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Wastewater Treatment > Introduction and Definitions > Flashcards

Introduction and Definitions Flashcards

1
Q

Why do we need biological wastewater treatment

A

Micro-organisms (predominantly bacteria) can grow on polluting substances in wastewater and they consume the pollutants

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2
Q

What does the breakdown of polluting components lead to

A

They lead to the growth of new microorganisms:
- Source of C, N and P as building blocks for new cell biomass (anabolism)
- source of energy to build these building blocks to make new biomass (catabolism)
> energy comes from breakdown of polluting substances
> most energy inferred from aerobic break down

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3
Q

What is the use of the catabolic process in biological wastewater treatment

A
  • Organic matter is broken down for ATP generation
  • Redox reaction occurs where carbon gets oxidised and oxygen gets reduced
  • The break down of pollutants breaks chemical bonds (in glucose) and releases this energy
  • The released energy is stored in ‘energy-rich’ molecules known as ATP
  • the breaking of the bonds lead to the electrons and hydrogen being shuttled to other molecules as the final electron acceptor:
    > O2 in aerobic conditions which produces 38 ATP
    > NO3- in anoxic conditions which produces 2-38 ATP
    > CHO in anaerobic conditions which produces 2 ATP
  • therefore the more oxygen available the more ATP that can be produced
  • the ATP molecules can afterwards be used to create new bonds to generate and build in the building blocks
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4
Q

What are the types of agents required to build oxygen atoms in the redox process

A
  • a chemical substance (strong oxidant) -> COD
  • a biochemical substance (activated sludge) -> BOD
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5
Q

General scheme for biological wastewater treatment

A

Raw sewage -> screening -> grit chamber -> primary clarifier -> aeration tank -> secondary clarifier (activated sludge returned to aeration tank or disposed) -> disinfectant -> discharge

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6
Q

What is the primary treatment process

A
  • Wastewater removed from the sewer or factory has to be pretreated to remove coarse particles (e.g. sand), oils and fats
  • Raw sewage -> screening -> grit chamber -> primary clarifier
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7
Q

What is the secondary treatment process

A
  • Carbon, nitrogen and phosphorous (nutrients) are removed by the activated sludge in the biodegradation tanks
  • Activated sludge is separated from the purified water in the sedimentation tanks
  • The largest part of the settled activated sludge is returned to the biodegradation tanks (recycle sludge) and the excess activated sludge is removed
    -aeration tank -> secondary clarifier (activated sludge returned to aeration tank or disposed)
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8
Q

What is the tertiary treatment process

A
  • The purified water is discharged into receiving water bodies or is further treated to upgrade its quality
  • disinfectant -> discharge
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9
Q

What is activated sludge

A
  • Activated sludge is not a pure culture but a very (unknown) mix of:
    > bacteria (morphologically dividable in floc and filament forming bacteria)
    > protozoa e.g. ciliates, flagellates (which eat up bacteria) and amoebes
    > metazoa e.g. rotifers, mematodes and worms
  • It is not an active culture
    > if active there is too much growth and no need to create extra biomass
    > When there is insufficient energy supply (starvation) oxidation of cell reserves occurs which leads to sludge mineralisation and lysis
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10
Q

What are the ideal demands for the activated sludge system

A
  • establishment of a microbial community that breaks down completely and at a fair rate compared to the incoming waste
  • little new biomass formed as possible
  • settles well in the sedimentation tank
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11
Q

What are the basic reactions that occur in an activated sludge system

A
  • Sorption of soluble, colloidal and suspended organics in or on sludge flocs
  • Biodegradation of the organics with end products
    > CO2, H2O and minerals
    > new microbial biomass
  • Ingestion of bacteria etc. by protozoa or other predators
  • Oxidation of ammonium to nitrite and nitrate by nitrifying bacteria
  • When there is insufficient energy supply (starvation) oxidation of cell reserves occurs which leads to sludge mineralisation and lysis
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12
Q

What are the measures of organic pollution in an effluent

A
  • Chemical Oxygen Demand
  • Biological Oxygen Demand
  • Total Organic Carbon
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13
Q

What is Chemical Oxygen Demand

A

The amount of oxygen required to oxidise organic carbon completely to CO2 by chemical means
- a strong oxidants is required such as K2Cr2O7
- the amount of dichromate oxygen used is determined and expressed as COD in mg O2/L
- for 1g of glucose we need 1.07g of COD to break it down (calcs)
- Oxygen to oxidise reduced nitrogen components to nitrate are not included
- Oxygen to oxidise reduced sulfur components to SO42- are not included

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14
Q

What is Biochemical Oxygen Demand

A

The amount of oxygen used by the non-photosynthetic micro-organisms at 20 degrees celcius to metabolise biologically degradable organic compounds
COD >= BOD
(graph)

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15
Q

What is the appropriate environmental conditions for measuring BOD

A
  • neutral pH
  • sufficiently large acclimatised microbial inoculum
  • appropriate amounts of necessary mineral nutrients = N, P, Ca, Mg, Fe and S
  • incubation in the dark -> avoiding interference with O2 producing organisms during daylight periods
  • no nitrification due to the addition of inhibitors
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16
Q

What is the measurement process for BOD

A
  • fill glass bottles with wastewater at different dilutions to make sure there is enough oxygen left at the end to measure
  • measure DO conc at T = 0
  • store bottles in the dark at 20 degrees C
  • measure the conc again after 5 days
  • calculate the difference between the two oxygen concentration values = BOD5
17
Q

