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Microbial Biotechnology (BIO-6004A) > Lecture 7: Wastewater treatment 3 > Flashcards

Lecture 7: Wastewater treatment 3 Flashcards

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1
Q

What are the two main goals of the activated sludge process

A
  1. Oxidation of biodegradeable organic matter in the aeration tank: soluble organic matter is converted to new cell biomass
  2. Flocculation: separation of newly formed biomass from treated effluent
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2
Q

What organisms are present in activated sludge flocs?:

A

bacteria, protozoa, viruses, fungi

Flocs contain aerobic, anoxic, and anaerobic zones

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3
Q

Explain bacteria in sludge flocs

A

make up bulk (~95%)

Bacterial EPS and filamentous bacteria play important roles in floc structure

Archaea (eg methanogens) are present but minor role

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4
Q

Explain protozoa in sludge flocs

A

Important in sludge ecology - graze on bacteria, role in maintaining floc size and density

Cillates - feed on bacteria, lrg numbers of stalked cillates indicate low BOD

Amoebae - feed on organic particles; lrg numbers suggest shock load

Flagellates - feed on organic particles; if lrg numbers remain in latter stages of treatment suggests high concentration of organic solutes remaining

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5
Q

What organisms other than protoza are present in activated sludge

A

Rotifers - help move free-living bacteria and ingest oocysts and contribute to floc formation

Worms: incl nematodes

Viruses: present in high numbers. Virus sludge ecology poorly understood. Some hazardous survive treatment process (rotaviruses)

Fungi: present, but not a major component

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6
Q

What is the content of a aeration tank called?

A

Mixed liquor

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7
Q

What are the units for mixed liquor suspended solids (MLSS)

A

mg/L

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8
Q

How can MLSS be determined

A

filtering known volume of mixed liquor, drying the filter at 105C, measuring the mass of dried solids and calc the concentration

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9
Q

What is “sludge age”

A

also known as ‘solids retention time (SRT)’ - residence time of microorganisms in the system - reciprocal of the microbial growth rate

Sludge age = MLSS x V / (SSe x Qe) + (SSw + Qw)
MLSS: mixed liquor suspended solids (mg/L)
V: volume of aeration tank (L)
SSe: suspended solids in wastewater effluent (mg/L)
Qe: quantity of wastewater effluent (L/day)
SSw: suspended solids in wasted sludge (mg/L)
Qw: quantity of wasted sludge (L/day)

TYPICAL SLUDGE AGE = 5 - 15 days (higher in winter)

MLSS x V = mass of solids in aeration tank
(SSe x Qe) = mass of solids leaving in effluent water per day
(SSe x Qw) = mass of solids leaving in wasted sludge per day

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10
Q

What is SRT equal to

A

SRT = mass of solids in system / mass of solids leaving system per day

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11
Q

What is HRT

A

hydraulic retention time - average time spent in influent liquid in the aeration tank

suitable contact time of wastewater with microbes is required for adequate degradation

Typical HRT ~4-12 hours

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12
Q

What is HRT equal too

A

HRT = V/Q = 1/D

HRT: (days)
V: volume of aeration tank (L)
Q: flow rate of influent wastewater (L/day)
D: dilution rate

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13
Q

What is cell yield (Y) in activated sluge)

A

“amount of biomass formed per unit of substrate removed”

Depends on:
Growth rates - faster growth rate generally higher Y
Electron acceptor - high yield with O2

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14
Q

What is the equation for cell yield in activated sludge

A

Y = X - X0 / S0 - S

Y: cell yield (mg/mg)
X: final [microbial] (mg/L)
X0: initial [microbial] (mg/L)
S0: inital [substrate] (mg/L)
S: final [substrate] (mg/L)

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15
Q

Are low growth rates desirable>

A
  1. less sludge too deal with but
  2. also means large and slow treatment facilities
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16
Q

What is the food/microorganism ratio (F/M)

A

indicates the organic load into the activated sludge system, controlled by rate of sludge wasting.

A low F/M ratio generally leads to more efficient wastewater treatment

Typical activated sludge requires F/M: 0.2 - 0.5kg BOD5/day/kg MLSS

17
Q

What is the equation for food/microorganism ratio (F/M)

A

F/M = Q x BOD5 / MLSS x V

Q = flow rate of influent wastewater (L/day)
BOD5 = 5-day biochemical oxygen demand (mg/L)
MLSS = mixed liquor suspended solids (mg/L)
V = volume of aeration tank (L)

18
Q

What are the biological processes which ake place in activated sludge?

A

Removal of carbonaceous substrate
Sludge settling
Nutrient removal (N and P)

19
Q

How does sludge settle an flocs form

A

In the clarifier (final settlement tank) solids settle and separate from the treated WW

settling requires flocculated and aggregated microbes

Growth in activated sludge must encourage floc forming organisms

20
Q

When is settling best seen?

