewage because, as mentioned, removal of nitrogen is
done in conjunction with the treatment of sewage. Again, the ammonium ion is
produced in the process, making it the source of nitrogen in the synthesis reaction.

Thus, modifying the ammonium synthesis reaction, using the s equivalents of cells,
produces the following reaction:
(15.33)
Denitrification is an anaerobic process; thus, it will not be using oxygen as the
electron acceptor. From Table 15.2, two possibilities exist for the electron acceptor:
nitrite or nitrate. In nitrate, the oxidation state of nitrogen is 5+; in nitrite, the oxidation
state of nitrogen is 3+. The nitrate ion is at the higher oxidation state, so it is easier
for it to be reduced than the nitrite ion. Thus, nitrate is the electron acceptor.
Let p, based on the acceptor reaction, be the equivalents of NO
3
–N utilized.
The revised acceptor reaction is
(15.34)
From Eqs. (15.33) and (15.34), the total e
−
needed from the donor reaction is
s + p electron moles. Thus, the donor reaction using sewage is
(15.35)
Adding Eqs. (15.33), (15.34), and (15.35), the overall reaction for denitrification
is produced,
(15.36)
From Equation (15.36), let Y
dn
be the cell yield in moles per unit moles of NO
3
–N.
Then,
(15.37)
s
5

CO
2
s
20

HCO
3
−
s
20

NH
4
+
sH
+
se
−
s
20

C
5
H
7
NO
2
9s
20

H
2
O+→++++
NO
3
−
p
5

NO
3
−
6 p
5

H
+
pe
−
++
p
10

N
2
3 p
5

H

2
O+→
sp+
50

C
10
H
19
NO
3
9 sp+()
25

H
2
O+
9 sp+()
50

CO
2
sp+
50

NH
4
+
+→

sp+
50

+ HCO
3
−
sp+()H
+
sp+()e
−
++
p
5

NO
3
−
p
5

H
+
sp+
50

C
10
H
19
NO

3
++
→
s
20

C
5
H
7
NO
2
p
10

N
2
2 p 3s–
100

NH
4
+
++
2 p 3s–
100

+ HCO
3
−

9 ps–
50

CO
2
24 p 9s+
100

H
2
O++
s 4 pY
dn
=
TX249_Frame_C15.fm Page 674 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
15.6.6 DENITRIFICATION: NO
2
-REDUCTION SIDE REACTION STAGE
Now, derive the overall reaction for the reduction of nitrite. can go only one
way: conversion to the nitrogen gas. Although right after cutting aeration, some
dissolved oxygen still remains in water, the concentration is not sufficient to oxidize
the nitrite to nitrate. As soon as the heterotrophic side reaction is complete, anaerobic
conditions set in and the environment becomes a reducing atmosphere. The
with nitrogen having an oxidation state of 3+ must then be reduced. The nitrogen
atom has three possible reduction products: reduction to nitrous oxide, N
2
O, reduction
to N
2

, and reduction to the ammonium ion.
The nitrogen in N
2
O has an oxidation state of 1+; that in N
2
has an oxidation
state of 0; and that in the ammonium ion has an oxidation state of 3−. Because the
environment is now severely reducing, the reduction process takes an extra step.
After reducing the nitrogen atom from 3+ in nitrite to 1+ in nitrous oxide, the process
continues one more step to the nitrogen molecule. In theory, it is still possible to
proceed with the reduction down to the formation of the ammonium ion. This,
however, will require the formation of bonds between the nitrogen atom and three
atoms of hydrogen (to form NH
3
), plus the hydrogen bond between the ammonia
molecule and the hydrogen proton (to form the ammonium ion, ). This needs
extra energy. Thus, the process stops at the liberation of the nitrogen gas. This whole
reduction process makes the nitrite ion the electron acceptor for it to be reduced.
Of course, the N
2
O could also be formed; but again, the atmosphere is severely
reduced and the nitrogen gas must be the one produced.
Let t, based on the acceptor reaction, be the equivalents of NO
2
–N utilized.
From Table 15.2, the revised acceptor reaction is
(15.38)
As in the normal anoxic process, sewage is used as the electron donor, because
it is the wastewater being treated. Again, as an electron donor, the ammonium ion
is produced serving as the nitrogen source for synthesis. Thus, let u, based on the

synthesis reaction, be the equivalents of cells formed. The synthesis reaction then
becomes
(15.39)
From Eqs. (15.38) and (15.39), the total moles of e
−
needed from the electron
donor is t + u. The modified donor reaction is
(15.40)
NO
2
−
NO
2
−
NH
4
+
NO
2
−
t
3

