Details about Biochemical Oxygen Demand(BOD) with solved examples. Extra examples are given for homework. You can contact me for details on pratik1516@gmail.com.

1. Carbonaceous Constituents
Biochemical Oxygen Demand
-Pratik Prakash Kulkarni
Department of Civil Engineering
IIT Guwahati

2. Effect of Oxygen Demanding Wastes
on Rivers
• Chhatrapati Rajaram Sugar mill
– Accidental release of molasses

3. 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.

4. Biochemical Oxygen Demand
• 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.

5. Biochemical Oxygen Demand
• 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.

6. Biochemical Oxygen Demand
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


7. Biochemical Oxygen Demand
• 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

9. 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.

10. Source Sawyer Mccarty and Parkin chemistry for environmental engineering

11. Biochemical Oxygen Demand
𝐵𝑂𝐷, 𝑚𝑔/𝐿 =
𝐷1 − 𝐷2
𝑃
When the dilution water is seeded
𝐵𝑂𝐷𝑚𝑔/𝐿, =
𝐷1 − 𝐷2 − 𝐵1 − 𝐵2 𝑓
𝑃
Where:
D1: Dissolved oxygen of diluted sample immediately after preparation, mg/L
D2: Dissolved oxygen of diluted sample immediately after 5day incubation at 20⁰C,
mg/L
B1: Dissolved oxygen of seed control before incubation, mg/L
B2: 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=290/300
P: Fraction of w/w sample to total combined volume i.e. 10/300

12. Biochemical Oxygen Demand
• As the microbial activity depends on
temperature, BOD test is highly depends on
temperature.
• Generally this temperature is chosen as 20⁰C
or 27⁰C.

13. Biochemical Oxygen Demand
60-70%
95-99%
Variation in DO profile during BOD test with duration of incubation at 20℃

14. Biochemical Oxygen Demand
• 𝐿𝑡 = 𝐿₀𝑒−𝑘1
𝑡
where, 𝐿₀: Ultimate BOD, mg/L
𝐿𝑡 : BOD remaining at any time, mg/L
𝑘1: 1st order reaction rate constant,1/d
t : time, d
• But 𝐿₀= 𝐿 + 𝑌
• ∴𝑌 = 𝐿₀(1- 𝑒−𝑘1
𝑡
)
• where, 𝑌: BOD Exerted at that time

15. Biochemical Oxygen Demand
• Rate constant 𝑘1 is dependent on
temperature, it can be calculated as,
• 𝑘1𝑡= 𝑘1 20⁰Ɵ 𝑇−20
where, Ɵ=1.047
• The value of Ɵ is temperature dependent

16. Biochemical Oxygen Demand
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) =
0.23𝑑−1
. What would have been the 5-day
BOD if the test have been conducted at 25⁰C?
Given : k1 (to the base e)=0.23/d
BOD5=200mg/L at 20℃
Calculate:
L0
BOD1 at 20℃
BOD5 at 25℃

17. Biochemical Oxygen Demand
Ans: Formula: 𝑌 = 𝐿₀(1 − 𝑒−𝑘1
𝑡
), 𝐿₀=
𝑌
(1−𝑒−𝑘1
𝑡)
, 𝑘1𝑇⁰=𝑘1 20
0
1.047 𝑇−20
1) Ultimate BOD: ∴ 𝐿₀=
𝑌
(1−𝑒−𝑘1
𝑡)
=
200
(1−𝑒−0.23∗5)
= 293𝑚𝑔/𝐿
2)Determine 1-day BOD:𝑌1 = 𝐿₀ 1 − 𝑒−𝑘1
∗1
= 293 1 − 𝑒−0.23∗1 = 60.1𝑚𝑔/𝐿
3)Determine 5day BOD at 25⁰C:
𝑘125⁰= 𝑘1 20
0
1.047 25−20
= 0.29𝑑−1
𝑌5 = 𝐿0 1 − 𝑒−𝑘1
∗5 =293 1 − 𝑒−0.29∗5 = 224𝑚𝑔/𝐿

18. 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/L
Final DO after 5 days D2=1.9mg/L
Initial DO of seeded dilution water B1=9.1mg/L
Final DO of seeded dilution water B2=7.9mg/L
Calculate
5-day BOD at 20⁰C

19. BOD
• Ans:Forrmula, 𝐵𝑂𝐷, 𝑚𝑔/𝐿 =
𝐷1
−𝐷2
− 𝐵1
−𝐵2
𝑓
𝑃
Where 𝑓 =
300−15
300
= 0.95
𝑃 =
15
300
= 0.05
𝐵𝑂𝐷, 𝑚𝑔/𝐿 =
8.8−1.9 − 9.1−7.9 0.95
0.05
= 115.2mg/L

20. BOD
Homework
3. 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 2000
4. 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 20⁰C
Ultimate Carbonaceous BOD
Remaining oxygen demand waste after 5-days
The 5 day BOD of this waste at 25⁰C assuming a temperature coefficient of 1.047
(Hint dilution factor=P)

21. 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 ⁰C
Ans: D. 35.0 ⁰C

22. 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

23. Why BOD test is conducted for 5th day

24. Why BOD test is conducted for 5th day
• 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.

25. BOD
• This test can be used to
1. Determine amount of oxygen required to
stabilize organic matter biologically
2. Determine size of treatment facility
3. Measure the efficiency of treatment facilities
4. Determine compliance with wastewater
discharge permits

26. Thank You