The following points will highlight the four main ways through which the demand of oxygen in sewage is expressed. The ways are: 1. Biochemical Oxygen Demand (BOD) 2. Chemical Oxygen Demand (COD) 3. Total Oxygen Demand (TOD) 4. Theoretical Oxygen Demand (ThOD).
Way # 1. Biochemical Oxygen Demand (BOD):
Biochemical oxygen demand (BOD) may be defined as the amount of oxygen required for the micro-organisms to carry out biological decomposition of organic matter in sewage under aerobic conditions at standard temperature. The amount of oxygen consumed in the process of decomposition is related to the amount of decomposable organic matter present in sewage.
Thus biochemical oxygen demand (BOD) may be taken as a measure of the amount of decomposable organic matter present in sewage. It is the most widely used parameter of organic pollution applied to both sewage and surface water. BOD is measured in milligrams of oxygen per litre of sewage, and it is expressed as mg/l or ppm.
The BOD test results are used to:
(1) Determine the approximate quantity of oxygen that will be required to biologically stabilize the organic matter present in sewage;
(2) Determine the size of sewage treatment facilities;
(3) Measure the efficiency of some treatment processes;
(4) Determine the strength of sewage; and
(5) Determine the amount of clear water required for the efficient disposal of sewage by dilution.
The organic matter present in sewage may belong to two groups:
(a) Carbonaceous matter and
(b) Nitrogenous matter.
Accordingly the biochemical decomposition (oxidation) proceeds in two stages. The first stage consists of breaking down carbonaceous matter with the production of carbon dioxide and water and simple amine compounds. The BOD corresponding to the first stage of decomposition (oxidation) is known as first-stage BOD or first-stage demand.
The first stage decomposition lasts for 7 to 10 days or more. In the second stage nitrogenous matter is decomposed (oxidized) into nitrites and nitrates. The BOD corresponding to the second stage of decomposition (oxidation) is known as second-stage BOD or nitrification demand. The second stage decomposition proceeds for a much longer period before the process gets stabilized.
Biochemical oxidation is a slow process and theoretically takes an infinite time to go to completion. Thus sewage will continue to absorb oxygen for many months, and it is impracticable to determine the complete oxygen demand.
However, by determining the oxygen used by a sample of sewage after a definite number of days of incubation at a given temperature, it is practically possible to arrive at a value very near to the total oxygen demand of the sample.
It is found that within a period of 5 days the decomposition or oxidation is about 60 to 70 percent complete, while within a period of 20 days the decomposition or oxidation is about 95 to 99 percent complete. The standard BOD test is usually carried out after 5 days of incubation of sample of sewage at a constant temperature of 20°C.
For this test 20°C temperature is used because the average value of temperature of the natural bodies of water is 20°C and it is easily duplicated in an incubator, and a period of 5 days is chosen because during this period a large percentage, about 60 to 70 percent, of the total BOD is satisfied (or about 60 to 70 percent decomposition or oxidation of organic matter is complete). The BOD value of 5-day incubation period is commonly written as BOD5 or 5-day BOD.
It may, however, be mentioned that since biochemical reaction rates are temperature dependent different values of BOD would be obtained at different temperatures. For instance the value of 5-day BOD at 20°C is to a reasonable extent comparable to 4-day BOD at 30°C and 3-day BOD at 35°C.
Further if the oxygen demand of the sample of sewage is greater than the available dissolved oxygen in the sample, a dilution of the sample is made. The amount of dilution depends upon the oxygen demand and must be such that an appreciable amount of dissolved oxygen (1.5 to 2.0 ppm) remains after the incubation period.
Limitations of BOD Test:
Despite the widespread use of the BOD test, it has a number of limitations as indicated below:
(1) In this test it is essential to have a high concentration of active bacteria present in the sample of sewage. In the absence of such bacteria, a high concentration of active, acclimated seed bacteria is required to be applied.
(2) Pretreatment is needed if the sample of sewage contains toxic wastes.
(3) The effects of nitrifying bacteria must be reduced or eliminated by pretreatment or by the use of inhibitory agents before the test is carried out.
(4) Only the biodegradable organics are measured in BOD test.
(5) The test does not have stoichiometric validity after the soluble organic matter present in the sample of sewage has been used or exhausted.
(6) In this test an arbitrary long period of time is required to obtain results.
(7) The 5-day period may or may not correspond to the point where the soluble organic matter that is present has been used. This reduces the usefulness of the test results.
