Question

PROBLEM A lake (reservoir) is connected to the rectangular channels as illustrated in the figure below. Details of the data for your design project is prepared and listed in a separate DATA TABLE you may find on Moodle. Therefore, Halid us epwall Ex fortuis reference), B, N, Sor, Mz and Saz values are given for your problem. ASKED 2. Check your slope condition (steep or mild slope) and proceed for calculations accordingly, b. Find out the discharge for your design problem, c. Find out the critical depth d. Find out normal depths of y, and >> c. If hydraulic jump exists, give the location of the jump (on steep slope channel or on mild slope channel), total length and, depths before and after the hydraulic jump. E. If there is no hydraulic jump expected, find out total length of the gradually varied flow curse using appropriate intervals and calculate flow depths for each interval. g. Imagine ys depth is required further down the So Set your water depth and design your channel accordingly (.. change widih, letowa). SKETCH HA yar A tresno Sar Stoper skye Nowy reach J yor S muilders Song reach NOTES i. Depending on the change of slope, hydraulic jump may or may not exist. If exists, there is no need to check for GVF calculations. GVF calculations are required for those works where hydraulic jump does not exist. Therefore, you should solve ONLY corf ii. Sketch abxwe is for illustrative purposes ONLY and may not represent the correct flow regimes, GVF curves, hydraulic jump occurrence, etc., so please check your values, draw your own sketch indicate flow regimes and comment on your calculations accordingly. ill. Formal submission is required! Your format will value your work, so you need to submit your work in an appropriate format

H is 1.32
B is 11.03
n1 is 0.022
n2 is 0.026
S1 is 0.00176
S2 is 0.00210

Answer 1

Sol) H- B.=11.03 52 = 0.00210 Specific Energy, E = y + 12 syt 82 Given data - 1.32 ni = 0.022 12= 0.026 SI = 0.00176 check for slope condition Si =0.00176 we need to find find normal depth for this w.k. T =yt 29 A2 a Here, Q² 29 2-39 2 نم q = 9 A = BY

neq (b) equate Let us specific energy at reservoir (section ③ - 0 f section @ @ H to a 92 a Yit 2.94,2 atx R$ xszxy - ®[ from man vingtee Area Hydraulic radius, R z welted Perimeter B.yi B +24 R = llo 034, 11.03+2y, Energy stope , S = s, = 0.00176[• Normal depth flow Let us substitite 1 qa 11.034, Х دمایہ) 0.022 11.03+2y x f 0.00176 xy

Now, substitute the value of q' in eq 1.32 to 9, + 2X 2x9.81xy, 11.039,) & (0.00178 0-022 X (11.03+2y) y, 0.61 m 1 q WIN (11.6 (11.03X6061 0.022 2 11.03+(2 x0.61)) X (0.00176) X O obl = لحة 0.783 m3/s.m Ar so, Q = 2 x B = 0.783 X 11.03 0 Discharge, @)= 8.63 m3/s

For critical depth, de dy =0 ie, Q2 T 9 AS - 1 q² 3 9. " co 9c 0o 7832 9.81 yo c = 0.40 m C critical depth, y e=0.40m Now, take s =S₂=0000210 & put it 0 in e qb Then , 0.783 - 11.03XY₂ 11.03+242 WIN X 0.026 x (0.0021)xyz Y q = 0.64 m

! slope condition check yu > Ус & Y2 > 4 is The slope is steep thorought Normal depths y = Oodlm y = 0.6um