Waste Water Engineering Questions and Answers – Chemical Clarification – 2

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This set of Waste Water Engineering test focuses 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 %
View Answer

Answer: d
Explanation: 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.
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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 %
View Answer

Answer: b
Explanation: 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 %
View Answer

Answer: a
Explanation: 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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: a
Explanation: 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.
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6. At what pH is phosphate precipitated out?
a) 8
b) 6
c) 9
d) 10
View Answer

Answer: d
Explanation: 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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: c
Explanation: The dosage rate is 220 Kg/hr. Dosage rate = 22000x 1000/1000. From m3/hr to convert it to kg/hr the flow is divided by 1000.

10. Calculate the volume of lime to be dosed for the below given data. Round it off to the nearest digit.
Flow rate: 2200m3/hr
Dosage: 25 ppm
Specific gravity: 1.12
Concentration: 10%
a) 491 lph
b) 450 lph
c) 475 lph
d) 400 lph
View Answer

Answer: a
Explanation: The volume of lime to be dosed is 491 lph. Dosage rate is calculated as (22000x 25/1000= 55 Kg/hr). Volume of lime to be dosed is calculated as 55/1.5/10%=491.1 lph.
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11. Calculate the size of the tank for the below given data. Round off to the nearest 1000 digit.
Flow rate: 440lph
Retention time: 17 hrs
a) 9000 L
b) 8000 L
c) 10000 L
d) 7000 L
View Answer

Answer: a
Explanation: A 9000 L tank is required. Volume is calculated as Q x Time. Therefore 440lph x 17hrs x 1.2 =8976 L. Also a 20% margin is considered for safety purpose.

12. What is the detention time assumed while designing a flash mixer?
a) 2 mins
b) 10 mins
c) >10 mins
d) 1min
View Answer

Answer: d
Explanation: The detention time considered would be 1 min for a flash mixer. A flash mixer is one in which chemical mixing is done to bring about coagulation. This is where coagulation occurs. It is prior to a flocculator.

13. Determine the size of a flash mixer for the following data. Round off to the nearest digit.
Flow rate: 2200 m3/hr
a) 37 m3
b) 73 m3
c) 367 m3
d) 183 m3
View Answer

Answer: a
Explanation: The size of the flash mixer would be 37 m3. Volume = 2200 x 60/3600. Thus a 37 m3 volume size flash mixer is required.

Sanfoundry Global Education & Learning Series – Waste Water Engineering.

To practice all areas of Waste Water Engineering for tests, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn