A \"seeded\" BOD (biochemical oxygen demand) test was conducted. Twelve ml of \"

Question

A "seeded" BOD (biochemical oxygen demand) test was conducted. Twelve ml of "seed" are added to 24 liters of dilution water. Thirty two milliliters of industrial wastewater are added to a 300-ml BOD bottle and the remaining volume consists of "seeded" dilution water. The average dissolved-oxygen concentration of the diluted wastewater samples and blanks, seeded dilution water, on the first day of the test is 7.4 mg/L and 9.0 mg/L, respectively. After incubating of separate BOD bottles at 20 degree C for 5 days, the average DO (dissolved oxygen) concentration of the diluted wastewater BOD bottles are seeded dilution water BOD bottles were 4.4 and 8.5 mg/L, respectively. Calculate the 5-day BOD of the industrial wastewater.

Explanation / Answer

BODt (mg/L)=(D1-D2)-(B1-B2)f/P

P= ml of sample/Volume of BOD bottle= 32ml/300ml=0.10

f=%see in D1/% seed in B1

% Seed in B1= 12ml/24L x (1000 ml/L)=0.0005

Volume of see in dilution water added to wastewater sample=0.0005x268ml=0.134

% Seed in D1=0.134/300ml=0.000447

f=%see in D1/% seed in B1=0.000447/0.0005=0.8934

BODt(mg/L)= (D1-D2)-(B1-B2)f/P=(7.4-4.4)-(9-8.5)*0.8934/0.1=3-0.4467/0.1=25.533 mg/L

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