1. If 8000m3/d of wastewater from an industry has a BOD5 of 190 mg/L and k = 0.17 per day (base 10)
A. How much oxygen (kg/day) is required to satisfy the BOD demand of this waste, assuming 1 kg of oxygen is to be supplied per kilogram of the ultimate BOD in the waste
B. What is the population equivalent of the waste (based on BOD5
Answer 1(a). BODt = BODu(1-10-kt) ..........................
Given,
BOD5 =190 mg/L
k = 0.17 per day
Q = 8000m3/day = 8x106 litres/day
Using BOD5 and k in equation (1) and using t =5,
We get ultimate BOD BODu = 221.25 mg/l
Now total ultimate BOD for 1 days wastewater = 221.25x8x106 mg/day =1770kg/day 1kg oxygen to be supplied for 1kg of BOD
Therefore total oxygen to be supplied is also equal to 1770kg/day
Answer 1(b). Population equivalent or number of people P...
2. A 5000-m3/day wastewater plant discharges to a river with a minimum drought flow of 137.7 m3/min (immediately upstream from the outfall)
A. What total allowable BOD5 (kg/day) could be discharged in the river after initial dilution must not exceed 1 mg/L and no upstream pollution is occurring?
B. What degree of treatment (% BOD satisfied) does this represent, assuming the plant influent to have a BOD5 of 250 mg/L
3. During recent winters there has been considerable criticism of the practice of dumping salt-laden snow from the streets into rivers and lakes for disposal. What problems are caused by this? (minimum of 4) Should the practice be prohibited? What alternatives are available? How would they resolve these problems
4. Calculate the surface area and depth of an "ideal" settling tank to remove all 10um spherical particles (SG = 2.65) from 1000 m3 of wastewater per day and provide a detention time.
5. From the information given, wastewater is made up of many different constituents, which one in your opinion create the greatest risks to human habitation? Natural habitation? Which ones are reasonably inert? Is treatment the only way to prevent these elements from entering the wastewater stream?