Activated-sludge math (MLSS, F:M, SVI, sludge age)

Problem set: 6 worked problems

Each problem is original, authored from the underlying non-copyrightable relationship, with the full worked solution and the exact slip that produces each wrong answer. Work the problem before opening the solution. For a timed, randomized run with a per-topic score, use the free practice test.

Watch for Every activated-sludge number rides on the pounds formula, so keep aeration and clarifier volumes in MILLION gallons in those steps. The food-to-microorganism ratio is pounds over pounds (unitless). The sludge volume index inverts easily: it is settled volume times 1,000 over MLSS, not the reverse.

1. A plant applies 3.0 MGD of 180 mg/L BOD to an aeration basin holding 0.6 MG of 2,500 mg/L MLVSS. What is the food-to-microorganism (F:M) ratio?

   • A 0.72
   • B 0.18
   • C 2.78
   • D 0.36
   Show the worked solution

   Correct answer: D. 0.36

   BOD load = 3.0 x 180 x 8.34 = 4,504 lb/day. MLVSS = 0.6 x 2,500 x 8.34 = 12,510 lb. F:M = 4,504 / 12,510 = 0.36. Inverting (MLVSS over BOD) gives 2.78; leaving out a factor gives the other distractors.

   Source: Public relationship: F:M = lb BOD applied per day / lb MLVSS under aeration, each via the pounds formula

2. A 30-minute settling test reads 200 mL/L and the MLSS is 2,000 mg/L. What is the sludge volume index (SVI)?

   • A 10 mL/g
   • B 40 mL/g
   • C 400 mL/g
   • D 100 mL/g
   Show the worked solution

   Correct answer: D. 100 mL/g

   SVI = settled volume (mL/L) x 1,000 / MLSS (mg/L) = 200 x 1,000 / 2,000 = 100 mL/g. An SVI near 100 settles well. Dropping the 1,000 gives 0.1; inverting the ratio gives 10.

   Source: Public relationship: SVI = (settled volume mL/L x 1,000) / MLSS mg/L

3. A settleometer reads 240 mL/L at an MLSS of 2,000 mg/L. What is the SVI, and what does it suggest?

   • A 83 mL/g, a dense sludge
   • B 100 mL/g, an ideal sludge
   • C 480 mL/g, a dense sludge
   • D 120 mL/g, a slightly bulking sludge
   Show the worked solution

   Correct answer: D. 120 mL/g, a slightly bulking sludge

   SVI = 240 x 1,000 / 2,000 = 120 mL/g. Above about 100 to 120 the sludge is settling more slowly and may be starting to bulk. Inverting the ratio gives 83.

   Source: Public relationship: SVI = (settled volume mL/L x 1,000) / MLSS mg/L; higher SVI settles slower

4. How many pounds of MLSS are in a 1.0-MG aeration basin holding 2,500 mg/L?

   • A 2,500 lb
   • B 208,500 lb
   • C 2,085 lb
   • D 20,850 lb
   Show the worked solution

   Correct answer: D. 20,850 lb

   Pounds = MG x mg/L x 8.34 = 1.0 x 2,500 x 8.34 = 20,850 lb. This is the pounds formula with volume in million gallons. A decimal slip on the volume produces the distractors.

   Source: Public relationship: lb = volume (MG) x concentration (mg/L) x 8.34

5. A system holds 20,000 pounds of MLSS and loses 2,000 pounds of suspended solids per day. What is the sludge age?

   • A 40,000 days
   • B 22,000 days
   • C 0.1 day
   • D 10 days
   Show the worked solution

   Correct answer: D. 10 days

   Sludge age = pounds of MLSS in the system / pounds of solids leaving per day = 20,000 / 2,000 = 10 days. Inverting gives 0.1. Sludge age (mean cell residence time) is a core process-control lever.

   Source: Public relationship: sludge age = lb MLSS in system / lb SS leaving per day

6. Two plants run at the same MLSS, but plant A has a much higher F:M ratio than plant B. What does that indicate about plant A?

   • A Plant A must have a lower BOD load
   • B F:M says nothing about loading
   • C Plant A has an older sludge with more organisms
   • D Plant A carries a higher food load per pound of organisms, a younger, higher-rate sludge
   Show the worked solution

   Correct answer: D. Plant A carries a higher food load per pound of organisms, a younger, higher-rate sludge

   A higher F:M means more BOD (food) per pound of MLVSS (microorganisms), which corresponds to a younger, higher-rate, less fully oxidized sludge. Low F:M plants run older sludge with extended aeration. F:M is the loading-per-biomass lever.

   Source: Public relationship: higher F:M = more food per unit biomass = younger, higher-rate sludge