Chapter ANAEROBIC TREATMENT BY METHANOGENESIS

Transcription

1 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

2 Ch. 13 Anaerobic Treatment by Methanogenesis Methanogenesis : anaerobic Organic matter (BOD L ) Reduced oxidation state (CH 4 ) At STP 1 g of BOD L generates 0.35 l of CH 4 Methanogenic process quite complex always contains methanogens Acetate Fermenters Hydrogen Oxidizers Electron Donor Acetate H 2 and formate Electron Acceptors Acetate CO 2 Carbon Sources Acetate CO 2 F 0 s Y 0.04 g VSS a /g Ac 0.45 g VSS a /g H 2 q^ (at 35 0 C) 7 g Ac/g VSS a -d 3 g H 2 /g VSSa-d K 400 mg Ac/l? B 0.03/d 0.03/d [ min x ] lim 3.6 d Slow growing 0.76 d S min 48 mg Ac/l?

3 13.1 Uses for Methanogenic Treatment 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

4 13.1 Uses for Methanogenic Treatment 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

5 Example 13.1 Energy Value of Methane Flow (Q) = 104 m 3 /d, BOD L = 20,000 mg/l Task : 1. Compute the volume of generated CH 4 if waste stabilization is 90% 2. Compute energy value of that methane BOD L loading rate = Q x BOD L = kg BOD L /d BOD L stabilization rate = waste stabilization x BOD L loading rate = kg BOD L /d Methane generation rate = BODL stabilization rate x 0.35 m 3 CH 4 (STP)/kg BOD L = m 3 CH 4 (STP)/d Energy rate = Methane generation rate x 35,800 kj/m 3 CH 4 (STP) = kj/d

6 13.2 Reactor Configurations

7 Completely Mixed Operational T = 35 0 C Td = days Earlier design : no mixing Solutions : operated at detention times, based on the ratio of tank volume to influent sludge flow rate 60 days Problems : 1. Fresh sludge & fermenting microbes are not brought together efficiently 2. Denser solids, tend to settle in the reactor

8 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 = % at loading rate kg/m 3 -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

9 Upflow and Downflow Packed Beds Upflow Commonly called anaerobic filter Trickling Filter Biofilm attached to a rock/plastic medium Used to treat soluble substrate with COD = ,000 mg/l T d = 4-36 h Problems : -clogging by biosolids -influent suspended solids -precipitated minerals Solution : occasional high-rate application into the bottom of the system turbulent fluid motion dislodge solids

10 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

11 Disadvantages 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 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

12 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 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

14 Miscellaneous Anaerobic Reactors 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

15 Miscellaneous Anaerobic Reactors 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

16 Miscellaneous Anaerobic Reactors Two-phase digestion Low ph accelerated hydrolysis Td = 1-2 d Td = 8-10 d Fermentation and methanogenesis takes place The overall volume is less than conventional CSTR Higher volumetric loading than a conventional CSTR

17 Miscellaneous Anaerobic Reactors 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

18 Miscellaneous Anaerobic Reactors Aerobic Polishing Remove the effluent from anaerobic system that may exceed the regulatory requirement

19 13.3 Process Chemistry and Microbiology Process Microbiology

20 Process Microbiology Complex organic Organic Acids + hydrogen Methane + Carbon dioxide Hydrolysis + fermentation Methane formation Microorganisms grow rapidly Sometimes being rate-limiting Should be balanced, by : -Proper seeding -ph control -Organic acid prod. Control -Enough amount of active seeding 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 Methanogens grow slowly Often being rate-limiting 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 ( )

21 Organic Acid and ph Process Microbiology 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 Volatile Acid Present in highest concentration : acetic, propionic, butyric, isobutyric acids at normal ph, present in the ionized form pk a (35 0 C) = should be monitored routinely, by GC, HPLC Microorganisms routine analyses using oligonucleotide probes (to be discussed later)

22 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 : CO 2, CH 4, H 2 O, biomass Other elements (nitrogen, sulfur) inorganic form (ammonium, sulfides) Conversion of acetate to methane : CH 3 COO - + H 2 O CH 4 + HCO 3 - (13.1) Using half-reactions : 1 3 R d CH CO H O CO HCO R a : CO H e CH 4 H 2 O R : CH3COO HCO H2O CH H e (13.2) (13.3) (13.4)

23 Process Chemistry Overall stoichiometric equations : 9dfs dfe CnH aob Nc 2n c b H 2O 20 4 dfe dfs dfe dfs CH 4 n c CO2 C5H 7O2N df df c s NH 4 c s HCO (13.5) Where d= 4n + a 2b 3c f s f 0 s 1 1 fd b x 1 b x f s = fraction of waste synthesized to cells f s 0 = common value Table 13.2 b = decay rate x = sludge retention time f d = percentage conversion to end product

