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newSchoolWorks.com › VoTechWorks › Agriculture, Environment & Public Service › Water / Wastewater Operator › Water and Wastewater Treatment Processes › Lesson 7
AGR-WATE71-03 · Lesson 7 · 12.5 clock hours · theory

Primary and Secondary Wastewater Treatment

Objective. Students will explain how primary sedimentation removes settleable solids and how biological secondary treatment—activated sludge and trickling filters—uses microbial metabolism to reduce BOD and suspended solids in wastewater.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by a licensed professional in the field or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

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Why Wastewater Must Be Treated

Wastewater leaving homes, businesses, and industries carries suspended solids, dissolved organic matter, nutrients, and pathogens. If discharged untreated, these contaminants deplete oxygen in receiving waters, harm aquatic life, and threaten public health. Treatment plants use a sequence of physical, biological, and chemical steps to reduce pollutants to permitted levels. This lesson covers the first two major steps: primary treatment (physical removal of settleable solids) and secondary treatment (biological reduction of dissolved and colloidal organic matter).

Primary Treatment: Screening, Grit Removal, and Primary Sedimentation

Wastewater enters the plant through bar screens and fine screens that catch rags, wipes, and large debris. Removed screenings are typically hauled to a landfill. Next, the flow passes through a grit chamber, where velocity is reduced enough for sand and grit to settle without capturing lighter organic solids. Removing grit protects downstream pumps and equipment from abrasion.

The flow then enters the primary clarifier, a large settling basin where retention time—commonly 1.5 to 2.5 hours—allows heavier organic solids (settleable solids) to sink to the bottom as primary sludge, while grease and scum float to the surface and are skimmed off. The clarified liquid flowing over the effluent weir is called primary effluent.

Primary sedimentation typically removes roughly 50–70% of total suspended solids (TSS) and 25–40% of five-day biochemical oxygen demand (BOD₅). These percentages vary with wastewater characteristics and clarifier design. Although significant, these removals are not enough to meet most National Pollutant Discharge Elimination System (NPDES) permit limits for BOD and TSS, which is why biological secondary treatment is required.

Secondary Treatment: Biological Processes

Secondary treatment uses microorganisms—primarily bacteria—to consume dissolved and colloidal organic matter. The bacteria use the organics as food, converting them into carbon dioxide, water, and new cell mass (biomass). Two common secondary processes are the activated-sludge system and the trickling filter.

Activated-Sludge Process

In an activated-sludge system, primary effluent enters an aeration basin where it mixes with a dense, living suspension of microorganisms called mixed liquor. Air blowers or mechanical aerators supply oxygen, which the bacteria need for aerobic metabolism. The concentration of microorganisms in the basin is measured as mixed-liquor suspended solids (MLSS), commonly maintained in the range of 1,500–4,000 mg/L in conventional systems.

From the aeration basin, mixed liquor flows to a secondary clarifier. There, the biomass settles, producing a clear secondary effluent that goes to disinfection and discharge. The settled biological solids are called return activated sludge (RAS). Most of the RAS is pumped back to the aeration basin to maintain the MLSS concentration—this recycled population of microorganisms is what makes the sludge "activated." A portion is removed daily as waste activated sludge (WAS) to control the population size and process performance.

Sludge retention time (SRT)—also called mean cell residence time or sludge age—is the average number of days microorganisms remain in the system. SRT is controlled by how much WAS is wasted. A longer SRT produces a more stable, nitrifying community; a shorter SRT keeps a younger, faster-growing population. Matching SRT to treatment goals is one of the operator's most important process control decisions.

Insufficient dissolved oxygen (DO) in the aeration basin is a common operational problem. When DO drops below about 1–2 mg/L, aerobic bacteria are stressed, BOD removal efficiency falls, and filamentous bacteria may proliferate, causing bulking sludge that settles poorly in the secondary clarifier. Operators respond by increasing aeration, checking blower operation, and verifying DO meter calibration.

