depth of flow (at the back of the suppressed weir) = 2.485 meters discharge = 0.84 m^3/s head = 0.285 meters What is the width (L) of the channel? *Used the Francis Formula in solving* *Consider the velocity of approach
depth of flow (at the back of the suppressed weir) = 2.485 meters discharge = 0.84 m^3/s head = 0.285 meters What is the width (L) of the channel? *Used the Francis Formula in solving* *Consider the velocity of approach
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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depth of flow (at the back of the suppressed weir) = 2.485 meters
discharge = 0.84 m^3/s
head = 0.285 meters
What is the width (L) of the channel?
*Used the Francis Formula in solving*
*Consider the velocity of approach
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