A rectangular channel of width 12 ft has an upstream slope of 0.02 ft/ft and a d

Question

A rectangular channel of width 12 ft has an upstream slope of 0.02 ft/ft and a downstream slope of 0.0005 ft/ft. Both reaches have lengths of 500 ft. The upstream reach has a Manning's roughness of 0.016 and the downstream reach has a roughness of 0.011. The design discharge is Q = 460 cfs, the low discharge is Q = 230 cfs, and the flood discharge is Q = 690 cfs. Using the direct step method by hand with 1 step, determine the location of the hydraulic jump relative to the slope break for the design discharge. Classify the jump type and calculate the energy lost.

Explanation / Answer

Hydraulic Jump In Sloping Channels

Figure (Bradley, 1961) indicates a method of delineating hydraulic jumps in horizontal and sloping channels. Horizontal channels (case A) were discussed in the previous section. Sloping channels are discussed in this section. If the channel bottom is selected as a datum, the momentum equation becomes:

(6.7)

where,

γ = unit weight of water, N/m3 (lb/ft3)

φ = angle of channel with the horizontal

B = channel bottom width (rectangular channel), m (ft)

w = weight of water in jump control volume, N (lb)

The momentum equation used for the horizontal channels cannot be applied directly to hydraulic jumps in sloping channels since the weight of water within the jump must be considered. The difficulty encountered is in defining the water surface profile to determine the volume of water within the jumps for various channel slopes. This volume may be neglected for slopes less than 10 percent and the jump analyzed as a horizontal channel.

The Bureau of Reclamation (Bradley, 1961) conducted extensive model tests on case B and C type jumps to define the length and depth relationships. This reference should be consulted if a hydraulic jump in a sloping rectangular channel is being considered. Model tests should be considered if other channel shapes are being considered.

Figure . Hydraulic Jump Types Sloping Channels (Bradley, 1961)

this how you will calulate your answer

i am posting a solved exapmle as well

Equations for hydraulic jump in horizontal rectangular channel (Chaudhry, 1993; Chow, 1959):

V=Q/(yB)      F=V/(gy)0.5     y2/y1 = 0.5 [(1+8F12 )0.5 - 1] 

L = 220 y1 tanh[(F1-1)/22]        h = (y2-y1)3 /(4y1y2)

where (subscript 1 indicates upstream of jump; subscript 2 indicates downstream of jump):
B=Channel width (m), F=Froude number (dimension-less), g=acceleration due to gravity (9.8066 m/s2), h=Head loss (m), L=Length of jump (m), Q=Discharge (m3/s), tanh=Hyperbolic tangent trigonometric function, V=Velocity (m/s), y=Water depth (m)

Need B>0."  Channel width must be a positive number.
"Need Q>0."  Discharge must be positive.
"Need y1>0."  Upstream depth must be positive.
"Need F1 >1."  Upstream flow must be supercritical.

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