Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 10 (Q.No. 29)
29.
The hydraulic radius for flow in a rectangular duct of cross-sectional dimension H, W is
Discussion:
9 comments Page 1 of 1.
Mr PEGU said:
1 year ago
According to me, the Correct answer is HW/2(H+W).
TANMAY SHAH said:
2 years ago
@All.
Equivalent diameter = 2HW/(H+W).
But, Hydraulic radius = HW/2(H+W).
Equivalent diameter = 2HW/(H+W).
But, Hydraulic radius = HW/2(H+W).
(2)
Chayanika said:
2 years ago
I think D is the correct answer.
Govind said:
3 years ago
D should be the correct answer.
Abhishek Kumar Bhargava said:
5 years ago
The hydraulic radius for rectangular duct.
4 HW/ (2H+2W) =2HW/H+W.
4 AREA of DUCT/WETTED PERIMETER.
4 HW/ (2H+2W) =2HW/H+W.
4 AREA of DUCT/WETTED PERIMETER.
Devesh said:
5 years ago
(4*H*W)/2(H+W) = HW/2(H+W).
ALEX said:
7 years ago
RH = cross sectional area/wetted perimeter = HW/2H+2W = HW/2(H+W).
(1)
M SUBBU said:
9 years ago
The correct answer is HW/2 (H+W).
Asmita said:
9 years ago
Can anyone explain it?
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