NO
2
−
4t
3

H

+
te
−
++
t
6

N
2
2t
3

H
2
O+→
u
5

CO
2
u
20

HCO
3
−
u
20

NH

4
+
uH
+
ue
−
++++
u
20

C
5
H
7
NO
2
9u
20

H
2
O+→
tu+
50

C
10
H
19
NO

3
9 tu+()
25

H
2
O
9 tu+()
50

CO
2
tu+
50

NH
4
+
+→+
tu+
50

+ HCO
3
−
tu+()H
+
tu+()e
−
++

TX249_Frame_C15.fm Page 675 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
Adding Eqs. (15.38), (15.39), and (15.40) produces the overall reaction for the
nitrite reduction,
(15.41)
Let Y
dc
be the cell yield in moles per unit moles of sewage. Then,
and,
(15.42)
Table 15.3 shows some values of Y
c
, Y
dn
, and Y
dc
. These values, however, may
only be used for very rough estimates. For an actual firm design, a laboratory or
pilot plant study for a given waste should be conducted.
TABLE 15.3
Values of Y
c
, Y
dn
, and Y
dc
Carbon Source Y
c
Y
dn

Y
dc
Domestic waste, mg VSS/mg BOD
5
0.40 — —
Domestic waste, mg VSS/mg COD 0.56 — —
Soft drink waste, mg VSS/mg COD 0.35 — —
Skim milk, mg VSS/mg BOD
5
0.38 — —
Pulp and paper, mg VSS/mg BOD
5
0.36 — —
Shrimp processing, mg VSS/mg BOD
5
0.35 — —
Methanol, mg VSS/mg NO
3
–N — 0.7–1.5 —
Domestic sludge, mg VSS/mg BOD
5
— — 0.040–0.100
Fatty acid, mg VSS/mg BOD
5
— — 0.040–0.070
Carbohydrate, mg VSS/mg BOD
5
— — 0.020–0.040
Protein, mg VSS/mg BOD
5

— — 0.050–0.090
Note: VSS = Volatile suspended solids
t
3

NO
2
−
t
3

H
+
tu+
50

C
10
H
19
NO
3
++
→
u
20

C
5
H

7
NO
2
2t 3u–
100

NH
4
+
+

2t 3u–
100

+ HCO
3
−
9tu–
50

CO
2
t
6

N
2
92t 27u+
300

H
2
O+++
Y
dc
u
20

tu+
50


5u
2 tu+()

==
u
2tY
dc
52Y
dc
–

=
TX249_Frame_C15.fm Page 676 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
15.7 TOTAL EFFLUENT NITROGEN
Although the idea behind denitrification is to remove nitrogen, due to the use of
sewage as a carbon source for synthesis in the heterotrophic side reaction, normal
anoxic denitrification, and nitrite reduction, some ammonia is produced. These

productions are indicated in Eqs. (15.31), (15.36), and (15.41). For convenience,
these reactions are reproduced next:
(15.43)
(15.44)
(15.45)
From Equation (15.43), the number of moles of ammonia nitrogen produced per
mole of oxygen is
The units of r and q are all in equivalents. In the derivation, although the unit of
equivalent was used, there is no restriction for using equivalents per liter instead.
Let r′ expressed in mmol/L units be the concentration corresponding to r, which is
now in meq/L. r′ is the concentration of dissolved oxygen left after the aeration is
cut off. For substitution into 2r − 3q/25r, these must be converted into units of
milligram equivalents per liter as follows:
rq+
50

