SME
Environmental Engineering · Problem type
BOD from a dilution test
Use this for “6 mL of wastewater in a 300 mL bottle drops the DO from 8.8 to 4.3 mg/L in 5 days. What is the BOD₅?”
Chart
8.8 mg/L
4.3 mg/L
This is an example. Type your own values to start a new problem.
P
=
Vₛ (mL)
Vb (mL)
BOD₅ (mg/L)
=
D₁ (mg/L)
−
D₂ (mg/L)
P
Vₛ is the wastewater put in the bottle and Vb the bottle’s volume, so P is the sample’s share of the bottle.
Formulas
P
=
Vₛ
Vb
0.02
=
6
300
BOD₅
=
D₁
−
D₂
P
225
=
8.8 − 4.3
0.02
Assumptions
•
DO (dissolved oxygen) is read at the start and after 5 days at 20 °C in the dark.
•
Unseeded: the dilution water itself uses no oxygen, so the whole drop is the sample’s.
•
A valid bottle uses at least 2 mg/L of DO and leaves at least 1 mg/L.
Step-by-step
WE KNOW
D₁ = 8.8 mg/L, D₂ = 4.3 mg/L, Vₛ = 6 mL, Vb = 300 mL
FIND
sample fraction (P), five-day BOD (BOD₅)
Step 1 · Find sample fraction (P)
Use P = Vₛ ÷ Vb
The sample’s share of the bottle; the rest is dilution water.
P
=
6
300
P
=
0.02
Step 2 · Find five-day BOD (BOD₅)
Use BOD₅ = (D₁ − D₂) ÷ P
The oxygen used in the bottle, scaled up from the diluted sample to full strength.
BOD₅
=
8.8 − 4.3
0.02
BOD₅
=
4.5
0.02
BOD₅ = 225 mg/L
Check
Put the numbers back in. Both sides should match.
0.02
=
6
300
✓
225
=
8.8 − 4.3
0.02
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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