ch. 13 anaerobic treatment by...

Chapter

13

ANAEROBIC TREATMENT

BY METHANOGENESIS

13.1 Uses for Methanogenic Treatment

13.2 Reactor Configuration

13.3 Process Chemistry and Microbiology

13.4 Process Kinetics

13.5 Special Factors for the Design of Anaerobic Sludge Digesters

Ch. 13 Anaerobic Treatment by Methanogenesis

Methanogenesis :

Organic matter (BODL) Reduced oxidation state (CH4)

anaerobic

At STP 1 g of BODL generates 0.35 l of CH4

Methanogenic process quite complex always contains methanogens

Acetate Fermenters Hydrogen Oxidizers

Electron Donor Acetate H2 and formate

Electron Acceptors Acetate CO2

Carbon Sources Acetate CO2

Fs0 0.05 0.08

Y 0.04 g VSSa/g Ac 0.45 g VSSa/g H2

q^ (at 350C) 7 g Ac/g VSSa-d 3 g H2/g VSSa-d

K 400 mg Ac/l ?

B 0.03/d 0.03/d

[xmin]lim 3.6 d 0.76 d

Smin 48 mg Ac/l ?

Slow growing


Anaerobic treatment municipal wastewater and solid waste

High strength industrial wastewater << - lack of experience with the process

- lack of process understanding

- presence of toxic compound

- concern over the need for high process

reliability in a strong regulatory

Anaerobic treatment popular in developing world & warm climate

More complex, need high skill & solid understanding

+

Enough nutrient, low pH problem and presence of toxicant

reliable

treatment

process


Advantages and disadvantages of anaerobic treatment

Advantages 1. Low production of waste biological solids

2. Low nutrient requirements

3. Methane is useful end products

4. Generally, a net energy producer

5. High organic loading is possible

Disadvantages 1. Low growth rate of microorganisms

2. Odor production

3. High buffer requirement for pH control

4. Poor removal efficiency with dilute wastes

Example 13.1 Energy Value of Methane

Flow (Q) = 104 m3/d, BODL = 20,000 mg/l

Task : 1. Compute the volume of generated CH4 if waste stabilization is 90%

2. Compute energy value of that methane

BODL loading rate = Q x BODL = 2.105 kg BODL/d

BODL stabilization rate = waste stabilization x BODL loading rate =

1.8 . 105 kg BODL/d

Methane generation rate = BODL stabilization rate x 0.35 m3 CH4 (STP)/kg BODL

= 6.3 . 104 m3 CH4 (STP)/d

Energy rate = Methane generation rate x 35,800 kJ/m3 CH4 (STP)

= 2.26 . 109 kJ/d

13.2 Reactor Configurations

13.2.1 Completely Mixed

Operational T = 350C

Td = 15 – 25 days

Earlier design : no mixing

Problems :

1. Fresh sludge & fermenting

microbes are not brought

together efficiently

2. Denser solids, tend to settle

in the reactor

Solutions : operated at detention times, based on the ratio

of tank volume to influent sludge flow rate

≥ 60 days

13.2.2 Anaerobic Contact

Anaerobic contact process

is analog to aerobic activated

sludge system

First : COD = 1,300 mg/l

Hydr. Td = 0.5 d

BOD removal = 91-95 %

at loading rate 2-2.5 kg/m3-d

PROBLEMS : - biosolids in the settling tank tends to rise due to bubble generation and

attachment in the settling tank

- biosolids loss to the effluent reduce x, affect process stability

SOLUTION : applying a vacuum in order to remove some of the gas supersaturation

13.2.3 Upflow and Downflow Packed Beds

Commonly called anaerobic filter

Trickling Filter

Biofilm attached to a rock/plastic

medium

Used to treat soluble substrate with

COD = 375-12,000 mg/l

Td = 4-36 h

Problems : -clogging by biosolids

-influent suspended solids

-precipitated minerals

Upflow

Solution : occasional high-rate application into

the bottom of the system

turbulent fluid motion

dislodge solids

13.2.3 Upflow and Downflow Packed Beds

Downflow

Advantages :

