Feb 26, 2016

Feb 26, 2016

Feb 26, 2016

# Open Channel Hydraulics (V.T Chow) Solved Example # 02

 Q.No. 02 Verify by computation the depth velocity relationships shown in figure below for the four flow regimes in a wide rectangular open channel. The temperature of the water is taken as 68°F. Depth Vs Velocity Chart Figure 1 SOLUTION: For verification of type of depth velocity relationship first we will assume a wide rectangular channel with a constant width then we will take values of V and Y from the given figure and by calculations we will verify our assumptions Channel width = 2 ft Flow range & type = Subcritical Laminar 1 Flow Velocity = 0.2 ft/s Flow depth = 0.02 ft Acc gravity = 32.2 ft/s Hydraulic radius = 0.0196 υ = 0.0000108 ft2/s @ 68F Froud No = 0.25 Rynold's No = 364 For Froud's No 0.25 and Rynold's No 364 it can be verified from Figure.1, that Flow is in Subcritical Laminar 1
 Channel width = 2 ft Flow range & type = Supercritical Laminar 2 Flow Velocity = 2 ft/s Flow depth = 0.0025 ft Acc gravity = 32.2 ft/s Hydraulic radius = 0.0025 ft υ = 0.0000108 ft2/s Froud No = 7.05 Rynold's No = 462 For Froud' No 7.05 and Rynold's No 462 it can be verified from Figure.1, that Flow is in Supercritical Laminar 2 Channel width = 2 ft Flow range & type = Supercritical Turbulent 3 Flow Velocity = 10 ft/s Flow depth = 0.1 ft Acc gravity = 32.2 ft/s Hydraulic radius = 0.0909 ft υ = 0.0000108 ft2/s Froud No = 5.57 Rynold's No = 84176 For Froud's No5.57 and Rynold's No 84176 it can be verified from Figure.1, that Flow is in Supercritical Turbulent 3
 Channel width = 2 ft Flow range & type = Subcritical Turbulent 4 Flow Velocity = 0.25 ft/s Flow depth = 0.1 ft Acc gravity = 32.2 ft/s Hydraulic radius = 0.0909 ft υ = 0.0000108 ft2/s Froud No = 0.14 Reynold's No = 2105 For Froud's No 0.14 and Rynold's No 2105 it can be verified from Figure.1, that Flow is in Subcritical Turbulent 4

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