I do not always get a reduction from plant influent BOD to CBOD. I was wondering if it was the volume of inhibitor compared to higher nitrogen content in sample or longer incubation. Any thoughts?
If I understand your question correctly, more than likely what you are observing is a lack of a sufficient nitrifying bacterial population and a test duration that is too short to see the “ultimate” biochemical oxygen demand required to completely degrade the organic and inorganic materials in the influent sample.
The Biochemical Oxygen Demand (BOD) test measures the consumption of oxygen by bacteria metabolizing both the organic and inorganic (ammonia) contaminants in the wastewater. Bacteria that metabolize organic materials are called heterotrophs, while those metabolizing the inorganic materials are called autotrophs. In raw wastewater, heterotrophs predominate. Autotrophs are but a small fraction of the total microbrial population. Furthermore, the biochemical conversion of ammonia to nitrate is a two-step process typically represented by the bacteria nitrosomona and nitrobacter in the literature (although other species have been identified as well). The rate limiting step appears to be the first one, (nitrosomona’s conversion of ammonia to nitrite), as only rarely does one
find a buildup of nitrite, in activated sludge plants with nitrification. The nitrite is then metabolized to nitrate by a second group of autotrophs represented by nitrobacter.
If sufficient oxygen is available, organic materials are readily acted upon by heterotrophs breaking down complex organic materials into smaller organic materials and ultimately carbon dioxide, water, creating additional biomass plus respiration. In an activated sludge plant this metabolic process requires on average ~ 1 – 3 pounds of oxygen per
pound of organic material stabilized.
The same activated sludge process requires an additional 4 – 5 pounds of oxygen to
convert 1 pound of Ammonia Nitrogen to Nitrate Nitrogen.
What might that look like in a comparison of BOD vs cBOD in an Influent sample?
First let us assume that your influent contains 240 mg/l of BOD and 24 mg/l of 5
Ammonia-Nitrogen. This could then mean that the total BOD (including nitrification demand is 240 mg/l) therefore, 90% is carbonaceous and 10% (24 mg/l) is from breakdown of ammonia nitrogen from autotrophs. Adding the inhibitor should yield a carbonaceous BOD or cBOD value of ~ 216 mg/l all things being equal. However, this assumes that there are sufficient bacterial populations naturally occurring in the raw influent (or seed material if added), to breakdown both the organic and inorganic
material, and that both materials breakdown to their end products using the same amount of oxygen. But, as mentioned above, it takes an activated sludge plant almost two times as much oxygen to metabolize ammonia nitrogen to nitrate nitrogen. Why would it be any different in a bottle? If that is the case, then the cBOD is only about 80% of the total BOD, while the BOD from Ammonia-nitrogen is 20%. So, when the inhibitor is added, there should be an even larger measurable decrease in influent cBOD values (192 mg/l) vs. total BOD (240 mg/l). But the observed results do not appear to reflect this.
The actual answer lies in the details. First, Biochemical Oxygen Demand, as the name implies, has a biological as well as a chemical component. The metabolic rate (often represented by the letter k) is a measure of how quickly a particular bacteria metabolizes a particular substrate. The higher the k rate the more quickly the bacteria can metabolize that particular substrate. The more complex the substrate - the lower the k rate. And of course, just to muck it up some more, different bacteria have different k rates for the same substance, not to mention that the k rate can be influenced by; incubation temperature, available oxygen, alkalinity, micronutrients, and inhibitory/toxic compounds or byproducts of bacterial decomposition in the raw wastewater sample. Second, the standard BOD analysis is a 5-day test. This time component is represented with a subscript: BOD. 5
Next look at how the BOD test is run. Typically, an aliquot of sample is placed into a 5
standard 300 ml BOD bottle, and oxygen saturated dilution water containing nutrients is 5
added to the remaining volume of water. An initial dissolved oxygen measurement with a dissolved oxygen probe is taken, and the bottle is capped so that there is no headspace obeneath the stopper. The sample is then placed in a darkened incubator at 20 C for 5
days. After 5 days the sample is removed from the incubator and a second dissolved oxygen measurement is taken. The net difference is then used to calculate the sample BOD. A second bottle containing only 300 ml of dilution water is run as control to 5
verify that dilution water itself does not contribute any BOD. 5
Keep in mind that Oxygen is only slightly soluble in water. Saturation is between 8.5 and 9.0 mg/l at 20 degrees C, depending on elevation and barometric pressure. For a BOD 5
test to be valid, the sample plus dilution water (or in the case of influent samples) the dilution water plus sample, must deplete a minimum of 2.0 mg/l over the five day period, and leave a minimum of 1.0 mg/l in the 300 ml sample bottle. That means to measure a BOD of 240 mg/l we need to select a sample volume that will not consume (deplete) 5
more than 7.5 mg/l of oxygen over the five days. If seeding the sample is required, the sample volume must be reduced further so that the depletion of the sample plus the seed is less than 7.5 mg/l. Assuming no seed is required, and the raw influent sample has a dissolved oxygen concentration of 0.0 mg/l, then a dilution water saturated at 8.7 mg/l should yield an initial dissolved oxygen concentration of ~8.5 mg/l if 7.5 ml of influent and 292.5 ml of dilution water are added to a 300 ml BOD bottle. At the maximum allowable depletion of 7.5 mg/l of oxygen occurred after 5 days, the BODwould be 240 5
mg/l. The bacteria only metabolized the organic and inorganic material contained within the 7.5 ml of influent sample. Stated another way, 2.5% of the total sample volume (7.5 ml/300ml) utilized 7.5 mg/l of oxygen. The BOD of the sample is then calculated by 5
dividing the oxygen consumed by the percent sample volume as a decimal, 7.5/0.25 = 240 mg/l.
While the cBOD may be about 75 - 90% of the ultimate total Biochemical Oxygen 5
Demand, during the 5-day test, it was closer to 98 % of the measured result. The ultimate BOD test typically lasts 20 or 30 days. It is sometimes represented: BOD 20.
By contrast Secondary Effluent samples typically have very little organic material left other than endogenous respiration from biomass so, nitrification inhibitor actually has a more pronounced effect in the result even though there is potentially the same 2-4 mg/l difference in cBOD and total BOD That’s why you would see a total BOD value of 55. 5
say 12 mg/l vs. a cBOD value of 8 mg/l for the same effluent sample. 5
The establishment of sufficient autotrophic bacterial populations and metabolic k rates of both the heterotrophic and autotrophic bacteria can significantly influence what is seen in the five-day test. The attached chart simulates the impacts these two factors could have and may help explain your observations.
By day 5 there is little difference between the cBOD (238 mg/l) and the total BOD (242 mg/l). The ultimate BOD is actually closer to 288 mg/l. 240 mg/l as cBOD and 48 mg/l from metabolizing 24 mg/l of Ammonia nitrogen (2x the amount of oxygen required).