Food-to-Microorganism Ratio

F / M = \frac{Q S _{0}}{V X}

F/M is the organic loading per day on the mixed-liquor solids. Conventional ~0.2–0.5 /d (BOD and MLVSS). High F/M: young sludge, possible bulking; low F/M: endogenous, better settling, nitrification.

$Q$ (m³/d): Influent flow
$S_{0}$ (kg/m³): Influent BOD or COD
$X$ (kg/m³): MLVSS

Step-by-step solved example

Q=4000 m³/d, S0=200 mg/L=0.20 kg/m³, V=500 m³, X=2.5 kg/m³. F/M?

Food Q S0 into a tank of solids V X. F/M is that loading.

1. Food
Q S0=4000×0.20=800 kg/d.
2. F/M
800/(500×2.5)=800/1250=0.64 /d.
$F / M = 0.64 d^{- 1}$

Answer: 0.64 /d

10 practice questions

0/10 correct

1.F/M units are typically

2.Raising X at fixed Q S0

3.Extended aeration has

4.S0 should be

5.If Q doubles and everything else is fixed, F/M

6.High F/M may cause

7.F/M and θc together specify

8.Use MLVSS not MLSS when

9.Q=2000 m³/d, S0=0.15 kg/m³, VX=600 kg. F/M=

10.Complete-mix vs plug-flow F/M is