F:M Ratio with MLVSS Derivation
Calculate the F:M ratio when you must derive MLVSS from MLSS and VSS%. Multi-step problem with primary effluent BOD, dual basins, and extraneous data.
Answer Choices
A) 0.08 Correct
Correct. Primary effluent BOD = 240 × (1 − 0.30) = 168 mg/L, so Food = 168 mg/L × 3.5 MGD × 8.34 = 4,904 lb/day. Total aeration volume = 2 × (200×80×15 ft³ × 7.48 gal/ft³) = 3.5904 MG, MLVSS = 2,800×0.75 = 2,100 mg/L, so MLVSS mass = 2,100 mg/L × 3.5904 MG × 8.34 = 62,882 lb. F:M = 4,904/62,882 = 0.078 ≈ 0.08; sanity check: this is a reasonable low F:M for a higher-solids/extended-aeration style operation.
B) 0.06
Wrong because it uses MLSS (2,800 mg/L) as if it were MLVSS instead of converting with the 75% volatile fraction. That makes the microorganism mass too large and drives the F:M too low (about 0.06).
C) 0.11
Wrong because it forgets the 30% primary BOD removal and uses 240 mg/L as the aeration influent BOD. That overstates the food load and makes F:M too high (about 0.11).
D) 0.16
Wrong because it uses only ONE aeration basin volume instead of both basins in service. Using half the correct MLVSS mass roughly doubles the F:M (about 0.16).
Step-by-Step Solution
- Primary effluent BOD (mg/L) = 240 mg/L × (1 − 0.30) = 168 mg/L
- Food (lb BOD/day) = 168 mg/L × 3.5 MGD × 8.34 (lb/MG per mg/L) = 4,903.92 lb/day
- One basin volume (ft³) = 200 ft × 80 ft × 15 ft = 240,000 ft³
- One basin volume (gal) = 240,000 ft³ × 7.48 gal/ft³ = 1,795,200 gal
- One basin volume (MG) = 1,795,200 gal ÷ 1,000,000 (gal/MG) = 1.7952 MG
- Total aeration volume (MG) = 1.7952 MG/basin × 2 basins = 3.5904 MG
- MLVSS concentration (mg/L) = MLSS × VSS fraction = 2,800 mg/L × 0.75 = 2,100 mg/L
- MLVSS mass in aeration (lb) = 2,100 mg/L × 3.5904 MG × 8.34 (lb/MG per mg/L) = 62,882.27 lb
- F:M (lb/day per lb) = 4,903.92 lb/day ÷ 62,882.27 lb = 0.0780 ≈ 0.08
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