CG Vyapam_CE Environmental Quiz-4

Determine the maximum upper limit of BOD of a glucose solution of concentration 540 mg/l.

Calculate the total 5 days BOD of sewage in kg per day, if the average sewage from the city is 75 x10⁶ litre per day and the average 5 day is 400 mg/litre.

When temperature of the waste increases

A waste water stream (flow = 3 m³/s, ultimate BOD = 80 mg/l) is joining a small river (flow = 12 m³/s, ultimate BOD = 10 mg/l). Both water streams get mixed up instantaneously. Cross-sectional area of the river is 30.0 m². Then the velocity of river flow will be:

‘if B.O.D. of waste water sample after 5 days incubation at 20° C is 100 mg/l, deoxygenation rate constant at 20° C is 0.1 per day, ultimate B.O.D. will be

A water sample has a . The initial DO in the BOD bottle is 8 mg/L and the dilution is 1:10. What is the final DO in the BOD bottle?

The bacterias which require oxygen for their survival is known as:

If initial Dissolved Oxygen (DO) and final DO after 5 days incubation at 20°C in 1.0% dilution sample are 5.0 mg/L and 4.0 mg/L respectively, what is 5-days BOD of the same sample in mg/L?

The following data pertains to a sewage sample:

Initial DO = 5.0 mg/l

Final DO = 0.5 mg/l

Dilution to 1.0%

The BOD of the given sewage sample is

Say a raw wastewater sample from AA WWTP has 5-day BOD equals to 2000 mg/L (reaction constant k = 0.23/day at 20°C). Calculate value of ultimate BOD?