Trickling Filters

A trickling filter is a bed of media (rock, plastic, or other material) over which primary effluent is distributed by rotating spray arms. A biofilm of microorganisms grows on the media surface and absorbs and metabolizes organics as wastewater trickles past. Oxygen is supplied by natural ventilation or forced air. Effluent from the filter passes to a secondary clarifier where sloughed biofilm settles out. Trickling filters are simpler to operate than activated sludge but generally achieve somewhat lower and less consistent BOD removal at high organic loads.

Safety Points

Key terms10

Primary clarifierA settling basin that uses gravity and retention time to remove settleable solids and scum from raw wastewater, producing primary sludge and primary effluent.
Biochemical oxygen demand (BOD₅)The amount of dissolved oxygen consumed by microorganisms breaking down organic matter in a water sample over five days at a standard temperature; a key measure of organic pollution strength.
Activated sludgeA secondary treatment process in which wastewater is mixed with a dense suspension of aerobic microorganisms in an aeration basin; the microorganisms consume dissolved organics, and the resulting biomass is settled and partly recycled.
Mixed-liquor suspended solids (MLSS)The concentration of suspended solids—active biomass plus inert material—in the aeration basin of an activated-sludge system, used to monitor and control the microbial population.
Return activated sludge (RAS)Settled biological solids from the secondary clarifier that are pumped back to the aeration basin to maintain the desired MLSS concentration.
Waste activated sludge (WAS)The portion of settled biological solids removed from the activated-sludge system daily to control the sludge age and microbial population size.
Sludge retention time (SRT)The average number of days microorganisms remain in the activated-sludge system before being wasted; also called sludge age or mean cell residence time. Operators adjust SRT by changing the WAS rate.
Trickling filterA secondary treatment unit consisting of a bed of media covered with a microbial biofilm over which wastewater is sprayed; organics are absorbed and metabolized as the liquid trickles through.
Bulking sludgeA condition in activated-sludge systems where filamentous microorganisms dominate, causing poor settling in the secondary clarifier and potential permit violations.
NPDES permitNational Pollutant Discharge Elimination System permit; a federal permit under the Clean Water Act that sets effluent limits a treatment plant must meet before discharging to waters of the United States.

Worked examples4

1. A primary clarifier receives an influent flow of 2.0 MGD with an influent TSS of 240 mg/L. The primary clarifier removes 60% of the TSS. What is the primary effluent TSS in mg/L?
  1. Identify the removal efficiency: 60%, or 0.60 as a decimal.
  2. Calculate TSS removed: 240 mg/L × 0.60 = 144 mg/L removed.
  3. Subtract TSS removed from influent TSS: 240 mg/L − 144 mg/L = 96 mg/L.
Answer: Primary effluent TSS = 96 mg/L
2. Using the same system (2.0 MGD, influent BOD = 200 mg/L), the primary clarifier removes 30% of the BOD. A secondary permit limit requires effluent BOD ≤ 30 mg/L. What is the primary effluent BOD, and does primary treatment alone meet the permit limit?
  1. Calculate BOD removed in the primary clarifier: 200 mg/L × 0.30 = 60 mg/L removed.
  2. Calculate primary effluent BOD: 200 mg/L − 60 mg/L = 140 mg/L.
  3. Compare to permit limit: 140 mg/L is far greater than 30 mg/L.
  4. Conclusion: Primary treatment alone does not meet the permit limit; biological secondary treatment is required to reduce BOD from 140 mg/L to 30 mg/L or less.
Answer: Primary effluent BOD = 140 mg/L; primary treatment alone does NOT meet the 30 mg/L permit limit.
3. An activated-sludge aeration basin holds 0.50 MG of mixed liquor at an MLSS of 2,400 mg/L. The plant wastes 12,000 lb of sludge per day. Calculate the SRT (sludge age) in days. (Use: lb = mg/L × MGD × 8.34)
  1. Calculate the total mass of solids in the aeration basin: MLSS (mg/L) × basin volume (MG) × 8.34 lb/gal = 2,400 × 0.50 × 8.34 = 10,008 lb. (Approximately 10,008 lb.)
  2. The mass of solids wasted per day (WAS + any solids leaving in effluent) is given as 12,000 lb/day. For this problem, use 12,000 lb/day as the total solids wasted.
  3. Calculate SRT: SRT = Total solids in system ÷ Solids wasted per day = 10,008 lb ÷ 12,000 lb/day ≈ 0.83 days.
  4. Interpret the result: An SRT of less than 1 day is extremely short and would result in a very young, poorly performing biomass. An operator would reduce WAS to raise SRT into a more appropriate operating range (commonly 5–15 days for conventional activated sludge).
Answer: SRT ≈ 0.83 days (well below the typical target; operator should reduce WAS rate to increase sludge age).
4. A primary clarifier treats 1.5 MGD of wastewater with influent TSS = 180 mg/L and removes 55% of TSS. Express the mass of TSS removed per day in pounds.
  1. Calculate the TSS concentration removed: 180 mg/L × 0.55 = 99 mg/L removed.
  2. Convert concentration removed to a daily mass using the water/wastewater conversion factor: lb/day = mg/L × flow (MGD) × 8.34 lb/gal.
  3. lb/day = 99 mg/L × 1.5 MGD × 8.34 = 1,238.49 lb/day.
  4. Round to appropriate significant figures: approximately 1,238 lb/day of TSS removed.
Answer: Approximately 1,238 lb/day of TSS are removed in the primary clarifier.

