1. CONCEPTUAL FRAMEWORK & WATER QUALITY THRESHOLDS

A. Definition & Standard Laboratory Test — BOD₅

Biological Oxygen Demand (BOD)

Biological Oxygen Demand (BOD) is the amount of dissolved oxygen (DO) required by aerobic microorganisms to biologically decompose biodegradable organic matter present in water or wastewater.

Standard BOD₅ Test

The complete decomposition of organic matter may take several weeks. Therefore, the standard laboratory test measures oxygen consumption over:

• 5 days

• 20°C

• In the dark

This is called BOD₅.

The test is kept in the dark to prevent algae from producing oxygen through photosynthesis, which would interfere with the measurement.

Remember:

BOD₅ = oxygen consumed biologically during 5 days at 20°C.

B. Dissolved Oxygen (DO) & Anoxic Conditions

When microorganisms decompose organic matter, they consume dissolved oxygen from the water.

Therefore:

More biodegradable organic matter → Higher BOD → Greater DO depletion

A high BOD generally indicates a greater oxygen demand on the receiving water body.

Important DO Levels

At very low DO levels, aerobic biological activity becomes severely limited and anaerobic decomposition can become dominant.

Anaerobic decomposition may produce compounds such as:

• Hydrogen sulfide (H₂S)

• Ammonia (NH₃)

• Methane (CH₄)

DO & Ecological Status

DISSOLVED OXYGEN (DO) & ECOLOGICAL STATUS

0 ───────── 1.5 ───────────── 4.0 ───────────── 8.0 ───────────── 14.0 mg/L
│            │                │                  │
▼            ▼                ▼                  ▼

Very Low     Critical         Highly             Well
DO           Oxygen           Polluted           Oxygenated
             Depletion

Anaerobic    Severe oxygen    Severe aquatic     Healthy
conditions   stress           stress             aquatic life

Possible formation of:
H₂S, NH₃, CH₄

Important: DO concentration alone should not be treated as a universal pollution-classification scale because natural DO varies with temperature, salinity, altitude, and water-body characteristics. For numerical problems, use the values specifically given in the question.

C. BOD₅ Water Quality Rating Scale

Exam logic:

Higher BOD → More biodegradable organic matter → More oxygen consumed → Greater pollution potential.

D. Effluent Standards & BOD vs COD

BOD Discharge Limit

For numerical problems, a commonly used reference value for treated wastewater discharged to surface waters is:

BOD ≤ 30 mg/L

However, actual regulatory limits depend on the applicable discharge category, location, consent conditions, and current standards.

Always use the standard specified in the question.

BOD vs COD

BOD

Measures oxygen required for biological oxidation of biodegradable organic matter.

COD

Measures oxygen required for chemical oxidation of oxidizable substances, including many compounds that may not be readily biodegradable.

Therefore, in most wastewater samples:

COD > BOD

The BOD/COD ratio is also useful for assessing wastewater biodegradability.

2. CORE MATHEMATICAL FORMULAS FOR BOD CALCULATIONS

                    BOD CALCULATION FORMULAS

                             BOD
                              │
       ┌──────────────────────┼──────────────────────┐
       │                      │                      │
       ▼                      ▼                      ▼
  Dilution Method       Reaction Kinetics      Temperature Correction
       │                      │                      │
       ▼                      ▼                      ▼
BOD₅ = (DOᵢ - DO𝒇) × DF   BODₜ = L₀(1-e⁻ᵏᵗ)   kT = k₂₀ × θ^(T-20)
       
                              │
                              ▼
                       Removal Efficiency
                              │
                              ▼
              % Removal = [(BODin-BODout)/BODin] × 100

FORMULA 1 — BOD DETERMINATION BY DILUTION METHOD

When raw wastewater is highly polluted, its dissolved oxygen may be completely consumed before 5 days.

Therefore, the wastewater is diluted with oxygen-saturated dilution water.

Formula

BOD₅ = (DOᵢ − DO𝒇) × Dilution Factor

Where:

• DOᵢ = Initial dissolved oxygen of diluted sample, mg/L

• DO𝒇 = Final dissolved oxygen after 5 days, mg/L

Dilution Factor

DF = Vtotal / Vsample

Therefore:

DF = (Vsample + Vdilution water) / Vsample

If P is the fraction of wastewater in the total mixture:

P = Vsample / Vtotal

Then:

DF = 1 / P

Basic Logic

More dilution
      ↓
More dilution factor
      ↓
Measured DO depletion is multiplied
      ↓
Original wastewater BOD is obtained

FORMULA 2 — FIRST-ORDER BOD REACTION KINETICS

The oxidation of biodegradable organic matter is commonly represented using first-order kinetics.