Way # 2. Chemical Oxygen Demand (COD):
The chemical oxygen demand (COD) test is used to measure the content of organic matter of both sewage and natural waters. In this test the oxygen equivalent of the organic matter that can be oxidized is measured by using a strong chemical oxidizing agent in an acidic medium. Potassium dichromate has been found to be an excellent oxidizing agent for this purpose and the same is used in COD test.
The COD test is carried out at a high temperature. Known quantities of standard potassium dichromate solution and concentrated H2SO4 are added to a known quantity of the sample of sewage suitably diluted with distilled water. The mixture is then boiled for 2 hours, cooled and the amount of dichromate remaining is measured by titration with standard ferrous ammonium sulphate solution.
A catalyst, usually silver sulphate, is used to aid the oxidation of certain types of organic compounds. The inorganic compounds likely to interfere with the test should be removed before the test is carried out.
The principal reaction using dichromate as the oxidizing agent may be represented in a general way by the following unbalanced equation:
The COD test is also used to measure the organic matter in industrial and municipal sewage that contain compounds which are toxic to biological life. The COD of a sewage is, in general, higher than the BOD because more compounds can be chemically oxidized than can be biologically oxidized.
No general correlation exists between BOD and COD, but for many types of wastes, it is possible to correlate COD with BOD. This can be very useful because the COD can be determined in 3 hours as against 5 days required for BOD.
Moreover, once the correlation between COD and BOD has been established at any treatment plant, the COD measurements can be used to good advantage for the control and operation of the treatment plant. For a typical untreated domestic sewage the COD/BOD5 ratio varies from 1.25 to 2.5.
If this ratio is greater than 3, the sewage is considered difficult to biodegrade and for non-biodegradable sewage the ratio exceeds 10. Further in general COD of sewage is approximately equal to 80% or 0.8 of the theoretical oxygen demand (ThOD) of the sewage. Table 8.1 gives some typical values of COD for domestic sewage.
Way # 3. Total Oxygen Demand (TOD):
The Total Oxygen Demand (TOD) test is another instrumental method used to measure the organic content of sewage. This test involves the quantitative measurement of the amount of oxygen use to burn the organic substances and, to a minor extent, inorganic substances. It is thus a direct measure of the oxygen demand of the sample of sewage.
The test is conducted in a platinum- catalyzed combustion chamber. The oxidizable components of the sample of sewage introduced into the combustion chamber are converted to their stable oxides by a reaction that disturbs the oxygen equilibrium in the nitrogen carrier gas stream.
The momentary depletion in the oxygen concentration in the carrier gas is detected by an oxygen recorder. The TOD of the sample of sewage is obtained by comparing this peak height with the peak height of standard TOD calibration solutions. This test can be carried out rapidly, and the results have been correlated with the COD.
Way # 4. Theoretical Oxygen Demand (ThOD):
This is a theoretical method of computing the oxygen demand of various constituents of the organic matter present in sewage. The organic matter present in sewage may be of animal or vegetable origin and it consists of principal groups such as carbohydrates, proteins, fats and products of their decomposition.
Each one of these is generally a combination of carbon, hydrogen, oxygen and nitrogen as indicated by its chemical formula. Thus if the chemical formulae of the constituents of the organic matter are known, ThOD can be easily computed. For example – glycerine, commonly present in sewage has a chemical formula [CH2(NH2)COOH].
Its ThOD can be analytically determined by assuming the following steps in the reactions:
(1) In the first step, carbon is converted to CO2 and nitrogen is converted to ammonia.
(2) In the second step, ammonia is oxidized to nitrite.
(3) In the third step, nitrite is oxidized to nitrate.
(4) The ThOD is the sum of the oxygen required for all the three steps of reactions as indicated below:
(1) Carbonaceous Demand:
(2) Nitrogenous Demand:
Total Organic Carbon TOC:
The Total Organic Carbon (TOC) test is another means for measuring the organic matter present in sewage or water. This test is especially applicable to small concentrations of organic matter. The test is performed by injecting a known quantity of sample of sewage into a high-temperature furnace.
The organic carbon is oxidized to carbon dioxide in the presence of catalyst. The carbon dioxide that is produced is quantitatively measured by means of an infrared analyzer. In order to eliminate errors due to the presence of inorganic carbon, the sample prior to its analysis is acidified and aerated.
This test can be performed very rapidly. However, certain resistant organic compounds may not be oxidized and hence the measured TOC value will be slightly less than the actual amount present in the sample. For a typical untreated domestic sewage the BOD5/TOC ratio varies from 1.0 to 1.6. Table 8.1 gives some typical values of TOC for domestic sewage.
