(b) A regulator was constructed to pass a Leliavsky discharge in a canal of depth equals to 2.2 m at u/s side and an allowable drop in water surface (Dw) equals to 0.1m. By considering velocity head (do not neglect it): 1. Derive a formula that can be used to calculate hy or AH directly in terms of yt Cd and Dw, equation, starting with Leliavsky q= Cd y√√2gAH. 2. What would be the discharge passing through the regulator? Hint: Cd-0.92; H-h+ hv; yt-h-Dw US WL ΔΗ Dw E.L. - AH=h+h, yt = Dw + hv yt D/S W.L.
(b) A regulator was constructed to pass a Leliavsky discharge in a canal of depth equals to 2.2 m at u/s side and an allowable drop in water surface (Dw) equals to 0.1m. By considering velocity head (do not neglect it): 1. Derive a formula that can be used to calculate hy or AH directly in terms of yt Cd and Dw, equation, starting with Leliavsky q= Cd y√√2gAH. 2. What would be the discharge passing through the regulator? Hint: Cd-0.92; H-h+ hv; yt-h-Dw US WL ΔΗ Dw E.L. - AH=h+h, yt = Dw + hv yt D/S W.L.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:(b) A regulator was constructed to pass a Leliavsky
discharge in a canal of depth equals to 2.2 m at
u/s side and an allowable drop in water surface
(Dw) equals to 0.1m.
By considering velocity head (do not neglect it):
1. Derive a formula that can be used to
calculate hy or AH directly in terms of yt
Cd and Dw,
equation,
starting with Leliavsky
q= Cd y√√2gAH.
2. What would be the discharge passing through the regulator?
Hint: Cd-0.92; H-h+ hv; yt-h-Dw
US WL
ΔΗ
Dw
E.L.
-
AH=h+h, yt = Dw + hv
yt
D/S W.L.
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