11 Anaerobic Processes F12
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ANAEROBIC PROCESSES (ME, p. 420; VH, p. 601) Definition of Anaerobic Preference order end of e- acceptor product (0) O2
Aerobic ....................
(-2)
H2O
(+5)
Anoxic (absence of O2) ....................
(0)
NO3- N2 (+6)
Anaerobic (complete absence of O2)
(Denitrification)
(-2)
SO42- H2S (+4)
(-4)
CH4
CO2
C2H4O2 (0)
(Sulfate reduction)
(Methane fermentation)
organics
(0)
CH3COOH
(-4)
CH4
(Methane fermentation)
- This route is not intuitively obvious but most of the CH4 comes from CH3COOH * We will be mainly concerned with the methane fermentation reactions. Organics
anaerobic bacteria
(-4)
(+4)
CH4 + CO2 + other end products (e.g., H2S)
Anaerobic Biological Treatment - difference in e- acceptor Example (K.M. Dumm, 2003) : Mead is a kind of wine produced from honey by yeast in the absence of oxygen. Honey is a concentrated solution of sugars, primarily glucose and fructose dissolved in water. A representative reaction for the synthesis of mead is C6 H12O6(aq) sugar (glucose)
2C2H5OH(aq) ethanol
+
2CO2 (g) carbon dioxide
The various types of mead can be produced in a fermentation reactor: If the reactor is sealed, carbonated mead will be produced since CO2 is trapped in the mead. If carbon dioxide is released from the reactor, dry unsweetened mead will be formed since sugar is completely consumed by the yeast. A sweet mead can be produced if a concentrated solution is used since sugar will still remain at the end of the fermentation reaction.
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In the presence of microorganisms (other than yeast) and oxygen in the reactor, alcohol (e.g., ethanol) will be converted to acid (e.g., acetic acid): C2 H5OH ethanol
O2 oxygen
+
CH3COOH acetic acid
+
H2 O water
Since vinegar consists of acetic acid, mead is spoiled to form vinegar, which tastes sour. Example: 1) The following three chemical reactions are redox reactions. Identify reduction and oxidation (mark with an arrow):
6(0) 12(1+) 6(2-)
2(0)
C6 H12 O6 + 6 O2 0
C 0
O
6(0) 12(1+) 6(2-)
0
(4+) 2(2-)
+
6 CO2 4+
C
O
C6 H12 O6
2(1+) (2-)
6 H2O
2-
2(2-) 5(1+) (2-) (1+)
0
2 CO2
4+
C
2(2-) 5(1+) (2-) (1+)
(0)
C2 H5 O H + O2
2-
C 0
O
Oxidation
2-
C
Reduction
(4+) 2(2-)
2 C2 H5 O H +
C
Oxidation
C
Reduction
(0) 3(1+) (0) (2-) (2-) (1+)
C H3 C O O H
2(1+) (2-)
+
H2 O
0
C
Oxidation 2-
O
2
Reduction
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2) How many grams of CO2 are produced when 200g of glucose (C6H12O6) are fermented anaerobically to ethanol (C2H5OH)? AW (g/mol): C= 12.011; O = 15.999; H= 1.008 MW (g/mol): C6H12O6: 6(12.011) + 12(1.008) + 6(15.999) = 180.156 C2H5OH: 2(12.011) + 6(1.008) + (15.999) = 46.069 CO2 : (12.011) + 2(15.999) = 44.009 Reaction:
C6H12O6
MW (g/mol):
180.156
2 C2H5OH
+
46.069
2 CO2 44.009
g CO2 2 mol CO2 44.009 g/mol 0.4886 g CO2 -------------- = ------------------- -------------------- = -----------------g C6H12O6 1 mol C6H12O6 180.156 g/mol g C6H12O6 0.4886 g CO2 ------------------ (200 g C6H12O6) = 97.713 g CO2 g C6H12O6 Thus, 97.713 g of CO2 are produced.
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History Septic tank Imhof tank Anaerobic digester 1) Seepage Systems (Cesspool, Septic Tank) Inspection and cleaning ports
Backfill covering trench Scum
Influent
Drain tile or perforated pipe
Liquid
Gravel bed
Seepage of wastewater
Solids
Settled Solids
Septic tank
Fig. A typical septic tank –adsorption field system for on-site disposal of household wastewater.
2) Imhof tank - The Imhoff tank, named for German engineer Karl Imhoff (1876–1965)
3) Anaerobic digester
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Process Configurations 1) Low rate or standard rate anerobic digester (single stage, not mixed, not heated) 2) High rate anaerobic digester (single-stage, two-stage, mixed, heated) 3) Anaerobic Contact Process 4) Anaerobic Filter 5) Anaerobic Fluidized (or Expanded) Bed 6) Anaerobic Upflow Sludge Blanket 7) Anaerobic Baffled Reactor 8) Anaerobic ponds 2. Purposes of Anaerobic Sludge Digestion:
Sludge volume reduction Stabilization of organic solids Disinfection
Applications of Anaerobic Processes a. Strong waste (high BOD, COD) - very high O2 demand - we want to avoid supply that much O2. 1) Domestic wastewater treatment Primary sludge (organic solids), Secondary sludge (biological solids); e.g. activated sludge, trickling filter sludge, etc. 2) Industrial waste treatment e.g., Wastes from meet packing, feed lots 3) Hazardous wastewater treatment (leachates, etc.) - new field of research, dechlorination of chlorinated compounds Advantages and Disadvantages of Anaerobic Treatment Advantages 1) Usable end products (CH4) digester gas (2/3 CH4, 1/3 CO2) natural gas
= 600 BTU/SCF = 1000 BTU/SCF
Energy (ME, p. 826) Digester gas (65% CH4 )
= 600 BTU/SCF (22,400 kJ/m3)
CH4
= 960 Btu/SCF (35,800 kJ/m3)
Natural gas (mixture of methane, propane butane)
= 1000 BTU/SCF (37,000 kJ/m3)
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2) 3) 4) 5)
No O2 or aeration costs Less nutrients are required - since less bacteria are produced Greater degree of waste stabilization Small amount of sludges synthesis
50 lbs of bacteria, sludge (VSS) 50 lbs of CO2 & H2O Aerobic
Yobs = 0.5 mg VSS / mg BOD
100 lb of BOD Anaerobic
Yobs = 0.05 mg VSS / mg BOD 5 lbs of bacteria, sludge (VSS) 95 lbs (CH4 & CO2)
Anaerobic Synthesis 0.04 - 0.05 mg VSS Yanaerobic = ---------------------------mg BOD . not much energy available . energy goes to maintenance rather than synthesis 0.5 mg VSS Yaerobic = ------------------mg BOD
Disadvantages 1) Slower growth - slow reaction 2) Sensitive - difficult to operate (Temperature, pH) 3) High tempeature (energy $) Methophilic 30 - 35°C Thermophilic 35 - 75°C 4) CH4 is explosive - safety (home near landfill)
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Process of Anaerobic Digestion Three Stage Processes: Stage 1 Hydrolysis Fermentation (Liquifaction)
Stage 2 Acetogenesis Acidgenesis (Acid formation)
Stage 3 . Methanogenesis (Gasification)
Partial oxidation Acid formers Non-methanogenic bacteria Complex Smaller organics soluble (sludge) organics (solids)
(liquid)
Methane formers Methanogenic bacteria
Simple organic acids* alchohols** (liquid)
No stabilization No BOD reduction in liquid Lipids Polysacchrides Protein DNA
CO2 CH4 (gas)
Large BOD reduction in liquid
Fatty acids Monosaccharides Acetic acid Amino acids Purines, pyrimidines
---------------------------------------------------------------* Simple acids ** Simple alcohols 1C Formic acid HFo 1C Methanol 2C Acetic acid HAc 2C Ethanol 3C Propionic acid HPr 3C Propanol 4C Buteric acid HBt 4C Butanol
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CH4 CO2
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Example: STAGE 1: HYDROLYSIS Polysacchride monosacchride
Starch
liquification
Starch
fermentation
glucose n C6H12O6 lactic acid
STAGE 2: ACETOGENESIS (acid formation)
C6H12O6 glucose lactic acid
fermentation
3 CH3COOH acetic acid HAc, HBt, HPr
Acetogenesis - glucose gives 3 acids (HAc, HBt, HPr) Types of bacteria ---------------------------pH < 5.5 C6H12O6 HBt + HPr + HAc + H2 HBt HAc + H2 HPr HAc + H2
Fermentative bacteria Hydrogen producing acetogenic bacteria Hydrogen producing acetogenic bacteria
STAGE 3: METHANOGENESIS (methane formation) CH3COOH CH4 + CO2 CO2 + 4 H2(g) CH4 + 2H2O Low energy, slow reduction Methanogenesis Types of bacteria ---------------------------pH >6.5 HAc CH4 + CO2 4H2 + CO2 CH4 + 2 H2O H2 + CO2 HAC
Aceticlatic methanogens CO2 reducing methanogens Hydrogen consuming acetogenic bacteria
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* Note that at least 5 different types of organisms are needed. 1) Fermentative bacteria 2) Hydrogen producing acetogenic bacteria 3) Hydrogen consuming acetogenic bacteria 4) CO2 reducing methanogens 5) Aceticlastic methanogens Consists of two distinct stages that occur simultaneously in digesting sludge.
