Answer:
a) i) [tex]v = 3.333\,\frac{m}{s}[/tex], ii) [tex]R_{h} = 0.06\,m[/tex], iii) [tex]E_{s} = 0.716\,m[/tex], b) [tex]Fr = 2.748[/tex] (Supercritical).
Explanation:
a) i) The flow velocity of water is:
[tex]v = \frac{\dot V}{w\cdot h}[/tex]
[tex]v = \frac{0.3\,\frac{m^{3}}{s} }{(0.6\,m)\cdot (0.15\,m)}[/tex]
[tex]v = 3.333\,\frac{m}{s}[/tex]
ii) The hydraulic radius of the channel:
[tex]R_{h} = \frac{w\cdot h}{2\cdot (w+h)}[/tex]
[tex]R_{h} = \frac{(0.6\,m)\cdot (0.15\,m)}{2\cdot (0.6\,m+0.15\,m)}[/tex]
[tex]R_{h} = 0.06\,m[/tex]
iii) Let assume that height is equal to zero. Then, the specific energy is:
[tex]E_{s} = 0.15\,m + \frac{(3.333\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s} )}[/tex]
[tex]E_{s} = 0.716\,m[/tex]
b) The characteristics of the flow in the open channel is inferred from the Froude number, whose formula is:
[tex]Fr = \frac{v}{\sqrt{g\cdot h} }[/tex]
[tex]Fr = \frac{3.333\,\frac{m}{s} }{\sqrt{(9.807\,\frac{m}{s^{2}} )\cdot (0.15\,m)} }[/tex]
[tex]Fr = 2.748[/tex]
As [tex]Fr > 1[/tex], then flow is supercritical.
