Standard Methods for Examination of Water & Wastewater_14 docx

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 gase

be effected by purely enzymatic means by providing the needed enzymes externallywithout ever using microorganisms

Similar to phosphorus, nitrogen is a very important element that has attractedmuch attention because of its ability to cause eutrophication in bodies of water Asstated in the chapter on phosphorus removal, the Chesapeake Bay in Maryland andVirginia is fed by tributaries from farmlands as far away as New York Because ofthe use of nitrogen in fertilizers for these farms, the bay receives an extraordinarilylarge amount of nitrogen input that has triggered excessive growths of algae in thewater body Presently, large portions of the bay are eutrophied

This chapter discusses removal of nitrogen using the unit process of nitrificationfollowed by denitrification Half reactions are utilized in the discussion of thechemical 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 arenormally tabulated in terms of free energies To understand the exact meaning offree energy as it relates to half reactions and thus to nitrogen removal, microbialthermodynamics 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]2s22p3 [He] means thatthe helium configuration is filled The valence configuration represented by the 2,

15

the L shell, shows five electrons in the orbitals: 2 electrons in the s orbitals and 3electrons in the p orbitals This means that, like phosphorus, nitrogen can have amaximum oxidation state of +5; its smallest oxidation state is 3− Examples arenitrous oxide (N2O, 1+); nitric oxide (NO, 2+); dinitrogen trioxide (N2O3, 3+);nitrogen dioxide (NO2, 4+); dinitrogen tetroxide (N2O4, 4+); and dinitrogen pentox-ide (N2O5, 5+) Our interest in nitrogen as it occurs in nature is in the form thatmakes it fertilizer to plants These forms are the nitrites, nitrates, and ammonia Thenitrogen 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, wewill, 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 (CH2O)106(NH3)16H3PO3 (Sincero and Sincero, 1996) Gleaningfrom this formula, to curtail its production in any water body such as the ChesapeakeBay, 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 bodyprevents 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, becausethere 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 showsvarious values of nitrogen and phosphorus concentrations in the water column andthe 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 leftalone 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

on the eutrophication potential of the water body should be investigated accuratelyand 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 namics may be used to evaluate life processes An example of a life process is thegrowth of bacteria when wastewater is fed to them to treat the waste Knowledge

thermody-of microbial thermodynamics is therefore important to prthermody-ofessionals involved incleaning 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 ornot 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 kineticenergies of the molecules, ionization energies of the electrons, bond energies, latticeenergies, etc The system possesses all these energies by virtue of its being and areall integral (that is, internal) with the system

Let us derive the relationship between enthalpy and the heat exchange during abiological reaction, where biological reaction is a chemical reaction mediated byorganisms Biological reactions are carried out at constant pressure; hence, the heatexchange 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 anywork W that the system is doing at the same time manifests itself in the form of an

V–

H = U+PV–

Because the biological reaction is at constant pressure, differentiate the enthalpyequation 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 constantpressure 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 systemcan be recovered back if the process is done in the opposite direction from that inwhich it is currently being done To perform a reversible process, the steps must beconducted very, very slowly, in an infinitesimal manner, and without friction Fromthe 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 steparound 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, theratio of the infinitesimal heat added to the temperature T during the infinitesimal timethat the heat is added defines the change in entropy If this addition is done around areversible cycle, the state or condition of the system at the end of the cycle will revertback 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 aroundthe 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 differentialstep, the heat added is dQ rev The differential change in entropy in every differentialstep 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 andsubscripts e and b refer to the end and the beginning of the cycle, respectively Ifthe process is not around a cycle, the previous subscripts simply mean the end and

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

S b

G = H–TS

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 asconstant 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 thechloroplast 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 aselectron movement is concerned, life processes have similarity to a battery cell Inthis cell, electrons move because of electrical pressure, the voltage difference By thesame 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 isreduced; that is, oxidation–reduction occurs in a battery cell Exactly the same processoccurs 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/Zn2+

has an electric pressure between them Now, take another couplesuch as Mg/Mg2+

whose half-cell reaction would be similar to that of Equation (15.10).The couples Zn/Zn2+

and Mg/Mg2+

do not possess the same voltage potential If thetwo couples are connected together, they form a cell Their voltage potentials arenot the same, so a voltage difference would be developed between their electrodesand electric current would flow

