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Answer :
To find the position of the apex of the V notch from the bed of the channel, we need to apply the concept of flow through a V-notch weir.
The formula for discharge [tex]Q[/tex] through a V-notch weir is given by:
[tex]Q = \frac{8}{15} C_d \sqrt{2g} \tan\left(\frac{\theta}{2}\right) H^{5/2}[/tex]
where:
- [tex]Q[/tex] is the discharge (0.2 \ \text{m}^3/\text{s}).
- [tex]C_d[/tex] is the coefficient of discharge (0.62).
- [tex]g[/tex] is the acceleration due to gravity, approximately [tex]9.81 \ \text{m/s}^2[/tex].
- [tex]\theta[/tex] is the angle of the notch which is 90°.
- [tex]H[/tex] is the head of the water over the notch we are trying to find.
Substitute the known values into the formula:
[tex]0.2 = \frac{8}{15} \times 0.62 \times \sqrt{2 \times 9.81} \cdot \tan\left(\frac{90°}{2}\right) \cdot H^{5/2}[/tex]
Simplify the expression:
[tex]\tan\left(\frac{90°}{2}\right) = \tan(45°) = 1[/tex]
The equation becomes:
[tex]0.2 = \frac{8}{15} \times 0.62 \times \sqrt{19.62} \cdot H^{5/2}[/tex]
Calculating further:
[tex]0.2 = 0.331 \cdot H^{5/2}[/tex]
Solve for [tex]H^{5/2}[/tex]:
[tex]H^{5/2} = \frac{0.2}{0.331} \approx 0.604[/tex]
To find [tex]H[/tex], raise both sides to the power of [tex]\frac{2}{5}[/tex]:
[tex]H = (0.604)^{\frac{2}{5}}[/tex]
Using a calculator, you'll find:
[tex]H \approx 0.78 \text{ m}[/tex]
This [tex]H[/tex] is the head of water over the V-notch, meaning the position of the apex of the notch from the bed of the channel is given by [tex]1 \text{ m} - 0.78 \text{ m}[/tex]. Therefore, the apex position:
[tex]\text{Position from bed} = 0.22 \text{ m}[/tex]
Thus, the apex of the notch should be 0.22 meters above the bed of the channel to not exceed the maximum water depth of 1 meter.
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