Wastewater Secondary Treatment Process

Diagnostic Exam

Fixed Film Reactors

• In fixed film reactors, microorganisms, snails, algae, and flies attach to media and feed on the organic material in the wastewater.

Wastewater Discharge

• Excessive wastewater discharge to receiving water can result in oxygen depletion.

Preliminary Wastewater Treatment

• Ineffective preliminary wastewater treatment can damage pumps and collector mechanisms and plug pipes.

Primary Settling Tank

• A typical weir overflow rate for a primary settling tank is 15,000 \text{ gpd/ft}. (gpd = gallons per day)

Gas in Primary Settling Tanks

• Gas produced in primary settling tanks is problematic because it can be toxic to microorganisms, cause short-circuiting of flow, buoy sludge and cause solids to rise, and reduce detention time.

Microorganism Clumping

• The process by which microorganisms clump together and become heavy enough to settle is called flocculation.

Activated Sludge Process

• A major advantage of the activated sludge process compared to a fixed film process is that the activated sludge process generally achieves a higher percent removal of BOD5 (Biochemical Oxygen Demand).

Anaerobic Lagoon

• A disadvantage of an anaerobic lagoon compared to an aerobic lagoon is odors.

Fixed Film Reactor Efficiency

• An important test in determining the efficiency of a fixed film reactor is BOD (Biochemical Oxygen Demand).

Growth on Fixed Media

• The growth of organisms on fixed media is called biological slime.

Biological Slime Composition

• In a fixed film reactor, biological slime generally consists of aerobic and anaerobic bacteria.

Disease-Producing Organisms

• Disease-producing organisms are called pathogens.

Reduction of Microorganisms

• The reduction of the number of microorganisms present in wastewater so that fewer pathogens are present is called disinfection.

Wastewater Disinfection

• Factors that influence the effectiveness of wastewater disinfection include the solids level of the wastewater, the amount of disinfectant used, and the degree of mixing, but NOT the dechlorination method utilized.

Cost Accounting Program

• A reasonable objective of a cost-accounting program for a wastewater utility is to identify methods for controlling increases in operating costs, provide data for budget development, and provide data for decisions about repairs versus replacement of equipment.

Preventative Maintenance Program

• An objective for a preventative maintenance program is to reduce emergency repairs and maintenance.

Temperature Conversion

• A temperature reading of 35 degrees Celsius is equal to 95 degrees Fahrenheit. The equation to convert Celsius to Fahrenheit is: F = (C \times \frac{9}{5}) + 32
  F = (35 \times \frac{9}{5}) + 32 = (35 \times 1.8) + 32 = 63 + 32 = 95

Parts Per Million

• Five milligrams per liter equals 5.0 parts per million.

Pump Types

• Centrifugal pumps work on the basis of inertia or mass moving in a circular motion.

Waterborne Disease

• One waterborne disease believed to be caused by a virus is Hepatitis.

Major Biological Treatment Processes in Wastewater Treatment

Secondary Treatment Process

• Objective: To remove the soluble and colloidal matter which remains after primary treatment.
• Requirements for treatment to be effective:
  • Suspended Growth process - maintain an adequate biological mass in suspension within the reactor by employing either natural or mechanical mixing.
  • Attached Growth process - utilize a solid medium on which bacterial solids are accumulated in order to maintain a high population.

Common Types of Secondary Biological Wastewater Treatment Processes

1. Suspended Growth Biological Treatment
   • Activated Sludge Process
2. Attached Growth Biological Treatment
   • Trickling Filter
   • Rotating Biological Contactor
3. Combined Processes
4. Lagoon Processes (more on industrial waste)

Aerobic, Anoxic, and Anaerobic Processes

Aerobic Processes

• Suspended Growth:
  • Activated Sludge Process(es)
  • Aerated Lagoons
  • Aerobic Digestion
• Attached Growth:
  • Trickling Filters
  • Rotating Biological Contactors
  • Packed Bed Reactors
• Hybrid (Combined): Trickling Filters/Activated Sludge Process
• Use: BOD removal, nitrification

Anoxic Processes

• Suspended Growth: Suspended Growth Denitrification
• Attached Growth: Attached Growth Denitrification
• Use: Denitrification

Anaerobic Processes

• Suspended Growth:
  • Anaerobic Contact Processes
  • Anaerobic Digestion
  • Upflow Anaerobic Sludge Blanket
• Attached Growth: Anaerobic Packed and Fluidized Bed
• Hybrid: Upflow Anaerobic Sludge Blanket/Attached Growth
• Use: Stabilization, Solid Destruction, BOD removal.

Combined Aerobic and Anaerobic Processes

• Suspended Growth: Single or Multi-Stage Processes
• Attached Growth: Single or Multi-Stage Processes
• Use: BOD removal, nitrification, denitrification, phosphorus removal

Lagoon Processes

• Aerobic Lagoons: Tertiary Lagoons, Maturation Lagoons
• Facultative Lagoons: Aerobic Lagoons, Facultative Lagoons
• Anaerobic Lagoon: Anaerobic Lagoons
• Use: BOD removal, waste stabilization streams

Activated Biosolids Treatment Process

• Manmade process that mimics the natural self-purification process that takes place in steams. In essence, we can state that the activated biosolids treatment process is a "stream in a container."
• Activated biosolids refers to biological treatment systems that use a suspended growth of organisms to remove BOD and suspended solids.
• The basic components of an activated biosolids sewage treatment system include an aeration tank and a secondary settling basin, or clarifier.
• Primary effluent is mixed with settled solids recycled from the secondary clarifier, and this mixture is then introduced into the aeration tank.
• Compressed air is injected continuously into the mixture through porous diffusers located at the bottom of the tank, usually along one side.
• Wastewater is fed continuously into an aerated tank, where the microorganisms metabolize and biologically flocculate the organics.
• Microorganisms (activated biosolids) are settled from the aerated mixed liquor under quiescent conditions in the final clarifier and are returned to the aeration tank.
• Left uncontrolled, the number of organisms would eventually become too great; therefore, some must be removed periodically (wasted).
• A portion of the concentrated solids from the bottom of the settling tank must be removed from the process (waste activated sludge, or WAS).
• Clear supernatant from the final settling tank is the plant effluent.

Aeration Tank

• Air diffusers, jet aerators, static aerators, and surface aerators are used to add water containing oxygen to decompose BOD

Secondary Clarifiers

• Settled wastewater and recycled activated sludge are introduced at different points in the aeration tank
• This is the most common method used today.
• Since the wastewater is completely mixed with bacteria and oxygen, the volatile suspended solids concentration and oxygen demand are the same throughout the tank.

Formulas

• \text{TS} = \text{VS} + \text{FS}
• \text{MLSS} = \text{MLVSS} + \text{Ash}
  • TS: total solids
  • VS: volatile suspended solids
  • FS: fixed solids
  • MLSS: mixed liquor suspended solids
  • MLVSS: mixed liquor volatile suspended solids

Conventional Activated Sludge System with Return Activated Sludge Stream

• The wastewater flows through as a plug and is treated as it winds its way through the tank.
• Settled wastewater and activated sludge enter the head end of the aeration tank and mixed by diffused air or mechanical aeration
• As the wastewater goes through the system, BOD and organics concentration are greatly reduced.

Contact Stabilization

• Microorganisms consume organics in the contact tank.
• Raw wastewater flows into the contact tank where it is aerated and mixed with bacteria.
• Soluble materials pass through bacterial cell walls, while insoluble materials stick to the outside.
• Solids settle out later and are wasted from the system or returned to a stabilization tank.
• Microbes digest organics in the stabilization tank and are then recycled back to the contact tank because they need more food.
• Detention time is minimized, so the size of the contact tank can be smaller.
• Volume requirements for the stabilization tank are also smaller because the basin receives only concentrated return sludge; there is no incoming raw wastewater.
• Often, there is no primary clarifier before the contact tank due to the rapid uptake of soluble and insoluble food.

Step-Aeration Activated Sludge

• Step feed-settled wastewater is introduced at several points in the aeration tank.
• With this arrangement, oxygen uptake requirements are relatively even, and the need for tapered aeration is limited.
• An advantage of step feed is that solids loading on the secondary clarifier can be decreased during periods of settling problems by moving the waste fed point toward the end of the aeration tank.
• Step feed plants can be designed with the flexibility to operate as contact stabilization or plug flow, and if fed at all ports evenly, under a condition of approximately complete mix.

Extended Aeration

• The extended aeration process uses the same flow scheme as the complete mix or plug flow process but retains the wastewater in the aeration tank for 18 hours or more.
• This process operates at a high MCRT (low FM ratio), resulting in a condition where the system lacks enough food to support all the microorganisms present. The microorganisms, therefore, compete for the remaining food and use their own cell mass for food. This competitive environment results in a highly treated effluent with low sludge production.
• The main disadvantages of this system are the high oxygen requirements per unit of waste entering the plant and the large tank volume needed to hold the waste for the extended period.

Oxidation Ditch

• The oxidation ditch is a variation of the extended aeration process
• The wastewater is typically pumped around a circular or oval pathway by a mechanical aerator or pumping device at one or more points along the flow pathway in the aeration tank; the mixed liquor velocity is maintained between 0.2 and 0.37 m/s in the channel to prevent solids from settling.
• Oxidation ditches typically use mechanical brush aerators, disk aerators, surface aerators, and jet aerators to aerate and pump the liquid flow. Combination diffused aeration and pumping devices are also used.

Kraus Process

• The anaerobically digested sludge and digester supernatant may be added to the return sludge, thus improving settling of the floc.

High Purity Oxygen

• The most common high purity oxygen activated sludge process uses a covered and stage aeration tank.
• The wastewater, return sludge, and oxygen feed gas enter the first stage of this system and flow concurrently through the tanks.
• The tanks in this system are covered to retain the oxygen gas and permit a high degree of oxygen use.
• A prime advantage of the stage reactor design of the oxygenation system is the system's ability to approximately match the biological activity with the amount of oxygen available.

High Rate Aeration

• High rate aeration is a modification combining low concentration of MLSS with high volumetric loadings.
• This combination facilities high F:M ratios and moderate MCRTs with short hydraulic detention times.

Important Acronyms

• MLSS = Mixed Liquor Suspended Solids
• BOD = Biological Oxygen Demand- utilized for design
• COD = Chemical Oxygen Demand - utilized for measurement

Example

• 550,000 \text{ gallons}= 2,750,000 \frac{\text{lbs}}{\text{day}}
• Conversion concentration -> concentration
• \frac{550,000 \text{ gal}}{x} = \frac{550 \text{ lbs}}{4.4}

Example 4

• If the MLSS concentration is 1200 \frac{\text{mg}}{\text{L}}, and the aeration tank has a volume of 550,000 gallons, how many pounds of suspended solids are in the aeration tank?
• \text{BOD} = \frac{\text{BOD} \times 10^6}{\text{MLVSS} \times \text{Vol}}

Typical F/M Ratios

• If the MLVSS concentration is not available, it can be calculated if the percent volatile matter (%VM) of the mixed liquor suspended solids (MLSS) is known.
• The food (F) value in the F/M ratio for computing loading to an activated biosolids process can be either BOD or COD. Remember, the reason for biosolids production in the activated biosolids process is to convert BOD to bacteria. One advantage of using COD over BOD for analysis of organic load is that COD is more accurate.

Example 5

• The aeration tank influent BOD is 145 \frac{\text{mg}}{\text{L}}, and the aeration tank influent flow rate is 16 MGD What is the F/M ratio if the MLVSS is 2300 \frac{\text{mg}}{\text{L}} and the aeration tank volume is 18 MG?
• \frac{145 \frac{\text{mg}}{\text{L}} \times 1.6 \text{MGD}}{2300 \frac{\text{mg}}{\text{L}} \times 1.8 \text{MG}} = \frac{0.6 \text{ BOD}}{ \text{MLVSS}}

Example 6

• The aeration tank contains 2885 \frac{\text{mg}}{\text{L}} of MLSS Lab tests indicate the MLSS is 66% volatile matter. What is the MLVSS concentration in the aeration tank?
• 2884 \frac{\text{mg}}{\text{L}} \times 0.66 = 1,904 \frac{\text{mg}}{\text{L}}

Waste Rates Using F/M Ratio

• Maintaining the desired F/M ratio is accomplished by controlling the MLVSS level in the aeration tank. This may be accomplished by adjustment of return rates; however, the most practical method is by proper control of the waste rate:
• Waste volatile solids (lb/day) = Actual MLVSS (lb) - desired MLVSS (lb)
• If the desired MLVSS is greater than the actual MLVSS, wasting is stopped until the desired level is achieved. Practical considerations require that the required waste quantity be converted to a required volume of waste per day.