Note: No waste stabilization occurs until CH4 is formed. You start with protein and form H2 & CH3COOH, and unless CH4 is formed the BOD is there in dissolved form.
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Characteristics of anaerobic digester organisms Acid formers Methane formers ====================================================================== 1. Facultative or Strict anaerobes 1. Strict anaerobes 2. Growth rate is high θc > 3 days
2. Growth rate is very low θc (35°C) > 10 days (optimum) θc (18°C) > 20 days (to 30 d)
θc = 1/ net 3. Produce acids and alcohols
3. Produce CH4 and CO2 CH4 = 2/3 CO2 = 1/3
4. Little waste stabilization BOD reduction is minimal
4. Large BOD reduction since CH4 gas leaves system CH4 has about the same energy as the alcohols and acids - Little energy conversion
5. Less pH sensitive (tolerance) - can take various environmental conditions
5. Very pH sensitive optimum 6.8-7.2 strict range 7.0-7.1 broad range 6.7-7.4 ======================================================================
General Conditions - consult Table 13.6 (VH, p.602) Anaerobic digester - symbiotic system Alkalinity: Volatile acid: Gas production: pH range:
2000-3500 mg/L as CaCO3 200-800 mg/L as HAc (max 2,000 mg/L) 2/3 CH4, 1/3 CO2 6.6 - 7.6 (6.8 - 7.2 optimum) 6.7 - 7.4 (7.1 - 7.2 optimum) … VH, p.13.6
Temperature
Low rate (ambient temperature) High rate (heated to 85 - 95 ºF or 30 - 35ºC) o Methophilic bacteria High rate (heated to 110-115 ºF or 50 -55ºC) o Thermophilic bacteria o too much energy input - energy is the hung up.
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Nutrients 1) must have N & P 2) less N & P are needed than for aerobic system For aerobic, BOD : N : P 100 : 5 : 1 3) nutrients are recycled by cell lysis Substrates 1) Domestic waste (biodegradable) – acceptable 2) Ring compounds - not acceptable (e.g., benzene, naphthalene) - Non biodegradable, perhaps toxic Toxics Odd organics metal > 1.0 mg/L total metals Cl> 5000 mg/L (industrial tannery) Ca 2+ > 2000 - 4000 mg/L NH4+ > 1500 mg/L S> 200 mg/L (sulfate reduction) Digester Failure Symptoms Sour If methane formers get sick - digester goes sour Any shift in environment adverse to the population of methane bacteria causes a buildup of organic acids. Large organic molecule
Acid formers --
Simple acids
fast growth
Methane formers -- CH4 + CO2 slow growth
a) pH drops due to: increase in volatile acid production and accumulation increase in alkalinity consumption (buffering capacity; ammonium carbonate) b) Gas production decrease in total gas production % CO2 will increase and % CH4 gas will decrease Causes Significant increase in organic loading Sharp decrease in digesting sludge volume Sudden increase in operating temperature Accumulation of a toxic or inhibiting substance 11
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pH pH 7 (6.8 - 7.2) Alk 2500 (2000-3000) Total gas, CH4 Volatile acids 250 (200-300)
Alk VA Total gas CH4 Time
Curing digester a. Cure by neutralize pH (control volatile acids) 1) Add lime Ca(OH)2 Ca(OH)2 + 2 CO2 ---> Ca(HCO3)2 - but not too much lime Ca(OH)2 + CO2
---> CaCO3 + H2O
2) Add NaHCO3
Alkalinity Balance CH3COOH + HCO3- CH3COO - + CO2 + H2O 1 moles alkalinity consumed/1 mole CH3COOH CH3COO - + H2O CH4 + HCO31 moles alkalinity produced/1 mole CH3COO * No net production and reduction of alkalinity.
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Types of Anaerobic Digestion Processes 1) By stage:
Single stage Two stage digester o heat only 1st digester, not 2nd digester.
2) By heating:
Standard rate digester
a) Cold temperature – unheated 0.02 - 0.05 lb VSS applied b) Organic loading rate = --------------------------------ft3.day
High rate digester
a) High temperature – heated 0.1 - 0.2 lb VSS applied b) Organic loading rate = -----------------------------ft3.day
3) By mixing: Table 13.7 (VH, p. 605) Loading and detention times for heated anaerobic digesters ------------------------------------------------------------------------------------------------------------------------Conventional Single-Stage First-Stage High-Rate (Unmixed) (Completely Mixed) ======================================================================== Organic Loading 0.02 - 0.05 0.1 - 0.2 (lb/ft3/day of VS) Detention time (days)
30 - 90
10 - 15
Volatile Acid reduction (%) 50 - 70 50 -------------------------------------------------------------------------------------------------------------------------
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Gas relief Gas Scum layer Supernatant Influent
Effluent Active layer Stabilized solids
Solid removal
Gas outlet
Gas outlet
Fixed cover
Floating cover
Gas storage
Gas storage
Sludge inlet
Scum layer Supernatant
Supernatant outlet
Influent Sludge heater Digested sludge Sludge outlet
First stage (Completely mixed)
Second stage (Stratified)
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1) Standard-Rate Anaerobic Digester (4th DC 517) - Single tank - No sludge mixing - Intermittent sludge feeding and withdrawal - Generally heated (SRT = 30 - 60 days for heated digester) - Organic loading rate = 0.48 -1.6 kg TVS / m3 of digester volume .day 2) High-Rate Anaerobic Digester (4th DC 518)
Continuous sludge feeding and withdrawal Organic loading rate = 1.6 -8.0 kg TVS/m3 of digester volume .day SRT = 10 - 15 days Two tanks in series (1st Tank and 2nd Tank) 1st Tank Fermentation
2nd Tank ____ gas/ Solid / Liquid separation
Thoroughly mixed No mixing Heated Not heated __________________________________
Anaerobic digesters, Pocatello, Idaho
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Gas Production a. Gas Composition (ME, p. 823) CH4 0.7 (2/3) 65-70% by volume CO2 0.3 (1/3) 25-30% by volume N2(g) trace H2S trace H2 trace Water vapor trace
b. Specific gravity of digester gas ~ 0.86 relative to air (ME, p. 825) Example: Modified from Example 8-5 (3rd ME 426) Conversion of BODL to methane gas. C6H12O6 3 CO2 + 3 CH4 3 CH4 + 6O2 3 CO2 + 6 H2O
.... (1) ..... (2)
Assuming that the starting compound is glucose, C6H12O6, determine the amount of methane produced per pound of BODL: 1) g CH4 produced / g BODL destroyed 2) m3 CH4 produced / kg BODL destroyed 3) ft3 CH4 produced / lb BODL destroyed (Solution) 1. 1) Write a balanced equation for the conversion of glucose to CH4 under anaerobic conditions. MW
C6H12O6 3 CO2 + 3 CH4 180 3(44) 3(16)
.... (1)
