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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 1
In a pipe of the sewage collection system of a town of 8000 inhabitants, a
device takes wastewater samples every two hours. It was installed to
optimize the depuration treatment taking into account the variability of theBiochemical Oxygen Demand BOD and the amount of Suspended Solids SS.
The table below shows the results obtained in the laboratory for these two
parameters as well as the flow rate Q during 24-hours: from 6 am to 4 am of
the next day.
1.1.Determine the average flow-rate in Ls-1
Averaging all the flow rates of the
values obtained during the period of24-hours, we obtained that the
average flow-rate in Ls-1is:
1.2.Estimate the mean 5-day Biochemical Oxygen Demand in mg O2L-1and the mean concentration of Suspended Solids in mgL-1
In order to calculate the mean 5-day
BOD and the concentration of
Suspended Solids, we have to apply theweighted arithmetic mean. As it is
shown in the table on the left, we
calculate how much represents the
amount of flow rate in each period of
time by dividing the flow rate in each
period over the sum of all the flow rates.
After that, we have to multiply these
values by the BOD5 and SS in every
range of time.
25,965Average flow (L/s):
Time (h)Q
(MLday-1
)
BODs
(mgO2L-1
)
SS
(mgL-1
)
6:00 1,06 30 348:00 3,09 180 98
10:00 5 300 250
12:00 3,55 380 292
14:00 3,21 260 281
16:00 3,51 195 205
18:00 3,36 205 208
20:00 2,12 130 122
22:00 1,06 45 42
24:00:00 0,37 33 15
2:00 0,22 20 144:00 0,37 16 13
% of QT BOD5 % of QT SS % of QT
3,94% 1,18127786 1,338781611,48% 20,66121842 11,248886
18,57% 55,72065379 46,433878
13,19% 50,11144131 38,506686
11,92% 31,00297177 33,507058
13,04% 25,42533432 26,729198
12,48% 25,58692422 25,961367
7,88% 10,23774146 9,6077266
3,94% 1,77191679 1,653789
1,37% 0,453566122 0,2061664
0,82% 0,163447251 0,11441311,37% 0,219910847 0,1786776
26,92Total flow QT(ML/day):
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Finally, adding all this final values of each column up, we obtain the
following results:
Mean BOD5 (mgO2/L): 222.54
Mean SS concentration (mg/L): 195.49
1.3.Calculate the contribution of each inhabitant to the daily BOD 5 and
SS in grams per day
Contribution to BOD5
26.9210 Ld x 222.54 10gO2
L8000hab =748.85
g O2
d hab
Contribution to SS 26.9210 Ld x 195.49 10
gO2L
8000hab =657.82g O2
d hab
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 2
In order to determine the oxygen demand of an industrial wastewater, 6 mL
of this industrial wastewater were added to a 500 mL bottle. The dissolved
oxygen concentration of the wastewater initially and after incubating it at20C for five days were 7 mg O2/L and 1 mg O2/L, respectively.
Determine the BOD5 of the industrial wastewater.
According to the table below that I have imported from the Excel document,
after entering the data of this concrete exercise, as the incubation period has been
5 days and the temperature is the one for which the formulae is applied, the result
obtained of the BOD is the BOD5that is required in this exercise.
Therefore, the BOD5 of the industrial wastewater is 500 mg O2/L (=ppm)
ml 6,00
ml 500,00
ml 494,00
[=] 0,0120
Initial diluted seeded wastewater dissolved oxygen (DO i) mg O2L-1
7,00
Final diluted seeded wastewater dissolved oxygen (DOf) mg O2L-1
1,00
Biochemical Oxygen Demand (BODt) mg O2L-1 500,00
ppm 500,00
volume BOD bottle
volume wastewater sample
volume dilution water
dilution factor
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 3
The 5-day Biochemical Oxygen Demand test performed on a wastewater
sample yielded 40 mg O2L-1. An aliquot of 40 mL of the wastewater sample
was taken and transferred to a 300 mL BOD bottle. The oxygen concentrationin the dilution water and after the incubation of the mixture were 9 mgL-1
and 2.74 mgL-1, in that order. Calculate the oxygen concentration in the
wastewater sample.
BOD= DO DOfp DO = B O D p + D Of
Using this expression, we can obtain the initial concentration of DO in the
BOD bottle, that is 8.07mg/L. As we also know that the oxygen concentration inthe dilution water is 9 mg/L, we can obtain the oxygen concentration in the
wastewater sample:
DO = 9 mgO2/L 10.1333 + x 0 . 1 3 3 3 x = [O]in the water sample = 2.05 mg/L
ml 40,00
ml 300,00
ml 260,00
[=] 0,1333
mg O2L-1 ?
mg O2L-1 2,74
mg O2L-1 9,00
mg O2L-1
40Biochemical Oxygen Demand (BODt)
Volume BOD bottle
Volume dilution water
Dilution factor
Initial diluted seeded wastewater dissolved oxygen (DO i)
Final diluted seeded wastewater dissolved oxygen (DO f)
Oxygen concentration in the dilution water
Volume wastewater sample
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 4
The 5-day Biochemical Oxygen Demand test performed on a waster sample
yielded 180 mg O2L-1. The k reaction rate constant at 20 C was about 0.21
days-1.
