FM Ratio Calculator
FM Ratio:
0.00
kg BOD/day per kg MLVSS
Food Mass:
0.00
kg BOD/day
Microorganism Mass:
0.00
kg MLVSS
About FM Ratio
What is FM Ratio?
The Food to Microorganism (FM) Ratio is a critical parameter in biological wastewater treatment that indicates the balance between organic loading (food) and the microorganisms available to process it.
FM Ratio Formula:
FM Ratio = (BOD × Flow Rate) ÷ (MLVSS × Aeration Tank Volume)
Where:
- BOD = Biochemical Oxygen Demand (mg/L)
- Flow Rate = Influent flow rate (m³/hr)
- MLVSS = Mixed Liquor Volatile Suspended Solids (mg/L)
- Volume = Aeration tank volume (m³)
Example Calculation:
For BOD = 350 mg/L, Flow = 600 m³/hr, MLVSS = 3000 mg/L, Volume = 6700 m³:
- Food Mass = (350 mg/L × 600 m³/hr × 24 hr/day) ÷ 1,000,000 = 5040 kg BOD/day
- Microorganism Mass = (3000 mg/L × 6700 m³) ÷ 1,000,000 = 20,100 kg MLVSS
- FM Ratio = 5040 ÷ 20,100 = 0.25 kg BOD/day per kg MLVSS
Typical FM Ratio Ranges:
| Process Type | FM Ratio Range (kg BOD/day/kg MLVSS) |
|---|---|
| Conventional Activated Sludge | 0.2 - 0.6 |
| Extended Aeration | 0.05 - 0.15 |
| Contact Stabilization | 0.2 - 0.6 |
| High Rate Treatment | 0.6 - 1.5 |
Importance in Wastewater Treatment:
- Process Control: Maintains optimal biological activity in aeration tanks
- Sludge Quality: Affects sludge settling characteristics and effluent quality
- Oxygen Requirements: Determines aeration needs and energy consumption
- System Design: Guides sizing of aeration tanks and return sludge rates
Interpretation Guidelines:
High FM Ratio (>0.6)
Indicates insufficient microorganisms for the organic load. May cause poor treatment, high effluent BOD, and sludge bulking.
Optimal Range (0.2-0.6)
Provides balanced microbial growth and efficient organic matter removal. Produces good quality sludge.
Low FM Ratio (<0.2)
Suggests excess microorganisms. May lead to endogenous respiration, pin floc, and increased sludge age.