2) Write a balanced equation for the oxidation of methane to CO2 and H2O. MW
3 CH4 + 6O2 3 CO2 3(16) 6(32) 3(44)
+ 6 H2O
..... (2)
c. Combine these two reactions (1) and (2): C6H12O6 3 CO2 + 3 CH4 ..... (1) 3 CH4 + 6O2 3 CO2 + 6 H2O ..... (2) ------------------------------------------------------------------C6H12O6 + 6O2 6CO2 + 6 H2O ..... (3)
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d. Calculate g BODL / g glucose g BODL 6 moles (32 g/mole) 192 g BODL ------------ = -------------------------- = -----------------g glucose 1 mole (180 g/mole) 180 g glucose 1.067 g BODL = -----------------g glucose
e. Calculate g CH4 /g BODL - use Reaction (2) 3 CH4 + 6O2 3 CO2
+ 6 H2O
..... (2)
g CH4 3 moles (16 g/mole) 48 g CH4 ------------ = ------------------------- ----------------g BODL 6 moles (32 g /mole) 192 g BODL 0.25 g CH4 produced 0.25 lb CH4 produced = ----------------------------- = ----------------------------g BODL destroyed lb BODL destroyed 2. Calculate the volume equivalent of the 0.25 g of CH4. 0.25 g CH4 1 mole CH4 22.4 L m3 103 g BODL z = --------------- ---------------- ---------- -------- ---------------g BODL 16 g CH4 1 mole 103 L kg BODL m3 CH4 produced z = 0.35 -------------------------kg BODL destroyed 0.35 m3 of CH4 is produced per kg of BODL converted (at 20ºC, 1 atm). 3. Calculate the volume equivalent of the 0.25 lb of CH4. 0.25 lb CH4 454 g 1 mole 22.4 L ft3 z = ----------------- -------- ---------- --------- ----------lb BODL lb 16 g 1 mole 28.32 L 5.62 ft3 CH4 = ----------------lb BODL
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5.62 ft of CH4 is produced per pound of BODL converted (at 20ºC, 1 atm). ft3 CH4 produced Common values: z = 5.6 - 5.7 ----------------------lb BODL destroyed c. Total Gas Yield, Ygas 8 - 12 SCF Ygas = --------------------lb VSS applied
(VH, p. 602, Table 13.6)
12 - 18 SCF = ----------------------lb VSS destroyed
(ME, p. 825)
16 - 18 SCF = ----------------------lb VSS destroyed
(VH, Table 13.6)
0.75 – 1.12 m3 = -----------------------kg VSS destroyed
(ME, p. 825)
For Primary sludge :|ME, p.825) 0.6 - 0.8 ft3 gas ------------------person-day
or
15 - 22 m3 gas ---------------------1000 person-day
or
28 m3 gas ---------------------1000 person-day
For Secondary sludge: 1.0 ft3 gas ----------------person-day
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c. CH4 Gas Production
dS VCH 4 z 142 . Px dt
VCH4 z Qo So Eff 142 . Px Px
Y ( So S ) 1 c kd
Y (So-S) Px = -------------1 + 0c kd 1.42 = O2 equivalent to biomass 0.25 lb CH4 produced z = -----------------------------lb BODL destroyed
or
5.6 - 5.7 ft3 CH4 produced z = ---------------------------------- = 5.62 lb BODL destroyed
Example: Determine ft2 CH4 produced per lb BODL destroyed. CH4 gram MW: 16
+ 2 O2 ------> CO2 + 2 H2O 2(32) 44 2(18)
2 moles of O2 2(32) g 4.0 lb O2 ------------------- = -------------- = -------------1 mole of CH4 16 g lb CH4
lb CH4 16 g 0.25 lb CH4 0.25 lb CH4 ---------- = ------------- = ----------------- = --------------------------lb O2 2(32) g lb O2 lb BODL destroyed
SCF CH4 produced 0.25 g CH4 1 mole CH4 22.414 L ft3 ------------------------- = ------------------ ----------- ------ ------------ ---------lb BODL destroyed g BODL 16 g CH4 1 mole 28.3 L
5.61 ft3 CH4 produced = -----------------------------lb BODL destroyed
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454 g BODL ------ ---------lb BODL
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ft3 CH4 produced = 5.6 - 5.7 ---------------------------lb BODL destroyed
m3 CH4 produced 0.25 g CH4 1 mole CH4 22.414 L m3 103 g BODL -------------------------- = --------------- ----------------- ---- ------ --------- ---------------kg BODL destroyed g BODL 16 g CH4 1 mole 103 L kg BODL 0.35 m3 CH4 produced = --------------------------------kg BODL destroyed
CH4
+ 2 O2 ------> CO2 + 2 H2O
1 mole gram MW: volume
2 moles
16 22.4 L
2(32)
44
2(18)
m3 CH4 1 mol (22.4 L CH4/mol) 1 m3 103 g 0.35 m3 CH4 produced ---------- = ---------------------------------- --------- -------- = --------------------------------kg BOD 2 mol (32 g/mol) BODL 103 L kg kg BODL destroyed
L CH4 1 mol (22.4 L CH4/mol) 0.35 L CH4 ------------ = ----------------------------------- ≈ -----------------g COD 2 mol (32 g/mol) BODL g COD
Example: BOD loading to a digester = 1500 lb BOD/day (680 kg BOD/d) and gas production = 5600 ft3 CH4/day (160 m3 CH4/d). Determine: (1) the BOD removed, and (2) VSS reduced. 3
5.6 ft CH4 ------------------------ or lb BOD destroyed
0.35 m3 CH4 -------------------------kg BODL removed
15 SCF total gas -------------------------lb VSS destroyed 20
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from
Gas:
16-18 SCF total gas --------------------------lb VSS destroyed 2/3 = 0.66 CH4 1/3 = 0.34 CO2
(Solution) Assume that: 1)
5600 ft3 CH4/day BOD removed = ---------------------------------- = 1000 lb BODL/day 5.6 ft3 CH4 ---------------------------lb BOD destroyed 5600 m3 CH4/day BOD removed = ----------------------------- = 457 kg BODL/day 0.35 m3 CH4 -------------------------kg BODL removed
2) 15 SCF gas 2/3 CH4 10 SCF CH4 CH4 production = (------------------------) (--------------) = ------------------------lb VSS destroyed gas lb VSS destroyed 5600 SCF CH4/day VSS destroyed = ------------------------------------------ = 560 lb VSS destroyed/ day 10 SCF CH4/lb VSS destroyed
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Example 12-4 (Modified from 3rd ME 819). Estimation of digester volume and performance Estimate the size of digester required to treat the sludge from a wastewater treatment plant. (a) Check the volumetric loading, and estimate (b) the percent stabilization and (c) the amount of gas produced per day. For the sludge to be treated, it has been found that the loadings of dry solids and BODL is 5455 kg/day and 5215 kg/day, respectively. Assume that the sludge contains about 95 percent moisture (95% water, 5% solids) and has a specific gravity (SG) of 1.02. Other pertinent design assumptions are as follows: a) The hydraulic regime of the reactor is complete mix. b) The sludge contains adequate nitrogen and phosphorus for biological growth. c) Sludge age, SRT = 10 days at 35°C (see Table 12-18, ME, p. 818) d) Efficiency of waste utilization = BODL removal efficiency E = 0.60 (ME., p. 818) e) Anaerobic cell yield Y = 0.05 mg cells/mg BODL utilized and f) Endogenous decay coefficient, kd = 0.03 d-1. g) Constants are for a temperature of 35°C.
