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Math problem set Grade 11 Science

Bramfield Water Recovery Plant: F:M Ratio and Solids Retention Time

A grade 11 problem set on activated sludge control: students compute BOD loading, an F:M ratio, solids retention time, and the wasting rate a ten-day target requires.

Objectives

  1. Students will apply the pounds formula to convert concentrations and flows into pounds per day.
  2. Students will calculate an F:M ratio and a solids retention time from plant operating data.
  3. Students will determine the wasting rate required to hold a target solids retention time.

Vocabulary

Activated sludge
Activated sludge is a living brown mixture of helpful bacteria and other microbes that wastewater plants grow on purpose. The microbes clean the water by eating the waste in it.

Also: mixed liquor, volatile suspended solids, F:M ratio, solids retention time, wasting

Teacher overview

An activated sludge plant is a managed microbial culture, and the operator's main levers are how much organic matter the bacteria receive and how long the bacteria stay in the system. This set works through one shift's data from the fictional Bramfield Water Recovery Plant, moving from influent loading to the food-to-microorganism ratio, then to solids retention time and the wasting rate a target retention time demands. Two errors show up repeatedly in this math. Students substitute mixed liquor suspended solids where the F:M ratio calls for the volatile fraction, which overstates the active biomass, and they divide a pounds inventory by a pounds-per-day rate without noticing that the answer comes out in days. Work the units out loud on the first two problems.

Bramfield Water Recovery Plant: F:M Ratio and Solids Retention Time — Grade 11 Science
Name: ____________________________ Date: ______________

Problems

Every problem describes the Bramfield Water Recovery Plant, a fictional facility. Use pounds per day = concentration in mg/L x flow in million gallons per day x 8.34 throughout.

Operating value Reading
Plant flow 2.40 MGD
Influent BOD 190 mg/L
Aeration basin volume 1.10 million gallons
Mixed liquor suspended solids 2,600 mg/L
Volatile fraction of the mixed liquor 78 percent
Waste sludge flow 22,000 gallons per day
Waste sludge concentration 7,800 mg/L
Effluent suspended solids 12 mg/L
Effluent BOD 8 mg/L
  1. Bramfield treats 2.40 MGD with an influent BOD of 190 mg/L. Calculate the BOD loading in pounds per day.
  2. The mixed liquor holds 2,600 mg/L of suspended solids, and 78 percent of that is volatile. Calculate the mixed liquor volatile suspended solids concentration in mg/L.
  3. The aeration basin holds 1.10 million gallons. Using your item 2 concentration, calculate the pounds of volatile suspended solids in the basin.
  4. The F:M ratio is the BOD loading in pounds per day divided by the pounds of volatile suspended solids in the basin. Using your item 1 and item 3 answers, calculate the F:M ratio, rounded to the nearest hundredth.
  5. Calculate the pounds of mixed liquor suspended solids in the 1.10-million-gallon aeration basin at 2,600 mg/L.
  6. Bramfield wastes 22,000 gallons per day of sludge at 7,800 mg/L. Calculate the pounds of solids wasted per day.
  7. The plant discharges 2.40 MGD of effluent carrying 12 mg/L of suspended solids. Calculate the pounds of solids leaving in the effluent each day.
  8. Solids retention time is the pounds of mixed liquor suspended solids in the basin divided by the total pounds of solids leaving the system each day. Using your item 5, 6, and 7 answers, calculate the solids retention time in days, rounded to the nearest tenth.
  9. Bramfield's operator wants to hold a solids retention time of 10.0 days. Using the item 5 inventory, the item 7 effluent solids, and a waste sludge concentration of 7,800 mg/L, calculate the waste sludge flow in gallons per day that would produce a 10.0-day retention time.
  10. Bramfield's influent BOD is 190 mg/L and its effluent BOD is 8 mg/L. Calculate the BOD removal efficiency as a percent, rounded to the nearest tenth, and state whether the effluent meets the secondary treatment standard of 30 mg/L.

