250+ TOP MCQs on Sand Filters – 2 and Answers

Waste Water Engineering Quiz on “Sand Filters – 2”.

1. Calculate the diameter of a dual media filter with the following data.
Flow: 2200 m3/hr
Velocity: 15 m/hr
a) 15.5 m
b) 13.6 m
c) 13.9 m
d) 16.5 m
Answer: b
Clarification: The calculated diameter is 13.6 m. Area is first calculated by dividing the flowrate with velocity. That is 2200/15 =147 m2. Then square root (147 x (4/3.14)) = 13.6 m.

2. Calculate the diameter of a multimedia filter with the following data.
Flow: 2200 m3/hr
Velocity: 18 m/hr
a) 13.3 m
b) 14.5 m
c) 12.5 m
d) 12.9 m
Answer: c
Clarification: The calculated diameter is 12.5 m. Area is first calculated by dividing the flowrate with velocity. That is 2200/18 =122 m2. Then square root (122 x (4/3.14)) = 12.5 m.

3. While designing a blower for backwash for sand filters without air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: a
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for sand filters without air scouring is considered as 36m/hr. In case air scouring is there the velocity considered would be different.

4. While designing a blower for backwash for sand filters with air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: c
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for sand filters with air scouring is considered as 24 m/hr. In case air scouring is not there the velocity considered would be different.

5. While designing a blower for backwash for dual media filters without air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: b
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for dual media filters without air scouring is considered as 30m/hr. In case air scouring is there the velocity considered would be different.

6. While designing a blower for backwash for multimedia filters without air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: c
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for multimedia filters without air scouring is considered as 24 m/hr. In case air scouring is there the velocity considered would be different.

7. While designing a blower for backwash for multimedia filters with air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: d
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for multimedia filters with air scouring is considered as 20 m/hr. In case air scouring is not there the velocity considered would be different.

8. While designing a pump for backwash for a filter what will be the velocity assumed?
a) 24 m/h
b) 36m/h
c) 30 m/h
d) 20 m/h
Answer: a
Clarification: The velocity of the backwash pump is considered as 24m/h. This is in case of Dual media filter or Pressure sand filter. A horizontal centrifugal pump would be used as a backwash pump.

9. Calculate the flow rate of the backwash pump to be used for a dual media filter with the following data.
Flowrate: 2200 m3/h
Rise velocity of DMF: 15 m/h
a) 4500 m3/ h
b) 3520 m3/ h
c) 3750 m3/ h
d) 4920 m3/ h
Answer: b
Clarification: First the area of the filter is found out. Area = 2200/15=147 m2. Then by multiplying area x the velocity of the backwash pump assumed we derive at the flowrate. Flowrate= 147 m2 x 24 m/h = 3520 m3/h.

10. Calculate the flow rate of the backwash pump to be used for a pressure sand filter with the following data.
Flowrate: 2200 m3/h
Rise velocity of PSF (Pressure Sand filter): 10 m/h
a) 4500 m3/ h
b) 5280 m3/ h
c) 6480 m3/ h
d) 7580 m3/ h
Answer: b
Clarification: First the area of the filter is found out. Area = 2200/10=220 m2. Then by multiplying area x the velocity of the backwash pump assumed we derive at the flowrate. Flowrate= 220 m2 x 24 m/h = 5280 m3/h.

11. Calculate the flow rate of the backwash pump to be used for a multimedia filter with the following data.
Flowrate: 2200 m3/h
Rise velocity of multimedia filter: 18 m/h
a) 3700 m3/ h
b) 2900 m3/ h
c) 5300 m3/ h
d) 3800 m3/ h
Answer: b
Clarification: First the area of the filter is found out. Area = 2200/18=122 m2. Then by multiplying area x the velocity of the backwash pump assumed we derive at the flowrate. Flowrate= 122 m2 x 24 m/h = 2900 m3/h.

12. What is the bulk density of pebbles assumed while designing a Pressure sand filter?
a) 1650 kg/m3
b) 1600 kg/m3
c) 650 kg/m3
d) 550 kg/m3
Answer: b
Clarification: The bulk density of the pebble used is generally assumed as 1600 kg/m3. This value is used to calculate the amount of pebble required. Generally the amount of pebble is calculated by number of bags required. Each bag constitutes around 140 Kg of pebble.

13. What is the bulk density of sand assumed while designing the amount required for a sand filter?
a) 1750 Kg/m3
b) 1300 Kg/m3
c) 1000Kg/m3
d) 790 Kg/m3
Answer: a
Clarification: While designing a pressure sand filter the bulk density of the sand is considered as 1750 kg/ m3. The sand is generally measured in bags. Each bag weighs around 140 Kg.

14. Calculate the amount of pebbles required for the following data.
Flow: 200 m3/h
Velocity of dual media filter: 15m/h
Height: 0.6 m
a) 12668 Kg
b) 13792 Kg
c) 35000 Kg
d) 15792 Kg
Answer: a
Clarification: First the diameter is calculated by finding the area. Area= 200/15. Diameter= square root (1.27 × 13.33) = 4.11. Amount of pebbles equals πd2/4. This equals 0.785 x 4.11 x 4.11 x 0.6 x 1600 = 12668 Kg.

15. Calculate the amount of sand required for the following data.
Flow: 200 m3/h
Velocity of dual media filter: 15m/h
Height: 0.4 m
a) 12668 Kg
b) 9237 Kg
c) 1020 Kg
d) 12752 Kg
Answer: b
Clarification: First the diameter is calculated by finding the area. Area= 200/15. Diameter= square root (1.27 × 13.33) = 4.11. Amount of sand equals πd2/4. This equals 0.785 x 4.11 x 4.11 x 0.4 x 1750 = 9237 Kg.

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