Example 7

• Given the following information, determine the required waste rate in gallons per minute to maintain an F/M ratio of 0.17 COD (\frac{\text{lb}}{\text{MLVSS} \text{ (lb)}})

• Primary effluent COD = 140 \frac{\text{mg}}{\text{L}}

• Primary effluent flow = 2.2 MGD

• MLVSS = 3549 \frac{\text{mg}}{\text{L}}

• Aeration tank volume = 0.75 MG

• Waste volatile concentrations = 4440 \frac{\text{mg}}{\text{L}} (volatile solids)

Gould Biosolids Age

• Biosolids age refers to the average number of days a particle of suspended solids remains under aeration. It is a calculation used to maintain the proper amount of activated biosolids in the aeration tank.
• This calculation is sometimes referred to as the Gould biosolids age so that it is not confused with similar calculations, such as solids retention time (or mean cell residence time). When considering sludge age, in effect, we are calculating how many days of suspended solids are in the aeration tank.
• For example, if 3000 lb of SS enter an aeration tank daily and the aeration tank contains 12,000 lb of SS, then when 4 days of solids are in the aeration tank, we have a sludge age of 4 days

Example 8

• A total of 2740 lb/day suspended solids enters an aeration tank in the primary effluent flow. If the aeration tank has a total of 13,800 lb of mixed liquor suspended solids, what is the biosolids age in the aeration tank?
• \frac{13800}{2740} = 5 \text{ days}

Mean Cell Residence Time (MCRT)

• Mean cell residence time (MCRT), sometimes called sludge retention time, is another process control calculation used for activated biosolids systems. MCRT represents the average length of time an activated biosolids particle remains in the activated biosolids system. It can also be defined as the length of time required at the current removal rate to remove all the solids in the system.
• MCRT can be calculated using only the aeration tank solids inventory.
• When comparing plant operational levels to reference materials, you must determine which calculation the reference manual uses to obtain its example values.

Example 9

• Given the following data, what is the MCRT?
  • Aeration volume = 1,000,000 gal
  • Clarifier volume = 600,000 gal
  • Flow 5.0 MGD
  • Waste rate = 0.085 MGD
  • MLSS = 2500 \frac{\text{mg}}{\text{L}}
  • Waste = 6400 \frac{\text{mg}}{\text{L}}
  • Effluent TSS = 14 \frac{\text{mg}}{\text{L}}

Sludge Volume Index (SVI)

• The sludge volume index (SVI) is a measure (an indicator) of the settling quality (a quality indicator) of the activated biosolids
• As the SVI increases, the biosolids settle more slowly, do not compact as well, and are likely to result in an increase in effluent suspended solids
• As the SVI decreases, the biosolids become denser, settling is more rapid, and the biosolids age SVI is the volume in milliliters occupied by 1 gram of activated biosolids
• For the settled biosolids volume (mL/L) and the mixed liquor suspended solids (MLSS) calculation, milligrams per liter are required. The proper SVI range for any plant must be determined by comparing SVI values with plant effluent quality

Example 10

• The SSV30 is 365 \frac{\text{mL}}{\text{L}}, and the MLSS is 2365 \frac{\text{mg}}{\text{L}}. What is the SVI?

• Other methods are available to determine the clarifier solids concentrations; however, the simplest method assumes that the average suspended solids concentration is equal to the solids concentration of the aeration tank.

Example 11

• Determine the size of SBR tank and the number of diffusers required using the following data:

1. Average flow rate = 800 \frac{\text{m}^3}{\text{day}}
2. Influent BOD5 = 300 \frac{\text{mg}}{\text{L}}
3. MLSS = 4,000 \frac{\text{mg}}{\text{L}} (4 kg/d)
4. F/M ratio = 0.03 \frac{\text{mg}-\text{BOD}}{\text{mg MLSS \times d}}
5. No. of tanks = One (1)
6. Primary treatment = None
7. Total draw & settle time = 3 hours
8. Total react time = 2 hours
9. Total idle time = 0.5 hrs