The rate of oxidation is proportional to the amount of biodegradable organic matter remaining.

Base-e Form

BODₜ = L₀(1 − e⁻ᵏᵗ)

Where:

• BODₜ = BOD exerted at time t, mg/L

• L₀ = Ultimate Carbonaceous BOD, mg/L

• k = First-order reaction rate constant, day⁻¹

• t = Time, days

The remaining organic matter is:

Lₜ = L₀e⁻ᵏᵗ

Therefore:

BODₜ = L₀ − Lₜ

Base-10 Form

The same relationship can also be written as:

BODₜ = L₀(1 − 10⁻ᴷᵗ)

Where K is the base-10 reaction-rate constant.

The relationship between the two rate constants is:

k = 2.303K

Important

• k → base-e constant

• K → base-10 constant

Do not mix the two forms in a calculation.

FORMULA 3 — TEMPERATURE CORRECTION OF REACTION RATE CONSTANT

Microbial activity generally increases with temperature within the normal biological operating range.

Therefore, the reaction rate constant changes with temperature.

Formula

kT = k₂₀ × θ^(T − 20)

Where:

• kT = reaction rate constant at temperature T, day⁻¹

• k₂₀ = reaction rate constant at 20°C, day⁻¹

• θ = temperature coefficient

• T = temperature, °C

Commonly used values in this context:

• θ = 1.047 for approximately 20–30°C

• θ = 1.056 for approximately 15–20°C

Meaning

If temperature increases:

Temperature ↑ → microbial activity ↑ → k ↑ → BOD is exerted faster

FORMULA 4 — BOD REMOVAL EFFICIENCY

Treatment plants reduce BOD through different treatment stages.

Formula

% BOD Removal = [(BODin − BODout) / BODin] × 100

Where:

• BODin = BOD entering the treatment stage

• BODout = BOD leaving the treatment stage

Typical Treatment Performance Used in Numerical Problems

Primary Treatment

Physical processes such as screening, grit removal and sedimentation can remove a portion of organic matter associated with suspended solids.

A typical numerical assumption may be:

BOD removal ≈ 35%

Secondary Treatment

Biological treatment processes such as:

• Activated Sludge Process

• Trickling Filters

• Oxidation Ponds

can achieve substantial BOD removal.

For numerical problems, the assumed removal efficiency should always be taken from the question.

FORMULA 5 — DAILY BOD MASS LOADING

Daily BOD load is the total mass of BOD entering or leaving a system per day.

Formula

BOD Load (kg/day) = Q × BOD × 10⁻³

Where:

• Q = wastewater flow, m³/day

• BOD = concentration, mg/L

Why 10⁻³?

Because:

1 m³ = 1,000 L

and:

1,000 mg = 0.001 kg

Therefore:

1 mg/L × 1 m³ = 0.001 kg

Hence:

Load = Q × concentration × 10⁻³

POPULATION EQUIVALENT (PE)

Population Equivalent expresses an industrial pollution load as the equivalent pollution load produced by a population.

Formula

PE = Total Industrial BOD Load (g/day) / Per-Capita BOD Load (g/person/day)

For the numerical problems in this chapter:

Per-capita domestic BOD contribution = 54 g/person/day

or:

0.054 kg/person/day

Meaning

If an industry produces 2,700 kg BOD/day:

PE = 2,700 / 0.054 = 50,000 persons

Therefore, its BOD load is equivalent to the domestic BOD load of approximately 50,000 people.

3. SOLVED NUMERICAL PROBLEMS

PROBLEM 1 — BOD₅ BY DILUTION METHOD

Question

A 10 mL sample of raw domestic sewage is diluted to 300 mL in a standard BOD bottle using aerated water.

Initial DO = 9.0 mg/L

Final DO after 5 days = 3.0 mg/L

Calculate the BOD₅ of the raw sewage.

Solution

Step 1 — Calculate Dilution Factor

DF = Vtotal / Vsample

DF = 300 / 10

DF = 30

Step 2 — Calculate BOD₅

BOD₅ = (DOᵢ − DO𝒇) × DF

BOD₅ = (9.0 − 3.0) × 30

BOD₅ = 6.0 × 30

BOD₅ = 180 mg/L

Answer

BOD₅ = 180 mg/L

PROBLEM 2 — FIRST-ORDER KINETICS & ULTIMATE BOD

Question

The 5-day BOD of wastewater at 20°C is 200 mg/L.

