Question

A channel has a bottom width of 350 m, depth 9 m and side slopes 1:2....

A channel has a bottom width of 350 m, depth 9 m and side slopes 1:2. If the depth is increased to 12 m by dredging, determine the percentage increase in velocity of flow in the channel. For the same increase in cross sectional area, if the channel is widened (instead of deepening), what is the percentage increase in the velocity of flow.
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Answer #1

Sol We know V=Yn R2/5/2 V= kR²/3 where K=sh n When depth y=am (B+ my) xy B+ 2y Jitm? 350+ 2x 9 )x9 R = Ale 8.486 350+ 2x9x554200 m2 Area = 350x12= Same increase in Now we have to keep the area by increasing the the width and beeping y=am B= 4200% 46

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