4.1. Determine the Ultimate Biochemical Oxygen Demand
According to the table above, the Ultimate Biochemical Oxygen Demand in
mg of O2per liter is 276.90.
4.2. Calculate the Biochemical Oxygen Demand and the percentage of
organic matter degraded after 7 days and after 15 days
Entering the date obtained in the previous section in the following table, we
obtain the BOD curve, representing the oxygen demand varying with time. Besides,
as it is correlated with the amount of organic matter that is biodegraded at that
time, we can also estimate the percentage of organic matter degraded after thenumber of days that are required.
However, the Biochemical Oxygen Demand and the percentage of
organic matter degraded after 7 days are 213.23 mg O2/L and 77.01%,
respectively. Lastly, this values after 15 days would be 265.03 mg O2/L and
95.71%.
Ultimate Biochemical oxygen demand (L) mg O2L-1
276,90
Time days 5
k reaction rate constant days-1 0,21
Biochemical Oxygen Demand (BODt) mg O2L-1 180,00
Ultimate Biochemical Oxygen Demand (L) mg O2L-1
276,90
k reaction rate constant days-1 0,21
time BODt degradation
days mg O2L-1
%
5 180,00 65,017 213,23 77,01
15 265,03 95,71
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Lab 2Biological Oxygen Demand
4.3. Represent the Biochemical Oxygen Demand curve of this sample
and describe it
The BOD curve of this sample is described by the graphic above. It shows
how the oxygen demand approaches to the BODmax=276.90mg/L as the time
increases. However, due to the fact that biochemical oxidation is a slow
process and it takes an infinite time to go to completion, it does not actuallyreach it. We can also see how after starting incubation, the oxygen demand
increases exponentially, but after some point, the rise occurs really slowly.
After 5 days, the BOD5 is found at 180 mg/L, being the 65% of the total,
which is within the common range. Also, as the reaction rate constant is
quite high (0.21 days-1), we can find the 95% of the aerobic oxidation even
sooner than 15 days, which is a rather rapid transformation.
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 5
The 5-day Biochemical Oxygen Demands of three different industrial
wastewaters were determined in the laboratory. The results were quite
similar for the three samples: their BOD5was around 250 mg O2L-1.
5.1. Considering that their k BOD reaction rate constants at 20C are0.13, 0.15 and 0.17 days-1, respectively, which wastewater sample
contains the higher concentration of easily biodegradable
compounds? Justify your answer.
The higher concentration of easily biodegradable compounds, the higher
the reaction rate constant is. Therefore, the wastewater sample containing the
higher concentration of easily biodegradable compounds must be the last one, the
one whose k reaction rate constant is 0.17 days-1, the highest one.
5.2. Determine the Ultimate Biochemical Oxygen Demand of the three
wastewaters as well as the percentage of organic matter degraded
after five days. Compare the results.
5
.
3
.
According to this two tables imported from the Excel document, the BODu
and the percentage of organic matter degraded after five days are 523.06 mg O2/L
and 47.5%, respectively, for the wastewater with a k reaction rate constant equal
to 0.13 days-1
.
Just changing the input of the k, we can easily obtain the results for the two
other wastewater samples:
k reaction rate constant days-1 0,15
Time days 5
Biochemical Oxygen Demand (BODt) mg O2L-1 250,00
Ultimate Biochemical oxygen demand (L) mg O2L-1
473,81
time BODt degradation
days mg O2L-1
%
5 250,00 57,26
k reaction rate constant days-1 0,13
Time days 5
Biochemical Oxygen Demand (BODt) mg O2L-1 250,00
Ultimate Biochemical oxygen demand (L) mg O2L-1
523,06
time BODt degradation
days mg O2L-1
%
5 250,00 47,80
time BODt degradation
days mg O2L-1
%
5 250,00 52,76
k reaction rate constant days-1 0,17
Time days 5Biochemical Oxygen Demand (BODt) mg O2L
-1 250,00
Ultimate Biochemical oxygen demand (L) mg O2L-1
436,62
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
For k=0.15 days-1BODu= 473.81 mg O2/L and degradation % = 52.76%
For k=0.17 days-1BODu= 436.62 mg O2/L and degradation % = 57.26%
5.3. Represent the BOD curve of each wastewater sample. How much
time does each wastewater sample require to get to the Ultimate
Biochemical Oxygen Demand?
The Ultimate Biochemical Oxygen Demand is reached in an infinite time.
Therefore, I will give the time it takes to transform the 95% of the organic matter.
For k = 0.13 days-1, the time to get it is approximately 24 days.
For k = 0.15 days-1, the time to get it is approximately 20 days.
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
For k = 0.17 days-1, the time to get it is approximately 18 days.
The results obtained in this section make sense. As we already knew, the
time to get to the Ultimate Biochemical Oxygen Demand decreases when the
reaction rate constant increases, as it happens in this case.
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 6
6.1. Calculate the 5-day Biochemical Oxygen Demand of the previous exercise
second wastewater sample (k =0.15) at three different temperatures: 4C,12C and 28C. Does the oxygen consumption vary with different
temperatures? Is it higher or lower than 250 mg O2L-1? Why?
As the wastewater sample is the same and the BODmax only depends on
the sample itself and not on the environmental conditions, for every different
temperature the BODmax is going to be the same and equal to 473.81 mg O2/L, the
value that we calculated in the previous exercise. What can vary is the time to get
to it, as well as the BOD5.
T = 4C k = 0.020 days-1BOD5= 45.09 mg O2/L.
T = 12C k = 0.054 days-1BOD5= 112.11 mg O2/L.
T = 28C k = 0.232 days-1BOD5= 325.28 mg O2/L.
The Biological Oxygen Demand 5 days after incubation has changed when
we change the temperature of the wastewater. As we already know, the reaction
rate constant increases with temperature. So, the k and, therefore, the BOD5 are
only higher than in the previous case in the laboratory at 20C in the last situation,
whose temperature is 28C, higher than the previous one. These are exactly the
only reasonable results that we could expect.
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
6.2. Represent the BOD curve at 4C, 12 C, 20 C and 28 C. What is the effect of the
temperature on the ultimate BOD?
As I have already explained, the temperature does not affect the ultimate BOD for
the reasons mentioned before.
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
6.3. How much time does this wastewater sample require to reach the UltimateBiochemical Oxygen Demand at these four different temperatures?
For T = 20C, the time required to reach the Ultimate Biochemical Oxygen Demand
has already been calculated in the exercise 5, being this value 18 days. At the other
different temperatures, this time can be estimated looking at the graphics above or taking
the value from the Excel document.
Furthermore, as we have mentioned before, the higher the temperature, the higher
the value of k, and lastly, the higher the time needed to get to the BODmax. Hence, the only
temperature that allows us to get to the BODmax quicker than in the laboratory is the
highest one, 28C.
For T = 4C, the time required to reach the Ultimate Biochemical Oxygen
Demand is 150 days.
For T = 12C, the time required to reach the Ultimate Biochemical Oxygen
Demand is 56 days.
For T = 28C, the time required to reach the Ultimate Biochemical Oxygen
Demand is 13 days.
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 7
The result of the Ultimate Biochemical Oxygen Demand test performed on a
sample of a sewage effluent was 30 g O2m-3
. The BOD reaction rate constantwas estimated to be 0.22 days-1at a temperature of 20C in previous studies.
As average, 0.5 m3s-1of this effluent are discharged into a river reducing the
levels of dissolved oxygen and causing of high fecal coliform counts. During
dry seasons, the temperature of the rivers water increases to 25 C and its
flow rate slows down to 6 m3s-1 intensifying the aforementioned impacts.
Determine the BOD5at these conditions.
Entering the datum of the k at 20C in the formulae, we obtain the reaction
rate constant at 25C, which is 0.289 days-1 = 1.056.The BODmax does not change with environmental conditions. However, it
changes because it is added another stream of water, the clean river, and when
they meet each other, the BOD of the water decreases. However, the rate of oxygen
(g O2/s) must be equal in both cases because the BOD of the up-rivers water is
considered null. Thus, we can obtain the BODmaxof the river after discharging into
it the effluent:
30 gO2
0.5s =
0.5+6
s =2.3077 gO2/
Once we know the Ultimate Biochemical Oxygen Demand, we must take into
account that the BOD5 actually varies with environmental conditions, because the
higher the reaction rate constant is, the higher the amount of organic matter
degraded is.
= 1 =2.3077 gO2 1 . = 1.7636 gO2/
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Ander Rangil Nieto G-61 C-2
Lab 2Biological Oxygen Demand
Exercise 8
The average flow rate of a small river during the dry season is 100 L/s.
Considering that the BOD5of the effluent of a wastewater treatment plant isequal or below 20 mg O2/L and that the limit value for the BOD5is 4 mg O2/L
in the receiving water, to how many inhabitants can serve this treatment
plant assuring that water quality standards are met in the rivers water?
Data:
- Suppose that the Biochemical Oxygen Demand of the up-rivers water is
null and that this river is not the source of drinking water for the
inhabitants of this village.
- The wastewater generation rate is 400 Lhab-1d-1.
In order to calculate the number of inhabitants to whom this treatment
plant can serve (assuring the water quality standards), we have to obtain first the
flow rate of the effluent stream going through the plant.
When this stream meets the river, the BOD of the water decreases.
However, the rate of oxygen (mg O2/s) must be equal in both cases because the
BOD of the up-rivers water is null.
Therefore, taking it into consideration, we can estimate the flow rate from
the treatment plant (Q):
20 mgO2L Q
= 4 mgO2L (Q +100 Ls)Q
=25L/s
As the wastewater generation rate is 400 L/(habd), the number of
inhabitants can be obtained in the following way:
25Ls 3600s1h 24h1d 1habd400L = 5400 inhabitants