Solids tend to accumulate more near the top surface remove solids by gas

recirculation

Sulfide produced through sulfate reduction may be stripped from the liquid

Disadvantages :

•May have a greater tendency to lose biosolids to the effluent

13.2.4 Fluidized and Expanded Beds

General

• The fluidized bed reactor contains small media

(sand, granular activated carbon)

• High upflow velocity of WW causes biofilm

carriers to rise to balanced point

• A portion of effluent is recycled back to the

influent

Advantages

• high flow rate creates good mass transfer of

DOM from bulk liquid to particle surface

• clogging and short-circuiting of flow through

the reactor less than in the packed-bed

• the small carrier size gives a very high specific

surface area

Disadvantages

• difficulty in developing strongly attached biosolids containing the correct blend of methanogens

• abrasion between particles and fluid shear stress can increase detachment rates of microbes

• high recycle rates can increase energy cost due to head loss in the recycle piping

• defluidization for an extended period can cause process instability

13.2.5 Upflow Anaerobic Sludge Blanket

The UASBR is similar to clarigester (Stander, 1966)

but different in the method for separating biosolids

from the effluent

Biosolids form granules contain acetate utilizing

methanogens (Methanothrix and Methanosaeta)

The formation of granules depends on :

-Characteristics of the waste stream

-Substrate loading

-Operational details (ex. upward fluid velocity)

Problems :

-Formation of granules that float

-Lack of granule formation

Loss of biomass from the system

13.2.6 Miscellaneous Anaerobic Reactors

Baffled Reactor

Wastewater alternately moves upward and downward,

upward movement

UASBR-like

biosolids from one chamber enter the next chamber

Prevent biosolids loss


Horizontal sludge-blanket reactor

Vol. loading is similar to UASBR

Advantages :

-Simple and lack of structures

minimize the costs

Disadvantages :

-Solids accumulation short circuiting

poor performance

Solids need to be removed regularly


Two-stage leaching bed leachate filter

Biomass (lignocellulosic materials) methane

Limiting step : hydrolysis of cellulose

By separating acid formation from methanogenesis,

the more sensitive methanogenic system can be

operated as a high-rate reactor


Two-phase digestion

Td = 1-2 d

Td = 8-10 d

Low pH accelerated hydrolysis

Fermentation and methanogenesis takes place

•The overall volume is less than conventional CSTR

•Higher volumetric loading than a conventional CSTR


Membrane Bioreactor

Pass the treated liquid, but retains the solids

Advantages : -Eliminates the possibility of uncontrolled biomass loss

to the effluent

-Effluent quality is improved, because of no suspended BOD,

and also removes some fraction of SMP

-The volumetric loading can be increased to very high levels,

since loss of biomass is impossible

Disadvantages : Adding costs : capital cost, energy cost, replacement/cleaning cost


Aerobic Polishing

Remove the effluent from anaerobic system that

may exceed the regulatory requirement

13.3 Process Chemistry and Microbiology

13.3.1 Process Microbiology


Complex

organic

Organic

Acids +

hydrogen

Methane +

Carbon

dioxide

Hydrolysis +

fermentation

Methane

formation

Organics acids of significance

Volatile acids

Formic acid

Acetic acid

Propionic acid

n-Butyric acid

Isobutyric acid

n-Valeric acid

Isovaleric acid

Caproic acid

Heptanoic acid

Octanoic acid

Non-volatile acid

Lactic acid

Pyruvic acid

Succinic acid

Microorganisms grow rapidly

Sometimes being rate-limiting

Methanogens grow slowly

Often being rate-limiting

Should be balanced, by :

-Proper seeding

-pH control

-Organic acid prod. Control

-Enough amount of active seeding

The crucial steps during start-up :

