Introduction on biological wastewater treatment

Question 1

what are the major sources of wastewater?

Answer 1

human sewage and process waste from manufacturing industries

Question 2

when was the method of biological treatment devised?

Answer 2

In the early years of the twentieth century

Question 3

Give short notes about the biological wastewater treatment idea?

Answer 3

It simply involves confining naturally occurring bacteria at very much higher concentrations in tanks. These bacteria, together with some protozoa and other microbes, are collectively referred to as activated sludge. The concept of treatment is very simple. The bacteria remove small organic carbon molecules by ‘eating’ them. As a result, the bacteria grow, and the wastewater is cleansed. The treated wastewater or effluent can then be discharged to receiving waters – normally a river or the sea.

Question 4

why controlling biological wastewater treatment is very complex?

Answer 4

because of the large number of variables that can affect it. These include changes in the composition of the bacterial flora of the treatment tanks, and changes in the sewage passing into the plant.
The influent can show variations in flow rate, chemical composition and pH, and temperature. Many municipal plants also have to contend with surge flows of rainwater following storms. Those plants receiving industrial wastewater have to cope with recalcitrant chemicals that the bacteria can degrade only very
slowly, and with toxic chemicals that inhibit the functioning of the activated sludge bacteria. High concentrations of toxic chemicals can produce a toxic shock that kills the bacteria. When this happens the plant may pass untreated effluent directly to the environment, until the dead bacteria have been removed from the tanks and new bacterial ‘seed’ introduced.

Question 5

what are the regulatory legislation agencies concerned with?

Answer 5

The legislation is concerned with the prevention of pollution and therefore sets concentration limits
on dissolved organic carbon (as BOD or COD), nitrogen and phosphates – which cause eutrophication in receiving waters. It also attempts to limit the discharge of known toxic chemicals by setting allowable concentration limits in the effluent.

Question 6

what are Direct Toxicity Assessment (DTA) tests and Whole Effluent Toxicity (WET) tests are used for?

Answer 6

are used to measure the toxic effects of effluents on
representative organisms from the receiving waters. Any toxicity detected in the effluents will obviously have been present in the sewage entering the plant.

Question 7

what is the composition of domestic sewage?

Answer 7

Domestic sewage is made up largely of organic carbon, either in solution or as particulate matter. About 60% is in particulate form, and of this, slightly under a half is large enough to settle out of suspension. Particles of 1nm to 100µm remain in colloidal suspension and during treatment become adsorbed on to the flocs of the activated sludge.

Question 8

what does the bulk of the organic matter contain?

Answer 8

proteins, amino acids, peptides, carbohydrates, fats, and fatty acids. The average carbon to nitrogen to phosphorus ratio (or C : N : P ratio) is variously stated as approx 100: 17 : 5 or 100 : 19 : 6. This is close to the ideal for the growth of the activated sludge bacteria. However, industrial wastewaters are very much more variable in composition. Those produced by the brewing, and pulp and paper industries, for example, are deficient in nitrogen and phosphate. These nutrients need to be added therefore to achieve the correct ratio for microbial growth and to allow treatment to proceed optimally.

Question 9

what are the measures used to know the organic

strength, or organic load, of the influent wastewater?

Answer 9

The Total Organic Carbon (TOC), the Chemical Oxygen

Demand (COD), and the Biological Oxygen Demand (BOD5)

Question 10

Describe The Total Organic Carbon (TOC) measure?

Answer 10

it involves oxidation by combustion at very high temperatures and measurement of the resultant CO2. However, TOC values include those stable organic carbon compounds that cannot be broken down biologically.

Question 11

Describe chemical oxidation?

Answer 11

It involves heating the sample in strong sulphuric acid containing potassium dichromate, and the carbon oxidized is determined by the amount of dichromate used up in the reaction. The result is expressed in units of oxygen, rather than carbon, and the procedure is referred to as the Chemical Oxygen Demand (COD).
The weakness of the is measure is that a number of recalcitrant organic carbon compounds that are not biologically oxidizable are included in the value obtained. Conversely, some aromatic
compounds, including benzene, toluene, and some pyridines, which can be broken down by bacteria, are only partly oxidized in the COD procedure. Overall, however, COD will overestimate the carbon that can be removed by the activated
sludge

Question 12

Describe the 5-day Biological Oxygen Demand (BOD5) measure?