Let the voltage between the electrodes of the above cell be measured by a tiometer Designate this voltage difference by ∆E In potentiometric measurements,

poten-no electrons are allowed to flow, but only the voltage tendency of the electrons toflow is measured No electrons are allowed to flow, therefore, no energy is dissipateddue to friction of electrons “rubbing along the wire.” Thus, any energy associated withthis no-electron-to-flow process represents the maximum energy available Becausevoltage is energy in joules per coulomb of charge, the energy associated can be

dG = dH–TdS = dH–Q rev

Zn→Zn2++2e−

calculated from the voltage difference This associated energy corresponds to a friction loss process; it is therefore a maximum energy—the change in free energy—after the entropic loss has been deducted.

no-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 measurementwas done with no energy loss, by definition, this associated energy represents thefree energy change of the cell (i.e., the maximum energy change in the cell) Insymbols,

(15.11)

The sign has been used A convention used in chemistry is that if the sign isnegative, the process is spontaneous and if the sign is positive, the process is notspontaneous

As mentioned, the battery cell process is analogous to the living cell process ofthe mitochondrion, the chloroplast, and the electron-transport system in the cyto-plasmic membrane of the procaryotes Thus, Equation (15.11) can represent the basicthermodynamics of a microbial system

In the living cell, organic materials are utilized for both energy (oxidation) andsynthesis (reduction) Microorganisms that utilize organic materials for energy are

called heterotrophs Those that utilize inorganics for energy are called autotrophs.

Autotrophs utilize CO2 and for the carbon needed for cell synthesis; the erotrophs utilize organic materials for their carbon source Autotrophs that use inorganic

het-chemicals for energy are called chemotrophs; those that use sunlight are phototrophs.

The bacteria that consumes the nitrogen species in the biological removal of nitrogenare chemotrophic autotrophs Algae are phototrophic autotrophs

Somehow, in life processes, the production of energy from release of electronsdoes not occur automatically but through a series of steps that produce a high energy-containing compound This high energy-containing compound is ATP (adenosinetriphosphate) 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 thatthe cell relies upon for energy supply

The energy function of ATP is explained as follows: ATP contains two energy bonds To form these bonds, energy must be obtained from an energy sourcethrough electron transfer The energy released is captured and stored in these bonds

high-On demand, hydrolysis of the bond releases the stored energy which the cell canthen use for synthesis and cell maintenance

ATP is produced from ADP (adenosine diphosphate) by coupling the release ofelectrons to the reaction of organic phosphates and ADP producing ATP ATP has

two modes of production: substrate-level phosphorylation and oxidative rylation In the former, the electrons released by the energy source are absorbed by

phospho-an intermediate product within the system The electron absorption is accompphospho-anied

by an energy release and ATP is formed The electron-transport system is simple

∆G = ±nF ∆E

±

HCO3−

Fermentation is an example of a substrate-level phosphorylation process that usesintermediate absorbers such as formaldehyde Substrate-level phosphorylation isinefficient and produces only a few molecules of ATP.

In the oxidative phosphorylation mode, the electron moves from one electroncarrier to another in a series of complex reduction and oxidation steps The differencebetween 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 fromthe molecule of the source The hydrogen carries with it the electron it shared withthe original source molecule, moving this electron through a series of intermediatecarriers such as NAD (nicotinamide adenine dinucleotide) The intermediate thatreceives the electron-carrying hydrogen becomes reduced The reduction of NAD,for example, produces NADH2 The series continues on with further reduction andoxidation steps The whole line of reduction and oxidation constitutes the electron-transport system At strategic points of the transport system, ATP is produced fromADP and inorganic phosphates

The other version of oxidative phosphorylation used by autotrophs involves therelease of electrons from an inorganic energy source An example of this is therelease of electrons from , oxidizing to , and the release of electronsfrom , oxidizing to

