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Quiz & test questions Grade 12 Science

Woldmere Regional: A Wastewater and Receiving Water Test

A grade 12 science unit test worth 50 points on wastewater unit processes, loading and detention calculations, and how a discharge affects oxygen in the receiving stream.

Objectives

  1. Students will identify the sequence and purpose of wastewater treatment unit processes.
  2. Students will calculate loading, removal efficiency, detention time, and surface overflow rate.
  3. Students will explain how organic and nutrient loading affect oxygen in a receiving stream.

Vocabulary

Biochemical oxygen demand (BOD)
Biochemical oxygen demand, or BOD, measures how much oxygen tiny microbes will use up as they eat the waste in water. Dirtier water has a higher BOD.
Total suspended solids (TSS)
Total suspended solids is the amount of solid material floating or suspended in water, measured by filtering a sample and weighing what gets left behind. High TSS makes water look cloudy and can smother fish habitat and clog treatment equipment.
Detention time
Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.

Also: surface overflow rate, removal efficiency, oxygen sag

Teacher overview

This unit test covers wastewater treatment processes and their effect on a receiving stream, using the fictional Woldmere Regional Water Reclamation Facility as its single scenario. It is worth 50 points across sixteen items, mixing multiple choice, short answer, computation, and one extended response.

The computations draw on the pounds formula, detention time, removal efficiency, and surface overflow rate, all of which are supplied on the test. Students who have practiced these on a worksheet will find the arithmetic familiar, while the items that separate performance are the ones asking why a number matters to the stream rather than how to produce it.

Woldmere Regional: A Wastewater and Receiving Water Test — Grade 12 Science
Name: ____________________________ Date: ______________

Assessment

All values below describe the fictional Woldmere Regional Water Reclamation Facility, which discharges to Calbourne Brook. Use these values for every item.

  • Plant flow: 5.5 MGD
  • Influent BOD: 210 mg/L; influent TSS: 240 mg/L
  • Effluent BOD: 18 mg/L; effluent TSS: 15 mg/L
  • Aeration basin volume: 1.2 million gallons
  • Final clarifier: circular, 70 feet in diameter
  • Secondary treatment standard: monthly average BOD and TSS at or below 30 mg/L
  • Most fish need dissolved oxygen above 5 mg/L
  • Pounds formula: pounds per day = concentration (mg/L) x flow (MGD) x 8.34
  • Detention time (hours) = basin volume (MG) / flow (MGD) x 24
  • Removal efficiency (percent) = (influent - effluent) / influent x 100
  • Surface overflow rate = flow (gpd) / clarifier surface area (sq ft); circular area = 0.785 x diameter squared

  1. (2 points) Which sequence correctly orders the plant's unit processes?
    A. Screening, primary clarification, aeration, final clarification, disinfection
    B. Primary clarification, screening, disinfection, aeration, final clarification
    C. Aeration, screening, primary clarification, disinfection, final clarification
    D. Screening, aeration, disinfection, primary clarification, final clarification

  2. (2 points) A primary clarifier chiefly removes which of the following?
    A. Dissolved nitrogen and phosphorus
    B. Settleable solids and floating material
    C. Bacteria and viruses
    D. Dissolved organic matter measured as BOD

  3. (3 points) Define biochemical oxygen demand and state what a high value in a discharge tells an operator about its likely effect on a stream.

  4. (3 points) Calculate the BOD load entering the plant, in pounds per day.

  5. (3 points) Calculate the plant's BOD removal efficiency as a percent. Round to one decimal place.

  6. (2 points) The 30 mg/L monthly average secondary standard applies to which pair of parameters?
    A. Chlorine residual and pH
    B. Nitrogen and phosphorus
    C. BOD and TSS
    D. Turbidity and dissolved oxygen

  7. (3 points) Calculate the detention time in the aeration basin, in hours, at the plant flow. Round to one decimal place.

  8. (3 points) Calculate the surface overflow rate of the final clarifier, in gallons per day per square foot. Round to the nearest whole number.

  9. (4 points) Explain why dissolved oxygen falls in a stream for some distance below a discharge carrying a high BOD load, and why it recovers farther downstream.

  10. (3 points) Calculate the mass of total suspended solids the plant discharges each day, in pounds per day.

  11. (3 points) Over one week, effluent TSS rises sharply while effluent BOD stays low. Which explanation best fits that pattern?
    A. The disinfection dose was increased
    B. Solids are washing out of the final clarifier faster than they settle
    C. The influent flow dropped by half
    D. Dissolved oxygen in the aeration basin fell to zero

  12. (4 points) The plant meets its BOD and TSS limits, yet Calbourne Brook develops thick algae growth downstream and its dissolved oxygen swings widely between night and day. Explain how an effluent that meets those limits can still produce that result.

  13. (3 points) Calculate the plant's TSS removal efficiency as a percent. Round to one decimal place.

  14. (3 points) Explain why the plant monitors both BOD and TSS instead of relying on either one alone.

  15. (3 points) During a wet-weather event, flow through the plant reaches 11.0 MGD. Calculate the detention time in the aeration basin at that flow, in hours. Round to one decimal place.

  16. (6 points) Recommend one change the fictional Woldmere plant should make to protect Calbourne Brook during wet weather. Support your recommendation with two specific pieces of evidence drawn from your answers on this test, and name one cost or trade-off the recommendation carries.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: A.
    Wastewater moves from screening, which protects the pumps, to primary clarification, which settles solids, to aeration, where bacteria consume dissolved organic matter, to final clarification, which settles those bacteria, and finally to disinfection before discharge.

