CEE 361: Introduction to Environmental Engineering

Spring 2019

Lab Title: Bod Determination of Wastewater

Date: 4/2/2019

Name:
INTRODUCTION

The objective of this lab is to learn how to conduct BOD5 tests. Biochemical Oxygen

Demand (BOD) i s t h e a m o u n t o f o x y g e n a b a c t e r i u m n e e d s t o a e r o b i c a l l y

o x i d i z e o r g a n i c w a s t e s . B O D i s a m o n g t h e m e t h o d s u s e d in quantifying water

pollution. W h e n o rganic wastes are disposed into water, they may either be suspended or totally

dissolved (Jing L. et. Al, 2002). Presence of oxygen and organic matter facilitate the live and

continuous growth of bacteria in the water because of provision of oxidation and food respectively.

Animals and bacteria oxidized organic matter by breaking them down with existence of oxygen

in their systems either by digestion or biochemical action of bacteria. Usually, to determine the

water sample BOD, a calculation of the amount of oxygen used by the bacteria is performed by

comparing the amount left at the final fifth day with the initial amount at the first day. At the end

of the fifth day, bacteria will have oxidized mostly the soluble organic matter available while a

very slight oxidation of the insoluble matter will take place (Michael R. et al, 2009).

The degree of water pollution is usually determined basically means BOD measurements.

This is the most vital measurement taken in managing a sewage treatment plant. Essentially, the

effectiveness and efficiency of sewage treatment can be studied by making a comparison of the

BOD of sewage into the plant and the BOD of treated water leaving the plant. If water with a high

content of BOD gets into a river, it will enable the bacteria in the river to oxidize the organic

matter, the oxygen from the river will be consumed faster than the rate it dissolves back in from the

river causing aquatic life to die out due to lack of oxygen (Jun H. et al, 2000). It is essential for

sewage treatment plants eliminate as much BOD as possible from the sewage water.

Wastewater typically contains a BOD of 110-440mg/L which must be reduced to 20mg/L in

wastewater treatment plants.

𝑚𝑔 𝐷𝑂𝑖𝑛𝑖𝑡𝑖𝑎𝑙 −𝐷𝑂5 𝐷𝑎𝑦𝑠

𝐵𝑂𝐷𝑠 ( )=
𝐿 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑡𝑖𝑜𝑛

𝑆𝑎𝑚𝑝𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒(𝑚𝐿)
𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =
𝐵𝑜𝑡𝑡𝑙𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒(300 𝑚𝐿)
METHODS

The materials used in this lab were; 300mL BOD bottles, large plastic bottles with siphons, DO meter

and graduated cylinders. The prepared nutrients solutions provided include; Phosphate buffer, Magnesium

sulfate solution, Calcium chloride solution, Ferric chloride solution and Ammonium Chloride solution.

Firstly, the desired volume of DI water was added to a 20-L on a plastic 20 L carboy. Water was

aerated overnight by bubbling in ambient air compressed air. The saturated O2 concentration in the water

was around 8.4 mg/L DO. 1 mL per liter of the five nutrients solutions were added to the aerated water.

The pH od the dilution water was adjusted to 7.

The appropriate volume of well mixed diluted sample was transferred into a well cleaned and rinsed

300mL BOD bottles using a tipped, volumetric pipette. BOD was slowly filled ensuring there was no air

bubble trapped inside. Triplicate bottles were also filled with dilution water. The bottles were carefully

labelled. The calibrated DO meter was used to measure the initial DO of the dilution water blanks and the

measure recorded. BOD bottles were placed in a cabinet at room temperature for 5 days. The calibrated DO

meter was again used to measure the final DO of the dilution water blanks and the sample bottles and the

measure recorded.
.

RESULTS

For the determination of biochemical oxygen demand (BOD) of wastewater, the BOD is

computed subtracting the final DO from the initial DO and dividing by the dilution fraction. The

test measures the ability of bacteria to digest organic matter after five days at room temperature

by analyzing the depletion of oxygen. The following analysis are done.

𝑚𝑔 𝐷𝑂𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐷𝑂5 𝐷𝑎𝑦𝑠

𝐵𝑂𝐷𝑠 ( ) =
𝐿 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑡𝑖𝑜𝑛

𝑆𝑎𝑚𝑝𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒(𝑚𝐿)
𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =
𝐵𝑜𝑡𝑡𝑙𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒(300 𝑚𝐿)

These formulas were used in excel to calculate BOD5 and the following results were obtained.
 Initial DO Final DO dilution BOD5
(mg/L) (mg/L) fraction (mg/L)
2.6 0 390
2.4 0 360
volume 2mL 0.006666667
1.9 0 285
2 0 300
7.6 5.7 190
9.4 6 340
volume 3mL 0.01
7.6 5 260
8.6 4.9 370
8.2 4.7 262.5
8.8 4.3 337.5
volume 4mL 0.013333333
8.2 4.7 262.5
8.6 5 270
8.1 2.9 312
7.9 2.3 336
volume 5mL 0.016666667
8.1 2.7 324
7.9 3 294
7 3.7 132
volume 6.8 0.5 252
0.025
7.5mL 6.7 2.5 168