(Solution) Given: SRT = θc = θ =10 d BOD removal efficiency, E = (So – S) / S = 0.6 Y = 0.05 kg VSS / kg BODL kd = 0.03 d-1 SG of sludge = ρs / ρ = 1.02 a = 1.42 g O2 / g VSS z =0.35 m3 CH4 generated / kg BODL destroyed
Given: Solid loading = 5455 kg/d dry solids Water content = 95%; SG = 1.02 BOD loading = 5215 kg/d BODL So
1. Calculate the daily sludge volume loading, Qd V
Ws Ws s 100 p S S 100 100
where V = volume of sludge, m3 Ws = weight of dry solids, kg Ws (daily base) = 5,455 kg/d dry solids s = solid content, % = (100 – p) = 100 – 95 = 5% P = water content, % = 95% γ = unit weight of water = 1000 kg/m3 S = S.G. of wet sludge = 1.02 22
Digester
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V 5455 kg/d Qd = ------- = ------------------------------- = 107 m3/d Time 5 1000 kg ----- ---------- (1.02) 100 m3
2. Calculated the digester volume, V Noting θ = θc = V/Qd, V = Qd θ = (107 m3/d) (10 d) = 1070 m3 3. Calculate BODL concentration, So (mg/L) BOD Loading = Qd So, So = BODL Loading / Q where Qd So = BODL Loading = 5,215 kg/d 5,215 kg/d (1,000 g/kg) So = ------------------------------ = 48,645 g/ m3 = 48,645 mg/L 107 m3/d 4. Calculate BODL conc., S (mg/L) Since BODL removal efficiency, E = (So – S) / S = 0.6 S = So (1– E) = 48,645 mg/L (1 – 0.6) = 19,458 mg/L BODL = (19,458 mg/L) (107 m3/d) (103 L/m3)( kg/106 mg) = 2082 kg/d BODL 5. Calculate the volatile solids produced per day, Px Px
Y ( So S ) Yobs So S 1 c kd
0.05 mgVSS 48, 645 mgBODL 19, 458 mgBODL mgBODL L L Px 1123 mgVSS / L 0.03 1 10d d
mg VSS 107 m3 103 L kg 120 kgVSS Px Yobs So S Qd 1123 L d m3 106 mg d 23
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6. Calculate BODL for Px BODL from Xbiodeg (Xbiodeg in terms of O2) = (1.42 mg O2/mg VSS) (1123 mg VSS / L) = 1595 mg O2/L BODL of Px = 1595 mg /L BODL or BODL from Xbiodeg (Xbiodeg in terms of O2) = (1.42 kg O2/kg VSS) (120 kg VSS / d) = 170 kg O2/d = 170 kg BODL/d BOD loading = 5215 kg/d BODL
Summary: Px = 120 kg VSS/d = 1123 mg VSS/L = 170 kg/d BODL = 1595 mg/L BODL
BOD in effluent = 2,082 kg/d BODL
So =48,645 mg/L BODL
S =19,458 mg/L BODL Digester
Qd = 107 m3/d
V = 1070 m3
170 kg/d BODL
8. Calculate BODL removed in the digester. Accum = Inputs – Outputs ± Rxns = (BODL)in – (BODL)out – (BODL) converted to cells (VSS) – (BODL) converted to CH4 and CO2 0 = 5215 kg/d – 2082 kg/d – 170 kg/d – BOD converted to CH4 and CO2. BOD converted to CH4 and CO2 = 2965 kg/d BOD removed = 2965 kg/d
Organics
CH4 + CO2 + Other end products
BODL anaerobic bacteria
new anaerobic bacteria
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9. Calculate % BODL stabilized BODL removed % BODL stabilized = --------------------- (100 ) BODL applied 2965 kg/d % BODL stabilized = --------------- (100 ) = 56.9% 5215 kg/d 10. Calculate the volume of methane produced per day, VCH4 0.35 m3 CH4 produced Use z = -----------------------------kg BODL destroyed 0.35 m3 CH4 produced 2965 kg BODL destroyed 1038 m3 CH4 produced z = ------------------------------ ---------------------------------- = ---------------------------kg BODL destroyed d d or VCH4 z So Se Qd a Px
where a = (1.42 mg O2/mg VSS) = (1.42 mg BODL/mg VSS)
VCH 4
0.35 m3CH 4 5215 2082 170 kgBODL 1037m3 / d kgBODL d
10. Estimate the total gas production (2/3)(volume of total gas) = volume of CH4 volume of CH4 1038 m3 CH4 /d (total gas volume) = --------------------- = --------------------- = 1556 m3 total gas /d 2/3 2/3 11. Calculate the volumetric BOD loading (kg/ m3 ∙d) BOD loading 5215 kg/d 4.87 kg ------------------ = --------------- = ---------V 1070 m3 m3 ∙d
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11-Anaerobic Processes_F12
Digester Hating Requirements The heat loss through the digester side wall, floor (bottom), roof (top area) can be computed using the following equation: q = UA ∆T where q = heat loss, J/d, Btu / h, W (Note: 1 W = 1 J/s = 3.413 Btu/h) U = overall coefficient of heat transfer, Btu/ft2 ºF hr (W/ m2 ºC) A = cross-sectional area through which the heat loss is occurring, ft2 (m2) ∆T = temperature drop across the surface in question, ºF (ºC) The heat loss from sludge, q, can be computed by q = c W ∆T where c = specific heat of sludge, 1 Btu/lb ºF (J/kg °C) W = weight of wet sludge per day, lb/d (kg/d) ∆T = change in temperature, ºC or q = c m (∆T /∆t) where m = mass of sludge, lb (kg) ∆t = duration of time Example 14-7 (4th ME 1527): A digester with a capacity of 45,000 kg/d (100,000 lb/d) of sludge is to be treated by the circulation of the sludge through an external hot water heat exchanger. Assuming that the following conditions apply, find the heat required to maintain the required digester temperature. If all heat sources were shut off for 24 h, what would be the average drop in temperature of the tank contents? Conditions: 1. Concrete digester dimensions: Diameter = 18 m (60 ft) Side depth = 6 m (20 ft) Mid depth = 9 m (30 ft) 2. Heat transfer coefficients: Dry earth embanked for entire depth Floor of digester in groundwater Roof exposed to air
U (W/m2 ºC) 0.68 0.85 0.91
U (Btu/ft2 .ºF . h) 0.12 0.15 0.16
3. Temperatures: Air Earth next to wall Incoming sludge Earth below floor Sludge contents in digester
( ºC) -5 0 10 5 32
ºF 23 32 50 42 90
26
11-Anaerobic Processes_F12
4. Specific heat of sludge = 4200 J/kg ºC = 1 Btu/ft2 ·ºF · h Assume the specific gravity of sludge = 1.02 and the density of water is 995.7 kg/m3 at 30ºC.
D = 18 m U = 0.91, -5ºC q
h1 = 6 m U = 0.68 0ºC q
H=9m 32ºC h2 = 3 m q U = 0.85, 5ºC
(Solution) 1. Preliminary calculations 1) The area of the walls, roof, and floor. a) The wall area = A1 = 2π r h1 = π D h1 = π (18 m)(6 m) = 339.29 m2
A2 r b) The floor area =
9m
r 2 h2 2
9m
2
2 3m 268.2m 2
c) The roof area = A3 = π r2 = π (9 m)2 = 254.5 m2 2) The digester volume The volume of circular cylinder, V1 = π r2 h1 The volume of circular cone, V2 = (1/3) π r2 h2 h 3 2 The volume of the digester = V1 V2 r 2 h1 2 9m 6 m 1781.3m3 3 3
27
11-Anaerobic Processes_F12
2. Compute the heat requirement for the sludge. q = c W ∆T = (4200 J/kg °C) (45000 kg/d) (32 – 10) ºC = 41.6 x 108 J/d where c = specific heat of sludge = 4200 J/kg °C W = weight of wet sludge per day = 45000 kg/d ∆T = change in temperature = (32 – 10) ºC 3. Compute the heat loss by conduction q = UA ∆T where q = heat loss, J/d, Btu / h, W (Note: 1 W = 1 J/s = 3.413 Btu/h) U = overall coefficient of heat transfer, Btu/ft2 ºF hr (W/ m2 ºC) A = cross-sectional area through which the heat loss is occurring, ft2 (m2) ∆T = temperature drop across the surface in question, ºF (ºC) 1) Walls: q = (0.68 W/m2 °C) (339.3 m2) (32 – 0) ºC (86400 s/d) = 6.38 x 108 J/d 2) Floor: q = (0.85 W/m2 °C) (268.2 m2) (32 – 5) ºC (86400 s/d) = 5.32 x 108 J/d 3) Roof: q = (0.91 W/m2 °C) (254.5 m2) (32 – (-5)) ºC (86400 s/d) = 7.40 x 108 J/d Total heat loss, qT = (6.38 x 108) + (5.32 x 108) + (7.40 x 108) = 19.1 x 108 J/d 4. Compute the required heat exchanger capacity Heat exchanger capacity = heat required for sludge + heat required for digester = (41.6 x 108 J/d) + (19.1 x 108 J/d) = 60.7 x 108 J/d 5. Determine the effect of heat shtoff. 1) Determine weight of sludge Since S.G. = ρsludge / ρH2O = 1.02 ρsludge = 1.02 (ρH2O) = 1.02 (995.7 kg/m3 at 30ºC) = 1.016 x 103 kg/m3 Weight of sludge = (1781.3 m3) (1.016 x 103 kg/m3) = 1.81 x 106 kg
2) Determine the temperature drop q = c m (∆T /∆t) where m = mass of sludge, lb (kg) ∆t = duration of time
T q 60.7 x108 J / d 0.80C / d t m c 1.81x106 kg 4200 J / kg C
28
ANAEROBIC PROCESSES (ME, p. 420; VH, p. 601) Definition of Anaerobic Preference order end of e- acceptor product (0) O2
Aerobic ....................
(-2)
H2O
(+5)
Anoxic (absence of O2) ....................
