Question

3. Water flows in a rectangular channel of width, w. In some cases, the water may suddenly increase in elevation as shown in the figure (from h to ha) through a highly disturbed region called a hydraulic jump. Assume the velocity is uniform at stations 1 and 2, and assume the pressures are distributed hydrostatically. Neglect friction on the channel bottom and sides. Hydraulic jump V2 Figure for problem 3. (a) Discuss the assumptions made in the problem statement. How likely are these to be valid assumptions? (b) Using the conservation of mass and momentum, find the equation (cubic) for the downstream depth, h, in terms of h, Vi, and g = 7.6 m/s, find the downstream height, h2. (Note that there are two (e) For hi 1 m and Vi valid solutions - explain.) What is the downstream velocity?

Answer 1

Assumptions made form across section ame and to : since it is given that flow is highly disturbed so there will be momentum tromsfer by mhing and assumption is valid, U Pressures are distributed by drostatically: Hydrostatic pressure distribution is valid if either fluid is steady slatie o stream lineschre shrought lines instead of being curred in a flow which is case here so assumption is valid (11) friction on channel bottom side is negligible because of laminar sublayer, forection at bottom combe noglected and assumption is valid (6 assuming width be b at both cross sections. hu fight fg ha conservation om ervation of mass Cat steady state) Si 1 h = B V2 102 assuming f1 = 12 vihi=12h2 12= vihi - 0 onservation of momentum EFx = change in momentum in x-direction r where Etx=net force I acting in a direction exb) = bP V2 h 2x 12 blvinive gh 12 gh2² - Bha-vihi - -)

substituting value of 12 in equation from P ghite_ghe = (Why Doba – weni hở - Vi bị vệ - - the year in ba tuwe co b2 Substituting hiram &vi= 7.6 m/s 1- 19 2x12 + 7.627 i 7 bat 7-6312=0 40905 h 2² - 62.665 b2 +57.76=0 h2= 3.432m or ho 3.432 m os h2= -3.97 m or h2=2.97 m or ha=lm Valid solutions: h2=2.97 m . ha = 1 me corresponding to no hydraullic Jump In case of hychraullic jump, h2=2.97m using equation o 1D = 147.62|= 355m/s 2.97