7.8 1.1 201

volume 8.4 1.2 216
0.033333333
10mL 8.2 1 216
8.2 1 216
10 0.1 237.6
volume 10.1 0.2 237.6
0.041666667
12.5mL 9.9 0.1 235.2
10 0 240
Table 1: calculated BOD5 results
Sample Avg Initial DO Avg Final DO dilution Avg
Volume (mg/L) (mg/L) fraction BOD5(mg/L)
2 2.225 0 0.006666667 333.75
3 8.3 5.4 0.01 290
4 8.45 4.675 0.013333333 283.125
5 8 2.725 0.016666667 316.5
7.5 6.833333333 2.233333333 0.025 184
10 8.15 1.075 0.033333333 212.25
12.5 10 0.1 0.041666667 237.6
Table 2: Average calculated BOD5 results

A graph showing variation in BOD5 in different

volume of sample water
400
350
300
BOD5 (mg/L)

250
200
150
100
50
0
0 2 4 6 8 10 12 14
Sample water volume (mL)

Fig 1 A graph showing variation in BOD5 in different volume of sample water

DISCUSSION

A neutral pH value of the sample should be achieved s i n c e microorganisms usually

need an environment that is suitable to survival. The dissolved oxygen in all the samples

decreased at the end of the five days. It is observed that the concentration of dissolved oxygen

decreases a s the dilution of the samples increased. This is because there is a high demand of

oxygen from bacteria. After fifth day, the dissolved oxygen became less as compared to the first

day. A high d e m a n d of oxygen from bacteria can be proven with dissolved oxygen

readings being low, which leads to a c o n c l u s i o n t h a t t h e r e a r e possible

contamination sources.

The calculations for BOD5 content was p e r f o r m e d as shown in Table 1 and

Table 2. From the graph in Fig 1, it is observed that the BOD for the samples generally reduced

with the increasing dilution of volume of the samples. The lesser the BOD value, the lower the

sample is polluted. High amount of BOD shows a high amount of organic material in the sample

that is easily degradable. While, a low BOD shows less amount of organic materials matter, that

are difficult to dissolve. The dilutions according to the BOD5 range provided in the lab manual

indicate the results obtained in this experiment are realistic since the fall in that range.

The quality of wastewater is very polluted when the BOD5 level is greater than 100, while

the wastewater quality is somewhat polluted when the BOD5 level is between 6 to 9. Hence, the

BOD5 level of all dilutions were above 100 ppm showing that all the dilution was very polluted.

Nitrification is the by which specific bacteria oxidize nitrogen from ammonium to nitrate.

The nitrification inhibitor prevents the specific bacteria that are accountable for the deprivation of

the ammonium without damaging the microorganisms that degrade carbon compounds.

Therefore, an inhibitor is supplemented to the dilution solution to stop the alteration of

ammonium to nitrate. In this experimental, the objective of learning how to conduct BOD5 tests

was successfully achieved. BOD of all wastewater samples were all determined.
REFERENCES

Jing Liu and Bo Mattiasson. (2002). Microbial BOD Sensors for Wastewater Analysis. Journal of

Water Research. Vol 36, issue 15 pg 3786-3802

Michael R. Penn, James J. Pauer and James R. Mihelcic. (2009). Biochemical Oxygen Demand.

Journal of Environmental and Ecological Chemistry. Vol II

Jun Ho. Lee and Ki Woong Bang. (2000). Characterization of Urban Stormwater Runoff. Journal

of Water Research. Vol 34, issue 6 pg 1773-1780

APPENDIX

Raw data table

Initial DO Final DO
(mg/L) (mg/L)
2.6 0
2.4 0
volume 2mL
1.9 0
2 0
7.6 5.7
9.4 6
volume 3mL
7.6 5
8.6 4.9
8.2 4.7
8 4.3
volume 3mL
8.2 4.7
8.6 5
8.1 2.9
7.9 2.3
volume 5mL
8.1 2.7
7.9 3
7 3.7
volume 6.8 0.5
7.5mL 6.7 2.5

7.8 1.1
volume 8.4 1.2
10mL 8.2 1
8.3 1
10 0.1
volume 10.1 0.2
12.5mL 9.9 0.1
10 0

Sample calculation, for volume 2mL

𝑆𝑎𝑚𝑝𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒(𝑚𝐿) 2 𝑚𝐿
𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = = = 0.006666667
𝐵𝑜𝑡𝑡𝑙𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒(300 𝑚𝐿) 300 𝑚𝐿

𝑚𝑔 𝐷𝑂𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐷𝑂5 𝐷𝑎𝑦𝑠 2.6 − 0

𝐵𝑂𝐷𝑠 ( )= = = 390
𝐿 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑡𝑖𝑜𝑛 0.00666667

𝑚𝑔 𝐷𝑂𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐷𝑂5 𝐷𝑎𝑦𝑠 2.4 − 0

𝐵𝑂𝐷𝑠 ( )= = = 360
𝐿 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑡𝑖𝑜𝑛 0.00666667
 𝑚𝑔 𝐷𝑂𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐷𝑂5 𝐷𝑎𝑦𝑠 1.9 − 0
𝐵𝑂𝐷𝑠 ( )= = = 285
𝐿 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑡𝑖𝑜𝑛 0.00666667

𝑚𝑔 𝐷𝑂𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝐷𝑂5 𝐷𝑎𝑦𝑠 2−0

𝐵𝑂𝐷𝑠 ( )= = = 300
𝐿 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑟𝑎𝑡𝑖𝑜𝑛 0.00666667