(0)
NO3- N2 (+6)
Anaerobic (complete absence of O2)
(Denitrification)
(-2)
SO42- H2S (+4)
(-4)
CH4
CO2
C2H4O2 (0)
(Sulfate reduction)
(Methane fermentation)
organics
(0)
CH3COOH
(-4)
CH4
(Methane fermentation)
- This route is not intuitively obvious but most of the CH4 comes from CH3COOH * We will be mainly concerned with the methane fermentation reactions. Organics
anaerobic bacteria
(-4)
(+4)
CH4 + CO2 + other end products (e.g., H2S)
Anaerobic Biological Treatment - difference in e- acceptor Example (K.M. Dumm, 2003) : Mead is a kind of wine produced from honey by yeast in the absence of oxygen. Honey is a concentrated solution of sugars, primarily glucose and fructose dissolved in water. A representative reaction for the synthesis of mead is C6 H12O6(aq) sugar (glucose)
2C2H5OH(aq) ethanol
+
2CO2 (g) carbon dioxide
The various types of mead can be produced in a fermentation reactor: If the reactor is sealed, carbonated mead will be produced since CO2 is trapped in the mead. If carbon dioxide is released from the reactor, dry unsweetened mead will be formed since sugar is completely consumed by the yeast. A sweet mead can be produced if a concentrated solution is used since sugar will still remain at the end of the fermentation reaction.
1
11-Anaerobic Processes_F12
In the presence of microorganisms (other than yeast) and oxygen in the reactor, alcohol (e.g., ethanol) will be converted to acid (e.g., acetic acid): C2 H5OH ethanol
O2 oxygen
+
CH3COOH acetic acid
+
H2 O water
Since vinegar consists of acetic acid, mead is spoiled to form vinegar, which tastes sour. Example: 1) The following three chemical reactions are redox reactions. Identify reduction and oxidation (mark with an arrow):
6(0) 12(1+) 6(2-)
2(0)
C6 H12 O6 + 6 O2 0
C 0
O
6(0) 12(1+) 6(2-)
0
(4+) 2(2-)
+
6 CO2 4+
C
O
C6 H12 O6
2(1+) (2-)
6 H2O
2-
2(2-) 5(1+) (2-) (1+)
0
2 CO2
4+
C
2(2-) 5(1+) (2-) (1+)
(0)
C2 H5 O H + O2
2-
C 0
O
Oxidation
2-
C
Reduction
(4+) 2(2-)
2 C2 H5 O H +
C
Oxidation
C
Reduction
(0) 3(1+) (0) (2-) (2-) (1+)
C H3 C O O H
2(1+) (2-)
+
H2 O
0
C
Oxidation 2-
O
2
Reduction
11-Anaerobic Processes_F12
2) How many grams of CO2 are produced when 200g of glucose (C6H12O6) are fermented anaerobically to ethanol (C2H5OH)? AW (g/mol): C= 12.011; O = 15.999; H= 1.008 MW (g/mol): C6H12O6: 6(12.011) + 12(1.008) + 6(15.999) = 180.156 C2H5OH: 2(12.011) + 6(1.008) + (15.999) = 46.069 CO2 : (12.011) + 2(15.999) = 44.009 Reaction:
C6H12O6
MW (g/mol):
180.156
2 C2H5OH
+
46.069
2 CO2 44.009
g CO2 2 mol CO2 44.009 g/mol 0.4886 g CO2 -------------- = ------------------- -------------------- = -----------------g C6H12O6 1 mol C6H12O6 180.156 g/mol g C6H12O6 0.4886 g CO2 ------------------ (200 g C6H12O6) = 97.713 g CO2 g C6H12O6 Thus, 97.713 g of CO2 are produced.
3
11-Anaerobic Processes_F12
History Septic tank Imhof tank Anaerobic digester 1) Seepage Systems (Cesspool, Septic Tank) Inspection and cleaning ports
Backfill covering trench Scum
Influent
Drain tile or perforated pipe
Liquid
Gravel bed
Seepage of wastewater
Solids
Settled Solids
Septic tank
Fig. A typical septic tank –adsorption field system for on-site disposal of household wastewater.
2) Imhof tank - The Imhoff tank, named for German engineer Karl Imhoff (1876–1965)
3) Anaerobic digester
4
11-Anaerobic Processes_F12
Process Configurations 1) Low rate or standard rate anerobic digester (single stage, not mixed, not heated) 2) High rate anaerobic digester (single-stage, two-stage, mixed, heated) 3) Anaerobic Contact Process 4) Anaerobic Filter 5) Anaerobic Fluidized (or Expanded) Bed 6) Anaerobic Upflow Sludge Blanket 7) Anaerobic Baffled Reactor 8) Anaerobic ponds 2. Purposes of Anaerobic Sludge Digestion:
Sludge volume reduction Stabilization of organic solids Disinfection
Applications of Anaerobic Processes a. Strong waste (high BOD, COD) - very high O2 demand - we want to avoid supply that much O2. 1) Domestic wastewater treatment Primary sludge (organic solids), Secondary sludge (biological solids); e.g. activated sludge, trickling filter sludge, etc. 2) Industrial waste treatment e.g., Wastes from meet packing, feed lots 3) Hazardous wastewater treatment (leachates, etc.) - new field of research, dechlorination of chlorinated compounds Advantages and Disadvantages of Anaerobic Treatment Advantages 1) Usable end products (CH4) digester gas (2/3 CH4, 1/3 CO2) natural gas
= 600 BTU/SCF = 1000 BTU/SCF
Energy (ME, p. 826) Digester gas (65% CH4 )
= 600 BTU/SCF (22,400 kJ/m3)
CH4
= 960 Btu/SCF (35,800 kJ/m3)
Natural gas (mixture of methane, propane butane)
= 1000 BTU/SCF (37,000 kJ/m3)
5
11-Anaerobic Processes_F12
2) 3) 4) 5)
No O2 or aeration costs Less nutrients are required - since less bacteria are produced Greater degree of waste stabilization Small amount of sludges synthesis
50 lbs of bacteria, sludge (VSS) 50 lbs of CO2 & H2O Aerobic
Yobs = 0.5 mg VSS / mg BOD
100 lb of BOD Anaerobic
Yobs = 0.05 mg VSS / mg BOD 5 lbs of bacteria, sludge (VSS) 95 lbs (CH4 & CO2)
Anaerobic Synthesis 0.04 - 0.05 mg VSS Yanaerobic = ---------------------------mg BOD . not much energy available . energy goes to maintenance rather than synthesis 0.5 mg VSS Yaerobic = ------------------mg BOD
Disadvantages 1) Slower growth - slow reaction 2) Sensitive - difficult to operate (Temperature, pH) 3) High tempeature (energy $) Methophilic 30 - 35°C Thermophilic 35 - 75°C 4) CH4 is explosive - safety (home near landfill)
6
11-Anaerobic Processes_F12
Process of Anaerobic Digestion Three Stage Processes: Stage 1 Hydrolysis Fermentation (Liquifaction)
Stage 2 Acetogenesis Acidgenesis (Acid formation)
Stage 3 . Methanogenesis (Gasification)
Partial oxidation Acid formers Non-methanogenic bacteria Complex Smaller organics soluble (sludge) organics (solids)
(liquid)
Methane formers Methanogenic bacteria
Simple organic acids* alchohols** (liquid)
No stabilization No BOD reduction in liquid Lipids Polysacchrides Protein DNA
CO2 CH4 (gas)
Large BOD reduction in liquid
Fatty acids Monosaccharides Acetic acid Amino acids Purines, pyrimidines
---------------------------------------------------------------* Simple acids ** Simple alcohols 1C Formic acid HFo 1C Methanol 2C Acetic acid HAc 2C Ethanol 3C Propionic acid HPr 3C Propanol 4C Buteric acid HBt 4C Butanol
7
CH4 CO2
11-Anaerobic Processes_F12
Example: STAGE 1: HYDROLYSIS Polysacchride monosacchride
Starch
liquification
Starch
fermentation
glucose n C6H12O6 lactic acid
STAGE 2: ACETOGENESIS (acid formation)
C6H12O6 glucose lactic acid
fermentation
3 CH3COOH acetic acid HAc, HBt, HPr
Acetogenesis - glucose gives 3 acids (HAc, HBt, HPr) Types of bacteria ---------------------------pH < 5.5 C6H12O6 HBt + HPr + HAc + H2 HBt HAc + H2 HPr HAc + H2
Fermentative bacteria Hydrogen producing acetogenic bacteria Hydrogen producing acetogenic bacteria
STAGE 3: METHANOGENESIS (methane formation) CH3COOH CH4 + CO2 CO2 + 4 H2(g) CH4 + 2H2O Low energy, slow reduction Methanogenesis Types of bacteria ---------------------------pH >6.5 HAc CH4 + CO2 4H2 + CO2 CH4 + 2 H2O H2 + CO2 HAC
Aceticlatic methanogens CO2 reducing methanogens Hydrogen consuming acetogenic bacteria
8
11-Anaerobic Processes_F12
* Note that at least 5 different types of organisms are needed. 1) Fermentative bacteria 2) Hydrogen producing acetogenic bacteria 3) Hydrogen consuming acetogenic bacteria 4) CO2 reducing methanogens 5) Aceticlastic methanogens Consists of two distinct stages that occur simultaneously in digesting sludge.
