Laboratory Exercise 7: DO and BOD
Contents
Objective
The objective of this experiment was to determine the Dissolved Oxygen and Biochemical Oxygen Demand (BOD5) for raw and treated wastewater samples.
Calculations
BOD Calculations.
To calculate the BOD exerted, which is denoted by “y” and at a particular time “t” the formula is:
Y = BODt = (BOo – BOt) – ∆DOcontrol * (Dilution factor)
y = BOD exerted at time “t”
DOo = Initial concentration of dissolved oxygen in a BOD bottle
DOt = Concentration of dissolved oxygen in the same BOD bottle after “t” days
Dilution Factor = Total volume / sample volume = (dilution water volume + sample volume) / (sample volume)
ΔDOcontrol = DO0,control – DOt, control Accounting for BOD in dilution water and/or drift in probe measurement Don't use plagiarised sources.Get your custom essay just from $11/page
Therefore, taking
BOD = 174
The Dilution factor, therefore
174 = (10.1 – 7.1) – o.1 * Dilution factor
Dilution factor = 60
Results and Discussion.
Table 1: Data for the recorded value of DO and BOD
Time | Initial DO | Final DO | Change in Control DO | BODt | |||||
Set I | Set II | Control | Set I | Set II | Control | Set I | Set II | ||
0.8819 | 10.1 | 9.9 | 9.1 | 7.1 | 6.8 | 9.0 | 0.1 | 174 | 180 |
4.0625 | 10.8 | 9.5 | 9.2 | 4.9 | 4.2 | 8.9 | 0.3 | 336 | 300 |
5.0104 | 9.7 | 9.6 | 8.4 | 3.5 | 4.1 | 8.2 | 0.2 | 360 | 318 |
5.9583 | 9.2 | 10.0 | 8.9 | 1.7 | 4.3 | 8.5 | 0.4 | 426 | 318 |
7.0208 | 9.9 | 9.4 | 8.6 | 3.2 | 1.4 | 8.3 | 0.3 | 384 | 462 |
Dissolved Oxygen.
The levels of Dissolved Oxygen fluctuate seasonally over an estimated period of 24 hours. This is the reason the sampling time was taken from the first day to the seventh day. The difference in the initial and final DO is what determines the amount of oxygen in a specific sample. This can then be used to determine the amount of oxygen required to stabilize organic matter biologically.
Biochemical Oxygen Demand.
Theoretically, the BOD takes an infinite time to go to completion. This is so because the amount of organic matter is assumed to be directly proportional to the rate of oxidation. The calculated dilution factor was 60, which, when multiplied by the difference in the initial, final, and control DO, gives the values for the BOD as indicated in the table above.
Graph1: For BOD (in mg/L) against time (in days)
In the graph, they, for the BOD, is almost linear until the 5th day when it forms a curve. This is proof of the theory that by the fifth day, the oxidation of carbonaceous matter is about 60-70%. After this period to the eight-day and beyond, the nitrifying bacteria start to using oxygen to convert ammonia nitrogen to nitrite then to nitrate. Being concerned only with the carbonaceous BOD, the monitored results are for the first five days, where the BODt is directly proportional to the time (in days).
Conclusion.
To sum up, the experiment was a success since all objectives were met. The values for the DO and the BOD for the collected samples were conclusively determined. There were a few errors due to oxygen that would have penetrated the samples during recording. This may have altered the results to a certain extend.
Answers
- Differentiating between CBOD and NBOD.
From the graph, it is evident that the amount of CBOD, which is the dissolved oxygen, increases with time. On the other hand, the NBOD levels are decreasing with an increase in the time since nitrifying bacteria starts to using oxygen to convert ammonia nitrogen to nitrites and nitrates. The DO thus gets depleted over time.
- Computing BOD5 for the sample.
Assuming:
Test Temp = 200C
Time, t = 5 days
BOD5 value = 20mg/l
Reaction rate = 0.18/day
Calculation :
At 20 O C, BOD 5 = ( L O ) [ 1 – 30 ( – k 1 ) ( 5 ) ]
thus ;
L O ( 20 O C ) = ( 20.0 ) / ( 1 – 30 – 0.9 ) = 22.9 mg/l
and L O at 30 O C ;
L O ( 30 O C ) = ( 22.9 ) [ 1 + ( 0.02 ) ( – 30 ) ] = 54.9 mg/l
k 1 at 30 O C ;
k 1 ( 30 O C ) = ( 0.18 ) ( 1.047 10 – 20 ) = 0.33 / day
Thus ;
BOD 5 ( 30O C ) = ( 54.9) [ 1 – 30( – 0.11 ) ( 5 ) ] = 39.3mg/l
- Recommend dilution.
COD value = 1200 mg/L
Since the COD: BOD ratio is 30:4, I would, therefore, recommend a dilution of 550mg/l. This because the COD has to be approximately 0.5 more than the BOD levels.
References.
Davis, Mackenzie L., and A. David. Introduction to environmental engineering. McGraw-Hill Companies, 2008.
Rice, Eugene W., et al. “Standard methods for the examination of water and wastewater.” American Public Health Association, Washington, DC 541 (2012).