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: 3,803 pounds per day.
    190 mg/L x 2.40 MGD x 8.34 = 3,803.04 pounds of BOD per day. Multiply 190 x 2.40 = 456 first, then 456 x 8.34 = 3,803.04.
  2. Answer: 2,028 mg/L.
    2,600 mg/L x 0.78 = 2,028 mg/L of volatile suspended solids. The remaining 572 mg/L is inert material that does not eat the incoming BOD.
  3. Answer: 18,605 pounds.
    2,028 mg/L x 1.10 million gallons x 8.34 = 18,604.87 pounds. Working stepwise: 2,028 x 1.10 = 2,230.8, and 2,230.8 x 8.34 = 18,604.87.
  4. Answer: 0.20 pounds of BOD per pound of volatile solids per day.
    3,803.04 pounds per day / 18,604.87 pounds = 0.2044, which rounds to 0.20. The pounds cancel, leaving a per-day ratio, and this value sits in the conventional range for a plant run for good settling.
  5. Answer: 23,852 pounds.
    2,600 mg/L x 1.10 million gallons x 8.34 = 23,852.4 pounds of mixed liquor suspended solids. Note this uses the full solids concentration, not the volatile fraction from item 2.
  6. Answer: 1,431 pounds per day.
    22,000 gallons per day is 0.022 million gallons per day. 7,800 mg/L x 0.022 MGD x 8.34 = 1,431.14 pounds per day. Working stepwise: 7,800 x 0.022 = 171.6, and 171.6 x 8.34 = 1,431.14.
  7. Answer: 240.2 pounds per day.
    12 mg/L x 2.40 MGD x 8.34 = 240.19 pounds per day. Even a clear effluent carries solids, and process control has to count them.
  8. Answer: 14.3 days.
    Total solids leaving per day = 1,431.14 + 240.19 = 1,671.33 pounds per day. 23,852.4 pounds / 1,671.33 pounds per day = 14.27 days, which rounds to 14.3 days. The pounds cancel and days remain.
  9. Answer: 32,974 gallons per day.
    A 10.0-day retention time requires removing 23,852.4 / 10.0 = 2,385.24 pounds of solids per day. The effluent already carries 240.19 pounds per day, so wasting must supply 2,385.24 - 240.19 = 2,145.05 pounds per day. One million gallons of 7,800 mg/L sludge holds 7,800 x 1.0 x 8.34 = 65,052 pounds, so the flow is 2,145.05 / 65,052 = 0.0329744 million gallons, or 32,974 gallons per day. That is an increase of about 11,000 gallons per day over the current wasting rate.
  10. Answer: 95.8 percent.
    (190 - 8) / 190 x 100 = 182 / 190 x 100 = 95.79, which rounds to 95.8 percent. An effluent BOD of 8 mg/L is well under the 30 mg/L secondary treatment standard, so the plant meets it with room to spare.

Differentiation

Support

Give students a filled-in units table for the pounds formula, with columns for concentration, flow in millions of gallons per day, and the constant, so each problem becomes a matter of placing three numbers. Work problem 6 with the class first, since converting 22,000 gallons per day into millions of gallons per day is where most students stall. Provide the item 5 inventory as a given for students who need to reach problems 8 and 9.

Extension

Bramfield thickens the wasted solids to 5.0 percent before hauling. Have students calculate the daily volume of thickened sludge produced at the item 9 wasting rate, using the fact that a gallon of water weighs 8.34 pounds, and check that the volume falls near 5,144 gallons per day. Ask them to explain why thickening matters to hauling cost even though the pounds of solids do not change.

ELL supports

Pre-teach mixed liquor, volatile, wasting, and retention time with a labeled flow diagram of the aeration basin, clarifier, return line, and waste line, marking where each measured value is taken. Provide the frame "The pounds of solids in the tank divided by the pounds leaving each day gives ___" so students can state the meaning of item 8 before computing it.