Note: No waste stabilization occurs until CH4 is formed. You start with protein and form H2 & CH3COOH, and unless CH4 is formed the BOD is there in dissolved form.
9
11-Anaerobic Processes_F12
Characteristics of anaerobic digester organisms Acid formers Methane formers ====================================================================== 1. Facultative or Strict anaerobes 1. Strict anaerobes 2. Growth rate is high θc > 3 days
2. Growth rate is very low θc (35°C) > 10 days (optimum) θc (18°C) > 20 days (to 30 d)
θc = 1/ net 3. Produce acids and alcohols
3. Produce CH4 and CO2 CH4 = 2/3 CO2 = 1/3
4. Little waste stabilization BOD reduction is minimal
4. Large BOD reduction since CH4 gas leaves system CH4 has about the same energy as the alcohols and acids - Little energy conversion
5. Less pH sensitive (tolerance) - can take various environmental conditions
5. Very pH sensitive optimum 6.8-7.2 strict range 7.0-7.1 broad range 6.7-7.4 ======================================================================
General Conditions - consult Table 13.6 (VH, p.602) Anaerobic digester - symbiotic system Alkalinity: Volatile acid: Gas production: pH range:
2000-3500 mg/L as CaCO3 200-800 mg/L as HAc (max 2,000 mg/L) 2/3 CH4, 1/3 CO2 6.6 - 7.6 (6.8 - 7.2 optimum) 6.7 - 7.4 (7.1 - 7.2 optimum) … VH, p.13.6
Temperature
Low rate (ambient temperature) High rate (heated to 85 - 95 ºF or 30 - 35ºC) o Methophilic bacteria High rate (heated to 110-115 ºF or 50 -55ºC) o Thermophilic bacteria o too much energy input - energy is the hung up.
10
11-Anaerobic Processes_F12
Nutrients 1) must have N & P 2) less N & P are needed than for aerobic system For aerobic, BOD : N : P 100 : 5 : 1 3) nutrients are recycled by cell lysis Substrates 1) Domestic waste (biodegradable) – acceptable 2) Ring compounds - not acceptable (e.g., benzene, naphthalene) - Non biodegradable, perhaps toxic Toxics Odd organics metal > 1.0 mg/L total metals Cl> 5000 mg/L (industrial tannery) Ca 2+ > 2000 - 4000 mg/L NH4+ > 1500 mg/L S> 200 mg/L (sulfate reduction) Digester Failure Symptoms Sour If methane formers get sick - digester goes sour Any shift in environment adverse to the population of methane bacteria causes a buildup of organic acids. Large organic molecule
Acid formers --
Simple acids
fast growth
Methane formers -- CH4 + CO2 slow growth
a) pH drops due to: increase in volatile acid production and accumulation increase in alkalinity consumption (buffering capacity; ammonium carbonate) b) Gas production decrease in total gas production % CO2 will increase and % CH4 gas will decrease Causes Significant increase in organic loading Sharp decrease in digesting sludge volume Sudden increase in operating temperature Accumulation of a toxic or inhibiting substance 11
11-Anaerobic Processes_F12
pH pH 7 (6.8 - 7.2) Alk 2500 (2000-3000) Total gas, CH4 Volatile acids 250 (200-300)
Alk VA Total gas CH4 Time
Curing digester a. Cure by neutralize pH (control volatile acids) 1) Add lime Ca(OH)2 Ca(OH)2 + 2 CO2 ---> Ca(HCO3)2 - but not too much lime Ca(OH)2 + CO2
---> CaCO3 + H2O
2) Add NaHCO3
Alkalinity Balance CH3COOH + HCO3- CH3COO - + CO2 + H2O 1 moles alkalinity consumed/1 mole CH3COOH CH3COO - + H2O CH4 + HCO31 moles alkalinity produced/1 mole CH3COO * No net production and reduction of alkalinity.
12
11-Anaerobic Processes_F12
Types of Anaerobic Digestion Processes 1) By stage:
Single stage Two stage digester o heat only 1st digester, not 2nd digester.
2) By heating:
Standard rate digester
a) Cold temperature – unheated 0.02 - 0.05 lb VSS applied b) Organic loading rate = --------------------------------ft3.day
High rate digester
a) High temperature – heated 0.1 - 0.2 lb VSS applied b) Organic loading rate = -----------------------------ft3.day
3) By mixing: Table 13.7 (VH, p. 605) Loading and detention times for heated anaerobic digesters ------------------------------------------------------------------------------------------------------------------------Conventional Single-Stage First-Stage High-Rate (Unmixed) (Completely Mixed) ======================================================================== Organic Loading 0.02 - 0.05 0.1 - 0.2 (lb/ft3/day of VS) Detention time (days)
30 - 90
10 - 15
Volatile Acid reduction (%) 50 - 70 50 -------------------------------------------------------------------------------------------------------------------------
13
11-Anaerobic Processes_F12
Gas relief Gas Scum layer Supernatant Influent
Effluent Active layer Stabilized solids
Solid removal
Gas outlet
Gas outlet
Fixed cover
Floating cover
Gas storage
Gas storage
Sludge inlet
Scum layer Supernatant
Supernatant outlet
Influent Sludge heater Digested sludge Sludge outlet
First stage (Completely mixed)
Second stage (Stratified)
14
11-Anaerobic Processes_F12
1) Standard-Rate Anaerobic Digester (4th DC 517) - Single tank - No sludge mixing - Intermittent sludge feeding and withdrawal - Generally heated (SRT = 30 - 60 days for heated digester) - Organic loading rate = 0.48 -1.6 kg TVS / m3 of digester volume .day 2) High-Rate Anaerobic Digester (4th DC 518)
Continuous sludge feeding and withdrawal Organic loading rate = 1.6 -8.0 kg TVS/m3 of digester volume .day SRT = 10 - 15 days Two tanks in series (1st Tank and 2nd Tank) 1st Tank Fermentation
2nd Tank ____ gas/ Solid / Liquid separation
Thoroughly mixed No mixing Heated Not heated __________________________________
Anaerobic digesters, Pocatello, Idaho
15
11-Anaerobic Processes_F12
Gas Production a. Gas Composition (ME, p. 823) CH4 0.7 (2/3) 65-70% by volume CO2 0.3 (1/3) 25-30% by volume N2(g) trace H2S trace H2 trace Water vapor trace
b. Specific gravity of digester gas ~ 0.86 relative to air (ME, p. 825) Example: Modified from Example 8-5 (3rd ME 426) Conversion of BODL to methane gas. C6H12O6 3 CO2 + 3 CH4 3 CH4 + 6O2 3 CO2 + 6 H2O
.... (1) ..... (2)
Assuming that the starting compound is glucose, C6H12O6, determine the amount of methane produced per pound of BODL: 1) g CH4 produced / g BODL destroyed 2) m3 CH4 produced / kg BODL destroyed 3) ft3 CH4 produced / lb BODL destroyed (Solution) 1. 1) Write a balanced equation for the conversion of glucose to CH4 under anaerobic conditions. MW
C6H12O6 3 CO2 + 3 CH4 180 3(44) 3(16)
.... (1)