C
10
H
19
NO
3
r
4

O
2
q
20

C
5
H
7
NO
2
9rq–
50

CO
2
+→+
+
2r 3q–
100

NH
4
+
2r 3q–
100

HCO
3
−
14r 9q+
100

H
2

O++
p
5

NO
3
−
p
5

H
+
sp+
50

C
10
H
19
NO
3
s
20

C
5
H
7
NO
2

p
10

N
2
2 p 3s–
100

NH
4
+
++→++
2 p 3s–
100

+ HCO
3
−
9 ps–
50

CO
2
24 p 9s+
100

H
2
O++
t

3

NO
2
−
t
3

H
+
tu+
50

C
10
H
19
NO
3
u
20

C
5
H
7
NO
2
2t 3u–
100

NH
4
+
+→++

2t 3u–
100

+ HCO
3
−
9tu–
50

CO
2
t
6

N
2
92t 27u+
300

H
2
O+++
2r 3q–
100

r
4


2r 3q–
25r

=
r
r′ 20()
r
4

O
2
r


4r′==
TX249_Frame_C15.fm Page 677 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
(r/4)O
2
/r is the equivalent mass of oxygen. This r may be substituted into q = 2rY
c
/(5 −
2Y
c
) such that q = 8r′Y

c
/(5 − 2Y
c
). Substituting these newfound values of r and q
into (2r − 3q)/25r, we have
(15.46)
2(1 − Y
c
)/5(5 − 2Y
c
) = (2r − 3q)/25r is the number of mmol/L of ammonia nitrogen
produced per mmol/L of dissolved oxygen. We have said the r′ is the concentration
of dissolved oxygen left after the aeration is cut off. This may be assumed as all
consumed, since the process is now going toward being completely anaerobic. There-
fore, the mmol/L of ammonia nitrogen produced in the heterotrophic side reaction is
(15.47)
A similar procedure is applied to find the ammonia nitrogen produced from the
normal anoxic denitrification. From Equation (15.44), the number of moles of
ammonia nitrogen produced per mole of nitrate nitrogen destroyed is
The units of p and s are all in equivalents. Again, in the derivation, although the
unit of equivalent was used, there is no restriction for using equivalents per liter
instead. Let p′ expressed in mmol/L units be the concentration corresponding to p,
which is now in meq/L. p′ is the concentration of nitrate nitrogen destroyed in the
normal anoxic denitrification stage. For substitution into 2p − 3s/20p, the conversion
into units of milligram equivalents per liter is as follows:
This p may be substituted into s = 4pY
dn
such that s = 20p′Y
dn
.

Substituting these newfound values of p and s into 2p − 3s/20p, we have
(15.48)
(1 − 6Y
dn
)/10 = (2p − 3s)/20p is the number of mmol/L of ammonia nitrogen
produced per mmol/L of nitrate nitrogen destroyed. The p′ mmol/L of nitrate nitrogen
2r 3q–
25r

8r′ 3
8r′Y
c
52Y
c
–

–
100r′

21 Y
c
–()
55 2Y
c
–()

==
milligram moles per liter of ammonia nitrogen produced
r′ milligram moles per liter of oxygen used

21 Y
c
–()
55 2Y
c
–()

r′=
2 p 3s–
100

p
5


2 p 3s–
20 p

=
p
p′ N()
p
5

N
P


5 p′==
2 p 3s–

20 p

16Y
dn
–
10

=
TX249_Frame_C15.fm Page 678 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
constitutes a fraction of the total concentration of nitrate nitrogen produced during
the nitrification process. Let this concentration be mmol/L and let f
p″
be the
fraction of destroyed. Then, p′ = f
p″
Therefore, the mmol/L of ammonia
nitrogen produced during the normal anoxic denitrification is
(15.49)
The equation to calculate the amount of ammonia nitrogen produced from the
nitrite-reduction side reaction may be derived in a similar manner. We will no longer
go through the steps but simply write the answer at once:
(15.50)
t′ is the mmol/L of nitrite nitrogen that corresponds to the t meq/L of nitrite
nitrogen destroyed or used during the nitrite-reduction stage of the denitrification
process. It is assumed that all of the nitrites that appeared after the aeration is shut
off are being converted to the nitrogen gas.
Adding Eqs. (15.47), (15.49), and (15.50) gives the concentration of NH
4
–Ν