• No primary treatment required
• \text{Food, F} = (800 \frac{\text{m}^3}{\text{day}} \times 300 \frac{\text{mg}}{\text{L}})/1000 = 240 \frac{\text{kg}}{\text{day}}
• \text{Sludge Mass} = \frac{F}{(F/M)} = \frac{240 \frac{\text{kg}}{\text{d}}}{(0.03)} = 8,000 \text{kg MLSS}
• \text{Volume} = \frac{8,000 \text{kg MLSS}}{4 \frac{\text{kg MLSS}}{\text{L}}} = 2,000 \text{ L} = 2 \text{ m}^3
• Use 1 tank
• Determine total time per cycle
• Use 8 hrs per cycle per day with the following:
  • Time to fill = 25% x 8hrs = 2.0 hrs
  • Time to React = 35% x 8hrs = 2.8 hrs
  • Time to Settle = 20% x 8 hrs = 1.6 hrs
  • Time to Draw = 15% x 8 hrs = 1.2 hrs
  • Time to Idle = 5% x 8 hrs = 0.4 hrs
  • TOTAL = 8.0 HRS
• For the settled biosolids volume (mL/L) and the mixed liquor suspended solids (MLSS) calculation, milligrams per liter are required. The proper SVI range for any plant must be determined by comparing SVI values with plant effluent quality
• Calculate the volume of liquid per tank per decant
• Number of cycle = \frac{24 \text{ hrs}}{8 \text{ hrs}} = 3 \frac{\text{cycles}}{\text{day}}
• V_d = \frac{800 \frac{\text{m}^3}{\text{d}}}{3 \frac{\text{cycles}}{\text{day}}} = 267 \frac{\text{m}^3}{\text{day}}
• Volume per tank per decant = \frac{267}{1 \text{ tank}} = 267 \text{ m}^3
• Calculate tank size = Volume of Mixed Liquor + Volume of decant
• =2 \text{ m}^3 + 267 \text{ m}^3
• =269 \text{ m}^3
• Considering that the water depth = 4.5m and free board 1.0m
• Area = \frac{269 \text{ m}^3}{4.5 \text{ m}} = 59.78 \text{ m}^2 \text{ vs } 5.5 \text{ Total depth}
• Use 6m x 10m x 5.5 m SBR tank
• Air Required = \frac{[(300 \times 800) \times 125 \text{ m}^3]}{1000 \times 1440} = 20.83 \frac{\text{ m}^3}{\text{min}}
• No. of diffusers = \frac{20.83 \frac{\text{ m}^3}{\text{min}}}{0.08 \frac{\text{ m}^3}{\text{min}}} = 260.4 \approx 262 \text{ pcs}

Environmental Engineering Questions

Question 1

• The rise in Earth's temperature is called global warming.

Question 2

• RA 7942 is a.k.a. Philippine Mining Act of 1995.

Question 3

• A pitot tube is used to measure the velocity of flow.

Question 4

• The Coriolis Effect is related to the Earth's rotation.

Question 5

• It is a function of the distance between the speaker and the listener and the frequency components of spoken words. None of the above are correct.

Question 6

• Catalytic Reactors are used to control NOx emissions arising from the burning of fossil fuels in industrial processes.

Question 8

• A leaching chamber is a sewage disposal system dug on the ground, with stones or bricks laid to allow contaminated sewage to leach into the under surface of the ground.

Question 9

• Health is the state of complete physical, mental, and social well-being and not merely the absence of disease or infirmity.

Question 10

• Atoms with the same atomic number but different mass number are called isotopes.

Question 11

• Flocculation is a process of contact and adhesion whereby the particles of a dispersion form larger-size clusters.

Question 12

• Recycled material refers to post-consumer material that has been recycled and returned to the economy.

Question 13

• Water with organics that has been chlorinated is typically associated with trihalomethanes.

Question 14

• Surge tanks are used to guard against water hammer.

Question 15

• The ratio of lime to copper sulfate for controlling algae growth on basin walls is 2 parts lime to 1 part copper sulfate.

Question 16

• Chlorofluorocarbons deplete the ozone layer.

Question 17

• Chapter XVII is the "Refuse Disposal" chapter on the Code of the Sanitation of the Philippines.

Question 18

• Bernoulli's equation assumes all of the above.
  • fluid is non-viscous
  • fluid is homogeneous
  • flow is steady
  • flow is along the stream line

Question 19

• Nitrate found in groundwater is most likely associated with agricultural land use.

Question 20

• The permissible noise exposure limit in a day for three (3) hours exposure is 97 dBA.

Question 21

• All of the above are types of dust collectors
  • Inertial separators
  • Fabric filters and Wet Scrubbers
  • Unit Collectors

Question 22

• Clostridia is/are Gram-positive bacteria, responsible for food poisoning.

Question 23

• Pathology is the study of the essential nature of diseases and especially of the structural and functional changes produced by them.

Question 24

• In the titration of a weak acid with a strong base, Phenolphthalein is the best indicator to be used.

Question 25

• TSS (Total Suspended Solids) is determined by regular monitoring & comparing of effluent quality

Question 26

• Materials recovery facility shall include solid waste transfer station or sorting station, drop-off center, a composting facility, and a recycling facility.

Question 27

• Corrosion of plumbing systems is the primary source of lead in drinking water.

Question 28

• Acidity in water is caused due to All of the above

• Mineral acids

• Free CO2

• Iron sulphate

• Aluminiumsulphate

Question 29

• Copper sulfate is used in surface water reservoirs to control algae.

Question 30

• Privy is that sewage disposal system where a hole is dug on the ground and a concrete vault constructed for the collection of raw sewage sealed within a wooden shelter.