Question

A rectangular channel having a width of 8m carries a discharge of 18 m3/s at the normal depth of 2.25m. (a) What should be the minimum width to which the channel can be contracted smoothly without causing any choking of flow assuming that the bed of the channel is level all through. (b) Compute the critical slope of this new section if n = 0.012. (c) How long should this channel be if there is a drop of 4m. at the end of its length.

Answer 1

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given 10x + 63.246 ²+ 10x = intet x=CF + P.I. on for Cof (103?+ 63,246D +10) 2:0 D--0.16 Henle --6:16 CO.KE Ae <6.16 + Ве 26 For Pita (10D? + +63:2V6 D tlo):1-dét i te-t (109?+62.246] Ho) T 20.2 to l-te-t 10 CD +0:16) (D+5:45) ĪT X Exió [ Dota16 04616]\-tot) =) 60X6.16 5 2. so yoola ['+eml]'C-tet) - (1+2) Cr-tet) 0.104 C1-P) C-ret) -0.0027 (-2)Ci-tet) 0.104 -0.104 tét -0.194 30+ 0-104-06tét). 0.0027 +0.00-27 tét + 5.002A 0019 = ) X- 0.16 0.16 8. 2027 LD (het) 6.16 G.16 a coloqot 0.1d3 -0.1013 det 70.65(e*- tét) 'Ix 0.1013 – 6-7513 tét X- P.I. +0.65 e Hence. x = C.F. +P.I xo Ае. -olst -6.kt + Be 10.1013-2.7513te torse 0 at too, to given than A + B +01013 - 34 0.65 $30+ 0.65 =) HH AAB-0 -0.7513 at odio xo given) ac -0.16€ -frit -0.16 A e 6.16Be. 5- 6,7513 otto.siz tét -0. -0.6set 0 - -0.16A-6-168 -0.7513 -0.65 ne b.16 A+ 6.16B = -1.4013 As, A.-0-5378, B=-22135 Hence -oilot 0.53780 70.1013 – 0.7913Fe*+0.65 ét 0. 213526.164