Given:

k₂₀ = 0.23 day⁻¹

Calculate:

1. Ultimate Carbonaceous BOD (L₀)

2. 3-day BOD (BOD₃)

Solution

Step 1 — Calculate Ultimate BOD

Use:

BODₜ = L₀(1 − e⁻ᵏᵗ)

At t = 5 days:

200 = L₀(1 − e⁻⁽⁰·²³ × ⁵⁾)

200 = L₀(1 − e⁻¹·¹⁵)

Since:

e⁻¹·¹⁵ ≈ 0.3166

Therefore:

200 = L₀(1 − 0.3166)

200 = L₀ × 0.6834

Therefore:

L₀ = 200 / 0.6834

L₀ ≈ 292.65 mg/L

Step 2 — Calculate BOD₃

Use:

BOD₃ = L₀(1 − e⁻ᵏᵗ)

BOD₃ = 292.65 × (1 − e⁻⁽⁰·²³ × ³⁾)

BOD₃ = 292.65 × (1 − e⁻⁰·⁶⁹)

Since:

e⁻⁰·⁶⁹ ≈ 0.5016

Therefore:

BOD₃ = 292.65 × (1 − 0.5016)

BOD₃ = 292.65 × 0.4984

BOD₃ ≈ 145.86 mg/L

Answer

• Ultimate BOD, L₀ ≈ 292.65 mg/L

• BOD₃ ≈ 145.86 mg/L

PROBLEM 3 — TEMPERATURE CORRECTION

Question

A wastewater sample has a base-e reaction rate constant:

k₂₀ = 0.10 day⁻¹ at 20°C

Determine the reaction rate constant at 30°C.

Given:

θ = 1.047

Solution

Use:

kT = k₂₀ × θ^(T − 20)

At 30°C:

k₃₀ = 0.10 × 1.047^(30 − 20)

k₃₀ = 0.10 × 1.047¹⁰

Since:

1.047¹⁰ ≈ 1.583

Therefore:

k₃₀ = 0.10 × 1.583

k₃₀ = 0.1583 day⁻¹

Answer

k₃₀ ≈ 0.1583 day⁻¹

The reaction rate constant increases by approximately 58.3%.

PROBLEM 4 — MULTI-STAGE SEWAGE TREATMENT

Question

Raw municipal sewage has:

BOD₅ = 250 mg/L

Primary sedimentation removes 35% of BOD.

Secondary activated sludge treatment removes 85% of the remaining BOD.

Calculate:

1. BOD after primary treatment

2. Final BOD after secondary treatment

3. Overall BOD removal efficiency

Step 1 — Primary Treatment

Remaining BOD:

BODprimary = 250 × (1 − 0.35)

BODprimary = 250 × 0.65

BODprimary = 162.5 mg/L

Step 2 — Secondary Treatment

Secondary treatment removes 85% of the remaining BOD.

Therefore, 15% remains:

BODfinal = 162.5 × (1 − 0.85)

BODfinal = 162.5 × 0.15

BODfinal = 24.375 mg/L

Approximately:

BODfinal = 24.38 mg/L

Step 3 — Compliance Check

Using the 30 mg/L reference value specified in this problem:

Final BOD = 24.38 mg/L

24.38 < 30 mg/L

Therefore, it satisfies the assumed discharge limit.

Step 4 — Overall Removal Efficiency

Overall removal = [(250 − 24.38) / 250] × 100

Overall removal ≈ 90.25%

Answer

• Primary effluent BOD = 162.5 mg/L

• Final effluent BOD = 24.38 mg/L

• Overall BOD removal = 90.25%

Important Concept

The removal efficiencies of successive treatment units are not simply added.

Wrong:

35% + 85% = 120%

Correct:

First remove 35%, then remove 85% of what remains.