The transported electrons emerge from the system to reduce a final externalelectron acceptor The type of the final acceptor depends upon the environment onwhich the electron transport is transpiring and may be one of the following: foraerobic environments, the acceptor is O2; for anaerobic environments, the possibleacceptors are , , and CO2 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, thehydrogen 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 asthe electron donor and the substance reduced as the electron acceptor The electrondonor is normally considered as food In the context of nitrogen removal, the foodsare the nitrites, nitrates, and ammonia Equation (15.10) is an example of an electrondonor reaction Zn is the donor of the electrons portrayed on the right-hand side ofthe 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 donorsand acceptors utilized in a bacterial systems The ones specific for the nitrogenspecies removal are shown in Table 15.2

∆G′ o = ±nF ∆E o′

15.4.1 C RITERION FOR S PONTANEOUS P ROCESS

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 temperatureequals absolute zero, is always disordered.* The energy required to maintain thisdisorder, we have found, is called entropy As mentioned, any system possesses freeenergy at any instant, this energy being the net energy remaining after the entropyrequired to maintain the current disorder has been subtracted from the enthalpy (energycontent)

When the system goes from state 1 (current state) to state 2, its free energy atthe 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:

20 - NH4+ H+ e−

+ + + + 20 - C1 5H 7 NO 2

9 20 - H2O +

+ + + 28 - C1 5H 7 NO 2

11 28 - H2O +

+ + 10 - N1 2 3

5 - H2O +

+ + 16 - N2 2

3 - H2O +

1 50 - NH4+ 150 - HCO3− H+ e−

1

6

- NH4+ 1

3 - H2O + 16 - NO2− 4

3 - H+ e−

→

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 beenadded 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 mol and 7.85 mol, respectively, has been added to the system; these values are indicated by theplus signs External sources of energy are being required, so these half-cell reactions

kcal/electron-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, someenergy 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 tablewith a negative sign This energy has been released “voluntarily” by the systemwithout 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 electrontransport system are finally accepted by oxygen, a large amount of energy equal to18.68 kcal/electron-mol is liberated This liberated energy is then captured in thebonds of ATP The same statement holds for the others whose free energy changeshave negative signs Thus, any material in wastewater, edible by organisms, willrelease 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 sed in a previous chapter The present chapter concerns only the removal of nitrogen

discus-by biochemical means, as mediated discus-by microorganisms The technique of the unitprocess is to release the nitrogen in the form of the gas N2 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, convertthe 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, N2 The

oxidation steps are mediated by Nitrosomonas and by Nitrobacter, as mentioned

previously The reduction step is mediated by the normal heterotrophic bacteria Wewill now discuss the chemical reactions involved in these oxidations and reduction.

15.6.1 N ITRIFICATION: N ITROSOMONAS S TAGE

From Table 15.2, the generalized donor reaction, acceptor reaction, and synthesis

reaction mediated by Nitrosomonas are, respectively, shown by the following

If the previous equations are simply added without modifications, electrons willremain free to roam in solution; this is not possible In actual reactions, the previousequations must be modified to make the electrons given up by Equation (15.13) bebalanced by the electrons accepted by Equations (15.14) and (15.15) Starting withone gram-equivalent of NH4–N, based on Equation (15.15), assume m equivalents are incorporated into the cell of Nitrosomonas, C5H7NO2 Thus, the NH4–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 NH4–N.] Thus, (1 − m) moles of NH4–N isavailable for the donor reaction to donate electrons By Equation (15.13), the mod-ified donor reaction is then

6 - NO2− 4

3 - H+ e−

→+

1

4

-O2 H+ e− 1

2 - H2O

20 - NH4+ H+ e−

→ 20 - C1 5H7NO2 9

20 - H2O+

 

  1 m( – )H2O+

→ 11 -/6/6 1

20 -

 

  1 m( – )H+ 1 -1/6 1

20 -

 

  1 m( – )e−

From Equation (15.15), the m equivalents of NH4–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− aregone 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 fromTable 15.2:

 

  HCO3− 1/20

1/20

- m20 -

 

1/20

- m20 -

 

  H++ 1

1/20

- m20 -

 

  H2O+

1

1/6

- 120 -

 

  1 m( – ) 1

1/20

- m20 -

 

 

10 -

=

14 - 3–13m10 -

10 -H+ 3–13m

10

-e−

10 -

20 -HCO3− m

5 CO2

m

20 -C5H7NO2+ 20 - 11( –m)NO2

− 1–2m

20 - H2O+

1

4

-O2+H++e− 1

2 -H2O

20 -NH4+ H+ e−

20 -C5H7NO2 9

20 -H2O+

In the BNR process, the Nitrobacter reaction follows the Nitrosomonas reaction.

