Question

Problem 5: Discharge Integral II Consider steady flow in a stream d=0.50 m deep and w = 3,0 m 0.50 m deep and w = 3.0 m wide ("into the page"): Part A Part B The velocity distribution is u = Umw(1-h2/d2), where m shown in the following table: "), where Umax = 0.20 m/s. A few sample values are = 0.20 m/s. A few sam h [m] u [m/s] 0 0.20 0.25 0.15 (a) on the figure, sketch the velocity profile using the equation. No need to calculate points. une ligure, sketch an approximation of the velocity profile. Split the cross-sectional area into two segments, and consider the average velocity for each. (C) Using your approximation, calculate the discharge 0 [m/s]. (d) Using your approximation, calculate the average velocity V [m/s]. (e) Calculate the exact discharge 0 [m/s].

Answer 1

Ceiven, stream width depth d=0.5m W=3.om. Pasto) Omeya 0-2ml Parto U=0.20m/sy d-om parbolic - 015164 Segmento seegmento Uz umy (i 0 him) 0 U(myse) :25 015 0.20 A Unax = 0.20 msec. 0-2 +0.15 0.175 m/s parte) Using approximation discharge for segment 1 - au hot so dischaye Q = AV for = 0.25) (3.4)(0-125) = 0.13125 m/sec Segment @ =) Ar velocity V₂ = 0.15 - 20.075 m/s t so discharge O₂ = AzV2 = ( w/vz = 0-25 X(3-) (0978) 0.05625 m/sec So total dischanze a= 0-13125 +0+05625 Appoox.clische = 0-1675 ms 1 Partedt. Average velocity - Average dischay 0.1875 m/se 0.50X3.0 Naug = 0.125 m/se .

e) exact discharje. Q = ? taking a strip of ch thickness at W..om h depth Couss section of strean So elemental area da = wath. elemental discharge d& - UDA de = Umaylland) DA da Jale- lung (1 - i min = Umywe en ben = Um W. [d- Q = ungu wel =) (020w/se)(3.0m) (o.com) exact dischaye 0 – 0.2 m/sec
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