Question

Q1) Consider a rectangular channel with a constant width b, a) Obtain the equation of water surface profile (dy/dx) as function of Froude number and channel bottom slope in a channel transition assuming that over a short distance Ax, the energy losses can be neglected. b) Using the equation you have obtained, draw the water surface profile if flow is supercritical and if there is a downward step at the channel bed and show your results on the graph of specific energy curve (E versus y curve). Q2) a)Derive the equation of critical flow. 8 T c) If channel cross section is rectangular, obtain the minimum specific energy and critical flow depth as a function of unit discharge rate (9) Q3) There occurs no change in the state of fluid flow in a rectangular channel having a constant channel width b=3 m. in a channel transition with an upward step of Az=0.2 m. The constant discharge rate in the channel is Q=18 m/s. Calculate the followings; a) Minimum specific energy Ec and y b) The water depth after the step yz, if water depth before the step is yı=2 m. c) The water depth after the step yz, if water depth before the step is yı=1 m. d) Draw the graph of specific energy curve (E versus y curve) for band c, yc, y, and y2 on the graph e) Draw the water surface profile for band c, indicating yc, yi and y2 on the graph Q=18 m/s - y2=? b =3 m TIMITTUNTUITITITITITITIT

Answer 1

Ans (b) Area = A= byc T= b to b * → Fog Gritical Flow condition I 92 T = 1 g AB y = g 63 43 vitical depth as a function

Now, Specific Energy at Ye Ec=yet ve ² 29 Ес-ус+ - 29 y 3 를 specific Energy as a function of q

Ques: 3 Q = 18 m3/s AZ= 0.am b=3m q= Q = 18 = 6 m/s Part(a) lo Colhei cel depth yea ) * 1% = 1.54 m Critical Energy / spoific Energy + Ec= 3 % Ec= 3 x 1.54 AE c = 2.31 m Paart (6) y = 2m V = q = 6 = 3 m/s. y, 2 V = 3 m/s = F = v= 3 = 0,677 Jay 4.43 4.43 Now E,= V, ² + y 29 2 F = 0.677 Froude no < II 7 Suboutical Flow E = 2+ 3 2x9.8) E, 2.45m Hence the Transition cause a drop in water Surface elevation - Applying Energy Conservation E = Eg+ AZ Eg= E, AZ Eg=2.45-0.2 Eg = 2.25m |

2.83 = yg + q + AZ 294 2083 - 0.2= y + 62 2x9.81 4 2.63 = Yg + 1.83 Yg 1 9. - 2.63 y +1.83=0 by Hit and Trial method. * yg = 2.3m