In water dissolved oxygen (DO) place an important role in maintaining a balance as enumerated above, for instance, when oil or for that matter any substance that is a ‘’stranger’ is introduced into a water body, some biodegradation will take place to break down the hydrocarbon molecules, usually oxygen is the fuel that is used up or consumed to carry out this breakdown of the hydrocarbon molecules. Biodegradation can be chemical reaction or biological that is caused by living organisms like; bacteria, fungi, sulphate reducing bacteria etc.
The amount of oxygen required to completely breakdown the hydrocarbon molecules by chemical reaction is called the Chemical Oxygen Demand (COD), this is a measure of the amount of oxygen required to breakdown the molecules. Also the amount of oxygen required to breakdown completely the hydrocarbon molecule by biological activity is called the Biochemical Oxygen demand (BOD). However, in these report Biochemical Oxygen Demand (BOD) will be focused on rather than Chemical Oxygen Demand (COD).
Biochemical Oxygen Demand (BOD) can simply be defined as the amount of oxygen used up by microorganism (e.g. aerobic bacteria) in water. The rate at which oxygen is used up is perhaps more important than the determination of dissolved oxygen (Pierce J.J et al… 1997). How much clean a water is can well be determined by the amount of BOD, this is because the amount of oxygen present is determined and also the amount available for both plants and organisms to use up and maintain a balance ecosystem. Increase of biochemical oxygen demand in a water body can be caused by;
- If there is high level of organic pollutant
- High level of nutrients.
Increase in BOD causes species or microroganisms that are sensitive to lower dissolved oxygen to be replaced by organisms that are more tolerant to low dissolved oxygen, these results in a shift in the ecosystem.
Organic and nitrogenous compounds are responsible for high levels of BOD, basically biochemical oxygen demand is divide into carbonaceous and nitrogenous oxygen demand. Aerobic organisms utilize organic and nitrogenous nutrients and these processes require high amount of dissolved oxygen. This is represented thus;
Organic
C6H2O6 + 6O2(aq) + bacteria 6CO2 (aq) + H2O (5)
Nitrogenous
COHNS + O2 + bacteria CO2 + NH3 + energy (6)
From equations 5 and 6, it is obvious that oxygen is consumed to the detriment of the water body.
Factors that causes high Biochemical Oxygen Demand
Organic matters are the source of high biochemical oxygen demand, the sources of these contaminant are, industrial effluents, leaves, dead plants, animal manure, sewage treatment works, urban runoff and agricultural runoff. However, in Nigeria the most significant contribution of organic compounds in water are the petroleum and petrochemical industries, oil exploration activities are being carried out. Formation water is one of the major pollutants of aquatic environment (Obiere O. et al..2003). In United Kingdom and developed countries, sewage treatment works are a major contributor. In this report/experiment the discharged water from a sewage treatment works in England was considered.
Sewage
A sewage treatment plant treats most of our domestic sewage, industrial discharges, inflow and infiltration. Domestic sewage varies substantially over time and no two municipal wastes are the same (Pierce J.J et al… 1998). Sewage undergoes primary, secondary and in exceptional cases tertiary treatment. After proper treatment, the wastewater is dumped into the nearest water body.
Sewage contains organic pollutants, microorganisms, and nutrient enriching compounds (e.g. nitrogen and phosphorus). If not well treated and dumped in a water body, these constituents will cause significant problem on the water as enumerated above. Therefore, most if not all of these constituents are removed before discharge into water. Sewage constituents are known to contribute significantly to high biochemical oxygen demand in water, treated sewage effluent has a BOD value raging from 20 – 100mg/l (). Over 80% of the consented discharge to for water in England and Wales are from sewage treatment works (). Before sewage is discharged into a water body from a sewage treatment work, various levels of pollutants are determined. Water modelling can be used to determine the quality or quantity of BOD that is discharged into a water body.
Efforts have been made by various governmental bodies to regulate and reduce the amount of pollutants (that is biochemical oxygen demand) discharged into water. In Nigeria, the Federal Environmental Protection Agency set a uniform effluent limits for all categories of industries.
