Question

The figure below shows a channel with an overflow dam section in place. The depth just upstream of the dam is 3.0 metres, . The energy grade line at the dam is located 1.125 m above the crest of the dam. The channel slope is 0.001, Manning's roughness coefficient is 0.02, and the width of the rectangular channel section is 5.0 m. EGL normal depth om 3.0 m Calculate the distance upstream of the dam at which the flow depth will be equal to the normal flow depth, using one step (i.e. you should have two depth values). Hint: It can be shown that Ec 3yc 2

Answer 1

Solution :- Data given :- + Depth just upstream of dam ( 81 ) = 3.0 m Energy Grade line over the dam ( y) = 1:1250 $ channel slope (so) = 0.001 s Manning roughness coefficient (m) -0.02 → width of Channel (b)-5.0m " - Normal depth = In EGIL y & Yc Yn 3m M

-) over For the flow to be possible minimum depth the dam should be critical depth (Yc) v² Yet = y = Ec 29 =) Yet V = 1.125 29 = 1.125 = yet qe 29 G² Y3 ag? = 1.125 =) Yct 3 ye = 1.125 ñ -> Yc =0.75m _j%3 3 0.75 ł g = q = 2.034 m?/ 3) 0 = 2.03 4 m/sx b =2.034x5 m3) =/0=10.17 m3/ & for normal depth of flow apply manning equation. q²m2 플 So = yo وار ) 7 0.001s 2.034x0.02% b=5m (you 110/3 =) Ym = 1.163 m

slope at 3m depth 13 5m 413 Q = 12 ARS x 15x [ 15 ſs1 * 10.17 oor 0.02 S1 = 0.0001216 Sz = so =0.001 - En = Ynt 5m 2 Von ag lym 1 7 1 1 63m 1 ♡ 10.17 Am = 1.75mis 5x1.263 - Vm = 1.75mis -2 1.75 = 1.32m = Vm = - Em = 1-163+ 19.62 » Ei = yit vi? ag 10.17 Iy Зот + Vi = -0.678 mis 5 15 =) El=3+ 0.6782 = 3.0234 m 19.62 AE A 2a unit step method of integration So - S DE: E1-Em =(3.0234 – 1:34)mm = 1.7034

SE = Si tso 2 0.000147 0001 I 2 SE = 0.00056 1.7034 AR 0.001 -0.00056 Aaa 3871-364 m