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Biochemical Oxygen Demand Calculator

BOD at time t equals ultimate BOD times 1 minus 10 to the power of negative k times t

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BOD Equation

Biochemical Oxygen Demand measures how much dissolved oxygen microorganisms consume while breaking down organic matter in water. The standard 5-day BOD test (BOD₅) is the most common water-quality benchmark.

BODt = L(1 − 10⁻ᵏᵗ)

Unseeded BOD

The unseeded BOD test measures dissolved oxygen depletion in a sample that contains enough native bacteria for decomposition.

BOD = (D₁ − D₂) / P

Seeded BOD

For samples with few native bacteria, a seeded BOD test adds a known microbial culture. The seed’s own oxygen consumption is subtracted to isolate the sample’s demand.

BOD = ((D₁ − D₂) − f(B₁ − B₂)) / P

How It Works

Biochemical Oxygen Demand (BOD) measures how much dissolved oxygen microorganisms consume while breaking down organic matter in water. The BOD equation relates oxygen demand at any time to the ultimate BOD and a deoxygenation rate constant. The standard 5-day BOD test (BOD₅) is the most common water-quality benchmark. For samples with few native bacteria (e.g., industrial effluent), a seeded BOD test adds a known microbial culture. The seed’s own oxygen consumption is subtracted to isolate the sample’s demand.

Example Problem

A wastewater sample has an ultimate BOD (L) of 300 mg/L and a deoxygenation rate constant (k) of 0.23 per day. What is BOD₅?

  1. Identify the knowns. Ultimate BOD L = 300 mg/L, deoxygenation rate constant k = 0.23 day⁻¹, elapsed time t = 5 days (the standard BOD₅ window).
  2. Identify what we're solving for. We want BODt at t = 5 days — the cumulative oxygen demand exerted by microbial decomposition during the 5-day test.
  3. Write the first-order BOD equation: BODt = L × (1 − 10⁻ᵏᵗ). The factor (1 − 10⁻ᵏᵗ) is the fraction of the ultimate demand that has been exerted by time t.
  4. Substitute the values: BOD₅ = 300 × (1 − 10⁻⁰·²³ × ⁵) = 300 × (1 − 10⁻¹·¹⁵).
  5. Simplify the arithmetic: 10⁻¹·¹⁵ ≈ 0.0708, so (1 − 0.0708) = 0.9292, and 300 × 0.9292 ≈ 278.8.
  6. **BOD₅ ≈ 278.8 mg/L** — about 93% of the ultimate oxygen demand is consumed by microbes during the standard 5-day test at this rate.

About 93% of the ultimate oxygen demand is exerted within 5 days at this rate.

When to Use Each Variable

  • Solve for BOD at time t — when you know the ultimate BOD, deoxygenation rate, and elapsed time, e.g., predicting oxygen demand after 5 days for a wastewater sample.
  • Solve for Ultimate BOD (L) — when you have a BOD measurement at a known time and rate constant, e.g., estimating total oxygen demand from a BOD5 test result.
  • Solve for Deoxygenation Rate (k) — when you have BOD and ultimate BOD measurements and need the rate constant, e.g., characterizing decomposition kinetics of an industrial effluent.
  • Solve for Time (t) — when you know the BOD, ultimate BOD, and rate constant and need to find elapsed time, e.g., determining how long until a stream reaches a target oxygen demand.
  • Solve for Unseeded BOD — when the sample has sufficient native bacteria and you measured initial and final dissolved oxygen with a known dilution ratio.
  • Solve for Seeded BOD — when the sample lacks native bacteria and you added a seed culture, e.g., testing chlorinated or high-temperature industrial waste.

Key Concepts

BOD quantifies the biodegradable organic load in water by measuring how much dissolved oxygen microorganisms consume during decomposition. The first-order deoxygenation model uses rate constant k to describe how quickly oxygen is consumed relative to the ultimate demand L. The unseeded test works for samples with enough native bacteria, while the seeded test adds a known microbial culture and subtracts the seed's own oxygen consumption to isolate the sample's demand.