Relative stability of sewage is defined as the ratio of the amount of oxygen available in sewage to the amount of oxygen required to satisfy the first-stage BOD of sewage. Thus relative stability indicates the oxygen available in sewage as compared to the oxygen required for complete stability of sewage. The oxygen available in sewage will include both dissolved oxygen (DO) as well as oxygen present as nitrite or nitrate. Relative stability is generally expressed in percent.
The test for relative stability of sewage is carried out as follows:
(i) The sample of sewage is filled in a glass-stoppered bottle.
(ii) A small quantity of methylene blue solution is added to the sample of sewage.
(iii) The mixture is then incubated either at a temperature of 20°C or at 37°C. In hot countries like India, a temperature of 37°C is preferred.
(iv) During incubation when the available dissolved oxygen in the mixture is consumed, the anaerobic bacteria start their function and produce hydrogen sulphide which bleaches the blue colour and decolourizes the mixture.
(v) The period in days required for bleaching the blue colour is noted.
The relative stability is worked out from the following expressions:
SR = 100 (1 -0.794t20) … (8.36a)
SR = 100(1 -0.605 t37) … (8.36b)
SR = relative stability in percent;
t20 = period in days of incubation at 20°C; and
t37 = period in days of incubation at 37°C
The above noted expressions for relative stability may be derived as indicated below:
Table 8.7 gives the values of relative stability of sewage at 20°C and 37°C as given by equations 8.36 (a) and 8.36(b) for different periods in days required for decolorization.
The sooner the decolorization takes place (i.e., the value of t comes out to be less), the earlier the anaerobic conditions develop, which indicates lesser availability of oxygen in sewage, and hence the sewage is relatively unstable. Thus if decolorization takes place in a period of less than 4 days, corresponding to which since the value of relative stability of sewage at 20°C is less than 60%, the sewage may be considered as relatively unstable and hence not safe for being discharged into a water body such as natural stream or river.
However, if t ≥ 5 days, corresponding to which the value of relative stability of sewage at 20°C is more than 60%, the sewage may be considered as relatively stable and hence it can be safely discharged into a water body.
The relative stability test is not suitable for raw sewage because the colour is precipitated out due to the presence of some dissolved and colloidal solids. The test is, however, suitable for studying the quality of polluted stream waters and effluents coming out of sewage treatment plants.
The population equivalent of a sewage is the expression of some characteristic such as BOD, total solids, etc., of the per capita flow of the sewage in terms of the same characteristic of the per capita flow of some standard sewage. A normal domestic sewage of a separate system is usually taken as a standard sewage.
Since out of all the tests, the BOD test is very important and reliable in the analysis of sewage, the BOD test result is generally used to convert a sewage of particular strength into equivalent population. Thus the population equivalent of any sewage would be the number of persons who could be responsible for the sewage which would have the same characteristics of BOD as the standard sewage. Generally industrial sewage is compared with per capita domestic sewage through the concept of population equivalent PE using per capita BOD value as the basis. Thus, we have-
For a normal domestic sewage 5-day BOD (or BOD5) at 20°C varies from 73 to 82 gm (or 0.073 to 0.082 kg) per capita per day and it may be taken as 80 gm (or 0.08 kg) per capita per day. Thus for example, if 5-day BOD (or BOD5) at 20°C of an industrial sewage is 548 kg per day, then
Population equivalent, PE = 548/0.08 = 6850
The uses of population equivalent are as indicated below:
(i) Charge for Industrial Sewage Treatment:
With the help of population equivalent, the sewage obtained from various industries is converted into equivalent population and each industry is then accordingly charged for the treatment of its sewage. This method is more logical than the ordinary method of charging on the basis of volume of sewage.
(ii) Strength of Industrial Sewage:
Population equivalent serves as an index of the strength of industrial sewage for the purpose of treatment at the municipal sewage treatment plant.
Sometimes population equivalent is also computed by considering total industrial sewage per day, and domestic sewage per capita per day, instead of their BOD values. The population equivalent so computed is known as hydraulic population equivalent (or hydraulic equivalent population) PE(H)y Thus, we have-
The population equivalent computed by considering BOD values as indicated in equation 8.37 is, therefore, sometimes known as BOD population equivalent (or BOD equivalent population) in order to distinguish it from hydraulic population equivalent (or hydraulic equivalent population).