Biochemical Oxygen Demand (BOD)

Carbonaceous Constituents

Biochemical Oxygen Demand

-Pratik Prakash KulkarniDepartment of Civil Engineering

IIT Guwahati

Effect of Oxygen Demanding Wastes on Rivers

• Chhatrapati Rajaram Sugar mill– Accidental release of molasses

Carbonaceous Constituents

• Carbonaceous constituents are measured in terms of BOD, COD or TOC analyses.

• While the BOD has been the common parameter to characterize carbonaceous material in wastewater, COD is becoming more common in most current comprehensive computer simulation design models.


• Sewage contains Organic matter– Fats, Proteins and Carbohydrates

• Requires common basis for quantification

• The BOD test gives a measure of the oxygen utilized by bacteria during the oxidation of organic material contained in a wastewater sample.


• The test is based on the premise that all the biodegradable organic material contained in the wastewater sample will be oxidized to CO2 and H2O, using molecular oxygen as the electron acceptor.

• It is a direct measure of oxygen requirements and an indirect measure of biodegradable organic matter.


60-70%60-70%

95-99%

Variation in DO profile during BOD test with duration of incubation at 20

2 2 2Organic matter + O CO + H O + New Cells + Stable ProductsMicroorganisms


• To perform this test on industrial waste generally there is requirement to add microbial seed.

• This seed is added as sewage or the water from river or stream or soil sample where the industry discharges its waste

1.The sample is taken into a BOD bottle of volume 300mL. Say 10 mL

BOD Bottle

Aspirator Bottles

Dilution water

BOD Bottle

2. Diluted it with sufficient amount of dilution water if required. Say 290mL

3. Initial DO is to be calculated and final DO after 3days @27⁰C or 5days @20⁰C is measured

4. Difference between these reading multiplied by dilution factor gives BOD value.

Source Sawyer Mccarty and Parkin chemistry for environmental engineering

Biochemical Oxygen Demand𝐵𝑂𝐷 ,𝑚𝑔 /𝐿=

𝐷1−𝐷2𝑃

When the dilution water is seeded

𝐵𝑂𝐷𝑚𝑔 /𝐿 ,= (𝐷1−𝐷2 )− (𝐵1−𝐵2 ) 𝑓𝑃

Where:D1: Dissolved oxygen of diluted sample immediately after preparation, mg/LD2: Dissolved oxygen of diluted sample immediately after 5day incubation at 20⁰C, mg/LB1: Dissolved oxygen of seed control before incubation, mg/LB2: Dissolved oxygen of seed control after incubation, mg/f : Fraction of seeded dilution water volume in sample to volume of seeded dilution water in seed control=1-P=290/300P: Fraction of w/w sample to total combined volume i.e. 10/300


• As the microbial activity depends on temperature, BOD test is highly depends on temperature.

• Generally this temperature is chosen as 20⁰C or 27⁰C.


60-70%

95-99%

Variation in DO profile during BOD test with duration of incubation at 20


where, : Ultimate BOD, mg/L : BOD remaining at any time, mg/L : 1st order reaction rate constant,1/d : time, d• But = • ∴(1- )• where, : BOD Exerted at that time


• Rate constant is dependent on temperature, it can be calculated as,• = where, =1.047 Ɵ• The value of is temperature Ɵdependent


1. Determine 1-day BOD and ultimate first stage BOD for a w/w whose 5- day BOD is 200mg/L at 20. The reaction constant k (base e) = 0.23. What would have been the 5-day BOD if the test have been conducted at 25⁰C?

Given : k1 (to the base e)=0.23/dBOD5=200mg/L at 20

Calculate: L0

BOD1 at 20BOD5 at 25

Biochemical Oxygen DemandAns: Formula: , = , =1) Ultimate BOD: ∴ = 2)Determine 1-day BOD:

3)Determine 5day BOD at 25⁰C: = =

BOD

2. A BOD test was conducted at 20⁰C in which 15mL of waste sample was diluted with dilution water to 300mL. Given: Initial DO of diluted sample D1 =8.8mg/LFinal DO after 5 days D2=1.9mg/LInitial DO of seeded dilution water B1=9.1mg/LFinal DO of seeded dilution water B2=7.9mg/L

Calculate 5-day BOD at 20⁰C

BOD

• Ans:Where

mg/L

BODHomework3. The BOD5of a wastewater is determined to be 150mg/L at 20⁰C. The k

is 0.23 per day. What would the BOD8 be if the test were run at 15⁰C?Gate 20004. The DO in an unseeded sample of diluted waste having an initial DO of

9.5 mg/L is measured to be 3.5mg/L after 5 days. The dilution factor is 0.03 and reaction rate constant, k=0.22/day (to the base e at 20⁰C) Estimate

The 5 day BOD of the waste at 20⁰CUltimate Carbonaceous BODRemaining oxygen demand waste after 5-daysThe 5 day BOD of this waste at 25⁰C assuming a temperature coefficient of 1.047

(Hint dilution factor=P)

Gate 2004: 5. A portion of w/w sample was subjected to standard

BOD test(5d,20⁰C), yielding a value of 180mg/L. The reaction rate constant (to the base ‘e’) at 20⁰C was taken as 0.18/d. temperature coefficient is 1.047. The temperature at which the other portion of the sample should be tested, to exert the same BOD in 2.5days, is…

A. 4.9⁰C B. 24.9 ⁰C C. 31.7 ⁰C D. 35.0 ⁰CAns: D. 35.0 ⁰C

Why BOD test is conducted for 5th day

• BOD theoretically takes infinite time to go to completion because the rate of oxidation is assumed to be proportional to the amount of organic matter remaining.

• Within 20day period the oxidation of carbonaceous matter is about 95-99% and in 5days it is 60-70%

• 20days is long time to wait for test

• After 8-10 days nitrifying bacteria start using oxygen to convert ammonia nitrogen to nitrite and then to nitrate.

• We are interested only in first stage BOD or carbonaceous BOD so we have chosen 5days for the test



• Carbonaceous demand can be obtained by supressing nitrifying bacteria by adding growth inhibiters of nitrifying bacteria.

• Also this test is conducted on 5th day so as to ensure less population of nitrifying bacteria.

BOD

• This test can be used to1. Determine amount of oxygen required to

stabilize organic matter biologically2. Determine size of treatment facility3. Measure the efficiency of treatment facilities4. Determine compliance with wastewater

discharge permits

Thank You