Prob. # S (a): In a partially flowing channel having a isosceles- ngular cross section as shown below (Figure 5a), the rate of flow can be expressed as Q=KAR 23, where k is a constant and A and R are respectively cross sectional area and the hydraulic radius. Determine the depth at which the discharge is maximum. For the channel, A= [B- (h/3)]h, and P= B + (4h/3). A Figure 5a-Triangular channel cross section.

Structural Analysis
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Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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Prob. # 5 (a): In a partially flowing channel having a isosceles-triangular cross section as shown
below (Figure 5a), the rate of flow can be expressed as Q=KAR 2³, where k is a constant and A
and R are respectively cross sectional area and the hydraulic radius. Determine the depth at which
the discharge is maximum. For the channel, A= [B - (h/√3)]h, and P= B + (4h/3).
60
60°
Figure
section.
5a-Triangular channel cross
Transcribed Image Text:Prob. # 5 (a): In a partially flowing channel having a isosceles-triangular cross section as shown below (Figure 5a), the rate of flow can be expressed as Q=KAR 2³, where k is a constant and A and R are respectively cross sectional area and the hydraulic radius. Determine the depth at which the discharge is maximum. For the channel, A= [B - (h/√3)]h, and P= B + (4h/3). 60 60° Figure section. 5a-Triangular channel cross
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