module 4: fundamentals of wastewater treatment wastewater treatment plant operator training

Module 4: Fundamentals of Wastewater Treatment

Wastewater Treatment Plant Operator Training

Learning Objectives• Explain the general purpose of preliminary treatment.• Explain the purpose of screening, grit removal, and

pre-aeration.• Differentiate between manually and mechanically

cleaned racks and screens.• Define differences between screening and

comminution.• Describe safe disposal of screenings and grit.

Unit 1–Preliminary Treatment Unit 1–Preliminary Treatment

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Manually Cleaned Bar ScreenManually Cleaned Bar Screen

3Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

AERATION TANK

Mechanically Cleaned Bar Screen

4

AERATION TANK

Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

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Aerated Grit ChamberAerated Grit Chamber

AERATION TANK


Cyclone SeparatorCyclone Separator

6

AERATION TANK


Grit WasherGrit Washer

7

AERATION TANK


ComminutorComminutor

8

AERATION TANK


Comminutor with By-Pass ScreenComminutor with By-Pass Screen

9

AERATION TANK


BarminutorBarminutor

10

AERATION TANK


Learning Objectives• Explain sedimentation and flotation

principles.• List factors that indicate a clarifier is not

performing properly.• Use mathematical formulas to solve for

detention time, weir overflow, surface loading, and solids loading.

Unit 2–Primary Treatment: Sedimentation & Flotation Unit 2–Primary Treatment: Sedimentation & Flotation

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Detention Time FormulasDetention Time Formulas

12

Formulas

Detention Time, hr. = Tank Volume, cu ft x 7.5 gal/cu ft x 24 hr/day Flow, gal/day

Rectangular Tank Volume, cu ft = Length, ft x Width, ft x Height (Depth), ft

Area of Circle, ft2 = (0.785)(Diameter2) or ()(Radius2)Note: = 3.14

Circular Tank Volume, cu ft = Area x Height (Depth), ft

Circular Tank Volume, cu ft = (0.785)(Diameter, ft)2 x Height (Depth), ft or (3.14)(Radius, ft)2 x Height (Depth), ft

What is the detention time when…

• The flow is 3.0 million gallons per day (MGD) or 3,000,000 gal/day, and

• Tank dimensions are 60 feet long by 30 feet wide by 10 feet deep?

Detention Time: Sample 2.1Detention Time: Sample 2.1

13

Given:

Volume = 60 feet by 30 feet by 10 feet

= 18,000 cu ft

Step #1 - VolumeStep #1 - Volume

14

Given:

Flow = 3.0 million gallons per day (MGD) or 3,000,000 gal/day

Step #2 - FlowStep #2 - Flow

15

Detention = Tank Volume, cu ft x 7.5 gal/cu ft x 24 hr/dayTime, hr Flow, gal/day

= 18,000 cu ft X 7.5 gal/cu ft x 24 hr/day3,000,000 gal/day

= 3,240,000 gal/hr/day3.0 MGD

= 1.08 hours

Step #3 - CalculationStep #3 - Calculation

16

What is the detention time when…

• The flow is 2.5 million gallons per day (MGD), and

• Circular clarifier is 60 ft in diameter with a depth of 12 ft?

Detention Time: Sample 2.2Detention Time: Sample 2.2

17

Reminder:

• Circular Tank Volume, cu ft = 0.785 x (Diameter, ft)2 x Depth, ft

• Volume = (0.785) x (60 ft)2 x 12 ft deep

= 33,912 cu ft

Step #1 - VolumeStep #1 - Volume

18

Given:

Flow = 2.5 million gallons per day (MGD) or 2,500,000 gal/day

Step #2 - FlowStep #2 - Flow

19

DetentionTime, hr= 33,912 cu ft x 7.48 gal/cu ft x 24 hr/day

2.5 MGD

= 6,087,882 gal/hr/day 2,500,000 gal/day

= 2.44 hours

Step #3 - CalculationStep #3 - Calculation

20

Formulas

• Weir Overflow, gpd/ft = Flow Rate, GPD Length of Weir,

ft

• Length of Circular Weir = 3.14 x Weir Diameter, ft

Weir Overflow RateWeir Overflow Rate

21

What is the weir overflow rate when…

• The flow rate into the unit is 3.5 MGD, and

• The circular clarifier has a 75 foot diameter overflow weir

Weir Overflow Rate: Problem 2.1Weir Overflow Rate: Problem 2.1

22

Weir overflow rate,gpd/ft

= 3,500,000 gallons/day 3.14 x 75 feet

= 3,500,000 gal/day 235.5 feet

= 14,862 gpd/ft

Weir Overflow Rate: Problem 2.1Weir Overflow Rate: Problem 2.1

23

What is the surface loading rate when…

• The flow into a rectangular clarifier is 5.0 MGD

• The clarifier is 40 feet wide by 110 feet long by 12 feet deep

– Reminder:Surface loading rate , gpd/ft2 = Flow Rate, gpd Surface area, ft2

Surface Loading Rate: Problem 2.2Surface Loading Rate: Problem 2.2

24

Surface loading rate, gpd/ft2

= 5,000,000 gallons/day 40 ft x 110 feet

= 5,000,000 gal/day 4400 ft2

= 1,136 gpd/ft2

Surface Loading Rate: Problem 2.2Surface Loading Rate: Problem 2.2

25

Pa DEP - Domestic Wastewater Facilities

Manual

Operation of Wastewater Treatment Plants, Vol I

Primary Clarifiers Not considered Not generally considered

Conventional Activated SludgeSecondary Clarifiers

40 #/day/sq ft average, 50 peak

12 to 30 lbs/day/sq ft

Extended Aeration Secondary Clarifiers

30 #/day/sq ft average, 50 peak

N/A

Nitrification Secondary Clarifiers:Separate Nitrification Stage

30 average, 50 peak N/A

Carbonaceous Stage 45 average, 50 peak N/A

Dissolved-Air Flotation 40 #/day/sq ft (w/o polymer addition),20 #/day sq ft (w polymer addition)

