SME
Environmental Engineering · Problem type
F/M ratio and detention time of an aeration tank
Use this for “10,000 m³/d at 200 mg/L BOD enters a 3000 m³ tank held at 2500 mg/L MLVSS. What are F/M and the HRT?”
Diagram
This is an example. Type your own values to start a new problem.
F/M (day⁻¹)
=
Q (m³/day)
×
S₀ (mg/L)
V (m³)
×
X (mg/L)
τ (h)
=
V (m³)
Q (m³/day)
Formulas
F/M
=
Q
×
S₀
V
×
X
0.2667
=
10,000
×
200
3,000 × 2,500
τ
=
24
×
V
Q
7.2
=
24
×
3,000
10,000
Assumptions
•
MLVSS (mixed-liquor volatile suspended solids) stands for the live biomass X in the tank.
•
Steady flow; the tank is well mixed, so X is the same throughout.
•
F/M is the BOD fed each day over the biomass held; the HRT τ is V ÷ Q, with 24 h a day.
Step-by-step
WE KNOW
Q = 10,000 m³/day, S₀ = 200 mg/L, V = 3,000 m³, X = 2,500 mg/L
FIND
food-to-microorganism ratio (F/M), hydraulic retention time (τ)
Step 1 · Find food-to-microorganism ratio (F/M)
Use F/M = Q × S₀ ÷ (V × X)
The BOD fed each day over the biomass in the tank; the mg/L cancel.
F/M
=
10,000
×
200
3,000 × 2,500
F/M
=
10,000
×
200
7,500,000
F/M
=
2,000,000
7,500,000
F/M = 0.2667 day⁻¹
Step 2 · Find hydraulic retention time (τ)
Use τ = 24 × V ÷ Q
V
Q
is
in
days;
24
turns
it
into
hours.
τ
=
24
×
3,000
10,000
τ
=
72,000
10,000
τ
=
7.2
h
Check
Put the numbers back in. Both sides should match.
0.2667
=
10,000
×
200
3,000 × 2,500
✓
7.2
=
24
×
3,000
10,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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