in milligram moles per liter, [NH
4
–Ν]
mmol
, that will appear in the effluent of the
nitrification–denitrification process:
(15.51)
The total nitrogen concentration that will appear at the effluent of the nitrification–
denitrification process as a result of the consumption of the residual oxygen from
nitrification, destruction of nitrate, and destruction of nitrite in the denitrification step
is the sum of the ammonia nitrogen in the above equation plus the nitrate nitrogen
not destroyed in the denitrification step. Let [TN]
mmol
be the milligram moles per
liter of total nitrogen in the effluent. Thus,
(15.52)
Again, t′ = mmol/L of nitrite nitrogen appearing after the nitrification step,
assumed totally destroyed; = mmol/L of nitrate nitrogen appearing after the
nitrification step, with f
p″
fraction destroyed; and r′ = mmol/L of dissolved oxygen
remaining right after aeration is cut off, assumed totally consumed.
Note that the production of effluent nitrogen depends upon the values of the cell
yields. For Y
c
equal to or greater than 1, no effluent ammonia nitrogen is produced
from the heterotrophic side of the denitrification; for 6Y
dn
equal to or greater than
1, no effluent ammonia nitrogen is produced from the normal anoxic denitrification;

p″
p″ p″.
milligram moles per liter of ammonia nitrogen produced
p′milligram moles per liter of nitrate nitrogen destroyed

16Y
dn
–
10

p′
16Y
dn
–
10

f
p″
p″==
milligram moles per liter of ammonia nitrogen produced
t′milligram moles per liter of nitrite nitrogen destroyed

=
31 Y
dc
–()
10 5 2Y
dc
–()

t′
NH
4
–N[]
mmol
21 Y
c
–()
55 2Y
c
–()

r′
16Y
dn
–
10

f
p″
p″
31 Y
dc
–()
10 5 2Y
dc
–()

t′++=
TN[]

mmol
NH
4
–N[]
mmol
1 f
p″
–()p″+=
21 Y
c
–()
55 2Y
c
–()

r′
16Y
dn
–
10

f
p″
p″
31 Y
dc
–()
10 5 2Y
dc
–()

t′ 1 f
p″
–()p″+++=
p″
TX249_Frame_C15.fm Page 679 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
and for Y
dc
equal to or greater than 1, no effluent ammonia nitrogen is produced
from the nitrite-reduction side of the denitrification. These facts should be considered
in the calculation by setting the values equal to zero when the respective conditions
are met.
15.7.1 UNITS OF CELL YIELDS
Three types of cell yields are used in the previous derivations: Y
c
, Y
dn
, and Y
dc
. The units
of Y
c
and Y
dc
are in terms of moles of organisms per unit mole of sewage. The units of
Y
dn
are in terms of moles of organism per unit mole of nitrate nitrogen. The units
normally used in practice for Y

c
and Y
dc
, however, are either in terms of mass of organisms
or cells (approximated by the volatile suspended solids value, VSS) per unit mass
of BOD
5
or mass of organisms or cells per unit mass of COD. For the case of Y
dn
,
the units used in practice are in terms of mass of the organisms or cells per unit of
mass of the nitrate nitrogen. Unlike the conversion of Y
c
and Y
dc
from mass basis to
mole basis which is harder, the conversion of Y
dn
from mass basis to mole basis is
very straightforward; thus, we will address the conversion of the former.
Let the cells yielding Y
c
and Y
dc
be designated collectively as when
expressed in terms of mass cells per mass BOD
5
and let them be designated collec-
tively as Y
COD