5 to 40 lbs/day/sq ft

Sludge Thickening 5 to 12 #/day/sq ft 5 to 20 lbs/day/sq ft

Solids Loading GuidelinesSolids Loading Guidelines

26

Formulas

• Solids Loading, lbs/day/ft2 = Solids Applied, lbs/day Surface Area, ft2

– Solids Applied, lbs/day = Flow, MGD x Conc., mg/L x 8.34 lbs/gal

Solids LoadingSolids Loading

27

Calculate the solids loading at which a clarifier is operating given the following…

•The circular clarifier has a diameter of 125 feet

•Forward flow is 6.0 MGD and the return sludge flow is 2.0 MGD

•MLSS is 4.000 mg/L

Solids Loading: Problem 2.3Solids Loading: Problem 2.3

28

Reminder:

Solids Applied, lbs/day

= Flow, MGD x MLSS conc., mg/L x 8.34 lbs/gal

Solids Applied = 8 MGD x 4,000 mg/L x 8.34 lbs/gal

= 266,880 lbs/day

Step # 1 - Solids AppliedStep # 1 - Solids Applied

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Given: The circular clarifier has a diameter of 125 feet

Area of Circle, ft2 = (0.785)(Diameter)2

= (0.785)(125 ft)2

= (0.785)(15,625 ft2)

= 12,266 ft2

Step # 2 – Surface AreaStep # 2 – Surface Area

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Solids Loading,lbs/day/ft2

= Solids Applied, lbs/day Surface Area, ft2

= 266,880 lbs/day 12,266 ft2

= 22 lbs/day/ft2

Step # 3 – Solids LoadingStep # 3 – Solids Loading

31

Learning Objectives• List four biological secondary treatment processes.

• Explain the principles of the trickling filter process.

• Identify the different types of trickling filters.

• Explain the principles of the rotating biological contactor (RBC) process.

• Explain the principles of the activated sludge process.

• List the three waste treatment pond classifications and explain the principles of each.

Unit 3–Overview of Biological Secondary TreatmentUnit 3–Overview of Biological Secondary Treatment

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Trickling Filter ProcessTrickling Filter Process

33

AERATION TANK

Prelim inaryTreatm ent

SolidsHandling

Wet PitTrickling

Filter

Distributor

FilterPum p

Station

SecondaryClarifier

ChlorineContact

Basin

WasteSolids

RawWastewater

PlantEffluent

TricklingFilter Process

Prim aryEffluent

Recirculation

FilterUnderflow

SecondaryClarifierInfluent

Prim aryClarifier

TRICKLING FILTER PROCESS FLOW SCHEMATIC

Trickling FilterTrickling Filter

34Diagram excerpted from Chapter 6: Trickling Filters. In Operation of Wastewater Treatment Plants Volume I.

AERATION TANK

Rotating Biological Contactor ProcessRotating Biological Contactor Process

35

AERATION TANK

No. 1 No. 2 No. 3 No. 4

ROTATING BIOLOGICAL CONTRACTOR PROCESS FLOW SCHEMATIC

RawWastewater

Prelim inaryTreatment

PrimaryClarifier

PrimaryEffluent

SolidsHandling

WasteSolids

SecondaryClarifier

SecondaryClarifierInfluent

RBC Process

RBC ReactorsChlorineContact

Basin

PlantEffluent

Shafts

Baffle (Typ)

Tank (Typ)

Media (Typ)

CONVENTIONAL ACTIVATED SLUDGE SCHEMATIC(COMPLETE MIX)

RawWastewater


PrimaryClarifier

PrimaryEffluent

SolidsHandling

WasteSolids

SecondaryClarifierAeration Tank

or Reactor

ChlorineContact

Basin

PlantEffluent

Return Activated Sludge

Waste Activated Sludge

Complete Mix Activated SludgeComplete Mix Activated Sludge

36

37

Contact Stabilization SchematicContact Stabilization Schematic

AERATION TANK

Aeration Contact Tankor Reactor

CONTACT STABILIZATION ACTIVATED SLUDGE SCHEMATIC

SolidsHandling

RawWastewater


ChlorineContact

Basin

WasteActivated

Sludge

WasteSolids

SecondaryClarifier

PlantEffluent

Reaeration Tankor Reactor Return

ActivatedSludge

38

AERATION TANK

Return Activated Sludge

Aeration Tankor Reactor

EXTENDED AERATION ACTIVATED SLUDGE SCHEMATIC

RawWastewater


ChlorineContact

Basin

Waste Activated Sludge

ClarifierPlant

Effluent

WasteSolids

SolidsHandling

Extended Aeration SchematicExtended Aeration Schematic

Two Unit SBR Time ChartTwo Unit SBR Time Chart

39

Time Period (hrs) Unit #1 Unit #20 - 1 Fill Aeration1 - 2 Fill Settle2 - 3 Fill / Aeration Settle3 - 4 Fill / Aeration Draw4 - 5 Aeration Fill5 - 6 Settle Fill6 - 7 Settle Fill / Aeration7 - 8 Draw Fill / Aeration

Oxidation Ditch Schematic

40

AERATION TANK

OXIDATION DITCH SCHEMATIC

RawWastewater

SolidsHandling

WasteSludge


WasteSolids

Return Sludge

FinalClarifier

ChlorineContact

Basin

PlantEffluent

Rotor (Typ)Level

ControlWeir

Oxidation Ditch

ReturnSludge

Pum ping

module 4: fundamentals of wastewater treatment wastewater treatment plant operator training

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