Table (2) source; FEPA 1991a
In England and Wales, the mount of sewage discharged into the water body has reduced significantly in the past 20 years as shown in table (3) and figure (1) below. All sewage treatment works are monitored for compliance with water quality targets. ()
Sewage treatment works pollution discharges in England and Wales, 1990 to 2002
Units: thousand tonnes/year
Table (3) Source: Environment Agency
Figure (1)
The effect of high biochemical oxygen demand in water cannot be overemphasised, that is why the European Union set up legislation or directives. The Urban Waste Water Treatment Directive (91/271/EEC) sets a BOD limit for net discharge at 25mg/l. Over 80% of rivers in Europe have a BOD less than 2mg O2/l ()
Table (4) source: European Environmental Agency
Table (4) indicates a drop on the mount of BOD discharged into European rivers.
The Winkler method was used to determine the concentration of dissolved oxygen present in the water sample. Winkler method is long established (Fifield F.W el al…1996). From several literatures the Winkler method was used in the determination of DO concentrations in water body. The recommended test method for DO on fresh water by European Environmental Agency and Environmental Protection Agency is the Winkler method.
This is a test method based upon an initial reaction between dissolved oxygen and precipitated manganese oxide, to produce higher oxidation of state of manganese (Fifield F.W el al…1996).
The biochemical oxygen demand (BOD) can be determined by subtracting the initial dissolved oxygen concentration (Do) from the dissolved solution concentration after 5 days (D1). However, in these experiment BOD7 was used instead of BOD5.
Therefore, the equation is given thus;
BOD (mg/l) = (Do – D1)/F (a)
Where
F = dilution factor
Aim
The advantage of biochemical oxygen demand measurement is that it parallel to natural processes closely (Fifield F.W el al…1996). In these report the biochemical oxygen demand (BOD7) concentration of a water sample from a sewage treatment works that has undergone primary treatment was checked, using the Winkler method to determine the dissolved oxygen present in the water sample. The class result was analysed, the mean value and standard deviation was also checked with confidence intervals.
The experiment also aims at comparing the BOD7 value with set guidelines or typical values and determines if the effluent is good for disposal into a water body.
Errors associated with the dilution and other possible errors were checked. The experiment aims at familiarizing each group with the expected quantity of BOD that should be discharged into a water body.
Methodology
Sample preparation
The sample was aerated to nearly saturation point, however, over saturation was avoided in other not to have excess oxygen in the sample. The water sample had a potential of high BOD, therefore, the sample was diluted to a ratio of 1:50.
5ml of the sample was added to two incubation bottles and were made up with 495ml of distilled water. The incubation bottles were completely filled and stoppers were immediately to ensure there was no air trapped in the bottles. The samples were inverted several times to ensure thorough mixing. The sample were marked 1 and 2,the sample marked 2 was placed in a dark incubator at 20oC for 7 days, the other sample was then checked for dissolved oxygen.
Dissolved oxygen determination
The Winkler method was used to determine dissolve oxygen concentration of the water sample in the incubation bottles.
Sample marked 1 was then checked for dissolved oxygen concentration, 2ml of manganese sulphate solution was added using a pipette. Immediately after 2ml of alkaline iodine – azide solution was added. The pipette tips were well below the surface of the sample bottle. The sample was inclined and the stopper carefully added to avoid air bubble.
The sample bottle was then inverted 20 times to allow for proper mixture and the orange - brown precipitate to settle. The mixing was repeated again and allowed to settle to give a clear supernatant liquid.
2ml of concentrated sulphuric acid was added to the supernatant liquid and the stopper was replaced. The sample was then mixed thoroughly by inverting the bottle until all the precipitate was dissolved.
200ml of the sample was transferred into a conical flask, 0.0125M thiosulphate solution was then titrated against. The sample brown colour due to iodine changed to pale straw, a drop of starch was added which changed the sample colour to blue. 0.0125M thiosulpate solution was titrated further until the blue colour changed to colourless. The volume of the thiosulphate was noted.
The sample marked 2 after 7 days of incubation was tested for dissolved oxygen, the process enumerated above was repeated. However, bubbles were found on retrieval from the incubator.
Result/calculation
In sample marked 1, 15.5ml of thiosulphate solution was titrated against the water sample.
In sample marked 2 (BOD7), 10.1ml of thiosulphate was titrated against the sample.
Table (5) Group 1 titration values.
Table (6) class result.
Where; Primary = 1 and Secondary = 2
Calculation
Sample 1
1ml of 0.0125M thiosulphate = 0.1mg of dissolve oxygen
15.5ml of 0.0125M thiosulphate = 1.55mg of DO
but there is 200ml of sample solution, therefore,
1.55mg of DO = 200ml of sample
in 1000 ml
therefore,
1.55 x 1000/200 = 7.75mg/l of dissolved oxygen.