From Equation (15.19), 1/20(1 − m) moles of NO2–N have been produced from theoriginal 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)

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

1/2

1/2

- 120 - 1( –m) NO3

− 1

1/2 -

20 - 1( –m) H+ 1

1/2

- 120 - 1( –m ) e−

+

→

n

C 5 H 7 O 2 N 20 -

 

  HCO3− 1/20

1/20

- n20 -

 

1/20

- n20 -

 

1/20

- n20 -

 

  C5H7O2N 9/20

1/20

- n20 -

 

  H2O+

10

- e− 1

2 - 1–m–10n10 - H2O

5 -CO2 n

20 -HCO3− n

20 -H2O 1–m–10n

40 - O2

→20 -Cn 5H7O2N 1–m

20 -NO3−+

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 NH4–N destroyed and cell yields for

Nitrobacter of 0.02 to 0.084 milligrams of the bacteria per milligrams of NO2–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 BOD5/TKN ratios should be less than 3 At theseratios, the nitrifier population is about 10% and higher For BOD5/TKN ratios ofgreater than 5.0, the process may be considered combined carbonaceous-nitrificationreaction At these ratios, the nitrifier population is less than 4% In addition, to ensurecomplete nitrification, the dissolved oxygen concentration should be, at least, about2.0 mg/L

15.6.4 D ENITRIFICATION : H ETEROTROPHIC S IDE R EACTION S TAGE

Aside from the normal anoxic reaction, two side reactions must be considered indenitrification: the continued oxidation using the leftover dissolved oxygen from thenitrification reaction stage, and nitrite reduction Immediately after nitrification isstopped, a large concentration of dissolved oxygen still exists in the reactor—thiswould 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 balancedduring nitrification and the two types of bacteria grow together As soon as the oxygensupply is cut off, however, the nitrifiers cannot compete against the heterotrophs forthe ever decreasing concentration of dissolved oxygen Thus, the activities of thenitrifiers “fade away,” and the heterotrophs predominate in the last stage of oxidationafter aeration is cut off

The other side reaction is the reduction of nitrite to the nitrogen gas Althoughthe 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 theremaining 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 O2 (based on the oxygen acceptor reaction) used at this stage The removal

1+n

20

-NH4+ m+n

20 -HCO3− m+n

5 -CO2 2–7m–5n

20 -O2

m+n

20 -C5H7NO2 1–m

10 -H+ 1–m

20 -NO3− 1–2m–n

20 -H2O

→

of nitrogen is done in conjunction with the treatment of sewage Thus, sewage(C10H19NO3) must be the electron donor Using sewage, the ammonium ion isproduced, see Table 15.2 As mentioned before, given and in solution,organisms prefer to use first, before for synthesis Thus, of the twocompeting 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

From these equations, the total electron moles needed from the donor is r + q.

Thus, the donor reaction is modified as follows:

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.

20 -NH4+ qH+ qe− q

20 -C5H7NO2 9q

20 -H2O+

→

r

4 -O2+rH++re− r

2 -H2O

→

r+q

50 -C10H19NO3 9 r( +q)

25 -H2O

50 -CO2 r+q

50 -NH4++

→

r+q

50 -HCO3− (r+q)H+

4 -O2 q20 -C5H7NO2 9r–q

50 -CO2+

→+

2r–3q

100 -

100 -HCO3− 14r+9q

100 -H2O

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 erotrophic, which must use sewage because, as mentioned, removal of nitrogen isdone in conjunction with the treatment of sewage Again, the ammonium ion isproduced in the process, making it the source of nitrogen in the synthesis reaction

het-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 theelectron 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 oxidationstate of nitrogen is 3+ The nitrate ion is at the higher oxidation state, so it is easierfor 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 NO3–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

