Biochemical Oxygen Demand
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BOD: Biochemical Oxygen Demand
BOD5 is the total amount of oxygen consumed by microorganisms during the first five days of biodegradation
BOD5 = (DOi - Dof)/(P) this is the five-day BOD of a diluted sample
Where DOi = the initial dissolved oxygen (DO) of the diluted wastewater
DOf = the final DO of the diluted wastewater, 5 days later
P = the dilution fraction = (volume of wastewater)/(volume of wastewater plus dilution water)
A standard BOD bottle holds 300 mL, so P is just the volume of wastewater divided by 300 mL
5.12)
Some wastewater has a BOD5 of 150 mg/L at 20 degrees Centigrade. The reaction rate k at that temperature has been determined to be 0.23/day.
a) Find the ultimate carbonaceous BOD
b) Find the reaction rate coefficient at 15 degrees Centigrade
c) Find BOD5 at 15 degrees Centigrade
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Solution Summary
Assistance with finding the ultimate carbonaceous BOD, reaction rate coefficient, and BOD5 at 15 degrees Centigrade given a BOD5 of 150 mg/L at 20 degrees Centigrade and reaction rate k at that temperature of 0.23/day.
Solution Preview
As you are probably aware, BrainMass policy precludes me from providing you with a complete answer. I will give you some pointers to help you get there.
a. Find the ultimate carbonaceous BOD
You can use this equation:
BODt ...
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