Answer 12

This is a measure of the oxygen uptake over a 5-day period by a small ‘seed’ of bacteria when confined, in the dark, in a bottle containing the wastewater. During this time the biodegradable organic carbon is taken up, and there is a corresponding decrease in the dissolved oxygen, as some of the carbon is used for the respiration of the bacteria. Respiration is a form
of biological oxidation and will be explained later. Rather unhelpfully, the biodegradable carbon, as in the COD test, is expressed in oxygen units. This is because the test was originally introduced to measure the oxygen depletion in receiving waters caused by the residual degradable carbon in the effluent. Its main value is in regulating the composition of effluents from the treatment water. For process management, where knowledge of the organic loading of the influent is
required, BOD5 is of limited value, because of the 5 days required to make the measurement.

Question 13

what measure is suggested to replace the use of BOD5?

Answer 13

The short-term test (BODST), which can be carried out

over a timescale of 30 minutes to several hours.

Question 14

Why are the values obtained for BOD5 are always lower than those for COD?

Answer 14

Because:
1- Activated sludge bacteria cannot degrade some of the compounds oxidized chemically in the COD test.
2- Some of the carbon removed during the BOD test is not oxidized, but ends up in new bacterial biomass. So the BOD is only measuring the biodegradable carbon that is actually oxidized by the bacteria.

Question 15

Describe the dependency of The ratio of BOD5/COD on the composition of the wastewater?

Answer 15

For domestic sewage, wastewaters from the slaughterhouse, dairy, distillery, and rubber industries, the ratio is about 0.5 - 0.6. However, for effluent leaving the treatment plant, it is closer to 0.2. This is because the readily biodegradable organic carbon has been removed during treatment, leaving behind the compounds that are not readily broken down by the bacteria – ‘hard’ BOD. These will be readily measured by chemical oxidation, but will not be readily degraded and removed by the bacteria in the BOD bottle.

Question 16

what is meant by ‘‘soft BOD’’?

Answer 16

It is the group of small molecular weight compounds that will start to be removed from the sewage immediately after it has entered the activated sludge tanks. Their removal may be completed in 1 – 2 hours.

Question 17

what is meant by ‘‘hard BOD’’?

Answer 17

It is the group of compounds that are more recalcitrant to degradation and may still be present after several days.

Question 18

what are the dominant organisms in activated sludge?

Answer 18

they are bacteria of which there may be 300 species present. Each comprises a single cell varying in size from about 0.5 – 2 µm

Question 19

what is the most common form of bacteria?

Answer 19

Spherical, but some may be rod-shaped or spiral.
Filamentous bacteria have long chains of small bacterial cells, sometimes surrounded by a tubular sheath and can reach lengths of 100µm.

Question 20

what is secretion?

Answer 20

In this process, small molecular weight compounds diffuse into the bacteria (ingestion) through the cell wall. At the same time, some larger complex molecules that have been synthesized within the bacteria, pass outwards.
The secretions include slimes and gels, that may bond the bacteria together, and also enzymes. The enzymes break down large organic molecules into smaller monomers that are small enough to be ingested.

Question 21

How do the bacteria grow?

Answer 21

The bacteria use the ingested molecules for the synthesis of new molecules, in the process of growth. When they have reached normal size, the bacterium divides into two, and the process is repeated. If nutrient molecules are not limiting, this results in exponential growth in the numbers of bacteria.

Question 22

What types of bacteria are involved in wastewater treatment?

Answer 22

heterotrophs and autotrophs
The heterotrophic or carbonaceous bacteria are the predominant group of organisms. They are characterized by feeding mainly on organic carbon molecules rather than inorganic ones. By contrast, the autotrophs take in inorganic chemicals and use these in the synthesis of organic compounds. The nitrifying bacteria that remove ammonia from the wastewater are the most important of this group. There are relatively few species of autotrophs, and since they have low growth rates, they tend to be out-competed by the faster-growing heterotrophs.

Question 23

where are the bacteria are mainly concentrated?

Answer 23

In the flocculent material of the activated sludge.

Question 24

what are the flocs are formed from?

Answer 24

aggregates of non-living organic polymers

that are probably secreted by bacteria.

