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A channel has a bottom width of 300 m, depth 8 m and side slopes 1:1.5....

A channel has a bottom width of 300 m, depth 8 m and side slopes 1:1.5. 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

case-I Z 14.8m lis 3oom top width, Ts 300 + 8x15x2 = 324m & ArealA) 1/2 x8*( 300+324) = 2496m2 Perimeter ID- 300 + 2x8* JitiCase - I Area, A= 3816m2 fy=sm (B+24) then y=sm 1s A = 1 / 8x8 ( B+B+24) -42B+24) Y sol 412B+ 24 ) = 3816 Ba 465m) new width

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