Civil Engineering - Waste Water Engineering - Discussion

Discussion Forum : Waste Water Engineering - Section 5 (Q.No. 44)
44.
The width of a settling tank with 2 hour detention period for treating sewage 378 cu m per hour, is
5 m
5.5 m
6 m
6.5 m
7 m.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

Anirban Roy said:   5 years ago
For Sedimentation tank :- l:b= 4 and b:d=2.
Vol=V= 378*2=756.
So, l*b*d= 756.
4b*b*(b/2)= 756; b= 7.23m ~ 7m.
(4)

Dheeraj kataria said:   6 years ago
@Lalit you are right. And @Naveen You are wrong.

This is not a formula that you are solving the question.

The original formula is:
Detention time(dt) = volume(Lxbxh)/Discharge(Q).
(1)

Lalit pathade said:   6 years ago
Q = Volume/Time.
L = 4B, D=4m.
Q = LBH/Time.
378= 4B*B*4/2.
B = 6.87m~7m.
(5)

Jinish said:   6 years ago
Depth of settling tank should d = 3 meter, not 4 meter and lenght= 4 times width, volume = q*t = 378*2.

So the width comes = 7.93.
(1)

Lunaf said:   6 years ago
How come 4 here? Please explain.

Manju said:   6 years ago
@Naveen.

V = Q/(B*L).
V= settling velocity.

Rakesh chouhan said:   7 years ago
@Naveen.

This is not a Formula.

You can compare the dimensions both sides, One side is sec. And other side is m/sec.
(1)

Dany said:   7 years ago
Q=378,
T=2hr,
Vol=QT=756m3.
Let depth D=4m.
Area B*L=Vol/D=189m2.
We know L=4B.
A=L*B=4B2=189m2.
So, B=6.87m~7m.
(2)

Abc said:   8 years ago
I think Q=AV and V considered 2m/hr.

Uru said:   8 years ago
Here T = Q/BxL.


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