20 -NH4+ sH+ se− s

20 -C5H7NO2 9s

20 -H2O+

5 -H+ pe−

5 -H2O+

50 -CO2 s+p

50 -NH4++

→

s+p

50 -+ HCO3− (s+p)H+

5 -H+ s+p50 -C10H19NO3

→ 20 -Cs 5H7NO2 p

10 -N2 2 p–3s

100 -NH4+

2 p–3s

100 -

− 9 p–s

50 -CO2 24 p+9s

100 -H2O

s = 4 pY dn

15.6.6 DENITRIFICATION: NO 2 -REDUCTION SIDE REACTION STAGE

Now, derive the overall reaction for the reduction of nitrite can go only oneway: conversion to the nitrogen gas Although right after cutting aeration, somedissolved oxygen still remains in water, the concentration is not sufficient to oxidizethe nitrite to nitrate As soon as the heterotrophic side reaction is complete, anaerobicconditions set in and the environment becomes a reducing atmosphere The with nitrogen having an oxidation state of 3+ must then be reduced The nitrogenatom has three possible reduction products: reduction to nitrous oxide, N2O, reduction

to N2, and reduction to the ammonium ion

The nitrogen in N2O has an oxidation state of 1+; that in N2 has an oxidationstate of 0; and that in the ammonium ion has an oxidation state of 3− Because theenvironment 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 processcontinues one more step to the nitrogen molecule In theory, it is still possible toproceed with the reduction down to the formation of the ammonium ion This,however, will require the formation of bonds between the nitrogen atom and threeatoms of hydrogen (to form NH3), plus the hydrogen bond between the ammoniamolecule and the hydrogen proton (to form the ammonium ion, ) This needsextra energy Thus, the process stops at the liberation of the nitrogen gas This wholereduction process makes the nitrite ion the electron acceptor for it to be reduced

Of course, the N2O could also be formed; but again, the atmosphere is severelyreduced and the nitrogen gas must be the one produced

Let t, based on the acceptor reaction, be the equivalents of NO2–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 thenbecomes

(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

3 -H+ te−

3 -H2O+

20 -NH4+ uH+ ue−

20 -C5H7NO2 9u

20 -H2O+

50 -CO2 t+u

50 -NH4++

→+

t+u

50 -+ HCO3− (t+u)H+

t+u

Adding Eqs (15.38), (15.39), and (15.40) produces the overall reaction for thenitrite reduction,

TABLE 15.3

Values of Y c, Ydn, and Ydc

Domestic waste, mg VSS/mg BOD5 0.40 — —

Soft drink waste, mg VSS/mg COD 0.35 — —

Pulp and paper, mg VSS/mg BOD5 0.36 — —

Shrimp processing, mg VSS/mg BOD 5 0.35 — —

Methanol, mg VSS/mg NO3–N — 0.7–1.5 —

Domestic sludge, mg VSS/mg BOD 5 — — 0.040–0.100

Fatty acid, mg VSS/mg BOD5 — — 0.040–0.070

→ 20 -Cu 5H7NO2 2t–3u

100 -NH4++

2t–3u

100 -

50 -CO2 t

6 -N2 92t+27u

300 -H2O

- 5u

2 t( +u) -

15.7 TOTAL EFFLUENT NITROGEN

Although the idea behind denitrification is to remove nitrogen, due to the use ofsewage as a carbon source for synthesis in the heterotrophic side reaction, normalanoxic denitrification, and nitrite reduction, some ammonia is produced Theseproductions are indicated in Eqs (15.31), (15.36), and (15.41) For convenience,these reactions are reproduced next:

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:

r+q

50 -C10H19NO3 r

4 -O2 q20 -C5H7NO2 9r–q

50 -CO2+

→++ 2r–3q100 -NH4+ 2r–3q

100 -HCO3− 14r+9q

100 -H2O

20 -C5H7NO2 p

10 -N2 2 p–3s

100 -NH4+

50 -CO2 24 p+9s

100 -H2O

20 -C5H7NO2 2t–3u

100 -NH4++

→

2t–3u

100 -

50 -CO2 t

6 -N2 92t+27u

300 -H2O

2r–3q

100 -

r

- 4r′

(r/4)O2/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

(15.47)

