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Math problem set Grade 12 Careers & CTE

Ludderby Creek Reclamation Plant: Operator Loading Math

Grade 12 operator-exam problem set on BOD and TSS loading, removal efficiency, basin volume, and detention time at a fictional water reclamation facility.

Objectives

  1. Students will calculate influent and effluent BOD loading in pounds per day using concentration, flow, and the 8.34 pounds-per-gallon constant.
  2. Students will compute percent removal efficiency for BOD and TSS from influent and effluent concentrations.
  3. Students will find basin volumes in gallons and convert them to detention times in hours at a stated plant flow.

Vocabulary

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.
Clarifier
A clarifier is a big round or rectangular tank where water sits calmly so solids can sink to the bottom and be scraped away. Treatment plants use clarifiers at more than one stage of cleaning.

Also: pounds formula, removal efficiency, mixed liquor, BOD, TSS

Teacher overview

The Ludderby Creek Water Reclamation Facility is a fictional plant serving the town of Ludderby, and every problem in this set draws on a single day of its operating data. Students work the loading, efficiency, volume, and detention-time calculations that appear on Class II and Class III wastewater operator certification exams, carrying units through each step the way a licensed operator is required to. The misconception this set is designed to correct is the habit of treating the pounds formula as an unlabeled multiplication of three numbers, because students routinely enter flow in gallons per day instead of million gallons per day and inflate every loading answer by a factor of one million. Require students to write the unit beside each factor before they multiply, and check problems 1 and 2 for that specific error.

Ludderby Creek Reclamation Plant: Operator Loading Math — Grade 12 Careers & CTE
Name: ____________________________ Date: ______________

Problems

Use the following operating data from one day at the Ludderby Creek Water Reclamation Facility, which serves the town of Ludderby in this scenario. Average daily flow is 2.4 MGD. Influent BOD is 240 mg/L and effluent BOD is 18 mg/L. Influent TSS is 210 mg/L and effluent TSS is 21 mg/L. The aeration basin is a rectangular tank 90 feet long, 30 feet wide, and 15 feet deep. The final clarifier is a cylinder 60 feet in diameter with a side water depth of 12 feet. Use 8.34 pounds per gallon for the weight of water and 7.48 gallons per cubic foot.

  1. Calculate the influent BOD loading in pounds per day, using the formula loading (lbs/day) = concentration (mg/L) x flow (MGD) x 8.34.
  2. Calculate the effluent BOD loading in pounds per day at the same flow.
  3. Calculate the plant's BOD removal efficiency as a percent, using removal efficiency (%) = (influent - effluent) / influent x 100. Round to one decimal place.
  4. Calculate the plant's TSS removal efficiency as a percent. Round to one decimal place.
  5. Find the aeration basin's volume in gallons. Give the volume in cubic feet first, then convert.
  6. Find the aeration basin's detention time in hours at the plant's average daily flow. Round to one decimal place.
  7. Find the final clarifier's volume in gallons, using cylindrical volume (cu ft) = 0.785 x diameter squared x depth.
  8. Find the final clarifier's detention time in hours at the plant's average daily flow. Round to one decimal place.
  9. Convert the plant's average daily flow to gallons per minute, using 1 MGD = 694 gpm. Round to the nearest whole gallon per minute.
  10. Ludderby's discharge permit sets a monthly-average effluent BOD limit of 30 mg/L, matching the federal secondary treatment standard. Calculate the pounds per day of BOD the plant would discharge if its effluent sat exactly at that limit.

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: 4,804 lbs/day. Loading = 240 mg/L x 2.4 MGD x 8.34 lb/gal. First, 240 x 2.4 = 576. Then 576 x 8.34 = 4,803.84, which rounds to 4,804 pounds per day. Students who enter 2,400,000 instead of 2.4 will report a figure one million times too large.
  2. Answer: 360.3 lbs/day. Loading = 18 mg/L x 2.4 MGD x 8.34 lb/gal. First, 18 x 2.4 = 43.2. Then 43.2 x 8.34 = 360.288, which rounds to 360.3 pounds per day.
  3. Answer: 92.5 percent. Removal efficiency = (240 - 18) / 240 x 100. The difference is 222 mg/L, and 222 / 240 = 0.925, which is 92.5 percent.
  4. Answer: 90.0 percent. Removal efficiency = (210 - 21) / 210 x 100. The difference is 189 mg/L, and 189 / 210 = 0.900, which is 90.0 percent.
  5. Answer: 302,940 gallons. Volume = 90 ft x 30 ft x 15 ft = 40,500 cubic feet. Converting, 40,500 cu ft x 7.48 gal/cu ft = 302,940 gallons.
  6. Answer: 3.0 hours. Detention time in days = 302,940 gal / 2,400,000 gal per day = 0.126225 day. Multiplying by 24 hours per day gives 3.029 hours, which rounds to 3.0 hours.
  7. Answer: 253,662 gallons. Volume = 0.785 x 60 ft x 60 ft x 12 ft. First, 60 x 60 = 3,600, and 0.785 x 3,600 = 2,826. Then 2,826 x 12 = 33,912 cubic feet. Converting, 33,912 x 7.48 = 253,661.76 gallons, which rounds to 253,662 gallons.
  8. Answer: 2.5 hours. Detention time in days = 253,662 gal / 2,400,000 gal per day = 0.105693 day. Multiplying by 24 gives 2.537 hours, which rounds to 2.5 hours.
  9. Answer: 1,666 gpm. Flow in gpm = 2.4 MGD x 694 gpm per MGD = 1,665.6, which rounds to 1,666 gallons per minute.
  10. Answer: 600.5 lbs/day. Loading = 30 mg/L x 2.4 MGD x 8.34 lb/gal. First, 30 x 2.4 = 72. Then 72 x 8.34 = 600.48, which rounds to 600.5 pounds per day. Comparing this figure with the answer to problem 2 shows the plant discharging well under its permitted load.

Differentiation

Support

Give students a printed formula card carrying the pounds formula, the removal-efficiency formula, the cylinder volume formula, and the two conversion constants used here, so recall is not the barrier. Work problem 1 aloud with the class, writing each factor with its unit label on the board, then let students attempt problem 2 independently before checking. For problems 5 through 8, provide a sketch of each tank with its dimensions labeled.

Extension

Ask students to calculate the sludge age or the food-to-microorganism ratio for the same plant after you supply mixed liquor suspended solids and waste sludge figures of your choosing. Alternatively, have students determine what flow would push the clarifier's detention time below two hours, then explain what an operator would expect to see in effluent quality at that point.

ELL supports

Pre-teach the terms influent, effluent, loading, detention time, and removal efficiency with a labeled process diagram that traces water from the headworks through aeration to the final clarifier. Allow students to record answers as a labeled calculation rather than in complete sentences, and pair the word "loading" with its unit, pounds per day, everywhere it appears.

Extensions

Invite an operator from a nearby water resource recovery facility to bring one month of daily monitoring reports and walk students through how those numbers become the discharge monitoring report submitted to the state. Students can also research the certification classes their own state uses, list the exam topics each class covers, and match the problems in this set to the exam sections they would appear under.