  2. Answer: B.
    Primary clarifiers work by gravity, so they capture what settles and what floats. Dissolved material passes through, which is why the aeration basin follows.

  3. Answer: BOD is the oxygen microbes consume while breaking down organic matter in a water sample over a set period.
    Award 2 points for a correct definition naming oxygen consumption by microorganisms and 1 point for the consequence: a high BOD discharge means the receiving stream's own microbes will draw heavily on dissolved oxygen, which can push the stream below the level fish need.

  4. Answer: 9,633 pounds per day.
    210 mg/L x 5.5 MGD x 8.34 = 1,155 x 8.34 = 9,632.7, which rounds to 9,633 pounds per day.

  5. Answer: 91.4 percent.
    (210 - 18) / 210 = 192 / 210 = 0.914286, which is 91.4 percent removal. That clears the 85 percent minimum removal the secondary treatment standard requires.

  6. Answer: C.
    The secondary treatment standard sets monthly average BOD and TSS at or below 30 mg/L. Chlorine, pH, and nutrients are governed by other parts of a permit.

  7. Answer: 5.2 hours.
    1.2 MG / 5.5 MGD = 0.21818 days. Multiplying by 24: 0.21818 x 24 = 5.236 hours, which rounds to 5.2 hours.

  8. Answer: 1,430 gallons per day per square foot.
    Area = 0.785 x 70 x 70 = 3,846.5 square feet. 5,500,000 gallons per day / 3,846.5 square feet = 1,429.9, which rounds to 1,430 gallons per day per square foot.

  9. Answer: Microbes in the stream consume oxygen as they break down the discharged organic matter, and reaeration restores it downstream.
    Award 2 points for the demand side: bacteria below the outfall metabolize the organic load and draw dissolved oxygen down faster than the stream replaces it, producing the lowest oxygen some distance downstream rather than at the pipe. Award 2 points for the recovery side: as the organic matter is consumed the demand falls, while oxygen keeps entering from the atmosphere and from photosynthesis, especially where the channel is turbulent, so the concentration climbs back toward saturation.

  10. Answer: 688 pounds per day.
    15 mg/L x 5.5 MGD x 8.34 = 82.5 x 8.34 = 688.05, which rounds to 688 pounds per day.

  11. Answer: B.
    Solids carryover raises TSS directly while leaving dissolved BOD largely unaffected, which matches the pattern. More chlorine would not add solids, a halved flow would not raise concentration this way, and a total loss of oxygen in the aeration basin would drive BOD up along with TSS.

  12. Answer: Nitrogen and phosphorus in the effluent fertilize algae, and the algae drive the oxygen swings.
    Award 2 points for identifying that BOD and TSS limits say nothing about nutrients, which pass through conventional secondary treatment. Award 2 points for the mechanism: nutrients support heavy algal growth, the algae add oxygen by day through photosynthesis and consume it all night through respiration, producing the wide daily swing, and when the bloom dies its decay consumes still more oxygen.

  13. Answer: 93.8 percent.
    (240 - 15) / 240 = 225 / 240 = 0.9375, which is 93.8 percent when rounded to one decimal place.

  14. Answer: The two measure different things, and a plant can fail one while passing the other.
    BOD tracks dissolved and suspended organic matter that will consume oxygen in the stream, while TSS tracks particles regardless of whether they are organic. Solids can wash out of a clarifier without a matching rise in dissolved organic matter, and a plant can also discharge oxygen-demanding dissolved material with very few particles. Award full credit for naming both parameters and giving one situation in which they diverge.

  15. Answer: 2.6 hours.
    1.2 MG / 11.0 MGD = 0.10909 days. Multiplying by 24: 0.10909 x 24 = 2.618 hours, which rounds to 2.6 hours. That is half the dry-weather detention time, leaving the bacteria far less contact time with the wastewater.

  16. Answer: Accept any recommendation supported by two calculations from this test and one named trade-off.
    Award 3 points for a specific, plausible recommendation, such as equalization storage to hold peak wet-weather flow and release it slowly, a larger or additional final clarifier, or separating stormwater from the collection system. Award 2 points for two pieces of evidence drawn from the students' own answers, such as the detention time falling from 5.2 hours to 2.6 hours in item 15, the surface overflow rate of 1,430 gallons per day per square foot in item 8 rising with flow, or the 688 pounds per day of solids in item 10 increasing if carryover occurs. Award 1 point for naming a genuine trade-off, such as capital cost, land required for a storage basin, or the years a sewer separation project would take.

Differentiation

Support

Give students a formula sheet with one worked example for each of the four computations, using a different fictional plant so the numbers cannot be copied. Allow the extended response in item 16 to be answered as a labeled list rather than a paragraph. For items 9 and 12, provide a blank oxygen-versus-distance sketch that students can annotate as part of the answer.

Extension

Ask students to calculate what effluent BOD concentration the fictional Woldmere plant would need in order to cut its daily BOD load in half, and to state whether that is achievable with the treatment described. A second option is to have them estimate how many hours of the 11.0 MGD wet-weather flow the plant could absorb if a 2.0 million gallon equalization basin were added.

ELL supports

Provide a bilingual glossary card for the six vocabulary terms, each paired with a small diagram of the unit process it belongs to. Allow extra time and permit students to answer items 3, 14, and 16 in short numbered sentences instead of paragraphs. Read the item stems aloud once at the start for students who benefit from hearing the language of the question.

Extensions

Students who finish early can sketch the dissolved oxygen profile of the fictional Calbourne Brook from one mile above the outfall to five miles below it, labeling the point of lowest oxygen and explaining what sets its position. A second option is to look up the discharge monitoring reports a real plant files with its state and describe which of the parameters on this test appear in those reports.