A similar procedure is applied to find the ammonia nitrogen produced from thenormal anoxic denitrification From Equation (15.44), the number of moles ofammonia 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

milligram moles per liter of ammonia nitrogen produced

r′ milligram moles per liter of oxygen used

p

5 -

=

constitutes a fraction of the total concentration of nitrate nitrogen produced duringthe nitrification process Let this concentration be mmol/L and let f p″ be thefraction of destroyed Then, p ′ = f p″ Therefore, the mmol/L of ammonianitrogen produced during the normal anoxic denitrification is

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 denitrificationprocess It is assumed that all of the nitrites that appeared after the aeration is shutoff are being converted to the nitrogen gas

Adding Eqs (15.47), (15.49), and (15.50) gives the concentration of NH4–Ν

in milligram moles per liter, [NH4–Ν]mmol, that will appear in the effluent of thenitrification–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 fromnitrification, 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 perliter of total nitrogen in the effluent Thus,

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″

milligram moles per liter of ammonia nitrogen produced

p′milligram moles per liter of nitrate nitrogen destroyed -

milligram moles per liter of ammonia nitrogen produced

t′milligram moles per liter of nitrite nitrogen destroyed

and for Y dc equal to or greater than 1, no effluent ammonia nitrogen is producedfrom 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 conditionsare 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 BOD5 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 isvery straightforward; thus, we will address the conversion of the former

Let the cells yielding Y c and Y dc be designated collectively as whenexpressed in terms of mass cells per mass BOD5 and let them be designated collec-

tively as YCOD when expressed in terms of mass cells per mass COD To make theconversion from or YCOD to Y c or Y dc, the electrons released by sewage as theelectron 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 ofthe electrons does not make the conversion The two half reactions are reproducedbelow 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 BOD5 to CBOD, which would be the same asthe mole ratio of BOD5 to oxygen The mole ratio of BOD5 to sewage is then (25/2)

50 -NH4+ 1

50 -HCO3− H+ e−

→

14 -O2+H++e− 1

2 -H2O

- implies YBOD

5

mass of cells

C5H7NO2 - mass of BOD 5

fBOD

5 32 ( )

fBOD5 25

2 -

 

 mol sewage

-=

5mol cells

fBOD5 25

2 -

- 1

113

-YCODmol cells

mol COD -

=

1113

-Y COD mol cells

25 2 -

fBOD5 = 0.67

Therefore,

Example 15.2 A domestic wastewater with a flow of 20,000 m3/d is to bedenitrified The effluent from the nitrification tank contains 30 mg/L of NO3–N,2.0 mg/L of dissolved oxygen, and 0.5 mg/L of NO2–N If the total nitrate nitrogen

in the effluent is limited by the state agency to 3.0 mg/L, calculate the percent tion of the nitrate nitrogen and the ammonia nitrogen concentration in the effluent

destruc-Solution:

Y c 7.08 10( −4) 0.35

0.67 -

mmol sewage - 3.7 10( −4) mmol cells

mmol sewage -

Y dc = 0.05 mg VSS/mg BOD5 from Table 15.3( )

Y dc 7.08 10( −4) 0.05

0.67 -

4 - 0.0357 mmol/L

TN

[ ]mmol = 3/14 = 0.214 mmol/L

0.214 = 5.0 10( −3)+1 - f–6 0.112(10 ) 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

15.8 CARBON REQUIREMENTS FOR DENITRIFICATION

As shown in the derivations, carbon is necessary during denitrification Because theorganisms responsible for this process are heterotrophs, sewage was provided as thesource The pertinent reactions for the heterotrophic side reaction, normal anoxicdenitrification reaction, and the nitrite-reduction side reaction are given, respectively,

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 duringthe heterotrophic side reaction is

r+q

50 -

milligram moles per liter of sewage needed

r′milligram moles per liter of oxygen used

p

5 -

10 -

From this equation, the total mmol/L of sewage needed during the normal anoxicdenitrification is

