Water & WW Treatment: Noteset 6 - BOD (Biochemical Oxygen Demand)

Biochemical Oxygen Demand (BOD)

Measures the oxygen requires by microorganisms to break down organic matter; indicator of biodegradable pollution.

Chemical Oxygen Demand (COD)

Measures the total oxidizable material (both biodegradable and non-biodegradable); broader than BOD.

Carbonaceous BOD (CBOD)

Oxygen demand from oxidation or organic carbon compounds, primary driver of BOD in untreated sewage.

Nitrogenous Oxygen Demand (NOD)

Oxygen required to biologically oxidize ammonia (NH₃) to nitrate (NO₃^-). Important for assessing total oxygen demand in nutrient-rich wastes.

ThOD (Theoretical Oxygen Demand)

Oxygen required if all organics were completely oxidized.

ThNOD

Theoretical oxygen demand for nitrification of ammonia.

BOD_L

Maximum biodegradable oxygen demand achievable with sufficient time

BOD_r

Concentration of organics (mg BOD/L) remaining at time=t

BOD_L

Initial concentration of organics at time = 0

BOD_t

O₂ consumed at time = t

BOD test setup

1. 300mL BOD Bottle (the volume added depends on the strength of the waste, it may need to be diluted)

2. If needed, add seed (a microbial inoculum)

3. Fill the top of the bottle with oxygenated buffer solution that contains nutrients and salts (ensures that initial DO starts at a standardized level)

4. Insert a glass stopper and incubate at a constant temperature (usually 20^oC) in the dark

5. Measure DO again at a specific time (e.g. t = 5 days)

6. Create seed controls in the same way, but omitting the sample

Why use seed?

To provide the microorganisms needed to break down the organic matter in the sample, ensuring the BOD test reflects the biochemical oxygen demand of the sample itself, not just the presence (or absence) of microbes.

Why is the bottle sealed?

So the only oxygen available to the microorganisms is already what’s dissolved in the water at the start of the tests. Atmospheric oxygen could dissolve into the water that would artificially inflate the DO.

BOD₅

Standard test, 5 days at 20^oC.

Calculate the COD’ of glucose (C₆H₁₂O₆) and present the answers in mole O₂ / mole compound and g O₂/g compound

C_nH_aO_bN_c

COD’ = (2n + 0.5a - 1.5c - b)/2

6(12) + 12(1) + 6(16) = 180 g/mol
COD’ = (2n + 0.5a - 1.5c - b)/2 = [2(6) + 0.5(12) - 1.5(0) - 6]/2 = 6 mol O₂/mol C₆H₁₂O₆

6 mol O₂/mol C₆H₁₂O₆ x [(32 g O₂)/1 mol O₂] x [(1 mol C₆H₁₂O₆)/(180 g/mol)] = 1.07 g O₂/g C₆H₁₂O₆

Calculate the ThNOD of glucose (C₆H₁₂O₆) and present the answers in mole O₂ / mole compound and g O₂/g compound

C_nH_aO_bN_c

ThNOD = 2( c )

ThNOD = 2(0) = 0 mol O₂/mol C₆H₁₂O₆ = 0 g O₂/g C₆H₁₂O₆

Calculate the ThOD of glucose (C₆H₁₂O₆) and present the answers in mole O₂ / mole compound and g O₂/g compound

ThOD = COD’ + ThNOD

6 mol O₂/C₆H₁₂O₁₆ + 0 O₂/C₆H₁₂O₂ = 6 O₂/C₆H₁₂O₂

6 mol O₂/mol C₆H₁₂O₆ x [(32 g O₂)/1 mol O₂] x [(1 mol C₆H₁₂O₆)/(180 g/mol)] = 1.07 g O₂/g C₆H₁₂O₆

Calculate the COD’ of a sludge (C₅H₇O₂N) and present the answers in mole O₂ / mole compound and g O₂/g compound

COD’ = (2n + 0.5a - 1.5c - b)/2

C_nH_aO_bN_c

5(12) + 1(7) + 2(16) + 1(14) = 113g/mol

COD’ = (2n + 0.5a - 1.5c - b)/2 = [2(5) + 0.5(7) - 1.5(1) - 2]/2 = 5 mol O₂/mol C₅H₇O₂N

5 mol O₂/mol C₅H₇O₂N x (32 g O₂/1 mol O₂) x (1 mol C₅H₇O₂N/113g) = 1.42 g O₂/g C₅H₇O₂N

Calculate the ThNOD of a sludge (C₅H₇O₂N) and present the answers in mole O₂ / mole compound and g O₂/g compound

ThNOD = 2( c )

C_nH_aO_bN_c

ThNOD = 2(1) = 2 mol O₂/mol C₅H₇O₂N

2 mol O₂/mol C₅H₇O₂N x (32 g O₂/1 mol O₂) x (1 mol C₅H₇O₂N/113g) = 1.42 g O₂/g C₅H₇O₂N = 0.57 g O₂/g C₅H₇O₂N

Calculate the ThOD of a sludge (C₅H₇O₂N) and present the answers in mole O₂ / mole compound and g O₂/g compound

ThOD = COD’ + ThNOD

ThOD = COD’ + ThNOD = 5 mol O₂/mol C₅H₇O₂N + 2 mol O₂/mol C₅H₇O₂N = 7 O₂/mol C₅H₇O₂N

7 O₂/mol C₅H₇O₂N x (32 g O₂/1 mol O₂) x (1 mol C₅H₇O₂N/113g) = 1.98 g O₂/g C₅H₇O₂N

If a candy making factory wastewater contains 150 mg glucose/L, what is the strength of the wastewater in units of COD’/L?

COD’ of glucose = 1.07 g COD’/g glucose

150 mg glucose/L x 1.07 mg COD’/mg glucose = 160.5 COD’/L

Remember: 1.07 g COD’/g glucose = 1.07 mg COD’/g glucose

The test was set up by pipetting 10 mL of the industrial waste sample into each 300-mL BOD bottle. A 0.5 mL sample of domestic settled sewage was then added as a bacterial inoculum (or seed). A separate seed control test was performed to determine the BOD contributed by the sewage seed by adding 10 mL of the settled sewage into a 300-mL bottle. Each bottle (seed control or wastewater) was then filled completely (to 300 mL) with appropriate dilution water.

Six bottles were set up for each test. On Day 0 three bottles were sacrificed to measure initial dissolved oxygen (DO). On Day 5 the remaining bottles were sacrificed for measurement of the final DO. The results are shown below.

a) Calculate the BOD₅ of the industrial wastewater

BOD_t = [(D1 - D2) - [(B1 - B2) * f]/P

First Take the Averages of Each Day:

[8.01 + 8.25 + 7.93]/3 = 8.06

[6.08 + 6.19 + 5.81]/3 = 6.03

[7.92 + 8.17 + 7.81]/3 = 7.97

[3.30 + 3.29 + 3.78]/3 = 3.46

BOD₅ = [(7.97 - 3.46) - (8.06 - 6.03) * (0.5/10)]/(10/300) = 132.3 mg BOD₅/L

The test was set up by pipetting 10 mL of the industrial waste sample into each 300-mL BOD bottle. A 0.5 mL sample of domestic settled sewage was then added as a bacterial inoculum (or seed). A separate seed control test was performed to determine the BOD contributed by the sewage seed by adding 10 mL of the settled sewage into a 300-mL bottle. Each bottle (seed control or wastewater) was then filled completely (to 300 mL) with appropriate dilution water.

Six bottles were set up for each test. On Day 0 three bottles were sacrificed to measure initial dissolved oxygen (DO). On Day 5 the remaining bottles were sacrificed for measurement of the final DO. The results are shown below.

b) If k is 0.1/d, calculate the BOD_L of the industrial wastewater

BOD_L = BOD_t/(1 - e^-kt)

BOD_L = 132.3 mg BOD₅/L/(1 - e^-(0.1)(5)) = 336.2 mg/L

The test was set up by pipetting 10 mL of the industrial waste sample into each 300-mL BOD bottle. A 0.5 mL sample of domestic settled sewage was then added as a bacterial inoculum (or seed). A separate seed control test was performed to determine the BOD contributed by the sewage seed by adding 10 mL of the settled sewage into a 300-mL bottle. Each bottle (seed control or wastewater) was then filled completely (to 300 mL) with appropriate dilution water.

Six bottles were set up for each test. On Day 0 three bottles were sacrificed to measure initial dissolved oxygen (DO). On Day 5 the remaining bottles were sacrificed for measurement of the final DO. The results are shown below.

c) Given the k and BOD_L, what is the BOD₇ of the industrial wastewater?

BOD_t = BOD_L(1 - e^-kt)

BOD₇ = 336(1 - e^-0.1×7)

BOD₇ = 167 mg/L

What is the relationship between COD’ and COD?

COD’ ≈ COD

What is COD’?

The carbonaceous portion of the COD i.e. the nitrogenous oxygen demand is excluded.

What is the general relationship between COD and BOD₅?

COD ≈ 2 x BOD₅