Applications

  • Wastewater treatment plant design: sizing aeration basins based on influent BOD loading to ensure adequate oxygen supply
  • Discharge permit compliance: demonstrating that treated effluent meets NPDES BOD5 limits (typically under 30 mg/L)
  • Stream water quality modeling: predicting dissolved oxygen sag downstream of a discharge using BOD and reaeration rates
  • Industrial pretreatment: assessing organic load from food processing, brewery, or paper mill waste before discharge to municipal sewers

Common Mistakes

  • Using a seeded test when the sample has adequate native bacteria — the seed correction introduces unnecessary uncertainty; use unseeded BOD when the sample can decompose on its own
  • Confusing BOD5 with ultimate BOD — BOD5 typically represents only 60-70% of the total oxygen demand; using BOD5 as the ultimate value underestimates the full organic load
  • Storing samples too long before testing — biological activity continues during storage, which alters the initial dissolved oxygen reading and skews results; test within 24 hours or refrigerate at 4 degrees C

Frequently Asked Questions

What is a normal BOD level for wastewater?

Untreated domestic wastewater typically has a BOD₅ of 200–300 mg/L. After secondary treatment, effluent BOD₅ usually drops below 20 mg/L to meet discharge permits.

What is the difference between BOD and COD?

BOD measures oxygen consumed by biological decomposition over days, while Chemical Oxygen Demand (COD) uses a strong oxidant to measure total oxidizable matter in a few hours. COD is always equal to or higher than BOD because it includes non-biodegradable organics.

Why is the BOD test done over 5 days?

Five days was chosen historically because it approximates the travel time of major English rivers from source to sea. It captures roughly 60–70% of the ultimate oxygen demand and provides a practical, reproducible measurement window.

When should you use a seeded BOD test?

Use a seeded test when the sample lacks sufficient microorganisms for decomposition, such as chlorinated effluent, industrial waste, or high-temperature discharges. The seed provides the bacteria needed for accurate results.

What deoxygenation rate constant is typical for domestic sewage?

For domestic sewage at 20 °C, k is typically 0.20–0.25 day⁻¹ (base-10), with 0.23 day⁻¹ being the textbook reference value. Industrial effluents with slow-to-degrade organics (food processing waste, paper mill liquors) often run lower — 0.05–0.15 day⁻¹ — because more of the BOD is exerted after the 5-day window.

What is the dilution factor P and how do I choose it?

P is the fraction of the BOD bottle volume that is sample water (typical 300 mL bottle). Choose P so the diluted sample uses 40–60% of its initial DO over 5 days. For high-strength wastewater (BOD₅ ≈ 250 mg/L), P ≈ 0.02; for cleaner samples, P ≈ 0.10–0.30. Run multiple dilutions in parallel to ensure at least one lands in the usable window.

How does temperature affect BOD progression?

The deoxygenation rate constant k roughly doubles for every 10 °C increase (Q10 ≈ 2). The BOD bottle is incubated at 20 °C as the regulatory standard so results are comparable across labs. Sample temperature in the field doesn't affect the BOD value reported — only the time the oxygen demand is exerted in nature.

Why use a seed correction factor f in the seeded BOD formula?

The seed organisms consume some oxygen on their own, separate from the sample. f scales the seed-control DO depletion (B₁ − B₂) by the volume ratio of seed used in the sample vs the seed control. Subtracting f × (B₁ − B₂) isolates the sample's own BOD from the seed's background respiration.

Worked Examples

Municipal Wastewater

What is the 5-day BOD of municipal wastewater with ultimate BOD 250 mg/L and k = 0.23 day⁻¹?

Domestic sewage entering a typical municipal treatment plant has ultimate first-stage BOD around 250 mg/L. With the textbook first-order rate constant k ≈ 0.23 day⁻¹ (base-10), compute the 5-day BOD₅ that the standard test would report.