Question 25

Describe bacterial flocs?

Answer 25

They have an open porous structure and are sufficiently robust to withstand the shear forces created by water movement, during aeration of the tanks. They vary in size from less than 10 µm up to 1mm (1000 µm).

Question 26

what advantage arises when the fine particulates, colloidal particles, and large molecules, become entangled with, and adsorbed to, the floc material?

Answer 26

The enzymes that are secreted by the bacteria into the water, will tend to be confined in the vicinity of the substrate, thereby facilitating their digestion.

Question 27

What is the problem for the bacteria living inside the floc?

Answer 27

Oxygen availability, because oxygen has to diffuse along a concentration gradient from the wastewater through the floc material to the inside.

Question 28

what is the minimum value of oxygen concentration of the mixed liquor required for the bacteria of disaggregated flocs to continue to grow?

Answer 28

only 0.6 mg O2/l

Question 29

what is the minimum value of oxygen concentration of the mixed liquor required for the bacteria inside a large floc?

Answer 29

1.2 – 2.0 mg O2/l

Question 30

what happens when the aeration tank is operated at below 2.0 mg O2/l?

Answer 30

the center of the flocs may become oxygen-depleted, and colonized by facultative anaerobic bacteria.

Question 31

what colonize the outer surface of the activated sludge flocs?

Answer 31

microorganisms of a higher trophic level, including

protozoa and rotifers, that feed on bacteria and particulate material in the wastewater.

Question 32

what is “adaptation” or acclimation?

Answer 32

It is the process of change of dominant bacterial species sometimes on a daily basis, in response to changes in the composition of the wastewater. Those species of
bacteria that have the ability to secrete the enzymes to break down a novel food source will grow more rapidly, thereby increasing in relative numbers.

Question 33

what are the major divisions of metabolism?

Answer 33

1- Catabolism or Energy Metabolism

2- Anabolism

Question 34

describe Catabolism or Energy Metabolism?

Answer 34

It comprises a series of reactions in which carbon compounds, are broken down to yield cellular energy. This is biological oxidation and involves oxygen uptake by the bacterium. This is also the basis of the process referred to as Respiration.

Question 35

describe Anabolism?

Answer 35

It is a series of biosynthetic reactions in which small molecules are joined together to form large molecular
weight macromolecules. This requires an input of energy from Catabolism and is the basis of the process of Growth.

Question 36

what are the three major processes in a bacterium?

Answer 36

1- Ingestion
2- Respiration
3- Growth and division
These processes are very highly integrated

Question 37

what is the purpose of respiration?

Answer 37

is to provide the energy that is required for growth and for the maintenance of the bacterium

Question 38

what is the implication of ingestion, respiration, and growth being highly integrated?

Answer 38

measuring the respiration rate means that you are indirectly also measuring the rate of growth and the rate of carbon ingestion.

Question 39

write short notes about ingestion?

Answer 39

It involves the passage of organic carbon compounds, other molecules and ions from the mixed liquor into the bacterium.
Small organic molecules similarly pass in along a concentration gradient or may be assisted in entry
by various mechanisms located in the inner membrane
In order to use large molecules for their nutrition and growth, the bacteria secrete enzymes into the water to digest them into small monomers, which can then pass into the cell.
Different species of bacteria are specific for what enzymes they secrete, and this determines which
chemicals they can exploit as a food source. The ability to secrete a particular enzyme may be latent. In other words, the bacterium requires the presence of the particular chemical compound in the water to switch on the genes for the synthesis of the enzyme required for its digestion

Question 40

How is the bacterial growth rate is measured?

Answer 40

it is measured as the increase in the number of cells with time.

Question 41

What factors affect the growth rate?

Answer 41

Genetic and environmental factors

Question 42

what are environmental factors affecting the growth rate of bacteria?

Answer 42

1- Substrate concentration
2- Availability of other nutrients
3- Oxygen
4- Temperature
5- Toxicity

Question 43

What happens when the substrate concentration increases?

Answer 43

the growth rate increases exponentially and then levels off. So with further increase in the concentration of substrate in the medium, there is no further increase in growth. The bacteria are at their maximum growth rate

Question 44

what happens to the substrate at very low concentrations?

Answer 44

It is being used for respiration simply to keep the bacterium alive.

