rectangular open channel 5m wide carrying a flow of 8 m3/s. Slope of the channel bed is 0.00652. Hydraulic jump occurs in the downstream part of a channel and produced a sequent depth of 2.5m. use n=0.015.
a. Determine the depth before the jump in
"F^2 = \\frac{Q^2T}{gA^3} \\\\\n\n= \\frac{8^2 \\times 5}{9.81 \\times (5 \\times 2.5)^3} \\\\\n\n= 0.0167 \\\\\n\n\\frac{y_2}{y_1} = \\frac{1}{2}(-1 + \\sqrt{1+QF^2}) \\\\\n\ny_2 = 2.5 \\times \\frac{1}{2}(-1 + \\sqrt{1 + 8 \\times 0.0167}) \\\\\n\n= 0.08 \\;m"
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