(15.60)

The relationship of t ′ (in mmol/L) and t was obtained as follows:

This t may be substituted into

milligram moles per liter of sewage needed

p′milligram moles per liter of nitrate nitrogen destroyed

milligram moles per liter of sewage needed

t′milligram moles per liter of nitrite nitrogen destroyed

- 3

10 5( –2Y dc)

-t′

=

Adding Eqs (15.58), (15.60), and (15.62) gives the overall milligram moles perliter of sewage, [SEWdenit]mmol, needed in the denitrification process:

(15.63)

[SEWdenit]mmol may be converted into units in terms of BOD5 in mg/L From Eqs.(15.53) and (15.54), the ratio of milligrams BOD5 to millimole of sewage is ((1/4)O2)/(1/50) = Therefore, the corresponding concentration of[SEWdenit]mmol in terms of BOD5 in mg/L, [BOD5denit]mg, is

(15.64)

Example 15.3 After cutting off the aeration to a nitrification plant, the solved oxygen concentration is 2.0 mg/L How much sewage is needed for thecarbonaceous reaction in this last stage of aerobic reaction before denitrification setsin? How much NH4–N is produced from this carbonaceous reaction?

dis-Solution:

Therefore,

Example 15.4 A domestic wastewater with a flow of 20,000 m3/d is to bedenitrified The effluent from the nitrification tank contains 30 mg / L of NO3–N,2.0 mg/L of dissolved oxygen, and 0.5 mg/L of NO2–N Assume that an effluent total

nitrogen not to exceed 3.0 mg/L is required by the permitting agency (a) Calculate

the quantity of domestic sewage to be provided to satisfy the carbon requirement in

mmoles per liter (b) What is the corresponding BOD5 to be provided?

10 5( –2Y dc) -

10 5( –2Y dc) -

=

milligram moles per liter of sewage needed

r′milligram moles per liter of oxygen used

- 2

5 5( –2Y c)

-r′

=

r′ = 32 -2 = 0.0625 mmol/L Y c = 3.7 10( −4)mmol sewage -mmol cells

milligram moles per liter of sewage needed

r′milligram moles per liter of oxygen used

- 2

5 5[ –2 3.7( ) 10( −4)] - 0.0625( )

=

milligram moles per liter of ammonia nitrogen produced

r′milligram moles per liter of oxygen used

3 0.0357( )

10 5[ –2 5.28 10( ( −5))] -

10 5( –2Y dr) -

1+n

20

- NH4+ m+n

20 - HCO3− m+n

5 - CO2 2–7m–5n

20 -O2

m+n

20 -

1–m

10 - H+ 1–m

20 - NO3 1–2m–n

20 - H2O

On the other hand, inspection of Table 15.2 for the half reaction of sewage as anelectron donor reproduced as Equation (15.66) shows that bicarbonate alkalinity isproduced from sewage.

of Nitrobacter per unit mole of nitrite nitrogen is

from which n = Y n(1 − Y m) Substituting these new values into , we have

Therefore, the mmol/L of alkalinity required, [A nit]mmol, for the mmol/L of nitratenitrogen produced in the nitrification stage is

(15.67) Equation (15.31) may be used to derive the production of alkalinity from sewage

To signify that the equation pertains to the regular heterotrophic secondary treatment

of sewage as opposed to the heterotrophic side reaction, change r and q to r s and q s,

respectively, where s stands for secondary in the regular heterotrophic secondary

treatment reaction The overall manipulations are indicated next Note that all the terms

of the equation are divided by the coefficient of and then multiplied by 1/20

50 -CO2 1

50 -NH4+ 1

50 -HCO3− H+ e−

→+

m+n

20 -

1–m

20 -

 

  C10H19NO3

r s

4

2r s–3q s

100 -

- 1

20 -

 

  O2+

q s

20

2r s–3q s

100 -

- 1

20 -

2r s–3q s

100 -

- 1

20 -

 

20 -

 

  NH4+ 1

20 - HCO3−

14r s+9q s

100 -

2r s–3q s

100 -

- 1

20 -

 

  H2O