Question 45

What is the optimum ratio of C:N:P in the mixed liquor?

Answer 45

It is generally thought to be 100: 5:1.

The ratio of these nutrients in settled domestic sewage is variously reported as 100 : 17: 5 or as 100:19 : 6.

Question 46

what are other minor components that are required in wastewater?

Answer 46

Trace components, which include S, Na, Ca, Mg,

K, and Fe and they are available in abundance in domestic sewage.

Question 47

what is the availability of N and P in wastewater from brewing, pulp and paper, and food-processing industries?

Answer 47

wastewater from these sources can be deficient in N and P. Nutrients, therefore, need to be added to the mixed liquor to obtain maximum bacterial growth and to optimize carbonaceous treatment.

Question 48

what is the viable temperature for most carbonaceous bacteria of the activated sludge?

Answer 48

From about 0 to 30°C.

However thermophyllic bacteria survive and grow between about 30°C and 60°C

Question 49

what does the rule of Arrhenius state?

Answer 49

it states that chemical reactions double in rate for a 10°C increase in temperature.

Question 50

what happens during respiration?

Answer 50

the energy is initially captured by the molecule adenosine diphosphate (ADP). This adds on another phosphate group to form adenosine triphosphate (ATP). The energy that is captured or transferred is stored in what is sometimes called a ‘high energy phosphate bond’.
The ATP then moves to another site within the cell and releases the energy to do work. At the same time, the phosphate group is released, regenerating ADP again

Question 51

How could the rate of respiration be measured?

Answer 51

by the rate of oxygen uptake, by the rate of CO2 production or by the rate of heat liberation. Carbon d be measured in a calorimeter, but the simplest measure of respiration rate is by measuring the oxygen uptake rate with an activated sludge respirometer.

Question 52

what is the main use of energy in bacteria?

Answer 52

It is for biosynthesis for growth, which involves joining together of small molecular weight compounds to form macromolecules, which may then be further modified and assembled to form structures such as membranes, cell walls, etc.

Question 53

what happens when bacteria grow, in terms of heat?

Answer 53

when they grow they release heat, and this causes the temperature of the aeration tank to be above ambient air temperature.

Question 54

Explain endogenous respiration?

Answer 54

When all of the biodegradable carbon in the mixed liquor has been used up, as may happen at the end of a plug flow reactor, growth ceases, and the bacterium is then starving. In order to remain alive it still requires energy for maintenance processes. It therefore starts
to metabolize its storage products (glycogen and poly-β-hydroxybutyrate (PHB)) to provide this energy

Question 55

what is endogenous respiration rate?

Answer 55

it is a low rate of respiration continues in order to provide the energy for maintenance.

Question 56

how is the respiration distributed in a growing bacterium in the aeration tanks?

Answer 56

part of its respiration is to provide the energy used in biosynthesis and growth, but there is still a small component that is being used in cell maintenance.

Question 57

compare filamentous bacteria to floc bacteria in terms of tolerance of low oxygen levels?

Answer 57

Filamentous bacteria have a greater tolerance

of low oxygen levels than floc bacteria.

Question 58

what happens to the solubility of oxygen with temperature increase?

Answer 58

It decreases with the increase in temperature.
One consequence of this is increase in the critical oxygen concentration value. Optimum aeration, therefore, becomes more and more difficult as the
temperature in the tanks rises. It is for this reason that most thermophilic plants, operating at 40-60ºC, have to use pure oxygen for aeration.

Question 59

Describe nitrifying bacteria?

Answer 59

They are autotrophs, requiring only inorganic chemicals as the starting point for their energy metabolism and growth.

Question 60

what do nitrifying bacteria do in wastewater?

Answer 60

they take up and oxidize ammonia to provide the energy required for growth. Carbon dioxide is used as the carbon source, and this is metabolized into organic carbon compounds inside the bacteria - a process which also requires energy.

Question 61

how is the nitrifying bacteria in terms of number and contribution to the total bacterial biomass?

Answer 61

There are relatively few species of nitrifiers, and their

contribution to the total bacterial biomass is small.

Question 62

what is nitrification?

Answer 62

It is the process of ammonia oxidation.

Question 63

how does nitrification take place?

