Worksheet Grade 12 Math
Deblin Creek Reclamation: F/M Ratio, Sludge Age, and Loading
A grade 12 wastewater worksheet on activated sludge process control: solids inventory, BOD loading, the F/M ratio, sludge age, and removal efficiency.
Objectives
- Students will calculate solids inventory and BOD loading in pounds per day using the pounds formula.
- Students will compute the F/M ratio and sludge age and judge whether an activated sludge plant is in its normal operating range.
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: MLSS, MLVSS, F/M ratio, sludge age
Teacher overview
An activated sludge plant is a living process: operators keep the right mass of microorganisms in the aeration tank relative to the amount of incoming waste, a balance measured by the food-to-microorganism, or F/M, ratio. This worksheet uses the fictional Deblin Creek Water Reclamation Facility to walk students through the process-control math an operator runs each shift. A common misconception to correct is that the "food" and the "microorganisms" in the F/M ratio use the same units; the food is a loading in pounds per day, while the microorganism mass is a standing inventory in pounds, so the ratio has units of per day.
Student Worksheet
Deblin Creek Water Reclamation Facility is a fictional plant. Use these readings for every item, and use the pounds formula (pounds per day equals concentration in mg/L times flow in MGD times 8.34) and the same 8.34 factor for the tank inventory.
- Plant flow: 2.5 MGD
- Influent BOD to the aeration tank: 160 mg/L
- Aeration tank volume: 0.60 MG
- Mixed liquor suspended solids (MLSS): 2,400 mg/L
- Volatile fraction: MLVSS is 70 percent of MLSS
- Solids entering the aeration tank: 150 mg/L
- Final effluent BOD: 12 mg/L
- Target F/M range for this process: 0.2 to 0.5 per day
- Calculate the MLSS inventory in the aeration tank, in pounds, using MLSS times tank volume times 8.34.
- Calculate the incoming BOD load (the food), in pounds per day, using the pounds formula.
- Find the MLVSS concentration in mg/L, which is 70 percent of the MLSS.
- Calculate the MLVSS inventory in the aeration tank, in pounds, using the MLVSS concentration, tank volume, and 8.34.
- Calculate the F/M ratio, in pounds of BOD per pound of MLVSS per day, by dividing the BOD load from item 2 by the MLVSS inventory from item 4. Round to the nearest hundredth.
- The target F/M range for this process is 0.2 to 0.5 per day. Is Deblin Creek inside that range?
- Calculate the sludge age in days: MLSS inventory from item 1 divided by the pounds of solids entering the aeration tank per day. Round to the nearest tenth of a day.
- Calculate the plant's BOD removal efficiency as a percent, using influent 160 mg/L and effluent 12 mg/L. Round to the nearest tenth.
- The secondary treatment limit for effluent BOD is a monthly average of 30 mg/L or less. Does the 12 mg/L effluent meet that limit?
- Suppose a high-strength industrial discharge doubles the influent BOD to 320 mg/L at the same 2.5 MGD flow, while MLVSS holds at your item 3 value. Calculate the new F/M ratio and state whether it is still inside the 0.2 to 0.5 range.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: 12,010 pounds. MLSS inventory equals 2,400 mg/L times 0.60 MG times 8.34: 2,400 times 0.60 is 1,440; 1,440 times 8.34 is 12,009.6, or about 12,010 pounds.
- Answer: 3,336 pounds per day. BOD load equals 160 times 2.5 times 8.34: 160 times 2.5 is 400; 400 times 8.34 is 3,336 pounds per day.
- Answer: 1,680 mg/L. MLVSS equals 0.70 times 2,400 mg/L, which is 1,680 mg/L.
- Answer: 8,407 pounds. MLVSS inventory equals 1,680 times 0.60 times 8.34: 1,680 times 0.60 is 1,008; 1,008 times 8.34 is 8,406.72, or about 8,407 pounds.
- Answer: 0.40 per day. F/M equals 3,336 pounds per day divided by 8,406.72 pounds, which is 0.397, rounding to 0.40 per day.
- Answer: Yes. An F/M of 0.40 per day falls between 0.2 and 0.5 per day, so the plant is inside its target range.
- Answer: 3.8 days. The solids entering per day equal 150 times 2.5 times 8.34, which is 3,127.5 pounds per day; sludge age is 12,009.6 divided by 3,127.5, which is 3.84, or about 3.8 days.
- Answer: 92.5 percent. Removal efficiency equals (160 minus 12) divided by 160, times 100: 148 divided by 160 is 0.925, or 92.5 percent.
- Answer: Yes. The effluent BOD of 12 mg/L is below the 30 mg/L secondary limit, so the plant meets the standard.
- Answer: 0.79 per day. The new BOD load is 320 times 2.5 times 8.34, which is 6,672 pounds per day; dividing by 8,406.72 pounds gives 0.79 per day, which is above the 0.5 upper limit, so the plant is out of range.
Differentiation
Support
Give students a worked version of item 1 and item 2 side by side so they can see that inventory uses tank volume in MG while loading uses flow in MGD. Provide a process diagram of the fictional aeration tank labeling where BOD enters and where mixed liquor is measured.
Extension
Ask students to find the daily pounds of solids the plant must waste to keep MLSS steady if the process produces about 0.6 pounds of new solids for every pound of BOD removed, then explain how wasting changes the sludge age.
ELL supports
Pre-teach "loading," "inventory," "food," and "microorganism" with paired visuals, and clarify that "food" here means incoming waste, not literal food. Provide a labeled formula card showing which values go on top and bottom of the F/M ratio.
Extensions
Have students track a week of fictional daily F/M values you provide and decide on which day the operator should have started wasting more sludge. Connect the exercise to how a real plant's process-control lab reports guide daily wasting decisions.