Waste Water Engineering test on “Chemical Clarification – 2”.
1. In percentage what is the amount of TSS removed by chemical clarification?
a) 50-60 %
b) 60-70 %
c) 70-80 %
d) 80-90 %
Answer: d
Clarification: Around 80-90% TSS is removed by chemical clarification. The degree of clarification depends on the amount of chemicals added to the waste water. Also the exact percentage removed depends on the amount of TSS present in the raw waste water.
2. In percentage what is the amount of BOD removed by chemical clarification?
a) 20-30 %
b) 50-80 %
c) 30-50 %
d) 80-90 %
Answer: b
Clarification: Around 50-80% BOD is removed by chemical clarification. The degree of clarification depends on the amount of chemicals added to the waste water. Also the exact percentage removed depends on the amount of BOD present in the raw waste water. In the case of primary sedimentation tanks around 25-40% of the BOD is removed.
3. In percentage, how much does the chemical sludge contribute to the total sludge formed?
a) 0.5 %
b) 1 %
c) 2 %
d) 3 %
Answer: a
Clarification: In percentage 0.5% is contributed from the chemical sludge to the total sludge formed. The handling and disposal of sludge resulting from these chemicals is one big problem. Sludge is produced in great volumes from most of the precipitation reactions.
4. Assume in a sample of waste water around 60% TSS is removed by primary sedimentation without chemical addition. Calculate the mass of TSS removed for the following data.
Amount of TSS present initially: 220 mg/L
Waste water flow: 1000 m3/d
a) 100 kg/d
b) 130 kg/d
c) 132 kg/d
d) 102 kg/d
Answer: c
Clarification: The amount of TSS removed is 132 kg/d. The mass of the TSS removed = 220kg/1000/m3 x 0.6×1000 m3/d =132 kg/d. 220 mg/L is converted into Kg/L.
5. Determine the sludge formed from the following set of data.
Specific gravity of sludge: 1.05
Moisture content: 92.5%
Total dry solids: 213.4 kg/1000 m3
a) 2.71 m3/d
b) 2.75 m3/d
c) 2.76 m3/d
d) 3.71 m3/d
Answer: a
Clarification: The sludge formed is 2.71 m3/d. The volume of sludge formed = 213.4 kg/d / (1.05 x 1000kg/m3 x 0.075). Thus the total sludge formed equals to 2.71 m3/d.
6. At what pH is phosphate precipitated out?
a) 8
b) 6
c) 9
d) 10
Answer: d
Clarification: At the pH 10 phosphate is precipitated out. This is brought about by adding lime. Lime increases the pH and brings about precipitation of phosphate.
7. Name the compound formed when lime is added in order to precipitate phosphate.
a) Hydroxylapatite
b) Ferric phosphate
c) Calcium phosphate
d) Calcium Carbonate
Answer: a
Clarification: When lime is added Hydroxylapatite is formed. This precipitate is formed only when the pH is above 10. To bring about this reaction lime is added.
8. At which point, chemical addition for phosphorous removal can’t be carried out?
a) Preliminary treatment
b) Prior to primary sedimentation tank
c) Prior to secondary treatment
d) After secondary treatment
Answer: a
Clarification: At the preliminary point chemicals can’t be added in order to remove phosphates. These chemicals can be added prior to the sedimentation tank. This can also be added prior to or after the secondary treatment.
9. Calculate the dosage rate for the following data.
Flow rate: 2200m3/hr
Dosage: 100 ppm
a) 200 Kg/hr
b) 400 Kg/hr
c) 220 Kg/hr
d) 440 Kg/hr