Answer:
a) [tex]y_{2} \approx 2.656\,m[/tex], b) [tex]Fr_{2} \approx 0.442[/tex], c) 20.5 % of initial energy is lost through the jump, d) Before the jump: Supercritical (Fr > 1), After the jump: Subcritical (Fr < 1).
Explanation:
a) The Froude number before the jump is:
[tex]Fr_{1} = \frac{v_{1}}{\sqrt{g\cdot y_{1}} }[/tex]
[tex]Fr_{1} = \frac{7.5\,\frac{m}{s} }{\sqrt{(9.807\,\frac{m}{s^{2}} )\cdot (0.8\,m)} }[/tex]
[tex]Fr_{1} \approx 2.678[/tex]
The depth after the jump is:
[tex]y_{2} = 0.5\cdot y_{1}\cdot \left(-1 + \sqrt{1 + 8\cdot Fr_{1}^{2}} \right)[/tex]
[tex]y_{2} = 0.5\cdot (0.8\,m)\cdot \left[-1 +\sqrt{1 +8\cdot (2.678)^{2}} \right][/tex]
[tex]y_{2} \approx 2.656\,m[/tex]
b) The speed after the hydraulic jump is derived from the continuity equation:
[tex]v_{1}\cdot y_{1} = v_{2}\cdot y_{2}[/tex]
[tex]v_{2} = \frac{y_{1}}{y_{2}}\cdot v_{1}[/tex]
[tex]v_{2} = \frac{0.8\,m}{2.656\,m} \cdot (7.5\,\frac{m}{s} )[/tex]
[tex]v_{2} = 2.259\,\frac{m}{s}[/tex]
The Froude number after the hydraulic jump is:
[tex]Fr_{2} = \frac{v_{2}}{\sqrt{g\cdot y_{2}} }[/tex]
[tex]Fr_{2} = \frac{2.259\,\frac{m}{s} }{\sqrt{(9.807\,\frac{m}{s^{2}} )\cdot (2.656\,m)} }[/tex]
[tex]Fr_{2} \approx 0.442[/tex]
c) The head loss is:
[tex]h_{L} = y_{1} - y_{2} + \frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g}[/tex]
[tex]h_{L} = 0.8\,m - 2.656\,m + \frac{(7.5\,\frac{m}{s} )^{2}-(2.259\,\frac{m}{s})^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]h_{L} = 0.752\,m[/tex]
The dissipation ratio is obtained afterwards:
[tex]DR = \frac{h_{L}}{y_{1}\cdot \left(1 + \frac{Fr_{1}^{2}}{2} \right)}[/tex]
[tex]DR = \frac{0.752\,m}{(0.8\,m)\cdot \left(1 + \frac{2.678^{2}}{2} \right)}[/tex]
[tex]DR = 0.205[/tex]
20.5 % of initial energy is lost through the jump.
d) The flow conditions are described below:
Before the jump: Supercritical (Fr > 1)
After the jump: Subcritical (Fr < 1)
