Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 2 (Q.No. 9)
9.
The settling velocity of the particles larger than 0.06 mm in a settling tank of depth 2.4 is 0.33 m per sec. The detention period recommended for the tank, is
Discussion:
11 comments Page 1 of 2.
Raj said:
4 years ago
Settling time = H/Vs,
So settling time =2 hours, and for settling, D.T >S.T and 3 hr is the only option which fulfils this condition.
So settling time =2 hours, and for settling, D.T >S.T and 3 hr is the only option which fulfils this condition.
(1)
Jatinder said:
7 years ago
Thanks @Rajesh.
Rajeshkumar J said:
7 years ago
DT = VOLUME / RATE OF FLOW,
VOLUME = AREA * DEPTH.
RATE OF FLOW,Q= V* A
therefore, DT = DEPTH/VELOCITY.
So, option D is correct.
VOLUME = AREA * DEPTH.
RATE OF FLOW,Q= V* A
therefore, DT = DEPTH/VELOCITY.
So, option D is correct.
(2)
Ravi mahawar said:
8 years ago
D.T = depth of tank / Vs of partical.
= 2.4m and .33 mm velocity,
( it's not .33 m )
= 2400/.33.
= 7272.72sec.
= 2 hour.
= 2.4m and .33 mm velocity,
( it's not .33 m )
= 2400/.33.
= 7272.72sec.
= 2 hour.
(5)
Kunti said:
8 years ago
@Shiva and @Gagnesh
Here 2.4m is depth, Not volume.
Here 2.4m is depth, Not volume.
(1)
Shiva said:
8 years ago
DT = lbh/q = lbh/av = h/v = 2400/.33.
(1)
Gagnesh said:
8 years ago
2.4m = 2400mm.
DT= 2400mm ÷.33mm/sec = 7272 sec.
7272sec ~ 2hours.
DT= 2400mm ÷.33mm/sec = 7272 sec.
7272sec ~ 2hours.
Adhi said:
9 years ago
Can anyone explain the solution?
Nagarajan said:
10 years ago
It is 181.82 sec so 3 min is the answer.
Rajmohan said:
10 years ago
2 hours for 0.33 mm /sec not m/sec.
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