BOD - Environmental Science & Technology

7/24/2019 BOD - Environmental Science & Technology

1/14

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):


2/14



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


3/14



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


4/14



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


5/14



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


Ultimate Biochemical Oxygen Demand (L) mg O2L-1

276,90


time BODt degradation

days mg O2L-1

%

5 180,00 65,017 213,23 77,01

15 265,03 95,71


6/14



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.


7/14



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:


Time days 5



473,81


days mg O2L-1

%

5 250,00 57,26


Time days 5



523,06


days mg O2L-1

%

5 250,00 47,80


days mg O2L-1

%

5 250,00 52,76


Time days 5Biochemical Oxygen Demand (BODt) mg O2L

-1 250,00


436,62


8/14



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.



9/14




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.


10/14



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.


11/14



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.


12/14



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.


Demand is 56 days.


Demand is 13 days.


13/14



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/


14/14



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

Documents

BOD - Environmental Science & Technology