Bod5

Water Quality

Last Updated on Mon, 02 Jan 2023

BOD5 refers to a particular empirical test, accepted as a standard, in which a specified volume of sample water is seeded with bacteria and nutrients (nitrogen and phosphorus) and then incubated for 5 days at 20°C in the dark. BOD5 is measured as the decrease in dissolved oxygen (in mg/L) after 5 days of incubation. The BOD5 test originated in England, where any river contaminant not decomposed within 5 days will have reached the ocean.

Water surface turbulence helps to dissolve oxygen from the atmosphere by increasing the water surface area. A BOD5 of 5 mg/L in a slow-moving stream might be enough to produce anaerobic conditions, while a turbulent mountain stream might be able to assimilate a BOD5 of 50 mg/L without appreciable oxygen depletion.

BOD Calculation Example 3.7

When a liter water sample is collected for analysis, an insect weighing 0.1g is accidentally trapped in the bottle. The initial DO is 10 mg/L. Assume that 10% of the insect’s fresh weight is readily

FIGURE 3.6 Dissolved oxygen sag curve caused by discharge of organic wastes into a river.

v Point of organic waste discharge Start of recovery zone

FIGURE 3.6 Dissolved oxygen sag curve caused by discharge of organic wastes into a river.

biodegradable and has the approximate unit formula CH2O.* Also, assume that microbes that will metabolize the insect are present. If the laboratory does not analyze the sample until biodegradation is complete, what DO will they measure?

Answer: The chemical reaction for oxidation of organic matter is

CH2O + O2 -> CO2 + H2O.

This equation shows that one mole of O2 oxidizes one mole of CH2O. Therefore, the moles of CH2O in the insect will equal the moles of O2 consumed during biodegradation. Find the moles of O2 initially present and the moles of CH2O in the insect.

Molecular weight of CH2O = 12 + 2 + 16 = 30.

Moles O2 initially present = 10 X10—^^ = 3.1 x 10-4 mol/L.

32 g/mol

Moles CH2O in roach = moles of O2 consumed = 0’01 g = 3.3 x 10-4 mol. 2 2 30 g/mol

Biodegradation of the insect will consume all of the DO present and there will be about 0.2 x 10-4 mol of insect tissue left undegraded or (0.2 x 10-4 mol)(30 g/mol) = 0.6 mg insect tissue left over. The laboratory will find the water anaerobic.

* Organic biomass contains carbon, hydrogen, and oxygen atoms in approximately the ratio of 1:2:1, so that CH2O serves as a convenient unit molecule of organic matter.

COD Calculation Example 3.8

COD levels of 60 mg/L were measured in groundwater. It is suspected that fuel contamination is the main cause. What concentration of hydrocarbons from fuel is necessary to account for all of the COD observed?

Answer: For simplicity, assume fuel hydrocarbons to have an average unit formula of CH2. The oxidation reaction is

CH2 + 1.5 O2 CO2 + H2O. For each carbon atom in the fuel, 1.5 oxygen molecules are consumed.

Weight of 1.5 mole of O2 = 1.5 x 32 = 48 g.

Weight ratio of oxygen to fuel is: = 3.4.

A COD of 60 mg/L requires: = 18 mg/L fuel hydrocarbons.

If dissolved fuel hydrocarbons in the groundwater are 18 mg/L or greater, the fuel alone could account for all the measured COD. If dissolved fuel hydrocarbons in the groundwater are less than 18 mg/L, then fuels could account for only a part of the COD; other organic substances, such as pesticides, fertilizers, solvents, PCBs, etc., must account for the rest.

Continue reading here: Nitrogen Ammonia Nh3 Nitrite No2 And Nitrate No3

Was this article helpful?

+1
0