Answer:
a)
, b)
, c) 20.5 % of initial energy is lost through the jump, d) Before the jump: Supercritical (Fr > 1), After the jump: Subcritical (Fr < 1).
Step-by-step explanation:
a) The Froude number before the jump is:
![Fr_(1) = \frac{v_(1)}{\sqrt{g\cdot y_(1)} }](https://img.qammunity.org/2021/formulas/engineering/college/71fl6jecywlzsrh1ueetno1lo3icpvj4gq.png)
![Fr_(1) = \frac{7.5\,(m)/(s) }{\sqrt{(9.807\,(m)/(s^(2)) )\cdot (0.8\,m)} }](https://img.qammunity.org/2021/formulas/engineering/college/qf5tvlwn2ejqahrib3r8s5hdtb3pjjn42v.png)
![Fr_(1) \approx 2.678](https://img.qammunity.org/2021/formulas/engineering/college/yubrqx30vhv5nsvdqnobq6k2wafaa5v41q.png)
The depth after the jump is:
![y_(2) = 0.5\cdot y_(1)\cdot \left(-1 + \sqrt{1 + 8\cdot Fr_(1)^(2)} \right)](https://img.qammunity.org/2021/formulas/engineering/college/qss3krt4y2j3z90fvtyaomfppwd51culyj.png)
![y_(2) = 0.5\cdot (0.8\,m)\cdot \left[-1 +\sqrt{1 +8\cdot (2.678)^(2)} \right]](https://img.qammunity.org/2021/formulas/engineering/college/f2wk03eo3ik6ft05zfhvte77pi4zhghlak.png)
![y_(2) \approx 2.656\,m](https://img.qammunity.org/2021/formulas/engineering/college/hhb258yhuh3qecqy0b5m6xu16t34kugpxo.png)
b) The speed after the hydraulic jump is derived from the continuity equation:
![v_(1)\cdot y_(1) = v_(2)\cdot y_(2)](https://img.qammunity.org/2021/formulas/engineering/college/xfhlub1qxe74yi06zv8z1dy3b3kgij9nh4.png)
![v_(2) = (y_(1))/(y_(2))\cdot v_(1)](https://img.qammunity.org/2021/formulas/engineering/college/9flxtwfzcv58klhvpwnt8ljo2mqwue6oq4.png)
![v_(2) = (0.8\,m)/(2.656\,m) \cdot (7.5\,(m)/(s) )](https://img.qammunity.org/2021/formulas/engineering/college/y6z4ty3lfa0g8n00riblmxpzsnuocy6pu5.png)
![v_(2) = 2.259\,(m)/(s)](https://img.qammunity.org/2021/formulas/engineering/college/i95d3xcw9dpxzla14rkoy0zzpob72mo8hb.png)
The Froude number after the hydraulic jump is:
![Fr_(2) = \frac{v_(2)}{\sqrt{g\cdot y_(2)} }](https://img.qammunity.org/2021/formulas/engineering/college/s0v0gv1jhw0qryaypnx5pue7016onvwhnm.png)
![Fr_(2) = \frac{2.259\,(m)/(s) }{\sqrt{(9.807\,(m)/(s^(2)) )\cdot (2.656\,m)} }](https://img.qammunity.org/2021/formulas/engineering/college/qvgojtv0rue8p2ujiyiteg8kkwhxuuvdrw.png)
![Fr_(2) \approx 0.442](https://img.qammunity.org/2021/formulas/engineering/college/ak62da8umtt4gif16vkp893pzxt3cgg95q.png)
c) The head loss is:
![h_(L) = y_(1) - y_(2) + (v_(1)^(2)-v_(2)^(2))/(2\cdot g)](https://img.qammunity.org/2021/formulas/engineering/college/emezea9cjp4q3g09zqrqot42jbck6me3ao.png)
![h_(L) = 0.8\,m - 2.656\,m + ((7.5\,(m)/(s) )^(2)-(2.259\,(m)/(s))^(2))/(2\cdot (9.807\,(m)/(s^(2)) ))](https://img.qammunity.org/2021/formulas/engineering/college/ox7lzzngp8p61e39yvsbmlj6yngp7m1ryd.png)
![h_(L) = 0.752\,m](https://img.qammunity.org/2021/formulas/engineering/college/gt5ibd0y0ug0mp9t1miienpdqtm77kya9r.png)
The dissipation ratio is obtained afterwards:
![DR = (h_(L))/(y_(1)\cdot \left(1 + (Fr_(1)^(2))/(2) \right))](https://img.qammunity.org/2021/formulas/engineering/college/2vgqihh5uwpb7yperhbdrhu0biucf2sjt5.png)
![DR = (0.752\,m)/((0.8\,m)\cdot \left(1 + (2.678^(2))/(2) \right))](https://img.qammunity.org/2021/formulas/engineering/college/ywj8wek7vx56iq1ifyoke3y2lzxfko4lr0.png)
![DR = 0.205](https://img.qammunity.org/2021/formulas/engineering/college/virv5n0udqazgqw6ebv6fxjvzrfw9tqcep.png)
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)