Please answer the question if you really know what is needed. All parts shown. D

Question

Please answer the question if you really know what is needed. All parts shown. Don't do one part. If you are unable please skip it for another person
.

A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. After 5 days, the DO is 4 mg/L. After 30 days, the DO in the bottle is 2 mg/L and it now longer seems to be dropping. At the beginning of the test, we added a nitrification inhibitor so we can assume that the nitrification is not occurring. Thus, the only BOD being measured is carbonaceous BOD. a) b) c) d) What is the BODs of the wastewater? What is the ultimate carbonaceous BOD? How much BOD remains after 5 days? Based on the data above, estimate the reaction rate constant k (1/day).

Explanation / Answer

a. Dilution rate (P)= actual sample volume/ dilution water volume = 30/270 = 1/9

BOD for first 5 days

BOD5 = (DOinitial - DOFinal )/ P = (9-4)/(1/9) = 45 mg/lit

b. After 30 days DO values not dropping, hence BOD after 30 days will be utilmate carbonaceous BOD

BODu = (DOinitial - DOU)/ P = (9-2)/(1/9) = 63 mg/lit

Utilmate carbonaceous BOD = 63 mg/lit

c. After 5 days, 45 mg/lit of oxygen demand out of total 63 mg/lit would be satisfied. Hence, the remaining oxygen demand would be 63 – 45 = 18 mg/lit

d. The amount of BOD that has been exerted at any time t is given by

BODt= BODultimate (1-exp (-kt))  

Above equation for 5 days is written as

BOD5= BODultimate (1-exp (-k*5))  

45=63 (1- exp(-5*k)

k= 0.25 (1/ day)

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