A
  1. microbes in endogenous state of growth (straved)
  2. sludge age is at least 3-4 days
  3. there is a low F/M ratio
21
Q

What is EPS and what does it do

A

Extracellular polymeric substance (EPS) consisting of sugars, uronic acids, and amino acids. IMPORTANT FOR FLOC FORMATION

produced by microorganisms

metals bridge carboxyl groups of the polymers and link bacteria together

22
Q

What is the sludge volume index

A

SVI - can be measured to monitor the ability of sludge to settle in WWTPs - volume occupied by 1g of sludge

SVI = Vs / V0 x MLSS x 0.001

SVI = sludge volume index (mL/g)
Vs = volume of settle sludge after 30 min (mL)
Vo = volume of sludge prior to settling (L)
MLSS = mixed liquor suspended solids (mg/L)
V = volume of aeration tank (L)

Typically SVI: ~ 50 - 150 mL/g

23
Q

What are problems of sludge settling

A
  1. Bulking and foaming - prevent effective settling in clarifier and thus cause increased solids in effluent
  2. slime/viscous bulking
  3. dispersed growth
  4. Pin floc
  5. raising sludge

SEE POWERPOINT FOR MORE INFO

24
Q

Why does bulking occur?

A

Overgrowth of filamentous bacteria in activated sludge. Filamentous bacteria are normal in flocs, but can out-compete other floc forming bacteria in certain conditions

Increases SVI = biomass fails to settle

poor sludge settling observed when total extended filament length exceeds 10^7 um/mg

25
Q

What factors influence filamentous bulking

A
  1. WW composition: high carb concentration in water
  2. F/M ratios: low ratio
  3. Sludge loading and age: generally increased age (ie decreased loading)
  4. pH: optimum = ~7 -7.5, <6 favours fungi which may lead to bulking
  5. sulphide concentration: high concentrations
  6. DO levels: certain filaments favoured by low DO
  7. Nutrient deficiency: deficiencies in P, N, Fe, or trace elements
  8. Temperature: increased temp + low DO

FILAMENTOUS BACERIA HAVE HIGHER SURFACE:VOLUME RATIOS THAN FLOC-FORMING BACTERIA = BETTER ABLE TO SURVIVE LOW DO AND NUTRIENT CONDITIONS

26
Q

What are common filamentous microbes which cause bulking

A

Sphaerotilus natans (Gram -) (B proteobacteria)
Microthrix parvicella (gram+)
Type 0041 (gram variable)
Nocardia sp (Gordonia) ( gram +) (high GC)

26
Q

What are common filamentous microbes which cause bulking

A

Sphaerotilus natans (Gram -) (B proteobacteria)
Microthrix parvicella (gram+)
Type 0041 (gram variable)
Nocardia sp (Gordonia) ( gram +) (high GC)

27
Q

How can filamentous bacteria in activated sludge be identified

A

Microscopy based techniques: based on gram stain, filaments: size, shape, branching, motility, attachment to other bacteria, sheath, S, PP, or PHB granules. Identification via dichotomous key

Fluorescent antibody and hybridisation techniques: bother fluorescent antibody and hybridisation (FISH) techniques. Fluoro antibodies can be used for identification of Sphaerotilus natans, Microthrix parvicella and Thiothrix spp. >20 FISH probes available which detect filamentous bacteria in sludge

16s rRNA sequence analysis: particuarly suitable for identifying uncurable microorganisms. DOes not rely on morphological characteristic or culture based techniques

28
Q

what is FISH

A

Fluorescence in-situ hybridisation

Uses fluorescently labelled DNA probes that complement target rRNA sequences.

Probes are designed to target specific microbial groups

29
Q

How can FISH be used to identify causal agents of bulking

A
  1. bulking smaple
  2. analyse DNA from sample and design probes for FISH
  3. FISH can confirm the common causal agent for bulking
30
Q

What is FISH-Microautoradiography (FISH-Mar)

A
  1. incubate sample with radioactive substrate
  2. fix samples and section
  3. fluorescently labelled probes used to identify target organisms
  4. FISH-probed samples are treated with radiolabelled substrate and a photographic emulsion
  5. Microscopy and image analysis
31
Q

How can you control bulking

A
  1. raise DO levels in aeration basin
  2. Treatment with oxidants (eg selective chlorination)
  3. raise F/M ratio
  4. Biological selectors ( a tank or compartment with controlled paramaters, eg aerobic, anoxic, anaerobic)
32
Q

Why is foaming undesirable?

A
  1. can overflow onto walkways = slippery
  2. may pass effluent into sludge effluent resulting in increased BOD and SS
  3. Interferes with anaerobic digestion of sludge
33
Q

What causes foaming

A
  1. undergraded surface-active organic compounds
  2. poorly degradedable detergents
  3. scums due to rising sludge and denitrification in the clarifier
  4. brown scumes due to excessive growth of filamentous bacteria

Mainly associated with mycolic acid-containing actinomycetes belonging to nocardiaceae (eg Nocardia) and Microthrix which may produce biosurfactants

Foams are often associated with long HRTs (>9dats) warm temps *>18C) and WW rich fats

34
Q

How can foaming be controlled

A

chlorination of RAS
increased sludge wasting
biological selectors
reduction of air flow in aerobic tank
physical removal

Microbial Biotechnology (BIO-6004A) (17 decks)

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