24 Process Chemistry

25 Example 13.2 Stoichiometry of Glucose Fermentation to Methane The wastewater contains 1.0 M glucose. Assume f s = 0.20, glucose is 100% consumed Task : estimate CH 4 prod., the mass of biological cells produced, concentration of NH 4 -N required for cell growth per m 3 of wastewater treated Answer : Molecular formula of glucose is C 6 H 12 O 6 Eq. 13.5, n=6, a=12, b=6, c=0, d=(4x6+12-2x6)=24, f e = 1-0.2=0.8 C 6 H 12 O NH HCO CH CO C 5 H 7 O 2 N H 2 O Thus, for each m 3 wastewater, 2.4 kmol CH 4 and 27.2 kg cells are produced NH 4 -N required = 3.36 kg/m 3 wastewater

26 Example 13.3 Stoichiometry of the Methanogenesis of an Organic Mixture Flow rate = 100 m 3 /d, COD = 5,000 mg/l, COD consists 50 % FAc. and 50 % protein x = 20 d, T = 35 0 C, Conversion = 80%. Protein = C 16 H 24 O 5 N 4. Fatty Acids = C 16 H 32 O 2 Estimate : CH 4 prod., biological cell prod. rate, relative percentage of CO 2 & CH 4 produced, bicarbonate alkalinity Answer : Reaction O-19 in Table 2.3, Fatty Acids : 4/23 CO 2 + H + + e - = 1/92 C 16 H 32 O /46 H 2 O Protein : 2/11 CO 2 + 2/33 NH /33 HCO H + + e - = 1/66 C 16 H 24 O 5 N /66 H 2 O CO NH HCO H + + e - = C 16 H 27.3 O 3.75 N H 2 O 50% + f s 0 in Table % adding to each other f s 0 = 0.07, b = 0.05 d (0.05)(20) fs (20) and f e = = 0.958

27 Example 13.3 Stoichiometry of the Methanogenesis of an Organic Mixture (cont.) Substituting to Eq overall raection : C 16 H 27.3 O 3.75 N H 2 O 9.20 CH CO C 5 H 7 O 2 N NH HCO 3 - COD removal rate = (S 0 -S)Q= [5, (5,000)] mg/l x 100 m 3 /d = 4 x 10 5 g/d At 35 0 C, 1 mol CH 4 = 25.3 l CH 4 prod. = 25.3 x 9.20 x 5 4(10 ) gcod / d 615gCOD / mol = 151 m 3 /d Cell production = x 113 x 5 4(10 ) gcod / d 615gCOD / mol = 11.8 kg/d 9.20 CH 4 + CO 2 = 100% CH 4 (100) 71% CO 2 = = 29% x5g COD 2.17 mol HCO3 mol 50,000mg alk Alk ( as CaCO3 )... 1 mol 615 g COD mol HCO = 706 mg/l as CaCO 3 3

28 Process Chemistry ph and Alkalinity Requirements ph = The main chemical species controlling ph in anaerobic treatment related to the carbonic acid system : [ CO2 ( g)] o K 38atm / mol (35 C) * H [ H CO ] 2 3 CO 2 (aq) = CO 2 (g) (13.6) CO 2 (aq) + H 2 O = H 2 CO 3 (13.7) H 2 CO 3 = H + + HCO - 3 (13.8) HCO - 3 = H + + CO 2-3 (13.9) H 2 O = H + + OH - (13.10) Equilibrium relationship among those species : [ 3 7 C H ][ HCO * [ H CO ] 2 3 H 2 CO 3 * = CO 2 (aq) = CO 2 (aq) + H 2 CO 3 (13.11) ] o Ka,1 510 (35 ) (13.12) 2 [ H ][ CO3 ] 11 o K 6 10 (35 [ ] ) a,2 C (13.13) HCO [ H ][ OH 14 o ] K 210 (35 C) (13.14) w 3

29 Process Chemistry Alkalinity can be quantified from a proton condition on the species of interest 2 [ H ] [ Alkalinity ] [ HCO3 ] 2[ CO3 ] [ OH ] (13.15) Alkalinity ( bicarb ) 50,000 [ HCO 3 ] (13.16) Taking log of Eq : ph pk a,1 [ HCO3 ] log * [ H CO ] 2 3 Subs. Eq & into Eq ph pk a, 1 log Alkalinity ( bicarb) 50,000 [ CO2 ( g)] K H

30 Process Chemistry

31 Example 13.4 Calculating the ph Calculate the ph for Ex. 13.3, T = 35 0 C, CO 2 partial pressure = 29% 0.29 atm Alkalinity was calculated as 706 mg/l. Eq ph log(5 x10 7 ) log 706 / 50, /

32 Process Chemistry Release of NH4+ alkanility is added to the water Biodegradation of organic matter can destroy alkalinity, by production of organic acid (HA) HA + HCO 3 - H 2 CO 3 * + A - (13.19) In the case when weak acids and bicarbonate are present together : [H + ] + [Alkalinity] = [A - ] + [HCO 3- ] + 2[CO 3 2- ] + [OH - ] (13.20) Where : [A - ] = summation of the molar conc. of all the weak acid salts present (except for bicarbonate and carbonate)

33 Example 13.5 Calculating the Effect of Volatile Acids on the ph Operating at 35 0 C, at sea level, 25% CO 2 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 : Alkalinity (bicarb) = 2,800 mg/l 2,500(50/60) = 717 mg/l Eq ph initial 2,800 / 50, log 0.25/ ph final 717 / 50, log 0.25/38 6.6