Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 3 (Q.No. 16)
16.
If the depth of flow in a circular sewer is 1/4th of its diameter D, the wetted perimeter is
πD.
Discussion:
7 comments Page 1 of 1.
Sudha said:
3 years ago
Yes, you are right, Thanks @Priya.
Zara said:
6 years ago
@Priya right answer. Thanks.
Priya said:
7 years ago
The depth of flow h=d/2(1-cosθ/2),
But h=d/4. Hence θ=120.
P=(120/360)*Π*d.
But h=d/4. Hence θ=120.
P=(120/360)*Π*d.
Ranjitha said:
8 years ago
Wetted perimeter(p) = cos-1(1-(h/r)) * d.
h = 0.5r; p= cos-1(0.5) * d.
p = 1.047d.
from given answer π/3 = 1.047.
So, p = πd/3,
h = 0.5r; p= cos-1(0.5) * d.
p = 1.047d.
from given answer π/3 = 1.047.
So, p = πd/3,
(1)
Pritesh Dabhade said:
8 years ago
Answer should be πD.
Because, R = A/P.
And R= always D/4.
Hence,
D/4=(π /4)*D2/P.
P=[π/4*D2]/(D ÷4)
P= π D.
Because, R = A/P.
And R= always D/4.
Hence,
D/4=(π /4)*D2/P.
P=[π/4*D2]/(D ÷4)
P= π D.
(2)
Avjot said:
9 years ago
How this D-D/4 came @Bharath.
Bharath said:
1 decade ago
(Area/Wetted perimeter) = (pi*D^2/4 / (D - (D/4)) = piD/3.
(2)
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