Answer 63

carried out by two different groups of nitrifiers. The
first group oxidizes ammonia to form nitrite. The most abundant genus is Nitrosomonas but there are other nitrifiers as well.
The oxidation of nitrite to nitrate is carried out by Nitrobacter and by other minor species

Question 64

can nitrification be referred to as respiration?

Answer 64

No, nitrification (ammonia oxidation) should not be referred to as respiration, although oxygen is consumed and the purpose-energy capture is the same. Respiration is the process found in carbonaceous bacteria and in plants and animals.

Question 65

How is Chemical oxidation by nitrifiers compared to the process of respiration in heterotrophs in terms of efficiency?

Answer 65

Chemical oxidation by nitrifiers is not as efficient as the process of respiration in heterotrophs. Relatively large amounts of oxygen are required per unit of energy produced. They, therefore, have a greater relative oxygen requirement and are slow growing.

Question 66

what is the temperature tolerance of nitrifiers ?

Answer 66

small one , from c 8°C - 30°C

Question 67

what is the metabolic rate of nitrifiers?

Answer 67

low metabolic rate below 15-20ºC.

Question 68

what is the critical oxygen concentration of nitrifiers?

Answer 68

high critical oxygen concentration of c 2.0 - 2.5 mg O2/l at 20°C

Question 69

what is the effect of toxicity on nitrifiers in comparison to carbonaceous bacteria?

Answer 69

They are also very much more susceptible to inhibition by toxic chemicals than carbonaceous bacteria, the nitrite-oxidizing species more so than the ammonia-oxidizing species.

Question 70

What are the types of aeration tanks?

Answer 70

There are many designs of the aeration tank, including:
1- plug-flow (the simplest one)
2- completely mixed
3- percolating filter
4- sequencing batch reactor (SBR), and so on

Question 71

what is mixed liquor?

Answer 71

It is a combination of wastewater and activated sludge.

Question 72

By the end of the aeration tanks, how much influent BOD should have been removed?

Answer 72

90-95% of the influent BOD should have been removed.
If it reaches this state before the point of discharge, there is an under-utilization of capacity. Conversely, if it has not reached this at the outlet, the effluent will be discharged to the environment with significant BOD still untreated and may cause eutrophication

Question 73

How much Return Activated Sludge is pumped back to the inlet

Answer 73

This normally represents about 25-50% of the flow through the aeration tank,

Question 74

what are the features of an ideal biological treatment plant?

Answer 74

1- Fast throughput of sewage
2- High rate of BOD removal
3- Good settlement of sludge in the clarifier
4- Low rate of sludge production
5- Minimal aeration costs
6- High quality effluent - low in BOD suspended solids, etc.

Question 75

what do high BOD concentrations result in?

Answer 75

They result in high growth rates and a high rate of biomass or sludge production. Operationally, it may cause aeration problems and will result in poor sludge settling characteristics.

Question 76

what is the problem with high sludge production?

Answer 76

it incurs high costs in dewatering, drying, and removal

Question 77

what do low BOD concentrations result in?

Answer 77

very little sludge will be produced, and it is characterized by having good setting characteristics.
The downside of this mode of operation, of course, is that the overall treatment time is increased and so is the total amount of aeration required.

Question 78

what are the main types of biological treatment plants?

Answer 78

1- High rate
2- Conventional
3- Low rate

Question 79

what is a high rate biological treatment plant?

Answer 79

It is used for pre-treatment or partial treatment of high BOD Sewage, as in some pharmaceutical and dairy wastes, etc.

Question 80

what is a conventional biological treatment plant?

Answer 80

It works with medium rate, characteristic of most municipal treatment works.

Question 81

what is a low rate biological treatment plant?

Answer 81

it works with Low BOD loading, and characteristic of small, extended aeration works and oxidation
ditches and lagoons, occupying a large land area.

Question 82

What is MLSS?

Answer 82

Mixed Liquor Suspended Solids
It is a measure of the biomass by filtering and drying
a sample of the suspended solids, and then weighing the dried residue.

Question 83

What is MLVSS?

Answer 83

Mixed Liquor Volatile Suspended Solids
It is a measure of the biomass by combusting the dried residue in a furnace at 500°C, reweighing, and obtaining the volatilized organic matter, by subtraction.

Question 84

what is the shortcoming of MLSS?

Answer 84

under some circumstances, a significant proportion of the MLSS may be an inorganic material.

Question 85

what is the shortcoming of MLVSS?

Answer 85

a significant part of the floc comprises inert organic matter.

Question 86

What are the values of MLSS?

Answer 86

They range from about 800 - 1,500 mg/l for extended-aeration and other low-rate systems to about 8,000 mg/l or more, for high-rate systems.

Question 87

what is “Hydraulic retention time” or “volumetric loading”?

Answer 87

This is the average time spent by the influent sewage in the aeration tank.
HRT = V/Q x 24 hours
where V is the tank volume m3
Q is flow rate m3/d
the higher the inflow rate Q, the sooner the sewage influent will reach the outlet and therefore the lower will be the residence time or hydraulic retention

Question 88

what are settlement problems?

Answer 88

1- high rate treatment causes poor settlement
2- nutritional imbalance
3- changes to the microbial components of the activated sludge
4- the presence of toxic compounds in the mixed liquor.

Question 89

how is the settlability of the sludge in the clarifier is measured?

Answer 89

by the sludge volume index (SVI) - the volume of sludge occupied by 1g of MLSS in a standard measuring cylinder, after 30 minutes of quiescent settling.
or by the sludge specific volume index (SSVI) in which the volume occupied is measured in a specially designed cylinder that incorporates a weak horizontal stirring, to mimic the nonquiescent conditions experienced in a clarifier.

Question 90

what is the value of HRT in a conventional activated sludge system?

Answer 90

between 5 and 14 hours

Question 91

what are the regular values of sludge age (SRT or ts)

Answer 91

they may vary from < 0.5days, in a very high-rate system, to 75 days in low growth-rate systems, such as extended aeration systems. In a conventional plant, SRT would normally be between 3 - 4 days.

Question 92

what is meant by “Sludge loading” or “f/m ratio”

Answer 92

It is calculated as the daily flow of BOD divided by the total MLSS in the aeration tank (derived from the product of the MLSS and the volume of the aeration tank)

Question 93

what are the values for f/m?

Answer 93

they range from about 0.5 to 1.0
For conventional plants, an f/m of between 0.2 and 0.5 is usually aimed for. At higher values, the rate of treatment increases, but at the cost of poor settlability of the sludge. f/m values below 0.2 are associated with slow BOD removal rates, but with very good sludge settlement

Question 94

what is instantaneous f/m, or BODST/B?

Answer 94

It is another measure of sludge loading, at any point in the tank. It is not yet in general use. Probably because of the lack of a rapid and effective way to measure BODST. However, with the availability of rapid respirometers such as Strathtox, this management tool may become more widespread.

Question 95

what are the causes of deflocculation?

Answer 95

1- failure of bacteria to form flocs
2- breakup of flocs due to the severe aeration turbulence
3- inadequate aeration (which causes low dissolved oxygen concentration/ high sludge loadings)
4- low pH
5- The presence of certain toxic chemicals in the influent.

Question 96

what is meant by “pin-point floc”?

Answer 96

It is the case where the flocs are very small and compact. Since they have little extracellular polymeric substance (EPS) to bond them together, they readily break into smaller units as they grow, due to the shear forces exerted by aeration.
It is often associated with long treatment times, e.g. in extended aeration systems where the
sludge age is above 5-6 days and the f/m ratio is very low. However, it can also appear in high-rate plants treating chemical or pharmaceutical waste, where it is thought to be a consequence of toxic chemicals inhibiting the growth of the filamentous bacteria that would otherwise assist floc formation.

Question 97

what causes foaming?

Answer 97

is sometimes associated with the presence of non-degradable detergents in the wastewater. This causes a light frothy foam

Question 98

what causes mousse formation,?

Answer 98

it is a more intractable type of foaming and it is caused by filamentous fungal hyphae of the genus Nocardia which bind the foam into a dense blanket. This can trap the activated sludge flocs. Thus foaming can result in a loss of MLSS from the return activated sludge.

Question 99

how can mousse formation be overcome?

Answer 99

the solution is to reduce the sludge age by increasing the waste sludge flow, thereby washing the Nocardia out of the system. Scum traps are usually unable to cope with the volume of foam produced, and anti-foaming chemical treatment and surface chlorination is sometimes used.