WASTEWATER TREATMENT-VIII

SOLUTION

1. 1 MLD of wastewater with influent soluble BOD ₅ (S _o ) = 75 mg/L is to be treated in an oxidation pond such that effluent soluble BOD ₅ (S _e ) is 10 mg/L. Determine the oxidation pond dimensions. Assume the depth of the pond to be 0.5 m or less.

Hint:

• K = 0.1 L/mg/d,
  where K is the first order microbial substrate utilization rate ( )
•   Y _T = 0.5 mg/mg; K _d = 0.05 /d
•   Assume only 50 percent of algal oxygen production should be utilized for microbial respiration
• Average intensity of solar radiation: 150 calories/cm ² /d
• Solar energy utilization efficiency for algae: 6 percent
• Energy content of algal bio-mass: 6000 calories/g algae
• Equation for algal photosynthesis:
  

Answer:

We know,    ;

Therefore,   

Also,          

Also,        , or,

Sludge Production

Volume of Oxidation Pond =

Assuming depth to be 0.3 m, Surface Area (A)   =

Algae production =

Total algal production = (15).(7400) = 111 kg/d

Assuming 1.3 Kg oxygen production per Kg algal production,

Oxygen Production = (1.3).(111) = 144.3 Kg/d

Since oxygen available is more that oxygen requirement, the design is adequate.

2   Explain why while oxidation ponds are efficient for BOD removal, aquatic plant ponds are inefficient for BOD removal and hence used only for nutrient removal

In aquatic plant ponds, the plants take carbon dioxide directly from the atmosphere for photosynthesis. Oxygen produced is directly released to the atmosphere. Hence no enhancement of aqueous oxygen concentration, which is required for microbial degradation of aqueous BOD, is achieved. Hence systems with such plants can only be utilized for nutrient removal.

In oxidation ponds, algae use dissolved carbon dioxide for photosynthesis and release oxygen into aqueous phase, which is used by microbes for degradation of aqueous BOD.