problem 8-9 Water slopc of 0.0017. Usc n 0.014. velocity of 1.2 m/s. Find the normal depth of flow if the channel is laid on a Water flows in a triangular V-notch steel channel, with vertex angle of 60°, at a flows in a triangular V-notch steel channel, with vertex angle of 60°, at a wanty of 1.2 m/S. Find the normal depth of flow if the channel is laid on a slopc Solution 1 R2/351/2 2d tan 30° A = 2(2d tan60°)d A = d2 tan60° %3D P 2d sec60° 30 30° R= A/P = d2 tan60°/2d sec60° R = 0,433d 600 %3D %3D 1.2 = 0Ở (0.433d)2/3(0.0017)1/2 d = 0.601 m (normal depth) d sec 30° d sec 30°

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Explain every step of the solution, especially why they used 60° rather than 30°. And how did they get R?
Water flows in a triangular V-notch steel channel, with vertex angle of 60°, at a
Problem 8-9
of 1.2 m/s. Find the normal depth of flow if the channel is laid on a
velocity
of 0.0017. Use n 0.014.
slope
Solution
v = -
R2/351/2
2d tan 30°
A = 2(2d tan60°)d
A = d? tan60°
P 2d sec60°
30%
30°
R = A/P
= d? tan60°/2d sec60°
R = 0,433d
60°
001a (0.433d)/3(0.0017)1/2
d = 0.601 m (normal depth)
1.2 =
d sec 30
d sec 30°
Transcribed Image Text:Water flows in a triangular V-notch steel channel, with vertex angle of 60°, at a Problem 8-9 of 1.2 m/s. Find the normal depth of flow if the channel is laid on a velocity of 0.0017. Use n 0.014. slope Solution v = - R2/351/2 2d tan 30° A = 2(2d tan60°)d A = d? tan60° P 2d sec60° 30% 30° R = A/P = d? tan60°/2d sec60° R = 0,433d 60° 001a (0.433d)/3(0.0017)1/2 d = 0.601 m (normal depth) 1.2 = d sec 30 d sec 30°
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