preliminary&secondary treatment_group 2.pptx

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    Wastewater TreatmenAran ~ Miranda ~ Riego ~ Santos ~ Sy

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    Why Treat Wastewater?

    To protect the wildlife that call those environments

    Causes a demand for dissolved oxygen (lower DO le

    streams)

    Increases suspended solids or sediments in stream

    (turbidity increase)

    In some cases, wastewater can be clean enough for

    particular purposes

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    Why Treat Wastewater?

    Wastewater treatment is important to remov

    constituents or contaminants that could harmor the environment.

    Common constituents in domestic wastewate Organics

    Solids

    Nutrients Pathogens

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    What happens to treated Wastew

    Treated wastewater (effluent) may bedischarged to a nearby water body suchstream or lake.

    It may also be reused, for example, for ir

    or industrial processes.

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    Preliminary Treatment

    Mechanical screens remove materials like rags, stic

    Grit removal processes remove grit, sand and granumaterials.

    Grit removal is the forced separation of gritty mater

    sand, egg shells, seeds, etc.) from wastewater.

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    Materials removed

    fats, oils, and greases (aka FOG)

    sand, gravels and rocks (aka grit)

    larger settleable solids including human

    and

    floating materials

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    Primary Treatment

    A physical process.

    Wastewater flow is slowed down and suspen

    solids settle to the bottom by gravity.

    The material that settles is called sludge or b

    Primary treatment reduces the suspended s

    the B.O.D. of the wastewater.

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    Primary Treatment

    Sludge from the primary sedimentation tankpumped to the sludge thickener.

    From the primary treatment tanks water is p

    the trickling filter for secondary treatment.

    Sludge is scraped from the bottom of the tank

    hopper, and is pumped from the tank for pro

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    Primary Sedimentation

    Primary sedimentation is the oldest and m

    used form of wastewater treatment.

    Sedimentation describes the tendency of floasuspended particles to settle to the bottom oby gravity.

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    SECONDARY TREATMENT PROCESS

    Activated sludge Trickling filter

    Non-electric secondary filtration (FilterPod)

    Oxidation ponds

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    Biological Wastewater Treatm

    Principal Goals

    Many of the microbes present in wastewater are befact, many wastewater treatment technologies are d

    on these beneficial microorganisms for remediation

    wastewater so that it won't detrimentally impact th

    environment. One of the primary goals of biological

    is the removal of organic material from wastewaterexcessive oxygen consumption won't become a prob

    it is released to the environment.

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    Waste

    DO

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    TRICKLING FILTER

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    TRICKLING FILTER

    A trickling filter consists of a bed of highly permeable media on whose

    mixed population of microorganisms is developed as a slime layer.

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    TRICKLING FILTER

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    TRICKLING FILTER

    Pros:

    Low construction cost

    Cheap oxygen delivery

    Non-electric systems available

    Cons:

    Temperature Dependent

    Vulnerable to congestion, which can lead to flooding and system failure, low oxygen

    restricted water flow

    Can occupy a larger area than activated sludge

    High Maintenance on some older types

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    COMMON PROBLEMS

    Pondingis normally the result of:

    (a) excessive organic loading without a correspondin

    recirculation rate,

    (b) use of media which is too small,

    (c) clogging of underdrain system,

    (d) non-uniform media size or breaking up of media,

    (e) trash or debris in filter voids

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    COMMON PROBLEMS

    Odors

    Since the trickling filter is an aerobic process, no seshould exist.

    If foul odors are present, anaerobic conditions arlikely cause.

    Anaerobic conditions usually predominate next tosurface.

    If the surface of the slime growth is aerobic, odorminimal.

    If odors are present, corrective action shouldimmediately or the condition could get worse.

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    COMMON PROBLEMS

    It can be eliminated by:

    Check the ventilation of the filter for clogging and stoppages.

    Check the underdrain system for clogging and stoppages.

    Increase recirculation rate; this usually provides added oxyg

    filter and may increase sloughing.

    Keep wastewater in filter; do not allow it to splash on expose

    weeds, or grass.

    Add odor-masking agents.

    Pre-chlorination at primary tank influent or at the dosing tan

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    COMMON PROBLEMS

    Weather Problems.