when expressed in terms of mass cells per mass COD. To make the
conversion from or Y
COD
to Y
c
or Y
dc
, the electrons released by sewage as the
electron donor must be assumed to be all taken up by the oxygen electron acceptor.
This is because we want the conversion of one to the other—partial taking up of
the electrons does not make the conversion. The two half reactions are reproduced
below for convenience.
(15.53)
(15.54)
Judging from these two equations, the mole ratio of oxygen to sewage is (1/4)/(1/50) =
(25/2). Oxygen is the same as the ultimate carbonaceous biochemical oxygen demand,
CBOD. Let be the mole ratio of BOD
5
to CBOD, which would be the same as
the mole ratio of BOD
5
to oxygen. The mole ratio of BOD
5
to sewage is then (25/2).
Y
BOD
5
Y
BOD
5

1
50

C
10
H
19
NO
3
9
25

H
2
O+
9
50

CO
2
1
50

NH
4
+
1
50

HCO

3
−
H
+
e
−
++ ++→
1
4

O
2
H
+
e
−
++
1
2

H
2
O→
f
BOD
5
f
BOD
5
Y

BOD
5
mass of cells
mass of BOD
5

implies Y
BOD
5
mass of cells
C
5
H
7
NO
2

mass of BOD
5
f
BOD
5
32()


1
113

Y
BOD

5
mol cells
mol BOD
5

=

1
113

Y
BOD
5
mol cells
f
BOD
5
25
2



mol sewage

=
TX249_Frame_C15.fm Page 680 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
(15.55)
Although not correct exactly, for practical purposes, COD may considered equal

to CBOD or oxygen:
Therefore,
(15.56)
Example 15.1 A domestic wastewater with a flow of 20,000 m
3
/d is to be den-
itrified. The effluent from the nitrification tank contains 30 mg/L of NO
3
–N, 2.0 mg/L
of dissolved oxygen, and 0.5 mg/L of NO
2
–N. If the total nitrate nitrogen is to be
destroyed by 95%, calculate the total ammonia nitrogen produced from the reactions
and total nitrogen in the effluent.
Solution:
Referring to Table 15.3, assume:
and assume
Y
c
or Y
dc
1
113

Y
BOD
5
mol cells
f
BOD

5
25
2



mol sewage

=
7.08 10
−4
()
Y
BOD
5
f
BOD
5



mol cells
mol sewage

=
Y
COD
mass of cells
mass of COD

implies Y
COD
mass of cells
C
5
H
7
NO
2

mass of COD
32()


1
113

Y
COD
mol cells
mol COD

=
1
113

Y COD
mol cells
25
2



mol sewage

=
Y
c
or Y
dc
7.08 10
−4
()Y
COD
mol cells
mol sewage

=
NH
4
–N[]
mmol
21 Y
c
–()
55 2Y
c
–()

r′

16Y
dn
–
10

f
p″
p″
31 Y
dc
–()
10 5 2Y
dc
–()

t′++=
TN[]
mmol
21 Y
c
–()
55 2Y
c
–()

r′
16Y
dn
–
10

f
p″
p″
31 Y
dc
–()
10 5 2Y
dc
–()

t′ 1 f
p″
–()p″+++=
Y
c
0.35 mg VSS/mg BOD
5
7.08 10
−4
()
Y
BOD
5
f
BOD
5




mmol cells
mmol sewage

==
f
BOD
5
0.67.=
TX249_Frame_C15.fm Page 681 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
Therefore,
Example 15.2 A domestic wastewater with a flow of 20,000 m
3
/d is to be
denitrified. The effluent from the nitrification tank contains 30 mg/L of NO
3
–N,
2.0 mg/L of dissolved oxygen, and 0.5 mg/L of NO
2
–N. If the total nitrate nitrogen
in the effluent is limited by the state agency to 3.0 mg/L, calculate the percent destruc-
tion of the nitrate nitrogen and the ammonia nitrogen concentration in the effluent.
Solution:
Y
c
7.08 10
−4
()
0.35