  • Knowns: L = 250 mg/L, k = 0.23 day⁻¹, t = 5 day
  • BOD_t = L × (1 − 10^(−k × t))
  • BOD_t = 250 × (1 − 10^(−0.23 × 5))
  • BOD_t = 250 × (1 − 10^(−1.15))
  • BOD_t = 250 × (1 − 0.0708)

BOD₅ ≈ 232 mg/L

Most of the ultimate BOD shows up in the first five days — that is why BOD₅ is the standard regulatory metric. After 20 days, residual oxygen demand is usually negligible for routine domestic sewage.

Drinking Water Source

How do you compute unseeded BOD for a clean surface-water sample diluted 30× in a BOD bottle?

A raw reservoir sample is run as an unseeded BOD test: 10 mL of sample is added to a 300 mL bottle and the initial dissolved oxygen reads 9.0 mg/L, then 4.5 mg/L after 5-day incubation at 20 °C. Compute the BOD from the standard dilution formula.

  • Knowns: D₁ = 9.0 mg/L, D₂ = 4.5 mg/L
  • Sample fraction P = 10 mL / 300 mL = 0.0333
  • BOD = (D₁ − D₂) / P
  • BOD = (9.0 − 4.5) / 0.0333
  • BOD = 4.5 / 0.0333

BOD ≈ 135 mg/L

For an unseeded test, the dilution water must be saturated with oxygen and free of organic seed. Typical drinking-water source BOD is well below 5 mg/L — a reading this high signals contamination, agricultural runoff, or sample handling error.

Industrial Effluent

What deoxygenation rate constant fits a food-processing wastewater with BOD₅ = 180 mg/L and ultimate BOD = 300 mg/L?

A cannery discharges process water with a measured BOD₅ of 180 mg/L; a 20-day BOD test confirms ultimate first-stage BOD L = 300 mg/L. Back out the first-order rate constant k so you can predict BOD at any incubation time.

  • Knowns: BOD_t = 180 mg/L, L = 300 mg/L, t = 5 day
  • k = −log₁₀(1 − BOD_t / L) / t
  • k = −log₁₀(1 − 180 / 300) / 5
  • k = −log₁₀(0.4) / 5
  • k = −(−0.39794) / 5

k ≈ 0.0796 day⁻¹

Industrial effluents with non-readily-biodegradable organics (sugars, fats, proteins) often show k below the 0.23 day⁻¹ value typical of domestic sewage. Lower k means more of the BOD is delivered after the 5-day window — relevant when sizing downstream secondary treatment.

Biochemical Oxygen Demand Formulas

BOD measurement uses three related equations: a first-order progression model for time-evolving oxygen demand, and two dilution-bottle formulas for unseeded and seeded laboratory tests.

BODt = L × (1 − 10−kt)First-order BOD progression at time t
BOD = (D₁ − D₂) / PUnseeded dilution-bottle BOD
BOD = ((D₁ − D₂) − f × (B₁ − B₂)) / PSeeded dilution-bottle BOD

Where:

  • BODt — oxygen demand exerted by time t (mg/L; BOD5 is the 5-day value)
  • L — ultimate first-stage BOD (mg/L) — the total carbonaceous oxygen demand
  • k — base-10 deoxygenation rate constant (day⁻¹; 0.23 typical for domestic sewage)
  • t — elapsed incubation time in days
  • D1, D2 — initial and final dissolved oxygen in the diluted sample bottle
  • B1, B2 — initial and final dissolved oxygen in the seed-control bottle
  • f — seed correction factor (volume ratio of seed in sample to seed in seed-control)
  • P — sample dilution fraction (mL of sample / total bottle volume in mL)

The first-order model assumes a single decay rate covers the carbonaceous BOD; the nitrogenous (second-stage) BOD is treated separately or suppressed with a nitrification inhibitor. Use the seeded form for chlorinated effluent, industrial wastes, or any sample lacking native microorganisms — otherwise the unseeded form is simpler and more precise.

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