1. Begin with as much good seed

2. Fill digester with this seed

3. Bring system to temperature

4. Add buffering material

5. Add small amount of substrate

6. Keep pH value (6.8 – 7.6)


• Organic acid concentration and pH should be checked daily

• Inhibition of the biological reaction or

overload with organic waste increase of organic acid concentration

• Buffering capacity is depleted pH buffer must be added prevent the death of methanogens

Organic Acid and pH

Volatile Acid

• Present in highest concentration : acetic, propionic, butyric, isobutyric acids

• at normal pH, present in the ionized form

• pKa (35 0C) = 3.8 – 4.9

• should be monitored routinely, by GC, HPLC

Microorganisms

• routine analyses using oligonucleotide probes (to be discussed later)

13.3.2 Process Chemistry

Important chemical aspects :

1. Reaction stoichiometry

2. pH and alkalinity requirements

3. Nutrient requirements

4. The effect of inhibitory materials on the process

Stoichiometry

Most of the organic is converted to main end products : CO2, CH4, H2O, biomass

Other elements (nitrogen, sulfur) inorganic form (ammonium, sulfides)

Conversion of acetate to methane :

CH3 ┇ COO- + H2O CH4 + HCO3- (13.1)

Using half-reactions : eHHCOCOOHCOCHRd 3223

8

1

8

1

8

3

8

1

OHCHeHCORa 2424

1

8

1

8

1:

34238

1

8

1

8

1

8

1: HCOCHOHCOOCHR

(13.2)

(13.3)

(13.4)


Overall stoichiometric equations :

34

27524

2

2020

20858

420

92

HCOdf

cNHdf

c

NOHCdf

COdfdf

cnCHdf

OHdfdf

bcnNOHC

ss

sese

escban

(13.5)

Where d= 4n + a – 2b – 3c

x

xdss

b

bfff

1

110fs = fraction of waste synthesized to cells

fs0 = common value Table 13.2

b = decay rate

x= sludge retention time

fd = percentage conversion to end product

Example 13.2 Stoichiometry of Glucose Fermentation to Methane

The wastewater contains 1.0 M glucose. Assume fs = 0.20, glucose is 100% consumed

Task : estimate CH4 prod., the mass of biological cells produced, concentration of NH4-N

required for cell growth per m3 of wastewater treated

Answer :

Molecular formula of glucose is C6H12O6 Eq. 13.5, n=6, a=12, b=6, c=0,

d=(4x6+12-2x6)=24, fe = 1-0.2=0.8

C6H12O6 + 0.24 NH4+ + 0.24 HCO3

- 2.4 CH4 + 2.64 CO2 + 0.24 C5H7O2N + 0.96 H2O

Thus, for each m3 wastewater, 2.4 kmol CH4 and 27.2 kg cells are produced

NH4-N required = 3.36 kg/m3 wastewater

Example 13.3 Stoichiometry of the Methanogenesis of an Organic

Mixture

Flow rate = 100 m3/d, COD = 5,000 mg/l, COD consists 50 % FAc. and 50 % protein

x = 20 d, T = 35 0C, Conversion = 80%. Protein = C16H24O5N4. Fatty Acids = C16H32O2

Estimate : CH4 prod., biological cell prod. rate, relative percentage of CO2 & CH4 produced,

bicarbonate alkalinity

Answer :

Reaction O-19 in Table 2.3,

Fatty Acids : 4/23 CO2 + H+ + e- = 1/92 C16H32O2 + 15/46 H2O

Protein : 2/11 CO2 + 2/33 NH4+ + 2/33 HCO3- + H+ + e- = 1/66 C16H24O5N4 + 31/66 H2O

50%

+

0.1779 CO2 + 0.0303 NH4+ + 0.0303 HCO3

- + H+ + e- = 0.013 C16H27.3O3.75N2.33 + 0.398 H2O

fs0 in Table 13.2 50% adding to each other fs

0 = 0.07, b = 0.05 d-1

042.0)20(05.01

)20)(05.0(2.0107.0

sf and fe = 1 - 0.042 = 0.958

Example 13.3 Stoichiometry of the Methanogenesis of an Organic

Mixture (cont.)