2) Write a balanced equation for the oxidation of methane to CO2 and H2O. MW
3 CH4 + 6O2 3 CO2 3(16) 6(32) 3(44)
+ 6 H2O
..... (2)
c. Combine these two reactions (1) and (2): C6H12O6 3 CO2 + 3 CH4 ..... (1) 3 CH4 + 6O2 3 CO2 + 6 H2O ..... (2) ------------------------------------------------------------------C6H12O6 + 6O2 6CO2 + 6 H2O ..... (3)
16
11-Anaerobic Processes_F12
d. Calculate g BODL / g glucose g BODL 6 moles (32 g/mole) 192 g BODL ------------ = -------------------------- = -----------------g glucose 1 mole (180 g/mole) 180 g glucose 1.067 g BODL = -----------------g glucose
e. Calculate g CH4 /g BODL - use Reaction (2) 3 CH4 + 6O2 3 CO2
+ 6 H2O
..... (2)
g CH4 3 moles (16 g/mole) 48 g CH4 ------------ = ------------------------- ----------------g BODL 6 moles (32 g /mole) 192 g BODL 0.25 g CH4 produced 0.25 lb CH4 produced = ----------------------------- = ----------------------------g BODL destroyed lb BODL destroyed 2. Calculate the volume equivalent of the 0.25 g of CH4. 0.25 g CH4 1 mole CH4 22.4 L m3 103 g BODL z = --------------- ---------------- ---------- -------- ---------------g BODL 16 g CH4 1 mole 103 L kg BODL m3 CH4 produced z = 0.35 -------------------------kg BODL destroyed 0.35 m3 of CH4 is produced per kg of BODL converted (at 20ºC, 1 atm). 3. Calculate the volume equivalent of the 0.25 lb of CH4. 0.25 lb CH4 454 g 1 mole 22.4 L ft3 z = ----------------- -------- ---------- --------- ----------lb BODL lb 16 g 1 mole 28.32 L 5.62 ft3 CH4 = ----------------lb BODL
17
11-Anaerobic Processes_F12 3
5.62 ft of CH4 is produced per pound of BODL converted (at 20ºC, 1 atm). ft3 CH4 produced Common values: z = 5.6 - 5.7 ----------------------lb BODL destroyed c. Total Gas Yield, Ygas 8 - 12 SCF Ygas = --------------------lb VSS applied
(VH, p. 602, Table 13.6)
12 - 18 SCF = ----------------------lb VSS destroyed
(ME, p. 825)
16 - 18 SCF = ----------------------lb VSS destroyed
(VH, Table 13.6)
0.75 – 1.12 m3 = -----------------------kg VSS destroyed
(ME, p. 825)
For Primary sludge :|ME, p.825) 0.6 - 0.8 ft3 gas ------------------person-day
or
15 - 22 m3 gas ---------------------1000 person-day
or
28 m3 gas ---------------------1000 person-day
For Secondary sludge: 1.0 ft3 gas ----------------person-day
18
11-Anaerobic Processes_F12
c. CH4 Gas Production
dS VCH 4 z 142 . Px dt
VCH4 z Qo So Eff 142 . Px Px
Y ( So S ) 1 c kd
Y (So-S) Px = -------------1 + 0c kd 1.42 = O2 equivalent to biomass 0.25 lb CH4 produced z = -----------------------------lb BODL destroyed
or
5.6 - 5.7 ft3 CH4 produced z = ---------------------------------- = 5.62 lb BODL destroyed
Example: Determine ft2 CH4 produced per lb BODL destroyed. CH4 gram MW: 16
+ 2 O2 ------> CO2 + 2 H2O 2(32) 44 2(18)
2 moles of O2 2(32) g 4.0 lb O2 ------------------- = -------------- = -------------1 mole of CH4 16 g lb CH4
lb CH4 16 g 0.25 lb CH4 0.25 lb CH4 ---------- = ------------- = ----------------- = --------------------------lb O2 2(32) g lb O2 lb BODL destroyed
SCF CH4 produced 0.25 g CH4 1 mole CH4 22.414 L ft3 ------------------------- = ------------------ ----------- ------ ------------ ---------lb BODL destroyed g BODL 16 g CH4 1 mole 28.3 L
5.61 ft3 CH4 produced = -----------------------------lb BODL destroyed
19
454 g BODL ------ ---------lb BODL
11-Anaerobic Processes_F12
ft3 CH4 produced = 5.6 - 5.7 ---------------------------lb BODL destroyed
m3 CH4 produced 0.25 g CH4 1 mole CH4 22.414 L m3 103 g BODL -------------------------- = --------------- ----------------- ---- ------ --------- ---------------kg BODL destroyed g BODL 16 g CH4 1 mole 103 L kg BODL 0.35 m3 CH4 produced = --------------------------------kg BODL destroyed
CH4
+ 2 O2 ------> CO2 + 2 H2O
1 mole gram MW: volume
2 moles
16 22.4 L
2(32)
44
2(18)
m3 CH4 1 mol (22.4 L CH4/mol) 1 m3 103 g 0.35 m3 CH4 produced ---------- = ---------------------------------- --------- -------- = --------------------------------kg BOD 2 mol (32 g/mol) BODL 103 L kg kg BODL destroyed
L CH4 1 mol (22.4 L CH4/mol) 0.35 L CH4 ------------ = ----------------------------------- ≈ -----------------g COD 2 mol (32 g/mol) BODL g COD
Example: BOD loading to a digester = 1500 lb BOD/day (680 kg BOD/d) and gas production = 5600 ft3 CH4/day (160 m3 CH4/d). Determine: (1) the BOD removed, and (2) VSS reduced. 3
5.6 ft CH4 ------------------------ or lb BOD destroyed
0.35 m3 CH4 -------------------------kg BODL removed
15 SCF total gas -------------------------lb VSS destroyed 20
11-Anaerobic Processes_F12
from
Gas:
16-18 SCF total gas --------------------------lb VSS destroyed 2/3 = 0.66 CH4 1/3 = 0.34 CO2
(Solution) Assume that: 1)
5600 ft3 CH4/day BOD removed = ---------------------------------- = 1000 lb BODL/day 5.6 ft3 CH4 ---------------------------lb BOD destroyed 5600 m3 CH4/day BOD removed = ----------------------------- = 457 kg BODL/day 0.35 m3 CH4 -------------------------kg BODL removed
2) 15 SCF gas 2/3 CH4 10 SCF CH4 CH4 production = (------------------------) (--------------) = ------------------------lb VSS destroyed gas lb VSS destroyed 5600 SCF CH4/day VSS destroyed = ------------------------------------------ = 560 lb VSS destroyed/ day 10 SCF CH4/lb VSS destroyed
21
11-Anaerobic Processes_F12
Example 12-4 (Modified from 3rd ME 819). Estimation of digester volume and performance Estimate the size of digester required to treat the sludge from a wastewater treatment plant. (a) Check the volumetric loading, and estimate (b) the percent stabilization and (c) the amount of gas produced per day. For the sludge to be treated, it has been found that the loadings of dry solids and BODL is 5455 kg/day and 5215 kg/day, respectively. Assume that the sludge contains about 95 percent moisture (95% water, 5% solids) and has a specific gravity (SG) of 1.02. Other pertinent design assumptions are as follows: a) The hydraulic regime of the reactor is complete mix. b) The sludge contains adequate nitrogen and phosphorus for biological growth. c) Sludge age, SRT = 10 days at 35°C (see Table 12-18, ME, p. 818) d) Efficiency of waste utilization = BODL removal efficiency E = 0.60 (ME., p. 818) e) Anaerobic cell yield Y = 0.05 mg cells/mg BODL utilized and f) Endogenous decay coefficient, kd = 0.03 d-1. g) Constants are for a temperature of 35°C.
