Given:-
The wetted perimeter of the channel = 8 m
The width of the channel = 6 m
To determine:-
The hydraulic radius of a channel for different depths
Considering the channel section as a rectangular channel section
So,
The hydraulic radius of a rectangular channel section can be calculated using the formula
${R}_{H}=\frac{{A}_{W}}{{P}_{W}}$
Here,
R_{H} = Hydraulic radius
A_{W} = Wetted cross-sectional area of the channel
P_{W} = Wetted perimeter
Now,
Putting, the width of the channel = 6 m and wetted perimeter of the channel = 8 m
For,
a) Depth of water = 0.80 m
The hydraulic radius will be
${R}_{H}=\frac{{A}_{W}}{{P}_{W}}\phantom{\rule{0ex}{0ex}}Putting,{A}_{w}=B\times y$
Here,
B = Width = 6 m
y = Depth = 0.80 m
${R}_{H}=\frac{6\times 0.80}{8}\phantom{\rule{0ex}{0ex}}{R}_{H}=\frac{4.8}{8}\phantom{\rule{0ex}{0ex}}{\mathit{R}}_{\mathbf{H}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathit{m}$
Similarly,
S.No | Width | Depth | Wetted perimeter | Hydraulic radius |
a) | 6 m | 0.80 m | 8 m | 0.6 m |
b) | 6 m | 0.65m | 8 m | 0.487 m |
c) | 6 m | 0.70 m | 8 m | 0.525 m |
d) | 6 m | 0.72 m | 8 m | 0.54 m |
e) | 6 m | 0.10 m | 8 m | 0.075 m |
f) | 6 m | 0.05 m | 8 m | 0.0375 m |
g) | 6 m | 1.55 m | 8 m | 1.1625 m |
h) | 6 m | 1.04 m | 8 m | 0.78 m |
i) | 6 m | 1.10 m | 8 m | 0.825 |
j) | 6 m | 0.80 m | 8 m | 0.6 m |
k) | 6 m | 0.10 m | 8 m | 0.075 m |
l) | 6 m | 0.15m | 8 m | 0.1125 m |
Hence,
The hydraulic radius of a channel for different depths are:-
Depth | Hydraulic radius |
0.80 m | 0.6 m |
0.65m | 0.487 m |
0.70 m | 0.525 m |
0.72 m | 0.54 m |
0.10 m | 0.075 m |
0.05 m | 0.0375 m |
1.55 m | 1.1625 m |
1.04 m | 0.78 m |
1.10 m | 0.825 |
0.80 m | 0.6 m |
0.10 m | 0.075 m |
0.15m | 0.1125 m |