The math in this lesson1

Learners apply 60% TSS removal to 240 mg/L influent to find 96 mg/L primary effluent.
7.RP.3
Brush up in the math halls at newSchoolWorks.com:
In class: Using an influent characteristics data set (flow, BOD, TSS), calculate primary clarifier removal efficiency, effluent BOD, and effluent TSS; compare the results to typical secondary treatment permit limits to understand why biological treatment is needed. · Draw and label the components of a conventional activated-sludge system (aeration basin, secondary clarifier, return activated sludge pump, waste activated sludge pump) and explain the role of the mixed-liquor suspended solids concentration and sludge retention time in process performance. · Read two short case studies—one describing a healthy activated-sludge system and one describing a system with low dissolved oxygen—and answer guided questions about what process adjustments an operator would make.
Materials: Wastewater operator study guide (state or WPCF/WEF manual), BOD/TSS calculation worksheets, activated-sludge process diagram handout, case study handouts, calculator.
Check: All three primary-treatment calculations correct; activated-sludge diagram labeled with all major components and flow paths; both case-study response sets earn ≥ 70% on the instructor rubric.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

Wastewater Treatment Processes
1. What is the primary purpose of a primary clarifier in a wastewater treatment plant?
Wastewater Treatment Calculations
2. A primary clarifier receives wastewater with an influent BOD of 220 mg/L and removes 35% of the BOD. What is the primary effluent BOD?
Wastewater Treatment Processes
3. In an activated-sludge system, what does return activated sludge (RAS) accomplish?
Activated-Sludge Process Control
4. An operator notices that dissolved oxygen (DO) in the aeration basin has dropped to 0.5 mg/L and the secondary clarifier effluent is turbid with rising TSS. What is the most likely cause and the appropriate first response?
Activated-Sludge Process Control
5. Which of the following best describes sludge retention time (SRT) in an activated-sludge system?
Wastewater Treatment Processes
6. How does a trickling filter achieve secondary treatment?
Safety and Health
7. Before entering a primary clarifier vault for inspection, which action is required by OSHA confined-space regulations?
Wastewater Treatment Calculations
8. A plant has an aeration basin volume of 0.40 MG and an MLSS of 2,000 mg/L. The total solids wasted daily is 6,000 lb. What is the approximate SRT?
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