(Solution) Given: SRT = θc = θ =10 d BOD removal efficiency, E = (So – S) / S = 0.6 Y = 0.05 kg VSS / kg BODL kd = 0.03 d-1 SG of sludge = ρs / ρ = 1.02 a = 1.42 g O2 / g VSS z =0.35 m3 CH4 generated / kg BODL destroyed
Given: Solid loading = 5455 kg/d dry solids Water content = 95%; SG = 1.02 BOD loading = 5215 kg/d BODL So
1. Calculate the daily sludge volume loading, Qd V
Ws Ws s 100 p S S 100 100
where V = volume of sludge, m3 Ws = weight of dry solids, kg Ws (daily base) = 5,455 kg/d dry solids s = solid content, % = (100 – p) = 100 – 95 = 5% P = water content, % = 95% γ = unit weight of water = 1000 kg/m3 S = S.G. of wet sludge = 1.02 22
Digester
Se
11-Anaerobic Processes_F12
V 5455 kg/d Qd = ------- = ------------------------------- = 107 m3/d Time 5 1000 kg ----- ---------- (1.02) 100 m3
2. Calculated the digester volume, V Noting θ = θc = V/Qd, V = Qd θ = (107 m3/d) (10 d) = 1070 m3 3. Calculate BODL concentration, So (mg/L) BOD Loading = Qd So, So = BODL Loading / Q where Qd So = BODL Loading = 5,215 kg/d 5,215 kg/d (1,000 g/kg) So = ------------------------------ = 48,645 g/ m3 = 48,645 mg/L 107 m3/d 4. Calculate BODL conc., S (mg/L) Since BODL removal efficiency, E = (So – S) / S = 0.6 S = So (1– E) = 48,645 mg/L (1 – 0.6) = 19,458 mg/L BODL = (19,458 mg/L) (107 m3/d) (103 L/m3)( kg/106 mg) = 2082 kg/d BODL 5. Calculate the volatile solids produced per day, Px Px
Y ( So S ) Yobs So S 1 c kd
0.05 mgVSS 48, 645 mgBODL 19, 458 mgBODL mgBODL L L Px 1123 mgVSS / L 0.03 1 10d d
mg VSS 107 m3 103 L kg 120 kgVSS Px Yobs So S Qd 1123 L d m3 106 mg d 23
11-Anaerobic Processes_F12
6. Calculate BODL for Px BODL from Xbiodeg (Xbiodeg in terms of O2) = (1.42 mg O2/mg VSS) (1123 mg VSS / L) = 1595 mg O2/L BODL of Px = 1595 mg /L BODL or BODL from Xbiodeg (Xbiodeg in terms of O2) = (1.42 kg O2/kg VSS) (120 kg VSS / d) = 170 kg O2/d = 170 kg BODL/d BOD loading = 5215 kg/d BODL
Summary: Px = 120 kg VSS/d = 1123 mg VSS/L = 170 kg/d BODL = 1595 mg/L BODL
BOD in effluent = 2,082 kg/d BODL
So =48,645 mg/L BODL
S =19,458 mg/L BODL Digester
Qd = 107 m3/d
V = 1070 m3
170 kg/d BODL
8. Calculate BODL removed in the digester. Accum = Inputs – Outputs ± Rxns = (BODL)in – (BODL)out – (BODL) converted to cells (VSS) – (BODL) converted to CH4 and CO2 0 = 5215 kg/d – 2082 kg/d – 170 kg/d – BOD converted to CH4 and CO2. BOD converted to CH4 and CO2 = 2965 kg/d BOD removed = 2965 kg/d
Organics
CH4 + CO2 + Other end products
BODL anaerobic bacteria
new anaerobic bacteria
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9. Calculate % BODL stabilized BODL removed % BODL stabilized = --------------------- (100 ) BODL applied 2965 kg/d % BODL stabilized = --------------- (100 ) = 56.9% 5215 kg/d 10. Calculate the volume of methane produced per day, VCH4 0.35 m3 CH4 produced Use z = -----------------------------kg BODL destroyed 0.35 m3 CH4 produced 2965 kg BODL destroyed 1038 m3 CH4 produced z = ------------------------------ ---------------------------------- = ---------------------------kg BODL destroyed d d or VCH4 z So Se Qd a Px
where a = (1.42 mg O2/mg VSS) = (1.42 mg BODL/mg VSS)
VCH 4
0.35 m3CH 4 5215 2082 170 kgBODL 1037m3 / d kgBODL d
10. Estimate the total gas production (2/3)(volume of total gas) = volume of CH4 volume of CH4 1038 m3 CH4 /d (total gas volume) = --------------------- = --------------------- = 1556 m3 total gas /d 2/3 2/3 11. Calculate the volumetric BOD loading (kg/ m3 ∙d) BOD loading 5215 kg/d 4.87 kg ------------------ = --------------- = ---------V 1070 m3 m3 ∙d
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Digester Hating Requirements The heat loss through the digester side wall, floor (bottom), roof (top area) can be computed using the following equation: q = UA ∆T where q = heat loss, J/d, Btu / h, W (Note: 1 W = 1 J/s = 3.413 Btu/h) U = overall coefficient of heat transfer, Btu/ft2 ºF hr (W/ m2 ºC) A = cross-sectional area through which the heat loss is occurring, ft2 (m2) ∆T = temperature drop across the surface in question, ºF (ºC) The heat loss from sludge, q, can be computed by q = c W ∆T where c = specific heat of sludge, 1 Btu/lb ºF (J/kg °C) W = weight of wet sludge per day, lb/d (kg/d) ∆T = change in temperature, ºC or q = c m (∆T /∆t) where m = mass of sludge, lb (kg) ∆t = duration of time Example 14-7 (4th ME 1527): A digester with a capacity of 45,000 kg/d (100,000 lb/d) of sludge is to be treated by the circulation of the sludge through an external hot water heat exchanger. Assuming that the following conditions apply, find the heat required to maintain the required digester temperature. If all heat sources were shut off for 24 h, what would be the average drop in temperature of the tank contents? Conditions: 1. Concrete digester dimensions: Diameter = 18 m (60 ft) Side depth = 6 m (20 ft) Mid depth = 9 m (30 ft) 2. Heat transfer coefficients: Dry earth embanked for entire depth Floor of digester in groundwater Roof exposed to air
U (W/m2 ºC) 0.68 0.85 0.91
U (Btu/ft2 .ºF . h) 0.12 0.15 0.16
3. Temperatures: Air Earth next to wall Incoming sludge Earth below floor Sludge contents in digester
( ºC) -5 0 10 5 32
ºF 23 32 50 42 90
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4. Specific heat of sludge = 4200 J/kg ºC = 1 Btu/ft2 ·ºF · h Assume the specific gravity of sludge = 1.02 and the density of water is 995.7 kg/m3 at 30ºC.
D = 18 m U = 0.91, -5ºC q
h1 = 6 m U = 0.68 0ºC q
H=9m 32ºC h2 = 3 m q U = 0.85, 5ºC
(Solution) 1. Preliminary calculations 1) The area of the walls, roof, and floor. a) The wall area = A1 = 2π r h1 = π D h1 = π (18 m)(6 m) = 339.29 m2
A2 r b) The floor area =
9m
r 2 h2 2
9m
2
2 3m 268.2m 2
c) The roof area = A3 = π r2 = π (9 m)2 = 254.5 m2 2) The digester volume The volume of circular cylinder, V1 = π r2 h1 The volume of circular cone, V2 = (1/3) π r2 h2 h 3 2 The volume of the digester = V1 V2 r 2 h1 2 9m 6 m 1781.3m3 3 3
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2. Compute the heat requirement for the sludge. q = c W ∆T = (4200 J/kg °C) (45000 kg/d) (32 – 10) ºC = 41.6 x 108 J/d where c = specific heat of sludge = 4200 J/kg °C W = weight of wet sludge per day = 45000 kg/d ∆T = change in temperature = (32 – 10) ºC 3. Compute the heat loss by conduction q = UA ∆T where q = heat loss, J/d, Btu / h, W (Note: 1 W = 1 J/s = 3.413 Btu/h) U = overall coefficient of heat transfer, Btu/ft2 ºF hr (W/ m2 ºC) A = cross-sectional area through which the heat loss is occurring, ft2 (m2) ∆T = temperature drop across the surface in question, ºF (ºC) 1) Walls: q = (0.68 W/m2 °C) (339.3 m2) (32 – 0) ºC (86400 s/d) = 6.38 x 108 J/d 2) Floor: q = (0.85 W/m2 °C) (268.2 m2) (32 – 5) ºC (86400 s/d) = 5.32 x 108 J/d 3) Roof: q = (0.91 W/m2 °C) (254.5 m2) (32 – (-5)) ºC (86400 s/d) = 7.40 x 108 J/d Total heat loss, qT = (6.38 x 108) + (5.32 x 108) + (7.40 x 108) = 19.1 x 108 J/d 4. Compute the required heat exchanger capacity Heat exchanger capacity = heat required for sludge + heat required for digester = (41.6 x 108 J/d) + (19.1 x 108 J/d) = 60.7 x 108 J/d 5. Determine the effect of heat shtoff. 1) Determine weight of sludge Since S.G. = ρsludge / ρH2O = 1.02 ρsludge = 1.02 (ρH2O) = 1.02 (995.7 kg/m3 at 30ºC) = 1.016 x 103 kg/m3 Weight of sludge = (1781.3 m3) (1.016 x 103 kg/m3) = 1.81 x 106 kg
2) Determine the temperature drop q = c m (∆T /∆t) where m = mass of sludge, lb (kg) ∆t = duration of time
T q 60.7 x108 J / d 0.80C / d t m c 1.81x106 kg 4200 J / kg C
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