Sample 2
1ml of 0.0125M thiosulphate = 0.1mg of dissolve oxygen
10.1ml of 0.0125M thiosulphate = 1.01mg of DO
but there is 200ml of sample solution, therefore,
1.01mg of DO = 200ml of sample
in 1000 ml
therefore,
- x 1000/200 = 5.05mg/l of dissolved oxygen
Biochemical oxygen demand (BOD) determination
From equation (a)
BOD (mg/l) = (Do – D1)/F (a)
Where
Do = 7.75mg/l
D1 = 5.05mg/l
F = 1:50 or 1/50
Therefore,
BOD = (7.75 – 5.05)/(1/50)
BOD = 135mgO2/l
Descriptive Statistics: BOD
Variable sample N N* Mean SE Mean StDev Minimum Q1 Median
BOD primary 4 0 137.50 8.29 16.58 120.00 123.75 135.00
secondary 5 0 9.20 1.04 2.32 7.25 7.50 8.25
Variable sample Q3 Maximum
BOD primary 153.75 160.00
secondary 11.38 13.00
One-Sample T
Test of mu = 25 vs not = 25
N Mean StDev SE Mean 95% CI T P
5 9.20000 2.32000 1.03754 (6.31934, 12.08066) -15.23 0.000
One-Sample T
Test of mu = 200 vs not = 200
N Mean StDev SE Mean 95% CI T P
4 137.500 16.580 8.290 (111.118, 163.882) -7.54 0.005
Discussion/conclusion
The result of the investigation shows that the BOD7 in the primary treatment stage is 135mgO2/l of the effluent water sample from the sewage treatment works. The BOD value at this point of treatment cannot be used to assess the effect on the water body, this is because the effluent from the primary treatment stage still flows to the secondary stage for further treatment by the sewage treatment works. However, there are set levels or quality of effluent that’s has to flow from the primary treatment to the secondary treatment and this is 200mg/l. Therefore, the BOD7 value meets the required level that can be further treated in the secondary phase.
Class result
On visual inspection of the values with the group that handled the primary effluent, it is obvious that the values are less than 200mg/l. A further statistical analysis was done on the values to corroborate this assumption. The t – sample test assumes that the values are equal, however, from the statistical analysis the null hypothesis can re rejected. The mean value from these groups is 137.5mg/l and the p – value is less than 0.05, therefore there is evidence against the null hypothesis. The primary effluent water has a BOD7 value that is less than 200mg/l. A 95% confidence interval was obtained, therefore, there is 95% confidence that the sample is less than the set value. Standard deviation is a measure of spread, that is, is the sample well spread. The mean value lies well within the variance, therefore, the result of the group can be assumed to be accurate and correct.
The secondary effluent water sample has a mean value of 9.2mg/l, the set guideline for a sewage treatment work is 25mg/l before discharge into the water or river. The value is lower than the set value, the t – sample shows clearly that there is very strong evidence against the null hypothesis. The mean value is less than the set limit of 25mg/l. confidence interval was also obtained and it shows that the mean value is well within the confidence intervals. There is 95% confidence that the sample is less than the set limit.
In an experiment errors can affect the degree of confidence of a result, error can occur as a result of the following;
- Lack of precision may arise from the many volumetric transfer of solution from one container to another.
- Air bubble noticed before the BOD7 test was carried out.
- Not being able to determine the actual end point during titration with 0.0125M thiosulphate solution.
- Interference from the atmosphere, for instance oxygen from the air.
- Error from dilution.
All the stated above errors can affect the result obtained, however, the significance or margin of error is quite minimal. From statistic the standard error values are quite small and could not have affected the final values obtained. It also shows that the dilution factor that was used was accurate as regards the level of dissolved oxygen present. An error of 0.1cm3 on a titration of 10cm3 is of 1% but on 100cm3 is reduced to 0.1% (Fifield F.W el al, 1996). Therefore, the larger the volume of sample the lesser the error that may occur within the experiment.
Conclusion
The biochemical oxygen demand concentration of water sample from a sewage treatment works was carefully monitored, the concentration value did not exceed the set guideline and regulation by the European Union for discharge into river by sewage treatment works (The Urban Waste Water Treatment Directive (91/271/EEC)). From figure (2) there is a marked difference between the primary and secondary values, therefore, it can be inferred that the treatment is efficient and the quality of effluent produced meets the European Union standard.
In conclusion the effluent water can be discharged into any receiving water body.
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