Question

Wastewater from a winery contains approximately 1.8 g/L glucose (C₆H₁₂O₆), 1.8 g/L fructose (C₆H₁₂O₆) and 250 mg/L acetic acid (CH₃COOH). What would be the BOD5 and ultimate BOD assuming the first order rate constant is 0.23 d^-1 (base e)?

Answer 1

The reaction will be

C6H12O6 + 6O2 --> 6CO2 + 6H2O

Theoretical oxygen demand = 6 X MO2 / Molecular weight of compound = 192 / 180 = 1.07

So BOD for glucose is 1.07 g / g of glucose

so for 1.8 grams of glucose + fructose BOD = 1.07 X 3.6 = 3.852

CH3COOH + 2O2 --> 2CO2 + 2H2O

Theoretical oxygen demand due to acetic acid =

so BOD = 2X 32 / 60 = 1.07 /g

So for 250 mg = 0.250 g the demand will be = 1.07 x 0.25 = 0.267 grams

Total BOD = 0.267 + 3.852 = 4.119

Rate constant = 0.23 d-1

Rate = K [4.119]

Rate = 0.23 X 4.119 = 0.947

so after five days = 0.947 / 5 = 0.189