Substituting to Eq. 13.5 overall raection :

C16H27.3O3.75N2.33 + 10.73 H2O 9.20 CH4 + 3.83 CO2 + 0.161 C5H7O2N + 2.17 NH4+ + 2.17 HCO3-

COD removal rate = (S0-S)Q= [5,000 – 0.2(5,000)] mg/l x 100 m3/d = 4 x 105 g/d

At 35 0C, 1 mol CH4 = 25.3 l

CH4 prod. = 25.3 x 9.20 x molgCOD

dgCOD

/615

/)10(4 5

= 151 m3/d

Cell production = 0.161 x 113 x molgCOD

dgCOD

/615

/)10(4 5

= 11.8 kg/d

CH4 + CO2 = 100% %71)100(83.320.9

20.94

CH CO2 = 100-71 = 29%

3

33

000,50.

615.

17.2.

1

58.0)(

HCOmol

alkmg

CODg

mol

mol

HCOmolCODgxCaCOasAlk = 706 mg/l as CaCO3


The main chemical species controlling pH in anaerobic treatment related to the

carbonic acid system :

CO2 (aq) = CO2 (g) (13.6)

CO2 (aq) + H2O = H2CO3 (13.7)

H2CO3 = H+ + HCO3- (13.8)

HCO3- = H+ + CO3

2- (13.9)

H2O = H+ + OH- (13.10)

)35(/38][

)]([*

32

2 CmolatmKCOH

gCO o

H

Equilibrium relationship among those species :

H2CO3* = CO2 (aq) = CO2 (aq) + H2CO3 (13.11)

)35(105][

]][[ 7

1,*

32

3 CKCOH

HCOH o

a

(13.12) )35(106][

]][[ 11

2,

3

2

3 CKHCO

COH o

a

(13.13)

pH and Alkalinity Requirements pH = 6.6 – 7.6

)35(102]][[ 14 CKOHH o

w

(13.14)


][000,50

)(3

HCObicarbAlkalinity

][

][log

*

32

31,

COH

HCOpKpH a

H

a

K

gCO

bicarbAlkalinity

pKpH)]([

000,50

)(

log2

1,

][][2][][][ 2

33

OHCOHCOAlkalinityH (13.15)

Alkalinity can be quantified from a proton condition on the species of interest

(13.16)

Taking log of Eq. 13.12 : Subs. Eq.13.11 & 13.16 into Eq. 13.17

Example 13.4 Calculating the pH

Calculate the pH for Ex. 13.3, T = 35 0C, CO2 partial pressure = 29% 0.29 atm

Alkalinity was calculated as 706 mg/l.

Eq. 13.18 6.6

38/29.0

000,50/706log)105log( 7 xpH


HA + HCO3- H2CO3

* + A- (13.19)

[H+] + [Alkalinity] = [A-] + [HCO3-] + 2[CO3

2-] + [OH-] (13.20)

Release of NH4+ alkanility is added to the water

Biodegradation of organic matter can destroy alkalinity, by production of organic acid (HA)

In the case when weak acids and bicarbonate are present together :

Where : [A-] = summation of the molar conc. of all the weak acid salts present

(except for bicarbonate and carbonate)

Example 13.5 Calculating the Effect of Volatile Acids on the pH

2.738/25.0

000,50/800,2log3.6 initialpH

6.638/25.0

000,50/717log3.6 finalpH

Alkalinity (bicarb) = 2,800 mg/l – 2,500(50/60) = 717 mg/l

Operating at 35 0C, at sea level, 25% CO2 in the gas phase, low volatile acid conc.,

Total alkalinity = 2,800 mg/l

Because of overload volatile acid. conc. increase to 2,500 mg/l (as acetic acid, MW=60)

Task : estimate pH before the volatile acid increase and after it.

Answer :

After the increase of volatile acids :

Eq. 13.18

ch. 13 anaerobic treatment by...

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