    Cold weather can cause an occasional build-up of ice on the

    distributor arms and orifices, resulting in operating problem

    efficiency.

    During cold temperatures, the organism's metabolic proces

    and as a result efficiency decreases.

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    COMMON PROBLEMS

    It can be eliminated by:

    Decrease the recirculation rate to prevent splashing at distrib

    but maintain sufficient flow to keep the filter working.

    Adjust orifices at splash plates to reduce the spraying effect.

    Construct wind screens or covers to reduce heat loss.

    Break up any ice build-up.

    Partially open flap gates at end of distributor arm to allow fo

    water rather than a spray of water.

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    Secondary Treatment- Aerati

    After leaving the primary clarifiers, the sewage g

    one of the aeration tanks. The contents of the aeration tanks, which require

    balance of food and oxygen, are commonly referremixed liquor suspended solids (MLSS) or activated activated sludge converts organic substances inproducts and a settleable floc which is settled

    secondary clarifiers. Raw sewage can be introducedlocations and be aerated and mixed for varying lengand intensity.

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    Aeration Tank

    Waste Water

    from 1stClarifier

    Microbes

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    PROCESS (ACTIVATED SLUDG

    Biochemical Oxygen Demand and

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    Biochemical Oxygen Demand andEutrophication

    Biochemical oxidation reactions involve the conver

    organic material using oxygen and nutrients into cadioxide, water and new cells. The equation that expis:

    Organic material + O2 + nutrients CO2 + H2O + newnutrients + energy

    It can be seen from this equation that organisms us

    breakdown carbon-based materials for assimilationcell mass and energy. A common measure of this oxbiochemical oxygen demand (BOD).

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    Soft and Hard BOD

    The time-course for the removal of the organic carbon vthe ability of the activated sludge bacteria to ingest it.

    Small molecular weight compounds will start to be remthe sewage immediately after it has entered theactivatetanks. Their removal may be completed in 1 2 hours. Tof compounds is often referred to as the readily biodegSoft BOD. Other, higher molecular weight compounds wseveral hours to be

    degraded and removed. Yet other compounds are morerecalcitrant, and may still be present after several daysreadily biodegradable BOD is often referred to as Hard

    The net result is that larger, complex organic carbon mobe not be degraded because the treatment time

    available (the hydraulic retention time) is not sufficientthey will therefore pass out in the effluent.

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    Calculations-Food to Microorganism

    The number of microorganisms which are used t

    aeration chamber is carefully controlled and isthe food to microorganism ratio (F/Microorganisms will most efficiently break down matter in water if they are present in the right prop

    If the appropriate food to microorganism ratio ithen there will be efficient B.O.D. removal in the a

    manufacturer, Clow, suggests that the bestmicroorganism ratio is about 0.6.

    C l l h F/M R

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    Calculating the F/M Ratio

    In order to calculate the proper amount of microorg

    be added to the aeration basin, you will need to usefollowing formula:

    (For the units of the formula to cancel out, you shou

    aware that the F/M ratio has no units and the constactually has the units "lb/(Million Gallons per Day(mg/L)".) The formula for the F/M ratio can also be as:

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    So let's calculate the pounds of microbes which need

    to the aeration basin of a plant over the span of a dahas a capacity (or flow) of 18,000 gallons per day. the water is 200 mg/L. You want to achieve a F/M radetermine the pounds of microbes, first you must mall of the units match. You should notice that you haof 18,000gallons/day while you need your flow

    units MGD. So you will need to convert the units:

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    Now all of your units match and you can simply pluginto the formula:

    So you will need to add 50.04 pounds of microbes tobasin for the day.

    R t d A ti t d Sl d

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    Returned Activated Sludge

    But how do you add 50.04 lbs of microbes to th

    basin? You do not have pure microbes available. Ihave a mixture of sludge and microbes from the clamixture is known as Returned Activated Sludgeo

    RAS has some typical properties. When dried, 10RAS weighs 5 grams. When cooked at 550C, 10RAS weighs 3 grams. And, most important

    calculations, 2% of RAS is microbes. So, if you haveof RAS, 2 grams of this is made up of microbes.

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    Thank You!