How can BOD me measured from COD and why

A

-COD is much faster (2 hrs)
- BOD5 = 0.65 x COD for municipal (non-industrial) wastewater if all COD is biodegradable (if not all biodegradable only relates to biodegradable part)
- Graph of biomass and time can be split into two phases
> 1-2 days of ‘feast; and growth on external substrate where 50% of COD is transformed to CO2 and energy and 50% is transformed into biomass; the amount of new biomass that is formed in those 2 days is 40% of the COD and is expressed in ‘dry weight’ of biomass
> the remaining 3 to 4 days the biomass is under starvation and there is no external substrate anymore so they start consuming themselves and there is a decline in biomass as oxygen consumptions is about 0.1 mg/LO2 per gram of dry weight of biomass per day

18
Q

What is BODu

A
  • after 20 days the ultimate BOD is measured
  • this is when all material that is biologically degradable has been biodegraded (bCOD)
  • BOD5 = COD x Fb (biodegradable fraction of COD) x 0.65 = bCOD x 0.65
19
Q

What is TOC

A
  • firstly acidification and purging of Inorganic Carbon (IC) measures the formed CO2
  • then combustion of total carbon (TC) at 800 degrees c measuring the formed CO2
  • TOC = TC - IC
  • of the formed CO2 only the amount of C is counted as TOC mg C/l
20
Q

What are the measures for suspended solids/activated sludge concentrations

A
  • TS (total solids)
  • SS (suspended solids)
21
Q

What are the total solids

A

The portion of wastewater dried at 105 degrees C to constant weight

22
Q

What are the suspended solids

A
  • the amount of particulate matter present in a sample
    -determined by separating the particulates out of the water by wither filtration or centrifugation and drying the residue at 105 degrees C to constant weight
  • expressed in dry weight
  • TS >= SS
23
Q

What are the types of suspended solids

A
  • Volatile Suspended Solids
  • Mixed Liquor Suspended Solids
  • Mixed Liquor Volatile Suspended Solids
  • Effluent Suspended Solids
24
Q

Volatile Suspended Solids

A
  • suspended solids, separated out of the sample, are dried and subsequently ashed at 600-650 degrees C
  • the amount of ash is subtracted from the total amount of SS; VSS = SS -ash
  • indicates the amount of activated sludge
25
Q

Mixed Liquor (Volatile) Suspended Solids

A
  • mixed liquor is the mixture of sludge and water in the mixed aeration basis
  • mixed liquor suspended solids is the total amount of sludge present in the mixed liquor
  • a better measure is the mixed liquor volatile suspended solids (65-75% of the SS are organic = VSS)
26
Q

Effluent Suspended Solids (ESS)

A

Amount (concentration) of SS in effluent

27
Q

sCOD

A
  • soluble COD (filtered COD) is the ESS
  • determined by filtering sample as for SS calculation
  • COD determination of the filtrate to avoid interference with COD of activated sludge sample
28
Q

What are the measures of nutrient concentration

A
  • Total nitrogen, Kjeldahl nitrogen,…
  • Total phosphorus, phsophates
29
Q

How is phosphorous content calculated

A
  • there are inorganic (e.g. orthophosphate PO43-) and organic components
  • analysis of orthophosphate concentration is calculated after oxidative destruction of an acid environment of the organic molecules to give the total phosphates
  • organic phosphates = total phosphates - original orthophosphate
30
Q

How is nitrogen content calculated

A
  • nitrogen comes in different inorganic forms such as ammonium, nitrite and nitrate
  • in an acidic environment organic nitrogen is transformed into NH4+
  • in alkaline environments the NH4+ is converted into NH3
  • the analysis of the NH3 results in the Kjeldahl-N
  • organic nitrogen = Kjeldahl-N - original NH4+
31
Q

Inhabitant Equivalent

A
  • the amount and specifications of ww that 1 inhabitant produced per day
  • loading:
    > 54g BOD/d or 300 mg/L
    > 135g COD/d or 750 mg/L
    > 90g SS/d or 500 mg/L
32
Q

Removal/Treatment Efficiency

A

Exxx = ((Co-Ce)/Co)x100
where
removal efficiency of parameter xxx in %
Co = influent amount/concentration
Ce = effluent amount/concentration
XXX = COD, BOD, SS, total N, total P,…

33
Q

Sludge Loading Rate (Bx)

A
  • the amount of substrate given each day to the bacteria to metabolise
  • the substrate is usually expressed as COD or BOD (bCOD = most meaningful)
  • the food/micro-organism ratio
  • Bx = +/- 0.15 kg BOD5/kg MLSS/d
    = +/- 0.25 kg bCOD/kg MLSS/d (biomass/activated sludge)
34
Q

Volumetric/Organic/Reactor loading rate (Bv)

A
  • the amount of substrate introduced per m3 reactor and per day
  • Bv = Bx * Cx where Cx is the activated sludge concentration
  • Bv = +/- 1 kg bCOD/m3/d
35
Q

Hydraulic Retention Time (HRT)

A
  • the time the water resides in the aeration system is
  • HRT = V reactor [m3] / Q [m3/h]
  • varies from 8 hours to several days
36
Q

Sludge Retention Time (SRT)

A
  • the average time the sludge resides in the biodegradation tanks
  • often kept around 20 days
37
Q

How can organisms be classified

A
  • Carbon Source
    > Autotrophic: C from CO2 in air or (bi)carbonates in the water
    > Heterotrophic: C from organic components
  • Energy Source
    > e-donors which can be phototrophic organism (i.e. e- from light) or heterotrophic (i.e. e- from biological redox) which is either chemo-litotrophic (inorganic components) or chemo-organotrophic (organic components)
    > e-acceptors which can be from aerobic respiration (O2), anoxic respiration (SO4-2 or NO3-) or fermentation (organic components)

Wastewater Treatment (10 decks)

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