PROBLEM 5 — DAILY BOD MASS LOAD & POPULATION EQUIVALENT

Question

An industrial food-processing plant discharges:

Flow = 2,000 m³/day

BOD₅ = 1,350 mg/L

Calculate:

1. Daily BOD mass load

2. Population Equivalent (PE)

Given:

Per-capita BOD contribution = 54 g/person/day

Step 1 — Calculate Daily BOD Load

Use:

Load (kg/day) = Q × BOD × 10⁻³

Therefore:

Load = 2,000 × 1,350 × 10⁻³

Load = 2,700,000 × 10⁻³

Load = 2,700 kg/day

Step 2 — Convert kg/day to g/day

2,700 kg/day × 1,000 g/kg

= 2,700,000 g/day

Step 3 — Calculate Population Equivalent

PE = Total BOD Load / Per-Capita BOD Load

PE = 2,700,000 / 54

PE = 50,000 persons

Answer

• BOD load = 2,700 kg/day

• Population Equivalent = 50,000 persons

Interpretation

The industry's BOD pollution load is approximately equivalent to the domestic BOD load generated by 50,000 people, using the assumed value of 54 g/person/day.

4. FORMULA SUMMARY & QUICK REFERENCE SHEET

5. EXAM-ORIENTED CALCULATION LOGIC

A. If DO values and dilution are given

Use:

BOD₅ = DO depletion × Dilution Factor

B. If BOD₅ and k are given and L₀ is asked

Use:

BOD₅ = L₀(1 − e⁻⁵ᵏ)

Rearrange:

L₀ = BOD₅ / (1 − e⁻⁵ᵏ)

C. If L₀, k and time are given

Use:

BODₜ = L₀(1 − e⁻ᵏᵗ)

D. If temperature changes

Use:

kT = k₂₀ × θ^(T−20)

E. If influent and effluent BOD are given

Use:

% Removal = [(BODin − BODout) / BODin] × 100

F. If flow and BOD concentration are given

Use:

BOD Load = Q × BOD × 10⁻³ kg/day

G. If industrial BOD load and per-capita load are given

Use:

PE = Industrial BOD Load / Per-Capita BOD Load

6. MUST REMEMBER

BOD

BOD = Biological Oxygen Demand

Measures oxygen required for biological oxidation of biodegradable organic matter.

BOD₅

5 days + 20°C + dark

Ultimate BOD

L₀ = total ultimate biodegradable oxygen demand represented by the kinetic model

First-Order Equation

BODₜ = L₀(1 − e⁻ᵏᵗ)

Base Conversion

k = 2.303K

Temperature Correction

kT = k₂₀ × θ^(T−20)

BOD Load

Load = Q × BOD × 10⁻³ kg/day

Population Equivalent

PE = Industrial BOD Load / Per-Capita BOD Load

Core Relationship

Organic pollution ↑ → BOD ↑ → DO depletion ↑ → Aquatic stress ↑

BOD vs COD

Usually COD > BOD

7. ONE-MINUTE MEMORY MAP

                         BOD CALCULATIONS
                               │
        ┌──────────────────────┼──────────────────────┐
        │                      │                      │
        ▼                      ▼                      ▼
   MEASUREMENT              KINETICS               TREATMENT
        │                      │                      │
        ▼                      ▼                      ▼
   DO depletion          BODₜ = L₀(1-e⁻ᵏᵗ)      BOD removal %
        │                      │                      │
        ▼                      ▼                      ▼
   Dilution Factor           L₀, k, t             BODin/BODout
        │                      │
        └──────────────┬───────┘
                       ▼
                 MASS LOADING
                       │
                       ▼
             Load = Q × BOD × 10⁻³
                       │
                       ▼
                Population
                Equivalent
                       │
                       ▼
                 PE = Load / 54

8. QUICK REVISION QUESTIONS

1. What does BOD measure?

2. Why is the standard BOD test conducted for 5 days?

3. Why is the BOD test conducted at 20°C?

4. Why is the BOD bottle kept in the dark?

5. What happens to DO when BOD increases?

6. What is the dilution factor?

7. What is ultimate BOD, L₀?

8. What is the first-order BOD equation?

9. What is the relationship between k and K?

10. How does temperature affect k?

11. What is BOD removal efficiency?

12. How do you calculate daily BOD load?

13. What is Population Equivalent?

14. Why is COD generally greater than BOD?

15. If treatment occurs in two stages, why can the removal percentages not simply be added?

CORE NUMERICAL FORMULAS

1. BOD₅ = (DOᵢ − DO𝒇) × DF

2. DF = Vtotal / Vsample

3. BODₜ = L₀(1 − e⁻ᵏᵗ)

4. Lₜ = L₀e⁻ᵏᵗ

5. k = 2.303K

6. kT = k₂₀ × θ^(T−20)

7. % Removal = [(BODin − BODout) / BODin] × 100

8. BOD Load = Q × BOD × 10⁻³ kg/day

9. PE = BOD Load (g/day) / 54

These nine formulas cover the main calculation patterns in this unit.