0.67



mmol cells
mmol sewage

3.7 10
−4
()
mmol cells
mmol sewage

==
r′
2
32

0.0625 mmol/L==
Y
dn
0.9 mg VSS/mg NO
3
–N (from Table 15.3)=
Y
dn
0.9/113
1/14

0.112 f

p″
0.95== = p″ 30/14 2.14 mmol/L==
Y
dc
0.05 mg VSS/mg BOD
5
from Table 15.3()=
Y
dc
7.08 10
−4
()
0.05
0.67



5.28 10
−5
() t′
0.5
4

0.0357 mmol/L====
NH
4
–N[]
mmol
2 1 3.7 10
−4

()–[]
55 23.7()10
−4
()–[]

0.0625()
1 6 0.112()–
10

0.95()2.14()+=
3 1 5.28 10
−5
()–[]
10 5 2 5.28()10
−5
()–[]

0.037()+
5.0 10
−3
()0.067 2.22 10
−3
()++ 0.074 mmol/L ⇒ 1.036 mg/L Ans==
TN[]
mmol
0.074 1 0.95–()2.14()+ 0.181 mmol/L ⇒ 2.53 mg/L Ans==
TN[]
mmol
21 Y
c

–()
55 2Y
c
–()

r′
16Y
dn
–
10

f
p″
p″
31 Y
dc
–()
10 5 2Y
dc
–()

t′ 1 f
p″
–()p″+++=
5.0 10
−3
()
1 6 0.112()–
10

f
p″
2.14()2.22 10
−3
()1 f
p″
–()2.14()+++=
TN[]
mmol
3/14 0.214 mmol/L==
0.214 5.0 10
−3
()
1 6 0.112()–
10

f
p″
2.14()2.22 10
−3
()1 f
p″
–()2.14()+++=
5.0 10
−3
()0.070 f
p″
2.22 10
−3
()2.14 2.14 f

p″
–++ +=
2.07 f
p″
1.93 f
p″
0.93 Ans==
TX249_Frame_C15.fm Page 682 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero
15.8 CARBON REQUIREMENTS FOR DENITRIFICATION
As shown in the derivations, carbon is necessary during denitrification. Because the
organisms responsible for this process are heterotrophs, sewage was provided as the
source. The pertinent reactions for the heterotrophic side reaction, normal anoxic
denitrification reaction, and the nitrite-reduction side reaction are given, respectively,
in Eqs. (15.43), (15.44), and (15.45).
From Equation (15.43), the number of moles of sewage needed per mole of
oxygen is
Substituting r = 4r′ and q = 8r′Y
c
/(5 − 2Y
c
), we have
(15.57)
2/5(5 − 2Y
c
) = 2(r + q)/25r is the number of mmol/L of sewage needed per
mmol/L of dissolved oxygen. Therefore, the total mmol/L of sewage needed during
the heterotrophic side reaction is
(15.58)
From Equation (15.44), the number of moles of sewage needed per mole of

nitrate nitrogen is
Substituting p = 5p′ and s = 20p′ Y
dn
, we have
(15.59)
NH
4
–N[]
mmol
21 Y
c
–()
55 2Y
c
–()

r′
16Y
dn
–
10

f
p″
p″
31 Y
dc
–()
10 5 2Y
dc

–()

t′++=
5.0 10
−3
()
1 6 0.112()–
10

0.93()2.14()2.22 10
−3
()++=
0.072 mmol/L ⇒1.008 mg/L Ans=
rq+
50

r
4


2 rq+()
25r

=
2 rq+()
25r

24r′
8r′Y
c

52Y
c
–

+()
25 4r′()

2
55 2Y
c
–()

==
milligram moles per liter of sewage needed
r′milligram moles per liter of oxygen used

2
55 2Y
c
–()

r′=
sp+
50

p
5


sp+

10 p

=
sp+
10 p

20 p′Y
dn
5 p′+
10 5 p′()

20Y
dn
5+
50

4Y
dn
1+
10

===
TX249_Frame_C15.fm Page 683 Friday, June 14, 2002 4:43 PM
© 2003 by A. P. Sincero and G. A. Sincero