CivilSolve

Environmental

BOD (Dilution Method)

Biochemical oxygen demand from the standard dilution bottle test: the oxygen depletion of the diluted mixture times the dilution factor, BOD = (DO_initial − DO_final)·(V_bottle/V_sample) — with the Standard Methods validity checks (at least 2 mg/L depletion and at least 1 mg/L residual DO). Use for: 'calculate the BOD5 of the wastewater', dilution-factor problems. Not for: seeded-dilution corrections or ultimate-BOD kinetics.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

standard 5/300 dilution (the motivating exam problem)

Given

sample volume
5 mL
bottle volume
300 mL
initial do
8.5
final do
3.5

Assumptions

  • Unseeded dilution test per Standard Methods; the dilution water exerts negligible oxygen demand.

Solution steps

  1. Oxygen depletion of the mixture

    The drop in dissolved oxygen over the incubation is the demand exerted by the DILUTED sample.

    ΔDO = DOi − DOf

    ΔDO = 8.5 − 3.5 = 5 mg/L

  2. Scale by the dilution factor

    The raw wastewater is stronger than the mixture by the ratio of bottle to sample volume.

    BOD = ΔDO · (V_bottle / V_sample)

    dilution factor = 60; BOD = 5 × 60 = 300 mg/L

    = 300

Results

BOD (5-day) in mg/L

300

DO depletion of the mixture (mg/L)

5

Dilution factor

60

Where this answer was checked
source
BOD = (DOi − DOf)·(V_bottle/V_sample) — 5 mL waste in a 300 mL bottle, 8.5 → 3.5 mg/L
verified by
hand-recomputed
derivation
ΔDO = 5 mg/L; dilution factor 300/5 = 60; BOD5 = 300 mg/L.

Example 2

10/300 dilution

Given

sample volume
10 mL
bottle volume
300 mL
initial do
9
final do
4.2

Assumptions

  • Unseeded dilution test per Standard Methods; the dilution water exerts negligible oxygen demand.

Solution steps

  1. Oxygen depletion of the mixture

    The drop in dissolved oxygen over the incubation is the demand exerted by the DILUTED sample.

    ΔDO = DOi − DOf

    ΔDO = 9 − 4.2 = 4.8 mg/L

  2. Scale by the dilution factor

    The raw wastewater is stronger than the mixture by the ratio of bottle to sample volume.

    BOD = ΔDO · (V_bottle / V_sample)

    dilution factor = 30; BOD = 4.8 × 30 = 144 mg/L

    = 144

Results

BOD (5-day) in mg/L

144

DO depletion of the mixture (mg/L)

4.8

Dilution factor

30

Where this answer was checked
source
Same formula, different fraction — 10 mL sample, 9.0 → 4.2 mg/L
verified by
hand-recomputed
derivation
ΔDO = 4.8; DF = 30; BOD = 144 mg/L.

Example 3

low residual DO triggers the validity warning

Given

sample volume
15 mL
bottle volume
300 mL
initial do
8
final do
0.5

Assumptions

  • Unseeded dilution test per Standard Methods; the dilution water exerts negligible oxygen demand.

Solution steps

  1. Oxygen depletion of the mixture

    The drop in dissolved oxygen over the incubation is the demand exerted by the DILUTED sample.

    ΔDO = DOi − DOf

    ΔDO = 8 − 0.5 = 7.5 mg/L

  2. Scale by the dilution factor

    The raw wastewater is stronger than the mixture by the ratio of bottle to sample volume.

    BOD = ΔDO · (V_bottle / V_sample)

    dilution factor = 20; BOD = 7.5 × 20 = 150 mg/L

    = 150

Results

BOD (5-day) in mg/L

150

DO depletion of the mixture (mg/L)

7.5

Dilution factor

20

Notes

  • Residual DO under 1 mg/L — the bottle may have gone anoxic and capped the demand; the true BOD could be higher (use a smaller sample fraction).
Where this answer was checked
source
Standard Methods: residual DO ≥ 1 mg/L — bottle nearly anoxic
verified by
hand-recomputed
derivation
ΔDO = 7.5; DF = 20; BOD = 150 mg/L, flagged unreliable (DOf = 0.5 < 1).